3.442 \(\int \frac {\tan (x)}{\sqrt [3]{a^3+b^3 \tan ^2(x)}} \, dx\)

Optimal. Leaf size=133 \[ \frac {\sqrt {3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{a^3+b^3 \tan ^2(x)}}{\sqrt [3]{a^3-b^3}}+1}{\sqrt {3}}\right )}{2 \sqrt [3]{a^3-b^3}}+\frac {3 \log \left (\sqrt [3]{a^3-b^3}-\sqrt [3]{a^3+b^3 \tan ^2(x)}\right )}{4 \sqrt [3]{a^3-b^3}}+\frac {\log (\cos (x))}{2 \sqrt [3]{a^3-b^3}} \]

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Rubi [A]  time = 0.16, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3670, 444, 55, 617, 204, 31} \[ \frac {\sqrt {3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{a^3+b^3 \tan ^2(x)}}{\sqrt [3]{a^3-b^3}}+1}{\sqrt {3}}\right )}{2 \sqrt [3]{a^3-b^3}}+\frac {3 \log \left (\sqrt [3]{a^3-b^3}-\sqrt [3]{a^3+b^3 \tan ^2(x)}\right )}{4 \sqrt [3]{a^3-b^3}}+\frac {\log (\cos (x))}{2 \sqrt [3]{a^3-b^3}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]/(a^3 + b^3*Tan[x]^2)^(1/3),x]

[Out]

(Sqrt[3]*ArcTan[(1 + (2*(a^3 + b^3*Tan[x]^2)^(1/3))/(a^3 - b^3)^(1/3))/Sqrt[3]])/(2*(a^3 - b^3)^(1/3)) + Log[C
os[x]]/(2*(a^3 - b^3)^(1/3)) + (3*Log[(a^3 - b^3)^(1/3) - (a^3 + b^3*Tan[x]^2)^(1/3)])/(4*(a^3 - b^3)^(1/3))

Rule 31

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

Rule 55

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[{q = Rt[(b*c - a*d)/b, 3]}, -Simp[L
og[RemoveContent[a + b*x, x]]/(2*b*q), x] + (Dist[3/(2*b), Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/
3)], x] - Dist[3/(2*b*q), Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] /; FreeQ[{a, b, c, d}, x] && PosQ
[(b*c - a*d)/b]

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 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

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 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 105, normalized size = 0.79 \[ \frac {2 \sqrt {3} \tan ^{-1}\left (\frac {\frac {2 \sqrt [3]{a^3+b^3 \tan ^2(x)}}{\sqrt [3]{a^3-b^3}}+1}{\sqrt {3}}\right )+3 \log \left (\sqrt [3]{a^3-b^3}-\sqrt [3]{a^3+b^3 \tan ^2(x)}\right )+2 \log (\cos (x))}{4 \sqrt [3]{a^3-b^3}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]/(a^3 + b^3*Tan[x]^2)^(1/3),x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan (x)}{\sqrt [3]{a^3+b^3 \tan ^2(x)}} \, dx \]

Verification is Not applicable to the result.

[In]

IntegrateAlgebraic[Tan[x]/(a^3 + b^3*Tan[x]^2)^(1/3),x]

[Out]

