\(\int (b \tan (c+d x))^{2/3} \, dx\) [18]

Optimal result

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

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.14 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.83 \[ \int (b \tan (c+d x))^{2/3} \, dx=\frac {\left (i \log \left (1-i \sqrt [6]{\tan ^2(c+d x)}\right )-i \log \left (1+i \sqrt [6]{\tan ^2(c+d x)}\right )+\sqrt [6]{-1} \left (\log \left (1-\sqrt [6]{-1} \sqrt [6]{\tan ^2(c+d x)}\right )-\log \left (1+\sqrt [6]{-1} \sqrt [6]{\tan ^2(c+d x)}\right )+(-1)^{2/3} \left (\log \left (1-(-1)^{5/6} \sqrt [6]{\tan ^2(c+d x)}\right )-\log \left (1+(-1)^{5/6} \sqrt [6]{\tan ^2(c+d x)}\right )\right )\right )\right ) (b \tan (c+d x))^{5/3}}{2 b d \tan ^2(c+d x)^{5/6}} \] 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)*(Log[1 - (-1)^(1/6)*(Tan[c + d*x]^2)^(1/6)] - Log[1 + (-1)^ 
(1/6)*(Tan[c + d*x]^2)^(1/6)] + (-1)^(2/3)*(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*Tan[c + d* 
x])^(5/3))/(2*b*d*(Tan[c + d*x]^2)^(5/6))
 

Rubi [A] (warning: unable to verify)

Time = 0.39 (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, 824, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3957

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

\(\Big \downarrow \) 266

\(\displaystyle \frac {3 b \int \frac {b^4 \tan ^4(c+d x)}{b^6 \tan ^6(c+d x)+b^2}d\sqrt [3]{b \tan (c+d x)}}{d}\)

\(\Big \downarrow \) 824

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 1142

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

(3*b*(ArcTan[b^(2/3)*Tan[c + d*x]]/(3*b^(1/3)) - (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)*Ta 
n[c + d*x] + b^2*Tan[c + d*x]^2])/2)/(6*b^(1/3)) - (-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^(1/3))))/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 824

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Module[{r = Numerator 
[Rt[a/b, n]], s = Denominator[Rt[a/b, n]], k, u}, Simp[u = Int[(r*Cos[(2*k 
- 1)*m*(Pi/n)] - s*Cos[(2*k - 1)*(m + 1)*(Pi/n)]*x)/(r^2 - 2*r*s*Cos[(2*k - 
 1)*(Pi/n)]*x + s^2*x^2), x] + Int[(r*Cos[(2*k - 1)*m*(Pi/n)] + s*Cos[(2*k 
- 1)*(m + 1)*(Pi/n)]*x)/(r^2 + 2*r*s*Cos[(2*k - 1)*(Pi/n)]*x + s^2*x^2), x] 
; 2*(-1)^(m/2)*(r^(m + 2)/(a*n*s^m))   Int[1/(r^2 + s^2*x^2), x] + 2*(r^(m 
+ 1)/(a*n*s^m))   Sum[u, {k, 1, (n - 2)/4}], x]] /; FreeQ[{a, b}, x] && IGt 
Q[(n - 2)/4, 0] && IGtQ[m, 0] && LtQ[m, n - 1] && 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 = 0.10 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.85

method	result	size
derivativedivides	\(\frac {3 b \left (-\frac {\sqrt {3}\, \left (b^{2}\right )^{\frac {5}{6}} \ln \left (\left (b \tan \left (d x +c \right )\right )^{\frac {2}{3}}+\sqrt {3}\, \left (b^{2}\right )^{\frac {1}{6}} \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}+\left (b^{2}\right )^{\frac {1}{3}}\right )}{12 b^{2}}+\frac {\arctan \left (\frac {2 \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{\left (b^{2}\right )^{\frac {1}{6}}}+\sqrt {3}\right )}{6 \left (b^{2}\right )^{\frac {1}{6}}}+\frac {\sqrt {3}\, \left (b^{2}\right )^{\frac {5}{6}} \ln \left (\left (b \tan \left (d x +c \right )\right )^{\frac {2}{3}}-\sqrt {3}\, \left (b^{2}\right )^{\frac {1}{6}} \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}+\left (b^{2}\right )^{\frac {1}{3}}\right )}{12 b^{2}}+\frac {\arctan \left (\frac {2 \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{\left (b^{2}\right )^{\frac {1}{6}}}-\sqrt {3}\right )}{6 \left (b^{2}\right )^{\frac {1}{6}}}+\frac {\arctan \left (\frac {\left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{\left (b^{2}\right )^{\frac {1}{6}}}\right )}{3 \left (b^{2}\right )^{\frac {1}{6}}}\right )}{d}\)	\(191\)
default	\(\frac {3 b \left (-\frac {\sqrt {3}\, \left (b^{2}\right )^{\frac {5}{6}} \ln \left (\left (b \tan \left (d x +c \right )\right )^{\frac {2}{3}}+\sqrt {3}\, \left (b^{2}\right )^{\frac {1}{6}} \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}+\left (b^{2}\right )^{\frac {1}{3}}\right )}{12 b^{2}}+\frac {\arctan \left (\frac {2 \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{\left (b^{2}\right )^{\frac {1}{6}}}+\sqrt {3}\right )}{6 \left (b^{2}\right )^{\frac {1}{6}}}+\frac {\sqrt {3}\, \left (b^{2}\right )^{\frac {5}{6}} \ln \left (\left (b \tan \left (d x +c \right )\right )^{\frac {2}{3}}-\sqrt {3}\, \left (b^{2}\right )^{\frac {1}{6}} \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}+\left (b^{2}\right )^{\frac {1}{3}}\right )}{12 b^{2}}+\frac {\arctan \left (\frac {2 \left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{\left (b^{2}\right )^{\frac {1}{6}}}-\sqrt {3}\right )}{6 \left (b^{2}\right )^{\frac {1}{6}}}+\frac {\arctan \left (\frac {\left (b \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{\left (b^{2}\right )^{\frac {1}{6}}}\right )}{3 \left (b^{2}\right )^{\frac {1}{6}}}\right )}{d}\)	\(191\)

