\(\int \frac {1}{(b \tan (c+d x))^{2/3}} \, dx\) [21]

Optimal result

Integrand size = 12, antiderivative size = 224 \[ \int \frac {1}{(b \tan (c+d x))^{2/3}} \, dx=\frac {\arctan \left (\frac {\sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{b^{2/3} d}-\frac {\arctan \left (\sqrt {3}-\frac {2 \sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{2 b^{2/3} d}+\frac {\arctan \left (\sqrt {3}+\frac {2 \sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{2 b^{2/3} d}-\frac {\sqrt {3} \log \left (b^{2/3}-\sqrt {3} \sqrt [3]{b} \sqrt [3]{b \tan (c+d x)}+(b \tan (c+d x))^{2/3}\right )}{4 b^{2/3} d}+\frac {\sqrt {3} \log \left (b^{2/3}+\sqrt {3} \sqrt [3]{b} \sqrt [3]{b \tan (c+d x)}+(b \tan (c+d x))^{2/3}\right )}{4 b^{2/3} d} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.14 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.84 \[ \int \frac {1}{(b \tan (c+d x))^{2/3}} \, dx=\frac {\left (i \log \left (1-i \sqrt [6]{\tan ^2(c+d x)}\right )-i \log \left (1+i \sqrt [6]{\tan ^2(c+d x)}\right )+\sqrt [6]{-1} \left ((-1)^{2/3} \log \left (1-\sqrt [6]{-1} \sqrt [6]{\tan ^2(c+d x)}\right )-(-1)^{2/3} \log \left (1+\sqrt [6]{-1} \sqrt [6]{\tan ^2(c+d x)}\right )+\log \left (1-(-1)^{5/6} \sqrt [6]{\tan ^2(c+d x)}\right )-\log \left (1+(-1)^{5/6} \sqrt [6]{\tan ^2(c+d x)}\right )\right )\right ) \sqrt [3]{b \tan (c+d x)}}{2 b d \sqrt [6]{\tan ^2(c+d x)}} \] Input:

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

Output:

((I*Log[1 - I*(Tan[c + d*x]^2)^(1/6)] - I*Log[1 + I*(Tan[c + d*x]^2)^(1/6) 
] + (-1)^(1/6)*((-1)^(2/3)*Log[1 - (-1)^(1/6)*(Tan[c + d*x]^2)^(1/6)] - (- 
1)^(2/3)*Log[1 + (-1)^(1/6)*(Tan[c + d*x]^2)^(1/6)] + Log[1 - (-1)^(5/6)*( 
Tan[c + d*x]^2)^(1/6)] - Log[1 + (-1)^(5/6)*(Tan[c + d*x]^2)^(1/6)]))*(b*T 
an[c + d*x])^(1/3))/(2*b*d*(Tan[c + d*x]^2)^(1/6))
 

Rubi [A] (warning: unable to verify)

Time = 0.38 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.88, 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, 753, 27, 216, 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.

\(\displaystyle \int \frac {1}{(b \tan (c+d x))^{2/3}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3957

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

\(\Big \downarrow \) 266

\(\displaystyle \frac {3 b \int \frac {1}{b^6 \tan ^6(c+d x)+b^2}d\sqrt [3]{b \tan (c+d x)}}{d}\)

