\(\int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx\) [411]

Optimal result

Integrand size = 33, antiderivative size = 448 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx=\frac {\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {\sqrt {a} \left (a^4 A b+18 a^2 A b^3-15 A b^5+3 a^5 B+6 a^3 b^2 B+35 a b^4 B\right ) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{4 b^{5/2} \left (a^2+b^2\right )^3 d}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (c+d x)}}{1+\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{2 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^2}-\frac {a \left (a^2 A b-7 A b^3+3 a^3 B+11 a b^2 B\right ) \sqrt {\tan (c+d x)}}{4 b^2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))} \] Output:

-1/2*(a^3*(A-B)-3*a*b^2*(A-B)+3*a^2*b*(A+B)-b^3*(A+B))*arctan(-1+2^(1/2)*t 
an(d*x+c)^(1/2))*2^(1/2)/(a^2+b^2)^3/d-1/2*(a^3*(A-B)-3*a*b^2*(A-B)+3*a^2* 
b*(A+B)-b^3*(A+B))*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)/(a^2+b^2)^3/ 
d+1/4*a^(1/2)*(A*a^4*b+18*A*a^2*b^3-15*A*b^5+3*B*a^5+6*B*a^3*b^2+35*B*a*b^ 
4)*arctan(b^(1/2)*tan(d*x+c)^(1/2)/a^(1/2))/b^(5/2)/(a^2+b^2)^3/d-1/2*(3*a 
^2*b*(A-B)-b^3*(A-B)-a^3*(A+B)+3*a*b^2*(A+B))*arctanh(2^(1/2)*tan(d*x+c)^( 
1/2)/(1+tan(d*x+c)))*2^(1/2)/(a^2+b^2)^3/d+1/2*a*(A*b-B*a)*tan(d*x+c)^(3/2 
)/b/(a^2+b^2)/d/(a+b*tan(d*x+c))^2-1/4*a*(A*a^2*b-7*A*b^3+3*B*a^3+11*B*a*b 
^2)*tan(d*x+c)^(1/2)/b^2/(a^2+b^2)^2/d/(a+b*tan(d*x+c))
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 4.68 (sec) , antiderivative size = 372, normalized size of antiderivative = 0.83 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx=\frac {2 \left (-\left ((A b+3 a B) \sqrt {\tan (c+d x)}\right )+\frac {\left (a^2 A b+4 A b^3+3 a^3 B\right ) \sqrt {\tan (c+d x)}}{4 \left (a^2+b^2\right )}-3 b B \tan ^{\frac {3}{2}}(c+d x)+\frac {(a+b \tan (c+d x)) \left (\frac {3}{8} a^{5/2} \sqrt {b} \left (a^2+b^2\right ) \left (a^3 A b+9 a A b^3+3 a^4 B+3 a^2 b^2 B+8 b^4 B\right ) \sqrt {\tan (c+d x)}+\left (\frac {3}{8} a^3 \left (a^4 A b+18 a^2 A b^3-15 A b^5+3 a^5 B+6 a^3 b^2 B+35 a b^4 B\right ) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )+\frac {3}{2} \sqrt [4]{-1} a^{5/2} b^{5/2} \left ((i a-b)^3 (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )-(i a+b)^3 (A+i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )\right )\right ) (a+b \tan (c+d x))\right )}{a^{5/2} \sqrt {b} \left (a^2+b^2\right )^3}\right )}{3 b^2 d (a+b \tan (c+d x))^2} \] Input:

Integrate[(Tan[c + d*x]^(5/2)*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^3 
,x]
 

Output:

(2*(-((A*b + 3*a*B)*Sqrt[Tan[c + d*x]]) + ((a^2*A*b + 4*A*b^3 + 3*a^3*B)*S 
qrt[Tan[c + d*x]])/(4*(a^2 + b^2)) - 3*b*B*Tan[c + d*x]^(3/2) + ((a + b*Ta 
n[c + d*x])*((3*a^(5/2)*Sqrt[b]*(a^2 + b^2)*(a^3*A*b + 9*a*A*b^3 + 3*a^4*B 
 + 3*a^2*b^2*B + 8*b^4*B)*Sqrt[Tan[c + d*x]])/8 + ((3*a^3*(a^4*A*b + 18*a^ 
2*A*b^3 - 15*A*b^5 + 3*a^5*B + 6*a^3*b^2*B + 35*a*b^4*B)*ArcTan[(Sqrt[b]*S 
qrt[Tan[c + d*x]])/Sqrt[a]])/8 + (3*(-1)^(1/4)*a^(5/2)*b^(5/2)*((I*a - b)^ 
3*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] - (I*a + b)^3*(A + I*B)* 
ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]]))/2)*(a + b*Tan[c + d*x])))/(a^(5/2 
)*Sqrt[b]*(a^2 + b^2)^3)))/(3*b^2*d*(a + b*Tan[c + d*x])^2)
 

