3.31 \(\int \frac {1}{(c \sin (a+b x))^{5/2}} \, dx\)

Optimal. Leaf size=77 \[ \frac {2 \sqrt {\sin (a+b x)} F\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{3 b c^2 \sqrt {c \sin (a+b x)}}-\frac {2 \cos (a+b x)}{3 b c (c \sin (a+b x))^{3/2}} \]

[Out]

-2/3*cos(b*x+a)/b/c/(c*sin(b*x+a))^(3/2)-2/3*(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)*Ell
ipticF(cos(1/2*a+1/4*Pi+1/2*b*x),2^(1/2))*sin(b*x+a)^(1/2)/b/c^2/(c*sin(b*x+a))^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 77, 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 = {2636, 2642, 2641} \[ \frac {2 \sqrt {\sin (a+b x)} F\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{3 b c^2 \sqrt {c \sin (a+b x)}}-\frac {2 \cos (a+b x)}{3 b c (c \sin (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*Sin[a + b*x])^(-5/2),x]

[Out]

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

Rule 2636

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

Rule 2641

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

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 55, normalized size = 0.71 \[ -\frac {2 \left (\cos (a+b x)+\sin ^{\frac {3}{2}}(a+b x) F\left (\left .\frac {1}{4} (-2 a-2 b x+\pi )\right |2\right )\right )}{3 b c (c \sin (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*Sin[a + b*x])^(-5/2),x]

[Out]

(-2*(Cos[a + b*x] + EllipticF[(-2*a + Pi - 2*b*x)/4, 2]*Sin[a + b*x]^(3/2)))/(3*b*c*(c*Sin[a + b*x])^(3/2))

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {c \sin \left (b x + a\right )}}{{\left (c^{3} \cos \left (b x + a\right )^{2} - c^{3}\right )} \sin \left (b x + a\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*sin(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

integral(-sqrt(c*sin(b*x + a))/((c^3*cos(b*x + a)^2 - c^3)*sin(b*x + a)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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 = 105, normalized size = 1.36 \[ -\frac {\sqrt {-\sin \left (b x +a \right )+1}\, \sqrt {2 \sin \left (b x +a \right )+2}\, \left (\sin ^{\frac {5}{2}}\left (b x +a \right )\right ) \EllipticF \left (\sqrt {-\sin \left (b x +a \right )+1}, \frac {\sqrt {2}}{2}\right )-2 \left (\sin ^{3}\left (b x +a \right )\right )+2 \sin \left (b x +a \right )}{3 c^{2} \sin \left (b x +a \right )^{2} \cos \left (b x +a \right ) \sqrt {c \sin \left (b x +a \right )}\, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*sin(b*x+a))^(5/2),x)

[Out]

-1/3/c^2*((-sin(b*x+a)+1)^(1/2)*(2*sin(b*x+a)+2)^(1/2)*sin(b*x+a)^(5/2)*EllipticF((-sin(b*x+a)+1)^(1/2),1/2*2^
(1/2))-2*sin(b*x+a)^3+2*sin(b*x+a))/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 \[ \int \frac {1}{\left (c \sin \left (b x + a\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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 \[ \int \frac {1}{{\left (c\,\sin \left (a+b\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*sin(a + b*x))^(5/2),x)

[Out]

int(1/(c*sin(a + b*x))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (c \sin {\left (a + b x \right )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*sin(b*x+a))**(5/2),x)

[Out]

Integral((c*sin(a + b*x))**(-5/2), x)

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