Integrand size = 18, antiderivative size = 287 \[ \int x^2 \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\frac {a x^3}{3}+\frac {1}{3} i b x^3-\frac {3 b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {12 i b x^{7/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {42 b x^2 \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {126 i b x^{5/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac {315 b x^{4/3} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}+\frac {630 i b x \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}-\frac {945 b x^{2/3} \operatorname {PolyLog}\left (7,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^7}-\frac {945 i b \sqrt [3]{x} \operatorname {PolyLog}\left (8,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^8}+\frac {945 b \operatorname {PolyLog}\left (9,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^9} \] Output:
1/3*a*x^3+1/3*I*b*x^3-3*b*x^(8/3)*ln(1+exp(2*I*(c+d*x^(1/3))))/d+12*I*b*x^ (7/3)*polylog(2,-exp(2*I*(c+d*x^(1/3))))/d^2-42*b*x^2*polylog(3,-exp(2*I*( c+d*x^(1/3))))/d^3-126*I*b*x^(5/3)*polylog(4,-exp(2*I*(c+d*x^(1/3))))/d^4+ 315*b*x^(4/3)*polylog(5,-exp(2*I*(c+d*x^(1/3))))/d^5+630*I*b*x*polylog(6,- exp(2*I*(c+d*x^(1/3))))/d^6-945*b*x^(2/3)*polylog(7,-exp(2*I*(c+d*x^(1/3)) ))/d^7-945*I*b*x^(1/3)*polylog(8,-exp(2*I*(c+d*x^(1/3))))/d^8+945/2*b*poly log(9,-exp(2*I*(c+d*x^(1/3))))/d^9
Time = 0.09 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.00 \[ \int x^2 \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\frac {a x^3}{3}+\frac {1}{3} i b x^3-\frac {3 b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {12 i b x^{7/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {42 b x^2 \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {126 i b x^{5/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac {315 b x^{4/3} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}+\frac {630 i b x \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}-\frac {945 b x^{2/3} \operatorname {PolyLog}\left (7,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^7}-\frac {945 i b \sqrt [3]{x} \operatorname {PolyLog}\left (8,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^8}+\frac {945 b \operatorname {PolyLog}\left (9,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^9} \] Input:
Integrate[x^2*(a + b*Tan[c + d*x^(1/3)]),x]
Output:
(a*x^3)/3 + (I/3)*b*x^3 - (3*b*x^(8/3)*Log[1 + E^((2*I)*(c + d*x^(1/3)))]) /d + ((12*I)*b*x^(7/3)*PolyLog[2, -E^((2*I)*(c + d*x^(1/3)))])/d^2 - (42*b *x^2*PolyLog[3, -E^((2*I)*(c + d*x^(1/3)))])/d^3 - ((126*I)*b*x^(5/3)*Poly Log[4, -E^((2*I)*(c + d*x^(1/3)))])/d^4 + (315*b*x^(4/3)*PolyLog[5, -E^((2 *I)*(c + d*x^(1/3)))])/d^5 + ((630*I)*b*x*PolyLog[6, -E^((2*I)*(c + d*x^(1 /3)))])/d^6 - (945*b*x^(2/3)*PolyLog[7, -E^((2*I)*(c + d*x^(1/3)))])/d^7 - ((945*I)*b*x^(1/3)*PolyLog[8, -E^((2*I)*(c + d*x^(1/3)))])/d^8 + (945*b*P olyLog[9, -E^((2*I)*(c + d*x^(1/3)))])/(2*d^9)
Time = 0.65 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, 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.
\(\displaystyle \int x^2 \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \int \left (a x^2+b x^2 \tan \left (c+d \sqrt [3]{x}\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a x^3}{3}+\frac {945 b \operatorname {PolyLog}\left (9,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^9}-\frac {945 i b \sqrt [3]{x} \operatorname {PolyLog}\left (8,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^8}-\frac {945 b x^{2/3} \operatorname {PolyLog}\left (7,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^7}+\frac {630 i b x \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}+\frac {315 b x^{4/3} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac {126 i b x^{5/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {42 b x^2 \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {12 i b x^{7/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {3 b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {1}{3} i b x^3\) |
Input:
Int[x^2*(a + b*Tan[c + d*x^(1/3)]),x]
Output:
(a*x^3)/3 + (I/3)*b*x^3 - (3*b*x^(8/3)*Log[1 + E^((2*I)*(c + d*x^(1/3)))]) /d + ((12*I)*b*x^(7/3)*PolyLog[2, -E^((2*I)*(c + d*x^(1/3)))])/d^2 - (42*b *x^2*PolyLog[3, -E^((2*I)*(c + d*x^(1/3)))])/d^3 - ((126*I)*b*x^(5/3)*Poly Log[4, -E^((2*I)*(c + d*x^(1/3)))])/d^4 + (315*b*x^(4/3)*PolyLog[5, -E^((2 *I)*(c + d*x^(1/3)))])/d^5 + ((630*I)*b*x*PolyLog[6, -E^((2*I)*(c + d*x^(1 /3)))])/d^6 - (945*b*x^(2/3)*PolyLog[7, -E^((2*I)*(c + d*x^(1/3)))])/d^7 - ((945*I)*b*x^(1/3)*PolyLog[8, -E^((2*I)*(c + d*x^(1/3)))])/d^8 + (945*b*P olyLog[9, -E^((2*I)*(c + d*x^(1/3)))])/(2*d^9)
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]]
\[\int x^{2} \left (a +b \tan \left (c +d \,x^{\frac {1}{3}}\right )\right )d x\]
Input:
int(x^2*(a+b*tan(c+d*x^(1/3))),x)
Output:
int(x^2*(a+b*tan(c+d*x^(1/3))),x)
\[ \int x^2 \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\int { {\left (b \tan \left (d x^{\frac {1}{3}} + c\right ) + a\right )} x^{2} \,d x } \] Input:
integrate(x^2*(a+b*tan(c+d*x^(1/3))),x, algorithm="fricas")
Output:
integral(b*x^2*tan(d*x^(1/3) + c) + a*x^2, x)
\[ \int x^2 \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\int x^{2} \left (a + b \tan {\left (c + d \sqrt [3]{x} \right )}\right )\, dx \] Input:
integrate(x**2*(a+b*tan(c+d*x**(1/3))),x)
Output:
Integral(x**2*(a + b*tan(c + d*x**(1/3))), x)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1119 vs. \(2 (222) = 444\).