\(\Big \downarrow \) 753

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {3 b \left (\frac {\int \frac {1}{b^2 \tan ^2(c+d x)+b^{2/3}}d\sqrt [3]{b \tan (c+d x)}}{3 b^{4/3}}+\frac {\int \frac {2 \sqrt [3]{b}-\sqrt {3} \sqrt [3]{b \tan (c+d x)}}{b^2 \tan ^2(c+d x)-\sqrt {3} b^{4/3} \tan (c+d x)+b^{2/3}}d\sqrt [3]{b \tan (c+d x)}}{6 b^{5/3}}+\frac {\int \frac {2 \sqrt [3]{b}+\sqrt {3} \sqrt [3]{b \tan (c+d x)}}{b^2 \tan ^2(c+d x)+\sqrt {3} b^{4/3} \tan (c+d x)+b^{2/3}}d\sqrt [3]{b \tan (c+d x)}}{6 b^{5/3}}\right )}{d}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {3 b \left (\frac {\int \frac {2 \sqrt [3]{b}-\sqrt {3} \sqrt [3]{b \tan (c+d x)}}{b^2 \tan ^2(c+d x)-\sqrt {3} b^{4/3} \tan (c+d x)+b^{2/3}}d\sqrt [3]{b \tan (c+d x)}}{6 b^{5/3}}+\frac {\int \frac {2 \sqrt [3]{b}+\sqrt {3} \sqrt [3]{b \tan (c+d x)}}{b^2 \tan ^2(c+d x)+\sqrt {3} b^{4/3} \tan (c+d x)+b^{2/3}}d\sqrt [3]{b \tan (c+d x)}}{6 b^{5/3}}+\frac {\arctan \left (b^{2/3} \tan (c+d x)\right )}{3 b^{5/3}}\right )}{d}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {3 b \left (\frac {\frac {1}{2} \sqrt [3]{b} \int \frac {1}{b^2 \tan ^2(c+d x)-\sqrt {3} b^{4/3} \tan (c+d x)+b^{2/3}}d\sqrt [3]{b \tan (c+d x)}-\frac {1}{2} \sqrt {3} \int -\frac {\sqrt {3} \sqrt [3]{b}-2 \sqrt [3]{b \tan (c+d x)}}{b^2 \tan ^2(c+d x)-\sqrt {3} b^{4/3} \tan (c+d x)+b^{2/3}}d\sqrt [3]{b \tan (c+d x)}}{6 b^{5/3}}+\frac {\frac {1}{2} \sqrt [3]{b} \int \frac {1}{b^2 \tan ^2(c+d x)+\sqrt {3} b^{4/3} \tan (c+d x)+b^{2/3}}d\sqrt [3]{b \tan (c+d x)}+\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3} \sqrt [3]{b}+2 \sqrt [3]{b \tan (c+d x)}}{b^2 \tan ^2(c+d x)+\sqrt {3} b^{4/3} \tan (c+d x)+b^{2/3}}d\sqrt [3]{b \tan (c+d x)}}{6 b^{5/3}}+\frac {\arctan \left (b^{2/3} \tan (c+d x)\right )}{3 b^{5/3}}\right )}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {3 b \left (\frac {\frac {1}{2} \sqrt [3]{b} \int \frac {1}{b^2 \tan ^2(c+d x)-\sqrt {3} b^{4/3} \tan (c+d x)+b^{2/3}}d\sqrt [3]{b \tan (c+d x)}+\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3} \sqrt [3]{b}-2 \sqrt [3]{b \tan (c+d x)}}{b^2 \tan ^2(c+d x)-\sqrt {3} b^{4/3} \tan (c+d x)+b^{2/3}}d\sqrt [3]{b \tan (c+d x)}}{6 b^{5/3}}+\frac {\frac {1}{2} \sqrt [3]{b} \int \frac {1}{b^2 \tan ^2(c+d x)+\sqrt {3} b^{4/3} \tan (c+d x)+b^{2/3}}d\sqrt [3]{b \tan (c+d x)}+\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3} \sqrt [3]{b}+2 \sqrt [3]{b \tan (c+d x)}}{b^2 \tan ^2(c+d x)+\sqrt {3} b^{4/3} \tan (c+d x)+b^{2/3}}d\sqrt [3]{b \tan (c+d x)}}{6 b^{5/3}}+\frac {\arctan \left (b^{2/3} \tan (c+d x)\right )}{3 b^{5/3}}\right )}{d}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {3 b \left (\frac {\frac {\int \frac {1}{-b^2 \tan ^2(c+d x)-\frac {1}{3}}d\left (1-\frac {2 b^{2/3} \tan (c+d x)}{\sqrt {3}}\right )}{\sqrt {3}}+\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3} \sqrt [3]{b}-2 \sqrt [3]{b \tan (c+d x)}}{b^2 \tan ^2(c+d x)-\sqrt {3} b^{4/3} \tan (c+d x)+b^{2/3}}d\sqrt [3]{b \tan (c+d x)}}{6 b^{5/3}}+\frac {\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3} \sqrt [3]{b}+2 \sqrt [3]{b \tan (c+d x)}}{b^2 \tan ^2(c+d x)+\sqrt {3} b^{4/3} \tan (c+d x)+b^{2/3}}d\sqrt [3]{b \tan (c+d x)}-\frac {\int \frac {1}{-b^2 \tan ^2(c+d x)-\frac {1}{3}}d\left (\frac {2 b^{2/3} \tan (c+d x)}{\sqrt {3}}+1\right )}{\sqrt {3}}}{6 b^{5/3}}+\frac {\arctan \left (b^{2/3} \tan (c+d x)\right )}{3 b^{5/3}}\right )}{d}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {3 b \left (\frac {\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3} \sqrt [3]{b}-2 \sqrt [3]{b \tan (c+d x)}}{b^2 \tan ^2(c+d x)-\sqrt {3} b^{4/3} \tan (c+d x)+b^{2/3}}d\sqrt [3]{b \tan (c+d x)}-\arctan \left (\sqrt {3} \left (1-\frac {2 b^{2/3} \tan (c+d x)}{\sqrt {3}}\right )\right )}{6 b^{5/3}}+\frac {\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3} \sqrt [3]{b}+2 \sqrt [3]{b \tan (c+d x)}}{b^2 \tan ^2(c+d x)+\sqrt {3} b^{4/3} \tan (c+d x)+b^{2/3}}d\sqrt [3]{b \tan (c+d x)}+\arctan \left (\sqrt {3} \left (\frac {2 b^{2/3} \tan (c+d x)}{\sqrt {3}}+1\right )\right )}{6 b^{5/3}}+\frac {\arctan \left (b^{2/3} \tan (c+d x)\right )}{3 b^{5/3}}\right )}{d}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {3 b \left (\frac {\arctan \left (b^{2/3} \tan (c+d x)\right )}{3 b^{5/3}}+\frac {-\arctan \left (\sqrt {3} \left (1-\frac {2 b^{2/3} \tan (c+d x)}{\sqrt {3}}\right )\right )-\frac {1}{2} \sqrt {3} \log \left (-\sqrt {3} b^{4/3} \tan (c+d x)+b^{2/3}+b^2 \tan ^2(c+d x)\right )}{6 b^{5/3}}+\frac {\arctan \left (\sqrt {3} \left (\frac {2 b^{2/3} \tan (c+d x)}{\sqrt {3}}+1\right )\right )+\frac {1}{2} \sqrt {3} \log \left (\sqrt {3} b^{4/3} \tan (c+d x)+b^{2/3}+b^2 \tan ^2(c+d x)\right )}{6 b^{5/3}}\right )}{d}\)

