3.1.47 \(\int x^2 (a+b \tan (c+d \sqrt [3]{x})) \, dx\) [47]

3.1.47.1 Optimal result

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
 

3.1.47.2 Mathematica [A] (verified)

Time = 0.10 (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)
 

3.1.47.3 Rubi [A] (verified)

Time = 0.63 (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)
 

3.1.47.3.1 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]]
 

3.1.47.4 Maple [F]

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

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)
 

3.1.47.5 Fricas [F]

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

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)
 

3.1.47.6 Sympy [F]

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

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)
 

3.1.47.7 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. 1119 vs. \(2 (222) = 444\).

Time = 0.49 (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...
 

3.1.47.8 Giac [F]

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

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)
 

3.1.47.9 Mupad [F(-1)]

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)