3.449 \(\int (1+2 \cos ^9(x))^{5/6} \tan (x) \, dx\)

Optimal. Leaf size=95 \[ -\frac{2}{15} \left (2 \cos ^9(x)+1\right )^{5/6}+\frac{\tan ^{-1}\left (\frac{1-\sqrt [3]{2 \cos ^9(x)+1}}{\sqrt{3} \sqrt [6]{2 \cos ^9(x)+1}}\right )}{3 \sqrt{3}}+\frac{1}{3} \tanh ^{-1}\left (\sqrt [6]{2 \cos ^9(x)+1}\right )-\frac{1}{9} \tanh ^{-1}\left (\sqrt{2 \cos ^9(x)+1}\right ) \]

[Out]

ArcTan[(1 - (1 + 2*Cos[x]^9)^(1/3))/(Sqrt[3]*(1 + 2*Cos[x]^9)^(1/6))]/(3*Sqrt[3]) + ArcTanh[(1 + 2*Cos[x]^9)^(
1/6)]/3 - ArcTanh[Sqrt[1 + 2*Cos[x]^9]]/9 - (2*(1 + 2*Cos[x]^9)^(5/6))/15

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Rubi [A]  time = 0.259556, antiderivative size = 162, normalized size of antiderivative = 1.71, number of steps used = 14, number of rules used = 10, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {3230, 266, 50, 63, 296, 634, 618, 204, 628, 206} \[ -\frac{2}{15} \left (2 \cos ^9(x)+1\right )^{5/6}-\frac{1}{18} \log \left (\sqrt [3]{2 \cos ^9(x)+1}-\sqrt [6]{2 \cos ^9(x)+1}+1\right )+\frac{1}{18} \log \left (\sqrt [3]{2 \cos ^9(x)+1}+\sqrt [6]{2 \cos ^9(x)+1}+1\right )+\frac{\tan ^{-1}\left (\frac{1-2 \sqrt [6]{2 \cos ^9(x)+1}}{\sqrt{3}}\right )}{3 \sqrt{3}}-\frac{\tan ^{-1}\left (\frac{2 \sqrt [6]{2 \cos ^9(x)+1}+1}{\sqrt{3}}\right )}{3 \sqrt{3}}+\frac{2}{9} \tanh ^{-1}\left (\sqrt [6]{2 \cos ^9(x)+1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 + 2*Cos[x]^9)^(5/6)*Tan[x],x]

[Out]

ArcTan[(1 - 2*(1 + 2*Cos[x]^9)^(1/6))/Sqrt[3]]/(3*Sqrt[3]) - ArcTan[(1 + 2*(1 + 2*Cos[x]^9)^(1/6))/Sqrt[3]]/(3
*Sqrt[3]) + (2*ArcTanh[(1 + 2*Cos[x]^9)^(1/6)])/9 - (2*(1 + 2*Cos[x]^9)^(5/6))/15 - Log[1 - (1 + 2*Cos[x]^9)^(
1/6) + (1 + 2*Cos[x]^9)^(1/3)]/18 + Log[1 + (1 + 2*Cos[x]^9)^(1/6) + (1 + 2*Cos[x]^9)^(1/3)]/18

Rule 3230

Int[((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Sin[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*(a + b*(c*ff*x)^n)^p)/(1 - ff^2*x^2)^(
(m + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && ILtQ[(m - 1)/2, 0]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 296

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Module[{r = Numerator[Rt[-(a/b), n]], s = Denominator[Rt
[-(a/b), n]], k, u}, Simp[u = Int[(r*Cos[(2*k*m*Pi)/n] - s*Cos[(2*k*(m + 1)*Pi)/n]*x)/(r^2 - 2*r*s*Cos[(2*k*Pi
)/n]*x + s^2*x^2), x] + Int[(r*Cos[(2*k*m*Pi)/n] + s*Cos[(2*k*(m + 1)*Pi)/n]*x)/(r^2 + 2*r*s*Cos[(2*k*Pi)/n]*x
+ s^2*x^2), x]; (2*r^(m + 2)*Int[1/(r^2 - s^2*x^2), x])/(a*n*s^m) + Dist[(2*r^(m + 1))/(a*n*s^m), Sum[u, {k,
1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] && IGtQ[m, 0] && LtQ[m, n - 1] && NegQ[a/b]

