\(\int \frac {1}{\sqrt [3]{b \tan (c+d x)}} \, dx\) [20]

Optimal result

Integrand size = 12, antiderivative size = 131 \[ \int \frac {1}{\sqrt [3]{b \tan (c+d x)}} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {b^{2/3}-2 (b \tan (c+d x))^{2/3}}{\sqrt {3} b^{2/3}}\right )}{2 \sqrt [3]{b} d}+\frac {\log \left (b^{2/3}+(b \tan (c+d x))^{2/3}\right )}{2 \sqrt [3]{b} d}-\frac {\log \left (b^{4/3}-b^{2/3} (b \tan (c+d x))^{2/3}+(b \tan (c+d x))^{4/3}\right )}{4 \sqrt [3]{b} d} \] Output:

-1/2*3^(1/2)*arctan(1/3*(b^(2/3)-2*(b*tan(d*x+c))^(2/3))*3^(1/2)/b^(2/3))/ 
b^(1/3)/d+1/2*ln(b^(2/3)+(b*tan(d*x+c))^(2/3))/b^(1/3)/d-1/4*ln(b^(4/3)-b^ 
(2/3)*(b*tan(d*x+c))^(2/3)+(b*tan(d*x+c))^(4/3))/b^(1/3)/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\sqrt [3]{b \tan (c+d x)}} \, dx=\frac {\left (2 \sqrt {3} \arctan \left (\frac {-1+2 \tan ^{\frac {2}{3}}(c+d x)}{\sqrt {3}}\right )+2 \log \left (1+\tan ^{\frac {2}{3}}(c+d x)\right )-\log \left (1-\tan ^{\frac {2}{3}}(c+d x)+\tan ^{\frac {4}{3}}(c+d x)\right )\right ) \sqrt [3]{\tan (c+d x)}}{4 d \sqrt [3]{b \tan (c+d x)}} \] Input:

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

Output:

((2*Sqrt[3]*ArcTan[(-1 + 2*Tan[c + d*x]^(2/3))/Sqrt[3]] + 2*Log[1 + Tan[c 
+ d*x]^(2/3)] - Log[1 - Tan[c + d*x]^(2/3) + Tan[c + d*x]^(4/3)])*Tan[c + 
d*x]^(1/3))/(4*d*(b*Tan[c + d*x])^(1/3))
 

Rubi [A] (warning: unable to verify)

Time = 0.29 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.85, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.917, Rules used = {3042, 3957, 266, 807, 750, 16, 1142, 25, 1082, 217, 1103}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sqrt [3]{b \tan (c+d x)}}dx\)

\(\Big \downarrow \) 3957

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

\(\Big \downarrow \) 266

\(\displaystyle \frac {3 b \int \frac {\sqrt [3]{b \tan (c+d x)}}{b^6 \tan ^6(c+d x)+b^2}d\sqrt [3]{b \tan (c+d x)}}{d}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {3 b \int \frac {1}{b^3 \tan ^3(c+d x)+b^2}d\left (b^2 \tan ^2(c+d x)\right )}{2 d}\)

