\(\int \frac {x}{(a+b \tan (c+d \sqrt [3]{x}))^2} \, dx\) [63]

Optimal result

Integrand size = 18, antiderivative size = 1155 \[ \int \frac {x}{\left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2} \, dx =\text {Too large to display} \] Output:

-6*I*b^2*x^(5/3)/(a^2+b^2)^2/d+6*b^2*x^(5/3)/(a+I*b)/(I*a+b)^2/d/(I*a-b+(I 
*a+b)*exp(2*I*(c+d*x^(1/3))))+1/2*x^2/(a-I*b)^2+2*b*x^2/(I*a-b)/(a-I*b)^2- 
2*b^2*x^2/(a^2+b^2)^2+15*b^2*x^(4/3)*ln(1+(a-I*b)*exp(2*I*(c+d*x^(1/3)))/( 
a+I*b))/(a^2+b^2)^2/d^2+6*b*x^(5/3)*ln(1+(a-I*b)*exp(2*I*(c+d*x^(1/3)))/(a 
+I*b))/(a-I*b)^2/(a+I*b)/d+45*I*b^2*x^(1/3)*polylog(4,-(a-I*b)*exp(2*I*(c+ 
d*x^(1/3)))/(a+I*b))/(a^2+b^2)^2/d^5-30*I*b^2*x*polylog(2,-(a-I*b)*exp(2*I 
*(c+d*x^(1/3)))/(a+I*b))/(a^2+b^2)^2/d^3+15*b*x^(4/3)*polylog(2,-(a-I*b)*e 
xp(2*I*(c+d*x^(1/3)))/(a+I*b))/(I*a-b)/(a-I*b)^2/d^2-15*b^2*x^(4/3)*polylo 
g(2,-(a-I*b)*exp(2*I*(c+d*x^(1/3)))/(a+I*b))/(a^2+b^2)^2/d^2+45*b^2*x^(2/3 
)*polylog(3,-(a-I*b)*exp(2*I*(c+d*x^(1/3)))/(a+I*b))/(a^2+b^2)^2/d^4+30*b* 
x*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 
+45*I*b^2*x^(1/3)*polylog(5,-(a-I*b)*exp(2*I*(c+d*x^(1/3)))/(a+I*b))/(a^2+ 
b^2)^2/d^5-30*I*b^2*x*polylog(3,-(a-I*b)*exp(2*I*(c+d*x^(1/3)))/(a+I*b))/( 
a^2+b^2)^2/d^3-45*b*x^(2/3)*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+45*b^2*x^(2/3)*polylog(4,-(a-I*b)*exp(2*I*(c+d* 
x^(1/3)))/(a+I*b))/(a^2+b^2)^2/d^4-45/2*b^2*polylog(5,-(a-I*b)*exp(2*I*(c+ 
d*x^(1/3)))/(a+I*b))/(a^2+b^2)^2/d^6-45*b*x^(1/3)*polylog(5,-(a-I*b)*exp(2 
*I*(c+d*x^(1/3)))/(a+I*b))/(a-I*b)^2/(a+I*b)/d^5-6*I*b^2*x^(5/3)*ln(1+(a-I 
*b)*exp(2*I*(c+d*x^(1/3)))/(a+I*b))/(a^2+b^2)^2/d+45/2*b*polylog(6,-(a-I*b 
)*exp(2*I*(c+d*x^(1/3)))/(a+I*b))/(I*a-b)/(a-I*b)^2/d^6-45/2*b^2*polylo...
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 2.99 (sec) , antiderivative size = 852, normalized size of antiderivative = 0.74 \[ \int \frac {x}{\left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2} \, dx=\frac {-\frac {i b \left (12 (a+i b) b (i a+b) d^5 x^{5/3}+4 a (a+i b) (i a+b) d^6 x^2+30 (a-i b) b d^4 \left (-i b \left (-1+e^{2 i c}\right )+a \left (1+e^{2 i c}\right )\right ) x^{4/3} \log \left (1+\frac {(a+i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )+12 a (a-i b) d^5 \left (-i b \left (-1+e^{2 i c}\right )+a \left (1+e^{2 i c}\right )\right ) x^{5/3} \log \left (1+\frac {(a+i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )+15 (a-i b) b \left (-i b \left (-1+e^{2 i c}\right )+a \left (1+e^{2 i c}\right )\right ) \left (4 i d^3 x \operatorname {PolyLog}\left (2,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )+6 d^2 x^{2/3} \operatorname {PolyLog}\left (3,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )-6 i d \sqrt [3]{x} \operatorname {PolyLog}\left (4,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )-3 \operatorname {PolyLog}\left (5,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )\right )+15 a (a-i b) \left (-i b \left (-1+e^{2 i c}\right )+a \left (1+e^{2 i c}\right )\right ) \left (2 i d^4 x^{4/3} \operatorname {PolyLog}\left (2,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )+4 d^3 x \operatorname {PolyLog}\left (3,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )-6 i d^2 x^{2/3} \operatorname {PolyLog}\left (4,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )-6 d \sqrt [3]{x} \operatorname {PolyLog}\left (5,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )+3 i \operatorname {PolyLog}\left (6,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )\right )\right )}{d^6 \left (b-b e^{2 i c}-i a \left (1+e^{2 i c}\right )\right )}+\frac {(a-i b)^2 (a+i b) x^2 (a \cos (c)-b \sin (c))}{a \cos (c)+b \sin (c)}+\frac {6 (a-i b)^2 (a+i b) b^2 x^{5/3} \sin \left (d \sqrt [3]{x}\right )}{d (a \cos (c)+b \sin (c)) \left (a \cos \left (c+d \sqrt [3]{x}\right )+b \sin \left (c+d \sqrt [3]{x}\right )\right )}}{2 (a-i b)^3 (a+i b)^2} \] Input:

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

Output:

(((-I)*b*(12*(a + I*b)*b*(I*a + b)*d^5*x^(5/3) + 4*a*(a + I*b)*(I*a + b)*d 
^6*x^2 + 30*(a - I*b)*b*d^4*((-I)*b*(-1 + E^((2*I)*c)) + a*(1 + E^((2*I)*c 
)))*x^(4/3)*Log[1 + (a + I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] + 12* 
a*(a - I*b)*d^5*((-I)*b*(-1 + E^((2*I)*c)) + a*(1 + E^((2*I)*c)))*x^(5/3)* 
Log[1 + (a + I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] + 15*(a - I*b)*b* 
((-I)*b*(-1 + E^((2*I)*c)) + a*(1 + E^((2*I)*c)))*((4*I)*d^3*x*PolyLog[2, 
(-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] + 6*d^2*x^(2/3)*PolyLog[ 
3, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] - (6*I)*d*x^(1/3)*Pol 
yLog[4, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] - 3*PolyLog[5, ( 
-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))]) + 15*a*(a - I*b)*((-I)*b 
*(-1 + E^((2*I)*c)) + a*(1 + E^((2*I)*c)))*((2*I)*d^4*x^(4/3)*PolyLog[2, ( 
-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] + 4*d^3*x*PolyLog[3, (-a 
- I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] - (6*I)*d^2*x^(2/3)*PolyLog[ 
4, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] - 6*d*x^(1/3)*PolyLog 
[5, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] + (3*I)*PolyLog[6, ( 
-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))])))/(d^6*(b - b*E^((2*I)*c 
) - I*a*(1 + E^((2*I)*c)))) + ((a - I*b)^2*(a + I*b)*x^2*(a*Cos[c] - b*Sin 
[c]))/(a*Cos[c] + b*Sin[c]) + (6*(a - I*b)^2*(a + I*b)*b^2*x^(5/3)*Sin[d*x 
^(1/3)])/(d*(a*Cos[c] + b*Sin[c])*(a*Cos[c + d*x^(1/3)] + b*Sin[c + d*x^(1 
/3)])))/(2*(a - I*b)^3*(a + I*b)^2)
 

Rubi [A] (verified)

Time = 2.28 (sec) , antiderivative size = 1215, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, 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.

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

\(\Big \downarrow \) 4234

\(\displaystyle 3 \int \frac {x^{5/3}}{\left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2}d\sqrt [3]{x}\)

\(\Big \downarrow \) 3042

\(\displaystyle 3 \int \frac {x^{5/3}}{\left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2}d\sqrt [3]{x}\)

\(\Big \downarrow \) 4217

\(\displaystyle 3 \int \left (-\frac {4 x^{5/3} b^2}{(i a+b)^2 \left (i a e^{2 i c+2 i d \sqrt [3]{x}} \left (1-\frac {i b}{a}\right )+i a \left (\frac {i b}{a}+1\right )\right )^2}+\frac {4 x^{5/3} b}{(a-i b)^2 \left (i a e^{2 i c+2 i d \sqrt [3]{x}} \left (1-\frac {i b}{a}\right )+i a \left (\frac {i b}{a}+1\right )\right )}+\frac {x^{5/3}}{(a-i b)^2}\right )d\sqrt [3]{x}\)

