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]

-3/8*arcsin(cos(b*x+a)-sin(b*x+a))/b+3/8*ln(cos(b*x+a)+sin(b*x+a)+sin(2*b*x+2*a)^(1/2))/b+1/2*sin(b*x+a)*sin(2
*b*x+2*a)^(3/2)/b-3/4*cos(b*x+a)*sin(2*b*x+2*a)^(1/2)/b

________________________________________________________________________________________

Rubi [A]  time = 0.10, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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]

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]

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 = 0.20, 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)

________________________________________________________________________________________

fricas [B]  time = 0.46, size = 281, normalized size = 2.55 \[ -\frac {8 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - 6 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) + 6 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) + 3 \, \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{32 \, b} \]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

maple [C]  time = 20.55, size = 243, normalized size = 2.21 \[ -\frac {8 \sqrt {-\frac {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1}}\, \left (\sqrt {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )+2}\, \sqrt {-\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}\, \EllipticF \left (\sqrt {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1}, \frac {\sqrt {2}}{2}\right ) \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-\sqrt {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1}\, \sqrt {-2 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )+2}\, \sqrt {-\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}\, \EllipticF \left (\sqrt {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1}, \frac {\sqrt {2}}{2}\right )+2 \left (\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+2 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b \sqrt {\tan \left (\frac {b x}{2}+\frac {a}{2}\right ) \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}\, \sqrt {\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )-\tan \left (\frac {b x}{2}+\frac {a}{2}\right )}} \]

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)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)*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)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________