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

[Out]

1/3*cos(x)^4*sin(x)/sin(2*x)^(5/2)+1/2*cos(x)^3*sin(x)^2/sin(2*x)^(5/2)-5/4*arctanh(1/2*tan(x)^(1/2)*2^(1/2))*
sin(x)^5/sin(2*x)^(5/2)*2^(1/2)/tan(x)^(5/2)

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Rubi [A]  time = 0.57, antiderivative size = 79, normalized size of antiderivative = 1.00, 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 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])

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 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]

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*}

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Mathematica [C]  time = 4.96, size = 183, normalized size = 2.32 \[ \frac {1}{96} \sqrt {\sin (2 x)} \sec (x) \left (2 \cot ^2(x)+6 \cot (x)+2 \csc ^2(x)+15 \sqrt {2} \sqrt {\frac {\cos (x)}{\cos (x)-1}} \sqrt {\tan \left (\frac {x}{2}\right )} F\left (\left .\sin ^{-1}\left (\frac {1}{\sqrt {\tan \left (\frac {x}{2}\right )}}\right )\right |-1\right )-15 \sqrt {2} \sqrt {\frac {\cos (x)}{\cos (x)-1}} \sqrt {\tan \left (\frac {x}{2}\right )} \Pi \left (-\frac {2}{-1+\sqrt {5}};\left .\sin ^{-1}\left (\frac {1}{\sqrt {\tan \left (\frac {x}{2}\right )}}\right )\right |-1\right )-15 \sqrt {2} \sqrt {\frac {\cos (x)}{\cos (x)-1}} \sqrt {\tan \left (\frac {x}{2}\right )} \Pi \left (\frac {1}{2} \left (-1+\sqrt {5}\right );\left .\sin ^{-1}\left (\frac {1}{\sqrt {\tan \left (\frac {x}{2}\right )}}\right )\right |-1\right )-2\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[x]*Sqrt[Sin[2*x]]*(-2 + 6*Cot[x] + 2*Cot[x]^2 + 2*Csc[x]^2 + 15*Sqrt[2]*Sqrt[Cos[x]/(-1 + Cos[x])]*Ellipt
icF[ArcSin[1/Sqrt[Tan[x/2]]], -1]*Sqrt[Tan[x/2]] - 15*Sqrt[2]*Sqrt[Cos[x]/(-1 + Cos[x])]*EllipticPi[-2/(-1 + S
qrt[5]), ArcSin[1/Sqrt[Tan[x/2]]], -1]*Sqrt[Tan[x/2]] - 15*Sqrt[2]*Sqrt[Cos[x]/(-1 + Cos[x])]*EllipticPi[(-1 +
Sqrt[5])/2, ArcSin[1/Sqrt[Tan[x/2]]], -1]*Sqrt[Tan[x/2]]))/96

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fricas [A]  time = 1.04, size = 120, normalized size = 1.52 \[ -\frac {4 \, \sqrt {2} \sqrt {\cos \relax (x) \sin \relax (x)} {\left (2 \, \cos \relax (x) + 3 \, \sin \relax (x)\right )} - 4 \, \cos \relax (x)^{2} - 15 \, {\left (\cos \relax (x)^{2} - 1\right )} \log \left (-\frac {1}{2} \, \sqrt {2} \sqrt {\cos \relax (x) \sin \relax (x)} {\left (4 \, \cos \relax (x) + 3 \, \sin \relax (x)\right )} + \frac {1}{2} \, \cos \relax (x)^{2} + \frac {7}{2} \, \cos \relax (x) \sin \relax (x) + \frac {1}{2}\right ) + 15 \, {\left (\cos \relax (x)^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \relax (x)^{2} + \frac {1}{2} \, \sqrt {2} \sqrt {\cos \relax (x) \sin \relax (x)} \sin \relax (x) - \frac {1}{2} \, \cos \relax (x) \sin \relax (x) + \frac {1}{2}\right ) + 4}{192 \, {\left (\cos \relax (x)^{2} - 1\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \relax (x)^{2} \sin \relax (x)}{{\left (\sin \relax (x)^{2} - \sin \left (2 \, x\right )\right )} \sin \left (2 \, x\right )^{\frac {5}{2}}}\,{d x} \]

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)

