3.112 \(\int \frac {\csc ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx\)

Optimal. Leaf size=77 \[ \frac {10 F\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{21 b}-\frac {10 \cos (2 a+2 b x)}{21 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {\csc ^2(a+b x)}{7 b \sin ^{\frac {3}{2}}(2 a+2 b x)} \]

[Out]

-10/21*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticF(cos(a+1/4*Pi+b*x),2^(1/2))/b-10/21*cos(2*b*x+2*
a)/b/sin(2*b*x+2*a)^(3/2)-1/7*csc(b*x+a)^2/b/sin(2*b*x+2*a)^(3/2)

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Rubi [A]  time = 0.05, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {4300, 2636, 2641} \[ \frac {10 F\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{21 b}-\frac {10 \cos (2 a+2 b x)}{21 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {\csc ^2(a+b x)}{7 b \sin ^{\frac {3}{2}}(2 a+2 b x)} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^2/Sin[2*a + 2*b*x]^(5/2),x]

[Out]

(10*EllipticF[a - Pi/4 + b*x, 2])/(21*b) - (10*Cos[2*a + 2*b*x])/(21*b*Sin[2*a + 2*b*x]^(3/2)) - Csc[a + b*x]^
2/(7*b*Sin[2*a + 2*b*x]^(3/2))

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 4300

Int[((e_.)*sin[(a_.) + (b_.)*(x_)])^(m_)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[((e*Sin[a + b
*x])^m*(g*Sin[c + d*x])^(p + 1))/(2*b*g*(m + p + 1)), x] + Dist[(m + 2*p + 2)/(e^2*(m + p + 1)), Int[(e*Sin[a
+ b*x])^(m + 2)*(g*Sin[c + d*x])^p, x], x] /; FreeQ[{a, b, c, d, e, g, p}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b,
2] &&  !IntegerQ[p] && LtQ[m, -1] && NeQ[m + 2*p + 2, 0] && NeQ[m + p + 1, 0] && IntegersQ[2*m, 2*p]

Rubi steps

\begin {align*} \int \frac {\csc ^2(a+b x)}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx &=-\frac {\csc ^2(a+b x)}{7 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {10}{7} \int \frac {1}{\sin ^{\frac {5}{2}}(2 a+2 b x)} \, dx\\ &=-\frac {10 \cos (2 a+2 b x)}{21 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {\csc ^2(a+b x)}{7 b \sin ^{\frac {3}{2}}(2 a+2 b x)}+\frac {10}{21} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}} \, dx\\ &=\frac {10 F\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{21 b}-\frac {10 \cos (2 a+2 b x)}{21 b \sin ^{\frac {3}{2}}(2 a+2 b x)}-\frac {\csc ^2(a+b x)}{7 b \sin ^{\frac {3}{2}}(2 a+2 b x)}\\ \end {align*}

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Mathematica [A]  time = 0.47, size = 66, normalized size = 0.86 \[ \frac {40 F\left (\left .a+b x-\frac {\pi }{4}\right |2\right )+\sqrt {\sin (2 (a+b x))} \left (-3 \csc ^4(a+b x)-13 \csc ^2(a+b x)+7 \sec ^2(a+b x)\right )}{84 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^2/Sin[2*a + 2*b*x]^(5/2),x]

[Out]

(40*EllipticF[a - Pi/4 + b*x, 2] + (-13*Csc[a + b*x]^2 - 3*Csc[a + b*x]^4 + 7*Sec[a + b*x]^2)*Sqrt[Sin[2*(a +
b*x)]])/(84*b)

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\csc \left (b x + a\right )^{2}}{{\left (\cos \left (2 \, b x + 2 \, a\right )^{2} - 1\right )} \sqrt {\sin \left (2 \, b x + 2 \, a\right )}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-csc(b*x + a)^2/((cos(2*b*x + 2*a)^2 - 1)*sqrt(sin(2*b*x + 2*a))), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 90.51, size = 154, normalized size = 2.00 \[ \frac {\sqrt {2}\, \left (-\frac {16 \sqrt {2}}{7 \sin \left (2 b x +2 a \right )^{\frac {7}{2}}}+\frac {8 \sqrt {2}\, \left (5 \sqrt {1+\sin \left (2 b x +2 a \right )}\, \sqrt {-2 \sin \left (2 b x +2 a \right )+2}\, \sqrt {-\sin \left (2 b x +2 a \right )}\, \EllipticF \left (\sqrt {1+\sin \left (2 b x +2 a \right )}, \frac {\sqrt {2}}{2}\right ) \left (\sin ^{3}\left (2 b x +2 a \right )\right )+10 \left (\sin ^{4}\left (2 b x +2 a \right )\right )-4 \left (\sin ^{2}\left (2 b x +2 a \right )\right )-6\right )}{21 \sin \left (2 b x +2 a \right )^{\frac {7}{2}} \cos \left (2 b x +2 a \right )}\right )}{16 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*2^(1/2)*(-16/7*2^(1/2)/sin(2*b*x+2*a)^(7/2)+8/21*2^(1/2)/sin(2*b*x+2*a)^(7/2)*(5*(1+sin(2*b*x+2*a))^(1/2)
*(-2*sin(2*b*x+2*a)+2)^(1/2)*(-sin(2*b*x+2*a))^(1/2)*EllipticF((1+sin(2*b*x+2*a))^(1/2),1/2*2^(1/2))*sin(2*b*x
+2*a)^3+10*sin(2*b*x+2*a)^4-4*sin(2*b*x+2*a)^2-6)/cos(2*b*x+2*a))/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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