\(\int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx\) [394]

Optimal result

Integrand size = 33, antiderivative size = 304 \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (3 a^2 b (A-B)-b^3 (A-B)+a^3 (A+B)-3 a b^2 (A+B)\right ) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)-3 a^2 b (A+B)+b^3 (A+B)\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (c+d x)}}{1+\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {2 b \left (15 a A b+14 a^2 B-5 b^2 B\right ) \sqrt {\tan (c+d x)}}{5 d}+\frac {2 b^2 (5 A b+9 a B) \tan ^{\frac {3}{2}}(c+d x)}{15 d}+\frac {2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d} \] Output:

1/2*(3*a^2*b*(A-B)-b^3*(A-B)+a^3*(A+B)-3*a*b^2*(A+B))*arctan(-1+2^(1/2)*ta 
n(d*x+c)^(1/2))*2^(1/2)/d+1/2*(3*a^2*b*(A-B)-b^3*(A-B)+a^3*(A+B)-3*a*b^2*( 
A+B))*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)/d+1/2*(a^3*(A-B)-3*a*b^2* 
(A-B)-3*a^2*b*(A+B)+b^3*(A+B))*arctanh(2^(1/2)*tan(d*x+c)^(1/2)/(1+tan(d*x 
+c)))*2^(1/2)/d+2/5*b*(15*A*a*b+14*B*a^2-5*B*b^2)*tan(d*x+c)^(1/2)/d+2/15* 
b^2*(5*A*b+9*B*a)*tan(d*x+c)^(3/2)/d+2/5*b*B*tan(d*x+c)^(1/2)*(a+b*tan(d*x 
+c))^2/d
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.00 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.50 \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\frac {-15 \sqrt [4]{-1} (a-i b)^3 (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )-15 \sqrt [4]{-1} (a+i b)^3 (A+i B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+2 b \sqrt {\tan (c+d x)} \left (15 \left (3 a A b+3 a^2 B-b^2 B\right )+5 b (A b+3 a B) \tan (c+d x)+3 b^2 B \tan ^2(c+d x)\right )}{15 d} \] Input:

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

Output:

(-15*(-1)^(1/4)*(a - I*b)^3*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]] 
] - 15*(-1)^(1/4)*(a + I*b)^3*(A + I*B)*ArcTanh[(-1)^(3/4)*Sqrt[Tan[c + d* 
x]]] + 2*b*Sqrt[Tan[c + d*x]]*(15*(3*a*A*b + 3*a^2*B - b^2*B) + 5*b*(A*b + 
 3*a*B)*Tan[c + d*x] + 3*b^2*B*Tan[c + d*x]^2))/(15*d)
 

Rubi [A] (verified)

Time = 1.17 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.03, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.576, Rules used = {3042, 4090, 27, 3042, 4120, 27, 3042, 4113, 3042, 4017, 27, 1482, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4090

\(\displaystyle \frac {2}{5} \int \frac {(a+b \tan (c+d x)) \left (b (5 A b+9 a B) \tan ^2(c+d x)+5 \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a (5 a A-b B)\right )}{2 \sqrt {\tan (c+d x)}}dx+\frac {2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \int \frac {(a+b \tan (c+d x)) \left (b (5 A b+9 a B) \tan ^2(c+d x)+5 \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a (5 a A-b B)\right )}{\sqrt {\tan (c+d x)}}dx+\frac {2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \int \frac {(a+b \tan (c+d x)) \left (b (5 A b+9 a B) \tan (c+d x)^2+5 \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a (5 a A-b B)\right )}{\sqrt {\tan (c+d x)}}dx+\frac {2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d}\)

\(\Big \downarrow \) 4120

\(\displaystyle \frac {1}{5} \left (\frac {2 b^2 (9 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}-\frac {2}{3} \int -\frac {3 \left ((5 a A-b B) a^2+b \left (14 B a^2+15 A b a-5 b^2 B\right ) \tan ^2(c+d x)+5 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)\right )}{2 \sqrt {\tan (c+d x)}}dx\right )+\frac {2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\int \frac {(5 a A-b B) a^2+b \left (14 B a^2+15 A b a-5 b^2 B\right ) \tan ^2(c+d x)+5 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx+\frac {2 b^2 (9 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\int \frac {(5 a A-b B) a^2+b \left (14 B a^2+15 A b a-5 b^2 B\right ) \tan (c+d x)^2+5 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx+\frac {2 b^2 (9 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d}\)

