3.452 \(\int \frac{\csc (x) \sec (x) (1+\sqrt [3]{1-8 \tan ^2(x)})}{(1-8 \tan ^2(x))^{2/3}} \, dx\)

Optimal. Leaf size=27 \[ \frac{3}{2} \log \left (1-\sqrt [3]{1-8 \tan ^2(x)}\right )-\log (\tan (x)) \]

[Out]

-Log[Tan[x]] + (3*Log[1 - (1 - 8*Tan[x]^2)^(1/3)])/2

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Rubi [A]  time = 0.967354, antiderivative size = 35, normalized size of antiderivative = 1.3, number of steps used = 15, number of rules used = 9, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.29, Rules used = {4366, 6725, 514, 444, 57, 618, 204, 31, 55} \[ \frac{3}{2} \log \left (1-\sqrt [3]{9-8 \sec ^2(x)}\right )-\frac{1}{2} \log \left (1-\sec ^2(x)\right ) \]

Antiderivative was successfully verified.

[In]

Int[(Csc[x]*Sec[x]*(1 + (1 - 8*Tan[x]^2)^(1/3)))/(1 - 8*Tan[x]^2)^(2/3),x]

[Out]

-Log[1 - Sec[x]^2]/2 + (3*Log[1 - (9 - 8*Sec[x]^2)^(1/3)])/2

Rule 4366

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^(n_), x_Symbol] :> With[{d = FreeFactors[Cos[c*(a + b*x)], x]}, -Dis
t[d/(b*c), Subst[Int[SubstFor[(1 - d^2*x^2)^((n - 1)/2), Cos[c*(a + b*x)]/d, u, x], x], x, Cos[c*(a + b*x)]/d]
, x] /; FunctionOfQ[Cos[c*(a + b*x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && IntegerQ[(n - 1)/2] && NonsumQ[u] &&
(EqQ[F, Sin] || EqQ[F, sin])

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 514

Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[x^(m - n*q)*
(a + b*x^n)^p*(d + c*x^n)^q, x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] |
|  !IntegerQ[p])

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 57

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

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]

Rubi steps

\begin{align*} \int \frac{\csc (x) \sec (x) \left (1+\sqrt [3]{1-8 \tan ^2(x)}\right )}{\left (1-8 \tan ^2(x)\right )^{2/3}} \, dx &=-\operatorname{Subst}\left (\int \frac{1+\sqrt [3]{9-\frac{8}{x^2}}}{\left (9-\frac{8}{x^2}\right )^{2/3} x \left (1-x^2\right )} \, dx,x,\cos (x)\right )\\ &=-\operatorname{Subst}\left (\int \left (-\frac{1}{\left (9-\frac{8}{x^2}\right )^{2/3} x \left (-1+x^2\right )}-\frac{1}{\sqrt [3]{9-\frac{8}{x^2}} x \left (-1+x^2\right )}\right ) \, dx,x,\cos (x)\right )\\ &=\operatorname{Subst}\left (\int \frac{1}{\left (9-\frac{8}{x^2}\right )^{2/3} x \left (-1+x^2\right )} \, dx,x,\cos (x)\right )+\operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{9-\frac{8}{x^2}} x \left (-1+x^2\right )} \, dx,x,\cos (x)\right )\\ &=\operatorname{Subst}\left (\int \frac{1}{\left (9-\frac{8}{x^2}\right )^{2/3} \left (1-\frac{1}{x^2}\right ) x^3} \, dx,x,\cos (x)\right )+\operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{9-\frac{8}{x^2}} \left (1-\frac{1}{x^2}\right ) x^3} \, dx,x,\cos (x)\right )\\ &=-\left (\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{(9-8 x)^{2/3} (1-x)} \, dx,x,\sec ^2(x)\right )\right )-\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{9-8 x} (1-x)} \, dx,x,\sec ^2(x)\right )\\ &=-\log (\tan (x))-2 \left (\frac{3}{4} \operatorname{Subst}\left (\int \frac{1}{1-x} \, dx,x,\sqrt [3]{9-8 \sec ^2(x)}\right )\right )\\ &=\frac{3}{2} \log \left (1-\sqrt [3]{9-8 \sec ^2(x)}\right )-\log (\tan (x))\\ \end{align*}