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maple [C]  time = 0.28, size = 397, normalized size = 5.03 \[ \frac {\sqrt {-\frac {\tan \left (\frac {x}{2}\right )}{\tan ^{2}\left (\frac {x}{2}\right )-1}}\, \left (140 \sqrt {1+\tan \left (\frac {x}{2}\right )}\, \sqrt {-\tan \left (\frac {x}{2}\right )}\, \sqrt {-2 \tan \left (\frac {x}{2}\right )+2}\, \EllipticF \left (\sqrt {1+\tan \left (\frac {x}{2}\right )}, \frac {\sqrt {2}}{2}\right ) \sqrt {\left (\tan ^{2}\left (\frac {x}{2}\right )-1\right ) \tan \left (\frac {x}{2}\right )}\, \tan \left (\frac {x}{2}\right )-240 \sqrt {1+\tan \left (\frac {x}{2}\right )}\, \sqrt {-\tan \left (\frac {x}{2}\right )}\, \sqrt {-2 \tan \left (\frac {x}{2}\right )+2}\, \EllipticE \left (\sqrt {1+\tan \left (\frac {x}{2}\right )}, \frac {\sqrt {2}}{2}\right ) \sqrt {\left (\tan ^{2}\left (\frac {x}{2}\right )-1\right ) \tan \left (\frac {x}{2}\right )}\, \tan \left (\frac {x}{2}\right )-\sqrt {\tan ^{3}\left (\frac {x}{2}\right )-\tan \left (\frac {x}{2}\right )}\, \sqrt {\left (\tan ^{2}\left (\frac {x}{2}\right )-1\right ) \tan \left (\frac {x}{2}\right )}\, \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (\textit {\_Z}^{4}+\textit {\_Z}^{3}+2 \textit {\_Z}^{2}-\textit {\_Z} +1\right )}{\sum }\frac {\left (14 \underline {\hspace {1.25 ex}}\alpha ^{3}+3 \underline {\hspace {1.25 ex}}\alpha ^{2}+14 \underline {\hspace {1.25 ex}}\alpha -11\right ) \left (\underline {\hspace {1.25 ex}}\alpha ^{3}+2 \underline {\hspace {1.25 ex}}\alpha -3\right ) \sqrt {1+\tan \left (\frac {x}{2}\right )}\, \sqrt {1-\tan \left (\frac {x}{2}\right )}\, \sqrt {-\tan \left (\frac {x}{2}\right )}\, \EllipticPi \left (\sqrt {1+\tan \left (\frac {x}{2}\right )}, -\frac {1}{4} \underline {\hspace {1.25 ex}}\alpha ^{3}-\frac {1}{2} \underline {\hspace {1.25 ex}}\alpha +\frac {3}{4}, \frac {\sqrt {2}}{2}\right )}{\sqrt {\left (\tan ^{2}\left (\frac {x}{2}\right )-1\right ) \tan \left (\frac {x}{2}\right )}}\right ) \tan \left (\frac {x}{2}\right )-40 \sqrt {\left (\tan ^{2}\left (\frac {x}{2}\right )-1\right ) \tan \left (\frac {x}{2}\right )}\, \left (\tan ^{4}\left (\frac {x}{2}\right )\right )-120 \left (\tan ^{3}\left (\frac {x}{2}\right )\right ) \sqrt {\tan ^{3}\left (\frac {x}{2}\right )-\tan \left (\frac {x}{2}\right )}+120 \sqrt {\tan ^{3}\left (\frac {x}{2}\right )-\tan \left (\frac {x}{2}\right )}\, \tan \left (\frac {x}{2}\right )+40 \sqrt {\left (\tan ^{2}\left (\frac {x}{2}\right )-1\right ) \tan \left (\frac {x}{2}\right )}\right )}{1920 \tan \left (\frac {x}{2}\right )^{2} \sqrt {\tan ^{3}\left (\frac {x}{2}\right )-\tan \left (\frac {x}{2}\right )}} \]

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*(1+tan(1/2*x))^(1/2)*(-tan(1/2*x))^(1/2)*(-2*tan
(1/2*x)+2)^(1/2)*EllipticF((1+tan(1/2*x))^(1/2),1/2*2^(1/2))*((tan(1/2*x)^2-1)*tan(1/2*x))^(1/2)*tan(1/2*x)-24
0*(1+tan(1/2*x))^(1/2)*(-tan(1/2*x))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*EllipticE((1+tan(1/2*x))^(1/2),1/2*2^(1/2))
*((tan(1/2*x)^2-1)*tan(1/2*x))^(1/2)*tan(1/2*x)-(tan(1/2*x)^3-tan(1/2*x))^(1/2)*((tan(1/2*x)^2-1)*tan(1/2*x))^
(1/2)*2^(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)^2-1)*tan(1/2*x))^(1/2)*EllipticPi((1+tan(1/2*x))^(1/2),-1/4*_alpha^
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)^2-1)*tan(1/2*x)
)^(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)*tan(
1/2*x)+40*((tan(1/2*x)^2-1)*tan(1/2*x))^(1/2))/(tan(1/2*x)^3-tan(1/2*x))^(1/2)

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ -\int \frac {{\cos \relax (x)}^2\,\sin \relax (x)}{{\sin \left (2\,x\right )}^{5/2}\,\left (\sin \left (2\,x\right )-{\sin \relax (x)}^2\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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