Input:

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

Output:

3/d*b*(-1/12/b^2*3^(1/2)*(b^2)^(5/6)*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)^(1/6)*arctan(2*(b*tan(d 
*x+c))^(1/3)/(b^2)^(1/6)+3^(1/2))+1/12/b^2*3^(1/2)*(b^2)^(5/6)*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)^(1/6)*arctan(2*(b*tan(d*x+c))^(1/3)/(b^2)^(1/6)-3^(1/2))+1/3/(b^2)^(1/6 
)*arctan((b*tan(d*x+c))^(1/3)/(b^2)^(1/6)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.40 \[ \int (b \tan (c+d x))^{2/3} \, dx=-\frac {1}{4} \, {\left (\sqrt {-3} - 1\right )} \left (-\frac {b^{4}}{d^{6}}\right )^{\frac {1}{6}} \log \left (\left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} b^{3} + \frac {1}{2} \, {\left (\sqrt {-3} d^{5} + d^{5}\right )} \left (-\frac {b^{4}}{d^{6}}\right )^{\frac {5}{6}}\right ) + \frac {1}{4} \, {\left (\sqrt {-3} - 1\right )} \left (-\frac {b^{4}}{d^{6}}\right )^{\frac {1}{6}} \log \left (\left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} b^{3} - \frac {1}{2} \, {\left (\sqrt {-3} d^{5} + d^{5}\right )} \left (-\frac {b^{4}}{d^{6}}\right )^{\frac {5}{6}}\right ) - \frac {1}{4} \, {\left (\sqrt {-3} + 1\right )} \left (-\frac {b^{4}}{d^{6}}\right )^{\frac {1}{6}} \log \left (\left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} b^{3} + \frac {1}{2} \, {\left (\sqrt {-3} d^{5} - d^{5}\right )} \left (-\frac {b^{4}}{d^{6}}\right )^{\frac {5}{6}}\right ) + \frac {1}{4} \, {\left (\sqrt {-3} + 1\right )} \left (-\frac {b^{4}}{d^{6}}\right )^{\frac {1}{6}} \log \left (\left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} b^{3} - \frac {1}{2} \, {\left (\sqrt {-3} d^{5} - d^{5}\right )} \left (-\frac {b^{4}}{d^{6}}\right )^{\frac {5}{6}}\right ) + \frac {1}{2} \, \left (-\frac {b^{4}}{d^{6}}\right )^{\frac {1}{6}} \log \left (d^{5} \left (-\frac {b^{4}}{d^{6}}\right )^{\frac {5}{6}} + \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} b^{3}\right ) - \frac {1}{2} \, \left (-\frac {b^{4}}{d^{6}}\right )^{\frac {1}{6}} \log \left (-d^{5} \left (-\frac {b^{4}}{d^{6}}\right )^{\frac {5}{6}} + \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} b^{3}\right ) \] Input:

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

Output:

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

Sympy [F]

