\(\int \frac {x^2}{(a+b \tan (c+d \sqrt [3]{x}))^2} \, dx\) [62]

Optimal result

Integrand size = 20, antiderivative size = 1691 \[ \int \frac {x^2}{\left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2} \, dx =\text {Too large to display} \] Output:

945*I*b^2*polylog(9,-(a-I*b)*exp(2*I*(c+d*x^(1/3)))/(a+I*b))/(a^2+b^2)^2/d 
^9+1/3*x^3/(a-I*b)^2+4/3*b*x^3/(I*a-b)/(a-I*b)^2+6*b^2*x^(8/3)/(a+I*b)/(I* 
a+b)^2/d/(I*a-b+(I*a+b)*exp(2*I*(c+d*x^(1/3))))-1890*b*x^(1/3)*polylog(8,- 
(a-I*b)*exp(2*I*(c+d*x^(1/3)))/(a+I*b))/(I*a-b)/(a-I*b)^2/d^8+1890*b*x^(2/ 
3)*polylog(7,-(a-I*b)*exp(2*I*(c+d*x^(1/3)))/(a+I*b))/(a-I*b)^2/(a+I*b)/d^ 
7+1260*b*x*polylog(6,-(a-I*b)*exp(2*I*(c+d*x^(1/3)))/(a+I*b))/(I*a-b)/(a-I 
*b)^2/d^6-630*b*x^(4/3)*polylog(5,-(a-I*b)*exp(2*I*(c+d*x^(1/3)))/(a+I*b)) 
/(a-I*b)^2/(a+I*b)/d^5-252*b*x^(5/3)*polylog(4,-(a-I*b)*exp(2*I*(c+d*x^(1/ 
3)))/(a+I*b))/(I*a-b)/(a-I*b)^2/d^4+84*b*x^2*polylog(3,-(a-I*b)*exp(2*I*(c 
+d*x^(1/3)))/(a+I*b))/(a-I*b)^2/(a+I*b)/d^3+24*b*x^(7/3)*polylog(2,-(a-I*b 
)*exp(2*I*(c+d*x^(1/3)))/(a+I*b))/(I*a-b)/(a-I*b)^2/d^2+6*b*x^(8/3)*ln(1+( 
a-I*b)*exp(2*I*(c+d*x^(1/3)))/(a+I*b))/(a-I*b)^2/(a+I*b)/d-1890*I*b^2*x^(2 
/3)*polylog(7,-(a-I*b)*exp(2*I*(c+d*x^(1/3)))/(a+I*b))/(a^2+b^2)^2/d^7-189 
0*I*b^2*x^(2/3)*polylog(6,-(a-I*b)*exp(2*I*(c+d*x^(1/3)))/(a+I*b))/(a^2+b^ 
2)^2/d^7-84*I*b^2*x^2*polylog(3,-(a-I*b)*exp(2*I*(c+d*x^(1/3)))/(a+I*b))/( 
a^2+b^2)^2/d^3-84*I*b^2*x^2*polylog(2,-(a-I*b)*exp(2*I*(c+d*x^(1/3)))/(a+I 
*b))/(a^2+b^2)^2/d^3-6*I*b^2*x^(8/3)*ln(1+(a-I*b)*exp(2*I*(c+d*x^(1/3)))/( 
a+I*b))/(a^2+b^2)^2/d+945*I*b^2*polylog(8,-(a-I*b)*exp(2*I*(c+d*x^(1/3)))/ 
(a+I*b))/(a^2+b^2)^2/d^9+630*I*b^2*x^(4/3)*polylog(5,-(a-I*b)*exp(2*I*(c+d 
*x^(1/3)))/(a+I*b))/(a^2+b^2)^2/d^5+630*I*b^2*x^(4/3)*polylog(4,-(a-I*b...
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 3.88 (sec) , antiderivative size = 1136, normalized size of antiderivative = 0.67 \[ \int \frac {x^2}{\left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

