3.5.32 \(\int \cos ^{\frac {3}{2}}(2 x) \sec ^3(x) \, dx\) [432]

Optimal. Leaf size=49 \[ 2 \sqrt {2} \sin ^{-1}\left (\sqrt {2} \sin (x)\right )-\frac {5}{2} \tan ^{-1}\left (\frac {\sin (x)}{\sqrt {\cos (2 x)}}\right )-\frac {1}{2} \sqrt {\cos (2 x)} \sec (x) \tan (x) \]

[Out]

-5/2*arctan(sin(x)/cos(2*x)^(1/2))+2*arcsin(sin(x)*2^(1/2))*2^(1/2)-1/2*sec(x)*cos(2*x)^(1/2)*tan(x)

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Rubi [A]
time = 0.08, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {4449, 424, 537, 222, 385, 209} \begin {gather*} 2 \sqrt {2} \text {ArcSin}\left (\sqrt {2} \sin (x)\right )-\frac {5}{2} \text {ArcTan}\left (\frac {\sin (x)}{\sqrt {\cos (2 x)}}\right )-\frac {1}{2} \sqrt {\cos (2 x)} \tan (x) \sec (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

2*Sqrt[2]*ArcSin[Sqrt[2]*Sin[x]] - (5*ArcTan[Sin[x]/Sqrt[Cos[2*x]]])/2 - (Sqrt[Cos[2*x]]*Sec[x]*Tan[x])/2

Rule 209

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

Rule 222

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

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 424

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a*d - c*b)*x*(a + b*x^n)^(
p + 1)*((c + d*x^n)^(q - 1)/(a*b*n*(p + 1))), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q
, x]

Rule 537

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 4449

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^(n_), x_Symbol] :> With[{d = FreeFactors[Sin[c*(a + b*x)], x]}, Dist
[d/(b*c), Subst[Int[SubstFor[(1 - d^2*x^2)^((n - 1)/2), Sin[c*(a + b*x)]/d, u, x], x], x, Sin[c*(a + b*x)]/d],
x] /; FunctionOfQ[Sin[c*(a + b*x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && IntegerQ[(n - 1)/2] && NonsumQ[u] && (
EqQ[F, Cos] || EqQ[F, cos])

Rubi steps

\begin {align*} \int \cos ^{\frac {3}{2}}(2 x) \sec ^3(x) \, dx &=\text {Subst}\left (\int \frac {\left (1-2 x^2\right )^{3/2}}{\left (1-x^2\right )^2} \, dx,x,\sin (x)\right )\\ &=-\frac {1}{2} \sqrt {\cos (2 x)} \sec (x) \tan (x)-\frac {1}{2} \text {Subst}\left (\int \frac {-3+8 x^2}{\sqrt {1-2 x^2} \left (1-x^2\right )} \, dx,x,\sin (x)\right )\\ &=-\frac {1}{2} \sqrt {\cos (2 x)} \sec (x) \tan (x)-\frac {5}{2} \text {Subst}\left (\int \frac {1}{\sqrt {1-2 x^2} \left (1-x^2\right )} \, dx,x,\sin (x)\right )+4 \text {Subst}\left (\int \frac {1}{\sqrt {1-2 x^2}} \, dx,x,\sin (x)\right )\\ &=2 \sqrt {2} \sin ^{-1}\left (\sqrt {2} \sin (x)\right )-\frac {1}{2} \sqrt {\cos (2 x)} \sec (x) \tan (x)-\frac {5}{2} \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sin (x)}{\sqrt {\cos (2 x)}}\right )\\ &=2 \sqrt {2} \sin ^{-1}\left (\sqrt {2} \sin (x)\right )-\frac {5}{2} \tan ^{-1}\left (\frac {\sin (x)}{\sqrt {\cos (2 x)}}\right )-\frac {1}{2} \sqrt {\cos (2 x)} \sec (x) \tan (x)\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 49, normalized size = 1.00 \begin {gather*} \frac {1}{2} \left (4 \sqrt {2} \sin ^{-1}\left (\sqrt {2} \sin (x)\right )-5 \tan ^{-1}\left (\frac {\sin (x)}{\sqrt {\cos (2 x)}}\right )-\sqrt {\cos (2 x)} \sec (x) \tan (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[2]*ArcSin[Sqrt[2]*Sin[x]] - 5*ArcTan[Sin[x]/Sqrt[Cos[2*x]]] - Sqrt[Cos[2*x]]*Sec[x]*Tan[x])/2