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

integrate((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 = 168, normalized size of antiderivative = 0.75 \[ \int (b \tan (c+d x))^{2/3} \, dx=-\frac {{\left (\frac {\sqrt {3} \log \left (\sqrt {3} \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} b^{\frac {1}{3}} + \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} + b^{\frac {2}{3}}\right )}{b^{\frac {1}{3}}} - \frac {\sqrt {3} \log \left (-\sqrt {3} \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} b^{\frac {1}{3}} + \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} + b^{\frac {2}{3}}\right )}{b^{\frac {1}{3}}} - \frac {2 \, \arctan \left (\frac {\sqrt {3} b^{\frac {1}{3}} + 2 \, \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}\right )}{b^{\frac {1}{3}}} - \frac {2 \, \arctan \left (-\frac {\sqrt {3} b^{\frac {1}{3}} - 2 \, \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}\right )}{b^{\frac {1}{3}}} - \frac {4 \, \arctan \left (\frac {\left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}}{b^{\frac {1}{3}}}\right )}{b^{\frac {1}{3}}}\right )} b}{4 \, d} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.92 \[ \int (b \tan (c+d x))^{2/3} \, dx=-\frac {\sqrt {3} {\left | b \right |}^{\frac {5}{3}} \log \left (\sqrt {3} \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} {\left | b \right |}^{\frac {1}{3}} + \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} + {\left | b \right |}^{\frac {2}{3}}\right )}{4 \, b d} + \frac {\sqrt {3} {\left | b \right |}^{\frac {5}{3}} \log \left (-\sqrt {3} \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}} {\left | b \right |}^{\frac {1}{3}} + \left (b \tan \left (d x + c\right )\right )^{\frac {2}{3}} + {\left | b \right |}^{\frac {2}{3}}\right )}{4 \, b d} + \frac {{\left | b \right |}^{\frac {5}{3}} \arctan \left (\frac {\sqrt {3} {\left | b \right |}^{\frac {1}{3}} + 2 \, \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}}{{\left | b \right |}^{\frac {1}{3}}}\right )}{2 \, b d} + \frac {{\left | b \right |}^{\frac {5}{3}} \arctan \left (-\frac {\sqrt {3} {\left | b \right |}^{\frac {1}{3}} - 2 \, \left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}}{{\left | b \right |}^{\frac {1}{3}}}\right )}{2 \, b d} + \frac {{\left | b \right |}^{\frac {5}{3}} \arctan \left (\frac {\left (b \tan \left (d x + c\right )\right )^{\frac {1}{3}}}{{\left | b \right |}^{\frac {1}{3}}}\right )}{b d} \] Input:

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

Output:

-1/4*sqrt(3)*abs(b)^(5/3)*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*d) + 1/4*sqrt(3)*abs(b)^(5/3)* 
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*d) + 1/2*abs(b)^(5/3)*arctan((sqrt(3)*abs(b)^(1/3) + 2* 
(b*tan(d*x + c))^(1/3))/abs(b)^(1/3))/(b*d) + 1/2*abs(b)^(5/3)*arctan(-(sq 
rt(3)*abs(b)^(1/3) - 2*(b*tan(d*x + c))^(1/3))/abs(b)^(1/3))/(b*d) + abs(b 
)^(5/3)*arctan((b*tan(d*x + c))^(1/3)/abs(b)^(1/3))/(b*d)
 

Mupad [B] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.16 \[ \int (b \tan (c+d x))^{2/3} \, dx=\frac {{\left (-1\right )}^{1/6}\,b^{2/3}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{2/3}\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}}{b^{1/3}}\right )\,1{}\mathrm {i}}{d}-\frac {{\left (-1\right )}^{1/6}\,b^{2/3}\,\ln \left (\frac {972\,b^9}{d^3}+\frac {486\,{\left (-1\right )}^{1/6}\,b^{26/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}}{d^3}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,d}-\frac {{\left (-1\right )}^{1/6}\,b^{2/3}\,\ln \left (\frac {972\,b^9}{d^3}+\frac {486\,{\left (-1\right )}^{1/6}\,b^{26/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}}{d^3}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,d}+\frac {{\left (-1\right )}^{1/6}\,b^{2/3}\,\ln \left (\frac {972\,b^9}{d^3}-\frac {486\,{\left (-1\right )}^{1/6}\,b^{26/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}}{d^3}\right )\,\left (-\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )}{d}+\frac {{\left (-1\right )}^{1/6}\,b^{2/3}\,\ln \left (\frac {972\,b^9}{d^3}-\frac {486\,{\left (-1\right )}^{1/6}\,b^{26/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{1/3}}{d^3}\right )\,\left (\frac {1}{4}+\frac {\sqrt {3}\,1{}\mathrm {i}}{4}\right )}{d} \] Input:

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

Output:

((-1)^(1/6)*b^(2/3)*atan(((-1)^(2/3)*(b*tan(c + d*x))^(1/3))/b^(1/3))*1i)/ 
d - ((-1)^(1/6)*b^(2/3)*log((972*b^9)/d^3 + (486*(-1)^(1/6)*b^(26/3)*(3^(1 
/2)*1i - 1)*(b*tan(c + d*x))^(1/3))/d^3)*((3^(1/2)*1i)/2 - 1/2))/(2*d) - ( 
(-1)^(1/6)*b^(2/3)*log((972*b^9)/d^3 + (486*(-1)^(1/6)*b^(26/3)*(3^(1/2)*1 
i + 1)*(b*tan(c + d*x))^(1/3))/d^3)*((3^(1/2)*1i)/2 + 1/2))/(2*d) + ((-1)^ 
(1/6)*b^(2/3)*log((972*b^9)/d^3 - (486*(-1)^(1/6)*b^(26/3)*(3^(1/2)*1i - 1 
)*(b*tan(c + d*x))^(1/3))/d^3)*((3^(1/2)*1i)/4 - 1/4))/d + ((-1)^(1/6)*b^( 
2/3)*log((972*b^9)/d^3 - (486*(-1)^(1/6)*b^(26/3)*(3^(1/2)*1i + 1)*(b*tan( 
c + d*x))^(1/3))/d^3)*((3^(1/2)*1i)/4 + 1/4))/d
 

Reduce [F]

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

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

Output:

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