\(\int x^2 (a+b \tan (c+d \sqrt [3]{x}))^2 \, dx\) [52]

Optimal result

Integrand size = 20, antiderivative size = 597 \[ \int x^2 \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2 \, dx=-\frac {3 i b^2 x^{8/3}}{d}+\frac {a^2 x^3}{3}+\frac {2}{3} i a b x^3-\frac {b^2 x^3}{3}+\frac {24 b^2 x^{7/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}-\frac {84 i b^2 x^2 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {24 i a b x^{7/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {252 b^2 x^{5/3} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {84 a b x^2 \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {630 i b^2 x^{4/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac {252 i a b x^{5/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {1260 b^2 x \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}+\frac {630 a b x^{4/3} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac {1890 i b^2 x^{2/3} \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^7}+\frac {1260 i a b x \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}+\frac {1890 b^2 \sqrt [3]{x} \operatorname {PolyLog}\left (7,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^8}-\frac {1890 a 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^2 \operatorname {PolyLog}\left (8,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^9}-\frac {1890 i a b \sqrt [3]{x} \operatorname {PolyLog}\left (8,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^8}+\frac {945 a b \operatorname {PolyLog}\left (9,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^9}+\frac {3 b^2 x^{8/3} \tan \left (c+d \sqrt [3]{x}\right )}{d} \] Output:

945*I*b^2*polylog(8,-exp(2*I*(c+d*x^(1/3))))/d^9+1/3*a^2*x^3+1260*I*a*b*x* 
polylog(6,-exp(2*I*(c+d*x^(1/3))))/d^6-1/3*b^2*x^3+24*b^2*x^(7/3)*ln(1+exp 
(2*I*(c+d*x^(1/3))))/d^2-6*a*b*x^(8/3)*ln(1+exp(2*I*(c+d*x^(1/3))))/d-252* 
I*a*b*x^(5/3)*polylog(4,-exp(2*I*(c+d*x^(1/3))))/d^4+2/3*I*a*b*x^3+252*b^2 
*x^(5/3)*polylog(3,-exp(2*I*(c+d*x^(1/3))))/d^4-84*a*b*x^2*polylog(3,-exp( 
2*I*(c+d*x^(1/3))))/d^3+630*I*b^2*x^(4/3)*polylog(4,-exp(2*I*(c+d*x^(1/3)) 
))/d^5-3*I*b^2*x^(8/3)/d-1260*b^2*x*polylog(5,-exp(2*I*(c+d*x^(1/3))))/d^6 
+630*a*b*x^(4/3)*polylog(5,-exp(2*I*(c+d*x^(1/3))))/d^5-1890*I*b^2*x^(2/3) 
*polylog(6,-exp(2*I*(c+d*x^(1/3))))/d^7-84*I*b^2*x^2*polylog(2,-exp(2*I*(c 
+d*x^(1/3))))/d^3+1890*b^2*x^(1/3)*polylog(7,-exp(2*I*(c+d*x^(1/3))))/d^8- 
1890*a*b*x^(2/3)*polylog(7,-exp(2*I*(c+d*x^(1/3))))/d^7+24*I*a*b*x^(7/3)*p 
olylog(2,-exp(2*I*(c+d*x^(1/3))))/d^2-1890*I*a*b*x^(1/3)*polylog(8,-exp(2* 
I*(c+d*x^(1/3))))/d^8+945*a*b*polylog(9,-exp(2*I*(c+d*x^(1/3))))/d^9+3*b^2 
*x^(8/3)*tan(c+d*x^(1/3))/d
 

Mathematica [A] (verified)

Time = 3.84 (sec) , antiderivative size = 828, normalized size of antiderivative = 1.39 \[ \int x^2 \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2 \, dx =\text {Too large to display} \] Input:

Integrate[x^2*(a + b*Tan[c + d*x^(1/3)])^2,x]
 

Output:

(((-I)*b*E^((2*I)*c)*((-18*b*d^8*x^(8/3))/E^((2*I)*c) + (4*a*d^9*x^3)/E^(( 
2*I)*c) + ((72*I)*b*d^7*(1 + E^((2*I)*c))*x^(7/3)*Log[1 + E^((-2*I)*(c + d 
*x^(1/3)))])/E^((2*I)*c) - ((18*I)*a*d^8*(1 + E^((2*I)*c))*x^(8/3)*Log[1 + 
 E^((-2*I)*(c + d*x^(1/3)))])/E^((2*I)*c) - 252*b*d^6*(1 + E^((-2*I)*c))*x 
^2*PolyLog[2, -E^((-2*I)*(c + d*x^(1/3)))] + 72*a*d^7*(1 + E^((-2*I)*c))*x 
^(7/3)*PolyLog[2, -E^((-2*I)*(c + d*x^(1/3)))] + ((756*I)*b*d^5*(1 + E^((2 
*I)*c))*x^(5/3)*PolyLog[3, -E^((-2*I)*(c + d*x^(1/3)))])/E^((2*I)*c) - ((2 
52*I)*a*d^6*(1 + E^((2*I)*c))*x^2*PolyLog[3, -E^((-2*I)*(c + d*x^(1/3)))]) 
/E^((2*I)*c) + 1890*b*d^4*(1 + E^((-2*I)*c))*x^(4/3)*PolyLog[4, -E^((-2*I) 
*(c + d*x^(1/3)))] - 756*a*d^5*(1 + E^((-2*I)*c))*x^(5/3)*PolyLog[4, -E^(( 
-2*I)*(c + d*x^(1/3)))] - ((3780*I)*b*d^3*(1 + E^((2*I)*c))*x*PolyLog[5, - 
E^((-2*I)*(c + d*x^(1/3)))])/E^((2*I)*c) + ((1890*I)*a*d^4*(1 + E^((2*I)*c 
))*x^(4/3)*PolyLog[5, -E^((-2*I)*(c + d*x^(1/3)))])/E^((2*I)*c) - 5670*b*d 
^2*(1 + E^((-2*I)*c))*x^(2/3)*PolyLog[6, -E^((-2*I)*(c + d*x^(1/3)))] + 37 
80*a*d^3*(1 + E^((-2*I)*c))*x*PolyLog[6, -E^((-2*I)*(c + d*x^(1/3)))] + (( 
5670*I)*b*d*(1 + E^((2*I)*c))*x^(1/3)*PolyLog[7, -E^((-2*I)*(c + d*x^(1/3) 
))])/E^((2*I)*c) - ((5670*I)*a*d^2*(1 + E^((2*I)*c))*x^(2/3)*PolyLog[7, -E 
^((-2*I)*(c + d*x^(1/3)))])/E^((2*I)*c) + 2835*b*(1 + E^((-2*I)*c))*PolyLo 
g[8, -E^((-2*I)*(c + d*x^(1/3)))] - 5670*a*d*(1 + E^((-2*I)*c))*x^(1/3)*Po 
lyLog[8, -E^((-2*I)*(c + d*x^(1/3)))] + ((2835*I)*a*(1 + E^((2*I)*c))*P...
 

Rubi [A] (verified)

Time = 1.06 (sec) , antiderivative size = 598, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4234, 3042, 4205, 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.

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

\(\Big \downarrow \) 4234

\(\displaystyle 3 \int x^{8/3} \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2d\sqrt [3]{x}\)

\(\Big \downarrow \) 3042

\(\displaystyle 3 \int x^{8/3} \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2d\sqrt [3]{x}\)

\(\Big \downarrow \) 4205

\(\displaystyle 3 \int \left (a^2 x^{8/3}+b^2 \tan ^2\left (c+d \sqrt [3]{x}\right ) x^{8/3}+2 a b \tan \left (c+d \sqrt [3]{x}\right ) x^{8/3}\right )d\sqrt [3]{x}\)

