3.113 \(\int \frac{\csc ^2(a+b x)}{\sin ^{\frac{7}{2}}(2 a+2 b x)} \, dx\)

Optimal. Leaf size=106 \[ -\frac{14 E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{15 b}-\frac{14 \cos (2 a+2 b x)}{45 b \sin ^{\frac{5}{2}}(2 a+2 b x)}-\frac{14 \cos (2 a+2 b x)}{15 b \sqrt{\sin (2 a+2 b x)}}-\frac{\csc ^2(a+b x)}{9 b \sin ^{\frac{5}{2}}(2 a+2 b x)} \]

[Out]

(-14*EllipticE[a - Pi/4 + b*x, 2])/(15*b) - (14*Cos[2*a + 2*b*x])/(45*b*Sin[2*a + 2*b*x]^(5/2)) - Csc[a + b*x]
^2/(9*b*Sin[2*a + 2*b*x]^(5/2)) - (14*Cos[2*a + 2*b*x])/(15*b*Sqrt[Sin[2*a + 2*b*x]])

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Rubi [A]  time = 0.0580496, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {4300, 2636, 2639} \[ -\frac{14 E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{15 b}-\frac{14 \cos (2 a+2 b x)}{45 b \sin ^{\frac{5}{2}}(2 a+2 b x)}-\frac{14 \cos (2 a+2 b x)}{15 b \sqrt{\sin (2 a+2 b x)}}-\frac{\csc ^2(a+b x)}{9 b \sin ^{\frac{5}{2}}(2 a+2 b x)} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^2/Sin[2*a + 2*b*x]^(7/2),x]

[Out]

(-14*EllipticE[a - Pi/4 + b*x, 2])/(15*b) - (14*Cos[2*a + 2*b*x])/(45*b*Sin[2*a + 2*b*x]^(5/2)) - Csc[a + b*x]
^2/(9*b*Sin[2*a + 2*b*x]^(5/2)) - (14*Cos[2*a + 2*b*x])/(15*b*Sqrt[Sin[2*a + 2*b*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]

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 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]

Rubi steps

\begin{align*} \int \frac{\csc ^2(a+b x)}{\sin ^{\frac{7}{2}}(2 a+2 b x)} \, dx &=-\frac{\csc ^2(a+b x)}{9 b \sin ^{\frac{5}{2}}(2 a+2 b x)}+\frac{14}{9} \int \frac{1}{\sin ^{\frac{7}{2}}(2 a+2 b x)} \, dx\\ &=-\frac{14 \cos (2 a+2 b x)}{45 b \sin ^{\frac{5}{2}}(2 a+2 b x)}-\frac{\csc ^2(a+b x)}{9 b \sin ^{\frac{5}{2}}(2 a+2 b x)}+\frac{14}{15} \int \frac{1}{\sin ^{\frac{3}{2}}(2 a+2 b x)} \, dx\\ &=-\frac{14 \cos (2 a+2 b x)}{45 b \sin ^{\frac{5}{2}}(2 a+2 b x)}-\frac{\csc ^2(a+b x)}{9 b \sin ^{\frac{5}{2}}(2 a+2 b x)}-\frac{14 \cos (2 a+2 b x)}{15 b \sqrt{\sin (2 a+2 b x)}}-\frac{14}{15} \int \sqrt{\sin (2 a+2 b x)} \, dx\\ &=-\frac{14 E\left (\left .a-\frac{\pi }{4}+b x\right |2\right )}{15 b}-\frac{14 \cos (2 a+2 b x)}{45 b \sin ^{\frac{5}{2}}(2 a+2 b x)}-\frac{\csc ^2(a+b x)}{9 b \sin ^{\frac{5}{2}}(2 a+2 b x)}-\frac{14 \cos (2 a+2 b x)}{15 b \sqrt{\sin (2 a+2 b x)}}\\ \end{align*}

Mathematica [A]  time = 0.814174, size = 85, normalized size = 0.8 \[ -\frac{336 E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )+\frac{(98 \cos (2 (a+b x))-28 \cos (4 (a+b x))-42 \cos (6 (a+b x))+21 \cos (8 (a+b x))-9) \csc ^2(a+b x)}{\sin ^{\frac{5}{2}}(2 (a+b x))}}{360 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^2/Sin[2*a + 2*b*x]^(7/2),x]

[Out]

