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

Optimal. Leaf size=110 \[ \frac{\sin (a+b x) \sin ^{\frac{3}{2}}(2 a+2 b x)}{2 b}-\frac{3 \sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{8 b}-\frac{3 \sqrt{\sin (2 a+2 b x)} \cos (a+b x)}{4 b}+\frac{3 \log \left (\sin (a+b x)+\sqrt{\sin (2 a+2 b x)}+\cos (a+b x)\right )}{8 b} \]

[Out]

(-3*ArcSin[Cos[a + b*x] - Sin[a + b*x]])/(8*b) + (3*Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*a + 2*b*x]]])
/(8*b) - (3*Cos[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(4*b) + (Sin[a + b*x]*Sin[2*a + 2*b*x]^(3/2))/(2*b)

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Rubi [A]  time = 0.095278, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4308, 4301, 4302, 4305} \[ \frac{\sin (a+b x) \sin ^{\frac{3}{2}}(2 a+2 b x)}{2 b}-\frac{3 \sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{8 b}-\frac{3 \sqrt{\sin (2 a+2 b x)} \cos (a+b x)}{4 b}+\frac{3 \log \left (\sin (a+b x)+\sqrt{\sin (2 a+2 b x)}+\cos (a+b x)\right )}{8 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*ArcSin[Cos[a + b*x] - Sin[a + b*x]])/(8*b) + (3*Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*a + 2*b*x]]])
/(8*b) - (3*Cos[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(4*b) + (Sin[a + b*x]*Sin[2*a + 2*b*x]^(3/2))/(2*b)

Rule 4308

Int[((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_)/sin[(a_.) + (b_.)*(x_)], x_Symbol] :> Dist[2*g, Int[Cos[a + b*x]*(g*S
in[c + d*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, g, p}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ
[p] && IntegerQ[2*p]

Rule 4301

Int[cos[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(2*Sin[a + b*x]*(g*Sin[c +
d*x])^p)/(d*(2*p + 1)), x] + Dist[(2*p*g)/(2*p + 1), Int[Sin[a + b*x]*(g*Sin[c + d*x])^(p - 1), x], x] /; Fre
eQ[{a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ[p] && GtQ[p, 0] && IntegerQ[2*p]

Rule 4302

Int[sin[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(-2*Cos[a + b*x]*(g*Sin[c
+ d*x])^p)/(d*(2*p + 1)), x] + Dist[(2*p*g)/(2*p + 1), Int[Cos[a + b*x]*(g*Sin[c + d*x])^(p - 1), x], x] /; Fr
eeQ[{a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ[p] && GtQ[p, 0] && IntegerQ[2*p]

Rule 4305

Int[cos[(a_.) + (b_.)*(x_)]/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> -Simp[ArcSin[Cos[a + b*x] - Sin[a + b*
x]]/d, x] + Simp[Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[c + d*x]]]/d, x] /; FreeQ[{a, b, c, d}, x] && EqQ[
b*c - a*d, 0] && EqQ[d/b, 2]

Rubi steps

\begin{align*} \int \csc (a+b x) \sin ^{\frac{5}{2}}(2 a+2 b x) \, dx &=2 \int \cos (a+b x) \sin ^{\frac{3}{2}}(2 a+2 b x) \, dx\\ &=\frac{\sin (a+b x) \sin ^{\frac{3}{2}}(2 a+2 b x)}{2 b}+\frac{3}{2} \int \sin (a+b x) \sqrt{\sin (2 a+2 b x)} \, dx\\ &=-\frac{3 \cos (a+b x) \sqrt{\sin (2 a+2 b x)}}{4 b}+\frac{\sin (a+b x) \sin ^{\frac{3}{2}}(2 a+2 b x)}{2 b}+\frac{3}{4} \int \frac{\cos (a+b x)}{\sqrt{\sin (2 a+2 b x)}} \, dx\\ &=-\frac{3 \sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{8 b}+\frac{3 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt{\sin (2 a+2 b x)}\right )}{8 b}-\frac{3 \cos (a+b x) \sqrt{\sin (2 a+2 b x)}}{4 b}+\frac{\sin (a+b x) \sin ^{\frac{3}{2}}(2 a+2 b x)}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.190355, size = 86, normalized size = 0.78 \[ \frac{3 \left (\log \left (\sin (a+b x)+\sqrt{\sin (2 (a+b x))}+\cos (a+b x)\right )-\sin ^{-1}(\cos (a+b x)-\sin (a+b x))\right )-2 \sqrt{\sin (2 (a+b x))} (2 \cos (a+b x)+\cos (3 (a+b x)))}{8 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(-ArcSin[Cos[a + b*x] - Sin[a + b*x]] + Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*(a + b*x)]]]) - 2*(2*C
os[a + b*x] + Cos[3*(a + b*x)])*Sqrt[Sin[2*(a + b*x)]])/(8*b)

