\(\int \frac {1}{(d \sec (e+f x))^{5/3} (a+b \tan (e+f x))} \, dx\) [643]

Optimal result

Integrand size = 25, antiderivative size = 581 \[ \int \frac {1}{(d \sec (e+f x))^{5/3} (a+b \tan (e+f x))} \, dx=\frac {3 b}{5 \left (a^2+b^2\right ) f (d \sec (e+f x))^{5/3}}+\frac {\sqrt {3} b^{8/3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt {3} \sqrt [6]{a^2+b^2}}\right ) \sec ^2(e+f x)^{5/6}}{2 \left (a^2+b^2\right )^{11/6} f (d \sec (e+f x))^{5/3}}-\frac {\sqrt {3} b^{8/3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt {3} \sqrt [6]{a^2+b^2}}\right ) \sec ^2(e+f x)^{5/6}}{2 \left (a^2+b^2\right )^{11/6} f (d \sec (e+f x))^{5/3}}-\frac {b^{8/3} \text {arctanh}\left (\frac {\sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt [6]{a^2+b^2}}\right ) \sec ^2(e+f x)^{5/6}}{\left (a^2+b^2\right )^{11/6} f (d \sec (e+f x))^{5/3}}+\frac {b^{8/3} \log \left (\sqrt [3]{a^2+b^2}-\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\sec ^2(e+f x)}+b^{2/3} \sqrt [3]{\sec ^2(e+f x)}\right ) \sec ^2(e+f x)^{5/6}}{4 \left (a^2+b^2\right )^{11/6} f (d \sec (e+f x))^{5/3}}-\frac {b^{8/3} \log \left (\sqrt [3]{a^2+b^2}+\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\sec ^2(e+f x)}+b^{2/3} \sqrt [3]{\sec ^2(e+f x)}\right ) \sec ^2(e+f x)^{5/6}}{4 \left (a^2+b^2\right )^{11/6} f (d \sec (e+f x))^{5/3}}+\frac {\operatorname {AppellF1}\left (\frac {1}{2},1,\frac {11}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{5/6} \tan (e+f x)}{a f (d \sec (e+f x))^{5/3}} \] Output:

3/5*b/(a^2+b^2)/f/(d*sec(f*x+e))^(5/3)-1/2*3^(1/2)*b^(8/3)*arctan(-1/3*3^( 
1/2)+2/3*b^(1/3)*(sec(f*x+e)^2)^(1/6)*3^(1/2)/(a^2+b^2)^(1/6))*(sec(f*x+e) 
^2)^(5/6)/(a^2+b^2)^(11/6)/f/(d*sec(f*x+e))^(5/3)-1/2*3^(1/2)*b^(8/3)*arct 
an(1/3*3^(1/2)+2/3*b^(1/3)*(sec(f*x+e)^2)^(1/6)*3^(1/2)/(a^2+b^2)^(1/6))*( 
sec(f*x+e)^2)^(5/6)/(a^2+b^2)^(11/6)/f/(d*sec(f*x+e))^(5/3)-b^(8/3)*arctan 
h(b^(1/3)*(sec(f*x+e)^2)^(1/6)/(a^2+b^2)^(1/6))*(sec(f*x+e)^2)^(5/6)/(a^2+ 
b^2)^(11/6)/f/(d*sec(f*x+e))^(5/3)+1/4*b^(8/3)*ln((a^2+b^2)^(1/3)-b^(1/3)* 
(a^2+b^2)^(1/6)*(sec(f*x+e)^2)^(1/6)+b^(2/3)*(sec(f*x+e)^2)^(1/3))*(sec(f* 
x+e)^2)^(5/6)/(a^2+b^2)^(11/6)/f/(d*sec(f*x+e))^(5/3)-1/4*b^(8/3)*ln((a^2+ 
b^2)^(1/3)+b^(1/3)*(a^2+b^2)^(1/6)*(sec(f*x+e)^2)^(1/6)+b^(2/3)*(sec(f*x+e 
)^2)^(1/3))*(sec(f*x+e)^2)^(5/6)/(a^2+b^2)^(11/6)/f/(d*sec(f*x+e))^(5/3)+A 
ppellF1(1/2,1,11/6,3/2,b^2*tan(f*x+e)^2/a^2,-tan(f*x+e)^2)*(sec(f*x+e)^2)^ 
(5/6)*tan(f*x+e)/a/f/(d*sec(f*x+e))^(5/3)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(6862\) vs. \(2(581)=1162\).