(((-I)*b*(18*(a + I*b)*b*(I*a + b)*d^8*x^(8/3) + 4*a*(a + I*b)*(I*a + b)*d 
^9*x^3 + 72*(a - I*b)*b*d^7*((-I)*b*(-1 + E^((2*I)*c)) + a*(1 + E^((2*I)*c 
)))*x^(7/3)*Log[1 + (a + I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] + 18* 
a*(a - I*b)*d^8*((-I)*b*(-1 + E^((2*I)*c)) + a*(1 + E^((2*I)*c)))*x^(8/3)* 
Log[1 + (a + I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] + 63*b*(I*a + b)* 
(b*(-1 + E^((2*I)*c)) + I*a*(1 + E^((2*I)*c)))*((-4*I)*d^6*x^2*PolyLog[2, 
(-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] - 12*d^5*x^(5/3)*PolyLog 
[3, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] + (15*I)*(2*d^4*x^(4 
/3)*PolyLog[4, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] - (4*I)*d 
^3*x*PolyLog[5, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] - 6*d^2* 
x^(2/3)*PolyLog[6, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] + (6* 
I)*d*x^(1/3)*PolyLog[7, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] 
+ 3*PolyLog[8, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))])) + 9*a*( 
a - I*b)*((-I)*b*(-1 + E^((2*I)*c)) + a*(1 + E^((2*I)*c)))*((8*I)*d^7*x^(7 
/3)*PolyLog[2, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] + 28*d^6* 
x^2*PolyLog[3, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] - (84*I)* 
d^5*x^(5/3)*PolyLog[4, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] - 
 105*(2*d^4*x^(4/3)*PolyLog[5, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/ 
3))))] - (4*I)*d^3*x*PolyLog[6, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1 
/3))))] - 6*d^2*x^(2/3)*PolyLog[7, (-a - I*b)/((a - I*b)*E^((2*I)*(c + ...
 

Rubi [A] (verified)

Time = 3.10 (sec) , antiderivative size = 1774, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4234, 3042, 4217, 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.

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

\(\Big \downarrow \) 4234

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4217

\(\displaystyle 3 \int \left (\frac {4 b x^{8/3}}{(a-i b)^2 \left (i a e^{2 i c+2 i d \sqrt [3]{x}} \left (1-\frac {i b}{a}\right )+i a \left (\frac {i b}{a}+1\right )\right )}+\frac {x^{8/3}}{(a-i b)^2}-\frac {4 b^2 x^{8/3}}{(i a+b)^2 \left (i a e^{2 i c+2 i d \sqrt [3]{x}} \left (1-\frac {i b}{a}\right )+i a \left (\frac {i b}{a}+1\right )\right )^2}\right )d\sqrt [3]{x}\)

\(\Big \downarrow \) 2009

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

Input:

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

Output:

3*(((-2*I)*b^2*x^(8/3))/((a^2 + b^2)^2*d) + (2*b^2*x^(8/3))/((a + I*b)*(I* 
a + b)^2*d*(I*a - b + (I*a + b)*E^((2*I)*c + (2*I)*d*x^(1/3)))) + x^3/(9*( 
a - I*b)^2) + (4*b*x^3)/(9*(I*a - b)*(a - I*b)^2) - (4*b^2*x^3)/(9*(a^2 + 
b^2)^2) + (8*b^2*x^(7/3)*Log[1 + ((a - I*b)*E^((2*I)*c + (2*I)*d*x^(1/3))) 
/(a + I*b)])/((a^2 + b^2)^2*d^2) + (2*b*x^(8/3)*Log[1 + ((a - I*b)*E^((2*I 
)*c + (2*I)*d*x^(1/3)))/(a + I*b)])/((a - I*b)^2*(a + I*b)*d) - ((2*I)*b^2 
*x^(8/3)*Log[1 + ((a - I*b)*E^((2*I)*c + (2*I)*d*x^(1/3)))/(a + I*b)])/((a 
^2 + b^2)^2*d) - ((28*I)*b^2*x^2*PolyLog[2, -(((a - I*b)*E^((2*I)*c + (2*I 
)*d*x^(1/3)))/(a + I*b))])/((a^2 + b^2)^2*d^3) + (8*b*x^(7/3)*PolyLog[2, - 
(((a - I*b)*E^((2*I)*c + (2*I)*d*x^(1/3)))/(a + I*b))])/((I*a - b)*(a - I* 
b)^2*d^2) - (8*b^2*x^(7/3)*PolyLog[2, -(((a - I*b)*E^((2*I)*c + (2*I)*d*x^ 
(1/3)))/(a + I*b))])/((a^2 + b^2)^2*d^2) + (84*b^2*x^(5/3)*PolyLog[3, -((( 
a - I*b)*E^((2*I)*c + (2*I)*d*x^(1/3)))/(a + I*b))])/((a^2 + b^2)^2*d^4) + 
 (28*b*x^2*PolyLog[3, -(((a - I*b)*E^((2*I)*c + (2*I)*d*x^(1/3)))/(a + I*b 
))])/((a - I*b)^2*(a + I*b)*d^3) - ((28*I)*b^2*x^2*PolyLog[3, -(((a - I*b) 
*E^((2*I)*c + (2*I)*d*x^(1/3)))/(a + I*b))])/((a^2 + b^2)^2*d^3) + ((210*I 
)*b^2*x^(4/3)*PolyLog[4, -(((a - I*b)*E^((2*I)*c + (2*I)*d*x^(1/3)))/(a + 
I*b))])/((a^2 + b^2)^2*d^5) - (84*b*x^(5/3)*PolyLog[4, -(((a - I*b)*E^((2* 
I)*c + (2*I)*d*x^(1/3)))/(a + I*b))])/((I*a - b)*(a - I*b)^2*d^4) + (84*b^ 
2*x^(5/3)*PolyLog[4, -(((a - I*b)*E^((2*I)*c + (2*I)*d*x^(1/3)))/(a + I...
 