Could not integrate

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [A]  time = 0.76, size = 186, normalized size = 1.40 \[ \frac {3 \, {\left (a^{3} - b^{3}\right )}^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b^{3} \tan \relax (x)^{2} + a^{3}\right )}^{\frac {1}{3}} + {\left (a^{3} - b^{3}\right )}^{\frac {1}{3}}\right )}}{3 \, {\left (a^{3} - b^{3}\right )}^{\frac {1}{3}}}\right )}{2 \, {\left (\sqrt {3} a^{3} - \sqrt {3} b^{3}\right )}} - \frac {\log \left ({\left (b^{3} \tan \relax (x)^{2} + a^{3}\right )}^{\frac {2}{3}} + {\left (b^{3} \tan \relax (x)^{2} + a^{3}\right )}^{\frac {1}{3}} {\left (a^{3} - b^{3}\right )}^{\frac {1}{3}} + {\left (a^{3} - b^{3}\right )}^{\frac {2}{3}}\right )}{4 \, {\left (a^{3} - b^{3}\right )}^{\frac {1}{3}}} + \frac {\log \left ({\left | {\left (b^{3} \tan \relax (x)^{2} + a^{3}\right )}^{\frac {1}{3}} - {\left (a^{3} - b^{3}\right )}^{\frac {1}{3}} \right |}\right )}{2 \, {\left (a^{3} - b^{3}\right )}^{\frac {1}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {\tan \relax (x )}{\left (a^{3}+b^{3} \left (\tan ^{2}\relax (x )\right )\right )^{\frac {1}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \relax (x)}{{\left (b^{3} \tan \relax (x)^{2} + a^{3}\right )}^{\frac {1}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(tan(x)/(b^3*tan(x)^2 + a^3)^(1/3), x)

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mupad [B]  time = 1.45, size = 250, normalized size = 1.88 \[ \frac {\ln \left (\frac {9\,{\left (a^3+b^3\,{\mathrm {tan}\relax (x)}^2\right )}^{1/3}}{4}-\frac {9\,a^3-9\,b^3}{4\,{\left (a-b\right )}^{2/3}\,{\left (a^2+a\,b+b^2\right )}^{2/3}}\right )}{2\,{\left (a-b\right )}^{1/3}\,{\left (a^2+a\,b+b^2\right )}^{1/3}}+\frac {\ln \left (\frac {9\,{\left (a^3+b^3\,{\mathrm {tan}\relax (x)}^2\right )}^{1/3}}{4}-\frac {{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (9\,a^3-9\,b^3\right )}{16\,{\left (a-b\right )}^{2/3}\,{\left (a^2+a\,b+b^2\right )}^{2/3}}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{4\,{\left (a-b\right )}^{1/3}\,{\left (a^2+a\,b+b^2\right )}^{1/3}}-\frac {\ln \left (\frac {9\,{\left (a^3+b^3\,{\mathrm {tan}\relax (x)}^2\right )}^{1/3}}{4}-\frac {{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (9\,a^3-9\,b^3\right )}{16\,{\left (a-b\right )}^{2/3}\,{\left (a^2+a\,b+b^2\right )}^{2/3}}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{4\,{\left (a-b\right )}^{1/3}\,{\left (a^2+a\,b+b^2\right )}^{1/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log((9*(b^3*tan(x)^2 + a^3)^(1/3))/4 - (9*a^3 - 9*b^3)/(4*(a - b)^(2/3)*(a*b + a^2 + b^2)^(2/3)))/(2*(a - b)^(
1/3)*(a*b + a^2 + b^2)^(1/3)) + (log((9*(b^3*tan(x)^2 + a^3)^(1/3))/4 - ((3^(1/2)*1i - 1)^2*(9*a^3 - 9*b^3))/(
16*(a - b)^(2/3)*(a*b + a^2 + b^2)^(2/3)))*(3^(1/2)*1i - 1))/(4*(a - b)^(1/3)*(a*b + a^2 + b^2)^(1/3)) - (log(
(9*(b^3*tan(x)^2 + a^3)^(1/3))/4 - ((3^(1/2)*1i + 1)^2*(9*a^3 - 9*b^3))/(16*(a - b)^(2/3)*(a*b + a^2 + b^2)^(2
/3)))*(3^(1/2)*1i + 1))/(4*(a - b)^(1/3)*(a*b + a^2 + b^2)^(1/3))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan {\relax (x )}}{\sqrt [3]{a^{3} + b^{3} \tan ^{2}{\relax (x )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(tan(x)/(a**3 + b**3*tan(x)**2)**(1/3), x)

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