\(\int \frac {1}{(a+b \tan (c+d \sqrt [3]{x}))^2} \, dx\) [64]

Optimal result

Integrand size = 16, antiderivative size = 610 \[ \int \frac {1}{\left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2} \, dx=-\frac {6 i b^2 x^{2/3}}{\left (a^2+b^2\right )^2 d}+\frac {6 b^2 x^{2/3}}{(a+i b) (i a+b)^2 d \left (i a-b+(i a+b) e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}+\frac {x}{(a-i b)^2}+\frac {4 b x}{(i a-b) (a-i b)^2}-\frac {4 b^2 x}{\left (a^2+b^2\right )^2}+\frac {6 b^2 \sqrt [3]{x} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac {6 b x^{2/3} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d}-\frac {6 i b^2 x^{2/3} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d}-\frac {3 i b^2 \operatorname {PolyLog}\left (2,-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^3}+\frac {6 b \sqrt [3]{x} \operatorname {PolyLog}\left (2,-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(i a-b) (a-i b)^2 d^2}-\frac {6 b^2 \sqrt [3]{x} \operatorname {PolyLog}\left (2,-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac {3 b \operatorname {PolyLog}\left (3,-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d^3}-\frac {3 i b^2 \operatorname {PolyLog}\left (3,-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^3} \] Output:

-6*I*b^2*x^(2/3)/(a^2+b^2)^2/d+6*b^2*x^(2/3)/(a+I*b)/(I*a+b)^2/d/(I*a-b+(I 
*a+b)*exp(2*I*(c+d*x^(1/3))))+x/(a-I*b)^2+4*b*x/(I*a-b)/(a-I*b)^2-4*b^2*x/ 
(a^2+b^2)^2+6*b^2*x^(1/3)*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^(2/3)*ln(1+(a-I*b)*exp(2*I*(c+d*x^(1/3)))/(a+I*b))/(a-I 
*b)^2/(a+I*b)/d-6*I*b^2*x^(2/3)*ln(1+(a-I*b)*exp(2*I*(c+d*x^(1/3)))/(a+I*b 
))/(a^2+b^2)^2/d-3*I*b^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*b*x^(1/3)*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^2*x^(1/3)*polylog(2,-(a-I*b)*exp(2*I*(c+d 
*x^(1/3)))/(a+I*b))/(a^2+b^2)^2/d^2+3*b*polylog(3,-(a-I*b)*exp(2*I*(c+d*x^ 
(1/3)))/(a+I*b))/(a-I*b)^2/(a+I*b)/d^3-3*I*b^2*polylog(3,-(a-I*b)*exp(2*I* 
(c+d*x^(1/3)))/(a+I*b))/(a^2+b^2)^2/d^3
                                                                                    
                                                                                    
 

Mathematica [A] (warning: unable to verify)

Time = 2.80 (sec) , antiderivative size = 538, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2} \, dx=\frac {\frac {b \left (\frac {6 b x^{2/3}}{a-i b}+\frac {4 a d x}{a-i b}+\frac {6 b \left (-i b \left (-1+e^{2 i c}\right )+a \left (1+e^{2 i c}\right )\right ) \sqrt [3]{x} \log \left (1+\frac {(a+i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )}{(a+i b) (i a+b) d}+\frac {6 a \left (-i b \left (-1+e^{2 i c}\right )+a \left (1+e^{2 i c}\right )\right ) x^{2/3} \log \left (1+\frac {(a+i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )}{(a+i b) (i a+b)}+\frac {3 b \left (-i b \left (-1+e^{2 i c}\right )+a \left (1+e^{2 i c}\right )\right ) \operatorname {PolyLog}\left (2,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )}{\left (a^2+b^2\right ) d^2}+\frac {3 a \left (-i b \left (-1+e^{2 i c}\right )+a \left (1+e^{2 i c}\right )\right ) \left (2 d \sqrt [3]{x} \operatorname {PolyLog}\left (2,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )-i \operatorname {PolyLog}\left (3,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )\right )}{\left (a^2+b^2\right ) d^2}\right )}{d \left (b-b e^{2 i c}-i a \left (1+e^{2 i c}\right )\right )}+\frac {x (a \cos (c)-b \sin (c))}{a \cos (c)+b \sin (c)}+\frac {3 b^2 x^{2/3} \sin \left (d \sqrt [3]{x}\right )}{d (a \cos (c)+b \sin (c)) \left (a \cos \left (c+d \sqrt [3]{x}\right )+b \sin \left (c+d \sqrt [3]{x}\right )\right )}}{a^2+b^2} \] Input:

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

Output:

((b*((6*b*x^(2/3))/(a - I*b) + (4*a*d*x)/(a - I*b) + (6*b*((-I)*b*(-1 + E^ 
((2*I)*c)) + a*(1 + E^((2*I)*c)))*x^(1/3)*Log[1 + (a + I*b)/((a - I*b)*E^( 
(2*I)*(c + d*x^(1/3))))])/((a + I*b)*(I*a + b)*d) + (6*a*((-I)*b*(-1 + E^( 
(2*I)*c)) + a*(1 + E^((2*I)*c)))*x^(2/3)*Log[1 + (a + I*b)/((a - I*b)*E^(( 
2*I)*(c + d*x^(1/3))))])/((a + I*b)*(I*a + b)) + (3*b*((-I)*b*(-1 + E^((2* 
I)*c)) + a*(1 + E^((2*I)*c)))*PolyLog[2, (-a - I*b)/((a - I*b)*E^((2*I)*(c 
 + d*x^(1/3))))])/((a^2 + b^2)*d^2) + (3*a*((-I)*b*(-1 + E^((2*I)*c)) + a* 
(1 + E^((2*I)*c)))*(2*d*x^(1/3)*PolyLog[2, (-a - I*b)/((a - I*b)*E^((2*I)* 
(c + d*x^(1/3))))] - I*PolyLog[3, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^ 
(1/3))))]))/((a^2 + b^2)*d^2)))/(d*(b - b*E^((2*I)*c) - I*a*(1 + E^((2*I)* 
c)))) + (x*(a*Cos[c] - b*Sin[c]))/(a*Cos[c] + b*Sin[c]) + (3*b^2*x^(2/3)*S 
in[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)])))/(a^2 + b^2)
 

Rubi [A] (verified)

Time = 1.67 (sec) , antiderivative size = 645, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4226, 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.

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

\(\Big \downarrow \) 4226

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4217

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

3*(((-2*I)*b^2*x^(2/3))/((a^2 + b^2)^2*d) + (2*b^2*x^(2/3))/((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*(a 
- I*b)^2) + (4*b*x)/(3*(I*a - b)*(a - I*b)^2) - (4*b^2*x)/(3*(a^2 + b^2)^2 
) + (2*b^2*x^(1/3)*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^(2/3)*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^(2/ 
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) - (I*b^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) + (2*b*x^(1/3)*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) - (2* 
b^2*x^(1/3)*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) + (b*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) - (I*b^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 
))
 

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 4226

Int[((a_.) + (b_.)*Tan[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Simp[1 
/n   Subst[Int[x^(1/n - 1)*(a + b*Tan[c + d*x])^p, x], x, x^n], x] /; FreeQ 
[{a, b, c, d, p}, x] && IGtQ[1/n, 0] && IntegerQ[p]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1187 vs. \(2 (491) = 982\).

