Integrand size = 25, antiderivative size = 552 \[ \int \frac {(d \sec (e+f x))^{5/3}}{a+b \tan (e+f x)} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt {3} \sqrt [6]{a^2+b^2}}\right ) (d \sec (e+f x))^{5/3}}{2 b^{2/3} \sqrt [6]{a^2+b^2} f \sec ^2(e+f x)^{5/6}}+\frac {\sqrt {3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt {3} \sqrt [6]{a^2+b^2}}\right ) (d \sec (e+f x))^{5/3}}{2 b^{2/3} \sqrt [6]{a^2+b^2} f \sec ^2(e+f x)^{5/6}}-\frac {\text {arctanh}\left (\frac {\sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt [6]{a^2+b^2}}\right ) (d \sec (e+f x))^{5/3}}{b^{2/3} \sqrt [6]{a^2+b^2} f \sec ^2(e+f x)^{5/6}}+\frac {\log \left (\sqrt [3]{a^2+b^2}-\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\sec ^2(e+f x)}+b^{2/3} \sqrt [3]{\sec ^2(e+f x)}\right ) (d \sec (e+f x))^{5/3}}{4 b^{2/3} \sqrt [6]{a^2+b^2} f \sec ^2(e+f x)^{5/6}}-\frac {\log \left (\sqrt [3]{a^2+b^2}+\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\sec ^2(e+f x)}+b^{2/3} \sqrt [3]{\sec ^2(e+f x)}\right ) (d \sec (e+f x))^{5/3}}{4 b^{2/3} \sqrt [6]{a^2+b^2} f \sec ^2(e+f x)^{5/6}}+\frac {\operatorname {AppellF1}\left (\frac {1}{2},1,\frac {1}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) (d \sec (e+f x))^{5/3} \tan (e+f x)}{a f \sec ^2(e+f x)^{5/6}} \] Output:
1/2*3^(1/2)*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))*(d*sec(f*x+e))^(5/3)/b^(2/3)/(a^2+b^2)^(1/6)/f/(sec(f*x+e) ^2)^(5/6)+1/2*3^(1/2)*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))*(d*sec(f*x+e))^(5/3)/b^(2/3)/(a^2+b^2)^(1/6)/f/(s ec(f*x+e)^2)^(5/6)-arctanh(b^(1/3)*(sec(f*x+e)^2)^(1/6)/(a^2+b^2)^(1/6))*( d*sec(f*x+e))^(5/3)/b^(2/3)/(a^2+b^2)^(1/6)/f/(sec(f*x+e)^2)^(5/6)+1/4*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))*(d*sec(f*x+e))^(5/3)/b^(2/3)/(a^2+b^2)^(1/6)/f/(sec(f*x+e )^2)^(5/6)-1/4*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))*(d*sec(f*x+e))^(5/3)/b^(2/3)/(a^2+b^2)^ (1/6)/f/(sec(f*x+e)^2)^(5/6)+AppellF1(1/2,1,1/6,3/2,b^2*tan(f*x+e)^2/a^2,- tan(f*x+e)^2)*(d*sec(f*x+e))^(5/3)*tan(f*x+e)/a/f/(sec(f*x+e)^2)^(5/6)
Result contains complex when optimal does not.
