\(\int \frac {\cos ^2(x) \sin (x)}{(\sin ^2(x)-\sin (2 x)) \sin ^{\frac {5}{2}}(2 x)} \, dx\) [635]

Optimal result

Integrand size = 28, antiderivative size = 79 \[ \int \frac {\cos ^2(x) \sin (x)}{\left (\sin ^2(x)-\sin (2 x)\right ) \sin ^{\frac {5}{2}}(2 x)} \, dx=\frac {\cos ^4(x) \sin (x)}{3 \sin ^{\frac {5}{2}}(2 x)}+\frac {\cos ^3(x) \sin ^2(x)}{2 \sin ^{\frac {5}{2}}(2 x)}-\frac {5 \text {arctanh}\left (\frac {\sqrt {\tan (x)}}{\sqrt {2}}\right ) \sin ^5(x)}{2 \sqrt {2} \sin ^{\frac {5}{2}}(2 x) \tan ^{\frac {5}{2}}(x)} \]

[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)

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4475, 912, 1276, 213} \[ \int \frac {\cos ^2(x) \sin (x)}{\left (\sin ^2(x)-\sin (2 x)\right ) \sin ^{\frac {5}{2}}(2 x)} \, dx=-\frac {5 \sin ^5(x) \text {arctanh}\left (\frac {\sqrt {\tan (x)}}{\sqrt {2}}\right )}{2 \sqrt {2} \sin ^{\frac {5}{2}}(2 x) \tan ^{\frac {5}{2}}(x)}+\frac {\sin (x) \cos ^4(x)}{3 \sin ^{\frac {5}{2}}(2 x)}+\frac {\sin ^2(x) \cos ^3(x)}{2 \sin ^{\frac {5}{2}}(2 x)} \]

[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 213

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

Rule 912

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 1276

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 4475

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*} \text {integral}& = \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) \text {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 ) \text {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 ) \text {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 ) \text {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 \text {arctanh}\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 [A] (verified)

Time = 2.23 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.54 \[ \int \frac {\cos ^2(x) \sin (x)}{\left (\sin ^2(x)-\sin (2 x)\right ) \sin ^{\frac {5}{2}}(2 x)} \, dx=\frac {\csc ^2\left (\frac {x}{2}\right ) \sqrt {\sin (2 x)} \left (-15 \arctan \left (\frac {\sqrt {\tan \left (\frac {x}{2}\right )}}{\sqrt {-1+\tan ^2\left (\frac {x}{2}\right )}}\right ) (-1+\cos (x))+\sqrt {2} \sqrt {-\frac {\cos (x)}{1+\cos (x)}} (2 \cos (x)+3 \sin (x)) \sqrt {\tan \left (\frac {x}{2}\right )}\right )}{96 (1+\cos (x)) \sqrt {\tan \left (\frac {x}{2}\right )} \sqrt {-1+\tan ^2\left (\frac {x}{2}\right )}} \]

[In]

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

[Out]

(Csc[x/2]^2*Sqrt[Sin[2*x]]*(-15*ArcTan[Sqrt[Tan[x/2]]/Sqrt[-1 + Tan[x/2]^2]]*(-1 + Cos[x]) + Sqrt[2]*Sqrt[-(Co
s[x]/(1 + Cos[x]))]*(2*Cos[x] + 3*Sin[x])*Sqrt[Tan[x/2]]))/(96*(1 + Cos[x])*Sqrt[Tan[x/2]]*Sqrt[-1 + Tan[x/2]^
2])

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.90 (sec) , antiderivative size = 396, normalized size of antiderivative = 5.01