Input:

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

Output:

(3*b*(ArcTan[b^(2/3)*Tan[c + d*x]]/(3*b^(5/3)) + (-ArcTan[Sqrt[3]*(1 - (2* 
b^(2/3)*Tan[c + d*x])/Sqrt[3])] - (Sqrt[3]*Log[b^(2/3) - Sqrt[3]*b^(4/3)*T 
an[c + d*x] + b^2*Tan[c + d*x]^2])/2)/(6*b^(5/3)) + (ArcTan[Sqrt[3]*(1 + ( 
2*b^(2/3)*Tan[c + d*x])/Sqrt[3])] + (Sqrt[3]*Log[b^(2/3) + Sqrt[3]*b^(4/3) 
*Tan[c + d*x] + b^2*Tan[c + d*x]^2])/2)/(6*b^(5/3))))/d
 

Defintions of rubi rules used

rule 25

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216

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

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 753

Int[((a_) + (b_.)*(x_)^(n_))^(-1), x_Symbol] :> Module[{r = Numerator[Rt[a/ 
b, n]], s = Denominator[Rt[a/b, n]], k, u, v}, Simp[u = Int[(r - s*Cos[(2*k 
 - 1)*(Pi/n)]*x)/(r^2 - 2*r*s*Cos[(2*k - 1)*(Pi/n)]*x + s^2*x^2), x] + Int[ 
(r + s*Cos[(2*k - 1)*(Pi/n)]*x)/(r^2 + 2*r*s*Cos[(2*k - 1)*(Pi/n)]*x + s^2* 
x^2), x]; 2*(r^2/(a*n))   Int[1/(r^2 + s^2*x^2), x] + 2*(r/(a*n))   Sum[u, 
{k, 1, (n - 2)/4}], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && PosQ[a 
/b]
 