\(\Big \downarrow \) 2009

\(\displaystyle 3 \left (\frac {a^2 x^3}{9}+\frac {315 a b \operatorname {PolyLog}\left (9,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^9}-\frac {630 i a b \sqrt [3]{x} \operatorname {PolyLog}\left (8,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^8}-\frac {630 a b x^{2/3} \operatorname {PolyLog}\left (7,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^7}+\frac {420 i a b x \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}+\frac {210 a b x^{4/3} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac {84 i a b x^{5/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {28 a b x^2 \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {8 i a b x^{7/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {2 a b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {2}{9} i a b x^3+\frac {315 i b^2 \operatorname {PolyLog}\left (8,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^9}+\frac {630 b^2 \sqrt [3]{x} \operatorname {PolyLog}\left (7,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^8}-\frac {630 i b^2 x^{2/3} \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^7}-\frac {420 b^2 x \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}+\frac {210 i b^2 x^{4/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}+\frac {84 b^2 x^{5/3} \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {28 i b^2 x^2 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {8 b^2 x^{7/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {b^2 x^{8/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}-\frac {i b^2 x^{8/3}}{d}-\frac {b^2 x^3}{9}\right )\)

Input:

Int[x^2*(a + b*Tan[c + d*x^(1/3)])^2,x]
 

Output:

3*(((-I)*b^2*x^(8/3))/d + (a^2*x^3)/9 + ((2*I)/9)*a*b*x^3 - (b^2*x^3)/9 + 
(8*b^2*x^(7/3)*Log[1 + E^((2*I)*(c + d*x^(1/3)))])/d^2 - (2*a*b*x^(8/3)*Lo 
g[1 + E^((2*I)*(c + d*x^(1/3)))])/d - ((28*I)*b^2*x^2*PolyLog[2, -E^((2*I) 
*(c + d*x^(1/3)))])/d^3 + ((8*I)*a*b*x^(7/3)*PolyLog[2, -E^((2*I)*(c + d*x 
^(1/3)))])/d^2 + (84*b^2*x^(5/3)*PolyLog[3, -E^((2*I)*(c + d*x^(1/3)))])/d 
^4 - (28*a*b*x^2*PolyLog[3, -E^((2*I)*(c + d*x^(1/3)))])/d^3 + ((210*I)*b^ 
2*x^(4/3)*PolyLog[4, -E^((2*I)*(c + d*x^(1/3)))])/d^5 - ((84*I)*a*b*x^(5/3 
)*PolyLog[4, -E^((2*I)*(c + d*x^(1/3)))])/d^4 - (420*b^2*x*PolyLog[5, -E^( 
(2*I)*(c + d*x^(1/3)))])/d^6 + (210*a*b*x^(4/3)*PolyLog[5, -E^((2*I)*(c + 
d*x^(1/3)))])/d^5 - ((630*I)*b^2*x^(2/3)*PolyLog[6, -E^((2*I)*(c + d*x^(1/ 
3)))])/d^7 + ((420*I)*a*b*x*PolyLog[6, -E^((2*I)*(c + d*x^(1/3)))])/d^6 + 
(630*b^2*x^(1/3)*PolyLog[7, -E^((2*I)*(c + d*x^(1/3)))])/d^8 - (630*a*b*x^ 
(2/3)*PolyLog[7, -E^((2*I)*(c + d*x^(1/3)))])/d^7 + ((315*I)*b^2*PolyLog[8 
, -E^((2*I)*(c + d*x^(1/3)))])/d^9 - ((630*I)*a*b*x^(1/3)*PolyLog[8, -E^(( 
2*I)*(c + d*x^(1/3)))])/d^8 + (315*a*b*PolyLog[9, -E^((2*I)*(c + d*x^(1/3) 
))])/d^9 + (b^2*x^(8/3)*Tan[c + d*x^(1/3)])/d)
 

Defintions of rubi rules used

rule 2009

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

rule 3042

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

rule 4205

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.) 
, x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (a + b*Tan[e + f*x])^n, x], 
 x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral(x**2*(a + b*tan(c + d*x**(1/3)))**2, 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. 4725 vs. \(2 (473) = 946\).