Rubi [A] (verified)

Time = 2.12 (sec) , antiderivative size = 466, normalized size of antiderivative = 1.04, number of steps used = 24, number of rules used = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.697, Rules used = {3042, 4088, 27, 3042, 4128, 27, 3042, 4136, 27, 3042, 4017, 27, 1482, 1476, 1082, 217, 1479, 25, 27, 1103, 4117, 73, 218}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)^{5/2} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3}dx\)

\(\Big \downarrow \) 4088

\(\displaystyle \frac {\int -\frac {\sqrt {\tan (c+d x)} \left (-\left (\left (3 B a^2+A b a+4 b^2 B\right ) \tan ^2(c+d x)\right )-4 b (A b-a B) \tan (c+d x)+3 a (A b-a B)\right )}{2 (a+b \tan (c+d x))^2}dx}{2 b \left (a^2+b^2\right )}+\frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {\int \frac {\sqrt {\tan (c+d x)} \left (-\left (\left (3 B a^2+A b a+4 b^2 B\right ) \tan ^2(c+d x)\right )-4 b (A b-a B) \tan (c+d x)+3 a (A b-a B)\right )}{(a+b \tan (c+d x))^2}dx}{4 b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {\int \frac {\sqrt {\tan (c+d x)} \left (-\left (\left (3 B a^2+A b a+4 b^2 B\right ) \tan (c+d x)^2\right )-4 b (A b-a B) \tan (c+d x)+3 a (A b-a B)\right )}{(a+b \tan (c+d x))^2}dx}{4 b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 4128

\(\displaystyle \frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {\frac {\int -\frac {-8 \left (A a^2+2 b B a-A b^2\right ) \tan (c+d x) b^2+\left (3 B a^4+A b a^3+3 b^2 B a^2+9 A b^3 a+8 b^4 B\right ) \tan ^2(c+d x)+a \left (3 B a^3+A b a^2+11 b^2 B a-7 A b^3\right )}{2 \sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{b \left (a^2+b^2\right )}+\frac {a \left (3 a^3 B+a^2 A b+11 a b^2 B-7 A b^3\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}}{4 b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {\frac {a \left (3 a^3 B+a^2 A b+11 a b^2 B-7 A b^3\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\int \frac {-8 \left (A a^2+2 b B a-A b^2\right ) \tan (c+d x) b^2+\left (3 B a^4+A b a^3+3 b^2 B a^2+9 A b^3 a+8 b^4 B\right ) \tan ^2(c+d x)+a \left (3 B a^3+A b a^2+11 b^2 B a-7 A b^3\right )}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{2 b \left (a^2+b^2\right )}}{4 b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {\frac {a \left (3 a^3 B+a^2 A b+11 a b^2 B-7 A b^3\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\int \frac {-8 \left (A a^2+2 b B a-A b^2\right ) \tan (c+d x) b^2+\left (3 B a^4+A b a^3+3 b^2 B a^2+9 A b^3 a+8 b^4 B\right ) \tan (c+d x)^2+a \left (3 B a^3+A b a^2+11 b^2 B a-7 A b^3\right )}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{2 b \left (a^2+b^2\right )}}{4 b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 4136

