3.2.6 \(\int \csc ^2(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx\) [106]

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

[Out]

-2/3*(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-2/5*cos(2*b*x+2*a)*s
in(2*b*x+2*a)^(5/2)/b+1/5*csc(b*x+a)^2*sin(2*b*x+2*a)^(9/2)/b-2/3*cos(2*b*x+2*a)*sin(2*b*x+2*a)^(1/2)/b

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Rubi [A]
time = 0.05, antiderivative size = 106, normalized size of antiderivative = 1.00, 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 = {4385, 2715, 2720} \begin {gather*} \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} \end {gather*}

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 2715

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] && Integ
erQ[2*n]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 4385

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 \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*}

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Mathematica [A]
time = 0.19, size = 76, normalized size = 0.72 \begin {gather*} \frac {20 F\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \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))}} \end {gather*}

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 = 14.15, size = 139, normalized size = 1.31

method	result	size
default	\(\frac {4 \sqrt {2}\, \left (\frac {\sqrt {2}\, \left (\sin ^{\frac {5}{2}}\left (2 x b +2 a \right )\right )}{20}+\frac {\sqrt {2}\, \left (\sqrt {\sin \left (2 x b +2 a \right )+1}\, \sqrt {-2 \sin \left (2 x b +2 a \right )+2}\, \sqrt {-\sin \left (2 x b +2 a \right )}\, \EllipticF \left (\sqrt {\sin \left (2 x b +2 a \right )+1}, \frac {\sqrt {2}}{2}\right )+2 \left (\sin ^{3}\left (2 x b +2 a \right )\right )-2 \sin \left (2 x b +2 a \right )\right )}{24 \cos \left (2 x b +2 a \right ) \sqrt {\sin \left (2 x b +2 a \right )}}\right )}{b}\)	\(139\)

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,method=_RETURNVERBOSE)

[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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\sin \left (2\,a+2\,b\,x\right )}^{7/2}}{{\sin \left (a+b\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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