\(\int \frac {x^3}{a+b \tan (c+d \sqrt {x})} \, dx\) [36]

Optimal result

Integrand size = 20, antiderivative size = 460 \[ \int \frac {x^3}{a+b \tan \left (c+d \sqrt {x}\right )} \, dx=\frac {x^4}{4 (a+i b)}+\frac {2 b x^{7/2} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d}-\frac {7 i b x^3 \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^2}+\frac {21 b x^{5/2} \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^3}+\frac {105 i b x^2 \operatorname {PolyLog}\left (4,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^4}-\frac {105 b x^{3/2} \operatorname {PolyLog}\left (5,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^5}-\frac {315 i b x \operatorname {PolyLog}\left (6,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^6}+\frac {315 b \sqrt {x} \operatorname {PolyLog}\left (7,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^7}+\frac {315 i b \operatorname {PolyLog}\left (8,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{4 \left (a^2+b^2\right ) d^8} \] Output:

x^4/(4*a+4*I*b)+2*b*x^(7/2)*ln(1+(a^2+b^2)*exp(2*I*(c+d*x^(1/2)))/(a+I*b)^ 
2)/(a^2+b^2)/d-7*I*b*x^3*polylog(2,-(a^2+b^2)*exp(2*I*(c+d*x^(1/2)))/(a+I* 
b)^2)/(a^2+b^2)/d^2+21*b*x^(5/2)*polylog(3,-(a^2+b^2)*exp(2*I*(c+d*x^(1/2) 
))/(a+I*b)^2)/(a^2+b^2)/d^3+105/2*I*b*x^2*polylog(4,-(a^2+b^2)*exp(2*I*(c+ 
d*x^(1/2)))/(a+I*b)^2)/(a^2+b^2)/d^4-105*b*x^(3/2)*polylog(5,-(a^2+b^2)*ex 
p(2*I*(c+d*x^(1/2)))/(a+I*b)^2)/(a^2+b^2)/d^5-315/2*I*b*x*polylog(6,-(a^2+ 
b^2)*exp(2*I*(c+d*x^(1/2)))/(a+I*b)^2)/(a^2+b^2)/d^6+315/2*b*x^(1/2)*polyl 
og(7,-(a^2+b^2)*exp(2*I*(c+d*x^(1/2)))/(a+I*b)^2)/(a^2+b^2)/d^7+315/4*I*b* 
polylog(8,-(a^2+b^2)*exp(2*I*(c+d*x^(1/2)))/(a+I*b)^2)/(a^2+b^2)/d^8
 

Mathematica [A] (verified)

Time = 1.42 (sec) , antiderivative size = 401, normalized size of antiderivative = 0.87 \[ \int \frac {x^3}{a+b \tan \left (c+d \sqrt {x}\right )} \, dx=\frac {a d^8 x^4+i b d^8 x^4+8 b d^7 x^{7/2} \log \left (1+\frac {(a+i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )+28 i b d^6 x^3 \operatorname {PolyLog}\left (2,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )+84 b d^5 x^{5/2} \operatorname {PolyLog}\left (3,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )-210 i b d^4 x^2 \operatorname {PolyLog}\left (4,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )-420 b d^3 x^{3/2} \operatorname {PolyLog}\left (5,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )+630 i b d^2 x \operatorname {PolyLog}\left (6,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )+630 b d \sqrt {x} \operatorname {PolyLog}\left (7,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )-315 i b \operatorname {PolyLog}\left (8,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt {x}\right )}}{a-i b}\right )}{4 \left (a^2+b^2\right ) d^8} \] Input:

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

Output:

(a*d^8*x^4 + I*b*d^8*x^4 + 8*b*d^7*x^(7/2)*Log[1 + (a + I*b)/((a - I*b)*E^ 
((2*I)*(c + d*Sqrt[x])))] + (28*I)*b*d^6*x^3*PolyLog[2, (-a - I*b)/((a - I 
*b)*E^((2*I)*(c + d*Sqrt[x])))] + 84*b*d^5*x^(5/2)*PolyLog[3, (-a - I*b)/( 
(a - I*b)*E^((2*I)*(c + d*Sqrt[x])))] - (210*I)*b*d^4*x^2*PolyLog[4, (-a - 
 I*b)/((a - I*b)*E^((2*I)*(c + d*Sqrt[x])))] - 420*b*d^3*x^(3/2)*PolyLog[5 
, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*Sqrt[x])))] + (630*I)*b*d^2*x*Poly 
Log[6, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*Sqrt[x])))] + 630*b*d*Sqrt[x] 
*PolyLog[7, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*Sqrt[x])))] - (315*I)*b* 
PolyLog[8, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*Sqrt[x])))])/(4*(a^2 + b^ 
2)*d^8)
 

Rubi [A] (verified)

Time = 1.98 (sec) , antiderivative size = 474, normalized size of antiderivative = 1.03, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {4234, 3042, 4215, 2620, 3011, 7163, 7163, 7163, 7163, 7163, 2720, 7143}

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.