\(\Big \downarrow \) 2009

\(\displaystyle 3 \left (-\frac {2 x^2 b^2}{3 \left (a^2+b^2\right )^2}-\frac {2 i x^{5/3} b^2}{\left (a^2+b^2\right )^2 d}+\frac {2 x^{5/3} b^2}{(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 {2 i x^{5/3} \log \left (\frac {e^{2 i c+2 i d \sqrt [3]{x}} (a-i b)}{a+i b}+1\right ) b^2}{\left (a^2+b^2\right )^2 d}+\frac {5 x^{4/3} \log \left (\frac {e^{2 i c+2 i d \sqrt [3]{x}} (a-i b)}{a+i b}+1\right ) b^2}{\left (a^2+b^2\right )^2 d^2}-\frac {5 x^{4/3} \operatorname {PolyLog}\left (2,-\frac {(a-i b) e^{2 i c+2 i d \sqrt [3]{x}}}{a+i b}\right ) b^2}{\left (a^2+b^2\right )^2 d^2}-\frac {10 i x \operatorname {PolyLog}\left (2,-\frac {(a-i b) e^{2 i c+2 i d \sqrt [3]{x}}}{a+i b}\right ) b^2}{\left (a^2+b^2\right )^2 d^3}-\frac {10 i x \operatorname {PolyLog}\left (3,-\frac {(a-i b) e^{2 i c+2 i d \sqrt [3]{x}}}{a+i b}\right ) b^2}{\left (a^2+b^2\right )^2 d^3}+\frac {15 x^{2/3} \operatorname {PolyLog}\left (3,-\frac {(a-i b) e^{2 i c+2 i d \sqrt [3]{x}}}{a+i b}\right ) b^2}{\left (a^2+b^2\right )^2 d^4}+\frac {15 x^{2/3} \operatorname {PolyLog}\left (4,-\frac {(a-i b) e^{2 i c+2 i d \sqrt [3]{x}}}{a+i b}\right ) b^2}{\left (a^2+b^2\right )^2 d^4}+\frac {15 i \sqrt [3]{x} \operatorname {PolyLog}\left (4,-\frac {(a-i b) e^{2 i c+2 i d \sqrt [3]{x}}}{a+i b}\right ) b^2}{\left (a^2+b^2\right )^2 d^5}+\frac {15 i \sqrt [3]{x} \operatorname {PolyLog}\left (5,-\frac {(a-i b) e^{2 i c+2 i d \sqrt [3]{x}}}{a+i b}\right ) b^2}{\left (a^2+b^2\right )^2 d^5}-\frac {15 \operatorname {PolyLog}\left (5,-\frac {(a-i b) e^{2 i c+2 i d \sqrt [3]{x}}}{a+i b}\right ) b^2}{2 \left (a^2+b^2\right )^2 d^6}-\frac {15 \operatorname {PolyLog}\left (6,-\frac {(a-i b) e^{2 i c+2 i d \sqrt [3]{x}}}{a+i b}\right ) b^2}{2 \left (a^2+b^2\right )^2 d^6}+\frac {2 x^2 b}{3 (i a-b) (a-i b)^2}+\frac {2 x^{5/3} \log \left (\frac {e^{2 i c+2 i d \sqrt [3]{x}} (a-i b)}{a+i b}+1\right ) b}{(a-i b)^2 (a+i b) d}+\frac {5 x^{4/3} \operatorname {PolyLog}\left (2,-\frac {(a-i b) e^{2 i c+2 i d \sqrt [3]{x}}}{a+i b}\right ) b}{(i a-b) (a-i b)^2 d^2}+\frac {10 x \operatorname {PolyLog}\left (3,-\frac {(a-i b) e^{2 i c+2 i d \sqrt [3]{x}}}{a+i b}\right ) b}{(a-i b)^2 (a+i b) d^3}-\frac {15 x^{2/3} \operatorname {PolyLog}\left (4,-\frac {(a-i b) e^{2 i c+2 i d \sqrt [3]{x}}}{a+i b}\right ) b}{(i a-b) (a-i b)^2 d^4}-\frac {15 \sqrt [3]{x} \operatorname {PolyLog}\left (5,-\frac {(a-i b) e^{2 i c+2 i d \sqrt [3]{x}}}{a+i b}\right ) b}{(a-i b)^2 (a+i b) d^5}+\frac {15 \operatorname {PolyLog}\left (6,-\frac {(a-i b) e^{2 i c+2 i d \sqrt [3]{x}}}{a+i b}\right ) b}{2 (i a-b) (a-i b)^2 d^6}+\frac {x^2}{6 (a-i b)^2}\right )\)

