\(\int x (a+b \tan (c+d \sqrt [3]{x})) \, dx\) [48]

Optimal result

Integrand size = 16, antiderivative size = 203 \[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\frac {a x^2}{2}+\frac {1}{2} i b x^2-\frac {3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {15 i b x^{4/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac {15 b x \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {45 i b x^{2/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^4}+\frac {45 b \sqrt [3]{x} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^5}+\frac {45 i b \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{4 d^6} \] Output:

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
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.00 \[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\frac {a x^2}{2}+\frac {1}{2} i b x^2-\frac {3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {15 i b x^{4/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac {15 b x \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {45 i b x^{2/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^4}+\frac {45 b \sqrt [3]{x} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^5}+\frac {45 i b \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{4 d^6} \] Input:

Integrate[x*(a + b*Tan[c + d*x^(1/3)]),x]
 

Output:

(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
 

Rubi [A] (verified)

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

\(\displaystyle \frac {a x^2}{2}+\frac {45 i b \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{4 d^6}+\frac {45 b \sqrt [3]{x} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^5}-\frac {45 i b x^{2/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^4}-\frac {15 b x \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {15 i b x^{4/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^2}-\frac {3 b x^{5/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {1}{2} i b x^2\)

Input:

Int[x*(a + b*Tan[c + d*x^(1/3)]),x]
 

Output:

(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
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2010

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]]
 

Maple [F]

\[\int x \left (a +b \tan \left (c +d \,x^{\frac {1}{3}}\right )\right )d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\int { {\left (b \tan \left (d x^{\frac {1}{3}} + c\right ) + a\right )} x \,d x } \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

Integral(x*(a + b*tan(c + d*x**(1/3))), 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. 618 vs. \(2 (150) = 300\).

Time = 0.20 (sec) , antiderivative size = 618, normalized size of antiderivative = 3.04 \[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

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
 

Giac [F]

\[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\int { {\left (b \tan \left (d x^{\frac {1}{3}} + c\right ) + a\right )} x \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\int x\,\left (a+b\,\mathrm {tan}\left (c+d\,x^{1/3}\right )\right ) \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int x \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\left (\int \tan \left (x^{\frac {1}{3}} d +c \right ) x d x \right ) b +\frac {a \,x^{2}}{2} \] Input:

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

Output:

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