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

Optimal. Leaf size=68 \[ -\frac{9 \cos (x)}{16 \sqrt{\sin (2 x)}}+\frac{\cos (x) \cot ^2(x)}{20 \sqrt{\sin (2 x)}}-\frac{5 \cos (x) \cot (x)}{24 \sqrt{\sin (2 x)}}+\frac{33}{32} \tanh ^{-1}\left (\frac{1}{2} \sqrt{\sin (2 x)} \sec (x)\right ) \]

[Out]

(33*ArcTanh[(Sec[x]*Sqrt[Sin[2*x]])/2])/32 - (9*Cos[x])/(16*Sqrt[Sin[2*x]]) - (5*Cos[x]*Cot[x])/(24*Sqrt[Sin[2
*x]]) + (Cos[x]*Cot[x]^2)/(20*Sqrt[Sin[2*x]])

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Rubi [A]  time = 0.855936, antiderivative size = 95, normalized size of antiderivative = 1.4, number of steps used = 6, number of rules used = 4, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.114, Rules used = {4390, 1619, 63, 207} \[ \frac{\cos ^5(x)}{5 \sin ^{\frac{5}{2}}(2 x)}-\frac{5 \sin (x) \cos ^4(x)}{6 \sin ^{\frac{5}{2}}(2 x)}-\frac{9 \sin ^2(x) \cos ^3(x)}{4 \sin ^{\frac{5}{2}}(2 x)}+\frac{33 \sin ^5(x) \tanh ^{-1}\left (\frac{\sqrt{\tan (x)}}{\sqrt{2}}\right )}{4 \sqrt{2} \sin ^{\frac{5}{2}}(2 x) \tan ^{\frac{5}{2}}(x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Cos[x]^5/(5*Sin[2*x]^(5/2)) - (5*Cos[x]^4*Sin[x])/(6*Sin[2*x]^(5/2)) - (9*Cos[x]^3*Sin[x]^2)/(4*Sin[2*x]^(5/2)
) + (33*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 1619

Int[((Px_)*((c_.) + (d_.)*(x_))^(n_.))/((a_.) + (b_.)*(x_)), x_Symbol] :> Int[ExpandIntegrand[1/Sqrt[c + d*x],
(Px*(c + d*x)^(n + 1/2))/(a + b*x), x], x] /; FreeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && ILtQ[n + 1/2, 0] &
& GtQ[Expon[Px, x], 2]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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)-3 \tan (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) (\cos (2 x)-3 \tan (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+3 x+x^2+3 x^3}{(2-x) x^{7/2}} \, dx,x,\tan (x)\right )}{\sin ^{\frac{5}{2}}(2 x) \tan ^{\frac{5}{2}}(x)}\\ &=\frac{\sin ^5(x) \operatorname{Subst}\left (\int \left (-\frac{1}{2 x^{7/2}}+\frac{5}{4 x^{5/2}}+\frac{9}{8 x^{3/2}}-\frac{33}{8 (-2+x) \sqrt{x}}\right ) \, dx,x,\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{5 \cos ^4(x) \sin (x)}{6 \sin ^{\frac{5}{2}}(2 x)}-\frac{9 \cos ^3(x) \sin ^2(x)}{4 \sin ^{\frac{5}{2}}(2 x)}-\frac{\left (33 \sin ^5(x)\right ) \operatorname{Subst}\left (\int \frac{1}{(-2+x) \sqrt{x}} \, dx,x,\tan (x)\right )}{8 \sin ^{\frac{5}{2}}(2 x) \tan ^{\frac{5}{2}}(x)}\\ &=\frac{\cos ^5(x)}{5 \sin ^{\frac{5}{2}}(2 x)}-\frac{5 \cos ^4(x) \sin (x)}{6 \sin ^{\frac{5}{2}}(2 x)}-\frac{9 \cos ^3(x) \sin ^2(x)}{4 \sin ^{\frac{5}{2}}(2 x)}-\frac{\left (33 \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{5 \cos ^4(x) \sin (x)}{6 \sin ^{\frac{5}{2}}(2 x)}-\frac{9 \cos ^3(x) \sin ^2(x)}{4 \sin ^{\frac{5}{2}}(2 x)}+\frac{33 \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 = 6.0357, size = 150, normalized size = 2.21 \[ \frac{\sqrt{\sin (2 x)} \cos (x) (\cos (2 x)-3 \tan (x)) \left (\frac{1}{15} \csc (x) \left (-50 \cot (x)+12 \csc ^2(x)-147\right )-33 \sqrt{\frac{\cos (x)}{2 \cos (x)-2}} \sqrt{\tan \left (\frac{x}{2}\right )} \sec (x) \left (\text{EllipticF}\left (\sin ^{-1}\left (\frac{1}{\sqrt{\tan \left (\frac{x}{2}\right )}}\right ),-1\right )+\Pi \left (-\frac{2}{-1+\sqrt{5}};\left .-\sin ^{-1}\left (\frac{1}{\sqrt{\tan \left (\frac{x}{2}\right )}}\right )\right |-1\right )+\Pi \left (\frac{1}{2} \left (-1+\sqrt{5}\right );\left .-\sin ^{-1}\left (\frac{1}{\sqrt{\tan \left (\frac{x}{2}\right )}}\right )\right |-1\right )\right )\right )}{16 (-6 \sin (x)+\cos (x)+\cos (3 x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[x]*Sqrt[Sin[2*x]]*((Csc[x]*(-147 - 50*Cot[x] + 12*Csc[x]^2))/15 - 33*Sqrt[Cos[x]/(-2 + 2*Cos[x])]*(Ellipt
icF[ArcSin[1/Sqrt[Tan[x/2]]], -1] + EllipticPi[-2/(-1 + Sqrt[5]), -ArcSin[1/Sqrt[Tan[x/2]]], -1] + EllipticPi[
(-1 + Sqrt[5])/2, -ArcSin[1/Sqrt[Tan[x/2]]], -1])*Sec[x]*Sqrt[Tan[x/2]])*(Cos[2*x] - 3*Tan[x]))/(16*(Cos[x] +
Cos[3*x] - 6*Sin[x]))