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(99\) vs. \(2(37)=74\).
time = 0.16, size = 100, normalized size = 2.04

method	result	size
default	\(-\frac {\sqrt {\left (2 \left (\cos ^{2}\left (x \right )\right )-1\right ) \left (\sin ^{2}\left (x \right )\right )}\, \left (4 \sqrt {2}\, \arcsin \left (4 \left (\cos ^{2}\left (x \right )\right )-3\right ) \left (\cos ^{2}\left (x \right )\right )-5 \arctan \left (\frac {3 \left (\cos ^{2}\left (x \right )\right )-2}{2 \sqrt {-2 \left (\sin ^{4}\left (x \right )\right )+\sin ^{2}\left (x \right )}}\right ) \left (\cos ^{2}\left (x \right )\right )+2 \sqrt {-2 \left (\sin ^{4}\left (x \right )\right )+\sin ^{2}\left (x \right )}\right )}{4 \cos \left (x \right )^{2} \sin \left (x \right ) \sqrt {2 \left (\cos ^{2}\left (x \right )\right )-1}}\)	\(100\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((2*cos(x)^2-1)*sin(x)^2)^(1/2)*(4*2^(1/2)*arcsin(4*cos(x)^2-3)*cos(x)^2-5*arctan(1/2*(3*cos(x)^2-2)/(-2*
sin(x)^4+sin(x)^2)^(1/2))*cos(x)^2+2*(-2*sin(x)^4+sin(x)^2)^(1/2))/cos(x)^2/sin(x)/(2*cos(x)^2-1)^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(cos(2*x)^(3/2)/cos(x)^3, x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 118 vs. \(2 (37) = 74\).
time = 1.62, size = 118, normalized size = 2.41 \begin {gather*} -\frac {2 \, \sqrt {2} \arctan \left (\frac {{\left (32 \, \sqrt {2} \cos \left (x\right )^{4} - 48 \, \sqrt {2} \cos \left (x\right )^{2} + 17 \, \sqrt {2}\right )} \sqrt {2 \, \cos \left (x\right )^{2} - 1}}{8 \, {\left (8 \, \cos \left (x\right )^{4} - 10 \, \cos \left (x\right )^{2} + 3\right )} \sin \left (x\right )}\right ) \cos \left (x\right )^{2} - 5 \, \arctan \left (\frac {3 \, \cos \left (x\right )^{2} - 2}{2 \, \sqrt {2 \, \cos \left (x\right )^{2} - 1} \sin \left (x\right )}\right ) \cos \left (x\right )^{2} + 2 \, \sqrt {2 \, \cos \left (x\right )^{2} - 1} \sin \left (x\right )}{4 \, \cos \left (x\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*sqrt(2)*arctan(1/8*(32*sqrt(2)*cos(x)^4 - 48*sqrt(2)*cos(x)^2 + 17*sqrt(2))*sqrt(2*cos(x)^2 - 1)/((8*c
os(x)^4 - 10*cos(x)^2 + 3)*sin(x)))*cos(x)^2 - 5*arctan(1/2*(3*cos(x)^2 - 2)/(sqrt(2*cos(x)^2 - 1)*sin(x)))*co
s(x)^2 + 2*sqrt(2*cos(x)^2 - 1)*sin(x))/cos(x)^2

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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)/cos(x)**3,x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(cos(2*x)^(3/2)/cos(x)^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\cos \left (2\,x\right )}^{3/2}}{{\cos \left (x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(cos(2*x)^(3/2)/cos(x)^3, x)

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