3.636 \(\int \frac{\cos ^3(x) \cos (2 x)}{(\sin ^2(x)-\sin (2 x)) \sin ^{\frac{5}{2}}(2 x)} \, dx\)

Optimal. Leaf size=95 \[ \frac{\cos ^5(x)}{5 \sin ^{\frac{5}{2}}(2 x)}+\frac{\sin (x) \cos ^4(x)}{6 \sin ^{\frac{5}{2}}(2 x)}-\frac{3 \sin ^2(x) \cos ^3(x)}{4 \sin ^{\frac{5}{2}}(2 x)}+\frac{3 \sin ^5(x) \tanh ^{-1}\left (\frac{\sqrt{\tan (x)}}{\sqrt{2}}\right )}{4 \sqrt{2} \sin ^{\frac{5}{2}}(2 x) \tan ^{\frac{5}{2}}(x)} \]

[Out]

Cos[x]^5/(5*Sin[2*x]^(5/2)) + (Cos[x]^4*Sin[x])/(6*Sin[2*x]^(5/2)) - (3*Cos[x]^3*Sin[x]^2)/(4*Sin[2*x]^(5/2))
+ (3*ArcTanh[Sqrt[Tan[x]]/Sqrt[2]]*Sin[x]^5)/(4*Sqrt[2]*Sin[2*x]^(5/2)*Tan[x]^(5/2))

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Rubi [A]  time = 0.575766, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {4390, 898, 1262, 207} \[ \frac{\cos ^5(x)}{5 \sin ^{\frac{5}{2}}(2 x)}+\frac{\sin (x) \cos ^4(x)}{6 \sin ^{\frac{5}{2}}(2 x)}-\frac{3 \sin ^2(x) \cos ^3(x)}{4 \sin ^{\frac{5}{2}}(2 x)}+\frac{3 \sin ^5(x) \tanh ^{-1}\left (\frac{\sqrt{\tan (x)}}{\sqrt{2}}\right )}{4 \sqrt{2} \sin ^{\frac{5}{2}}(2 x) \tan ^{\frac{5}{2}}(x)} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[x]^3*Cos[2*x])/((Sin[x]^2 - Sin[2*x])*Sin[2*x]^(5/2)),x]

[Out]

Cos[x]^5/(5*Sin[2*x]^(5/2)) + (Cos[x]^4*Sin[x])/(6*Sin[2*x]^(5/2)) - (3*Cos[x]^3*Sin[x]^2)/(4*Sin[2*x]^(5/2))
+ (3*ArcTanh[Sqrt[Tan[x]]/Sqrt[2]]*Sin[x]^5)/(4*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 ^3(x) \cos (2 x)}{\left (\sin ^2(x)-\sin (2 x)\right ) \sin ^{\frac{5}{2}}(2 x)} \, dx &=\frac{\sin ^5(x) \int \frac{\cos (2 x) \csc ^2(x)}{\left (\sin ^2(x)-\sin (2 x)\right ) \sqrt{\tan (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^{7/2}} \, 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^6 \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^6}-\frac{1}{4 x^4}+\frac{3}{8 x^2}-\frac{3}{8 \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 ^5(x)}{5 \sin ^{\frac{5}{2}}(2 x)}+\frac{\cos ^4(x) \sin (x)}{6 \sin ^{\frac{5}{2}}(2 x)}-\frac{3 \cos ^3(x) \sin ^2(x)}{4 \sin ^{\frac{5}{2}}(2 x)}-\frac{\left (3 \sin ^5(x)\right ) \operatorname{Subst}\left (\int \frac{1}{-2+x^2} \, dx,x,\sqrt{\tan (x)}\right )}{4 \sin ^{\frac{5}{2}}(2 x) \tan ^{\frac{5}{2}}(x)}\\ &=\frac{\cos ^5(x)}{5 \sin ^{\frac{5}{2}}(2 x)}+\frac{\cos ^4(x) \sin (x)}{6 \sin ^{\frac{5}{2}}(2 x)}-\frac{3 \cos ^3(x) \sin ^2(x)}{4 \sin ^{\frac{5}{2}}(2 x)}+\frac{3 \tanh ^{-1}\left (\frac{\sqrt{\tan (x)}}{\sqrt{2}}\right ) \sin ^5(x)}{4 \sqrt{2} \sin ^{\frac{5}{2}}(2 x) \tan ^{\frac{5}{2}}(x)}\\ \end{align*}

