3.26 \(\int (c \sin (a+b x))^{5/2} \, dx\)

Optimal. Leaf size=75 \[ \frac {6 c^2 E\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {c \sin (a+b x)}}{5 b \sqrt {\sin (a+b x)}}-\frac {2 c \cos (a+b x) (c \sin (a+b x))^{3/2}}{5 b} \]

[Out]

-2/5*c*cos(b*x+a)*(c*sin(b*x+a))^(3/2)/b-6/5*c^2*(sin(1/2*a+1/4*Pi+1/2*b*x)^2)^(1/2)/sin(1/2*a+1/4*Pi+1/2*b*x)
*EllipticE(cos(1/2*a+1/4*Pi+1/2*b*x),2^(1/2))*(c*sin(b*x+a))^(1/2)/b/sin(b*x+a)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2635, 2640, 2639} \[ \frac {6 c^2 E\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {c \sin (a+b x)}}{5 b \sqrt {\sin (a+b x)}}-\frac {2 c \cos (a+b x) (c \sin (a+b x))^{3/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[(c*Sin[a + b*x])^(5/2),x]

[Out]

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

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 66, normalized size = 0.88 \[ -\frac {(c \sin (a+b x))^{5/2} \left (\sqrt {\sin (a+b x)} \sin (2 (a+b x))+6 E\left (\left .\frac {1}{4} (-2 a-2 b x+\pi )\right |2\right )\right )}{5 b \sin ^{\frac {5}{2}}(a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*Sin[a + b*x])^(5/2),x]

[Out]

-1/5*((c*Sin[a + b*x])^(5/2)*(6*EllipticE[(-2*a + Pi - 2*b*x)/4, 2] + Sqrt[Sin[a + b*x]]*Sin[2*(a + b*x)]))/(b
*Sin[a + b*x]^(5/2))

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (c^{2} \cos \left (b x + a\right )^{2} - c^{2}\right )} \sqrt {c \sin \left (b x + a\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(c^2*cos(b*x + a)^2 - c^2)*sqrt(c*sin(b*x + a)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*sin(b*x + a))^(5/2), x)

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maple [A]  time = 0.06, size = 152, normalized size = 2.03 \[ -\frac {c^{3} \left (6 \sqrt {-\sin \left (b x +a \right )+1}\, \sqrt {2 \sin \left (b x +a \right )+2}\, \left (\sqrt {\sin }\left (b x +a \right )\right ) \EllipticE \left (\sqrt {-\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {-\sin \left (b x +a \right )+1}\, \sqrt {2 \sin \left (b x +a \right )+2}\, \left (\sqrt {\sin }\left (b x +a \right )\right ) \EllipticF \left (\sqrt {-\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \left (\sin ^{4}\left (b x +a \right )\right )+2 \left (\sin ^{2}\left (b x +a \right )\right )\right )}{5 \cos \left (b x +a \right ) \sqrt {c \sin \left (b x +a \right )}\, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*c^3*(6*(-sin(b*x+a)+1)^(1/2)*(2*sin(b*x+a)+2)^(1/2)*sin(b*x+a)^(1/2)*EllipticE((-sin(b*x+a)+1)^(1/2),1/2*
2^(1/2))-3*(-sin(b*x+a)+1)^(1/2)*(2*sin(b*x+a)+2)^(1/2)*sin(b*x+a)^(1/2)*EllipticF((-sin(b*x+a)+1)^(1/2),1/2*2
^(1/2))-2*sin(b*x+a)^4+2*sin(b*x+a)^2)/cos(b*x+a)/(c*sin(b*x+a))^(1/2)/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*sin(b*x + a))^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (c\,\sin \left (a+b\,x\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sin(a + b*x))^(5/2),x)

[Out]

int((c*sin(a + b*x))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c \sin {\left (a + b x \right )}\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c*sin(a + b*x))**(5/2), x)

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