3.5.30 \(\int \cos ^{\frac {3}{2}}(2 x) \sin (x) \, dx\) [430]

Optimal. Leaf size=55 \[ -\frac {3 \tanh ^{-1}\left (\frac {\sqrt {2} \cos (x)}{\sqrt {\cos (2 x)}}\right )}{8 \sqrt {2}}+\frac {3}{8} \cos (x) \sqrt {\cos (2 x)}-\frac {1}{4} \cos (x) \cos ^{\frac {3}{2}}(2 x) \]

[Out]

-1/4*cos(x)*cos(2*x)^(3/2)-3/16*arctanh(cos(x)*2^(1/2)/cos(2*x)^(1/2))*2^(1/2)+3/8*cos(x)*cos(2*x)^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {4442, 201, 223, 212} \begin {gather*} -\frac {1}{4} \cos (x) \cos ^{\frac {3}{2}}(2 x)+\frac {3}{8} \cos (x) \sqrt {\cos (2 x)}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {2} \cos (x)}{\sqrt {\cos (2 x)}}\right )}{8 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*ArcTanh[(Sqrt[2]*Cos[x])/Sqrt[Cos[2*x]]])/(8*Sqrt[2]) + (3*Cos[x]*Sqrt[Cos[2*x]])/8 - (Cos[x]*Cos[2*x]^(3/
2))/4

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 212

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

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 4442

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFactors[Cos[c*(a + b*x)], x]}, Dist[-d/(
b*c), Subst[Int[SubstFor[1, Cos[c*(a + b*x)]/d, u, x], x], x, Cos[c*(a + b*x)]/d], x] /; FunctionOfQ[Cos[c*(a
+ b*x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Sin] || EqQ[F, sin])

Rubi steps

\begin {align*} \int \cos ^{\frac {3}{2}}(2 x) \sin (x) \, dx &=-\text {Subst}\left (\int \left (-1+2 x^2\right )^{3/2} \, dx,x,\cos (x)\right )\\ &=-\frac {1}{4} \cos (x) \cos ^{\frac {3}{2}}(2 x)+\frac {3}{4} \text {Subst}\left (\int \sqrt {-1+2 x^2} \, dx,x,\cos (x)\right )\\ &=\frac {3}{8} \cos (x) \sqrt {\cos (2 x)}-\frac {1}{4} \cos (x) \cos ^{\frac {3}{2}}(2 x)-\frac {3}{8} \text {Subst}\left (\int \frac {1}{\sqrt {-1+2 x^2}} \, dx,x,\cos (x)\right )\\ &=\frac {3}{8} \cos (x) \sqrt {\cos (2 x)}-\frac {1}{4} \cos (x) \cos ^{\frac {3}{2}}(2 x)-\frac {3}{8} \text {Subst}\left (\int \frac {1}{1-2 x^2} \, dx,x,\frac {\cos (x)}{\sqrt {\cos (2 x)}}\right )\\ &=-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {2} \cos (x)}{\sqrt {\cos (2 x)}}\right )}{8 \sqrt {2}}+\frac {3}{8} \cos (x) \sqrt {\cos (2 x)}-\frac {1}{4} \cos (x) \cos ^{\frac {3}{2}}(2 x)\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 49, normalized size = 0.89 \begin {gather*} -\frac {1}{8} \sqrt {\cos (2 x)} (-2 \cos (x)+\cos (3 x))-\frac {3 \log \left (\sqrt {2} \cos (x)+\sqrt {\cos (2 x)}\right )}{8 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(Sqrt[Cos[2*x]]*(-2*Cos[x] + Cos[3*x])) - (3*Log[Sqrt[2]*Cos[x] + Sqrt[Cos[2*x]]])/(8*Sqrt[2])

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Maple [A]
time = 0.09, size = 55, normalized size = 1.00

method	result	size
default	\(-\frac {\left (\cos ^{3}\left (x \right )\right ) \sqrt {2 \left (\cos ^{2}\left (x \right )\right )-1}}{2}+\frac {5 \cos \left (x \right ) \sqrt {2 \left (\cos ^{2}\left (x \right )\right )-1}}{8}-\frac {3 \ln \left (\cos \left (x \right ) \sqrt {2}+\sqrt {2 \left (\cos ^{2}\left (x \right )\right )-1}\right ) \sqrt {2}}{16}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(2*x)^(3/2)*sin(x),x,method=_RETURNVERBOSE)

[Out]

-1/2*cos(x)^3*(2*cos(x)^2-1)^(1/2)+5/8*cos(x)*(2*cos(x)^2-1)^(1/2)-3/16*ln(cos(x)*2^(1/2)+(2*cos(x)^2-1)^(1/2)
)*2^(1/2)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 790 vs. \(2 (39) = 78\).
time = 3.22, size = 790, normalized size = 14.36 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(2*x)^(3/2)*sin(x),x, algorithm="maxima")

[Out]