Mathematica [C]  time = 15.3218, size = 188, normalized size = 1.98 \[ \frac{1}{960} \sqrt{\sin (2 x)} \sec (x) \left (-45 \sqrt{2} \sqrt{\frac{\cos (x)}{\cos (x)-1}} \sqrt{\tan \left (\frac{x}{2}\right )} \text{EllipticF}\left (\sin ^{-1}\left (\frac{1}{\sqrt{\tan \left (\frac{x}{2}\right )}}\right ),-1\right )+20 \cot ^2(x)-114 \cot (x)+24 \cot (x) \csc ^2(x)-45 \sqrt{2} \sqrt{\frac{\cos (x)}{\cos (x)-1}} \sqrt{\tan \left (\frac{x}{2}\right )} \Pi \left (-\frac{2}{-1+\sqrt{5}};\left .-\sin ^{-1}\left (\frac{1}{\sqrt{\tan \left (\frac{x}{2}\right )}}\right )\right |-1\right )-45 \sqrt{2} \sqrt{\frac{\cos (x)}{\cos (x)-1}} \sqrt{\tan \left (\frac{x}{2}\right )} \Pi \left (\frac{1}{2} \left (-1+\sqrt{5}\right );\left .-\sin ^{-1}\left (\frac{1}{\sqrt{\tan \left (\frac{x}{2}\right )}}\right )\right |-1\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[x]^3*Cos[2*x])/((Sin[x]^2 - Sin[2*x])*Sin[2*x]^(5/2)),x]

[Out]

(Sec[x]*Sqrt[Sin[2*x]]*(-114*Cot[x] + 20*Cot[x]^2 + 24*Cot[x]*Csc[x]^2 - 45*Sqrt[2]*Sqrt[Cos[x]/(-1 + Cos[x])]
*EllipticF[ArcSin[1/Sqrt[Tan[x/2]]], -1]*Sqrt[Tan[x/2]] - 45*Sqrt[2]*Sqrt[Cos[x]/(-1 + Cos[x])]*EllipticPi[-2/
(-1 + Sqrt[5]), -ArcSin[1/Sqrt[Tan[x/2]]], -1]*Sqrt[Tan[x/2]] - 45*Sqrt[2]*Sqrt[Cos[x]/(-1 + Cos[x])]*Elliptic
Pi[(-1 + Sqrt[5])/2, -ArcSin[1/Sqrt[Tan[x/2]]], -1]*Sqrt[Tan[x/2]]))/960

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Maple [C]  time = 0.187, size = 761, normalized size = 8. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^3*cos(2*x)/(sin(x)^2-sin(2*x))/sin(2*x)^(5/2),x)

[Out]

