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

Optimal. Leaf size=224 \[ \frac{\tan ^{-1}\left (\frac{\sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{b^{2/3} d}-\frac{\tan ^{-1}\left (\sqrt{3}-\frac{2 \sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{2 b^{2/3} d}+\frac{\tan ^{-1}\left (\frac{2 \sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}+\sqrt{3}\right )}{2 b^{2/3} d}-\frac{\sqrt{3} \log \left (b^{2/3}-\sqrt{3} \sqrt [3]{b} \sqrt [3]{b \tan (c+d x)}+(b \tan (c+d x))^{2/3}\right )}{4 b^{2/3} d}+\frac{\sqrt{3} \log \left (b^{2/3}+\sqrt{3} \sqrt [3]{b} \sqrt [3]{b \tan (c+d x)}+(b \tan (c+d x))^{2/3}\right )}{4 b^{2/3} d} \]

[Out]

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

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Rubi [A]  time = 0.326431, antiderivative size = 224, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {3476, 329, 209, 634, 618, 204, 628, 203} \[ \frac{\tan ^{-1}\left (\frac{\sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{b^{2/3} d}-\frac{\tan ^{-1}\left (\sqrt{3}-\frac{2 \sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{2 b^{2/3} d}+\frac{\tan ^{-1}\left (\frac{2 \sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}+\sqrt{3}\right )}{2 b^{2/3} d}-\frac{\sqrt{3} \log \left (b^{2/3}-\sqrt{3} \sqrt [3]{b} \sqrt [3]{b \tan (c+d x)}+(b \tan (c+d x))^{2/3}\right )}{4 b^{2/3} d}+\frac{\sqrt{3} \log \left (b^{2/3}+\sqrt{3} \sqrt [3]{b} \sqrt [3]{b \tan (c+d x)}+(b \tan (c+d x))^{2/3}\right )}{4 b^{2/3} d} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[c + d*x])^(-2/3),x]

[Out]

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

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[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]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 209

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*Int[1/(r^2 +
s^2*x^2), x])/(a*n) + Dist[(2*r)/(a*n), Sum[u, {k, 1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)
/4, 0] && PosQ[a/b]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(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 203

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

Rubi steps

\begin{align*} \int \frac{1}{(b \tan (c+d x))^{2/3}} \, dx &=\frac{b \operatorname{Subst}\left (\int \frac{1}{x^{2/3} \left (b^2+x^2\right )} \, dx,x,b \tan (c+d x)\right )}{d}\\ &=\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{b^2+x^6} \, dx,x,\sqrt [3]{b \tan (c+d x)}\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\sqrt [3]{b}-\frac{\sqrt{3} x}{2}}{b^{2/3}-\sqrt{3} \sqrt [3]{b} x+x^2} \, dx,x,\sqrt [3]{b \tan (c+d x)}\right )}{b^{2/3} d}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt [3]{b}+\frac{\sqrt{3} x}{2}}{b^{2/3}+\sqrt{3} \sqrt [3]{b} x+x^2} \, dx,x,\sqrt [3]{b \tan (c+d x)}\right )}{b^{2/3} d}+\frac{\operatorname{Subst}\left (\int \frac{1}{b^{2/3}+x^2} \, dx,x,\sqrt [3]{b \tan (c+d x)}\right )}{\sqrt [3]{b} d}\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{b^{2/3} d}-\frac{\sqrt{3} \operatorname{Subst}\left (\int \frac{-\sqrt{3} \sqrt [3]{b}+2 x}{b^{2/3}-\sqrt{3} \sqrt [3]{b} x+x^2} \, dx,x,\sqrt [3]{b \tan (c+d x)}\right )}{4 b^{2/3} d}+\frac{\sqrt{3} \operatorname{Subst}\left (\int \frac{\sqrt{3} \sqrt [3]{b}+2 x}{b^{2/3}+\sqrt{3} \sqrt [3]{b} x+x^2} \, dx,x,\sqrt [3]{b \tan (c+d x)}\right )}{4 b^{2/3} d}+\frac{\operatorname{Subst}\left (\int \frac{1}{b^{2/3}-\sqrt{3} \sqrt [3]{b} x+x^2} \, dx,x,\sqrt [3]{b \tan (c+d x)}\right )}{4 \sqrt [3]{b} d}+\frac{\operatorname{Subst}\left (\int \frac{1}{b^{2/3}+\sqrt{3} \sqrt [3]{b} x+x^2} \, dx,x,\sqrt [3]{b \tan (c+d x)}\right )}{4 \sqrt [3]{b} d}\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{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}+\frac{\operatorname{Subst}\left (\int \frac{1}{-\frac{1}{3}-x^2} \, dx,x,1-\frac{2 \sqrt [3]{b \tan (c+d x)}}{\sqrt{3} \sqrt [3]{b}}\right )}{2 \sqrt{3} b^{2/3} d}-\frac{\operatorname{Subst}\left (\int \frac{1}{-\frac{1}{3}-x^2} \, dx,x,1+\frac{2 \sqrt [3]{b \tan (c+d x)}}{\sqrt{3} \sqrt [3]{b}}\right )}{2 \sqrt{3} b^{2/3} d}\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )}{b^{2/3} d}-\frac{\tan ^{-1}\left (\frac{1}{3} \left (3 \sqrt{3}-\frac{6 \sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )\right )}{2 b^{2/3} d}+\frac{\tan ^{-1}\left (\frac{1}{3} \left (3 \sqrt{3}+\frac{6 \sqrt [3]{b \tan (c+d x)}}{\sqrt [3]{b}}\right )\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}\\ \end{align*}