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Maple [C]  time = 2.911, size = 243, normalized size = 2.2 \begin{align*} -{\frac{8}{3\,b}\sqrt{-{\tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \left ( \left ( \tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}-1 \right ) ^{-1}}} \left ( \sqrt{\tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) +1}\sqrt{-2\,\tan \left ( 1/2\,bx+a/2 \right ) +2}\sqrt{-\tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) }{\it EllipticF} \left ( \sqrt{\tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) +1},{\frac{\sqrt{2}}{2}} \right ) \left ( \tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}-\sqrt{\tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) +1}\sqrt{-2\,\tan \left ( 1/2\,bx+a/2 \right ) +2}\sqrt{-\tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) }{\it EllipticF} \left ( \sqrt{\tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) +1},{\frac{\sqrt{2}}{2}} \right ) +2\, \left ( \tan \left ( 1/2\,bx+a/2 \right ) \right ) ^{3}+2\,\tan \left ( 1/2\,bx+a/2 \right ) \right ){\frac{1}{\sqrt{\tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \left ( \left ( \tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{2}-1 \right ) }}}{\frac{1}{\sqrt{ \left ( \tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) \right ) ^{3}-\tan \left ({\frac{bx}{2}}+{\frac{a}{2}} \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/3/b*(-tan(1/2*b*x+1/2*a)/(tan(1/2*b*x+1/2*a)^2-1))^(1/2)*((tan(1/2*b*x+1/2*a)+1)^(1/2)*(-2*tan(1/2*b*x+1/2*
a)+2)^(1/2)*(-tan(1/2*b*x+1/2*a))^(1/2)*EllipticF((tan(1/2*b*x+1/2*a)+1)^(1/2),1/2*2^(1/2))*tan(1/2*b*x+1/2*a)
^2-(tan(1/2*b*x+1/2*a)+1)^(1/2)*(-2*tan(1/2*b*x+1/2*a)+2)^(1/2)*(-tan(1/2*b*x+1/2*a))^(1/2)*EllipticF((tan(1/2
*b*x+1/2*a)+1)^(1/2),1/2*2^(1/2))+2*tan(1/2*b*x+1/2*a)^3+2*tan(1/2*b*x+1/2*a))/(tan(1/2*b*x+1/2*a)*(tan(1/2*b*
x+1/2*a)^2-1))^(1/2)/(tan(1/2*b*x+1/2*a)^3-tan(1/2*b*x+1/2*a))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \csc \left (b x + a\right ) \sin \left (2 \, b x + 2 \, a\right )^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 0.55652, size = 768, normalized size = 6.98 \begin{align*} -\frac{8 \, \sqrt{2}{\left (4 \, \cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \sqrt{\cos \left (b x + a\right ) \sin \left (b x + a\right )} - 6 \, \arctan \left (-\frac{\sqrt{2} \sqrt{\cos \left (b x + a\right ) \sin \left (b x + a\right )}{\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) + 6 \, \arctan \left (-\frac{2 \, \sqrt{2} \sqrt{\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) + 3 \, \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt{2}{\left (4 \, \cos \left (b x + a\right )^{3} -{\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt{\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{32 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(8*sqrt(2)*(4*cos(b*x + a)^3 - cos(b*x + a))*sqrt(cos(b*x + a)*sin(b*x + a)) - 6*arctan(-(sqrt(2)*sqrt(c
os(b*x + a)*sin(b*x + a))*(cos(b*x + a) - sin(b*x + a)) + cos(b*x + a)*sin(b*x + a))/(cos(b*x + a)^2 + 2*cos(b
*x + a)*sin(b*x + a) - 1)) + 6*arctan(-(2*sqrt(2)*sqrt(cos(b*x + a)*sin(b*x + a)) - cos(b*x + a) - sin(b*x + a
))/(cos(b*x + a) - sin(b*x + a))) + 3*log(-32*cos(b*x + a)^4 + 4*sqrt(2)*(4*cos(b*x + a)^3 - (4*cos(b*x + a)^2
+ 1)*sin(b*x + a) - 5*cos(b*x + a))*sqrt(cos(b*x + a)*sin(b*x + a)) + 32*cos(b*x + a)^2 + 16*cos(b*x + a)*sin
(b*x + a) + 1))/b

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \csc \left (b x + a\right ) \sin \left (2 \, b x + 2 \, a\right )^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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