\(\displaystyle \frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {\frac {a \left (3 a^3 B+a^2 A b+11 a b^2 B-7 A b^3\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\frac {\int -\frac {8 \left (\left (-B a^3+3 A b a^2+3 b^2 B a-A b^3\right ) b^2+\left (A a^3+3 b B a^2-3 A b^2 a-b^3 B\right ) \tan (c+d x) b^2\right )}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}+\frac {a \left (3 a^5 B+a^4 A b+6 a^3 b^2 B+18 a^2 A b^3+35 a b^4 B-15 A b^5\right ) \int \frac {\tan ^2(c+d x)+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{2 b \left (a^2+b^2\right )}}{4 b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {\frac {a \left (3 a^3 B+a^2 A b+11 a b^2 B-7 A b^3\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\frac {a \left (3 a^5 B+a^4 A b+6 a^3 b^2 B+18 a^2 A b^3+35 a b^4 B-15 A b^5\right ) \int \frac {\tan ^2(c+d x)+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {8 \int \frac {\left (-B a^3+3 A b a^2+3 b^2 B a-A b^3\right ) b^2+\left (A a^3+3 b B a^2-3 A b^2 a-b^3 B\right ) \tan (c+d x) b^2}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}}{2 b \left (a^2+b^2\right )}}{4 b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {\frac {a \left (3 a^3 B+a^2 A b+11 a b^2 B-7 A b^3\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\frac {a \left (3 a^5 B+a^4 A b+6 a^3 b^2 B+18 a^2 A b^3+35 a b^4 B-15 A b^5\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {8 \int \frac {\left (-B a^3+3 A b a^2+3 b^2 B a-A b^3\right ) b^2+\left (A a^3+3 b B a^2-3 A b^2 a-b^3 B\right ) \tan (c+d x) b^2}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}}{2 b \left (a^2+b^2\right )}}{4 b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 4017

\(\displaystyle \frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {\frac {a \left (3 a^3 B+a^2 A b+11 a b^2 B-7 A b^3\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\frac {a \left (3 a^5 B+a^4 A b+6 a^3 b^2 B+18 a^2 A b^3+35 a b^4 B-15 A b^5\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {16 \int \frac {b^2 \left (-B a^3+3 A b a^2+3 b^2 B a-A b^3+\left (A a^3+3 b B a^2-3 A b^2 a-b^3 B\right ) \tan (c+d x)\right )}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}}{4 b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {\frac {a \left (3 a^3 B+a^2 A b+11 a b^2 B-7 A b^3\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\frac {a \left (3 a^5 B+a^4 A b+6 a^3 b^2 B+18 a^2 A b^3+35 a b^4 B-15 A b^5\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {16 b^2 \int \frac {-B a^3+3 A b a^2+3 b^2 B a-A b^3+\left (A a^3+3 b B a^2-3 A b^2 a-b^3 B\right ) \tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}}{4 b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 1482