Time = 0.12 (sec) , antiderivative size = 1187, normalized size of antiderivative = 1.95 \[ \int \frac {1}{\left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/2*(6*b^3*d^2*x^(2/3) - 2*(a^3 - a*b^2)*d^3*x + 2*(a^3 - a*b^2)*d^3 + 3* 
(-2*I*a^2*b*d*x^(1/3) - I*a*b^2 + (-2*I*a*b^2*d*x^(1/3) - I*b^3)*tan(d*x^( 
1/3) + c))*dilog(2*((I*a*b - b^2)*tan(d*x^(1/3) + c)^2 - a^2 - I*a*b + (I* 
a^2 - 2*a*b - I*b^2)*tan(d*x^(1/3) + c))/((a^2 + b^2)*tan(d*x^(1/3) + c)^2 
 + a^2 + b^2) + 1) + 3*(2*I*a^2*b*d*x^(1/3) + I*a*b^2 + (2*I*a*b^2*d*x^(1/ 
3) + I*b^3)*tan(d*x^(1/3) + c))*dilog(2*((-I*a*b - b^2)*tan(d*x^(1/3) + c) 
^2 - a^2 + I*a*b + (-I*a^2 - 2*a*b + I*b^2)*tan(d*x^(1/3) + c))/((a^2 + b^ 
2)*tan(d*x^(1/3) + c)^2 + a^2 + b^2) + 1) - 6*(a^2*b*d^2*x^(2/3) - a^2*b*c 
^2 + a*b^2*d*x^(1/3) + a*b^2*c + (a*b^2*d^2*x^(2/3) - a*b^2*c^2 + b^3*d*x^ 
(1/3) + b^3*c)*tan(d*x^(1/3) + c))*log(-2*((I*a*b - b^2)*tan(d*x^(1/3) + c 
)^2 - a^2 - I*a*b + (I*a^2 - 2*a*b - I*b^2)*tan(d*x^(1/3) + c))/((a^2 + b^ 
2)*tan(d*x^(1/3) + c)^2 + a^2 + b^2)) - 6*(a^2*b*d^2*x^(2/3) - a^2*b*c^2 + 
 a*b^2*d*x^(1/3) + a*b^2*c + (a*b^2*d^2*x^(2/3) - a*b^2*c^2 + b^3*d*x^(1/3 
) + b^3*c)*tan(d*x^(1/3) + c))*log(-2*((-I*a*b - b^2)*tan(d*x^(1/3) + c)^2 
 - a^2 + I*a*b + (-I*a^2 - 2*a*b + I*b^2)*tan(d*x^(1/3) + c))/((a^2 + b^2) 
*tan(d*x^(1/3) + c)^2 + a^2 + b^2)) - 6*(a^2*b*c^2 - a*b^2*c + (a*b^2*c^2 
- b^3*c)*tan(d*x^(1/3) + c))*log(((I*a*b + b^2)*tan(d*x^(1/3) + c)^2 - a^2 
 + I*a*b + (I*a^2 + I*b^2)*tan(d*x^(1/3) + c))/(tan(d*x^(1/3) + c)^2 + 1)) 
 - 6*(a^2*b*c^2 - a*b^2*c + (a*b^2*c^2 - b^3*c)*tan(d*x^(1/3) + c))*log((( 
I*a*b - b^2)*tan(d*x^(1/3) + c)^2 + a^2 + I*a*b + (I*a^2 + I*b^2)*tan(d...
 

Sympy [F]

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

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

Output:

Integral((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. 1732 vs. \(2 (491) = 982\).

Time = 0.46 (sec) , antiderivative size = 1732, normalized size of antiderivative = 2.84 \[ \int \frac {1}{\left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

(3*(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^2 + ((a^3 - I*a^2*b + a*b^2 - I*b^3)*(d*x^(1/3) + c)^3 - 3*(a^ 
3 - I*a^2*b + a*b^2 - I*b^3)*(d*x^(1/3) + c)^2*c - 6*((I*a*b^2 + b^3)*c*co 
s(2*d*x^(1/3) + 2*c) - (a*b^2 - I*b^3)*c*sin(2*d*x^(1/3) + 2*c) + (I*a*b^2 
 - b^3)*c)*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) - 6*((I*a^2*b 
- a*b^2)*(d*x^(1/3) + c)^2 + (I*a*b^2 - b^3 + 2*(-I*a^2*b + a*b^2)*c)*(d*x 
^(1/3) + c) + ((I*a^2*b + a*b^2)*(d*x^(1/3) + c)^2 + (I*a*b^2 + b^3 + 2*(- 
I*a^2*b - a*b^2)*c)*(d*x^(1/3) + c))*cos(2*d*x^(1/3) + 2*c) - ((a^2*b - I* 
a*b^2)*(d*x^(1/3) + c)^2 + (a*b^2 - I*b^3 - 2*(a^2*b - I*a*b^2)*c)*(d*x^(1 
/3) + c))*sin(2*d*x^(1/3) + 2*c))*arctan2((2*a*b*cos(2*d*x^(1/3) + 2*c) - 
(a^2 - b^2)*sin(2*d*x^(1/3) + 2*c))/(a^2 + b^2), (2*a*b*sin(2*d*x^(1/3) + 
2*c) + a^2 + b^2 + (a^2 - b^2)*cos(2*d*x^(1/3) + 2*c))/(a^2 + b^2)) + ((a^ 
3 - 3*I*a^2*b - 3*a*b^2 + I*b^3)*(d*x^(1/3) + c)^3 - 3*(2*I*a*b^2 + 2*b^3 
+ (a^3 - 3*I*a^2*b - 3*a*b^2 + I*b^3)*c)*(d*x^(1/3) + c)^2 - 12*(-I*a*b^2 
- b^3)*(d*x^(1/3) + c)*c)*cos(2*d*x^(1/3) + 2*c) - 3*(I*a*b^2 - b^3 + 2*(I 
*a^2*b - a*b^2)*(d*x^(1/3) + c) + 2*(-I*a^2*b + a*b^2)*c + (I*a*b^2 + b^3 
+ 2*(I*a^2*b + a*b^2)*(d*x^(1/3) + c) + 2*(-I*a^2*b - a*b^2)*c)*cos(2*d...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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