Time = 0.21 (sec) , antiderivative size = 1119, normalized size of antiderivative = 3.90 \[ \int x^2 \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\text {Too large to display} \] Input:
integrate(x^2*(a+b*tan(c+d*x^(1/3))),x, algorithm="maxima")
Output:
1/105*(35*(d*x^(1/3) + c)^9*a + 35*I*(d*x^(1/3) + c)^9*b - 315*(d*x^(1/3) + c)^8*a*c - 315*I*(d*x^(1/3) + c)^8*b*c + 1260*(d*x^(1/3) + c)^7*a*c^2 + 1260*I*(d*x^(1/3) + c)^7*b*c^2 - 2940*(d*x^(1/3) + c)^6*a*c^3 - 2940*I*(d* x^(1/3) + c)^6*b*c^3 + 4410*(d*x^(1/3) + c)^5*a*c^4 + 4410*I*(d*x^(1/3) + c)^5*b*c^4 - 4410*(d*x^(1/3) + c)^4*a*c^5 - 4410*I*(d*x^(1/3) + c)^4*b*c^5 + 2940*(d*x^(1/3) + c)^3*a*c^6 + 2940*I*(d*x^(1/3) + c)^3*b*c^6 - 1260*(d *x^(1/3) + c)^2*a*c^7 - 1260*I*(d*x^(1/3) + c)^2*b*c^7 + 315*(d*x^(1/3) + c)*a*c^8 + 315*b*c^8*log(sec(d*x^(1/3) + c)) + 12*(-420*I*(d*x^(1/3) + c)^ 8*b + 1920*I*(d*x^(1/3) + c)^7*b*c - 3920*I*(d*x^(1/3) + c)^6*b*c^2 + 4704 *I*(d*x^(1/3) + c)^5*b*c^3 - 3675*I*(d*x^(1/3) + c)^4*b*c^4 + 1960*I*(d*x^ (1/3) + c)^3*b*c^5 - 735*I*(d*x^(1/3) + c)^2*b*c^6 + 210*I*(d*x^(1/3) + c) *b*c^7)*arctan2(sin(2*d*x^(1/3) + 2*c), cos(2*d*x^(1/3) + 2*c) + 1) + 1260 *(16*I*(d*x^(1/3) + c)^7*b - 64*I*(d*x^(1/3) + c)^6*b*c + 112*I*(d*x^(1/3) + c)^5*b*c^2 - 112*I*(d*x^(1/3) + c)^4*b*c^3 + 70*I*(d*x^(1/3) + c)^3*b*c ^4 - 28*I*(d*x^(1/3) + c)^2*b*c^5 + 7*I*(d*x^(1/3) + c)*b*c^6 - I*b*c^7)*d ilog(-e^(2*I*d*x^(1/3) + 2*I*c)) - 6*(420*(d*x^(1/3) + c)^8*b - 1920*(d*x^ (1/3) + c)^7*b*c + 3920*(d*x^(1/3) + c)^6*b*c^2 - 4704*(d*x^(1/3) + c)^5*b *c^3 + 3675*(d*x^(1/3) + c)^4*b*c^4 - 1960*(d*x^(1/3) + c)^3*b*c^5 + 735*( d*x^(1/3) + c)^2*b*c^6 - 210*(d*x^(1/3) + c)*b*c^7)*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) + 793...
\[ \int x^2 \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\int { {\left (b \tan \left (d x^{\frac {1}{3}} + c\right ) + a\right )} x^{2} \,d x } \] Input:
integrate(x^2*(a+b*tan(c+d*x^(1/3))),x, algorithm="giac")
Output:
integrate((b*tan(d*x^(1/3) + c) + a)*x^2, x)
Timed out. \[ \int x^2 \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\int x^2\,\left (a+b\,\mathrm {tan}\left (c+d\,x^{1/3}\right )\right ) \,d x \] Input:
int(x^2*(a + b*tan(c + d*x^(1/3))),x)
Output:
int(x^2*(a + b*tan(c + d*x^(1/3))), x)
\[ \int x^2 \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\left (\int \tan \left (x^{\frac {1}{3}} d +c \right ) x^{2}d x \right ) b +\frac {a \,x^{3}}{3} \] Input:
int(x^2*(a+b*tan(c+d*x^(1/3))),x)
Output:
(3*int(tan(x**(1/3)*d + c)*x**2,x)*b + a*x**3)/3