3.1.22 \(\int \frac {1}{(c \csc (a+b x))^{3/2}} \, dx\) [22]

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

[Out]

-2/3*cos(b*x+a)/b/c/(c*csc(b*x+a))^(1/2)-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))*(c*csc(b*x+a))^(1/2)*sin(b*x+a)^(1/2)/b/c^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 = {3854, 3856, 2720} \begin {gather*} \frac {2 \sqrt {\sin (a+b x)} F\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {c \csc (a+b x)}}{3 b c^2}-\frac {2 \cos (a+b x)}{3 b c \sqrt {c \csc (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*Csc[a + b*x])^(-3/2),x]

[Out]

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

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]

Rule 3854

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

Rule 3856

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rubi steps

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

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Mathematica [A]
time = 0.09, size = 63, normalized size = 0.82 \begin {gather*} -\frac {\csc ^2(a+b x) \left (2 F\left (\left .\frac {1}{4} (-2 a+\pi -2 b x)\right |2\right ) \sqrt {\sin (a+b x)}+\sin (2 (a+b x))\right )}{3 b (c \csc (a+b x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c*Csc[a + b*x])^(-3/2),x]

[Out]

-1/3*(Csc[a + b*x]^2*(2*EllipticF[(-2*a + Pi - 2*b*x)/4, 2]*Sqrt[Sin[a + b*x]] + Sin[2*(a + b*x)]))/(b*(c*Csc[
a + b*x])^(3/2))

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Maple [C] Result contains complex when optimal does not.
time = 1.13, size = 189, normalized size = 2.45

method	result	size
default	\(-\frac {\left (i \sqrt {-\frac {i \cos \left (x b +a \right )-i-\sin \left (x b +a \right )}{\sin \left (x b +a \right )}}\, \EllipticF \left (\sqrt {\frac {i \cos \left (x b +a \right )-i+\sin \left (x b +a \right )}{\sin \left (x b +a \right )}}, \frac {\sqrt {2}}{2}\right ) \sin \left (x b +a \right ) \sqrt {-\frac {i \left (-1+\cos \left (x b +a \right )\right )}{\sin \left (x b +a \right )}}\, \sqrt {\frac {i \cos \left (x b +a \right )-i+\sin \left (x b +a \right )}{\sin \left (x b +a \right )}}+\left (\cos ^{2}\left (x b +a \right )\right ) \sqrt {2}-\cos \left (x b +a \right ) \sqrt {2}\right ) \sqrt {2}}{3 b \left (-1+\cos \left (x b +a \right )\right ) \left (\frac {c}{\sin \left (x b +a \right )}\right )^{\frac {3}{2}} \sin \left (x b +a \right )}\)	\(189\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*csc(b*x+a))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/3/b*(I*(-(I*cos(b*x+a)-I-sin(b*x+a))/sin(b*x+a))^(1/2)*EllipticF(((I*cos(b*x+a)-I+sin(b*x+a))/sin(b*x+a))^(
1/2),1/2*2^(1/2))*sin(b*x+a)*(-I*(-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((I*cos(b*x+a)-I+sin(b*x+a))/sin(b*x+a))^(1
/2)+cos(b*x+a)^2*2^(1/2)-cos(b*x+a)*2^(1/2))/(-1+cos(b*x+a))/(c/sin(b*x+a))^(3/2)/sin(b*x+a)*2^(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(1/(c*csc(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

integrate((c*csc(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 = 0.80, size = 85, normalized size = 1.10 \begin {gather*} -\frac {2 \, \sqrt {\frac {c}{\sin \left (b x + a\right )}} \cos \left (b x + a\right ) \sin \left (b x + a\right ) + i \, \sqrt {2 i \, c} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) - i \, \sqrt {-2 i \, c} {\rm weierstrassPInverse}\left (4, 0, \cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right )}{3 \, b c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*csc(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (c \csc {\left (a + b x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*csc(b*x+a))**(3/2),x)

[Out]

Integral((c*csc(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(1/(c*csc(b*x+a))^(3/2),x, algorithm="giac")

[Out]

integrate((c*csc(b*x + a))^(-3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (\frac {c}{\sin \left (a+b\,x\right )}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c/sin(a + b*x))^(3/2),x)

[Out]

int(1/(c/sin(a + b*x))^(3/2), x)

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