\(\int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx\) [404]

Optimal result

Integrand size = 33, antiderivative size = 364 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=\frac {\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}-\frac {\left (a^2 (A-B)-b^2 (A-B)+2 a b (A+B)\right ) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}+\frac {a^{3/2} \left (a^2 A b+5 A b^3-3 a^3 B-7 a b^2 B\right ) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{b^{5/2} \left (a^2+b^2\right )^2 d}-\frac {\left (2 a b (A-B)-a^2 (A+B)+b^2 (A+B)\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (c+d x)}}{1+\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^2 d}-\frac {\left (a A b-3 a^2 B-2 b^2 B\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right ) d}+\frac {a (A b-a B) \tan ^{\frac {3}{2}}(c+d x)}{b \left (a^2+b^2\right ) d (a+b \tan (c+d x))} \] Output:

-1/2*(a^2*(A-B)-b^2*(A-B)+2*a*b*(A+B))*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)) 
*2^(1/2)/(a^2+b^2)^2/d-1/2*(a^2*(A-B)-b^2*(A-B)+2*a*b*(A+B))*arctan(1+2^(1 
/2)*tan(d*x+c)^(1/2))*2^(1/2)/(a^2+b^2)^2/d+a^(3/2)*(A*a^2*b+5*A*b^3-3*B*a 
^3-7*B*a*b^2)*arctan(b^(1/2)*tan(d*x+c)^(1/2)/a^(1/2))/b^(5/2)/(a^2+b^2)^2 
/d-1/2*(2*a*b*(A-B)-a^2*(A+B)+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)^2/d-(A*a*b-3*B*a^2-2*B*b^2)*tan(d*x+c)^( 
1/2)/b^2/(a^2+b^2)/d+a*(A*b-B*a)*tan(d*x+c)^(3/2)/b/(a^2+b^2)/d/(a+b*tan(d 
*x+c))
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.60 (sec) , antiderivative size = 275, normalized size of antiderivative = 0.76 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=\frac {2 \left ((-A b+3 a B) \sqrt {\tan (c+d x)}+\frac {\left (a^2 A b+2 A b^3-3 a^3 B-4 a b^2 B\right ) \sqrt {\tan (c+d x)}}{2 \left (a^2+b^2\right )}+b B \tan ^{\frac {3}{2}}(c+d x)-\frac {\left (\sqrt [4]{-1} (a+i b)^2 b^{5/2} (i A+B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+a^{3/2} \left (-a^2 A b-5 A b^3+3 a^3 B+7 a b^2 B\right ) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )+(-1)^{3/4} b^{5/2} (i a+b)^2 (A+i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )\right ) (a+b \tan (c+d x))}{2 \sqrt {b} \left (a^2+b^2\right )^2}\right )}{b^2 d (a+b \tan (c+d x))} \] Input:

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

Output:

(2*((-(A*b) + 3*a*B)*Sqrt[Tan[c + d*x]] + ((a^2*A*b + 2*A*b^3 - 3*a^3*B - 
4*a*b^2*B)*Sqrt[Tan[c + d*x]])/(2*(a^2 + b^2)) + b*B*Tan[c + d*x]^(3/2) - 
(((-1)^(1/4)*(a + I*b)^2*b^(5/2)*(I*A + B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + 
d*x]]] + a^(3/2)*(-(a^2*A*b) - 5*A*b^3 + 3*a^3*B + 7*a*b^2*B)*ArcTan[(Sqrt 
[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]] + (-1)^(3/4)*b^(5/2)*(I*a + b)^2*(A + I*B 
)*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])*(a + b*Tan[c + d*x]))/(2*Sqrt[b] 
*(a^2 + b^2)^2)))/(b^2*d*(a + b*Tan[c + d*x]))
 

Rubi [A] (verified)

Time = 1.81 (sec) , antiderivative size = 379, 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, 4130, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4088

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4130

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4136

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4017

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1482

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

\(\Big \downarrow \) 4117

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

-1/2*(-(((2*a^(3/2)*(a^2*A*b + 5*A*b^3 - 3*a^3*B - 7*a*b^2*B)*ArcTan[(Sqrt 
[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(Sqrt[b]*(a^2 + b^2)*d) - (4*b^2*(((a^2* 
(A - B) - b^2*(A - B) + 2*a*b*(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 + ((2 
*a*b*(A - B) - a^2*(A + B) + 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]] + Ta 
n[c + d*x]]/(2*Sqrt[2])))/2))/((a^2 + b^2)*d))/b) + (2*(a*A*b - 3*a^2*B - 
2*b^2*B)*Sqrt[Tan[c + d*x]])/(b*d))/(b*(a^2 + b^2)) + (a*(A*b - a*B)*Tan[c 
 + d*x]^(3/2))/(b*(a^2 + b^2)*d*(a + b*Tan[c + d*x]))
 