Time = 103.42 (sec) , antiderivative size = 6862, normalized size of antiderivative = 11.81 \[ \int \frac {1}{(d \sec (e+f x))^{5/3} (a+b \tan (e+f x))} \, dx=\text {Result too large to show} \] Input:

Integrate[1/((d*Sec[e + f*x])^(5/3)*(a + b*Tan[e + f*x])),x]
 

Output:

Result too large to show
 

Rubi [A] (warning: unable to verify)

Time = 0.80 (sec) , antiderivative size = 415, normalized size of antiderivative = 0.71, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.640, Rules used = {3042, 3994, 504, 333, 353, 61, 73, 754, 27, 221, 1142, 25, 27, 1082, 217, 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.

\(\displaystyle \int \frac {1}{(d \sec (e+f x))^{5/3} (a+b \tan (e+f x))} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(d \sec (e+f x))^{5/3} (a+b \tan (e+f x))}dx\)

\(\Big \downarrow \) 3994

\(\displaystyle \frac {\sec ^2(e+f x)^{5/6} \int \frac {1}{(a+b \tan (e+f x)) \left (\tan ^2(e+f x)+1\right )^{11/6}}d(b \tan (e+f x))}{b f (d \sec (e+f x))^{5/3}}\)

\(\Big \downarrow \) 504

\(\displaystyle \frac {\sec ^2(e+f x)^{5/6} \left (a \int \frac {1}{\left (\tan ^2(e+f x)+1\right )^{11/6} \left (a^2-b^2 \tan ^2(e+f x)\right )}d(b \tan (e+f x))-\int \frac {b \tan (e+f x)}{\left (\tan ^2(e+f x)+1\right )^{11/6} \left (a^2-b^2 \tan ^2(e+f x)\right )}d(b \tan (e+f x))\right )}{b f (d \sec (e+f x))^{5/3}}\)

\(\Big \downarrow \) 333

\(\displaystyle \frac {\sec ^2(e+f x)^{5/6} \left (\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {11}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a}-\int \frac {b \tan (e+f x)}{\left (\tan ^2(e+f x)+1\right )^{11/6} \left (a^2-b^2 \tan ^2(e+f x)\right )}d(b \tan (e+f x))\right )}{b f (d \sec (e+f x))^{5/3}}\)

\(\Big \downarrow \) 353

\(\displaystyle \frac {\sec ^2(e+f x)^{5/6} \left (\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {11}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a}-\frac {1}{2} \int \frac {1}{\left (\frac {\tan (e+f x)}{b}+1\right )^{11/6} \left (a^2-b^2 \tan ^2(e+f x)\right )}d\left (b^2 \tan ^2(e+f x)\right )\right )}{b f (d \sec (e+f x))^{5/3}}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {\sec ^2(e+f x)^{5/6} \left (\frac {1}{2} \left (\frac {6 b^2}{5 \left (a^2+b^2\right ) \left (\frac {\tan (e+f x)}{b}+1\right )^{5/6}}-\frac {b^2 \int \frac {1}{\left (\frac {\tan (e+f x)}{b}+1\right )^{5/6} \left (a^2-b^2 \tan ^2(e+f x)\right )}d\left (b^2 \tan ^2(e+f x)\right )}{a^2+b^2}\right )+\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {11}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a}\right )}{b f (d \sec (e+f x))^{5/3}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\sec ^2(e+f x)^{5/6} \left (\frac {1}{2} \left (\frac {6 b^2}{5 \left (a^2+b^2\right ) \left (\frac {\tan (e+f x)}{b}+1\right )^{5/6}}-\frac {6 b^4 \int \frac {1}{-\tan ^6(e+f x) b^8+b^2+a^2}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{a^2+b^2}\right )+\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {11}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a}\right )}{b f (d \sec (e+f x))^{5/3}}\)