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 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 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]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \left (1+2 \cos ^9(x)\right )^{5/6} \tan (x) \, dx &=-\operatorname{Subst}\left (\int \frac{\left (1+2 x^9\right )^{5/6}}{x} \, dx,x,\cos (x)\right )\\ &=-\left (\frac{1}{9} \operatorname{Subst}\left (\int \frac{(1+2 x)^{5/6}}{x} \, dx,x,\cos ^9(x)\right )\right )\\ &=-\frac{2}{15} \left (1+2 \cos ^9(x)\right )^{5/6}-\frac{1}{9} \operatorname{Subst}\left (\int \frac{1}{x \sqrt [6]{1+2 x}} \, dx,x,\cos ^9(x)\right )\\ &=-\frac{2}{15} \left (1+2 \cos ^9(x)\right )^{5/6}-\frac{1}{3} \operatorname{Subst}\left (\int \frac{x^4}{-\frac{1}{2}+\frac{x^6}{2}} \, dx,x,\sqrt [6]{1+2 \cos ^9(x)}\right )\\ &=-\frac{2}{15} \left (1+2 \cos ^9(x)\right )^{5/6}+\frac{2}{9} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sqrt [6]{1+2 \cos ^9(x)}\right )+\frac{2}{9} \operatorname{Subst}\left (\int \frac{-\frac{1}{2}-\frac{x}{2}}{1-x+x^2} \, dx,x,\sqrt [6]{1+2 \cos ^9(x)}\right )+\frac{2}{9} \operatorname{Subst}\left (\int \frac{-\frac{1}{2}+\frac{x}{2}}{1+x+x^2} \, dx,x,\sqrt [6]{1+2 \cos ^9(x)}\right )\\ &=\frac{2}{9} \tanh ^{-1}\left (\sqrt [6]{1+2 \cos ^9(x)}\right )-\frac{2}{15} \left (1+2 \cos ^9(x)\right )^{5/6}-\frac{1}{18} \operatorname{Subst}\left (\int \frac{-1+2 x}{1-x+x^2} \, dx,x,\sqrt [6]{1+2 \cos ^9(x)}\right )+\frac{1}{18} \operatorname{Subst}\left (\int \frac{1+2 x}{1+x+x^2} \, dx,x,\sqrt [6]{1+2 \cos ^9(x)}\right )-\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{1-x+x^2} \, dx,x,\sqrt [6]{1+2 \cos ^9(x)}\right )-\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{1+x+x^2} \, dx,x,\sqrt [6]{1+2 \cos ^9(x)}\right )\\ &=\frac{2}{9} \tanh ^{-1}\left (\sqrt [6]{1+2 \cos ^9(x)}\right )-\frac{2}{15} \left (1+2 \cos ^9(x)\right )^{5/6}-\frac{1}{18} \log \left (1-\sqrt [6]{1+2 \cos ^9(x)}+\sqrt [3]{1+2 \cos ^9(x)}\right )+\frac{1}{18} \log \left (1+\sqrt [6]{1+2 \cos ^9(x)}+\sqrt [3]{1+2 \cos ^9(x)}\right )+\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+2 \sqrt [6]{1+2 \cos ^9(x)}\right )+\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1+2 \sqrt [6]{1+2 \cos ^9(x)}\right )\\ &=\frac{\tan ^{-1}\left (\frac{1-2 \sqrt [6]{1+2 \cos ^9(x)}}{\sqrt{3}}\right )}{3 \sqrt{3}}-\frac{\tan ^{-1}\left (\frac{1+2 \sqrt [6]{1+2 \cos ^9(x)}}{\sqrt{3}}\right )}{3 \sqrt{3}}+\frac{2}{9} \tanh ^{-1}\left (\sqrt [6]{1+2 \cos ^9(x)}\right )-\frac{2}{15} \left (1+2 \cos ^9(x)\right )^{5/6}-\frac{1}{18} \log \left (1-\sqrt [6]{1+2 \cos ^9(x)}+\sqrt [3]{1+2 \cos ^9(x)}\right )+\frac{1}{18} \log \left (1+\sqrt [6]{1+2 \cos ^9(x)}+\sqrt [3]{1+2 \cos ^9(x)}\right )\\ \end{align*}

Mathematica [A]  time = 0.130941, size = 154, normalized size = 1.62 \[ \frac{1}{90} \left (-12 \left (2 \cos ^9(x)+1\right )^{5/6}-5 \log \left (\sqrt [3]{2 \cos ^9(x)+1}-\sqrt [6]{2 \cos ^9(x)+1}+1\right )+5 \log \left (\sqrt [3]{2 \cos ^9(x)+1}+\sqrt [6]{2 \cos ^9(x)+1}+1\right )+10 \sqrt{3} \tan ^{-1}\left (\frac{1-2 \sqrt [6]{2 \cos ^9(x)+1}}{\sqrt{3}}\right )-10 \sqrt{3} \tan ^{-1}\left (\frac{2 \sqrt [6]{2 \cos ^9(x)+1}+1}{\sqrt{3}}\right )+20 \tanh ^{-1}\left (\sqrt [6]{2 \cos ^9(x)+1}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + 2*Cos[x]^9)^(5/6)*Tan[x],x]

[Out]

(10*Sqrt[3]*ArcTan[(1 - 2*(1 + 2*Cos[x]^9)^(1/6))/Sqrt[3]] - 10*Sqrt[3]*ArcTan[(1 + 2*(1 + 2*Cos[x]^9)^(1/6))/
Sqrt[3]] + 20*ArcTanh[(1 + 2*Cos[x]^9)^(1/6)] - 12*(1 + 2*Cos[x]^9)^(5/6) - 5*Log[1 - (1 + 2*Cos[x]^9)^(1/6) +
(1 + 2*Cos[x]^9)^(1/3)] + 5*Log[1 + (1 + 2*Cos[x]^9)^(1/6) + (1 + 2*Cos[x]^9)^(1/3)])/90

