3.419 \(\int (1+2 \cos ^2(x))^{5/2} \sin (x) \, dx\)

Optimal. Leaf size=73 \[ -\frac{1}{6} \cos (x) \left (2 \cos ^2(x)+1\right )^{5/2}-\frac{5}{24} \cos (x) \left (2 \cos ^2(x)+1\right )^{3/2}-\frac{5}{16} \cos (x) \sqrt{2 \cos ^2(x)+1}-\frac{5 \sinh ^{-1}\left (\sqrt{2} \cos (x)\right )}{16 \sqrt{2}} \]

[Out]

(-5*ArcSinh[Sqrt[2]*Cos[x]])/(16*Sqrt[2]) - (5*Cos[x]*Sqrt[1 + 2*Cos[x]^2])/16 - (5*Cos[x]*(1 + 2*Cos[x]^2)^(3
/2))/24 - (Cos[x]*(1 + 2*Cos[x]^2)^(5/2))/6

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Rubi [A]  time = 0.0465499, antiderivative size = 67, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3190, 195, 215} \[ -\frac{1}{6} \cos (x) (\cos (2 x)+2)^{5/2}-\frac{5}{24} \cos (x) (\cos (2 x)+2)^{3/2}-\frac{5}{16} \cos (x) \sqrt{\cos (2 x)+2}-\frac{5 \sinh ^{-1}\left (\sqrt{2} \cos (x)\right )}{16 \sqrt{2}} \]

Antiderivative was successfully verified.

[In]

Int[(1 + 2*Cos[x]^2)^(5/2)*Sin[x],x]

[Out]

(-5*ArcSinh[Sqrt[2]*Cos[x]])/(16*Sqrt[2]) - (5*Cos[x]*Sqrt[2 + Cos[2*x]])/16 - (5*Cos[x]*(2 + Cos[2*x])^(3/2))
/24 - (Cos[x]*(2 + Cos[2*x])^(5/2))/6

Rule 3190

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e +
f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rubi steps

\begin{align*} \int \left (1+2 \cos ^2(x)\right )^{5/2} \sin (x) \, dx &=-\operatorname{Subst}\left (\int \left (1+2 x^2\right )^{5/2} \, dx,x,\cos (x)\right )\\ &=-\frac{1}{6} \cos (x) (2+\cos (2 x))^{5/2}-\frac{5}{6} \operatorname{Subst}\left (\int \left (1+2 x^2\right )^{3/2} \, dx,x,\cos (x)\right )\\ &=-\frac{5}{24} \cos (x) (2+\cos (2 x))^{3/2}-\frac{1}{6} \cos (x) (2+\cos (2 x))^{5/2}-\frac{5}{8} \operatorname{Subst}\left (\int \sqrt{1+2 x^2} \, dx,x,\cos (x)\right )\\ &=-\frac{5}{16} \cos (x) \sqrt{2+\cos (2 x)}-\frac{5}{24} \cos (x) (2+\cos (2 x))^{3/2}-\frac{1}{6} \cos (x) (2+\cos (2 x))^{5/2}-\frac{5}{16} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+2 x^2}} \, dx,x,\cos (x)\right )\\ &=-\frac{5 \sinh ^{-1}\left (\sqrt{2} \cos (x)\right )}{16 \sqrt{2}}-\frac{5}{16} \cos (x) \sqrt{2+\cos (2 x)}-\frac{5}{24} \cos (x) (2+\cos (2 x))^{3/2}-\frac{1}{6} \cos (x) (2+\cos (2 x))^{5/2}\\ \end{align*}

Mathematica [A]  time = 0.138242, size = 61, normalized size = 0.84 \[ \frac{1}{96} \left (-2 \sqrt{\cos (2 x)+2} (92 \cos (x)+23 \cos (3 x)+2 \cos (5 x))-15 \sqrt{2} \log \left (\sqrt{2} \cos (x)+\sqrt{\cos (2 x)+2}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(1 + 2*Cos[x]^2)^(5/2)*Sin[x],x]

[Out]

(-2*Sqrt[2 + Cos[2*x]]*(92*Cos[x] + 23*Cos[3*x] + 2*Cos[5*x]) - 15*Sqrt[2]*Log[Sqrt[2]*Cos[x] + Sqrt[2 + Cos[2
*x]]])/96

