\(\int (1+2 \cos ^2(x))^{5/2} \sin (x) \, dx\) [419]

Optimal result

Integrand size = 15, antiderivative size = 73 \[ \int \left (1+2 \cos ^2(x)\right )^{5/2} \sin (x) \, dx=-\frac {5 \text {arcsinh}\left (\sqrt {2} \cos (x)\right )}{16 \sqrt {2}}-\frac {5}{16} \cos (x) \sqrt {1+2 \cos ^2(x)}-\frac {5}{24} \cos (x) \left (1+2 \cos ^2(x)\right )^{3/2}-\frac {1}{6} \cos (x) \left (1+2 \cos ^2(x)\right )^{5/2} \]

[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)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3269, 201, 221} \[ \int \left (1+2 \cos ^2(x)\right )^{5/2} \sin (x) \, dx=-\frac {5 \text {arcsinh}\left (\sqrt {2} \cos (x)\right )}{16 \sqrt {2}}-\frac {1}{6} \cos (x) (\cos (2 x)+2)^{5/2}-\frac {5}{24} \cos (x) (\cos (2 x)+2)^{3/2}-\frac {5}{16} \cos (x) \sqrt {\cos (2 x)+2} \]

[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 201

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 221

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]

Rule 3269

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]

Rubi steps \begin{align*} \text {integral}& = -\text {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} \text {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} \text {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} \text {Subst}\left (\int \frac {1}{\sqrt {1+2 x^2}} \, dx,x,\cos (x)\right ) \\ & = -\frac {5 \text {arcsinh}\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] (verified)

Time = 0.17 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.84 \[ \int \left (1+2 \cos ^2(x)\right )^{5/2} \sin (x) \, dx=\frac {1}{96} \left (-2 \sqrt {2+\cos (2 x)} (92 \cos (x)+23 \cos (3 x)+2 \cos (5 x))-15 \sqrt {2} \log \left (\sqrt {2} \cos (x)+\sqrt {2+\cos (2 x)}\right )\right ) \]

[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

Maple [A] (verified)

Time = 1.08 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.77

method	result	size
derivativedivides	\(-\frac {5 \cos \left (x \right ) {\left (1+2 \left (\cos ^{2}\left (x \right )\right )\right )}^{\frac {3}{2}}}{24}-\frac {\cos \left (x \right ) {\left (1+2 \left (\cos ^{2}\left (x \right )\right )\right )}^{\frac {5}{2}}}{6}-\frac {5 \,\operatorname {arcsinh}\left (\cos \left (x \right ) \sqrt {2}\right ) \sqrt {2}}{32}-\frac {5 \cos \left (x \right ) \sqrt {1+2 \left (\cos ^{2}\left (x \right )\right )}}{16}\)	\(56\)
default	\(-\frac {5 \cos \left (x \right ) {\left (1+2 \left (\cos ^{2}\left (x \right )\right )\right )}^{\frac {3}{2}}}{24}-\frac {\cos \left (x \right ) {\left (1+2 \left (\cos ^{2}\left (x \right )\right )\right )}^{\frac {5}{2}}}{6}-\frac {5 \,\operatorname {arcsinh}\left (\cos \left (x \right ) \sqrt {2}\right ) \sqrt {2}}{32}-\frac {5 \cos \left (x \right ) \sqrt {1+2 \left (\cos ^{2}\left (x \right )\right )}}{16}\)	\(56\)

[In]

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.48 \[ \int \left (1+2 \cos ^2(x)\right )^{5/2} \sin (x) \, dx=-\frac {1}{48} \, {\left (32 \, \cos \left (x\right )^{5} + 52 \, \cos \left (x\right )^{3} + 33 \, \cos \left (x\right )\right )} \sqrt {2 \, \cos \left (x\right )^{2} + 1} + \frac {5}{256} \, \sqrt {2} \log \left (2048 \, \cos \left (x\right )^{8} + 2048 \, \cos \left (x\right )^{6} + 640 \, \cos \left (x\right )^{4} + 64 \, \cos \left (x\right )^{2} - 8 \, {\left (128 \, \sqrt {2} \cos \left (x\right )^{7} + 96 \, \sqrt {2} \cos \left (x\right )^{5} + 20 \, \sqrt {2} \cos \left (x\right )^{3} + \sqrt {2} \cos \left (x\right )\right )} \sqrt {2 \, \cos \left (x\right )^{2} + 1} + 1\right ) \]

[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)

Sympy [F(-1)]

Timed out. \[ \int \left (1+2 \cos ^2(x)\right )^{5/2} \sin (x) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.75 \[ \int \left (1+2 \cos ^2(x)\right )^{5/2} \sin (x) \, dx=-\frac {1}{6} \, {\left (2 \, \cos \left (x\right )^{2} + 1\right )}^{\frac {5}{2}} \cos \left (x\right ) - \frac {5}{24} \, {\left (2 \, \cos \left (x\right )^{2} + 1\right )}^{\frac {3}{2}} \cos \left (x\right ) - \frac {5}{32} \, \sqrt {2} \operatorname {arsinh}\left (\sqrt {2} \cos \left (x\right )\right ) - \frac {5}{16} \, \sqrt {2 \, \cos \left (x\right )^{2} + 1} \cos \left (x\right ) \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.75 \[ \int \left (1+2 \cos ^2(x)\right )^{5/2} \sin (x) \, dx=-\frac {1}{48} \, {\left (4 \, {\left (8 \, \cos \left (x\right )^{2} + 13\right )} \cos \left (x\right )^{2} + 33\right )} \sqrt {2 \, \cos \left (x\right )^{2} + 1} \cos \left (x\right ) + \frac {5}{32} \, \sqrt {2} \log \left (-\sqrt {2} \cos \left (x\right ) + \sqrt {2 \, \cos \left (x\right )^{2} + 1}\right ) \]

[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))

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.59 \[ \int \left (1+2 \cos ^2(x)\right )^{5/2} \sin (x) \, dx=-\frac {5\,\sqrt {2}\,\mathrm {asinh}\left (\sqrt {2}\,\cos \left (x\right )\right )}{32}-\frac {\sqrt {2}\,\sqrt {{\cos \left (x\right )}^2+\frac {1}{2}}\,\left (\frac {4\,{\cos \left (x\right )}^5}{3}+\frac {13\,{\cos \left (x\right )}^3}{6}+\frac {11\,\cos \left (x\right )}{8}\right )}{2} \]

[In]

int(sin(x)*(2*cos(x)^2 + 1)^(5/2),x)

[Out]

- (5*2^(1/2)*asinh(2^(1/2)*cos(x)))/32 - (2^(1/2)*(cos(x)^2 + 1/2)^(1/2)*((11*cos(x))/8 + (13*cos(x)^3)/6 + (4
*cos(x)^5)/3))/2