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

\(\Big \downarrow \) 4234

\(\displaystyle 2 \int \frac {x^{7/2}}{a+b \tan \left (c+d \sqrt {x}\right )}d\sqrt {x}\)

\(\Big \downarrow \) 3042

\(\displaystyle 2 \int \frac {x^{7/2}}{a+b \tan \left (c+d \sqrt {x}\right )}d\sqrt {x}\)

\(\Big \downarrow \) 4215

\(\displaystyle 2 \left (2 i b \int \frac {e^{2 i \left (c+d \sqrt {x}\right )} x^{7/2}}{(a+i b)^2+\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}d\sqrt {x}+\frac {x^4}{8 (a+i b)}\right )\)

\(\Big \downarrow \) 2620

\(\displaystyle 2 \left (2 i b \left (\frac {7 i \int x^3 \log \left (\frac {e^{2 i \left (c+d \sqrt {x}\right )} \left (a^2+b^2\right )}{(a+i b)^2}+1\right )d\sqrt {x}}{2 d \left (a^2+b^2\right )}-\frac {i x^{7/2} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {x^4}{8 (a+i b)}\right )\)

\(\Big \downarrow \) 3011

\(\displaystyle 2 \left (2 i b \left (\frac {7 i \left (\frac {i x^3 \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}-\frac {3 i \int x^{5/2} \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )d\sqrt {x}}{d}\right )}{2 d \left (a^2+b^2\right )}-\frac {i x^{7/2} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {x^4}{8 (a+i b)}\right )\)

\(\Big \downarrow \) 7163

\(\displaystyle 2 \left (2 i b \left (\frac {7 i \left (\frac {i x^3 \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}-\frac {3 i \left (\frac {5 i \int x^2 \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )d\sqrt {x}}{2 d}-\frac {i x^{5/2} \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}\right )}{2 d \left (a^2+b^2\right )}-\frac {i x^{7/2} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {x^4}{8 (a+i b)}\right )\)

\(\Big \downarrow \) 7163

\(\displaystyle 2 \left (2 i b \left (\frac {7 i \left (\frac {i x^3 \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}-\frac {3 i \left (\frac {5 i \left (\frac {2 i \int x^{3/2} \operatorname {PolyLog}\left (4,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )d\sqrt {x}}{d}-\frac {i x^2 \operatorname {PolyLog}\left (4,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{2 d}-\frac {i x^{5/2} \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}\right )}{2 d \left (a^2+b^2\right )}-\frac {i x^{7/2} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {x^4}{8 (a+i b)}\right )\)

\(\Big \downarrow \) 7163

\(\displaystyle 2 \left (2 i b \left (\frac {7 i \left (\frac {i x^3 \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}-\frac {3 i \left (\frac {5 i \left (\frac {2 i \left (\frac {3 i \int x \operatorname {PolyLog}\left (5,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )d\sqrt {x}}{2 d}-\frac {i x^{3/2} \operatorname {PolyLog}\left (5,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}-\frac {i x^2 \operatorname {PolyLog}\left (4,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{2 d}-\frac {i x^{5/2} \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}\right )}{2 d \left (a^2+b^2\right )}-\frac {i x^{7/2} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {x^4}{8 (a+i b)}\right )\)

\(\Big \downarrow \) 7163

\(\displaystyle 2 \left (2 i b \left (\frac {7 i \left (\frac {i x^3 \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}-\frac {3 i \left (\frac {5 i \left (\frac {2 i \left (\frac {3 i \left (\frac {i \int \sqrt {x} \operatorname {PolyLog}\left (6,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )d\sqrt {x}}{d}-\frac {i x \operatorname {PolyLog}\left (6,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{2 d}-\frac {i x^{3/2} \operatorname {PolyLog}\left (5,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}-\frac {i x^2 \operatorname {PolyLog}\left (4,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{2 d}-\frac {i x^{5/2} \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}\right )}{2 d \left (a^2+b^2\right )}-\frac {i x^{7/2} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {x^4}{8 (a+i b)}\right )\)