Time = 14.60 (sec) , antiderivative size = 276, normalized size of antiderivative = 0.50 \[ \int \frac {(d \sec (e+f x))^{5/3}}{a+b \tan (e+f x)} \, dx=-\frac {24 d^2 \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{6},\frac {1}{6},\frac {4}{3},\frac {a-i b}{a+b \tan (e+f x)},\frac {a+i b}{a+b \tan (e+f x)}\right ) (a+b \tan (e+f x))}{b f \sqrt [3]{d \sec (e+f x)} \left ((a+i b) \operatorname {AppellF1}\left (\frac {4}{3},\frac {1}{6},\frac {7}{6},\frac {7}{3},\frac {a-i b}{a+b \tan (e+f x)},\frac {a+i b}{a+b \tan (e+f x)}\right )+(a-i b) \operatorname {AppellF1}\left (\frac {4}{3},\frac {7}{6},\frac {1}{6},\frac {7}{3},\frac {a-i b}{a+b \tan (e+f x)},\frac {a+i b}{a+b \tan (e+f x)}\right )+8 \operatorname {AppellF1}\left (\frac {1}{3},\frac {1}{6},\frac {1}{6},\frac {4}{3},\frac {a-i b}{a+b \tan (e+f x)},\frac {a+i b}{a+b \tan (e+f x)}\right ) (a+b \tan (e+f x))\right )} \] Input:
Integrate[(d*Sec[e + f*x])^(5/3)/(a + b*Tan[e + f*x]),x]
Output:
(-24*d^2*AppellF1[1/3, 1/6, 1/6, 4/3, (a - I*b)/(a + b*Tan[e + f*x]), (a + I*b)/(a + b*Tan[e + f*x])]*(a + b*Tan[e + f*x]))/(b*f*(d*Sec[e + f*x])^(1 /3)*((a + I*b)*AppellF1[4/3, 1/6, 7/6, 7/3, (a - I*b)/(a + b*Tan[e + f*x]) , (a + I*b)/(a + b*Tan[e + f*x])] + (a - I*b)*AppellF1[4/3, 7/6, 1/6, 7/3, (a - I*b)/(a + b*Tan[e + f*x]), (a + I*b)/(a + b*Tan[e + f*x])] + 8*Appel lF1[1/3, 1/6, 1/6, 4/3, (a - I*b)/(a + b*Tan[e + f*x]), (a + I*b)/(a + b*T an[e + f*x])]*(a + b*Tan[e + f*x])))
Time = 0.74 (sec) , antiderivative size = 379, normalized size of antiderivative = 0.69, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 3994, 504, 333, 353, 73, 825, 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.
\(\displaystyle \int \frac {(d \sec (e+f x))^{5/3}}{a+b \tan (e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(d \sec (e+f x))^{5/3}}{a+b \tan (e+f x)}dx\) |
\(\Big \downarrow \) 3994 |
\(\displaystyle \frac {(d \sec (e+f x))^{5/3} \int \frac {1}{(a+b \tan (e+f x)) \sqrt [6]{\tan ^2(e+f x)+1}}d(b \tan (e+f x))}{b f \sec ^2(e+f x)^{5/6}}\) |
\(\Big \downarrow \) 504 |
\(\displaystyle \frac {(d \sec (e+f x))^{5/3} \left (a \int \frac {1}{\sqrt [6]{\tan ^2(e+f x)+1} \left (a^2-b^2 \tan ^2(e+f x)\right )}d(b \tan (e+f x))-\int \frac {b \tan (e+f x)}{\sqrt [6]{\tan ^2(e+f x)+1} \left (a^2-b^2 \tan ^2(e+f x)\right )}d(b \tan (e+f x))\right )}{b f \sec ^2(e+f x)^{5/6}}\) |
\(\Big \downarrow \) 333 |
\(\displaystyle \frac {(d \sec (e+f x))^{5/3} \left (\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {1}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a}-\int \frac {b \tan (e+f x)}{\sqrt [6]{\tan ^2(e+f x)+1} \left (a^2-b^2 \tan ^2(e+f x)\right )}d(b \tan (e+f x))\right )}{b f \sec ^2(e+f x)^{5/6}}\) |
\(\Big \downarrow \) 353 |
\(\displaystyle \frac {(d \sec (e+f x))^{5/3} \left (\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {1}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a}-\frac {1}{2} \int \frac {1}{\sqrt [6]{\frac {\tan (e+f x)}{b}+1} \left (a^2-b^2 \tan ^2(e+f x)\right )}d\left (b^2 \tan ^2(e+f x)\right )\right )}{b f \sec ^2(e+f x)^{5/6}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(d \sec (e+f x))^{5/3} \left (\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {1}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a}-3 b^2 \int \frac {b^4 \tan ^4(e+f x)}{-\tan ^6(e+f x) b^8+b^2+a^2}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}\right )}{b f \sec ^2(e+f x)^{5/6}}\) |
