3.443 \(\int \tan (x) (1-7 \tan ^2(x))^{2/3} \, dx\)

Optimal. Leaf size=69 \[ 2 \sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{1-7 \tan ^2(x)}+1}{\sqrt{3}}\right )+\frac{3}{4} \left (1-7 \tan ^2(x)\right )^{2/3}+3 \log \left (2-\sqrt [3]{1-7 \tan ^2(x)}\right )+2 \log (\cos (x)) \]

[Out]

2*Sqrt[3]*ArcTan[(1 + (1 - 7*Tan[x]^2)^(1/3))/Sqrt[3]] + 2*Log[Cos[x]] + 3*Log[2 - (1 - 7*Tan[x]^2)^(1/3)] + (
3*(1 - 7*Tan[x]^2)^(2/3))/4

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Rubi [A]  time = 0.0871591, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.467, Rules used = {3670, 444, 50, 55, 618, 204, 31} \[ 2 \sqrt{3} \tan ^{-1}\left (\frac{\sqrt [3]{1-7 \tan ^2(x)}+1}{\sqrt{3}}\right )+\frac{3}{4} \left (1-7 \tan ^2(x)\right )^{2/3}+3 \log \left (2-\sqrt [3]{1-7 \tan ^2(x)}\right )+2 \log (\cos (x)) \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]*(1 - 7*Tan[x]^2)^(2/3),x]

[Out]

2*Sqrt[3]*ArcTan[(1 + (1 - 7*Tan[x]^2)^(1/3))/Sqrt[3]] + 2*Log[Cos[x]] + 3*Log[2 - (1 - 7*Tan[x]^2)^(1/3)] + (
3*(1 - 7*Tan[x]^2)^(2/3))/4

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]))

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 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, 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 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 31

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

Rubi steps

\begin{align*} \int \tan (x) \left (1-7 \tan ^2(x)\right )^{2/3} \, dx &=\operatorname{Subst}\left (\int \frac{x \left (1-7 x^2\right )^{2/3}}{1+x^2} \, dx,x,\tan (x)\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(1-7 x)^{2/3}}{1+x} \, dx,x,\tan ^2(x)\right )\\ &=\frac{3}{4} \left (1-7 \tan ^2(x)\right )^{2/3}+4 \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{1-7 x} (1+x)} \, dx,x,\tan ^2(x)\right )\\ &=2 \log (\cos (x))+\frac{3}{4} \left (1-7 \tan ^2(x)\right )^{2/3}-3 \operatorname{Subst}\left (\int \frac{1}{2-x} \, dx,x,\sqrt [3]{1-7 \tan ^2(x)}\right )+6 \operatorname{Subst}\left (\int \frac{1}{4+2 x+x^2} \, dx,x,\sqrt [3]{1-7 \tan ^2(x)}\right )\\ &=2 \log (\cos (x))+3 \log \left (2-\sqrt [3]{1-7 \tan ^2(x)}\right )+\frac{3}{4} \left (1-7 \tan ^2(x)\right )^{2/3}-12 \operatorname{Subst}\left (\int \frac{1}{-12-x^2} \, dx,x,2+2 \sqrt [3]{1-7 \tan ^2(x)}\right )\\ &=2 \sqrt{3} \tan ^{-1}\left (\frac{1+\sqrt [3]{1-7 \tan ^2(x)}}{\sqrt{3}}\right )+2 \log (\cos (x))+3 \log \left (2-\sqrt [3]{1-7 \tan ^2(x)}\right )+\frac{3}{4} \left (1-7 \tan ^2(x)\right )^{2/3}\\ \end{align*}

Mathematica [C]  time = 0.138839, size = 42, normalized size = 0.61 \[ -\frac{3}{4} \left (1-7 \tan ^2(x)\right )^{2/3} \left (\, _2F_1\left (\frac{2}{3},1;\frac{5}{3};\frac{1}{8} (4 \cos (2 x)-3) \sec ^2(x)\right )-1\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]*(1 - 7*Tan[x]^2)^(2/3),x]

[Out]

(-3*(-1 + Hypergeometric2F1[2/3, 1, 5/3, ((-3 + 4*Cos[2*x])*Sec[x]^2)/8])*(1 - 7*Tan[x]^2)^(2/3))/4