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Maple [F]  time = 0.11, size = 0, normalized size = 0. \begin{align*} \int \left ( 1+2\, \left ( \cos \left ( x \right ) \right ) ^{9} \right ) ^{{\frac{5}{6}}}\tan \left ( x \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+2*cos(x)^9)^(5/6)*tan(x),x)

[Out]

int((1+2*cos(x)^9)^(5/6)*tan(x),x)

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Maxima [B]  time = 1.41716, size = 196, normalized size = 2.06 \begin{align*} -\frac{1}{9} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \,{\left (2 \, \cos \left (x\right )^{9} + 1\right )}^{\frac{1}{6}} + 1\right )}\right ) - \frac{1}{9} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \,{\left (2 \, \cos \left (x\right )^{9} + 1\right )}^{\frac{1}{6}} - 1\right )}\right ) - \frac{2}{15} \,{\left (2 \, \cos \left (x\right )^{9} + 1\right )}^{\frac{5}{6}} + \frac{1}{18} \, \log \left ({\left (2 \, \cos \left (x\right )^{9} + 1\right )}^{\frac{1}{3}} +{\left (2 \, \cos \left (x\right )^{9} + 1\right )}^{\frac{1}{6}} + 1\right ) - \frac{1}{18} \, \log \left ({\left (2 \, \cos \left (x\right )^{9} + 1\right )}^{\frac{1}{3}} -{\left (2 \, \cos \left (x\right )^{9} + 1\right )}^{\frac{1}{6}} + 1\right ) + \frac{1}{9} \, \log \left ({\left (2 \, \cos \left (x\right )^{9} + 1\right )}^{\frac{1}{6}} + 1\right ) - \frac{1}{9} \, \log \left ({\left (2 \, \cos \left (x\right )^{9} + 1\right )}^{\frac{1}{6}} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*cos(x)^9)^(5/6)*tan(x),x, algorithm="maxima")

[Out]

-1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*(2*cos(x)^9 + 1)^(1/6) + 1)) - 1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*(2*cos(x)^
9 + 1)^(1/6) - 1)) - 2/15*(2*cos(x)^9 + 1)^(5/6) + 1/18*log((2*cos(x)^9 + 1)^(1/3) + (2*cos(x)^9 + 1)^(1/6) +
1) - 1/18*log((2*cos(x)^9 + 1)^(1/3) - (2*cos(x)^9 + 1)^(1/6) + 1) + 1/9*log((2*cos(x)^9 + 1)^(1/6) + 1) - 1/9
*log((2*cos(x)^9 + 1)^(1/6) - 1)

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Fricas [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((1+2*cos(x)^9)^(5/6)*tan(x),x, algorithm="fricas")

[Out]

Timed out

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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((1+2*cos(x)**9)**(5/6)*tan(x),x)

[Out]

Timed out

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Giac [B]  time = 1.14937, size = 197, normalized size = 2.07 \begin{align*} -\frac{1}{9} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \,{\left (2 \, \cos \left (x\right )^{9} + 1\right )}^{\frac{1}{6}} + 1\right )}\right ) - \frac{1}{9} \, \sqrt{3} \arctan \left (\frac{1}{3} \, \sqrt{3}{\left (2 \,{\left (2 \, \cos \left (x\right )^{9} + 1\right )}^{\frac{1}{6}} - 1\right )}\right ) - \frac{2}{15} \,{\left (2 \, \cos \left (x\right )^{9} + 1\right )}^{\frac{5}{6}} + \frac{1}{18} \, \log \left ({\left (2 \, \cos \left (x\right )^{9} + 1\right )}^{\frac{1}{3}} +{\left (2 \, \cos \left (x\right )^{9} + 1\right )}^{\frac{1}{6}} + 1\right ) - \frac{1}{18} \, \log \left ({\left (2 \, \cos \left (x\right )^{9} + 1\right )}^{\frac{1}{3}} -{\left (2 \, \cos \left (x\right )^{9} + 1\right )}^{\frac{1}{6}} + 1\right ) + \frac{1}{9} \, \log \left ({\left (2 \, \cos \left (x\right )^{9} + 1\right )}^{\frac{1}{6}} + 1\right ) - \frac{1}{9} \, \log \left ({\left |{\left (2 \, \cos \left (x\right )^{9} + 1\right )}^{\frac{1}{6}} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*cos(x)^9)^(5/6)*tan(x),x, algorithm="giac")

[Out]

-1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*(2*cos(x)^9 + 1)^(1/6) + 1)) - 1/9*sqrt(3)*arctan(1/3*sqrt(3)*(2*(2*cos(x)^
9 + 1)^(1/6) - 1)) - 2/15*(2*cos(x)^9 + 1)^(5/6) + 1/18*log((2*cos(x)^9 + 1)^(1/3) + (2*cos(x)^9 + 1)^(1/6) +
1) - 1/18*log((2*cos(x)^9 + 1)^(1/3) - (2*cos(x)^9 + 1)^(1/6) + 1) + 1/9*log((2*cos(x)^9 + 1)^(1/6) + 1) - 1/9
*log(abs((2*cos(x)^9 + 1)^(1/6) - 1))