\(\displaystyle \frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {\frac {a \left (3 a^3 B+a^2 A b+11 a b^2 B-7 A b^3\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\frac {a \left (3 a^5 B+a^4 A b+6 a^3 b^2 B+18 a^2 A b^3+35 a b^4 B-15 A b^5\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {16 b^2 \left (\frac {1}{2} \left (-\left (a^3 (A+B)\right )+3 a^2 b (A-B)+3 a b^2 (A+B)-b^3 (A-B)\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \left (a^3 (A-B)+3 a^2 b (A+B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \int \frac {\tan (c+d x)+1}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}}{4 b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {\frac {a \left (3 a^3 B+a^2 A b+11 a b^2 B-7 A b^3\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\frac {a \left (3 a^5 B+a^4 A b+6 a^3 b^2 B+18 a^2 A b^3+35 a b^4 B-15 A b^5\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {16 b^2 \left (\frac {1}{2} \left (-\left (a^3 (A+B)\right )+3 a^2 b (A-B)+3 a b^2 (A+B)-b^3 (A-B)\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \left (a^3 (A-B)+3 a^2 b (A+B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \left (\frac {1}{2} \int \frac {1}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \int \frac {1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )\right )}{d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}}{4 b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {\frac {a \left (3 a^3 B+a^2 A b+11 a b^2 B-7 A b^3\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\frac {a \left (3 a^5 B+a^4 A b+6 a^3 b^2 B+18 a^2 A b^3+35 a b^4 B-15 A b^5\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {16 b^2 \left (\frac {1}{2} \left (-\left (a^3 (A+B)\right )+3 a^2 b (A-B)+3 a b^2 (A+B)-b^3 (A-B)\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \left (a^3 (A-B)+3 a^2 b (A+B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \left (\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}}{4 b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {\frac {a \left (3 a^3 B+a^2 A b+11 a b^2 B-7 A b^3\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\frac {a \left (3 a^5 B+a^4 A b+6 a^3 b^2 B+18 a^2 A b^3+35 a b^4 B-15 A b^5\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {16 b^2 \left (\frac {1}{2} \left (-\left (a^3 (A+B)\right )+3 a^2 b (A-B)+3 a b^2 (A+B)-b^3 (A-B)\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \left (a^3 (A-B)+3 a^2 b (A+B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}}{4 b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {\frac {a \left (3 a^3 B+a^2 A b+11 a b^2 B-7 A b^3\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\frac {a \left (3 a^5 B+a^4 A b+6 a^3 b^2 B+18 a^2 A b^3+35 a b^4 B-15 A b^5\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {16 b^2 \left (\frac {1}{2} \left (-\left (a^3 (A+B)\right )+3 a^2 b (A-B)+3 a b^2 (A+B)-b^3 (A-B)\right ) \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (a^3 (A-B)+3 a^2 b (A+B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}}{4 b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {\frac {a \left (3 a^3 B+a^2 A b+11 a b^2 B-7 A b^3\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\frac {a \left (3 a^5 B+a^4 A b+6 a^3 b^2 B+18 a^2 A b^3+35 a b^4 B-15 A b^5\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {16 b^2 \left (\frac {1}{2} \left (-\left (a^3 (A+B)\right )+3 a^2 b (A-B)+3 a b^2 (A+B)-b^3 (A-B)\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (a^3 (A-B)+3 a^2 b (A+B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}}{4 b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {\frac {a \left (3 a^3 B+a^2 A b+11 a b^2 B-7 A b^3\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\frac {a \left (3 a^5 B+a^4 A b+6 a^3 b^2 B+18 a^2 A b^3+35 a b^4 B-15 A b^5\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {16 b^2 \left (\frac {1}{2} \left (-\left (a^3 (A+B)\right )+3 a^2 b (A-B)+3 a b^2 (A+B)-b^3 (A-B)\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (c+d x)}+1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )+\frac {1}{2} \left (a^3 (A-B)+3 a^2 b (A+B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}}{4 b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {\frac {a \left (3 a^3 B+a^2 A b+11 a b^2 B-7 A b^3\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\frac {a \left (3 a^5 B+a^4 A b+6 a^3 b^2 B+18 a^2 A b^3+35 a b^4 B-15 A b^5\right ) \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {16 b^2 \left (\frac {1}{2} \left (a^3 (A-B)+3 a^2 b (A+B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (-\left (a^3 (A+B)\right )+3 a^2 b (A-B)+3 a b^2 (A+B)-b^3 (A-B)\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}}{4 b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 4117

\(\displaystyle \frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {\frac {a \left (3 a^3 B+a^2 A b+11 a b^2 B-7 A b^3\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\frac {a \left (3 a^5 B+a^4 A b+6 a^3 b^2 B+18 a^2 A b^3+35 a b^4 B-15 A b^5\right ) \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}d\tan (c+d x)}{d \left (a^2+b^2\right )}-\frac {16 b^2 \left (\frac {1}{2} \left (a^3 (A-B)+3 a^2 b (A+B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (-\left (a^3 (A+B)\right )+3 a^2 b (A-B)+3 a b^2 (A+B)-b^3 (A-B)\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}}{4 b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {\frac {a \left (3 a^3 B+a^2 A b+11 a b^2 B-7 A b^3\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\frac {2 a \left (3 a^5 B+a^4 A b+6 a^3 b^2 B+18 a^2 A b^3+35 a b^4 B-15 A b^5\right ) \int \frac {1}{a+b \tan (c+d x)}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}-\frac {16 b^2 \left (\frac {1}{2} \left (a^3 (A-B)+3 a^2 b (A+B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (-\left (a^3 (A+B)\right )+3 a^2 b (A-B)+3 a b^2 (A+B)-b^3 (A-B)\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}}{4 b \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{2 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {\frac {a \left (3 a^3 B+a^2 A b+11 a b^2 B-7 A b^3\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {\frac {2 \sqrt {a} \left (3 a^5 B+a^4 A b+6 a^3 b^2 B+18 a^2 A b^3+35 a b^4 B-15 A b^5\right ) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {b} d \left (a^2+b^2\right )}-\frac {16 b^2 \left (\frac {1}{2} \left (a^3 (A-B)+3 a^2 b (A+B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (-\left (a^3 (A+B)\right )+3 a^2 b (A-B)+3 a b^2 (A+B)-b^3 (A-B)\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{2 b \left (a^2+b^2\right )}}{4 b \left (a^2+b^2\right )}\)