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Maple [F]  time = 0.05, size = 0, normalized size = 0. \begin{align*} \int \tan \left ( x \right ) \left ( 1-7\, \left ( \tan \left ( x \right ) \right ) ^{2} \right ) ^{{\frac{2}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)*(1-7*tan(x)^2)^(2/3),x)

[Out]

int(tan(x)*(1-7*tan(x)^2)^(2/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-7 \, \tan \left (x\right )^{2} + 1\right )}^{\frac{2}{3}} \tan \left (x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)*(1-7*tan(x)^2)^(2/3),x, algorithm="maxima")

[Out]

integrate((-7*tan(x)^2 + 1)^(2/3)*tan(x), x)

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Fricas [B]  time = 11.5915, size = 360, normalized size = 5.22 \begin{align*} 2 \, \sqrt{3} \arctan \left (\frac{7 \, \sqrt{3} \tan \left (x\right )^{2} + 4 \, \sqrt{3}{\left (-7 \, \tan \left (x\right )^{2} + 1\right )}^{\frac{2}{3}} - 16 \, \sqrt{3}{\left (-7 \, \tan \left (x\right )^{2} + 1\right )}^{\frac{1}{3}} - \sqrt{3}}{7 \, \tan \left (x\right )^{2} - 65}\right ) + \frac{3}{4} \,{\left (-7 \, \tan \left (x\right )^{2} + 1\right )}^{\frac{2}{3}} + \log \left (\frac{7 \, \tan \left (x\right )^{2} + 6 \,{\left (-7 \, \tan \left (x\right )^{2} + 1\right )}^{\frac{2}{3}} - 12 \,{\left (-7 \, \tan \left (x\right )^{2} + 1\right )}^{\frac{1}{3}} + 7}{\tan \left (x\right )^{2} + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)*(1-7*tan(x)^2)^(2/3),x, algorithm="fricas")

[Out]

2*sqrt(3)*arctan((7*sqrt(3)*tan(x)^2 + 4*sqrt(3)*(-7*tan(x)^2 + 1)^(2/3) - 16*sqrt(3)*(-7*tan(x)^2 + 1)^(1/3)
- sqrt(3))/(7*tan(x)^2 - 65)) + 3/4*(-7*tan(x)^2 + 1)^(2/3) + log((7*tan(x)^2 + 6*(-7*tan(x)^2 + 1)^(2/3) - 12
*(-7*tan(x)^2 + 1)^(1/3) + 7)/(tan(x)^2 + 1))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (1 - 7 \tan ^{2}{\left (x \right )}\right )^{\frac{2}{3}} \tan{\left (x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)*(1-7*tan(x)**2)**(2/3),x)

[Out]

Integral((1 - 7*tan(x)**2)**(2/3)*tan(x), x)

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Giac [A]  time = 1.10444, size = 107, normalized size = 1.55 \begin{align*} 2 \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left ({\left (-7 \, \tan \left (x\right )^{2} + 1\right )}^{\frac{1}{3}} + 1\right )}\right ) + \frac{3}{4} \,{\left (-7 \, \tan \left (x\right )^{2} + 1\right )}^{\frac{2}{3}} - \log \left ({\left (-7 \, \tan \left (x\right )^{2} + 1\right )}^{\frac{2}{3}} + 2 \,{\left (-7 \, \tan \left (x\right )^{2} + 1\right )}^{\frac{1}{3}} + 4\right ) + 2 \, \log \left ({\left |{\left (-7 \, \tan \left (x\right )^{2} + 1\right )}^{\frac{1}{3}} - 2 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)*(1-7*tan(x)^2)^(2/3),x, algorithm="giac")

[Out]

2*sqrt(3)*arctan(1/3*sqrt(3)*((-7*tan(x)^2 + 1)^(1/3) + 1)) + 3/4*(-7*tan(x)^2 + 1)^(2/3) - log((-7*tan(x)^2 +
1)^(2/3) + 2*(-7*tan(x)^2 + 1)^(1/3) + 4) + 2*log(abs((-7*tan(x)^2 + 1)^(1/3) - 2))