\(\Big \downarrow \) 825 |
\(\displaystyle \frac {(d \sec (e+f x))^{5/3} \left (\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {1}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a}-3 b^2 \left (\frac {\int \frac {1}{\sqrt [3]{a^2+b^2}-b^{8/3} \tan ^2(e+f x)}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{3 b^{4/3}}+\frac {\int -\frac {\tan (e+f x) b^{4/3}+\sqrt [6]{a^2+b^2}}{2 \left (\tan ^2(e+f x) b^{8/3}-\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}\right )}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{3 b^{4/3} \sqrt [6]{a^2+b^2}}+\frac {\int -\frac {\sqrt [6]{a^2+b^2}-b^{4/3} \tan (e+f x)}{2 \left (\tan ^2(e+f x) b^{8/3}+\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}\right )}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{3 b^{4/3} \sqrt [6]{a^2+b^2}}\right )\right )}{b f \sec ^2(e+f x)^{5/6}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(d \sec (e+f x))^{5/3} \left (\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {1}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a}-3 b^2 \left (\frac {\int \frac {1}{\sqrt [3]{a^2+b^2}-b^{8/3} \tan ^2(e+f x)}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{3 b^{4/3}}-\frac {\int \frac {\tan (e+f x) b^{4/3}+\sqrt [6]{a^2+b^2}}{\tan ^2(e+f x) b^{8/3}-\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{6 b^{4/3} \sqrt [6]{a^2+b^2}}-\frac {\int \frac {\sqrt [6]{a^2+b^2}-b^{4/3} \tan (e+f x)}{\tan ^2(e+f x) b^{8/3}+\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{6 b^{4/3} \sqrt [6]{a^2+b^2}}\right )\right )}{b f \sec ^2(e+f x)^{5/6}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {(d \sec (e+f x))^{5/3} \left (\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {1}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a}-3 b^2 \left (-\frac {\int \frac {\tan (e+f x) b^{4/3}+\sqrt [6]{a^2+b^2}}{\tan ^2(e+f x) b^{8/3}-\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{6 b^{4/3} \sqrt [6]{a^2+b^2}}-\frac {\int \frac {\sqrt [6]{a^2+b^2}-b^{4/3} \tan (e+f x)}{\tan ^2(e+f x) b^{8/3}+\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{6 b^{4/3} \sqrt [6]{a^2+b^2}}+\frac {\text {arctanh}\left (\frac {b^{4/3} \tan (e+f x)}{\sqrt [6]{a^2+b^2}}\right )}{3 b^{5/3} \sqrt [6]{a^2+b^2}}\right )\right )}{b f \sec ^2(e+f x)^{5/6}}\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {(d \sec (e+f x))^{5/3} \left (\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {1}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a}-3 b^2 \left (-\frac {\frac {3}{2} \sqrt [6]{a^2+b^2} \int \frac {1}{\tan ^2(e+f x) b^{8/3}-\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}+\frac {\int -\frac {\sqrt [3]{b} \left (\sqrt [6]{a^2+b^2}-2 b^{4/3} \tan (e+f x)\right )}{\tan ^2(e+f