Input:

Int[(Tan[c + d*x]^(5/2)*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^3,x]
 

Output:

(a*(A*b - a*B)*Tan[c + d*x]^(3/2))/(2*b*(a^2 + b^2)*d*(a + b*Tan[c + d*x]) 
^2) - (-1/2*((2*Sqrt[a]*(a^4*A*b + 18*a^2*A*b^3 - 15*A*b^5 + 3*a^5*B + 6*a 
^3*b^2*B + 35*a*b^4*B)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(Sqrt 
[b]*(a^2 + b^2)*d) - (16*b^2*(((a^3*(A - B) - 3*a*b^2*(A - B) + 3*a^2*b*(A 
 + B) - b^3*(A + B))*(-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]) + 
ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]))/2 + ((3*a^2*b*(A - B) - b 
^3*(A - B) - a^3*(A + B) + 3*a*b^2*(A + B))*(-1/2*Log[1 - Sqrt[2]*Sqrt[Tan 
[c + d*x]] + Tan[c + d*x]]/Sqrt[2] + Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + 
Tan[c + d*x]]/(2*Sqrt[2])))/2))/((a^2 + b^2)*d))/(b*(a^2 + b^2)) + (a*(a^2 
*A*b - 7*A*b^3 + 3*a^3*B + 11*a*b^2*B)*Sqrt[Tan[c + d*x]])/(b*(a^2 + b^2)* 
d*(a + b*Tan[c + d*x])))/(4*b*(a^2 + b^2))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

rule 1482

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
a*c, 2]}, Simp[(d*q + a*e)/(2*a*c)   Int[(q + c*x^2)/(a + c*x^4), x], x] + 
Simp[(d*q - a*e)/(2*a*c)   Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ[{a 
, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(- 
a)*c]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4017

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ 
)]], x_Symbol] :> Simp[2/f   Subst[Int[(b*c + d*x^2)/(b^2 + x^4), x], x, Sq 
rt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] & 
& NeQ[c^2 + d^2, 0]
 

rule 4088

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[(b*c - a*d)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x 
])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Simp[1/(d*(n + 1)*(c^2 + d^2)) 
  Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a*A*d* 
(b*d*(m - 1) - a*c*(n + 1)) + (b*B*c - (A*b + a*B)*d)*(b*c*(m - 1) + a*d*(n 
 + 1)) - d*((a*A - b*B)*(b*c - a*d) + (A*b + a*B)*(a*c + b*d))*(n + 1)*Tan[ 
e + f*x] - b*(d*(A*b*c + a*B*c - a*A*d)*(m + n) - b*B*(c^2*(m - 1) - d^2*(n 
 + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && 
 NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] & 
& LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
 

rule 4117

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) 
+ (f_.)*(x_)])^(n_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
 Simp[A/f   Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; 
FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]
 

rule 4128

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*d^2 + c*(c*C - B*d))*(a + b*Tan[e + 
 f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Sim 
p[1/(d*(n + 1)*(c^2 + d^2))   Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e 
 + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c*m + a*d* 
(n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b 
*(d*(B*c - A*d)*(m + n + 1) - C*(c^2*m - d^2*(n + 1)))*Tan[e + f*x]^2, x], 
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ 
[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
 

rule 4136

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) 
+ (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) 
+ (f_.)*(x_)]), x_Symbol] :> Simp[1/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^ 
n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Simp[ 
(A*b^2 - a*b*B + a^2*C)/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^n*((1 + Tan[ 
e + f*x]^2)/(a + b*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, 
 C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] & 
&  !GtQ[n, 0] &&  !LeQ[n, -1]
 