1/3840*(-tan(1/2*x)/(tan(1/2*x)^2-1))^(1/2)/tan(1/2*x)^3*(1772*(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))*(tan(1/2*x)*(tan(1/2*x)-1)*(1+tan(1/2*x)))^(1/2)*(1+t
an(1/2*x))^(1/2)*(-tan(1/2*x))^(1/2)*tan(1/2*x)^2-4464*(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))*(tan(1/2*x)*(tan(1/2*x)-1)*(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+24*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*(tan(1/2*x)*(tan(1/2*x)-1)*(1
+tan(1/2*x)))^(1/2)*tan(1/2*x)^6+3*2^(1/2)*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*(tan(1/2*x)^3-tan(1/2*x))^(1/2)
*(tan(1/2*x)*(tan(1/2*x)-1)*(1+tan(1/2*x)))^(1/2)*sum((6*_alpha^3+7*_alpha^2+6*_alpha+1)*(_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)*EllipticPi((1+tan(1/2*x))^(1/2),-1/4*_alpha^3-1
/2*_alpha+3/4,1/2*2^(1/2))/(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2),_alpha=RootOf(_Z^4+_Z^3+2*_Z^2-_Z+1))*tan(1/2*x
)^2-40*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*(tan(1/2*x)*(tan(1/2*x)-1)*(1+tan(1/2*x)))^(1/2)*tan(1/2*x)^5-1920*
tan(1/2*x)^4*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)-24*tan(1/2*x)^4*(tan(1/2*x)*(
tan(1/2*x)-1)*(1+tan(1/2*x)))^(1/2)*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)-1272*tan(1/2*x)^4*(tan(1/2*x)^3-tan(1/
2*x))^(1/2)*(tan(1/2*x)*(tan(1/2*x)-1)*(1+tan(1/2*x)))^(1/2)-24*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*(tan(1/2*x
)*(tan(1/2*x)-1)*(1+tan(1/2*x)))^(1/2)*tan(1/2*x)^2+1272*(tan(1/2*x)^3-tan(1/2*x))^(1/2)*(tan(1/2*x)*(tan(1/2*
x)-1)*(1+tan(1/2*x)))^(1/2)*tan(1/2*x)^2+40*tan(1/2*x)*(tan(1/2*x)*(tan(1/2*x)-1)*(1+tan(1/2*x)))^(1/2)*(tan(1
/2*x)*(tan(1/2*x)^2-1))^(1/2)+24*(tan(1/2*x)*(tan(1/2*x)-1)*(1+tan(1/2*x)))^(1/2)*(tan(1/2*x)*(tan(1/2*x)^2-1)
)^(1/2))/(tan(1/2*x)^3-tan(1/2*x))^(1/2)/(tan(1/2*x)*(tan(1/2*x)-1)*(1+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)^3*cos(2*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 = 2.63272, size = 501, normalized size = 5.27 \begin{align*} -\frac{45 \,{\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 ) \sin \left (x\right ) - 45 \,{\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 ) \sin \left (x\right ) + 4 \, \sqrt{2}{\left (57 \, \cos \left (x\right )^{2} + 10 \, \cos \left (x\right ) \sin \left (x\right ) - 45\right )} \sqrt{\cos \left (x\right ) \sin \left (x\right )} + 268 \,{\left (\cos \left (x\right )^{2} - 1\right )} \sin \left (x\right )}{1920 \,{\left (\cos \left (x\right )^{2} - 1\right )} \sin \left (x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(x)^3*cos(2*x)/(sin(x)^2-sin(2*x))/sin(2*x)^(5/2),x, algorithm="fricas")

[Out]

-1/1920*(45*(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)*sin(x) - 45*(cos(x)^2 - 1)*log(1/2*cos(x)^2 + 1/2*sqrt(2)*sqrt(cos(x)*sin(x))*sin(x) - 1/2*c
os(x)*sin(x) + 1/2)*sin(x) + 4*sqrt(2)*(57*cos(x)^2 + 10*cos(x)*sin(x) - 45)*sqrt(cos(x)*sin(x)) + 268*(cos(x)
^2 - 1)*sin(x))/((cos(x)^2 - 1)*sin(x))

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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)**3*cos(2*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 (2 \, x\right ) \cos \left (x\right )^{3}}{{\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)^3*cos(2*x)/(sin(x)^2-sin(2*x))/sin(2*x)^(5/2),x, algorithm="giac")

[Out]

integrate(cos(2*x)*cos(x)^3/((sin(x)^2 - sin(2*x))*sin(2*x)^(5/2)), x)