\(\Big \downarrow \) 754

\(\displaystyle \frac {\sec ^2(e+f x)^{5/6} \left (\frac {1}{2} \left (\frac {6 b^2}{5 \left (a^2+b^2\right ) \left (\frac {\tan (e+f x)}{b}+1\right )^{5/6}}-\frac {6 b^4 \left (\frac {\int \frac {1}{\sqrt [3]{a^2+b^2}-b^{8/3} \tan ^2(e+f x)}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{3 \left (a^2+b^2\right )^{2/3}}+\frac {\int \frac {2 \sqrt [6]{a^2+b^2}-b^{4/3} \tan (e+f x)}{2 \left (\tan ^2(e+f x) b^{8/3}-\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}\right )}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{3 \left (a^2+b^2\right )^{5/6}}+\frac {\int \frac {\tan (e+f x) b^{4/3}+2 \sqrt [6]{a^2+b^2}}{2 \left (\tan ^2(e+f x) b^{8/3}+\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}\right )}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{3 \left (a^2+b^2\right )^{5/6}}\right )}{a^2+b^2}\right )+\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {11}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a}\right )}{b f (d \sec (e+f x))^{5/3}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sec ^2(e+f x)^{5/6} \left (\frac {1}{2} \left (\frac {6 b^2}{5 \left (a^2+b^2\right ) \left (\frac {\tan (e+f x)}{b}+1\right )^{5/6}}-\frac {6 b^4 \left (\frac {\int \frac {1}{\sqrt [3]{a^2+b^2}-b^{8/3} \tan ^2(e+f x)}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{3 \left (a^2+b^2\right )^{2/3}}+\frac {\int \frac {2 \sqrt [6]{a^2+b^2}-b^{4/3} \tan (e+f x)}{\tan ^2(e+f x) b^{8/3}-\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{6 \left (a^2+b^2\right )^{5/6}}+\frac {\int \frac {\tan (e+f x) b^{4/3}+2 \sqrt [6]{a^2+b^2}}{\tan ^2(e+f x) b^{8/3}+\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{6 \left (a^2+b^2\right )^{5/6}}\right )}{a^2+b^2}\right )+\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {11}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a}\right )}{b f (d \sec (e+f x))^{5/3}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\sec ^2(e+f x)^{5/6} \left (\frac {1}{2} \left (\frac {6 b^2}{5 \left (a^2+b^2\right ) \left (\frac {\tan (e+f x)}{b}+1\right )^{5/6}}-\frac {6 b^4 \left (\frac {\int \frac {2 \sqrt [6]{a^2+b^2}-b^{4/3} \tan (e+f x)}{\tan ^2(e+f x) b^{8/3}-\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{6 \left (a^2+b^2\right )^{5/6}}+\frac {\int \frac {\tan (e+f x) b^{4/3}+2 \sqrt [6]{a^2+b^2}}{\tan ^2(e+f x) b^{8/3}+\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{6 \left (a^2+b^2\right )^{5/6}}+\frac {\text {arctanh}\left (\frac {b^{4/3} \tan (e+f x)}{\sqrt [6]{a^2+b^2}}\right )}{3 \sqrt [3]{b} \left (a^2+b^2\right )^{5/6}}\right )}{a^2+b^2}\right )+\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {11}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a}\right )}{b f (d \sec (e+f x))^{5/3}}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {\sec ^2(e+f x)^{5/6} \left (\frac {1}{2} \left (\frac {6 b^2}{5 \left (a^2+b^2\right ) \left (\frac {\tan (e+f x)}{b}+1\right )^{5/6}}-\frac {6 b^4 \left (\frac {\frac {3}{2} \sqrt [6]{a^2+b^2} \int \frac {1}{\tan ^2(e+f x) b^{8/3}-\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}-\frac {\int -\frac {\sqrt [3]{b} \left (\sqrt [6]{a^2+b^2}-2 b^{4/3} \tan (e+f x)\right )}{\tan ^2(e+f x) b^{8/3}-\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{2 \sqrt [3]{b}}}{6 \left (a^2+b^2\right )^{5/6}}+\frac {\frac {3}{2} \sqrt [6]{a^2+b^2} \int \frac {1}{\tan ^2(e+f x) b^{8/3}+\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}+\frac {\int \frac {\sqrt [3]{b} \left (2 \tan (e+f x) b^{4/3}+\sqrt [6]{a^2+b^2}\right )}{\tan ^2(e+f x) b^{8/3}+\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{2 \sqrt [3]{b}}}{6 \left (a^2+b^2\right )^{5/6}}+\frac {\text {arctanh}\left (\frac {b^{4/3} \tan (e+f x)}{\sqrt [6]{a^2+b^2}}\right )}{3 \sqrt [3]{b} \left (a^2+b^2\right )^{5/6}}\right )}{a^2+b^2}\right )+\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {11}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a}\right )}{b f (d \sec (e+f x))^{5/3}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\sec ^2(e+f x)^{5/6} \left (\frac {1}{2} \left (\frac {6 b^2}{5 \left (a^2+b^2\right ) \left (\frac {\tan (e+f x)}{b}+1\right )^{5/6}}-\frac {6 b^4 \left (\frac {\frac {3}{2} \sqrt [6]{a^2+b^2} \int \frac {1}{\tan ^2(e+f x) b^{8/3}-\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}+\frac {\int \frac {\sqrt [3]{b} \left (\sqrt [6]{a^2+b^2}-2 b^{4/3} \tan (e+f x)\right )}{\tan ^2(e+f x) b^{8/3}-\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{2 \sqrt [3]{b}}}{6 \left (a^2+b^2\right )^{5/6}}+\frac {\frac {3}{2} \sqrt [6]{a^2+b^2} \int \frac {1}{\tan ^2(e+f x) b^{8/3}+\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}+\frac {\int \frac {\sqrt [3]{b} \left (2 \tan (e+f x) b^{4/3}+\sqrt [6]{a^2+b^2}\right )}{\tan ^2(e+f x) b^{8/3}+\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{2 \sqrt [3]{b}}}{6 \left (a^2+b^2\right )^{5/6}}+\frac {\text {arctanh}\left (\frac {b^{4/3} \tan (e+f x)}{\sqrt [6]{a^2+b^2}}\right )}{3 \sqrt [3]{b} \left (a^2+b^2\right )^{5/6}}\right )}{a^2+b^2}\right )+\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {11}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a}\right )}{b f (d \sec (e+f x))^{5/3}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sec ^2(e+f x)^{5/6} \left (\frac {1}{2} \left (\frac {6 b^2}{5 \left (a^2+b^2\right ) \left (\frac {\tan (e+f x)}{b}+1\right )^{5/6}}-\frac {6 b^4 \left (\frac {\frac {3}{2} \sqrt [6]{a^2+b^2} \int \frac {1}{\tan ^2(e+f x) b^{8/3}-\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}+\frac {1}{2} \int \frac {\sqrt [6]{a^2+b^2}-2 b^{4/3} \tan (e+f x)}{\tan ^2(e+f x) b^{8/3}-\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{6 \left (a^2+b^2\right )^{5/6}}+\frac {\frac {3}{2} \sqrt [6]{a^2+b^2} \int \frac {1}{\tan ^2(e+f x) b^{8/3}+\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}+\frac {1}{2} \int \frac {2 \tan (e+f x) b^{4/3}+\sqrt [6]{a^2+b^2}}{\tan ^2(e+f x) b^{8/3}+\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{6 \left (a^2+b^2\right )^{5/6}}+\frac {\text {arctanh}\left (\frac {b^{4/3} \tan (e+f x)}{\sqrt [6]{a^2+b^2}}\right )}{3 \sqrt [3]{b} \left (a^2+b^2\right )^{5/6}}\right )}{a^2+b^2}\right )+\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {11}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a}\right )}{b f (d \sec (e+f x))^{5/3}}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {\sec ^2(e+f x)^{5/6} \left (\frac {1}{2} \left (\frac {6 b^2}{5 \left (a^2+b^2\right ) \left (\frac {\tan (e+f x)}{b}+1\right )^{5/6}}-\frac {6 b^4 \left (\frac {\frac {1}{2} \int \frac {\sqrt [6]{a^2+b^2}-2 b^{4/3} \tan (e+f x)}{\tan ^2(e+f x) b^{8/3}-\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}+\frac {3 \int \frac {1}{\frac {2 b^{4/3} \tan (e+f x)}{\sqrt [6]{a^2+b^2}}-4}d\left (1-\frac {2 b^{4/3} \tan (e+f x)}{\sqrt [6]{a^2+b^2}}\right )}{\sqrt [3]{b}}}{6 \left (a^2+b^2\right )^{5/6}}+\frac {\frac {1}{2} \int \frac {2 \tan (e+f x) b^{4/3}+\sqrt [6]{a^2+b^2}}{\tan ^2(e+f x) b^{8/3}+\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}-\frac {3 \int \frac {1}{-\frac {2 \tan (e+f x) b^{4/3}}{\sqrt [6]{a^2+b^2}}-4}d\left (\frac {2 \tan (e+f x) b^{4/3}}{\sqrt [6]{a^2+b^2}}+1\right )}{\sqrt [3]{b}}}{6 \left (a^2+b^2\right )^{5/6}}+\frac {\text {arctanh}\left (\frac {b^{4/3} \tan (e+f x)}{\sqrt [6]{a^2+b^2}}\right )}{3 \sqrt [3]{b} \left (a^2+b^2\right )^{5/6}}\right )}{a^2+b^2}\right )+\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {11}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a}\right )}{b f (d \sec (e+f x))^{5/3}}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\sec ^2(e+f x)^{5/6} \left (\frac {1}{2} \left (\frac {6 b^2}{5 \left (a^2+b^2\right ) \left (\frac {\tan (e+f x)}{b}+1\right )^{5/6}}-\frac {6 b^4 \left (\frac {\frac {1}{2} \int \frac {\sqrt [6]{a^2+b^2}-2 b^{4/3} \tan (e+f x)}{\tan ^2(e+f x) b^{8/3}-\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 b^{4/3} \tan (e+f x)}{\sqrt [6]{a^2+b^2}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{6 \left (a^2+b^2\right )^{5/6}}+\frac {\frac {1}{2} \int \frac {2 \tan (e+f x) b^{4/3}+\sqrt [6]{a^2+b^2}}{\tan ^2(e+f x) b^{8/3}+\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}+\frac {\sqrt {3} \arctan \left (\frac {\frac {2 b^{4/3} \tan (e+f x)}{\sqrt [6]{a^2+b^2}}+1}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{6 \left (a^2+b^2\right )^{5/6}}+\frac {\text {arctanh}\left (\frac {b^{4/3} \tan (e+f x)}{\sqrt [6]{a^2+b^2}}\right )}{3 \sqrt [3]{b} \left (a^2+b^2\right )^{5/6}}\right )}{a^2+b^2}\right )+\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {11}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a}\right )}{b f (d \sec (e+f x))^{5/3}}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {\sec ^2(e+f x)^{5/6} \left (\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {11}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a}+\frac {1}{2} \left (\frac {6 b^2}{5 \left (a^2+b^2\right ) \left (\frac {\tan (e+f x)}{b}+1\right )^{5/6}}-\frac {6 b^4 \left (\frac {-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 b^{4/3} \tan (e+f x)}{\sqrt [6]{a^2+b^2}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a^2+b^2}-b^{4/3} \sqrt [6]{a^2+b^2} \tan (e+f x)+b^{8/3} \tan ^2(e+f x)\right )}{2 \sqrt [3]{b}}}{6 \left (a^2+b^2\right )^{5/6}}+\frac {\frac {\sqrt {3} \arctan \left (\frac {\frac {2 b^{4/3} \tan (e+f x)}{\sqrt [6]{a^2+b^2}}+1}{\sqrt {3}}\right )}{\sqrt [3]{b}}+\frac {\log \left (\sqrt [3]{a^2+b^2}+b^{4/3} \sqrt [6]{a^2+b^2} \tan (e+f x)+b^{8/3} \tan ^2(e+f x)\right )}{2 \sqrt [3]{b}}}{6 \left (a^2+b^2\right )^{5/6}}+\frac {\text {arctanh}\left (\frac {b^{4/3} \tan (e+f x)}{\sqrt [6]{a^2+b^2}}\right )}{3 \sqrt [3]{b} \left (a^2+b^2\right )^{5/6}}\right )}{a^2+b^2}\right )\right )}{b f (d \sec (e+f x))^{5/3}}\)

