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

Optimal. Leaf size=106 \[ \frac{2 \text{EllipticF}\left (a+b x-\frac{\pi }{4},2\right )}{3 b}-\frac{2 \sin ^{\frac{5}{2}}(2 a+2 b x) \cos (2 a+2 b x)}{5 b}-\frac{2 \sqrt{\sin (2 a+2 b x)} \cos (2 a+2 b x)}{3 b}+\frac{\sin ^{\frac{9}{2}}(2 a+2 b x) \csc ^2(a+b x)}{5 b} \]

[Out]

(2*EllipticF[a - Pi/4 + b*x, 2])/(3*b) - (2*Cos[2*a + 2*b*x]*Sqrt[Sin[2*a + 2*b*x]])/(3*b) - (2*Cos[2*a + 2*b*
x]*Sin[2*a + 2*b*x]^(5/2))/(5*b) + (Csc[a + b*x]^2*Sin[2*a + 2*b*x]^(9/2))/(5*b)

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Rubi [A]  time = 0.059291, 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, 2635, 2641} \[ \frac{2 F\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{3 b}-\frac{2 \sin ^{\frac{5}{2}}(2 a+2 b x) \cos (2 a+2 b x)}{5 b}-\frac{2 \sqrt{\sin (2 a+2 b x)} \cos (2 a+2 b x)}{3 b}+\frac{\sin ^{\frac{9}{2}}(2 a+2 b x) \csc ^2(a+b x)}{5 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*EllipticF[a - Pi/4 + b*x, 2])/(3*b) - (2*Cos[2*a + 2*b*x]*Sqrt[Sin[2*a + 2*b*x]])/(3*b) - (2*Cos[2*a + 2*b*
x]*Sin[2*a + 2*b*x]^(5/2))/(5*b) + (Csc[a + b*x]^2*Sin[2*a + 2*b*x]^(9/2))/(5*b)

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 2635

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

Rubi steps

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

Mathematica [A]  time = 0.257858, size = 76, normalized size = 0.72 \[ \frac{20 \sqrt{\sin (2 (a+b x))} \text{EllipticF}\left (a+b x-\frac{\pi }{4},2\right )+9 \sin (2 (a+b x))-10 \sin (4 (a+b x))-3 \sin (6 (a+b x))}{30 b \sqrt{\sin (2 (a+b x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(20*EllipticF[a - Pi/4 + b*x, 2]*Sqrt[Sin[2*(a + b*x)]] + 9*Sin[2*(a + b*x)] - 10*Sin[4*(a + b*x)] - 3*Sin[6*(
a + b*x)])/(30*b*Sqrt[Sin[2*(a + b*x)]])

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Maple [A]  time = 5.161, size = 139, normalized size = 1.3 \begin{align*} 4\,{\frac{\sqrt{2}}{b} \left ( 1/20\,\sqrt{2} \left ( \sin \left ( 2\,bx+2\,a \right ) \right ) ^{5/2}+1/24\,{\frac{\sqrt{2} \left ( \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 ) }{\it EllipticF} \left ( \sqrt{\sin \left ( 2\,bx+2\,a \right ) +1},1/2\,\sqrt{2} \right ) +2\, \left ( \sin \left ( 2\,bx+2\,a \right ) \right ) ^{3}-2\,\sin \left ( 2\,bx+2\,a \right ) \right ) }{\cos \left ( 2\,bx+2\,a \right ) \sqrt{\sin \left ( 2\,bx+2\,a \right ) }}} \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]

4*2^(1/2)*(1/20*2^(1/2)*sin(2*b*x+2*a)^(5/2)+1/24*2^(1/2)*((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)*EllipticF((sin(2*b*x+2*a)+1)^(1/2),1/2*2^(1/2))+2*sin(2*b*x+2*a)^3-2*sin(2*b*x+2*a))
/cos(2*b*x+2*a)/sin(2*b*x+2*a)^(1/2))/b

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \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 (-{\left (\cos \left (2 \, b x + 2 \, a\right )^{2} - 1\right )} \csc \left (b x + a\right )^{2} \sin \left (2 \, b x + 2 \, a\right )^{\frac{3}{2}}, 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(-(cos(2*b*x + 2*a)^2 - 1)*csc(b*x + a)^2*sin(2*b*x + 2*a)^(3/2), 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(-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, algorithm="giac")

[Out]

Timed out