\(\Big \downarrow \) 4113

\(\displaystyle \frac {1}{5} \left (\int \frac {5 \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right )+5 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx+\frac {2 b \left (14 a^2 B+15 a A b-5 b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b^2 (9 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\int \frac {5 \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B\right )+5 \left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx+\frac {2 b \left (14 a^2 B+15 a A b-5 b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b^2 (9 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d}\)

\(\Big \downarrow \) 4017

\(\displaystyle \frac {1}{5} \left (\frac {2 \int \frac {5 \left (A a^3-3 b B a^2-3 A b^2 a+b^3 B+\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)\right )}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}+\frac {2 b \left (14 a^2 B+15 a A b-5 b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b^2 (9 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\frac {10 \int \frac {A a^3-3 b B a^2-3 A b^2 a+b^3 B+\left (B a^3+3 A b a^2-3 b^2 B a-A b^3\right ) \tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d}+\frac {2 b \left (14 a^2 B+15 a A b-5 b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b^2 (9 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d}\)

\(\Big \downarrow \) 1482

\(\displaystyle \frac {1}{5} \left (\frac {10 \left (\frac {1}{2} \left (a^3 (A-B)-3 a^2 b (A+B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \left (a^3 (A+B)+3 a^2 b (A-B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \int \frac {\tan (c+d x)+1}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d}+\frac {2 b \left (14 a^2 B+15 a A b-5 b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b^2 (9 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {1}{5} \left (\frac {10 \left (\frac {1}{2} \left (a^3 (A-B)-3 a^2 b (A+B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \left (a^3 (A+B)+3 a^2 b (A-B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \left (\frac {1}{2} \int \frac {1}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \int \frac {1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )\right )}{d}+\frac {2 b \left (14 a^2 B+15 a A b-5 b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b^2 (9 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {1}{5} \left (\frac {10 \left (\frac {1}{2} \left (a^3 (A-B)-3 a^2 b (A+B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \left (a^3 (A+B)+3 a^2 b (A-B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \left (\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}\right )\right )}{d}+\frac {2 b \left (14 a^2 B+15 a A b-5 b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b^2 (9 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{5} \left (\frac {10 \left (\frac {1}{2} \left (a^3 (A-B)-3 a^2 b (A+B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \left (a^3 (A+B)+3 a^2 b (A-B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {2 b \left (14 a^2 B+15 a A b-5 b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b^2 (9 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {1}{5} \left (\frac {10 \left (\frac {1}{2} \left (a^3 (A-B)-3 a^2 b (A+B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (a^3 (A+B)+3 a^2 b (A-B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {2 b \left (14 a^2 B+15 a A b-5 b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b^2 (9 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{5} \left (\frac {10 \left (\frac {1}{2} \left (a^3 (A-B)-3 a^2 b (A+B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} \left (a^3 (A+B)+3 a^2 b (A-B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {2 b \left (14 a^2 B+15 a A b-5 b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b^2 (9 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\frac {10 \left (\frac {1}{2} \left (a^3 (A-B)-3 a^2 b (A+B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (c+d x)}+1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )+\frac {1}{2} \left (a^3 (A+B)+3 a^2 b (A-B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d}+\frac {2 b \left (14 a^2 B+15 a A b-5 b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {2 b^2 (9 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {1}{5} \left (\frac {2 b \left (14 a^2 B+15 a A b-5 b^2 B\right ) \sqrt {\tan (c+d x)}}{d}+\frac {10 \left (\frac {1}{2} \left (a^3 (A+B)+3 a^2 b (A-B)-3 a b^2 (A+B)-b^3 (A-B)\right ) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} \left (a^3 (A-B)-3 a^2 b (A+B)-3 a b^2 (A-B)+b^3 (A+B)\right ) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d}+\frac {2 b^2 (9 a B+5 A b) \tan ^{\frac {3}{2}}(c+d x)}{3 d}\right )+\frac {2 b B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^2}{5 d}\)

Input:

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

Output:

(2*b*B*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^2)/(5*d) + ((10*(((3*a^2*b* 
(A - B) - b^3*(A - B) + a^3*(A + B) - 3*a*b^2*(A + B))*(-(ArcTan[1 - Sqrt[ 
2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/S 
qrt[2]))/2 + ((a^3*(A - B) - 3*a*b^2*(A - B) - 3*a^2*b*(A + B) + b^3*(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))/d + (2*b 
*(15*a*A*b + 14*a^2*B - 5*b^2*B)*Sqrt[Tan[c + d*x]])/d + (2*b^2*(5*A*b + 9 
*a*B)*Tan[c + d*x]^(3/2))/(3*d))/5
 

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 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 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 4090

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

rule 4113

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

rule 4120

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.03

method	result	size
derivativedivides	\(\frac {\frac {2 B \,b^{3} \tan \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {2 A \,b^{3} \tan \left (d x +c \right )^{\frac {3}{2}}}{3}+2 B a \,b^{2} \tan \left (d x +c \right )^{\frac {3}{2}}+6 \sqrt {\tan \left (d x +c \right )}\, A a \,b^{2}+6 \sqrt {\tan \left (d x +c \right )}\, B \,a^{2} b -2 \sqrt {\tan \left (d x +c \right )}\, B \,b^{3}+\frac {\left (A \,a^{3}-3 A a \,b^{2}-3 B \,a^{2} b +B \,b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {\left (3 A \,a^{2} b -A \,b^{3}+B \,a^{3}-3 B a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}}{d}\)	\(314\)
default	\(\frac {\frac {2 B \,b^{3} \tan \left (d x +c \right )^{\frac {5}{2}}}{5}+\frac {2 A \,b^{3} \tan \left (d x +c \right )^{\frac {3}{2}}}{3}+2 B a \,b^{2} \tan \left (d x +c \right )^{\frac {3}{2}}+6 \sqrt {\tan \left (d x +c \right )}\, A a \,b^{2}+6 \sqrt {\tan \left (d x +c \right )}\, B \,a^{2} b -2 \sqrt {\tan \left (d x +c \right )}\, B \,b^{3}+\frac {\left (A \,a^{3}-3 A a \,b^{2}-3 B \,a^{2} b +B \,b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {\left (3 A \,a^{2} b -A \,b^{3}+B \,a^{3}-3 B a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}}{d}\)	\(314\)
parts	\(\frac {\left (A \,b^{3}+3 B a \,b^{2}\right ) \left (\frac {2 \tan \left (d x +c \right )^{\frac {3}{2}}}{3}-\frac {\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}\right )}{d}+\frac {\left (3 A a \,b^{2}+3 B \,a^{2} b \right ) \left (2 \sqrt {\tan \left (d x +c \right )}-\frac {\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}\right )}{d}+\frac {\left (3 A \,a^{2} b +B \,a^{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 d}+\frac {A \,a^{3} \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 d}+\frac {B \,b^{3} \left (\frac {2 \tan \left (d x +c \right )^{\frac {5}{2}}}{5}-2 \sqrt {\tan \left (d x +c \right )}+\frac {\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}\right )}{d}\)	\(537\)

Input:

int((a+b*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),x,method=_RETURNV 
ERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2051 vs. \(2 (275) = 550\).

Time = 0.15 (sec) , antiderivative size = 2051, normalized size of antiderivative = 6.75 \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\text {Too large to display} \] Input:

integrate((a+b*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),x, algorith 
m="fricas")
 

Output:

1/30*(30*sqrt(1/2)*d*sqrt(((A^2 + 2*A*B + B^2)*a^6 + 6*(A^2 - B^2)*a^5*b + 
 3*(A^2 - 10*A*B + B^2)*a^4*b^2 - 20*(A^2 - B^2)*a^3*b^3 + 3*(A^2 + 10*A*B 
 + B^2)*a^2*b^4 + 6*(A^2 - B^2)*a*b^5 + (A^2 - 2*A*B + B^2)*b^6)/d^2)*arct 
an(-(2*sqrt(1/2)*((A - B)*a^3 - 3*(A + B)*a^2*b - 3*(A - B)*a*b^2 + (A + B 
)*b^3)*d*sqrt(((A^2 + 2*A*B + B^2)*a^6 + 6*(A^2 - B^2)*a^5*b + 3*(A^2 - 10 
*A*B + B^2)*a^4*b^2 - 20*(A^2 - B^2)*a^3*b^3 + 3*(A^2 + 10*A*B + B^2)*a^2* 
b^4 + 6*(A^2 - B^2)*a*b^5 + (A^2 - 2*A*B + B^2)*b^6)/d^2)*sqrt(tan(d*x + c 
)) + d^2*sqrt(((A^2 + 2*A*B + B^2)*a^6 + 6*(A^2 - B^2)*a^5*b + 3*(A^2 - 10 
*A*B + B^2)*a^4*b^2 - 20*(A^2 - B^2)*a^3*b^3 + 3*(A^2 + 10*A*B + B^2)*a^2* 
b^4 + 6*(A^2 - B^2)*a*b^5 + (A^2 - 2*A*B + B^2)*b^6)/d^2)*sqrt(((A^2 - 2*A 
*B + B^2)*a^6 - 6*(A^2 - B^2)*a^5*b + 3*(A^2 + 10*A*B + B^2)*a^4*b^2 + 20* 
(A^2 - B^2)*a^3*b^3 + 3*(A^2 - 10*A*B + B^2)*a^2*b^4 - 6*(A^2 - B^2)*a*b^5 
 + (A^2 + 2*A*B + B^2)*b^6)/d^2))/(12*A*B*a^5*b - 40*A*B*a^3*b^3 + 12*A*B* 
a*b^5 - (A^2 - B^2)*a^6 + 15*(A^2 - B^2)*a^4*b^2 - 15*(A^2 - B^2)*a^2*b^4 
+ (A^2 - B^2)*b^6)) + 30*sqrt(1/2)*d*sqrt(((A^2 + 2*A*B + B^2)*a^6 + 6*(A^ 
2 - B^2)*a^5*b + 3*(A^2 - 10*A*B + B^2)*a^4*b^2 - 20*(A^2 - B^2)*a^3*b^3 + 
 3*(A^2 + 10*A*B + B^2)*a^2*b^4 + 6*(A^2 - B^2)*a*b^5 + (A^2 - 2*A*B + B^2 
)*b^6)/d^2)*arctan(-(2*sqrt(1/2)*((A - B)*a^3 - 3*(A + B)*a^2*b - 3*(A - B 
)*a*b^2 + (A + B)*b^3)*d*sqrt(((A^2 + 2*A*B + B^2)*a^6 + 6*(A^2 - B^2)*a^5 
*b + 3*(A^2 - 10*A*B + B^2)*a^4*b^2 - 20*(A^2 - B^2)*a^3*b^3 + 3*(A^2 +...
 

Sympy [F]

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

integrate((a+b*tan(d*x+c))**3*(A+B*tan(d*x+c))/tan(d*x+c)**(1/2),x)
 

Output:

Integral((A + B*tan(c + d*x))*(a + b*tan(c + d*x))**3/sqrt(tan(c + d*x)), 
x)
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.08 \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\frac {24 \, B b^{3} \tan \left (d x + c\right )^{\frac {5}{2}} + 30 \, \sqrt {2} {\left ({\left (A + B\right )} a^{3} + 3 \, {\left (A - B\right )} a^{2} b - 3 \, {\left (A + B\right )} a b^{2} - {\left (A - B\right )} b^{3}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 30 \, \sqrt {2} {\left ({\left (A + B\right )} a^{3} + 3 \, {\left (A - B\right )} a^{2} b - 3 \, {\left (A + B\right )} a b^{2} - {\left (A - B\right )} b^{3}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 15 \, \sqrt {2} {\left ({\left (A - B\right )} a^{3} - 3 \, {\left (A + B\right )} a^{2} b - 3 \, {\left (A - B\right )} a b^{2} + {\left (A + B\right )} b^{3}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - 15 \, \sqrt {2} {\left ({\left (A - B\right )} a^{3} - 3 \, {\left (A + B\right )} a^{2} b - 3 \, {\left (A - B\right )} a b^{2} + {\left (A + B\right )} b^{3}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 40 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \tan \left (d x + c\right )^{\frac {3}{2}} + 120 \, {\left (3 \, B a^{2} b + 3 \, A a b^{2} - B b^{3}\right )} \sqrt {\tan \left (d x + c\right )}}{60 \, d} \] Input:

integrate((a+b*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),x, algorith 
m="maxima")
 

Output:

1/60*(24*B*b^3*tan(d*x + c)^(5/2) + 30*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)))) + 30*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)))) + 
 15*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) - 15*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) + 40*(3*B*a*b^2 + A*b^3)*tan(d*x + c)^(3 
/2) + 120*(3*B*a^2*b + 3*A*a*b^2 - B*b^3)*sqrt(tan(d*x + c)))/d
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+b*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),x, algorith 
m="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [B] (verification not implemented)