x) b^{8/3}-\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{2 \sqrt [3]{b}}}{6 b^{4/3} \sqrt [6]{a^2+b^2}}-\frac {\frac {3}{2} \sqrt [6]{a^2+b^2} \int \frac {1}{\tan ^2(e+f x) b^{8/3}+\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}-\frac {\int \frac {\sqrt [3]{b} \left (2 \tan (e+f x) b^{4/3}+\sqrt [6]{a^2+b^2}\right )}{\tan ^2(e+f x) b^{8/3}+\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{2 \sqrt [3]{b}}}{6 b^{4/3} \sqrt [6]{a^2+b^2}}+\frac {\text {arctanh}\left (\frac {b^{4/3} \tan (e+f x)}{\sqrt [6]{a^2+b^2}}\right )}{3 b^{5/3} \sqrt [6]{a^2+b^2}}\right )\right )}{b f \sec ^2(e+f x)^{5/6}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {(d \sec (e+f x))^{5/3} \left (\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {1}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a}-3 b^2 \left (-\frac {\frac {3}{2} \sqrt [6]{a^2+b^2} \int \frac {1}{\tan ^2(e+f x) b^{8/3}-\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}-\frac {\int \frac {\sqrt [3]{b} \left (\sqrt [6]{a^2+b^2}-2 b^{4/3} \tan (e+f x)\right )}{\tan ^2(e+f x) b^{8/3}-\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{2 \sqrt [3]{b}}}{6 b^{4/3} \sqrt [6]{a^2+b^2}}-\frac {\frac {3}{2} \sqrt [6]{a^2+b^2} \int \frac {1}{\tan ^2(e+f x) b^{8/3}+\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}-\frac {\int \frac {\sqrt [3]{b} \left (2 \tan (e+f x) b^{4/3}+\sqrt [6]{a^2+b^2}\right )}{\tan ^2(e+f x) b^{8/3}+\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{2 \sqrt [3]{b}}}{6 b^{4/3} \sqrt [6]{a^2+b^2}}+\frac {\text {arctanh}\left (\frac {b^{4/3} \tan (e+f x)}{\sqrt [6]{a^2+b^2}}\right )}{3 b^{5/3} \sqrt [6]{a^2+b^2}}\right )\right )}{b f \sec ^2(e+f x)^{5/6}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(d \sec (e+f x))^{5/3} \left (\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {1}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a}-3 b^2 \left (-\frac {\frac {3}{2} \sqrt [6]{a^2+b^2} \int \frac {1}{\tan ^2(e+f x) b^{8/3}-\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}-\frac {1}{2} \int \frac {\sqrt [6]{a^2+b^2}-2 b^{4/3} \tan (e+f x)}{\tan ^2(e+f x) b^{8/3}-\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{6 b^{4/3} \sqrt [6]{a^2+b^2}}-\frac {\frac {3}{2} \sqrt [6]{a^2+b^2} \int \frac {1}{\tan ^2(e+f x) b^{8/3}+\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}-\frac {1}{2} \int \frac {2 \tan (e+f x) b^{4/3}+\sqrt [6]{a^2+b^2}}{\tan ^2(e+f x) b^{8/3}+\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{6 b^{4/3} \sqrt [6]{a^2+b^2}}+\frac {\text {arctanh}\left (\frac {b^{4/3} \tan (e+f x)}{\sqrt [6]{a^2+b^2}}\right )}{3 b^{5/3} \sqrt [6]{a^2+b^2}}\right )\right )}{b f \sec ^2(e+f x)^{5/6}}\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {(d \sec (e+f x))^{5/3} \left (\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {1}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a}-3 b^2 \left (-\frac {\frac {3 \int \frac {1}{\frac {2 b^{4/3} \tan (e+f x)}{\sqrt [6]{a^2+b^2}}-4}d\left (1-\frac {2 b^{4/3} \tan (e+f x)}{\sqrt [6]{a^2+b^2}}\right )}{\sqrt [3]{b}}-\frac {1}{2} \int \frac {\sqrt [6]{a^2+b^2}-2 