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Maple [A]  time = 0.012, size = 56, normalized size = 0.8 \begin{align*} -{\frac{5\,\cos \left ( x \right ) }{24} \left ( 1+2\, \left ( \cos \left ( x \right ) \right ) ^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{\cos \left ( x \right ) }{6} \left ( 1+2\, \left ( \cos \left ( x \right ) \right ) ^{2} \right ) ^{{\frac{5}{2}}}}-{\frac{5\,{\it Arcsinh} \left ( \cos \left ( x \right ) \sqrt{2} \right ) \sqrt{2}}{32}}-{\frac{5\,\cos \left ( x \right ) }{16}\sqrt{1+2\, \left ( \cos \left ( x \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1+2*cos(x)^2)^(5/2)*sin(x),x)

[Out]

-5/24*cos(x)*(1+2*cos(x)^2)^(3/2)-1/6*cos(x)*(1+2*cos(x)^2)^(5/2)-5/32*arcsinh(cos(x)*2^(1/2))*2^(1/2)-5/16*co
s(x)*(1+2*cos(x)^2)^(1/2)

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Maxima [A]  time = 1.46665, size = 74, normalized size = 1.01 \begin{align*} -\frac{1}{6} \,{\left (2 \, \cos \left (x\right )^{2} + 1\right )}^{\frac{5}{2}} \cos \left (x\right ) - \frac{5}{24} \,{\left (2 \, \cos \left (x\right )^{2} + 1\right )}^{\frac{3}{2}} \cos \left (x\right ) - \frac{5}{32} \, \sqrt{2} \operatorname{arsinh}\left (\sqrt{2} \cos \left (x\right )\right ) - \frac{5}{16} \, \sqrt{2 \, \cos \left (x\right )^{2} + 1} \cos \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*cos(x)^2 + 1)^(5/2)*cos(x) - 5/24*(2*cos(x)^2 + 1)^(3/2)*cos(x) - 5/32*sqrt(2)*arcsinh(sqrt(2)*cos(x))
- 5/16*sqrt(2*cos(x)^2 + 1)*cos(x)

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Fricas [A]  time = 3.39366, size = 352, normalized size = 4.82 \begin{align*} -\frac{1}{48} \,{\left (32 \, \cos \left (x\right )^{5} + 52 \, \cos \left (x\right )^{3} + 33 \, \cos \left (x\right )\right )} \sqrt{2 \, \cos \left (x\right )^{2} + 1} + \frac{5}{256} \, \sqrt{2} \log \left (2048 \, \cos \left (x\right )^{8} + 2048 \, \cos \left (x\right )^{6} + 640 \, \cos \left (x\right )^{4} + 64 \, \cos \left (x\right )^{2} - 8 \,{\left (128 \, \sqrt{2} \cos \left (x\right )^{7} + 96 \, \sqrt{2} \cos \left (x\right )^{5} + 20 \, \sqrt{2} \cos \left (x\right )^{3} + \sqrt{2} \cos \left (x\right )\right )} \sqrt{2 \, \cos \left (x\right )^{2} + 1} + 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1+2*cos(x)^2)^(5/2)*sin(x),x, algorithm="fricas")

[Out]

-1/48*(32*cos(x)^5 + 52*cos(x)^3 + 33*cos(x))*sqrt(2*cos(x)^2 + 1) + 5/256*sqrt(2)*log(2048*cos(x)^8 + 2048*co
s(x)^6 + 640*cos(x)^4 + 64*cos(x)^2 - 8*(128*sqrt(2)*cos(x)^7 + 96*sqrt(2)*cos(x)^5 + 20*sqrt(2)*cos(x)^3 + sq
rt(2)*cos(x))*sqrt(2*cos(x)^2 + 1) + 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((1+2*cos(x)**2)**(5/2)*sin(x),x)

[Out]

Timed out

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Giac [A]  time = 1.08495, size = 74, normalized size = 1.01 \begin{align*} -\frac{1}{48} \,{\left (4 \,{\left (8 \, \cos \left (x\right )^{2} + 13\right )} \cos \left (x\right )^{2} + 33\right )} \sqrt{2 \, \cos \left (x\right )^{2} + 1} \cos \left (x\right ) + \frac{5}{32} \, \sqrt{2} \log \left (-\sqrt{2} \cos \left (x\right ) + \sqrt{2 \, \cos \left (x\right )^{2} + 1}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(4*(8*cos(x)^2 + 13)*cos(x)^2 + 33)*sqrt(2*cos(x)^2 + 1)*cos(x) + 5/32*sqrt(2)*log(-sqrt(2)*cos(x) + sqr
t(2*cos(x)^2 + 1))