3.635 \(\int \frac{\cos ^2(x) \sin (x)}{(\sin ^2(x)-\sin (2 x)) \sin ^{\frac{5}{2}}(2 x)} \, dx\)

Optimal. Leaf size=79 \[ \frac{\sin ^2(x) \cos ^3(x)}{2 \sin ^{\frac{5}{2}}(2 x)}+\frac{\sin (x) \cos ^4(x)}{3 \sin ^{\frac{5}{2}}(2 x)}-\frac{5 \sin ^5(x) \tanh ^{-1}\left (\frac{\sqrt{\tan (x)}}{\sqrt{2}}\right )}{2 \sqrt{2} \sin ^{\frac{5}{2}}(2 x) \tan ^{\frac{5}{2}}(x)} \]

[Out]

(Cos[x]^4*Sin[x])/(3*Sin[2*x]^(5/2)) + (Cos[x]^3*Sin[x]^2)/(2*Sin[2*x]^(5/2)) - (5*ArcTanh[Sqrt[Tan[x]]/Sqrt[2
]]*Sin[x]^5)/(2*Sqrt[2]*Sin[2*x]^(5/2)*Tan[x]^(5/2))

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Rubi [A]  time = 0.567016, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4390, 898, 1262, 207} \[ \frac{\sin ^2(x) \cos ^3(x)}{2 \sin ^{\frac{5}{2}}(2 x)}+\frac{\sin (x) \cos ^4(x)}{3 \sin ^{\frac{5}{2}}(2 x)}-\frac{5 \sin ^5(x) \tanh ^{-1}\left (\frac{\sqrt{\tan (x)}}{\sqrt{2}}\right )}{2 \sqrt{2} \sin ^{\frac{5}{2}}(2 x) \tan ^{\frac{5}{2}}(x)} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[x]^2*Sin[x])/((Sin[x]^2 - Sin[2*x])*Sin[2*x]^(5/2)),x]

[Out]

(Cos[x]^4*Sin[x])/(3*Sin[2*x]^(5/2)) + (Cos[x]^3*Sin[x]^2)/(2*Sin[2*x]^(5/2)) - (5*ArcTanh[Sqrt[Tan[x]]/Sqrt[2
]]*Sin[x]^5)/(2*Sqrt[2]*Sin[2*x]^(5/2)*Tan[x]^(5/2))

Rule 4390

Int[(u_)*((c_.)*sin[v_])^(m_), x_Symbol] :> With[{w = FunctionOfTrig[(u*Sin[v/2]^(2*m))/(c*Tan[v/2])^m, x]}, D
ist[((c*Sin[v])^m*(c*Tan[v/2])^m)/Sin[v/2]^(2*m), Int[(u*Sin[v/2]^(2*m))/(c*Tan[v/2])^m, x], x] /;  !FalseQ[w]
&& FunctionOfQ[NonfreeFactors[Tan[w], x], (u*Sin[v/2]^(2*m))/(c*Tan[v/2])^m, x]] /; FreeQ[c, x] && LinearQ[v,
x] && IntegerQ[m + 1/2] &&  !SumQ[u] && InverseFunctionFreeQ[u, x]

Rule 898

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> With[{q = De
nominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 + a*e^2)/e^2 - (2*c
*d*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*
g, 0] && NeQ[c*d^2 + a*e^2, 0] && IntegersQ[n, p] && FractionQ[m]

Rule 1262

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegra
nd[(f*x)^m*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a, c, d, e, f, m, q}, x] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 207