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Maple [C]  time = 0.215, size = 761, normalized size = 11.2 \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)-3*tan(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*(932*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*(1+tan(1/2*x
))^(1/2)*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(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)*tan(1/2*x)^2-3024*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*(1+tan(1/2*x))^(1/2)
*(-2*tan(1/2*x)+2)^(1/2)*(-tan(1/2*x))^(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)*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*(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((34*_alpha^3+13*_alpha^2+34*_alpha-21)*(_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*_alpha^3-1/2*_a
lpha+3/4,1/2*2^(1/2)),_alpha=RootOf(_Z^4+_Z^3+2*_Z^2-_Z+1))*(tan(1/2*x)*(tan(1/2*x)^2-1))^(1/2)*2^(1/2)*tan(1/
2*x)^2+200*(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)^5-5
52*(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)^4-1920*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)+552*(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)^2-24*(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)^2-200*(tan(1/2*x)*(tan(1/2*x)-1)*(1+tan(1/2*x)))^(1/2)*(tan(1/2*x)*(t
an(1/2*x)^2-1))^(1/2)*tan(1/2*x)+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)-3*tan(x))/(sin(x)^2-sin(2*x))/sin(2*x)^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 2.11844, size = 506, normalized size = 7.44 \begin{align*} -\frac{495 \,{\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 ) - 495 \,{\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 (147 \, \cos \left (x\right )^{2} - 50 \, \cos \left (x\right ) \sin \left (x\right ) - 135\right )} \sqrt{\cos \left (x\right ) \sin \left (x\right )} + 388 \,{\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)-3*tan(x))/(sin(x)^2-sin(2*x))/sin(2*x)^(5/2),x, algorithm="fricas")

[Out]

-1/1920*(495*(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*co
s(x)*sin(x) + 1/2)*sin(x) - 495*(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)*sin(x) + 4*sqrt(2)*(147*cos(x)^2 - 50*cos(x)*sin(x) - 135)*sqrt(cos(x)*sin(x)) + 388*(co
s(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)-3*tan(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{{\left (\cos \left (2 \, x\right ) - 3 \, \tan \left (x\right )\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)-3*tan(x))/(sin(x)^2-sin(2*x))/sin(2*x)^(5/2),x, algorithm="giac")

[Out]

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