\(\Big \downarrow \) 7163

\(\displaystyle 2 \left (2 i b \left (\frac {7 i \left (\frac {i x^3 \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}-\frac {3 i \left (\frac {5 i \left (\frac {2 i \left (\frac {3 i \left (\frac {i \left (\frac {i \int \operatorname {PolyLog}\left (7,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )d\sqrt {x}}{2 d}-\frac {i \sqrt {x} \operatorname {PolyLog}\left (7,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}-\frac {i x \operatorname {PolyLog}\left (6,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{2 d}-\frac {i x^{3/2} \operatorname {PolyLog}\left (5,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}-\frac {i x^2 \operatorname {PolyLog}\left (4,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{2 d}-\frac {i x^{5/2} \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}\right )}{2 d \left (a^2+b^2\right )}-\frac {i x^{7/2} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {x^4}{8 (a+i b)}\right )\)

\(\Big \downarrow \) 2720

\(\displaystyle 2 \left (2 i b \left (\frac {7 i \left (\frac {i x^3 \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}-\frac {3 i \left (\frac {5 i \left (\frac {2 i \left (\frac {3 i \left (\frac {i \left (\frac {\int \frac {\operatorname {PolyLog}\left (7,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{\sqrt {x}}de^{2 i \left (c+d \sqrt {x}\right )}}{4 d^2}-\frac {i \sqrt {x} \operatorname {PolyLog}\left (7,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}-\frac {i x \operatorname {PolyLog}\left (6,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{2 d}-\frac {i x^{3/2} \operatorname {PolyLog}\left (5,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}-\frac {i x^2 \operatorname {PolyLog}\left (4,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{2 d}-\frac {i x^{5/2} \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}\right )}{2 d \left (a^2+b^2\right )}-\frac {i x^{7/2} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {x^4}{8 (a+i b)}\right )\)

\(\Big \downarrow \) 7143

\(\displaystyle 2 \left (2 i b \left (\frac {7 i \left (\frac {i x^3 \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}-\frac {3 i \left (\frac {5 i \left (\frac {2 i \left (\frac {3 i \left (\frac {i \left (\frac {\operatorname {PolyLog}\left (8,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{4 d^2}-\frac {i \sqrt {x} \operatorname {PolyLog}\left (7,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}-\frac {i x \operatorname {PolyLog}\left (6,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{2 d}-\frac {i x^{3/2} \operatorname {PolyLog}\left (5,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}-\frac {i x^2 \operatorname {PolyLog}\left (4,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{2 d}-\frac {i x^{5/2} \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}\right )}{2 d \left (a^2+b^2\right )}-\frac {i x^{7/2} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt {x}\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {x^4}{8 (a+i b)}\right )\)

Input:

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

Output:

2*(x^4/(8*(a + I*b)) + (2*I)*b*(((-1/2*I)*x^(7/2)*Log[1 + ((a^2 + b^2)*E^( 
(2*I)*(c + d*Sqrt[x])))/(a + I*b)^2])/((a^2 + b^2)*d) + (((7*I)/2)*(((I/2) 
*x^3*PolyLog[2, -(((a^2 + b^2)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b)^2)])/d 
 - ((3*I)*(((-1/2*I)*x^(5/2)*PolyLog[3, -(((a^2 + b^2)*E^((2*I)*(c + d*Sqr 
t[x])))/(a + I*b)^2)])/d + (((5*I)/2)*(((-1/2*I)*x^2*PolyLog[4, -(((a^2 + 
b^2)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b)^2)])/d + ((2*I)*(((-1/2*I)*x^(3/ 
2)*PolyLog[5, -(((a^2 + b^2)*E^((2*I)*(c + d*Sqrt[x])))/(a + I*b)^2)])/d + 
 (((3*I)/2)*(((-1/2*I)*x*PolyLog[6, -(((a^2 + b^2)*E^((2*I)*(c + d*Sqrt[x] 
)))/(a + I*b)^2)])/d + (I*(((-1/2*I)*Sqrt[x]*PolyLog[7, -(((a^2 + b^2)*E^( 
(2*I)*(c + d*Sqrt[x])))/(a + I*b)^2)])/d + PolyLog[8, -(((a^2 + b^2)*E^((2 
*I)*(c + d*Sqrt[x])))/(a + I*b)^2)]/(4*d^2)))/d))/d))/d))/d))/d))/((a^2 + 
b^2)*d)))
 

