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

Optimal. Leaf size=77 \[ \frac {6 E\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{5 b c^2 \sqrt {\sin (a+b x)} \sqrt {c \csc (a+b x)}}-\frac {2 \cos (a+b x)}{5 b c (c \csc (a+b x))^{3/2}} \]

[Out]

-2/5*cos(b*x+a)/b/c/(c*csc(b*x+a))^(3/2)-6/5*(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
ipticE(cos(1/2*a+1/4*Pi+1/2*b*x),2^(1/2))/b/c^2/(c*csc(b*x+a))^(1/2)/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 = {3769, 3771, 2639} \[ \frac {6 E\left (\left .\frac {1}{2} \left (a+b x-\frac {\pi }{2}\right )\right |2\right )}{5 b c^2 \sqrt {\sin (a+b x)} \sqrt {c \csc (a+b x)}}-\frac {2 \cos (a+b x)}{5 b c (c \csc (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*Csc[a + b*x])^(-5/2),x]

[Out]

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

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

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Mathematica [A]  time = 0.14, size = 60, normalized size = 0.78 \[ \frac {-2 \sin (2 (a+b x))-\frac {12 E\left (\left .\frac {1}{4} (-2 a-2 b x+\pi )\right |2\right )}{\sqrt {\sin (a+b x)}}}{10 b c^2 \sqrt {c \csc (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*Csc[a + b*x])^(-5/2),x]

[Out]

((-12*EllipticE[(-2*a + Pi - 2*b*x)/4, 2])/Sqrt[Sin[a + b*x]] - 2*Sin[2*(a + b*x)])/(10*b*c^2*Sqrt[c*Csc[a + b
*x]])

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c \csc \left (b x + a\right )}}{c^{3} \csc \left (b x + a\right )^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(c*csc(b*x + a))/(c^3*csc(b*x + a)^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*csc(b*x + a))^(-5/2), x)

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maple [C]  time = 0.85, size = 534, normalized size = 6.94 \[ \frac {\left (-6 \cos \left (b x +a \right ) \sqrt {-\frac {i \left (-1+\cos \left (b x +a \right )\right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-i \cos \left (b x +a \right )+\sin \left (b x +a \right )+i}{\sin \left (b x +a \right )}}\, \EllipticE \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 ) \sqrt {\frac {i \cos \left (b x +a \right )-i+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}+3 \cos \left (b x +a \right ) \sqrt {-\frac {i \left (-1+\cos \left (b x +a \right )\right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-i \cos \left (b x +a \right )+\sin \left (b x +a \right )+i}{\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 ) \sqrt {\frac {i \cos \left (b x +a \right )-i+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}+\left (\cos ^{3}\left (b x +a \right )\right ) \sqrt {2}-6 \sqrt {-\frac {i \left (-1+\cos \left (b x +a \right )\right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-i \cos \left (b x +a \right )+\sin \left (b x +a \right )+i}{\sin \left (b x +a \right )}}\, \EllipticE \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 ) \sqrt {\frac {i \cos \left (b x +a \right )-i+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}+3 \sqrt {-\frac {i \left (-1+\cos \left (b x +a \right )\right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-i \cos \left (b x +a \right )+\sin \left (b x +a \right )+i}{\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 ) \sqrt {\frac {i \cos \left (b x +a \right )-i+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}-4 \cos \left (b x +a \right ) \sqrt {2}+3 \sqrt {2}\right ) \sqrt {2}}{5 b \left (\frac {c}{\sin \left (b x +a \right )}\right )^{\frac {5}{2}} \sin \left (b x +a \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5/b*(-6*cos(b*x+a)*(-I*(-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((-I*cos(b*x+a)+sin(b*x+a)+I)/sin(b*x+a))^(1/2)*Ell
ipticE(((I*cos(b*x+a)-I+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((I*cos(b*x+a)-I+sin(b*x+a))/sin(b*x+a))^(1
/2)+3*cos(b*x+a)*(-I*(-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((-I*cos(b*x+a)+sin(b*x+a)+I)/sin(b*x+a))^(1/2)*Ellipti
cF(((I*cos(b*x+a)-I+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((I*cos(b*x+a)-I+sin(b*x+a))/sin(b*x+a))^(1/2)+
cos(b*x+a)^3*2^(1/2)-6*(-I*(-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((-I*cos(b*x+a)+sin(b*x+a)+I)/sin(b*x+a))^(1/2)*E
llipticE(((I*cos(b*x+a)-I+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((I*cos(b*x+a)-I+sin(b*x+a))/sin(b*x+a))^
(1/2)+3*(-I*(-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((-I*cos(b*x+a)+sin(b*x+a)+I)/sin(b*x+a))^(1/2)*EllipticF(((I*co
s(b*x+a)-I+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((I*cos(b*x+a)-I+sin(b*x+a))/sin(b*x+a))^(1/2)-4*cos(b*x
+a)*2^(1/2)+3*2^(1/2))/(c/sin(b*x+a))^(5/2)/sin(b*x+a)^3*2^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*csc(b*x + a))^(-5/2), x)

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

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 \[ \int \frac {1}{\left (c \csc {\left (a + b x \right )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c*csc(a + b*x))**(-5/2), x)

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