\(\Big \downarrow \) 750

\(\displaystyle \frac {3 b \left (\frac {\int \frac {1}{b^2 \tan ^2(c+d x)+b^{2/3}}d\left (b^2 \tan ^2(c+d x)\right )}{3 b^{4/3}}+\frac {\int \frac {2 b^{2/3}-b^2 \tan ^2(c+d x)}{b^2 \tan ^2(c+d x)-b^{5/3} \tan (c+d x)+b^{4/3}}d\left (b^2 \tan ^2(c+d x)\right )}{3 b^{4/3}}\right )}{2 d}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {3 b \left (\frac {\int \frac {2 b^{2/3}-b^2 \tan ^2(c+d x)}{b^2 \tan ^2(c+d x)-b^{5/3} \tan (c+d x)+b^{4/3}}d\left (b^2 \tan ^2(c+d x)\right )}{3 b^{4/3}}+\frac {\log \left (b^{2/3}+b^2 \tan ^2(c+d x)\right )}{3 b^{4/3}}\right )}{2 d}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {3 b \left (\frac {\frac {3}{2} b^{2/3} \int \frac {1}{b^2 \tan ^2(c+d x)-b^{5/3} \tan (c+d x)+b^{4/3}}d\left (b^2 \tan ^2(c+d x)\right )-\frac {1}{2} \int -\frac {b^{2/3}-2 b^2 \tan ^2(c+d x)}{b^2 \tan ^2(c+d x)-b^{5/3} \tan (c+d x)+b^{4/3}}d\left (b^2 \tan ^2(c+d x)\right )}{3 b^{4/3}}+\frac {\log \left (b^{2/3}+b^2 \tan ^2(c+d x)\right )}{3 b^{4/3}}\right )}{2 d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {3 b \left (\frac {\frac {3}{2} b^{2/3} \int \frac {1}{b^2 \tan ^2(c+d x)-b^{5/3} \tan (c+d x)+b^{4/3}}d\left (b^2 \tan ^2(c+d x)\right )+\frac {1}{2} \int \frac {b^{2/3}-2 b^2 \tan ^2(c+d x)}{b^2 \tan ^2(c+d x)-b^{5/3} \tan (c+d x)+b^{4/3}}d\left (b^2 \tan ^2(c+d x)\right )}{3 b^{4/3}}+\frac {\log \left (b^{2/3}+b^2 \tan ^2(c+d x)\right )}{3 b^{4/3}}\right )}{2 d}\)

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1103

\(\displaystyle \frac {3 b \left (\frac {-\sqrt {3} \arctan \left (\frac {1-2 \sqrt [3]{b} \tan (c+d x)}{\sqrt {3}}\right )-\frac {1}{2} \log \left (-b^{5/3} \tan (c+d x)+b^{4/3}+b^2 \tan ^2(c+d x)\right )}{3 b^{4/3}}+\frac {\log \left (b^{2/3}+b^2 \tan ^2(c+d x)\right )}{3 b^{4/3}}\right )}{2 d}\)

Input:

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

Output:

(3*b*(Log[b^(2/3) + b^2*Tan[c + d*x]^2]/(3*b^(4/3)) + (-(Sqrt[3]*ArcTan[(1 
 - 2*b^(1/3)*Tan[c + d*x])/Sqrt[3]]) - Log[b^(4/3) - b^(5/3)*Tan[c + d*x] 
+ b^2*Tan[c + d*x]^2]/2)/(3*b^(4/3))))/(2*d)
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 750

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Simp[1/(3*Rt[a, 3]^2)   Int[1/ 
(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]^2)   Int[(2*Rt[a, 3] - 
 Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x] /; 
 FreeQ[{a, b}, x]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1142

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 3042

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

rule 3957

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d   Subst[Int 
[x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && 
!IntegerQ[n]
 

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.82

method	result	size
derivativedivides	\(\frac {3 b \left (\frac {\ln \left (\left (b \tan \left (d x +c \right )\right )^{\frac {2}{3}}+\left (b^{2}\right )^{\frac {1}{3}}\right )}{6 \left (b^{2}\right )^{\frac {2}{3}}}-\frac {\ln \left (\left (b \tan \left (d x +c \right )\right )^{\frac {4}{3}}-\left (b \tan \left (d x +c \right )\right )^{\frac {2}{3}} \left (b^{2}\right )^{\frac {1}{3}}+\left (b^{2}\right )^{\frac {2}{3}}\right )}{12 \left (b^{2}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{\left (b^{2}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{6 \left (b^{2}\right )^{\frac {2}{3}}}\right )}{d}\)	\(108\)
default	\(\frac {3 b \left (\frac {\ln \left (\left (b \tan \left (d x +c \right )\right )^{\frac {2}{3}}+\left (b^{2}\right )^{\frac {1}{3}}\right )}{6 \left (b^{2}\right )^{\frac {2}{3}}}-\frac {\ln \left (\left (b \tan \left (d x +c \right )\right )^{\frac {4}{3}}-\left (b \tan \left (d x +c \right )\right )^{\frac {2}{3}} \left (b^{2}\right )^{\frac {1}{3}}+\left (b^{2}\right )^{\frac {2}{3}}\right )}{12 \left (b^{2}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \left (b \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{\left (b^{2}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{6 \left (b^{2}\right )^{\frac {2}{3}}}\right )}{d}\)	\(108\)