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 4130

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[C*(a + b*Tan[e + f*x])^m*((c + d*Tan[ 
e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a 
+ b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C 
*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f*x] - (C* 
m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, 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[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[ 
c, 0] && NeQ[a, 0])))
 

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 = 0.12 (sec) , antiderivative size = 351, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {\frac {2 \sqrt {\tan \left (d x +c \right )}\, B}{b^{2}}+\frac {2 a^{2} \left (\frac {\left (-\frac {1}{2} A \,a^{2} b -\frac {1}{2} A \,b^{3}+\frac {1}{2} B \,a^{3}+\frac {1}{2} B a \,b^{2}\right ) \sqrt {\tan \left (d x +c \right )}}{a +b \tan \left (d x +c \right )}+\frac {\left (A \,a^{2} b +5 A \,b^{3}-3 B \,a^{3}-7 B a \,b^{2}\right ) \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{b^{2} \left (a^{2}+b^{2}\right )^{2}}+\frac {\frac {\left (-2 A a b +B \,a^{2}-B \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {\left (-A \,a^{2}+A \,b^{2}-2 B a b \right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}}{\left (a^{2}+b^{2}\right )^{2}}}{d}\)	\(351\)
default	\(\frac {\frac {2 \sqrt {\tan \left (d x +c \right )}\, B}{b^{2}}+\frac {2 a^{2} \left (\frac {\left (-\frac {1}{2} A \,a^{2} b -\frac {1}{2} A \,b^{3}+\frac {1}{2} B \,a^{3}+\frac {1}{2} B a \,b^{2}\right ) \sqrt {\tan \left (d x +c \right )}}{a +b \tan \left (d x +c \right )}+\frac {\left (A \,a^{2} b +5 A \,b^{3}-3 B \,a^{3}-7 B a \,b^{2}\right ) \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{b^{2} \left (a^{2}+b^{2}\right )^{2}}+\frac {\frac {\left (-2 A a b +B \,a^{2}-B \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {\left (-A \,a^{2}+A \,b^{2}-2 B a b \right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}}{\left (a^{2}+b^{2}\right )^{2}}}{d}\)	\(351\)

Input:

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

Output:

1/d*(2*tan(d*x+c)^(1/2)*B/b^2+2*a^2/b^2/(a^2+b^2)^2*((-1/2*A*a^2*b-1/2*A*b 
^3+1/2*B*a^3+1/2*B*a*b^2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))+1/2*(A*a^2*b+5 
*A*b^3-3*B*a^3-7*B*a*b^2)/(a*b)^(1/2)*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/2 
)))+2/(a^2+b^2)^2*(1/8*(-2*A*a*b+B*a^2-B*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^2+A*b^2-2*B*a*b)*2^(1/2)*(ln((tan(d*x+c)-2^(1/2)*tan(d*x+c)^(1/2)+1)/(t 
an(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)))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2436 vs. \(2 (335) = 670\).

Time = 33.24 (sec) , antiderivative size = 4897, normalized size of antiderivative = 13.45 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 377, normalized size of antiderivative = 1.04 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=-\frac {\frac {4 \, {\left (3 \, B a^{5} - A a^{4} b + 7 \, B a^{3} b^{2} - 5 \, A a^{2} b^{3}\right )} \arctan \left (\frac {b \sqrt {\tan \left (d x + c\right )}}{\sqrt {a b}}\right )}{{\left (a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}\right )} \sqrt {a b}} - \frac {4 \, {\left (B a^{3} - A a^{2} b\right )} \sqrt {\tan \left (d x + c\right )}}{a^{3} b^{2} + a b^{4} + {\left (a^{2} b^{3} + b^{5}\right )} \tan \left (d x + c\right )} + \frac {2 \, \sqrt {2} {\left ({\left (A - B\right )} a^{2} + 2 \, {\left (A + B\right )} a b - {\left (A - B\right )} b^{2}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left ({\left (A - B\right )} a^{2} + 2 \, {\left (A + B\right )} a b - {\left (A - B\right )} b^{2}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left ({\left (A + B\right )} a^{2} - 2 \, {\left (A - B\right )} a b - {\left (A + B\right )} b^{2}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left ({\left (A + B\right )} a^{2} - 2 \, {\left (A - B\right )} a b - {\left (A + B\right )} b^{2}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {8 \, B \sqrt {\tan \left (d x + c\right )}}{b^{2}}}{4 \, d} \] Input:

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

Output:

-1/4*(4*(3*B*a^5 - A*a^4*b + 7*B*a^3*b^2 - 5*A*a^2*b^3)*arctan(b*sqrt(tan( 
d*x + c))/sqrt(a*b))/((a^4*b^2 + 2*a^2*b^4 + b^6)*sqrt(a*b)) - 4*(B*a^3 - 
A*a^2*b)*sqrt(tan(d*x + c))/(a^3*b^2 + a*b^4 + (a^2*b^3 + b^5)*tan(d*x + c 
)) + (2*sqrt(2)*((A - B)*a^2 + 2*(A + B)*a*b - (A - B)*b^2)*arctan(1/2*sqr 
t(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 2*sqrt(2)*((A - B)*a^2 + 2*(A + B 
)*a*b - (A - B)*b^2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) 
 - sqrt(2)*((A + B)*a^2 - 2*(A - B)*a*b - (A + B)*b^2)*log(sqrt(2)*sqrt(ta 
n(d*x + c)) + tan(d*x + c) + 1) + sqrt(2)*((A + B)*a^2 - 2*(A - B)*a*b - ( 
A + B)*b^2)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1))/(a^4 + 2* 
a^2*b^2 + b^4) - 8*B*sqrt(tan(d*x + c))/b^2)/d
 

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument Type
 

Mupad [B] (verification not implemented)

Time = 35.53 (sec) , antiderivative size = 18313, normalized size of antiderivative = 50.31 \[ \int \frac {\tan ^{\frac {5}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

(log(((((((((128*b^3*tan(c + d*x)^(1/2)*(a^2 - b^2)*(a^2 + b^2)^2*((4*(-B^ 
4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) - 16*B^2*a*b^3*d^2 + 16*B^2*a^3*b*d 
^2)/(d^4*(a^2 + b^2)^4))^(1/2) - (128*B*a*b^2*(7*a^4 + b^4 + 8*a^2*b^2))/d 
)*((4*(-B^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) - 16*B^2*a*b^3*d^2 + 16*B 
^2*a^3*b*d^2)/(d^4*(a^2 + b^2)^4))^(1/2))/4 + (64*B^2*a*tan(c + d*x)^(1/2) 
*(18*a^10 - 15*b^10 + 17*a^2*b^8 - a^4*b^6 + 97*a^6*b^4 + 84*a^8*b^2))/(b^ 
2*d^2*(a^2 + b^2)^2))*((4*(-B^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) - 16* 
B^2*a*b^3*d^2 + 16*B^2*a^3*b*d^2)/(d^4*(a^2 + b^2)^4))^(1/2))/4 + (32*B^3* 
a^4*(127*a^2*b^6 - 112*b^8 - 9*a^8 + 173*a^4*b^4 + 21*a^6*b^2))/(b^3*d^3*( 
a^2 + b^2)^3))*((4*(-B^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) - 16*B^2*a*b 
^3*d^2 + 16*B^2*a^3*b*d^2)/(d^4*(a^2 + b^2)^4))^(1/2))/4 - (16*B^4*tan(c + 
 d*x)^(1/2)*(9*a^12 + 2*b^12 + 4*a^2*b^10 + 2*a^4*b^8 - 49*a^6*b^6 + 7*a^8 
*b^4 + 33*a^10*b^2))/(b^3*d^4*(a^2 + b^2)^4))*((4*(-B^4*d^4*(a^4 + b^4 - 6 
*a^2*b^2)^2)^(1/2) - 16*B^2*a*b^3*d^2 + 16*B^2*a^3*b*d^2)/(d^4*(a^2 + b^2) 
^4))^(1/2))/4 - (16*B^5*a^3*b^2*(3*a^2 + 7*b^2))/(d^5*(a^2 + b^2)^4))*(((1 
92*B^4*a^2*b^6*d^4 - 16*B^4*b^8*d^4 - 16*B^4*a^8*d^4 - 608*B^4*a^4*b^4*d^4 
 + 192*B^4*a^6*b^2*d^4)^(1/2) - 16*B^2*a*b^3*d^2 + 16*B^2*a^3*b*d^2)/(a^8* 
d^4 + b^8*d^4 + 4*a^2*b^6*d^4 + 6*a^4*b^4*d^4 + 4*a^6*b^2*d^4))^(1/2))/4 + 
 (log(((((((((128*b^3*tan(c + d*x)^(1/2)*(a^2 - b^2)*(a^2 + b^2)^2*(-(4*(- 
B^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) + 16*B^2*a*b^3*d^2 - 16*B^2*a^...
 

Reduce [F]

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

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

Output:

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