Mathematica [C]  time = 0.0281756, size = 38, normalized size = 0.17 \[ \frac{3 \sqrt [3]{b \tan (c+d x)} \, _2F_1\left (\frac{1}{6},1;\frac{7}{6};-\tan ^2(c+d x)\right )}{b d} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[c + d*x])^(-2/3),x]

[Out]

(3*Hypergeometric2F1[1/6, 1, 7/6, -Tan[c + d*x]^2]*(b*Tan[c + d*x])^(1/3))/(b*d)

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Maple [A]  time = 0.04, size = 211, normalized size = 0.9 \begin{align*}{\frac{\sqrt{3}}{4\,bd}\sqrt [6]{{b}^{2}}\ln \left ( \left ( b\tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}+\sqrt{3}\sqrt [6]{{b}^{2}}\sqrt [3]{b\tan \left ( dx+c \right ) }+\sqrt [3]{{b}^{2}} \right ) }+{\frac{1}{2\,bd}\sqrt [6]{{b}^{2}}\arctan \left ( 2\,{\frac{\sqrt [3]{b\tan \left ( dx+c \right ) }}{\sqrt [6]{{b}^{2}}}}+\sqrt{3} \right ) }-{\frac{\sqrt{3}}{4\,bd}\sqrt [6]{{b}^{2}}\ln \left ( \sqrt{3}\sqrt [6]{{b}^{2}}\sqrt [3]{b\tan \left ( dx+c \right ) }- \left ( b\tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}-\sqrt [3]{{b}^{2}} \right ) }+{\frac{1}{2\,bd}\sqrt [6]{{b}^{2}}\arctan \left ( 2\,{\frac{\sqrt [3]{b\tan \left ( dx+c \right ) }}{\sqrt [6]{{b}^{2}}}}-\sqrt{3} \right ) }+{\frac{1}{bd}\sqrt [6]{{b}^{2}}\arctan \left ({\sqrt [3]{b\tan \left ( dx+c \right ) }{\frac{1}{\sqrt [6]{{b}^{2}}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.60274, size = 1494, normalized size = 6.67 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(3)*(1/(b^4*d^6))^(1/6)*log(b^2*d^2*(1/(b^4*d^6))^(1/3) + sqrt(3)*b*d*(b*sin(d*x + c)/cos(d*x + c))^(1
/3)*(1/(b^4*d^6))^(1/6) + (b*sin(d*x + c)/cos(d*x + c))^(2/3)) - 1/4*sqrt(3)*(1/(b^4*d^6))^(1/6)*log(b^2*d^2*(
1/(b^4*d^6))^(1/3) - sqrt(3)*b*d*(b*sin(d*x + c)/cos(d*x + c))^(1/3)*(1/(b^4*d^6))^(1/6) + (b*sin(d*x + c)/cos
(d*x + c))^(2/3)) - (1/(b^4*d^6))^(1/6)*arctan(2*sqrt(b^2*d^2*(1/(b^4*d^6))^(1/3) + sqrt(3)*b*d*(b*sin(d*x + c
)/cos(d*x + c))^(1/3)*(1/(b^4*d^6))^(1/6) + (b*sin(d*x + c)/cos(d*x + c))^(2/3))*b^3*d^5*(1/(b^4*d^6))^(5/6) -
2*b^3*d^5*(b*sin(d*x + c)/cos(d*x + c))^(1/3)*(1/(b^4*d^6))^(5/6) - sqrt(3)) - (1/(b^4*d^6))^(1/6)*arctan(2*s
qrt(b^2*d^2*(1/(b^4*d^6))^(1/3) - sqrt(3)*b*d*(b*sin(d*x + c)/cos(d*x + c))^(1/3)*(1/(b^4*d^6))^(1/6) + (b*sin
(d*x + c)/cos(d*x + c))^(2/3))*b^3*d^5*(1/(b^4*d^6))^(5/6) - 2*b^3*d^5*(b*sin(d*x + c)/cos(d*x + c))^(1/3)*(1/
(b^4*d^6))^(5/6) + sqrt(3)) - 2*(1/(b^4*d^6))^(1/6)*arctan(sqrt(b^2*d^2*(1/(b^4*d^6))^(1/3) + (b*sin(d*x + c)/
cos(d*x + c))^(2/3))*b^3*d^5*(1/(b^4*d^6))^(5/6) - b^3*d^5*(b*sin(d*x + c)/cos(d*x + c))^(1/3)*(1/(b^4*d^6))^(
5/6))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \tan{\left (c + d x \right )}\right )^{\frac{2}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*tan(c + d*x))**(-2/3), x)

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Giac [A]  time = 1.34661, size = 282, normalized size = 1.26 \begin{align*} \frac{1}{4} \, b{\left (\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^{2} d} - \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^{2} d} + \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^{2} d} + \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^{2} d} + \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^{2} d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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