Input:

int(1/(b*tan(d*x+c))^(1/3),x,method=_RETURNVERBOSE)
 

Output:

3/d*b*(1/6/(b^2)^(2/3)*ln((b*tan(d*x+c))^(2/3)+(b^2)^(1/3))-1/12/(b^2)^(2/ 
3)*ln((b*tan(d*x+c))^(4/3)-(b*tan(d*x+c))^(2/3)*(b^2)^(1/3)+(b^2)^(2/3))+1 
/6/(b^2)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2*(b*tan(d*x+c))^(2/3)/(b^2)^(1 
/3)-1)))
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 299, normalized size of antiderivative = 2.28 \[ \int \frac {1}{\sqrt [3]{b \tan (c+d x)}} \, dx=\left [\frac {\sqrt {3} b \sqrt {-\frac {1}{b^{\frac {2}{3}}}} \log \left (\frac {2 \, \sqrt {3} \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} b \sqrt {-\frac {1}{b^{\frac {2}{3}}}} \tan \left (d x + c\right ) + 2 \, b \tan \left (d x + c\right )^{2} - \sqrt {3} b^{\frac {4}{3}} \sqrt {-\frac {1}{b^{\frac {2}{3}}}} + \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} {\left (\sqrt {3} b^{\frac {2}{3}} \sqrt {-\frac {1}{b^{\frac {2}{3}}}} - 3 \, b^{\frac {1}{3}}\right )} - b}{\tan \left (d x + c\right )^{2} + 1}\right ) - b^{\frac {2}{3}} \log \left (\left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} b \tan \left (d x + c\right ) - \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} b^{\frac {2}{3}} + b^{\frac {4}{3}}\right ) + 2 \, b^{\frac {2}{3}} \log \left (\left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} + b^{\frac {2}{3}}\right )}{4 \, b d}, \frac {2 \, \sqrt {3} b^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} b^{\frac {2}{3}} - b^{\frac {4}{3}}\right )}}{3 \, b^{\frac {4}{3}}}\right ) - b^{\frac {2}{3}} \log \left (\left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} b \tan \left (d x + c\right ) - \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} b^{\frac {2}{3}} + b^{\frac {4}{3}}\right ) + 2 \, b^{\frac {2}{3}} \log \left (\left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} + b^{\frac {2}{3}}\right )}{4 \, b d}\right ] \] Input:

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

Output:

[1/4*(sqrt(3)*b*sqrt(-1/b^(2/3))*log((2*sqrt(3)*(b*tan(d*x + c))^(1/3)*b*s 
qrt(-1/b^(2/3))*tan(d*x + c) + 2*b*tan(d*x + c)^2 - sqrt(3)*b^(4/3)*sqrt(- 
1/b^(2/3)) + (b*tan(d*x + c))^(2/3)*(sqrt(3)*b^(2/3)*sqrt(-1/b^(2/3)) - 3* 
b^(1/3)) - b)/(tan(d*x + c)^2 + 1)) - b^(2/3)*log((b*tan(d*x + c))^(1/3)*b 
*tan(d*x + c) - (b*tan(d*x + c))^(2/3)*b^(2/3) + b^(4/3)) + 2*b^(2/3)*log( 
(b*tan(d*x + c))^(2/3) + b^(2/3)))/(b*d), 1/4*(2*sqrt(3)*b^(2/3)*arctan(1/ 
3*sqrt(3)*(2*(b*tan(d*x + c))^(2/3)*b^(2/3) - b^(4/3))/b^(4/3)) - b^(2/3)* 
log((b*tan(d*x + c))^(1/3)*b*tan(d*x + c) - (b*tan(d*x + c))^(2/3)*b^(2/3) 
 + b^(4/3)) + 2*b^(2/3)*log((b*tan(d*x + c))^(2/3) + b^(2/3)))/(b*d)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\sqrt [3]{b \tan (c+d x)}} \, dx=\frac {2 \, \sqrt {3} b^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} - b^{\frac {2}{3}}\right )}}{3 \, b^{\frac {2}{3}}}\right ) - b^{\frac {2}{3}} \log \left (\left (b \tan \left (d x + c\right )\right )^{\frac {4}{3}} - \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} b^{\frac {2}{3}} + b^{\frac {4}{3}}\right ) + 2 \, b^{\frac {2}{3}} \log \left (\left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} + b^{\frac {2}{3}}\right )}{4 \, b d} \] Input:

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

Output:

1/4*(2*sqrt(3)*b^(2/3)*arctan(1/3*sqrt(3)*(2*(b*tan(d*x + c))^(2/3) - b^(2 
/3))/b^(2/3)) - b^(2/3)*log((b*tan(d*x + c))^(4/3) - (b*tan(d*x + c))^(2/3 
)*b^(2/3) + b^(4/3)) + 2*b^(2/3)*log((b*tan(d*x + c))^(2/3) + b^(2/3)))/(b 
*d)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.95 \[ \int \frac {1}{\sqrt [3]{b \tan (c+d x)}} \, dx=\frac {\sqrt {3} {\left | b \right |}^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} - {\left | b \right |}^{\frac {2}{3}}\right )}}{3 \, {\left | b \right |}^{\frac {2}{3}}}\right )}{2 \, b d} - \frac {{\left | b \right |}^{\frac {2}{3}} \log \left (\left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} b \tan \left (d x + c\right ) - \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} {\left | b \right |}^{\frac {2}{3}} + {\left | b \right |}^{\frac {4}{3}}\right )}{4 \, b d} + \frac {{\left | b \right |}^{\frac {2}{3}} \log \left (\left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} + {\left | b \right |}^{\frac {2}{3}}\right )}{2 \, b d} \] Input:

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

Output:

1/2*sqrt(3)*abs(b)^(2/3)*arctan(1/3*sqrt(3)*(2*(b*tan(d*x + c))^(2/3) - ab 
s(b)^(2/3))/abs(b)^(2/3))/(b*d) - 1/4*abs(b)^(2/3)*log((b*tan(d*x + c))^(1 
/3)*b*tan(d*x + c) - (b*tan(d*x + c))^(2/3)*abs(b)^(2/3) + abs(b)^(4/3))/( 
b*d) + 1/2*abs(b)^(2/3)*log((b*tan(d*x + c))^(2/3) + abs(b)^(2/3))/(b*d)
 

Mupad [B] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.98 \[ \int \frac {1}{\sqrt [3]{b \tan (c+d x)}} \, dx=\frac {\ln \left ({\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{2/3}+b^{2/3}\right )}{2\,b^{1/3}\,d}+\frac {\ln \left (\frac {81\,b^{11/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{d^3}+\frac {162\,b^3\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{2/3}}{d^3}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{4\,b^{1/3}\,d}-\frac {\ln \left (\frac {81\,b^{11/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{d^3}-\frac {162\,b^3\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{2/3}}{d^3}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{4\,b^{1/3}\,d} \] Input:

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

Output:

log((b*tan(c + d*x))^(2/3) + b^(2/3))/(2*b^(1/3)*d) + (log((81*b^(11/3)*(3 
^(1/2)*1i - 1))/d^3 + (162*b^3*(b*tan(c + d*x))^(2/3))/d^3)*(3^(1/2)*1i - 
1))/(4*b^(1/3)*d) - (log((81*b^(11/3)*(3^(1/2)*1i + 1))/d^3 - (162*b^3*(b* 
tan(c + d*x))^(2/3))/d^3)*(3^(1/2)*1i + 1))/(4*b^(1/3)*d)
 

Reduce [F]

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

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

Output:

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