Input:

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

Output:

3*(((-2*I)*b^2*x^(5/3))/((a^2 + b^2)^2*d) + (2*b^2*x^(5/3))/((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^2/(6*( 
a - I*b)^2) + (2*b*x^2)/(3*(I*a - b)*(a - I*b)^2) - (2*b^2*x^2)/(3*(a^2 + 
b^2)^2) + (5*b^2*x^(4/3)*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^(5/3)*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^(5/3)*Log[1 + ((a - I*b)*E^((2*I)*c + (2*I)*d*x^(1/3)))/(a + I*b)])/((a 
^2 + b^2)^2*d) - ((10*I)*b^2*x*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) + (5*b*x^(4/3)*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) - (5*b^2*x^(4/3)*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) + (15*b^2*x^(2/3)*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) + ( 
10*b*x*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) - ((10*I)*b^2*x*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) + ((15*I)*b^2*x 
^(1/3)*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) - (15*b*x^(2/3)*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) + (15*b^2*x^(2/ 
3)*PolyLog[4, -(((a - I*b)*E^((2*I)*c + (2*I)*d*x^(1/3)))/(a + I*b))])/...
 

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral(x/(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. 4345 vs. \(2 (928) = 1856\).

Time = 0.99 (sec) , antiderivative size = 4345, normalized size of antiderivative = 3.76 \[ \int \frac {x}{\left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/10*(30*(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^5 - (5*(a^3 - I*a^2*b + a*b^2 - I*b^3)*(d*x^(1/3) + c)^ 
6 - 30*(a^3 - I*a^2*b + a*b^2 - I*b^3)*(d*x^(1/3) + c)^5*c + 75*(a^3 - I*a 
^2*b + a*b^2 - I*b^3)*(d*x^(1/3) + c)^4*c^2 - 100*(a^3 - I*a^2*b + a*b^2 - 
 I*b^3)*(d*x^(1/3) + c)^3*c^3 + 75*(a^3 - I*a^2*b + a*b^2 - I*b^3)*(d*x^(1 
/3) + c)^2*c^4 - 150*((-I*a*b^2 - b^3)*c^4*cos(2*d*x^(1/3) + 2*c) + (a*b^2 
 - I*b^3)*c^4*sin(2*d*x^(1/3) + 2*c) + (-I*a*b^2 + b^3)*c^4)*arctan2(-b*co 
s(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) - 4*(48*(I*a^2*b - a*b^2)*(d*x^(1/3) + 
 c)^5 + 75*(I*a*b^2 - b^3 + 2*(-I*a^2*b + a*b^2)*c)*(d*x^(1/3) + c)^4 + 20 
0*((I*a^2*b - a*b^2)*c^2 + (-I*a*b^2 + b^3)*c)*(d*x^(1/3) + c)^3 + 75*(2*( 
-I*a^2*b + a*b^2)*c^3 + 3*(I*a*b^2 - b^3)*c^2)*(d*x^(1/3) + c)^2 + 75*((I* 
a^2*b - a*b^2)*c^4 + 2*(-I*a*b^2 + b^3)*c^3)*(d*x^(1/3) + c) + (48*(I*a^2* 
b + a*b^2)*(d*x^(1/3) + c)^5 + 75*(I*a*b^2 + b^3 + 2*(-I*a^2*b - a*b^2)*c) 
*(d*x^(1/3) + c)^4 + 200*((I*a^2*b + a*b^2)*c^2 + (-I*a*b^2 - b^3)*c)*(d*x 
^(1/3) + c)^3 + 75*(2*(-I*a^2*b - a*b^2)*c^3 + 3*(I*a*b^2 + b^3)*c^2)*(d*x 
^(1/3) + c)^2 + 75*((I*a^2*b + a*b^2)*c^4 + 2*(-I*a*b^2 - b^3)*c^3)*(d*x^( 
1/3) + c))*cos(2*d*x^(1/3) + 2*c) - (48*(a^2*b - I*a*b^2)*(d*x^(1/3) + ...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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