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

Optimal. Leaf size=131 \[ -\frac{\sqrt{3} \tan ^{-1}\left (\frac{b^{2/3}-2 (b \tan (c+d x))^{2/3}}{\sqrt{3} b^{2/3}}\right )}{2 \sqrt [3]{b} d}+\frac{\log \left (b^{2/3}+(b \tan (c+d x))^{2/3}\right )}{2 \sqrt [3]{b} d}-\frac{\log \left (-b^{2/3} (b \tan (c+d x))^{2/3}+b^{4/3}+(b \tan (c+d x))^{4/3}\right )}{4 \sqrt [3]{b} d} \]

[Out]

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

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Rubi [A]  time = 0.0996872, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {3476, 329, 275, 200, 31, 634, 617, 204, 628} \[ -\frac{\sqrt{3} \tan ^{-1}\left (\frac{b^{2/3}-2 (b \tan (c+d x))^{2/3}}{\sqrt{3} b^{2/3}}\right )}{2 \sqrt [3]{b} d}+\frac{\log \left (b^{2/3}+(b \tan (c+d x))^{2/3}\right )}{2 \sqrt [3]{b} d}-\frac{\log \left (-b^{2/3} (b \tan (c+d x))^{2/3}+b^{4/3}+(b \tan (c+d x))^{4/3}\right )}{4 \sqrt [3]{b} d} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[c + d*x])^(-1/3),x]

[Out]

-(Sqrt[3]*ArcTan[(b^(2/3) - 2*(b*Tan[c + d*x])^(2/3))/(Sqrt[3]*b^(2/3))])/(2*b^(1/3)*d) + Log[b^(2/3) + (b*Tan
[c + d*x])^(2/3)]/(2*b^(1/3)*d) - Log[b^(4/3) - b^(2/3)*(b*Tan[c + d*x])^(2/3) + (b*Tan[c + d*x])^(4/3)]/(4*b^
(1/3)*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 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[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 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[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 31

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

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 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[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])] /;
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]

Rubi steps

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

Mathematica [A]  time = 0.133476, size = 100, normalized size = 0.76 \[ \frac{\sqrt [3]{\tan (c+d x)} \left (2 \sqrt{3} \tan ^{-1}\left (\frac{2 \tan ^{\frac{2}{3}}(c+d x)-1}{\sqrt{3}}\right )+2 \log \left (\tan ^{\frac{2}{3}}(c+d x)+1\right )-\log \left (\tan ^{\frac{4}{3}}(c+d x)-\tan ^{\frac{2}{3}}(c+d x)+1\right )\right )}{4 d \sqrt [3]{b \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[c + d*x])^(-1/3),x]

[Out]

((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))

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Maple [A]  time = 0.013, size = 114, normalized size = 0.9 \begin{align*}{\frac{b}{2\,d}\ln \left ( \left ( b\tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}+\sqrt [3]{{b}^{2}} \right ) \left ({b}^{2} \right ) ^{-{\frac{2}{3}}}}-{\frac{b}{4\,d}\ln \left ( \left ( b\tan \left ( dx+c \right ) \right ) ^{{\frac{4}{3}}}-\sqrt [3]{{b}^{2}} \left ( b\tan \left ( dx+c \right ) \right ) ^{{\frac{2}{3}}}+ \left ({b}^{2} \right ) ^{{\frac{2}{3}}} \right ) \left ({b}^{2} \right ) ^{-{\frac{2}{3}}}}+{\frac{b\sqrt{3}}{2\,d}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{\frac{ \left ( b\tan \left ( dx+c \right ) \right ) ^{2/3}}{\sqrt [3]{{b}^{2}}}}-1 \right ) } \right ) \left ({b}^{2} \right ) ^{-{\frac{2}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.4241, size = 167, normalized size = 1.27 \begin{align*} \frac{\frac{2 \, \sqrt{3} b^{2} \arctan \left (\frac{\sqrt{3}{\left (2 \, \left (b \tan \left (d x + c\right )\right )^{\frac{2}{3}} -{\left (b^{2}\right )}^{\frac{1}{3}}\right )}}{3 \,{\left (b^{2}\right )}^{\frac{1}{3}}}\right )}{{\left (b^{2}\right )}^{\frac{2}{3}}} - \frac{b^{2} \log \left (\left (b \tan \left (d x + c\right )\right )^{\frac{4}{3}} -{\left (b^{2}\right )}^{\frac{1}{3}} \left (b \tan \left (d x + c\right )\right )^{\frac{2}{3}} +{\left (b^{2}\right )}^{\frac{2}{3}}\right )}{{\left (b^{2}\right )}^{\frac{2}{3}}} + \frac{2 \, b^{2} \log \left (\left (b \tan \left (d x + c\right )\right )^{\frac{2}{3}} +{\left (b^{2}\right )}^{\frac{1}{3}}\right )}{{\left (b^{2}\right )}^{\frac{2}{3}}}}{4 \, b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.38658, size = 898, normalized size = 6.85 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(sqrt(3)*b*sqrt(-1/b^(2/3))*log((2*sqrt(3)*(b*tan(d*x + c))^(1/3)*b*sqrt(-1/b^(2/3))*tan(d*x + c) + 2*b*t
an(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)*arcta
n(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)
]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*tan(c + d*x))**(-1/3), x)

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Giac [A]  time = 1.35768, size = 173, normalized size = 1.32 \begin{align*} \frac{1}{4} \, b{\left (\frac{2 \, \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 )}{b^{2} 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 )}{b^{2} d} + \frac{2 \,{\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 )}{b^{2} d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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