Input:

Int[1/((d*Sec[e + f*x])^(5/3)*(a + b*Tan[e + f*x])),x]
 

Output:

((Sec[e + f*x]^2)^(5/6)*((b*AppellF1[1/2, 1, 11/6, 3/2, (b^2*Tan[e + f*x]^ 
2)/a^2, -Tan[e + f*x]^2]*Tan[e + f*x])/a + ((-6*b^4*(ArcTanh[(b^(4/3)*Tan[ 
e + f*x])/(a^2 + b^2)^(1/6)]/(3*b^(1/3)*(a^2 + b^2)^(5/6)) + (-((Sqrt[3]*A 
rcTan[(1 - (2*b^(4/3)*Tan[e + f*x])/(a^2 + b^2)^(1/6))/Sqrt[3]])/b^(1/3)) 
- Log[(a^2 + b^2)^(1/3) - b^(4/3)*(a^2 + b^2)^(1/6)*Tan[e + f*x] + b^(8/3) 
*Tan[e + f*x]^2]/(2*b^(1/3)))/(6*(a^2 + b^2)^(5/6)) + ((Sqrt[3]*ArcTan[(1 
+ (2*b^(4/3)*Tan[e + f*x])/(a^2 + b^2)^(1/6))/Sqrt[3]])/b^(1/3) + Log[(a^2 
 + b^2)^(1/3) + b^(4/3)*(a^2 + b^2)^(1/6)*Tan[e + f*x] + b^(8/3)*Tan[e + f 
*x]^2]/(2*b^(1/3)))/(6*(a^2 + b^2)^(5/6))))/(a^2 + b^2) + (6*b^2)/(5*(a^2 
+ b^2)*(1 + Tan[e + f*x]/b)^(5/6)))/2))/(b*f*(d*Sec[e + f*x])^(5/3))
 

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 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, 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 221

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

rule 333

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F 
reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 
0]) && (IntegerQ[q] || GtQ[c, 0])
 

rule 353

Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] 
 :> Simp[1/2   Subst[Int[(a + b*x)^p*(c + d*x)^q, x], x, x^2], x] /; FreeQ[ 
{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0]
 

rule 504

Int[((a_) + (b_.)*(x_)^2)^(p_)/((c_) + (d_.)*(x_)), x_Symbol] :> Simp[c   I 
nt[(a + b*x^2)^p/(c^2 - d^2*x^2), x], x] - Simp[d   Int[x*((a + b*x^2)^p/(c 
^2 - d^2*x^2)), x], x] /; FreeQ[{a, b, c, d, p}, x]
 

rule 754

Int[((a_) + (b_.)*(x_)^(n_))^(-1), x_Symbol] :> Module[{r = Numerator[Rt[-a 
/b, n]], s = Denominator[Rt[-a/b, n]], k, u}, Simp[u = Int[(r - s*Cos[(2*k* 
Pi)/n]*x)/(r^2 - 2*r*s*Cos[(2*k*Pi)/n]*x + s^2*x^2), x] + Int[(r + s*Cos[(2 
*k*Pi)/n]*x)/(r^2 + 2*r*s*Cos[(2*k*Pi)/n]*x + s^2*x^2), x]; 2*(r^2/(a*n)) 
 Int[1/(r^2 - s^2*x^2), x] + 2*(r/(a*n))   Sum[u, {k, 1, (n - 2)/4}], x]] / 
; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && NegQ[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 1142