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

Rubi steps

\begin{align*} \int \frac{\cos ^2(x) \sin (x)}{\left (\sin ^2(x)-\sin (2 x)\right ) \sin ^{\frac{5}{2}}(2 x)} \, dx &=\frac{\sin ^5(x) \int \frac{\csc ^2(x) \sqrt{\tan (x)}}{\sin ^2(x)-\sin (2 x)} \, dx}{\sin ^{\frac{5}{2}}(2 x) \tan ^{\frac{5}{2}}(x)}\\ &=\frac{\sin ^5(x) \operatorname{Subst}\left (\int \frac{-1-x^2}{(2-x) x^{5/2}} \, dx,x,\tan (x)\right )}{\sin ^{\frac{5}{2}}(2 x) \tan ^{\frac{5}{2}}(x)}\\ &=\frac{\left (2 \sin ^5(x)\right ) \operatorname{Subst}\left (\int \frac{-1-x^4}{x^4 \left (2-x^2\right )} \, dx,x,\sqrt{\tan (x)}\right )}{\sin ^{\frac{5}{2}}(2 x) \tan ^{\frac{5}{2}}(x)}\\ &=\frac{\left (2 \sin ^5(x)\right ) \operatorname{Subst}\left (\int \left (-\frac{1}{2 x^4}-\frac{1}{4 x^2}+\frac{5}{4 \left (-2+x^2\right )}\right ) \, dx,x,\sqrt{\tan (x)}\right )}{\sin ^{\frac{5}{2}}(2 x) \tan ^{\frac{5}{2}}(x)}\\ &=\frac{\cos ^4(x) \sin (x)}{3 \sin ^{\frac{5}{2}}(2 x)}+\frac{\cos ^3(x) \sin ^2(x)}{2 \sin ^{\frac{5}{2}}(2 x)}+\frac{\left (5 \sin ^5(x)\right ) \operatorname{Subst}\left (\int \frac{1}{-2+x^2} \, dx,x,\sqrt{\tan (x)}\right )}{2 \sin ^{\frac{5}{2}}(2 x) \tan ^{\frac{5}{2}}(x)}\\ &=\frac{\cos ^4(x) \sin (x)}{3 \sin ^{\frac{5}{2}}(2 x)}+\frac{\cos ^3(x) \sin ^2(x)}{2 \sin ^{\frac{5}{2}}(2 x)}-\frac{5 \tanh ^{-1}\left (\frac{\sqrt{\tan (x)}}{\sqrt{2}}\right ) \sin ^5(x)}{2 \sqrt{2} \sin ^{\frac{5}{2}}(2 x) \tan ^{\frac{5}{2}}(x)}\\ \end{align*}

Mathematica [C]  time = 4.80679, size = 139, normalized size = 1.76 \[ -\frac{\sqrt{\sin (2 x)} \csc (x) (2 \cos (x)-\sin (x)) \left (-5 \sqrt{\frac{\cos (x)}{2 \cos (x)-2}} \sqrt{\tan \left (\frac{x}{2}\right )} \sec (x) \left (\text{EllipticF}\left (\sin ^{-1}\left (\frac{1}{\sqrt{\tan \left (\frac{x}{2}\right )}}\right ),-1\right )+\Pi \left (-\frac{2}{-1+\sqrt{5}};\left .-\sin ^{-1}\left (\frac{1}{\sqrt{\tan \left (\frac{x}{2}\right )}}\right )\right |-1\right )+\Pi \left (\frac{1}{2} \left (-1+\sqrt{5}\right );\left .-\sin ^{-1}\left (\frac{1}{\sqrt{\tan \left (\frac{x}{2}\right )}}\right )\right |-1\right )\right )-\frac{1}{3} (2 \cot (x)+3) \csc (x)\right )}{16 (2 \cot (x)-1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Csc[x]*(2*Cos[x] - Sin[x])*Sqrt[Sin[2*x]]*(-((3 + 2*Cot[x])*Csc[x])/3 - 5*Sqrt[Cos[x]/(-2 + 2*Cos[x])]*(Elli
pticF[ArcSin[1/Sqrt[Tan[x/2]]], -1] + EllipticPi[-2/(-1 + Sqrt[5]), -ArcSin[1/Sqrt[Tan[x/2]]], -1] + EllipticP
i[(-1 + Sqrt[5])/2, -ArcSin[1/Sqrt[Tan[x/2]]], -1])*Sec[x]*Sqrt[Tan[x/2]]))/(16*(-1 + 2*Cot[x]))