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 451, normalized size of antiderivative = 1.01

method	result	size
derivativedivides	\(\frac {\frac {2 a \left (\frac {\frac {\left (A \,a^{4} b +10 A \,a^{2} b^{3}+9 A \,b^{5}-5 B \,a^{5}-18 B \,a^{3} b^{2}-13 B a \,b^{4}\right ) \tan \left (d x +c \right )^{\frac {3}{2}}}{8 b}-\frac {a \left (A \,a^{4} b -6 A \,a^{2} b^{3}-7 A \,b^{5}+3 B \,a^{5}+14 B \,a^{3} b^{2}+11 B a \,b^{4}\right ) \sqrt {\tan \left (d x +c \right )}}{8 b^{2}}}{\left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {\left (A \,a^{4} b +18 A \,a^{2} b^{3}-15 A \,b^{5}+3 B \,a^{5}+6 B \,a^{3} b^{2}+35 B a \,b^{4}\right ) \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right )}{8 b^{2} \sqrt {a b}}\right )}{\left (a^{2}+b^{2}\right )^{3}}+\frac {\frac {\left (-3 A \,a^{2} b +A \,b^{3}+B \,a^{3}-3 B a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {\left (-A \,a^{3}+3 A a \,b^{2}-3 B \,a^{2} b +B \,b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}}{\left (a^{2}+b^{2}\right )^{3}}}{d}\)	\(451\)
default	\(\frac {\frac {2 a \left (\frac {\frac {\left (A \,a^{4} b +10 A \,a^{2} b^{3}+9 A \,b^{5}-5 B \,a^{5}-18 B \,a^{3} b^{2}-13 B a \,b^{4}\right ) \tan \left (d x +c \right )^{\frac {3}{2}}}{8 b}-\frac {a \left (A \,a^{4} b -6 A \,a^{2} b^{3}-7 A \,b^{5}+3 B \,a^{5}+14 B \,a^{3} b^{2}+11 B a \,b^{4}\right ) \sqrt {\tan \left (d x +c \right )}}{8 b^{2}}}{\left (a +b \tan \left (d x +c \right )\right )^{2}}+\frac {\left (A \,a^{4} b +18 A \,a^{2} b^{3}-15 A \,b^{5}+3 B \,a^{5}+6 B \,a^{3} b^{2}+35 B a \,b^{4}\right ) \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right )}{8 b^{2} \sqrt {a b}}\right )}{\left (a^{2}+b^{2}\right )^{3}}+\frac {\frac {\left (-3 A \,a^{2} b +A \,b^{3}+B \,a^{3}-3 B a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {\left (-A \,a^{3}+3 A a \,b^{2}-3 B \,a^{2} b +B \,b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}}{\left (a^{2}+b^{2}\right )^{3}}}{d}\)	\(451\)

Input:

int(tan(d*x+c)^(5/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x,method=_RETURNV 
ERBOSE)
 

Output:

1/d*(2*a/(a^2+b^2)^3*((1/8*(A*a^4*b+10*A*a^2*b^3+9*A*b^5-5*B*a^5-18*B*a^3* 
b^2-13*B*a*b^4)/b*tan(d*x+c)^(3/2)-1/8*a*(A*a^4*b-6*A*a^2*b^3-7*A*b^5+3*B* 
a^5+14*B*a^3*b^2+11*B*a*b^4)/b^2*tan(d*x+c)^(1/2))/(a+b*tan(d*x+c))^2+1/8* 
(A*a^4*b+18*A*a^2*b^3-15*A*b^5+3*B*a^5+6*B*a^3*b^2+35*B*a*b^4)/b^2/(a*b)^( 
1/2)*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/2)))+2/(a^2+b^2)^3*(1/8*(-3*A*a^2* 
b+A*b^3+B*a^3-3*B*a*b^2)*2^(1/2)*(ln((tan(d*x+c)+2^(1/2)*tan(d*x+c)^(1/2)+ 
1)/(tan(d*x+c)-2^(1/2)*tan(d*x+c)^(1/2)+1))+2*arctan(1+2^(1/2)*tan(d*x+c)^ 
(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))+1/8*(-A*a^3+3*A*a*b^2-3*B*a^ 
2*b+B*b^3)*2^(1/2)*(ln((tan(d*x+c)-2^(1/2)*tan(d*x+c)^(1/2)+1)/(tan(d*x+c) 
+2^(1/2)*tan(d*x+c)^(1/2)+1))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arcta 
n(-1+2^(1/2)*tan(d*x+c)^(1/2)))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3664 vs. \(2 (408) = 816\).