-1/128*sqrt(2)*(4*(cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1)^(1/4)*(((cos(4*x) - 2)*cos(1/2*arctan2(sin(4*x),
cos(4*x))) + sin(4*x)*sin(1/2*arctan2(sin(4*x), cos(4*x))) + cos(4*x) - 2)*cos(1/2*arctan2(sin(4*x), cos(4*x)
+ 1)) - (cos(1/2*arctan2(sin(4*x), cos(4*x)))*sin(4*x) - (cos(4*x) - 2)*sin(1/2*arctan2(sin(4*x), cos(4*x))) -
sin(4*x))*sin(1/2*arctan2(sin(4*x), cos(4*x) + 1))) + 3*log(sqrt(cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1)*co
s(1/2*arctan2(sin(4*x), cos(4*x) + 1))^2 + sqrt(cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1)*sin(1/2*arctan2(sin(
4*x), cos(4*x) + 1))^2 + 2*(cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1)^(1/4)*cos(1/2*arctan2(sin(4*x), cos(4*x)
+ 1)) + 1) - 3*log(sqrt(cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1)*cos(1/2*arctan2(sin(4*x), cos(4*x) + 1))^2
+ sqrt(cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1)*sin(1/2*arctan2(sin(4*x), cos(4*x) + 1))^2 - 2*(cos(4*x)^2 +
sin(4*x)^2 + 2*cos(4*x) + 1)^(1/4)*cos(1/2*arctan2(sin(4*x), cos(4*x) + 1)) + 1) + 3*log(((cos(1/2*arctan2(sin
(4*x), cos(4*x)))^2 + sin(1/2*arctan2(sin(4*x), cos(4*x)))^2)*cos(1/2*arctan2(sin(4*x), cos(4*x) + 1))^2 + (co
s(1/2*arctan2(sin(4*x), cos(4*x)))^2 + sin(1/2*arctan2(sin(4*x), cos(4*x)))^2)*sin(1/2*arctan2(sin(4*x), cos(4
*x) + 1))^2)*sqrt(cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1) + 2*(cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1)^(1/
4)*(cos(1/2*arctan2(sin(4*x), cos(4*x) + 1))*cos(1/2*arctan2(sin(4*x), cos(4*x))) + sin(1/2*arctan2(sin(4*x),
cos(4*x) + 1))*sin(1/2*arctan2(sin(4*x), cos(4*x)))) + 1) - 3*log(((cos(1/2*arctan2(sin(4*x), cos(4*x)))^2 + s
in(1/2*arctan2(sin(4*x), cos(4*x)))^2)*cos(1/2*arctan2(sin(4*x), cos(4*x) + 1))^2 + (cos(1/2*arctan2(sin(4*x),
cos(4*x)))^2 + sin(1/2*arctan2(sin(4*x), cos(4*x)))^2)*sin(1/2*arctan2(sin(4*x), cos(4*x) + 1))^2)*sqrt(cos(4
*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1) - 2*(cos(4*x)^2 + sin(4*x)^2 + 2*cos(4*x) + 1)^(1/4)*(cos(1/2*arctan2(sin
(4*x), cos(4*x) + 1))*cos(1/2*arctan2(sin(4*x), cos(4*x))) + sin(1/2*arctan2(sin(4*x), cos(4*x) + 1))*sin(1/2*
arctan2(sin(4*x), cos(4*x)))) + 1))

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 103 vs. \(2 (39) = 78\).
time = 1.52, size = 103, normalized size = 1.87 \begin {gather*} -\frac {1}{8} \, {\left (4 \, \cos \left (x\right )^{3} - 5 \, \cos \left (x\right )\right )} \sqrt {2 \, \cos \left (x\right )^{2} - 1} + \frac {3}{128} \, \sqrt {2} \log \left (2048 \, \cos \left (x\right )^{8} - 2048 \, \cos \left (x\right )^{6} + 640 \, \cos \left (x\right )^{4} - 64 \, \cos \left (x\right )^{2} - 8 \, {\left (128 \, \sqrt {2} \cos \left (x\right )^{7} - 96 \, \sqrt {2} \cos \left (x\right )^{5} + 20 \, \sqrt {2} \cos \left (x\right )^{3} - \sqrt {2} \cos \left (x\right )\right )} \sqrt {2 \, \cos \left (x\right )^{2} - 1} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(4*cos(x)^3 - 5*cos(x))*sqrt(2*cos(x)^2 - 1) + 3/128*sqrt(2)*log(2048*cos(x)^8 - 2048*cos(x)^6 + 640*cos(
x)^4 - 64*cos(x)^2 - 8*(128*sqrt(2)*cos(x)^7 - 96*sqrt(2)*cos(x)^5 + 20*sqrt(2)*cos(x)^3 - sqrt(2)*cos(x))*sqr
t(2*cos(x)^2 - 1) + 1)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(2*x)**(3/2)*sin(x),x)

[Out]

Timed out

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Giac [A]
time = 0.98, size = 48, normalized size = 0.87 \begin {gather*} -\frac {1}{8} \, {\left (4 \, \cos \left (x\right )^{2} - 5\right )} \sqrt {2 \, \cos \left (x\right )^{2} - 1} \cos \left (x\right ) + \frac {3}{16} \, \sqrt {2} \log \left ({\left | -\sqrt {2} \cos \left (x\right ) + \sqrt {2 \, \cos \left (x\right )^{2} - 1} \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(2*x)^(3/2)*sin(x),x, algorithm="giac")

[Out]

-1/8*(4*cos(x)^2 - 5)*sqrt(2*cos(x)^2 - 1)*cos(x) + 3/16*sqrt(2)*log(abs(-sqrt(2)*cos(x) + sqrt(2*cos(x)^2 - 1
)))

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Mupad [B]
time = 0.42, size = 29, normalized size = 0.53 \begin {gather*} -\frac {{\cos \left (2\,x\right )}^{3/2}\,\cos \left (x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {1}{2};\ \frac {3}{2};\ \cos \left (2\,x\right )+1\right )}{{\left (-\cos \left (2\,x\right )\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(cos(2*x)^(3/2)*cos(x)*hypergeom([-3/2, 1/2], 3/2, cos(2*x) + 1))/(-cos(2*x))^(3/2)

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