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

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

[Out]

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

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Rubi [A]  time = 0.103936, 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, 292, 31, 634, 617, 204, 628} \[ -\frac{\sqrt{3} \sqrt [3]{b} \tan ^{-1}\left (\frac{b^{2/3}-2 (b \tan (c+d x))^{2/3}}{\sqrt{3} b^{2/3}}\right )}{2 d}-\frac{\sqrt [3]{b} \log \left (b^{2/3}+(b \tan (c+d x))^{2/3}\right )}{2 d}+\frac{\sqrt [3]{b} \log \left (-b^{2/3} (b \tan (c+d x))^{2/3}+b^{4/3}+(b \tan (c+d x))^{4/3}\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[(b*Tan[c + d*x])^(1/3),x]

[Out]

-(Sqrt[3]*b^(1/3)*ArcTan[(b^(2/3) - 2*(b*Tan[c + d*x])^(2/3))/(Sqrt[3]*b^(2/3))])/(2*d) - (b^(1/3)*Log[b^(2/3)
+ (b*Tan[c + d*x])^(2/3)])/(2*d) + (b^(1/3)*Log[b^(4/3) - b^(2/3)*(b*Tan[c + d*x])^(2/3) + (b*Tan[c + d*x])^(
4/3)])/(4*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 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(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 \sqrt [3]{b \tan (c+d x)} \, dx &=\frac{b \operatorname{Subst}\left (\int \frac{\sqrt [3]{x}}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{d}\\ &=\frac{(3 b) \operatorname{Subst}\left (\int \frac{x^3}{b^2+x^6} \, dx,x,\sqrt [3]{b \tan (c+d x)}\right )}{d}\\ &=\frac{(3 b) \operatorname{Subst}\left (\int \frac{x}{b^2+x^3} \, dx,x,(b \tan (c+d x))^{2/3}\right )}{2 d}\\ &=-\frac{\sqrt [3]{b} \operatorname{Subst}\left (\int \frac{1}{b^{2/3}+x} \, dx,x,(b \tan (c+d x))^{2/3}\right )}{2 d}+\frac{\sqrt [3]{b} \operatorname{Subst}\left (\int \frac{b^{2/3}+x}{b^{4/3}-b^{2/3} x+x^2} \, dx,x,(b \tan (c+d x))^{2/3}\right )}{2 d}\\ &=-\frac{\sqrt [3]{b} \log \left (b^{2/3}+(b \tan (c+d x))^{2/3}\right )}{2 d}+\frac{\sqrt [3]{b} \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 d}+\frac{(3 b) \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{\sqrt [3]{b} \log \left (b^{2/3}+(b \tan (c+d x))^{2/3}\right )}{2 d}+\frac{\sqrt [3]{b} \log \left (b^{4/3}-b^{2/3} (b \tan (c+d x))^{2/3}+(b \tan (c+d x))^{4/3}\right )}{4 d}+\frac{\left (3 \sqrt [3]{b}\right ) \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 d}\\ &=-\frac{\sqrt{3} \sqrt [3]{b} \tan ^{-1}\left (\frac{1-\frac{2 (b \tan (c+d x))^{2/3}}{b^{2/3}}}{\sqrt{3}}\right )}{2 d}-\frac{\sqrt [3]{b} \log \left (b^{2/3}+(b \tan (c+d x))^{2/3}\right )}{2 d}+\frac{\sqrt [3]{b} \log \left (b^{4/3}-b^{2/3} (b \tan (c+d x))^{2/3}+(b \tan (c+d x))^{4/3}\right )}{4 d}\\ \end{align*}

Mathematica [C]  time = 0.0397503, size = 40, normalized size = 0.31 \[ \frac{3 (b \tan (c+d x))^{4/3} \, _2F_1\left (\frac{2}{3},1;\frac{5}{3};-\tan ^2(c+d x)\right )}{4 b d} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Tan[c + d*x])^(1/3),x]

[Out]

(3*Hypergeometric2F1[2/3, 1, 5/3, -Tan[c + d*x]^2]*(b*Tan[c + d*x])^(4/3))/(4*b*d)

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Maple [A]  time = 0.015, 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 ){\frac{1}{\sqrt [3]{{b}^{2}}}}}+{\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 ){\frac{1}{\sqrt [3]{{b}^{2}}}}}+{\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 ){\frac{1}{\sqrt [3]{{b}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d*b/(b^2)^(1/3)*ln((b*tan(d*x+c))^(2/3)+(b^2)^(1/3))+1/4/d*b/(b^2)^(1/3)*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*3^(1/2)/(b^2)^(1/3)*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 = 1.44522, size = 166, 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{1}{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{1}{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{1}{3}}}}{4 \, b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]  time = 1.34673, size = 358, normalized size = 2.73 \begin{align*} \frac{2 \, \sqrt{3} \left (-b\right )^{\frac{1}{3}} \arctan \left (\frac{2 \, \sqrt{3} \left (b \tan \left (d x + c\right )\right )^{\frac{2}{3}} \left (-b\right )^{\frac{1}{3}} + \sqrt{3} b}{3 \, b}\right ) - \left (-b\right )^{\frac{1}{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{1}{3}} b\right ) + 2 \, \left (-b\right )^{\frac{1}{3}} \log \left (\left (b \tan \left (d x + c\right )\right )^{\frac{2}{3}} + \left (-b\right )^{\frac{2}{3}}\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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 = 1.42799, size = 171, normalized size = 1.31 \begin{align*} \frac{1}{4} \, b{\left (\frac{2 \, \sqrt{3}{\left | b \right |}^{\frac{4}{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{4}{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{4}{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((b*tan(d*x+c))^(1/3),x, algorithm="giac")

[Out]

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