Time = 101.41 (sec) , antiderivative size = 7355, normalized size of antiderivative = 16.42 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx=\text {Too large to display} \] Input:

integrate(tan(d*x+c)^(5/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x, algorith 
m="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate(tan(d*x+c)**(5/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 551, normalized size of antiderivative = 1.23 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx=\frac {\frac {{\left (3 \, B a^{6} + A a^{5} b + 6 \, B a^{4} b^{2} + 18 \, A a^{3} b^{3} + 35 \, B a^{2} b^{4} - 15 \, A a b^{5}\right )} \arctan \left (\frac {b \sqrt {\tan \left (d x + c\right )}}{\sqrt {a b}}\right )}{{\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} \sqrt {a b}} - \frac {2 \, \sqrt {2} {\left ({\left (A - B\right )} a^{3} + 3 \, {\left (A + B\right )} a^{2} b - 3 \, {\left (A - B\right )} a b^{2} - {\left (A + B\right )} b^{3}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left ({\left (A - B\right )} a^{3} + 3 \, {\left (A + B\right )} a^{2} b - 3 \, {\left (A - B\right )} a b^{2} - {\left (A + B\right )} b^{3}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left ({\left (A + B\right )} a^{3} - 3 \, {\left (A - B\right )} a^{2} b - 3 \, {\left (A + B\right )} a b^{2} + {\left (A - B\right )} b^{3}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left ({\left (A + B\right )} a^{3} - 3 \, {\left (A - B\right )} a^{2} b - 3 \, {\left (A + B\right )} a b^{2} + {\left (A - B\right )} b^{3}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {{\left (5 \, B a^{4} b - A a^{3} b^{2} + 13 \, B a^{2} b^{3} - 9 \, A a b^{4}\right )} \tan \left (d x + c\right )^{\frac {3}{2}} + {\left (3 \, B a^{5} + A a^{4} b + 11 \, B a^{3} b^{2} - 7 \, A a^{2} b^{3}\right )} \sqrt {\tan \left (d x + c\right )}}{a^{6} b^{2} + 2 \, a^{4} b^{4} + a^{2} b^{6} + {\left (a^{4} b^{4} + 2 \, a^{2} b^{6} + b^{8}\right )} \tan \left (d x + c\right )^{2} + 2 \, {\left (a^{5} b^{3} + 2 \, a^{3} b^{5} + a b^{7}\right )} \tan \left (d x + c\right )}}{4 \, d} \] Input:

integrate(tan(d*x+c)^(5/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x, algorith 
m="maxima")
 

Output:

1/4*((3*B*a^6 + A*a^5*b + 6*B*a^4*b^2 + 18*A*a^3*b^3 + 35*B*a^2*b^4 - 15*A 
*a*b^5)*arctan(b*sqrt(tan(d*x + c))/sqrt(a*b))/((a^6*b^2 + 3*a^4*b^4 + 3*a 
^2*b^6 + b^8)*sqrt(a*b)) - (2*sqrt(2)*((A - B)*a^3 + 3*(A + B)*a^2*b - 3*( 
A - B)*a*b^2 - (A + B)*b^3)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + 
 c)))) + 2*sqrt(2)*((A - B)*a^3 + 3*(A + B)*a^2*b - 3*(A - B)*a*b^2 - (A + 
 B)*b^3)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) - sqrt(2)*( 
(A + B)*a^3 - 3*(A - B)*a^2*b - 3*(A + B)*a*b^2 + (A - B)*b^3)*log(sqrt(2) 
*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + sqrt(2)*((A + B)*a^3 - 3*(A - B) 
*a^2*b - 3*(A + B)*a*b^2 + (A - B)*b^3)*log(-sqrt(2)*sqrt(tan(d*x + c)) + 
tan(d*x + c) + 1))/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - ((5*B*a^4*b - A*a 
^3*b^2 + 13*B*a^2*b^3 - 9*A*a*b^4)*tan(d*x + c)^(3/2) + (3*B*a^5 + A*a^4*b 
 + 11*B*a^3*b^2 - 7*A*a^2*b^3)*sqrt(tan(d*x + c)))/(a^6*b^2 + 2*a^4*b^4 + 
a^2*b^6 + (a^4*b^4 + 2*a^2*b^6 + b^8)*tan(d*x + c)^2 + 2*(a^5*b^3 + 2*a^3* 
b^5 + a*b^7)*tan(d*x + c)))/d
 