b^{4/3} \tan (e+f x)}{\tan ^2(e+f x) b^{8/3}-\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{6 b^{4/3} \sqrt [6]{a^2+b^2}}-\frac {-\frac {1}{2} \int \frac {2 \tan (e+f x) b^{4/3}+\sqrt [6]{a^2+b^2}}{\tan ^2(e+f x) b^{8/3}+\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}-\frac {3 \int \frac {1}{-\frac {2 \tan (e+f x) b^{4/3}}{\sqrt [6]{a^2+b^2}}-4}d\left (\frac {2 \tan (e+f x) b^{4/3}}{\sqrt [6]{a^2+b^2}}+1\right )}{\sqrt [3]{b}}}{6 b^{4/3} \sqrt [6]{a^2+b^2}}+\frac {\text {arctanh}\left (\frac {b^{4/3} \tan (e+f x)}{\sqrt [6]{a^2+b^2}}\right )}{3 b^{5/3} \sqrt [6]{a^2+b^2}}\right )\right )}{b f \sec ^2(e+f x)^{5/6}}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {(d \sec (e+f x))^{5/3} \left (\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {1}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a}-3 b^2 \left (-\frac {-\frac {1}{2} \int \frac {\sqrt [6]{a^2+b^2}-2 b^{4/3} \tan (e+f x)}{\tan ^2(e+f x) b^{8/3}-\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 b^{4/3} \tan (e+f x)}{\sqrt [6]{a^2+b^2}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{6 b^{4/3} \sqrt [6]{a^2+b^2}}-\frac {\frac {\sqrt {3} \arctan \left (\frac {\frac {2 b^{4/3} \tan (e+f x)}{\sqrt [6]{a^2+b^2}}+1}{\sqrt {3}}\right )}{\sqrt [3]{b}}-\frac {1}{2} \int \frac {2 \tan (e+f x) b^{4/3}+\sqrt [6]{a^2+b^2}}{\tan ^2(e+f x) b^{8/3}+\sqrt [6]{a^2+b^2} \tan (e+f x) b^{4/3}+\sqrt [3]{a^2+b^2}}d\sqrt [6]{\frac {\tan (e+f x)}{b}+1}}{6 b^{4/3} \sqrt [6]{a^2+b^2}}+\frac {\text {arctanh}\left (\frac {b^{4/3} \tan (e+f x)}{\sqrt [6]{a^2+b^2}}\right )}{3 b^{5/3} \sqrt [6]{a^2+b^2}}\right )\right )}{b f \sec ^2(e+f x)^{5/6}}\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {(d \sec (e+f x))^{5/3} \left (\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},1,\frac {1}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a}-3 b^2 \left (-\frac {\frac {\log \left (\sqrt [3]{a^2+b^2}-b^{4/3} \sqrt [6]{a^2+b^2} \tan (e+f x)+b^{8/3} \tan ^2(e+f x)\right )}{2 \sqrt [3]{b}}-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 b^{4/3} \tan (e+f x)}{\sqrt [6]{a^2+b^2}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{6 b^{4/3} \sqrt [6]{a^2+b^2}}-\frac {\frac {\sqrt {3} \arctan \left (\frac {\frac {2 b^{4/3} \tan (e+f x)}{\sqrt [6]{a^2+b^2}}+1}{\sqrt {3}}\right )}{\sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a^2+b^2}+b^{4/3} \sqrt [6]{a^2+b^2} \tan (e+f x)+b^{8/3} \tan ^2(e+f x)\right )}{2 \sqrt [3]{b}}}{6 b^{4/3} \sqrt [6]{a^2+b^2}}+\frac {\text {arctanh}\left (\frac {b^{4/3} \tan (e+f x)}{\sqrt [6]{a^2+b^2}}\right )}{3 b^{5/3} \sqrt [6]{a^2+b^2}}\right )\right )}{b f \sec ^2(e+f x)^{5/6}}\) |
Input:
Int[(d*Sec[e + f*x])^(5/3)/(a + b*Tan[e + f*x]),x]
Output:
((d*Sec[e + f*x])^(5/3)*(-3*b^2*(ArcTanh[(b^(4/3)*Tan[e + f*x])/(a^2 + b^2 )^(1/6)]/(3*b^(5/3)*(a^2 + b^2)^(1/6)) - (-((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*b^(4/3)*(a^2 + b^2)^(1/6)) - ((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*b^(4/3)*(a^2 + b^2)^(1/6))) + (b*AppellF1[1/2, 1, 1/6, 3/2, (b^ 2*Tan[e + f*x]^2)/a^2, -Tan[e + f*x]^2]*Tan[e + f*x])/a))/(b*f*(Sec[e + f* x]^2)^(5/6))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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])