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Maple [C]  time = 0.116, size = 396, normalized size = 5. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1920*(-tan(1/2*x)/(tan(1/2*x)^2-1))^(1/2)/tan(1/2*x)^2*(-140*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*(-2*tan(1/
2*x)+2)^(1/2)*EllipticF((1+tan(1/2*x))^(1/2),1/2*2^(1/2))*(1+tan(1/2*x))^(1/2)*(-tan(1/2*x))^(1/2)*tan(1/2*x)+
240*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*EllipticE((1+tan(1/2*x))^(1/2),1/2*2^(1/2))*(1
+tan(1/2*x))^(1/2)*(-tan(1/2*x))^(1/2)*tan(1/2*x)+2^(1/2)*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*(tan(1/2*x)^3-ta
n(1/2*x))^(1/2)*sum((14*_alpha^3+3*_alpha^2+14*_alpha-11)*(_alpha^3+2*_alpha-3)*(1+tan(1/2*x))^(1/2)*(1-tan(1/
2*x))^(1/2)*(-tan(1/2*x))^(1/2)/(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*EllipticPi((1+tan(1/2*x))^(1/2),-1/4*_alph
a^3-1/2*_alpha+3/4,1/2*2^(1/2)),_alpha=RootOf(_Z^4+_Z^3+2*_Z^2-_Z+1))*tan(1/2*x)+40*(tan(1/2*x)*(tan(1/2*x)^2-
1))^(1/2)*tan(1/2*x)^4+120*tan(1/2*x)^3*(tan(1/2*x)^3-tan(1/2*x))^(1/2)-120*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*ta
n(1/2*x)-40*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2))/(tan(1/2*x)^3-tan(1/2*x))^(1/2)

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Maxima [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(cos(x)^2*sin(x)/(sin(x)^2-sin(2*x))/sin(2*x)^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.65727, size = 432, normalized size = 5.47 \begin{align*} -\frac{4 \, \sqrt{2} \sqrt{\cos \left (x\right ) \sin \left (x\right )}{\left (2 \, \cos \left (x\right ) + 3 \, \sin \left (x\right )\right )} - 4 \, \cos \left (x\right )^{2} - 15 \,{\left (\cos \left (x\right )^{2} - 1\right )} \log \left (-\frac{1}{2} \, \sqrt{2} \sqrt{\cos \left (x\right ) \sin \left (x\right )}{\left (4 \, \cos \left (x\right ) + 3 \, \sin \left (x\right )\right )} + \frac{1}{2} \, \cos \left (x\right )^{2} + \frac{7}{2} \, \cos \left (x\right ) \sin \left (x\right ) + \frac{1}{2}\right ) + 15 \,{\left (\cos \left (x\right )^{2} - 1\right )} \log \left (\frac{1}{2} \, \cos \left (x\right )^{2} + \frac{1}{2} \, \sqrt{2} \sqrt{\cos \left (x\right ) \sin \left (x\right )} \sin \left (x\right ) - \frac{1}{2} \, \cos \left (x\right ) \sin \left (x\right ) + \frac{1}{2}\right ) + 4}{192 \,{\left (\cos \left (x\right )^{2} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*(4*sqrt(2)*sqrt(cos(x)*sin(x))*(2*cos(x) + 3*sin(x)) - 4*cos(x)^2 - 15*(cos(x)^2 - 1)*log(-1/2*sqrt(2)*
sqrt(cos(x)*sin(x))*(4*cos(x) + 3*sin(x)) + 1/2*cos(x)^2 + 7/2*cos(x)*sin(x) + 1/2) + 15*(cos(x)^2 - 1)*log(1/
2*cos(x)^2 + 1/2*sqrt(2)*sqrt(cos(x)*sin(x))*sin(x) - 1/2*cos(x)*sin(x) + 1/2) + 4)/(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(cos(x)**2*sin(x)/(sin(x)**2-sin(2*x))/sin(2*x)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (x\right )^{2} \sin \left (x\right )}{{\left (\sin \left (x\right )^{2} - \sin \left (2 \, x\right )\right )} \sin \left (2 \, x\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(cos(x)^2*sin(x)/((sin(x)^2 - sin(2*x))*sin(2*x)^(5/2)), x)