-(336*EllipticE[a - Pi/4 + b*x, 2] + ((-9 + 98*Cos[2*(a + b*x)] - 28*Cos[4*(a + b*x)] - 42*Cos[6*(a + b*x)] +
21*Cos[8*(a + b*x)])*Csc[a + b*x]^2)/Sin[2*(a + b*x)]^(5/2))/(360*b)

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Maple [B]  time = 30.493, size = 240, normalized size = 2.3 \begin{align*}{\frac{\sqrt{2}}{32\,b} \left ( -{\frac{32\,\sqrt{2}}{9} \left ( \sin \left ( 2\,bx+2\,a \right ) \right ) ^{-{\frac{9}{2}}}}+{\frac{16\,\sqrt{2}}{45\,\cos \left ( 2\,bx+2\,a \right ) } \left ( 42\,\sqrt{\sin \left ( 2\,bx+2\,a \right ) +1}\sqrt{-2\,\sin \left ( 2\,bx+2\,a \right ) +2}\sqrt{-\sin \left ( 2\,bx+2\,a \right ) } \left ( \sin \left ( 2\,bx+2\,a \right ) \right ) ^{4}{\it EllipticE} \left ( \sqrt{\sin \left ( 2\,bx+2\,a \right ) +1},1/2\,\sqrt{2} \right ) -21\,\sqrt{\sin \left ( 2\,bx+2\,a \right ) +1}\sqrt{-2\,\sin \left ( 2\,bx+2\,a \right ) +2}\sqrt{-\sin \left ( 2\,bx+2\,a \right ) } \left ( \sin \left ( 2\,bx+2\,a \right ) \right ) ^{4}{\it EllipticF} \left ( \sqrt{\sin \left ( 2\,bx+2\,a \right ) +1},1/2\,\sqrt{2} \right ) +42\, \left ( \sin \left ( 2\,bx+2\,a \right ) \right ) ^{6}-28\, \left ( \sin \left ( 2\,bx+2\,a \right ) \right ) ^{4}-4\, \left ( \sin \left ( 2\,bx+2\,a \right ) \right ) ^{2}-10 \right ) \left ( \sin \left ( 2\,bx+2\,a \right ) \right ) ^{-{\frac{9}{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)^2/sin(2*b*x+2*a)^(7/2),x)

[Out]

1/32*2^(1/2)*(-32/9*2^(1/2)/sin(2*b*x+2*a)^(9/2)+16/45*2^(1/2)/sin(2*b*x+2*a)^(9/2)*(42*(sin(2*b*x+2*a)+1)^(1/
2)*(-2*sin(2*b*x+2*a)+2)^(1/2)*(-sin(2*b*x+2*a))^(1/2)*sin(2*b*x+2*a)^4*EllipticE((sin(2*b*x+2*a)+1)^(1/2),1/2
*2^(1/2))-21*(sin(2*b*x+2*a)+1)^(1/2)*(-2*sin(2*b*x+2*a)+2)^(1/2)*(-sin(2*b*x+2*a))^(1/2)*sin(2*b*x+2*a)^4*Ell
ipticF((sin(2*b*x+2*a)+1)^(1/2),1/2*2^(1/2))+42*sin(2*b*x+2*a)^6-28*sin(2*b*x+2*a)^4-4*sin(2*b*x+2*a)^2-10)/co
s(2*b*x+2*a))/b

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (b x + a\right )^{2}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^2/sin(2*b*x+2*a)^(7/2),x, algorithm="maxima")

[Out]

integrate(csc(b*x + a)^2/sin(2*b*x + 2*a)^(7/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\csc \left (b x + a\right )^{2} \sqrt{\sin \left (2 \, b x + 2 \, a\right )}}{\cos \left (2 \, b x + 2 \, a\right )^{4} - 2 \, \cos \left (2 \, b x + 2 \, a\right )^{2} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^2/sin(2*b*x+2*a)^(7/2),x, algorithm="fricas")

[Out]

integral(csc(b*x + a)^2*sqrt(sin(2*b*x + 2*a))/(cos(2*b*x + 2*a)^4 - 2*cos(2*b*x + 2*a)^2 + 1), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)**2/sin(2*b*x+2*a)**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (b x + a\right )^{2}}{\sin \left (2 \, b x + 2 \, a\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^2/sin(2*b*x+2*a)^(7/2),x, algorithm="giac")

[Out]

integrate(csc(b*x + a)^2/sin(2*b*x + 2*a)^(7/2), x)