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]
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])
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]
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]
Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Module[{r = Numerator [Rt[-a/b, n]], s = Denominator[Rt[-a/b, n]], k, u}, Simp[u = Int[(r*Cos[2*k *m*(Pi/n)] - s*Cos[2*k*(m + 1)*(Pi/n)]*x)/(r^2 - 2*r*s*Cos[2*k*(Pi/n)]*x + s^2*x^2), x] + Int[(r*Cos[2*k*m*(Pi/n)] + s*Cos[2*k*(m + 1)*(Pi/n)]*x)/(r^2 + 2*r*s*Cos[2*k*(Pi/n)]*x + s^2*x^2), x]; 2*(r^(m + 2)/(a*n*s^m)) Int[1/ (r^2 - s^2*x^2), x] + 2*(r^(m + 1)/(a*n*s^m)) Sum[u, {k, 1, (n - 2)/4}], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && IGtQ[m, 0] && LtQ[m, n - 1 ] && NegQ[a/b]
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]
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]
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]
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]
\[\int \frac {\left (d \sec \left (f x +e \right )\right )^{\frac {5}{3}}}{a +b \tan \left (f x +e \right )}d x\]
Input:
int((d*sec(f*x+e))^(5/3)/(a+b*tan(f*x+e)),x)
Output:
int((d*sec(f*x+e))^(5/3)/(a+b*tan(f*x+e)),x)
Timed out. \[ \int \frac {(d \sec (e+f x))^{5/3}}{a+b \tan (e+f x)} \, dx=\text {Timed out} \] Input:
integrate((d*sec(f*x+e))^(5/3)/(a+b*tan(f*x+e)),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {(d \sec (e+f x))^{5/3}}{a+b \tan (e+f x)} \, dx=\int \frac {\left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{3}}}{a + b \tan {\left (e + f x \right )}}\, dx \] Input:
integrate((d*sec(f*x+e))**(5/3)/(a+b*tan(f*x+e)),x)
Output:
Integral((d*sec(e + f*x))**(5/3)/(a + b*tan(e + f*x)), x)
\[ \int \frac {(d \sec (e+f x))^{5/3}}{a+b \tan (e+f x)} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {5}{3}}}{b \tan \left (f x + e\right ) + a} \,d x } \] Input:
integrate((d*sec(f*x+e))^(5/3)/(a+b*tan(f*x+e)),x, algorithm="maxima")
Output:
integrate((d*sec(f*x + e))^(5/3)/(b*tan(f*x + e) + a), x)
\[ \int \frac {(d \sec (e+f x))^{5/3}}{a+b \tan (e+f x)} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {5}{3}}}{b \tan \left (f x + e\right ) + a} \,d x } \] Input:
integrate((d*sec(f*x+e))^(5/3)/(a+b*tan(f*x+e)),x, algorithm="giac")
Output:
integrate((d*sec(f*x + e))^(5/3)/(b*tan(f*x + e) + a), x)
Timed out. \[ \int \frac {(d \sec (e+f x))^{5/3}}{a+b \tan (e+f x)} \, dx=\int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/3}}{a+b\,\mathrm {tan}\left (e+f\,x\right )} \,d x \] Input:
int((d/cos(e + f*x))^(5/3)/(a + b*tan(e + f*x)),x)
Output:
int((d/cos(e + f*x))^(5/3)/(a + b*tan(e + f*x)), x)
\[ \int \frac {(d \sec (e+f x))^{5/3}}{a+b \tan (e+f x)} \, dx=d^{\frac {5}{3}} \left (\int \frac {\sec \left (f x +e \right )^{\frac {5}{3}}}{\tan \left (f x +e \right ) b +a}d x \right ) \] Input:
int((d*sec(f*x+e))^(5/3)/(a+b*tan(f*x+e)),x)
Output:
d**(2/3)*int((sec(e + f*x)**(2/3)*sec(e + f*x))/(tan(e + f*x)*b + a),x)*d