Giac [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx=\text {Timed out} \] Input:

integrate(tan(d*x+c)^(5/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x, algorith 
m="giac")
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 52.53 (sec) , antiderivative size = 26614, normalized size of antiderivative = 59.41 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx=\text {Too large to display} \] Input:

int((tan(c + d*x)^(5/2)*(A + B*tan(c + d*x)))/(a + b*tan(c + d*x))^3,x)
 

Output:

(log((((((((((64*A*a*b^3*(11*a^2 - 13*b^2))/d + 128*b^3*tan(c + d*x)^(1/2) 
*(a^2 - b^2)*(a^2 + b^2)^2*((4*(-A^4*d^4*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4* 
b^2)^2)^(1/2) + 80*A^2*a^3*b^3*d^2 - 24*A^2*a*b^5*d^2 - 24*A^2*a^5*b*d^2)/ 
(d^4*(a^2 + b^2)^6))^(1/2))*((4*(-A^4*d^4*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4 
*b^2)^2)^(1/2) + 80*A^2*a^3*b^3*d^2 - 24*A^2*a*b^5*d^2 - 24*A^2*a^5*b*d^2) 
/(d^4*(a^2 + b^2)^6))^(1/2))/4 + (8*A^2*a*tan(c + d*x)^(1/2)*(a^10 - 184*b 
^10 + 833*a^2*b^8 - 812*a^4*b^6 + 262*a^6*b^4 + 44*a^8*b^2))/(d^2*(a^2 + b 
^2)^4))*((4*(-A^4*d^4*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2)^2)^(1/2) + 80* 
A^2*a^3*b^3*d^2 - 24*A^2*a*b^5*d^2 - 24*A^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^6) 
)^(1/2))/4 + (2*A^3*a^2*(5*a^10 - 1199*b^10 + 5017*a^2*b^8 - 5142*a^4*b^6 
+ 1106*a^6*b^4 + 181*a^8*b^2))/(d^3*(a^2 + b^2)^6))*((4*(-A^4*d^4*(a^6 - b 
^6 + 15*a^2*b^4 - 15*a^4*b^2)^2)^(1/2) + 80*A^2*a^3*b^3*d^2 - 24*A^2*a*b^5 
*d^2 - 24*A^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^6))^(1/2))/4 + (A^4*tan(c + d*x) 
^(1/2)*(a^14 - 32*b^14 + 97*a^2*b^12 - 2082*a^4*b^10 + 3631*a^6*b^8 - 2300 
*a^8*b^6 + 79*a^10*b^4 + 30*a^12*b^2))/(b*d^4*(a^2 + b^2)^8))*((4*(-A^4*d^ 
4*(a^6 - b^6 + 15*a^2*b^4 - 15*a^4*b^2)^2)^(1/2) + 80*A^2*a^3*b^3*d^2 - 24 
*A^2*a*b^5*d^2 - 24*A^2*a^5*b*d^2)/(d^4*(a^2 + b^2)^6))^(1/2))/4 + (A^5*a* 
(a^10 - 120*b^10 + 249*a^2*b^8 - 388*a^4*b^6 + 302*a^6*b^4 + 36*a^8*b^2))/ 
(2*b*d^5*(a^2 + b^2)^8))*(((480*A^4*a^2*b^10*d^4 - 16*A^4*b^12*d^4 - 16*A^ 
4*a^12*d^4 - 4080*A^4*a^4*b^8*d^4 + 7232*A^4*a^6*b^6*d^4 - 4080*A^4*a^8...
 

Reduce [F]

\[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx=\int \frac {\sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{2} b^{2}+2 \tan \left (d x +c \right ) a b +a^{2}}d x \] Input:

int(tan(d*x+c)^(5/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x)
 

Output:

int((sqrt(tan(c + d*x))*tan(c + d*x)**2)/(tan(c + d*x)**2*b**2 + 2*tan(c + 
 d*x)*a*b + a**2),x)