\(\int (a+b \tan (c+d \sqrt [3]{x}))^2 \, dx\) [54]

Optimal result

Integrand size = 16, antiderivative size = 206 \[ \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2 \, dx=-\frac {3 i b^2 x^{2/3}}{d}+a^2 x+2 i a b x-b^2 x+\frac {6 b^2 \sqrt [3]{x} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{2/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}-\frac {3 i b^2 \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {6 i a b \sqrt [3]{x} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {3 a b \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {3 b^2 x^{2/3} \tan \left (c+d \sqrt [3]{x}\right )}{d} \] Output:

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

Mathematica [A] (verified)

Time = 1.51 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.90 \[ \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2 \, dx=\frac {b \left (\frac {6 i b d^2 x^{2/3}-4 i a d^3 x}{1+e^{2 i c}}+6 d \left (b-a d \sqrt [3]{x}\right ) \sqrt [3]{x} \log \left (1+e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )+3 i \left (b-2 a d \sqrt [3]{x}\right ) \operatorname {PolyLog}\left (2,-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )-3 a \operatorname {PolyLog}\left (3,-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )\right )}{d^3}+\frac {3 b^2 x^{2/3} \sec (c) \sec \left (c+d \sqrt [3]{x}\right ) \sin \left (d \sqrt [3]{x}\right )}{d}+x \left (a^2-b^2+2 a b \tan (c)\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {4226, 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.

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

\(\Big \downarrow \) 4226

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4205

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

3*(((-I)*b^2*x^(2/3))/d + (a^2*x)/3 + ((2*I)/3)*a*b*x - (b^2*x)/3 + (2*b^2 
*x^(1/3)*Log[1 + E^((2*I)*(c + d*x^(1/3)))])/d^2 - (2*a*b*x^(2/3)*Log[1 + 
E^((2*I)*(c + d*x^(1/3)))])/d - (I*b^2*PolyLog[2, -E^((2*I)*(c + d*x^(1/3) 
))])/d^3 + ((2*I)*a*b*x^(1/3)*PolyLog[2, -E^((2*I)*(c + d*x^(1/3)))])/d^2 
- (a*b*PolyLog[3, -E^((2*I)*(c + d*x^(1/3)))])/d^3 + (b^2*x^(2/3)*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 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]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.55 \[ \int \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2 \, dx=\frac {6 \, b^{2} d^{2} x^{\frac {2}{3}} \tan \left (d x^{\frac {1}{3}} + c\right ) + 2 \, {\left (a^{2} - b^{2}\right )} d^{3} x - 3 \, a b {\rm polylog}\left (3, \frac {\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 2 i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1}\right ) - 3 \, a b {\rm polylog}\left (3, \frac {\tan \left (d x^{\frac {1}{3}} + c\right )^{2} - 2 i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1}\right ) - 3 \, {\left (2 i \, a b d x^{\frac {1}{3}} - i \, b^{2}\right )} {\rm Li}_2\left (\frac {2 \, {\left (i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1\right )}}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1} + 1\right ) - 3 \, {\left (-2 i \, a b d x^{\frac {1}{3}} + i \, b^{2}\right )} {\rm Li}_2\left (\frac {2 \, {\left (-i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1\right )}}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1} + 1\right ) - 6 \, {\left (a b d^{2} x^{\frac {2}{3}} - b^{2} d x^{\frac {1}{3}}\right )} \log \left (-\frac {2 \, {\left (i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1\right )}}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1}\right ) - 6 \, {\left (a b d^{2} x^{\frac {2}{3}} - b^{2} d x^{\frac {1}{3}}\right )} \log \left (-\frac {2 \, {\left (-i \, \tan \left (d x^{\frac {1}{3}} + c\right ) - 1\right )}}{\tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 1}\right )}{2 \, d^{3}} \] Input:

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

Output:

1/2*(6*b^2*d^2*x^(2/3)*tan(d*x^(1/3) + c) + 2*(a^2 - b^2)*d^3*x - 3*a*b*po 
lylog(3, (tan(d*x^(1/3) + c)^2 + 2*I*tan(d*x^(1/3) + c) - 1)/(tan(d*x^(1/3 
) + c)^2 + 1)) - 3*a*b*polylog(3, (tan(d*x^(1/3) + c)^2 - 2*I*tan(d*x^(1/3 
) + c) - 1)/(tan(d*x^(1/3) + c)^2 + 1)) - 3*(2*I*a*b*d*x^(1/3) - I*b^2)*di 
log(2*(I*tan(d*x^(1/3) + c) - 1)/(tan(d*x^(1/3) + c)^2 + 1) + 1) - 3*(-2*I 
*a*b*d*x^(1/3) + I*b^2)*dilog(2*(-I*tan(d*x^(1/3) + c) - 1)/(tan(d*x^(1/3) 
 + c)^2 + 1) + 1) - 6*(a*b*d^2*x^(2/3) - b^2*d*x^(1/3))*log(-2*(I*tan(d*x^ 
(1/3) + c) - 1)/(tan(d*x^(1/3) + c)^2 + 1)) - 6*(a*b*d^2*x^(2/3) - b^2*d*x 
^(1/3))*log(-2*(-I*tan(d*x^(1/3) + c) - 1)/(tan(d*x^(1/3) + c)^2 + 1)))/d^ 
3
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

a^2*x + (6*b^2*x^(2/3)*sin(2*d*x^(1/3) + 2*c) - (b^2*d*cos(2*d*x^(1/3) + 2 
*c)^2 + b^2*d*sin(2*d*x^(1/3) + 2*c)^2 + 2*b^2*d*cos(2*d*x^(1/3) + 2*c) + 
b^2*d)*x - (d*cos(2*d*x^(1/3) + 2*c)^2 + d*sin(2*d*x^(1/3) + 2*c)^2 + 2*d* 
cos(2*d*x^(1/3) + 2*c) + d)*integrate(-4*(a*b*d*x*sin(2*d*x^(1/3) + 2*c) - 
 b^2*x^(2/3)*sin(2*d*x^(1/3) + 2*c))/((d*cos(2*d*x^(1/3) + 2*c)^2 + d*sin( 
2*d*x^(1/3) + 2*c)^2 + 2*d*cos(2*d*x^(1/3) + 2*c) + d)*x), x))/(d*cos(2*d* 
x^(1/3) + 2*c)^2 + d*sin(2*d*x^(1/3) + 2*c)^2 + 2*d*cos(2*d*x^(1/3) + 2*c) 
 + d)
 

Giac [F]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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