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 = 0.08 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(\frac {3 b \left (\frac {\sqrt {3}\, \left (b^{2}\right )^{\frac {1}{6}} \ln \left (\left (b \tan \left (d x +c \right )\right )^{\frac {2}{3}}+\sqrt {3}\, \left (b^{2}\right )^{\frac {1}{6}} \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}+\left (b^{2}\right )^{\frac {1}{3}}\right )}{12 b^{2}}+\frac {\left (b^{2}\right )^{\frac {1}{6}} \arctan \left (\frac {2 \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{\left (b^{2}\right )^{\frac {1}{6}}}+\sqrt {3}\right )}{6 b^{2}}-\frac {\sqrt {3}\, \left (b^{2}\right )^{\frac {1}{6}} \ln \left (-\left (b \tan \left (d x +c \right )\right )^{\frac {2}{3}}+\sqrt {3}\, \left (b^{2}\right )^{\frac {1}{6}} \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\left (b^{2}\right )^{\frac {1}{3}}\right )}{12 b^{2}}+\frac {\left (b^{2}\right )^{\frac {1}{6}} \arctan \left (\frac {2 \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{\left (b^{2}\right )^{\frac {1}{6}}}-\sqrt {3}\right )}{6 b^{2}}+\frac {\left (b^{2}\right )^{\frac {1}{6}} \arctan \left (\frac {\left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{\left (b^{2}\right )^{\frac {1}{6}}}\right )}{3 b^{2}}\right )}{d}\)	\(203\)
default	\(\frac {3 b \left (\frac {\sqrt {3}\, \left (b^{2}\right )^{\frac {1}{6}} \ln \left (\left (b \tan \left (d x +c \right )\right )^{\frac {2}{3}}+\sqrt {3}\, \left (b^{2}\right )^{\frac {1}{6}} \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}+\left (b^{2}\right )^{\frac {1}{3}}\right )}{12 b^{2}}+\frac {\left (b^{2}\right )^{\frac {1}{6}} \arctan \left (\frac {2 \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{\left (b^{2}\right )^{\frac {1}{6}}}+\sqrt {3}\right )}{6 b^{2}}-\frac {\sqrt {3}\, \left (b^{2}\right )^{\frac {1}{6}} \ln \left (-\left (b \tan \left (d x +c \right )\right )^{\frac {2}{3}}+\sqrt {3}\, \left (b^{2}\right )^{\frac {1}{6}} \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}-\left (b^{2}\right )^{\frac {1}{3}}\right )}{12 b^{2}}+\frac {\left (b^{2}\right )^{\frac {1}{6}} \arctan \left (\frac {2 \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{\left (b^{2}\right )^{\frac {1}{6}}}-\sqrt {3}\right )}{6 b^{2}}+\frac {\left (b^{2}\right )^{\frac {1}{6}} \arctan \left (\frac {\left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{\left (b^{2}\right )^{\frac {1}{6}}}\right )}{3 b^{2}}\right )}{d}\)	\(203\)

Input:

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

Output:

3/d*b*(1/12/b^2*3^(1/2)*(b^2)^(1/6)*ln((b*tan(d*x+c))^(2/3)+3^(1/2)*(b^2)^ 
(1/6)*(b*tan(d*x+c))^(1/3)+(b^2)^(1/3))+1/6/b^2*(b^2)^(1/6)*arctan(2*(b*ta 
n(d*x+c))^(1/3)/(b^2)^(1/6)+3^(1/2))-1/12/b^2*3^(1/2)*(b^2)^(1/6)*ln(-(b*t 
an(d*x+c))^(2/3)+3^(1/2)*(b^2)^(1/6)*(b*tan(d*x+c))^(1/3)-(b^2)^(1/3))+1/6 
/b^2*(b^2)^(1/6)*arctan(2*(b*tan(d*x+c))^(1/3)/(b^2)^(1/6)-3^(1/2))+1/3/b^ 
2*(b^2)^(1/6)*arctan((b*tan(d*x+c))^(1/3)/(b^2)^(1/6)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.26 \[ \int \frac {1}{(b \tan (c+d x))^{2/3}} \, dx=\frac {1}{4} \, {\left (\sqrt {-3} + 1\right )} \left (-\frac {1}{b^{4} d^{6}}\right )^{\frac {1}{6}} \log \left (\frac {1}{2} \, {\left (\sqrt {-3} b d + b d\right )} \left (-\frac {1}{b^{4} d^{6}}\right )^{\frac {1}{6}} + \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}\right ) - \frac {1}{4} \, {\left (\sqrt {-3} + 1\right )} \left (-\frac {1}{b^{4} d^{6}}\right )^{\frac {1}{6}} \log \left (-\frac {1}{2} \, {\left (\sqrt {-3} b d + b d\right )} \left (-\frac {1}{b^{4} d^{6}}\right )^{\frac {1}{6}} + \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}\right ) + \frac {1}{4} \, {\left (\sqrt {-3} - 1\right )} \left (-\frac {1}{b^{4} d^{6}}\right )^{\frac {1}{6}} \log \left (\frac {1}{2} \, {\left (\sqrt {-3} b d - b d\right )} \left (-\frac {1}{b^{4} d^{6}}\right )^{\frac {1}{6}} + \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}\right ) - \frac {1}{4} \, {\left (\sqrt {-3} - 1\right )} \left (-\frac {1}{b^{4} d^{6}}\right )^{\frac {1}{6}} \log \left (-\frac {1}{2} \, {\left (\sqrt {-3} b d - b d\right )} \left (-\frac {1}{b^{4} d^{6}}\right )^{\frac {1}{6}} + \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}\right ) + \frac {1}{2} \, \left (-\frac {1}{b^{4} d^{6}}\right )^{\frac {1}{6}} \log \left (b d \left (-\frac {1}{b^{4} d^{6}}\right )^{\frac {1}{6}} + \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}\right ) - \frac {1}{2} \, \left (-\frac {1}{b^{4} d^{6}}\right )^{\frac {1}{6}} \log \left (-b d \left (-\frac {1}{b^{4} d^{6}}\right )^{\frac {1}{6}} + \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}\right ) \] Input:

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

Output:

1/4*(sqrt(-3) + 1)*(-1/(b^4*d^6))^(1/6)*log(1/2*(sqrt(-3)*b*d + b*d)*(-1/( 
b^4*d^6))^(1/6) + (b*tan(d*x + c))^(1/3)) - 1/4*(sqrt(-3) + 1)*(-1/(b^4*d^ 
6))^(1/6)*log(-1/2*(sqrt(-3)*b*d + b*d)*(-1/(b^4*d^6))^(1/6) + (b*tan(d*x 
+ c))^(1/3)) + 1/4*(sqrt(-3) - 1)*(-1/(b^4*d^6))^(1/6)*log(1/2*(sqrt(-3)*b 
*d - b*d)*(-1/(b^4*d^6))^(1/6) + (b*tan(d*x + c))^(1/3)) - 1/4*(sqrt(-3) - 
 1)*(-1/(b^4*d^6))^(1/6)*log(-1/2*(sqrt(-3)*b*d - b*d)*(-1/(b^4*d^6))^(1/6 
) + (b*tan(d*x + c))^(1/3)) + 1/2*(-1/(b^4*d^6))^(1/6)*log(b*d*(-1/(b^4*d^ 
6))^(1/6) + (b*tan(d*x + c))^(1/3)) - 1/2*(-1/(b^4*d^6))^(1/6)*log(-b*d*(- 
1/(b^4*d^6))^(1/6) + (b*tan(d*x + c))^(1/3))
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.76 \[ \int \frac {1}{(b \tan (c+d x))^{2/3}} \, dx=\frac {\sqrt {3} b^{\frac {1}{3}} \log \left (\sqrt {3} \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} b^{\frac {1}{3}} + \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} + b^{\frac {2}{3}}\right ) - \sqrt {3} b^{\frac {1}{3}} \log \left (-\sqrt {3} \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} b^{\frac {1}{3}} + \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} + b^{\frac {2}{3}}\right ) + 2 \, b^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} b^{\frac {1}{3}} + 2 \, \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}\right ) + 2 \, b^{\frac {1}{3}} \arctan \left (-\frac {\sqrt {3} b^{\frac {1}{3}} - 2 \, \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}\right ) + 4 \, b^{\frac {1}{3}} \arctan \left (\frac {\left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}\right )}{4 \, b d} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.88 \[ \int \frac {1}{(b \tan (c+d x))^{2/3}} \, dx=\frac {\frac {\sqrt {3} {\left | b \right |}^{\frac {1}{3}} \log \left (\sqrt {3} \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} {\left | b \right |}^{\frac {1}{3}} + \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} + {\left | b \right |}^{\frac {2}{3}}\right )}{b} - \frac {\sqrt {3} {\left | b \right |}^{\frac {1}{3}} \log \left (-\sqrt {3} \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} {\left | b \right |}^{\frac {1}{3}} + \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} + {\left | b \right |}^{\frac {2}{3}}\right )}{b} + \frac {2 \, {\left | b \right |}^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left | b \right |}^{\frac {1}{3}} + 2 \, \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}}{{\left | b \right |}^{\frac {1}{3}}}\right )}{b} + \frac {2 \, {\left | b \right |}^{\frac {1}{3}} \arctan \left (-\frac {\sqrt {3} {\left | b \right |}^{\frac {1}{3}} - 2 \, \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}}{{\left | b \right |}^{\frac {1}{3}}}\right )}{b} + \frac {4 \, {\left | b \right |}^{\frac {1}{3}} \arctan \left (\frac {\left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}}{{\left | b \right |}^{\frac {1}{3}}}\right )}{b}}{4 \, d} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.03 \[ \int \frac {1}{(b \tan (c+d x))^{2/3}} \, dx=\frac {{\left (-1\right )}^{1/6}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{5/6}\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}\,1{}\mathrm {i}}{b^{1/3}}\right )\,1{}\mathrm {i}}{b^{2/3}\,d}-\frac {{\left (-1\right )}^{1/6}\,\ln \left ({\left (-1\right )}^{1/6}\,b^{1/3}-2\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}+{\left (-1\right )}^{2/3}\,\sqrt {3}\,b^{1/3}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,b^{2/3}\,d}-\frac {{\left (-1\right )}^{1/6}\,\ln \left ({\left (-1\right )}^{1/6}\,b^{1/3}+2\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}-{\left (-1\right )}^{2/3}\,\sqrt {3}\,b^{1/3}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,b^{2/3}\,d}+\frac {{\left (-1\right )}^{1/6}\,\ln \left ({\left (-1\right )}^{1/6}\,b^{1/3}+2\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}+{\left (-1\right )}^{2/3}\,\sqrt {3}\,b^{1/3}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )}{b^{2/3}\,d}+\frac {{\left (-1\right )}^{1/6}\,\ln \left (2\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}-{\left (-1\right )}^{1/6}\,b^{1/3}+{\left (-1\right )}^{2/3}\,\sqrt {3}\,b^{1/3}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )}{b^{2/3}\,d} \] Input:

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

Output:

((-1)^(1/6)*atan(((-1)^(5/6)*(b*tan(c + d*x))^(1/3)*1i)/b^(1/3))*1i)/(b^(2 
/3)*d) - ((-1)^(1/6)*log((-1)^(1/6)*b^(1/3) - 2*(b*tan(c + d*x))^(1/3) + ( 
-1)^(2/3)*3^(1/2)*b^(1/3))*((3^(1/2)*1i)/2 + 1/2))/(2*b^(2/3)*d) - ((-1)^( 
1/6)*log((-1)^(1/6)*b^(1/3) + 2*(b*tan(c + d*x))^(1/3) - (-1)^(2/3)*3^(1/2 
)*b^(1/3))*((3^(1/2)*1i)/2 - 1/2))/(2*b^(2/3)*d) + ((-1)^(1/6)*log((-1)^(1 
/6)*b^(1/3) + 2*(b*tan(c + d*x))^(1/3) + (-1)^(2/3)*3^(1/2)*b^(1/3))*((3^( 
1/2)*1i)/4 + 1/4))/(b^(2/3)*d) + ((-1)^(1/6)*log(2*(b*tan(c + d*x))^(1/3) 
- (-1)^(1/6)*b^(1/3) + (-1)^(2/3)*3^(1/2)*b^(1/3))*((3^(1/2)*1i)/4 - 1/4)) 
/(b^(2/3)*d)
 

Reduce [F]

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

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

Output:

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