Time = 9.60 (sec) , antiderivative size = 6657, normalized size of antiderivative = 21.90 \[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\text {Too large to display} \] Input:

int(((A + B*tan(c + d*x))*(a + b*tan(c + d*x))^3)/tan(c + d*x)^(1/2),x)
 

Output:

atan((((8*(4*A*a^3*d^2 - 12*A*a*b^2*d^2)*((5*A^2*a^3*b^3)/d^2 - (30*A^4*a^ 
2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a 
^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (3 
*A^2*a*b^5)/(2*d^2) - (3*A^2*a^5*b)/(2*d^2))^(1/2))/d^3 - (16*tan(c + d*x) 
^(1/2)*(A^2*a^6 - A^2*b^6 + 15*A^2*a^2*b^4 - 15*A^2*a^4*b^2))/d^2)*((5*A^2 
*a^3*b^3)/d^2 - (30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A 
^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b 
^2*d^4)^(1/2)/(4*d^4) - (3*A^2*a*b^5)/(2*d^2) - (3*A^2*a^5*b)/(2*d^2))^(1/ 
2)*1i - ((8*(4*A*a^3*d^2 - 12*A*a*b^2*d^2)*((5*A^2*a^3*b^3)/d^2 - (30*A^4* 
a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4 
*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - 
(3*A^2*a*b^5)/(2*d^2) - (3*A^2*a^5*b)/(2*d^2))^(1/2))/d^3 + (16*tan(c + d* 
x)^(1/2)*(A^2*a^6 - A^2*b^6 + 15*A^2*a^2*b^4 - 15*A^2*a^4*b^2))/d^2)*((5*A 
^2*a^3*b^3)/d^2 - (30*A^4*a^2*b^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255 
*A^4*a^4*b^8*d^4 + 452*A^4*a^6*b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10 
*b^2*d^4)^(1/2)/(4*d^4) - (3*A^2*a*b^5)/(2*d^2) - (3*A^2*a^5*b)/(2*d^2))^( 
1/2)*1i)/((16*(3*A^3*a^8*b - A^3*b^9 + 6*A^3*a^4*b^5 + 8*A^3*a^6*b^3))/d^3 
 + ((8*(4*A*a^3*d^2 - 12*A*a*b^2*d^2)*((5*A^2*a^3*b^3)/d^2 - (30*A^4*a^2*b 
^10*d^4 - A^4*b^12*d^4 - A^4*a^12*d^4 - 255*A^4*a^4*b^8*d^4 + 452*A^4*a^6* 
b^6*d^4 - 255*A^4*a^8*b^4*d^4 + 30*A^4*a^10*b^2*d^4)^(1/2)/(4*d^4) - (3...
 

Reduce [F]

\[ \int \frac {(a+b \tan (c+d x))^3 (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\frac {2 \sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{2} b^{4}+60 \sqrt {\tan \left (d x +c \right )}\, a^{2} b^{2}-10 \sqrt {\tan \left (d x +c \right )}\, b^{4}+5 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}}{\tan \left (d x +c \right )}d x \right ) a^{4} d -30 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}}{\tan \left (d x +c \right )}d x \right ) a^{2} b^{2} d +5 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}}{\tan \left (d x +c \right )}d x \right ) b^{4} d +20 \left (\int \sqrt {\tan \left (d x +c \right )}d x \right ) a^{3} b d +20 \left (\int \sqrt {\tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{2}d x \right ) a \,b^{3} d}{5 d} \] Input:

int((a+b*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),x)
 

Output:

(2*sqrt(tan(c + d*x))*tan(c + d*x)**2*b**4 + 60*sqrt(tan(c + d*x))*a**2*b* 
*2 - 10*sqrt(tan(c + d*x))*b**4 + 5*int(sqrt(tan(c + d*x))/tan(c + d*x),x) 
*a**4*d - 30*int(sqrt(tan(c + d*x))/tan(c + d*x),x)*a**2*b**2*d + 5*int(sq 
rt(tan(c + d*x))/tan(c + d*x),x)*b**4*d + 20*int(sqrt(tan(c + d*x)),x)*a** 
3*b*d + 20*int(sqrt(tan(c + d*x))*tan(c + d*x)**2,x)*a*b**3*d)/(5*d)