Integrand size = 16, antiderivative size = 203 \[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\frac {a x^2}{2}+\frac {1}{2} i b x^2-\frac {3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {15 i b x^{4/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac {15 b x \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {45 i b x^{2/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^4}+\frac {45 b \sqrt [3]{x} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^5}+\frac {45 i b \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{4 d^6} \]
1/2*a*x^2+1/2*I*b*x^2-3*b*x^(5/3)*ln(1+exp(2*I*(c+d*x^(1/3))))/d+15/2*I*b* x^(4/3)*polylog(2,-exp(2*I*(c+d*x^(1/3))))/d^2-15*b*x*polylog(3,-exp(2*I*( c+d*x^(1/3))))/d^3-45/2*I*b*x^(2/3)*polylog(4,-exp(2*I*(c+d*x^(1/3))))/d^4 +45/2*b*x^(1/3)*polylog(5,-exp(2*I*(c+d*x^(1/3))))/d^5+45/4*I*b*polylog(6, -exp(2*I*(c+d*x^(1/3))))/d^6
Time = 0.03 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.00 \[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\frac {a x^2}{2}+\frac {1}{2} i b x^2-\frac {3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {15 i b x^{4/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac {15 b x \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {45 i b x^{2/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^4}+\frac {45 b \sqrt [3]{x} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^5}+\frac {45 i b \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{4 d^6} \]
(a*x^2)/2 + (I/2)*b*x^2 - (3*b*x^(5/3)*Log[1 + E^((2*I)*(c + d*x^(1/3)))]) /d + (((15*I)/2)*b*x^(4/3)*PolyLog[2, -E^((2*I)*(c + d*x^(1/3)))])/d^2 - ( 15*b*x*PolyLog[3, -E^((2*I)*(c + d*x^(1/3)))])/d^3 - (((45*I)/2)*b*x^(2/3) *PolyLog[4, -E^((2*I)*(c + d*x^(1/3)))])/d^4 + (45*b*x^(1/3)*PolyLog[5, -E ^((2*I)*(c + d*x^(1/3)))])/(2*d^5) + (((45*I)/4)*b*PolyLog[6, -E^((2*I)*(c + d*x^(1/3)))])/d^6
Time = 0.48 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2010, 2009}
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.
| \(\displaystyle \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx\) |
| \(\Big \downarrow \) 2010 |
| \(\displaystyle \int \left (a x+b x \tan \left (c+d \sqrt [3]{x}\right )\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a x^2}{2}+\frac {45 i b \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{4 d^6}+\frac {45 b \sqrt [3]{x} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^5}-\frac {45 i b x^{2/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^4}-\frac {15 b x \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {15 i b x^{4/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac {3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {1}{2} i b x^2\) |
(a*x^2)/2 + (I/2)*b*x^2 - (3*b*x^(5/3)*Log[1 + E^((2*I)*(c + d*x^(1/3)))]) /d + (((15*I)/2)*b*x^(4/3)*PolyLog[2, -E^((2*I)*(c + d*x^(1/3)))])/d^2 - ( 15*b*x*PolyLog[3, -E^((2*I)*(c + d*x^(1/3)))])/d^3 - (((45*I)/2)*b*x^(2/3) *PolyLog[4, -E^((2*I)*(c + d*x^(1/3)))])/d^4 + (45*b*x^(1/3)*PolyLog[5, -E ^((2*I)*(c + d*x^(1/3)))])/(2*d^5) + (((45*I)/4)*b*PolyLog[6, -E^((2*I)*(c + d*x^(1/3)))])/d^6
3.1.48.3.1 Defintions of rubi rules used
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
\[\int x \left (a +b \tan \left (c +d \,x^{\frac {1}{3}}\right )\right )d x\]
\[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\int { {\left (b \tan \left (d x^{\frac {1}{3}} + c\right ) + a\right )} x \,d x } \]
\[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\int x \left (a + b \tan {\left (c + d \sqrt [3]{x} \right )}\right )\, dx \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 618 vs. \(2 (150) = 300\).