Time = 0.66 (sec) , antiderivative size = 4725, normalized size of antiderivative = 7.91 \[ \int x^2 \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2 \, dx=\text {Too large to display} \] Input:

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

Output:

1/3*((d*x^(1/3) + c)^9*a^2 - 9*(d*x^(1/3) + c)^8*a^2*c + 36*(d*x^(1/3) + c 
)^7*a^2*c^2 - 84*(d*x^(1/3) + c)^6*a^2*c^3 + 126*(d*x^(1/3) + c)^5*a^2*c^4 
 - 126*(d*x^(1/3) + c)^4*a^2*c^5 + 84*(d*x^(1/3) + c)^3*a^2*c^6 - 36*(d*x^ 
(1/3) + c)^2*a^2*c^7 + 9*(d*x^(1/3) + c)*a^2*c^8 + 18*a*b*c^8*log(sec(d*x^ 
(1/3) + c)) - 9*(-315*I*(d*x^(1/3) + c)*b^2*c^8 - 35*(2*a*b + I*b^2)*(d*x^ 
(1/3) + c)^9 + 315*(2*a*b + I*b^2)*(d*x^(1/3) + c)^8*c - 1260*(2*a*b + I*b 
^2)*(d*x^(1/3) + c)^7*c^2 + 2940*(2*a*b + I*b^2)*(d*x^(1/3) + c)^6*c^3 - 4 
410*(2*a*b + I*b^2)*(d*x^(1/3) + c)^5*c^4 + 4410*(2*a*b + I*b^2)*(d*x^(1/3 
) + c)^4*c^5 - 2940*(2*a*b + I*b^2)*(d*x^(1/3) + c)^3*c^6 + 1260*(2*a*b + 
I*b^2)*(d*x^(1/3) + c)^2*c^7 - 630*b^2*c^8 + 24*(420*(d*x^(1/3) + c)^8*a*b 
 + 105*b^2*c^7 - 960*(2*a*b*c + b^2)*(d*x^(1/3) + c)^7 + 3920*(a*b*c^2 + b 
^2*c)*(d*x^(1/3) + c)^6 - 2352*(2*a*b*c^3 + 3*b^2*c^2)*(d*x^(1/3) + c)^5 + 
 3675*(a*b*c^4 + 2*b^2*c^3)*(d*x^(1/3) + c)^4 - 980*(2*a*b*c^5 + 5*b^2*c^4 
)*(d*x^(1/3) + c)^3 + 735*(a*b*c^6 + 3*b^2*c^5)*(d*x^(1/3) + c)^2 - 105*(2 
*a*b*c^7 + 7*b^2*c^6)*(d*x^(1/3) + c) + (420*(d*x^(1/3) + c)^8*a*b + 105*b 
^2*c^7 - 960*(2*a*b*c + b^2)*(d*x^(1/3) + c)^7 + 3920*(a*b*c^2 + b^2*c)*(d 
*x^(1/3) + c)^6 - 2352*(2*a*b*c^3 + 3*b^2*c^2)*(d*x^(1/3) + c)^5 + 3675*(a 
*b*c^4 + 2*b^2*c^3)*(d*x^(1/3) + c)^4 - 980*(2*a*b*c^5 + 5*b^2*c^4)*(d*x^( 
1/3) + c)^3 + 735*(a*b*c^6 + 3*b^2*c^5)*(d*x^(1/3) + c)^2 - 105*(2*a*b*c^7 
 + 7*b^2*c^6)*(d*x^(1/3) + c))*cos(2*d*x^(1/3) + 2*c) - (-420*I*(d*x^(1...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int x^2 \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2 \, dx=\frac {9 x^{\frac {8}{3}} \tan \left (x^{\frac {1}{3}} d +c \right ) b^{2}-24 \left (\int x^{\frac {5}{3}} \tan \left (x^{\frac {1}{3}} d +c \right )d x \right ) b^{2}+6 \left (\int \tan \left (x^{\frac {1}{3}} d +c \right ) x^{2}d x \right ) a b d +a^{2} d \,x^{3}-b^{2} d \,x^{3}}{3 d} \] Input:

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

Output:

(9*x**(2/3)*tan(x**(1/3)*d + c)*b**2*x**2 - 24*int(x**(2/3)*tan(x**(1/3)*d 
 + c)*x,x)*b**2 + 6*int(tan(x**(1/3)*d + c)*x**2,x)*a*b*d + a**2*d*x**3 - 
b**2*d*x**3)/(3*d)