Defintions of rubi rules used

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

rule 3042

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

rule 4215

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Sy 
mbol] :> Simp[(c + d*x)^(m + 1)/(d*(m + 1)*(a + I*b)), x] + Simp[2*I*b   In 
t[(c + d*x)^m*(E^Simp[2*I*(e + f*x), x]/((a + I*b)^2 + (a^2 + b^2)*E^Simp[2 
*I*(e + f*x), x])), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 + b^2 
, 0] && IGtQ[m, 0]
 

rule 4234

Int[(x_)^(m_.)*((a_.) + (b_.)*Tan[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Tan[c + d*x])^ 
p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[(m + 
 1)/n], 0] && IntegerQ[p]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

rule 7163

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_. 
)*(x_))))^(p_.)], x_Symbol] :> Simp[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a 
+ b*x)))^p]/(b*c*p*Log[F])), x] - Simp[f*(m/(b*c*p*Log[F]))   Int[(e + f*x) 
^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c 
, d, e, f, n, p}, x] && GtQ[m, 0]
 

Maple [F]

\[\int \frac {x^{3}}{a +b \tan \left (c +d \sqrt {x}\right )}d x\]

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(x^3/(b*tan(d*sqrt(x) + c) + a), x)
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1133 vs. \(2 (383) = 766\).

Time = 0.34 (sec) , antiderivative size = 1133, normalized size of antiderivative = 2.46 \[ \int \frac {x^3}{a+b \tan \left (c+d \sqrt {x}\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/420*(420*(2*(d*sqrt(x) + c)*a/(a^2 + b^2) + 2*b*log(b*tan(d*sqrt(x) + c 
) + a)/(a^2 + b^2) - b*log(tan(d*sqrt(x) + c)^2 + 1)/(a^2 + b^2))*c^7 - (1 
05*(d*sqrt(x) + c)^8*(a - I*b) - 840*(d*sqrt(x) + c)^7*(a - I*b)*c + 2940* 
(d*sqrt(x) + c)^6*(a - I*b)*c^2 - 5880*(d*sqrt(x) + c)^5*(a - I*b)*c^3 + 7 
350*(d*sqrt(x) + c)^4*(a - I*b)*c^4 - 5880*(d*sqrt(x) + c)^3*(a - I*b)*c^5 
 + 2940*(d*sqrt(x) + c)^2*(a - I*b)*c^6 - 8*(960*I*(d*sqrt(x) + c)^7*b - 3 
920*I*(d*sqrt(x) + c)^6*b*c + 7056*I*(d*sqrt(x) + c)^5*b*c^2 - 7350*I*(d*s 
qrt(x) + c)^4*b*c^3 + 4900*I*(d*sqrt(x) + c)^3*b*c^4 - 2205*I*(d*sqrt(x) + 
 c)^2*b*c^5 + 735*I*(d*sqrt(x) + c)*b*c^6)*arctan2((2*a*b*cos(2*d*sqrt(x) 
+ 2*c) - (a^2 - b^2)*sin(2*d*sqrt(x) + 2*c))/(a^2 + b^2), (2*a*b*sin(2*d*s 
qrt(x) + 2*c) + a^2 + b^2 + (a^2 - b^2)*cos(2*d*sqrt(x) + 2*c))/(a^2 + b^2 
)) - 420*(64*I*(d*sqrt(x) + c)^6*b - 224*I*(d*sqrt(x) + c)^5*b*c + 336*I*( 
d*sqrt(x) + c)^4*b*c^2 - 280*I*(d*sqrt(x) + c)^3*b*c^3 + 140*I*(d*sqrt(x) 
+ c)^2*b*c^4 - 42*I*(d*sqrt(x) + c)*b*c^5 + 7*I*b*c^6)*dilog((I*a + b)*e^( 
2*I*d*sqrt(x) + 2*I*c)/(-I*a + b)) + 4*(960*(d*sqrt(x) + c)^7*b - 3920*(d* 
sqrt(x) + c)^6*b*c + 7056*(d*sqrt(x) + c)^5*b*c^2 - 7350*(d*sqrt(x) + c)^4 
*b*c^3 + 4900*(d*sqrt(x) + c)^3*b*c^4 - 2205*(d*sqrt(x) + c)^2*b*c^5 + 735 
*(d*sqrt(x) + c)*b*c^6)*log(((a^2 + b^2)*cos(2*d*sqrt(x) + 2*c)^2 + 4*a*b* 
sin(2*d*sqrt(x) + 2*c) + (a^2 + b^2)*sin(2*d*sqrt(x) + 2*c)^2 + a^2 + b^2 
+ 2*(a^2 - b^2)*cos(2*d*sqrt(x) + 2*c))/(a^2 + b^2)) + 302400*I*b*polyl...
 

Giac [F]

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

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

Output:

integrate(x^3/(b*tan(d*sqrt(x) + c) + a), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(x**3/(tan(sqrt(x)*d + c)*b + a),x)