Mathematica [B]  time = 4.10294, size = 58, normalized size = 2.15 \[ \frac{1}{4} \left (5 \log \left (1-\sqrt [3]{1-8 \tan ^2(x)}\right )-\log \left (\left (1-8 \tan ^2(x)\right )^{2/3}+\sqrt [3]{1-8 \tan ^2(x)}+1\right )-2 \log (\tan (x))\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(Csc[x]*Sec[x]*(1 + (1 - 8*Tan[x]^2)^(1/3)))/(1 - 8*Tan[x]^2)^(2/3),x]

[Out]

(-2*Log[Tan[x]] + 5*Log[1 - (1 - 8*Tan[x]^2)^(1/3)] - Log[1 + (1 - 8*Tan[x]^2)^(1/3) + (1 - 8*Tan[x]^2)^(2/3)]
)/4

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Maple [F]  time = 0.485, size = 0, normalized size = 0. \begin{align*} \int{\frac{\cot \left ( x \right ) }{ \left ( \cos \left ( x \right ) \right ) ^{2}} \left ( 1+\sqrt [3]{1-8\, \left ( \tan \left ( x \right ) \right ) ^{2}} \right ) \left ( 1-8\, \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(cot(x)*(1+(1-8*tan(x)^2)^(1/3))/cos(x)^2/(1-8*tan(x)^2)^(2/3),x)

[Out]

int(cot(x)*(1+(1-8*tan(x)^2)^(1/3))/cos(x)^2/(1-8*tan(x)^2)^(2/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left ({\left (-8 \, \tan \left (x\right )^{2} + 1\right )}^{\frac{1}{3}} + 1\right )} \cot \left (x\right )}{{\left (-8 \, \tan \left (x\right )^{2} + 1\right )}^{\frac{2}{3}} \cos \left (x\right )^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)*(1+(1-8*tan(x)^2)^(1/3))/cos(x)^2/(1-8*tan(x)^2)^(2/3),x, algorithm="maxima")

[Out]

integrate(((-8*tan(x)^2 + 1)^(1/3) + 1)*cot(x)/((-8*tan(x)^2 + 1)^(2/3)*cos(x)^2), x)

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Fricas [B]  time = 16.9191, size = 281, normalized size = 10.41 \begin{align*} -\frac{1}{2} \, \log \left (\frac{16 \,{\left (145 \, \cos \left (x\right )^{4} - 200 \, \cos \left (x\right )^{2} + 3 \,{\left (11 \, \cos \left (x\right )^{4} - 8 \, \cos \left (x\right )^{2}\right )} \left (\frac{9 \, \cos \left (x\right )^{2} - 8}{\cos \left (x\right )^{2}}\right )^{\frac{2}{3}} + 3 \,{\left (19 \, \cos \left (x\right )^{4} - 16 \, \cos \left (x\right )^{2}\right )} \left (\frac{9 \, \cos \left (x\right )^{2} - 8}{\cos \left (x\right )^{2}}\right )^{\frac{1}{3}} + 64\right )}}{\cos \left (x\right )^{4} - 2 \, \cos \left (x\right )^{2} + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)*(1+(1-8*tan(x)^2)^(1/3))/cos(x)^2/(1-8*tan(x)^2)^(2/3),x, algorithm="fricas")

[Out]

-1/2*log(16*(145*cos(x)^4 - 200*cos(x)^2 + 3*(11*cos(x)^4 - 8*cos(x)^2)*((9*cos(x)^2 - 8)/cos(x)^2)^(2/3) + 3*
(19*cos(x)^4 - 16*cos(x)^2)*((9*cos(x)^2 - 8)/cos(x)^2)^(1/3) + 64)/(cos(x)^4 - 2*cos(x)^2 + 1))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)*(1+(1-8*tan(x)**2)**(1/3))/cos(x)**2/(1-8*tan(x)**2)**(2/3),x)

[Out]

Timed out

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Giac [A]  time = 1.13704, size = 55, normalized size = 2.04 \begin{align*} -\frac{1}{2} \, \log \left ({\left (-8 \, \tan \left (x\right )^{2} + 1\right )}^{\frac{2}{3}} +{\left (-8 \, \tan \left (x\right )^{2} + 1\right )}^{\frac{1}{3}} + 1\right ) + \log \left (-{\left (-8 \, \tan \left (x\right )^{2} + 1\right )}^{\frac{1}{3}} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)*(1+(1-8*tan(x)^2)^(1/3))/cos(x)^2/(1-8*tan(x)^2)^(2/3),x, algorithm="giac")

[Out]

-1/2*log((-8*tan(x)^2 + 1)^(2/3) + (-8*tan(x)^2 + 1)^(1/3) + 1) + log(-(-8*tan(x)^2 + 1)^(1/3) + 1)