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 4217

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), 
 x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (1/(a - I*b) - 2*I*(b/(a^2 + 
 b^2 + (a - I*b)^2*E^(2*I*(e + f*x)))))^(-n), x], x] /; FreeQ[{a, b, c, d, 
e, f}, x] && NeQ[a^2 + b^2, 0] && ILtQ[n, 0] && IGtQ[m, 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]

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

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]

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

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

Output:

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

Sympy [F]

\[ \int \frac {x^2}{\left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2} \, dx=\int \frac {x^{2}}{\left (a + b \tan {\left (c + d \sqrt [3]{x} \right )}\right )^{2}}\, dx \] 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. 8152 vs. \(2 (1362) = 2724\).

Time = 2.38 (sec) , antiderivative size = 8152, normalized size of antiderivative = 4.82 \[ \int \frac {x^2}{\left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/105*(315*(2*a*b*log(b*tan(d*x^(1/3) + c) + a)/(a^4 + 2*a^2*b^2 + b^4) - 
a*b*log(tan(d*x^(1/3) + c)^2 + 1)/(a^4 + 2*a^2*b^2 + b^4) + (a^2 - b^2)*(d 
*x^(1/3) + c)/(a^4 + 2*a^2*b^2 + b^4) - b/(a^3 + a*b^2 + (a^2*b + b^3)*tan 
(d*x^(1/3) + c)))*c^8 + (35*(a^3 - I*a^2*b + a*b^2 - I*b^3)*(d*x^(1/3) + c 
)^9 - 315*(a^3 - I*a^2*b + a*b^2 - I*b^3)*(d*x^(1/3) + c)^8*c + 1260*(a^3 
- I*a^2*b + a*b^2 - I*b^3)*(d*x^(1/3) + c)^7*c^2 - 2940*(a^3 - I*a^2*b + a 
*b^2 - I*b^3)*(d*x^(1/3) + c)^6*c^3 + 4410*(a^3 - I*a^2*b + a*b^2 - I*b^3) 
*(d*x^(1/3) + c)^5*c^4 - 4410*(a^3 - I*a^2*b + a*b^2 - I*b^3)*(d*x^(1/3) + 
 c)^4*c^5 + 2940*(a^3 - I*a^2*b + a*b^2 - I*b^3)*(d*x^(1/3) + c)^3*c^6 - 1 
260*(a^3 - I*a^2*b + a*b^2 - I*b^3)*(d*x^(1/3) + c)^2*c^7 - 2520*((I*a*b^2 
 + b^3)*c^7*cos(2*d*x^(1/3) + 2*c) - (a*b^2 - I*b^3)*c^7*sin(2*d*x^(1/3) + 
 2*c) + (I*a*b^2 - b^3)*c^7)*arctan2(-b*cos(2*d*x^(1/3) + 2*c) + a*sin(2*d 
*x^(1/3) + 2*c) + b, a*cos(2*d*x^(1/3) + 2*c) + b*sin(2*d*x^(1/3) + 2*c) + 
 a) - 24*(420*(I*a^2*b - a*b^2)*(d*x^(1/3) + c)^8 + 960*(I*a*b^2 - b^3 + 2 
*(-I*a^2*b + a*b^2)*c)*(d*x^(1/3) + c)^7 + 3920*((I*a^2*b - a*b^2)*c^2 + ( 
-I*a*b^2 + b^3)*c)*(d*x^(1/3) + c)^6 + 2352*(2*(-I*a^2*b + a*b^2)*c^3 + 3* 
(I*a*b^2 - b^3)*c^2)*(d*x^(1/3) + c)^5 + 3675*((I*a^2*b - a*b^2)*c^4 + 2*( 
-I*a*b^2 + b^3)*c^3)*(d*x^(1/3) + c)^4 + 980*(2*(-I*a^2*b + a*b^2)*c^5 + 5 
*(I*a*b^2 - b^3)*c^4)*(d*x^(1/3) + c)^3 + 735*((I*a^2*b - a*b^2)*c^6 + 3*( 
-I*a*b^2 + b^3)*c^5)*(d*x^(1/3) + c)^2 + 105*(2*(-I*a^2*b + a*b^2)*c^7 ...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

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

Output:

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