3.1.11 \(\int \cos ^{\frac {3}{2}}(a+b x) \, dx\) [11]

Optimal. Leaf size=42 \[ \frac {2 F\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{3 b}+\frac {2 \sqrt {\cos (a+b x)} \sin (a+b x)}{3 b} \]

[Out]

2/3*(cos(1/2*a+1/2*b*x)^2)^(1/2)/cos(1/2*a+1/2*b*x)*EllipticF(sin(1/2*a+1/2*b*x),2^(1/2))/b+2/3*sin(b*x+a)*cos
(b*x+a)^(1/2)/b

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Rubi [A]
time = 0.01, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2715, 2720} \begin {gather*} \frac {2 F\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{3 b}+\frac {2 \sin (a+b x) \sqrt {\cos (a+b x)}}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^(3/2),x]

[Out]

(2*EllipticF[(a + b*x)/2, 2])/(3*b) + (2*Sqrt[Cos[a + b*x]]*Sin[a + b*x])/(3*b)

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin {align*} \int \cos ^{\frac {3}{2}}(a+b x) \, dx &=\frac {2 \sqrt {\cos (a+b x)} \sin (a+b x)}{3 b}+\frac {1}{3} \int \frac {1}{\sqrt {\cos (a+b x)}} \, dx\\ &=\frac {2 F\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{3 b}+\frac {2 \sqrt {\cos (a+b x)} \sin (a+b x)}{3 b}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 36, normalized size = 0.86 \begin {gather*} \frac {2 \left (F\left (\left .\frac {1}{2} (a+b x)\right |2\right )+\sqrt {\cos (a+b x)} \sin (a+b x)\right )}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^(3/2),x]

[Out]

(2*(EllipticF[(a + b*x)/2, 2] + Sqrt[Cos[a + b*x]]*Sin[a + b*x]))/(3*b)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(178\) vs. \(2(62)=124\).
time = 0.03, size = 179, normalized size = 4.26

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, \left (4 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \cos \left (\frac {b x}{2}+\frac {a}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1}\, b}\)	\(179\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*((2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)*(4*sin(1/2*b*x+1/2*a)^4*cos(1/2*b*x+1/2*a)-2*sin(
1/2*b*x+1/2*a)^2*cos(1/2*b*x+1/2*a)+(sin(1/2*b*x+1/2*a)^2)^(1/2)*(2*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*EllipticF(co
s(1/2*b*x+1/2*a),2^(1/2)))/(-2*sin(1/2*b*x+1/2*a)^4+sin(1/2*b*x+1/2*a)^2)^(1/2)/sin(1/2*b*x+1/2*a)/(2*cos(1/2*
b*x+1/2*a)^2-1)^(1/2)/b

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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(b*x+a)^(3/2),x, algorithm="maxima")

[Out]

integrate(cos(b*x + a)^(3/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.09, size = 68, normalized size = 1.62 \begin {gather*} \frac {2 \, \sqrt {\cos \left (b x + a\right )} \sin \left (b x + a\right ) - i \, \sqrt {2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + i \, \sqrt {2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )}{3 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(2*sqrt(cos(b*x + a))*sin(b*x + a) - I*sqrt(2)*weierstrassPInverse(-4, 0, cos(b*x + a) + I*sin(b*x + a)) +
I*sqrt(2)*weierstrassPInverse(-4, 0, cos(b*x + a) - I*sin(b*x + a)))/b

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \cos ^{\frac {3}{2}}{\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**(3/2),x)

[Out]

Integral(cos(a + b*x)**(3/2), x)

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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(b*x+a)^(3/2),x, algorithm="giac")

[Out]

integrate(cos(b*x + a)^(3/2), x)

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Mupad [B]
time = 0.08, size = 35, normalized size = 0.83 \begin {gather*} \frac {2\,\mathrm {F}\left (\frac {a}{2}+\frac {b\,x}{2}\middle |2\right )}{3\,b}+\frac {2\,\sqrt {\cos \left (a+b\,x\right )}\,\sin \left (a+b\,x\right )}{3\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^(3/2),x)

[Out]

(2*ellipticF(a/2 + (b*x)/2, 2))/(3*b) + (2*cos(a + b*x)^(1/2)*sin(a + b*x))/(3*b)

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