Time = 0.40 (sec) , antiderivative size = 618, normalized size of antiderivative = 3.04 \[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\frac {5 \, {\left (d x^{\frac {1}{3}} + c\right )}^{6} a + 5 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{6} b - 30 \, {\left (d x^{\frac {1}{3}} + c\right )}^{5} a c - 30 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{5} b c + 75 \, {\left (d x^{\frac {1}{3}} + c\right )}^{4} a c^{2} + 75 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{4} b c^{2} - 100 \, {\left (d x^{\frac {1}{3}} + c\right )}^{3} a c^{3} - 100 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{3} b c^{3} + 75 \, {\left (d x^{\frac {1}{3}} + c\right )}^{2} a c^{4} + 75 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{2} b c^{4} - 30 \, {\left (d x^{\frac {1}{3}} + c\right )} a c^{5} - 30 \, b c^{5} \log \left (\sec \left (d x^{\frac {1}{3}} + c\right )\right ) + 2 \, {\left (-48 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{5} b + 150 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{4} b c - 200 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{3} b c^{2} + 150 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{2} b c^{3} - 75 i \, {\left (d x^{\frac {1}{3}} + c\right )} b c^{4}\right )} \arctan \left (\sin \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right ), \cos \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right ) + 1\right ) + 15 \, {\left (16 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{4} b - 40 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{3} b c + 40 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{2} b c^{2} - 20 i \, {\left (d x^{\frac {1}{3}} + c\right )} b c^{3} + 5 i \, b c^{4}\right )} {\rm Li}_2\left (-e^{\left (2 i \, d x^{\frac {1}{3}} + 2 i \, c\right )}\right ) - {\left (48 \, {\left (d x^{\frac {1}{3}} + c\right )}^{5} b - 150 \, {\left (d x^{\frac {1}{3}} + c\right )}^{4} b c + 200 \, {\left (d x^{\frac {1}{3}} + c\right )}^{3} b c^{2} - 150 \, {\left (d x^{\frac {1}{3}} + c\right )}^{2} b c^{3} + 75 \, {\left (d x^{\frac {1}{3}} + c\right )} b c^{4}\right )} \log \left (\cos \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right )^{2} + \sin \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x^{\frac {1}{3}} + 2 \, c\right ) + 1\right ) + 360 i \, b {\rm Li}_{6}(-e^{\left (2 i \, d x^{\frac {1}{3}} + 2 i \, c\right )}) + 90 \, {\left (8 \, {\left (d x^{\frac {1}{3}} + c\right )} b - 5 \, b c\right )} {\rm Li}_{5}(-e^{\left (2 i \, d x^{\frac {1}{3}} + 2 i \, c\right )}) + 60 \, {\left (-12 i \, {\left (d x^{\frac {1}{3}} + c\right )}^{2} b + 15 i \, {\left (d x^{\frac {1}{3}} + c\right )} b c - 5 i \, b c^{2}\right )} {\rm Li}_{4}(-e^{\left (2 i \, d x^{\frac {1}{3}} + 2 i \, c\right )}) - 30 \, {\left (16 \, {\left (d x^{\frac {1}{3}} + c\right )}^{3} b - 30 \, {\left (d x^{\frac {1}{3}} + c\right )}^{2} b c + 20 \, {\left (d x^{\frac {1}{3}} + c\right )} b c^{2} - 5 \, b c^{3}\right )} {\rm Li}_{3}(-e^{\left (2 i \, d x^{\frac {1}{3}} + 2 i \, c\right )})}{10 \, d^{6}} \]
1/10*(5*(d*x^(1/3) + c)^6*a + 5*I*(d*x^(1/3) + c)^6*b - 30*(d*x^(1/3) + c) ^5*a*c - 30*I*(d*x^(1/3) + c)^5*b*c + 75*(d*x^(1/3) + c)^4*a*c^2 + 75*I*(d *x^(1/3) + c)^4*b*c^2 - 100*(d*x^(1/3) + c)^3*a*c^3 - 100*I*(d*x^(1/3) + c )^3*b*c^3 + 75*(d*x^(1/3) + c)^2*a*c^4 + 75*I*(d*x^(1/3) + c)^2*b*c^4 - 30 *(d*x^(1/3) + c)*a*c^5 - 30*b*c^5*log(sec(d*x^(1/3) + c)) + 2*(-48*I*(d*x^ (1/3) + c)^5*b + 150*I*(d*x^(1/3) + c)^4*b*c - 200*I*(d*x^(1/3) + c)^3*b*c ^2 + 150*I*(d*x^(1/3) + c)^2*b*c^3 - 75*I*(d*x^(1/3) + c)*b*c^4)*arctan2(s in(2*d*x^(1/3) + 2*c), cos(2*d*x^(1/3) + 2*c) + 1) + 15*(16*I*(d*x^(1/3) + c)^4*b - 40*I*(d*x^(1/3) + c)^3*b*c + 40*I*(d*x^(1/3) + c)^2*b*c^2 - 20*I *(d*x^(1/3) + c)*b*c^3 + 5*I*b*c^4)*dilog(-e^(2*I*d*x^(1/3) + 2*I*c)) - (4 8*(d*x^(1/3) + c)^5*b - 150*(d*x^(1/3) + c)^4*b*c + 200*(d*x^(1/3) + c)^3* b*c^2 - 150*(d*x^(1/3) + c)^2*b*c^3 + 75*(d*x^(1/3) + c)*b*c^4)*log(cos(2* d*x^(1/3) + 2*c)^2 + sin(2*d*x^(1/3) + 2*c)^2 + 2*cos(2*d*x^(1/3) + 2*c) + 1) + 360*I*b*polylog(6, -e^(2*I*d*x^(1/3) + 2*I*c)) + 90*(8*(d*x^(1/3) + c)*b - 5*b*c)*polylog(5, -e^(2*I*d*x^(1/3) + 2*I*c)) + 60*(-12*I*(d*x^(1/3 ) + c)^2*b + 15*I*(d*x^(1/3) + c)*b*c - 5*I*b*c^2)*polylog(4, -e^(2*I*d*x^ (1/3) + 2*I*c)) - 30*(16*(d*x^(1/3) + c)^3*b - 30*(d*x^(1/3) + c)^2*b*c + 20*(d*x^(1/3) + c)*b*c^2 - 5*b*c^3)*polylog(3, -e^(2*I*d*x^(1/3) + 2*I*c)) )/d^6
\[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\int { {\left (b \tan \left (d x^{\frac {1}{3}} + c\right ) + a\right )} x \,d x } \]
Timed out. \[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\int x\,\left (a+b\,\mathrm {tan}\left (c+d\,x^{1/3}\right )\right ) \,d x \]