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 3042

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

rule 3994

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( 
x_)])^(n_), x_Symbol] :> Simp[d^(2*IntPart[m/2])*((d*Sec[e + f*x])^(2*FracP 
art[m/2])/(b*f*(Sec[e + f*x]^2)^FracPart[m/2]))   Subst[Int[(a + x)^n*(1 + 
x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, 
n}, x] && NeQ[a^2 + b^2, 0] &&  !IntegerQ[m] && IntegerQ[n]
 

Maple [F]

\[\int \frac {1}{\left (d \sec \left (f x +e \right )\right )^{\frac {5}{3}} \left (a +b \tan \left (f x +e \right )\right )}d x\]

Input:

int(1/(d*sec(f*x+e))^(5/3)/(a+b*tan(f*x+e)),x)
 

Output:

int(1/(d*sec(f*x+e))^(5/3)/(a+b*tan(f*x+e)),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{(d \sec (e+f x))^{5/3} (a+b \tan (e+f x))} \, dx=\text {Timed out} \] Input:

integrate(1/(d*sec(f*x+e))^(5/3)/(a+b*tan(f*x+e)),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {1}{(d \sec (e+f x))^{5/3} (a+b \tan (e+f x))} \, dx=\int \frac {1}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{3}} \left (a + b \tan {\left (e + f x \right )}\right )}\, dx \] Input:

integrate(1/(d*sec(f*x+e))**(5/3)/(a+b*tan(f*x+e)),x)
 

Output:

Integral(1/((d*sec(e + f*x))**(5/3)*(a + b*tan(e + f*x))), x)
 

Maxima [F]

\[ \int \frac {1}{(d \sec (e+f x))^{5/3} (a+b \tan (e+f x))} \, dx=\int { \frac {1}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{3}} {\left (b \tan \left (f x + e\right ) + a\right )}} \,d x } \] Input:

integrate(1/(d*sec(f*x+e))^(5/3)/(a+b*tan(f*x+e)),x, algorithm="maxima")
 

Output:

integrate(1/((d*sec(f*x + e))^(5/3)*(b*tan(f*x + e) + a)), x)
 

Giac [F]

\[ \int \frac {1}{(d \sec (e+f x))^{5/3} (a+b \tan (e+f x))} \, dx=\int { \frac {1}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{3}} {\left (b \tan \left (f x + e\right ) + a\right )}} \,d x } \] Input:

integrate(1/(d*sec(f*x+e))^(5/3)/(a+b*tan(f*x+e)),x, algorithm="giac")
 

Output:

integrate(1/((d*sec(f*x + e))^(5/3)*(b*tan(f*x + e) + a)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(d \sec (e+f x))^{5/3} (a+b \tan (e+f x))} \, dx=\int \frac {1}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/3}\,\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )} \,d x \] Input:

int(1/((d/cos(e + f*x))^(5/3)*(a + b*tan(e + f*x))),x)
 

Output:

int(1/((d/cos(e + f*x))^(5/3)*(a + b*tan(e + f*x))), x)
 

Reduce [F]

\[ \int \frac {1}{(d \sec (e+f x))^{5/3} (a+b \tan (e+f x))} \, dx=\frac {\int \frac {1}{\sec \left (f x +e \right )^{\frac {5}{3}} \tan \left (f x +e \right ) b +\sec \left (f x +e \right )^{\frac {5}{3}} a}d x}{d^{\frac {5}{3}}} \] Input:

int(1/(d*sec(f*x+e))^(5/3)/(a+b*tan(f*x+e)),x)
                                                                                    
                                                                                    
 

Output:

int(1/(sec(e + f*x)**(2/3)*sec(e + f*x)*tan(e + f*x)*b + sec(e + f*x)**(2/ 
3)*sec(e + f*x)*a),x)/(d**(2/3)*d)