3.10 \(\int \cos ^{\frac {5}{2}}(a+b x) \, dx\)

Optimal. Leaf size=42 \[ \frac {6 E\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{5 b}+\frac {2 \sin (a+b x) \cos ^{\frac {3}{2}}(a+b x)}{5 b} \]

[Out]

6/5*(cos(1/2*b*x+1/2*a)^2)^(1/2)/cos(1/2*b*x+1/2*a)*EllipticE(sin(1/2*b*x+1/2*a),2^(1/2))/b+2/5*cos(b*x+a)^(3/
2)*sin(b*x+a)/b

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Rubi [A]  time = 0.02, 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 = {2635, 2639} \[ \frac {6 E\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{5 b}+\frac {2 \sin (a+b x) \cos ^{\frac {3}{2}}(a+b x)}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^(5/2),x]

[Out]

(6*EllipticE[(a + b*x)/2, 2])/(5*b) + (2*Cos[a + b*x]^(3/2)*Sin[a + b*x])/(5*b)

Rule 2635

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] && Integer
Q[2*n]

Rule 2639

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

Rubi steps

\begin {align*} \int \cos ^{\frac {5}{2}}(a+b x) \, dx &=\frac {2 \cos ^{\frac {3}{2}}(a+b x) \sin (a+b x)}{5 b}+\frac {3}{5} \int \sqrt {\cos (a+b x)} \, dx\\ &=\frac {6 E\left (\left .\frac {1}{2} (a+b x)\right |2\right )}{5 b}+\frac {2 \cos ^{\frac {3}{2}}(a+b x) \sin (a+b x)}{5 b}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 40, normalized size = 0.95 \[ \frac {6 E\left (\left .\frac {1}{2} (a+b x)\right |2\right )+\sin (2 (a+b x)) \sqrt {\cos (a+b x)}}{5 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^(5/2),x]

[Out]

(6*EllipticE[(a + b*x)/2, 2] + Sqrt[Cos[a + b*x]]*Sin[2*(a + b*x)])/(5*b)

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fricas [F]  time = 0.40, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\cos \left (b x + a\right )^{\frac {5}{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^(5/2),x, algorithm="fricas")

[Out]

integral(cos(b*x + a)^(5/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \cos \left (b x + a\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^(5/2),x, algorithm="giac")

[Out]

integrate(cos(b*x + a)^(5/2), x)

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maple [B]  time = 0.08, size = 202, normalized size = 4.81 \[ -\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}\, \left (-8 \cos \left (\frac {b x}{2}+\frac {a}{2}\right ) \left (\sin ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+8 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \cos \left (\frac {b x}{2}+\frac {a}{2}\right )-3 \sqrt {\frac {1}{2}-\frac {\cos \left (b x +a \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {b x}{2}+\frac {a}{2}\right ), \sqrt {2}\right )-2 \left (\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \cos \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{5 \sqrt {-2 \left (\sin ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+\sin ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )}\, \sin \left (\frac {b x}{2}+\frac {a}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1}\, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^(5/2),x)

[Out]

-2/5*((2*cos(1/2*b*x+1/2*a)^2-1)*sin(1/2*b*x+1/2*a)^2)^(1/2)*(-8*cos(1/2*b*x+1/2*a)*sin(1/2*b*x+1/2*a)^6+8*sin
(1/2*b*x+1/2*a)^4*cos(1/2*b*x+1/2*a)-3*(sin(1/2*b*x+1/2*a)^2)^(1/2)*(2*sin(1/2*b*x+1/2*a)^2-1)^(1/2)*EllipticE
(cos(1/2*b*x+1/2*a),2^(1/2))-2*sin(1/2*b*x+1/2*a)^2*cos(1/2*b*x+1/2*a))/(-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 \[ \int \cos \left (b x + a\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^(5/2),x, algorithm="maxima")

[Out]

integrate(cos(b*x + a)^(5/2), x)

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mupad [B]  time = 0.17, size = 42, normalized size = 1.00 \[ -\frac {2\,{\cos \left (a+b\,x\right )}^{7/2}\,\sin \left (a+b\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (a+b\,x\right )}^2\right )}{7\,b\,\sqrt {{\sin \left (a+b\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)^(5/2),x)

[Out]

-(2*cos(a + b*x)^(7/2)*sin(a + b*x)*hypergeom([1/2, 7/4], 11/4, cos(a + b*x)^2))/(7*b*(sin(a + b*x)^2)^(1/2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**(5/2),x)

[Out]

Timed out

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