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

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

[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 = {3769, 3771, 2641} \[ \frac {2 \sqrt {\sin (a+b x)} F\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {c \csc (a+b x)}}{3 b c^2}-\frac {2 \cos (a+b x)}{3 b c \sqrt {c \csc (a+b x)}} \]

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 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 3769

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 3771

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 = 0.07, size = 63, normalized size = 0.82 \[ -\frac {\csc ^2(a+b x) \left (\sin (2 (a+b x))+2 \sqrt {\sin (a+b x)} F\left (\left .\frac {1}{4} (-2 a-2 b x+\pi )\right |2\right )\right )}{3 b (c \csc (a+b x))^{3/2}} \]

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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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c \csc \left (b x + a\right )}}{c^{2} \csc \left (b x + a\right )^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*csc(b*x + a))/(c^2*csc(b*x + a)^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (c \csc \left (b x + a\right )\right )^{\frac {3}{2}}}\,{d x} \]

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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maple [C]  time = 0.84, size = 186, normalized size = 2.42 \[ -\frac {\left (i \sqrt {\frac {-i \cos \left (b x +a \right )+\sin \left (b x +a \right )+i}{\sin \left (b x +a \right )}}\, \sqrt {\frac {i \cos \left (b x +a \right )-i+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {-\frac {i \left (-1+\cos \left (b x +a \right )\right )}{\sin \left (b x +a \right )}}\, \sin \left (b x +a \right ) \EllipticF \left (\sqrt {\frac {i \cos \left (b x +a \right )-i+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right )+\left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {2}-\cos \left (b x +a \right ) \sqrt {2}\right ) \sqrt {2}}{3 b \left (-1+\cos \left (b x +a \right )\right ) \left (\frac {c}{\sin \left (b x +a \right )}\right )^{\frac {3}{2}} \sin \left (b x +a \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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

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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