method	result	size
default	\(-\frac {\sqrt {-\frac {\tan \left (\frac {x}{2}\right )}{\tan \left (\frac {x}{2}\right )^{2}-1}}\, \left (-140 \sqrt {\tan \left (\frac {x}{2}\right ) \left (\tan \left (\frac {x}{2}\right )^{2}-1\right )}\, \sqrt {-2 \tan \left (\frac {x}{2}\right )+2}\, \operatorname {EllipticF}\left (\sqrt {\tan \left (\frac {x}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \sqrt {\tan \left (\frac {x}{2}\right )+1}\, \sqrt {-\tan \left (\frac {x}{2}\right )}\, \tan \left (\frac {x}{2}\right )+240 \sqrt {\tan \left (\frac {x}{2}\right ) \left (\tan \left (\frac {x}{2}\right )^{2}-1\right )}\, \sqrt {-2 \tan \left (\frac {x}{2}\right )+2}\, \operatorname {EllipticE}\left (\sqrt {\tan \left (\frac {x}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \sqrt {\tan \left (\frac {x}{2}\right )+1}\, \sqrt {-\tan \left (\frac {x}{2}\right )}\, \tan \left (\frac {x}{2}\right )+\sqrt {2}\, \sqrt {\tan \left (\frac {x}{2}\right ) \left (\tan \left (\frac {x}{2}\right )^{2}-1\right )}\, \sqrt {\tan \left (\frac {x}{2}\right )^{3}-\tan \left (\frac {x}{2}\right )}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {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 {\tan \left (\frac {x}{2}\right )+1}\, \sqrt {1-\tan \left (\frac {x}{2}\right )}\, \sqrt {-\tan \left (\frac {x}{2}\right )}\, \operatorname {EllipticPi}\left (\sqrt {\tan \left (\frac {x}{2}\right )+1}, -\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 {\tan \left (\frac {x}{2}\right ) \left (\tan \left (\frac {x}{2}\right )^{2}-1\right )}}\right ) \tan \left (\frac {x}{2}\right )+40 \sqrt {\tan \left (\frac {x}{2}\right ) \left (\tan \left (\frac {x}{2}\right )^{2}-1\right )}\, \tan \left (\frac {x}{2}\right )^{4}+120 \tan \left (\frac {x}{2}\right )^{3} \sqrt {\tan \left (\frac {x}{2}\right )^{3}-\tan \left (\frac {x}{2}\right )}-120 \sqrt {\tan \left (\frac {x}{2}\right )^{3}-\tan \left (\frac {x}{2}\right )}\, \tan \left (\frac {x}{2}\right )-40 \sqrt {\tan \left (\frac {x}{2}\right ) \left (\tan \left (\frac {x}{2}\right )^{2}-1\right )}\right )}{1920 \tan \left (\frac {x}{2}\right )^{2} \sqrt {\tan \left (\frac {x}{2}\right )^{3}-\tan \left (\frac {x}{2}\right )}}\)	\(396\)

[In]

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

[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((tan(1/2*x)+1)^(1/2),1/2*2^(1/2))*(tan(1/2*x)+1)^(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((tan(1/2*x)+1)^(1/2),1/2*2^(1/2))*(t
an(1/2*x)+1)^(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)*(tan(1/2*x)+1)^(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((tan(1/2*x)+1)^(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)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.52 \[ \int \frac {\cos ^2(x) \sin (x)}{\left (\sin ^2(x)-\sin (2 x)\right ) \sin ^{\frac {5}{2}}(2 x)} \, dx=-\frac {4 \, \sqrt {2} \sqrt {\cos \left (x\right ) \sin \left (x\right )} {\left (2 \, \cos \left (x\right ) + 3 \, \sin \left (x\right )\right )} - 4 \, \cos \left (x\right )^{2} - 15 \, {\left (\cos \left (x\right )^{2} - 1\right )} \log \left (-\frac {1}{2} \, \sqrt {2} \sqrt {\cos \left (x\right ) \sin \left (x\right )} {\left (4 \, \cos \left (x\right ) + 3 \, \sin \left (x\right )\right )} + \frac {1}{2} \, \cos \left (x\right )^{2} + \frac {7}{2} \, \cos \left (x\right ) \sin \left (x\right ) + \frac {1}{2}\right ) + 15 \, {\left (\cos \left (x\right )^{2} - 1\right )} \log \left (\frac {1}{2} \, \cos \left (x\right )^{2} + \frac {1}{2} \, \sqrt {2} \sqrt {\cos \left (x\right ) \sin \left (x\right )} \sin \left (x\right ) - \frac {1}{2} \, \cos \left (x\right ) \sin \left (x\right ) + \frac {1}{2}\right ) + 4}{192 \, {\left (\cos \left (x\right )^{2} - 1\right )}} \]

[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)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [F]

\[ \int \frac {\cos ^2(x) \sin (x)}{\left (\sin ^2(x)-\sin (2 x)\right ) \sin ^{\frac {5}{2}}(2 x)} \, dx=\int { \frac {\cos \left (x\right )^{2} \sin \left (x\right )}{{\left (\sin \left (x\right )^{2} - \sin \left (2 \, x\right )\right )} \sin \left (2 \, x\right )^{\frac {5}{2}}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^2(x) \sin (x)}{\left (\sin ^2(x)-\sin (2 x)\right ) \sin ^{\frac {5}{2}}(2 x)} \, dx=-\int \frac {{\cos \left (x\right )}^2\,\sin \left (x\right )}{{\sin \left (2\,x\right )}^{5/2}\,\left (\sin \left (2\,x\right )-{\sin \left (x\right )}^2\right )} \,d x \]

[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)