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

Optimal. Leaf size=136 \[ \frac {5 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{24 b}-\frac {5 \sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{32 b}-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (a+b x)}{6 b}-\frac {5 \sqrt {\sin (2 a+2 b x)} \cos (a+b x)}{16 b}+\frac {5 \log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{32 b} \]

[Out]

-5/32*arcsin(cos(b*x+a)-sin(b*x+a))/b+5/32*ln(cos(b*x+a)+sin(b*x+a)+sin(2*b*x+2*a)^(1/2))/b+5/24*sin(b*x+a)*si
n(2*b*x+2*a)^(3/2)/b-1/6*cos(b*x+a)*sin(2*b*x+2*a)^(5/2)/b-5/16*cos(b*x+a)*sin(2*b*x+2*a)^(1/2)/b

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Rubi [A]  time = 0.09, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4302, 4301, 4305} \[ \frac {5 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{24 b}-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (a+b x)}{6 b}-\frac {5 \sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{32 b}-\frac {5 \sqrt {\sin (2 a+2 b x)} \cos (a+b x)}{16 b}+\frac {5 \log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{32 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*ArcSin[Cos[a + b*x] - Sin[a + b*x]])/(32*b) + (5*Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*a + 2*b*x]]]
)/(32*b) - (5*Cos[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(16*b) + (5*Sin[a + b*x]*Sin[2*a + 2*b*x]^(3/2))/(24*b) - (
Cos[a + b*x]*Sin[2*a + 2*b*x]^(5/2))/(6*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]

Rubi steps

\begin {align*} \int \sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx &=-\frac {\cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b}+\frac {5}{6} \int \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx\\ &=\frac {5 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{24 b}-\frac {\cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b}+\frac {5}{8} \int \sin (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx\\ &=-\frac {5 \cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{16 b}+\frac {5 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{24 b}-\frac {\cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b}+\frac {5}{16} \int \frac {\cos (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx\\ &=-\frac {5 \sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{32 b}+\frac {5 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{32 b}-\frac {5 \cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{16 b}+\frac {5 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{24 b}-\frac {\cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b}\\ \end {align*}

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Mathematica [A]  time = 0.30, size = 98, normalized size = 0.72 \[ \frac {15 \left (\log \left (\sin (a+b x)+\sqrt {\sin (2 (a+b x))}+\cos (a+b x)\right )-\sin ^{-1}(\cos (a+b x)-\sin (a+b x))\right )-2 \sqrt {\sin (2 (a+b x))} (14 \cos (a+b x)+3 \cos (3 (a+b x))-2 \cos (5 (a+b x)))}{96 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*(-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*(14
*Cos[a + b*x] + 3*Cos[3*(a + b*x)] - 2*Cos[5*(a + b*x)])*Sqrt[Sin[2*(a + b*x)]])/(96*b)

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fricas [B]  time = 0.62, size = 291, normalized size = 2.14 \[ \frac {8 \, \sqrt {2} {\left (32 \, \cos \left (b x + a\right )^{5} - 52 \, \cos \left (b x + a\right )^{3} + 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 30 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) - 30 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) - 15 \, \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{384 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/384*(8*sqrt(2)*(32*cos(b*x + a)^5 - 52*cos(b*x + a)^3 + 5*cos(b*x + a))*sqrt(cos(b*x + a)*sin(b*x + a)) + 30
*arctan(-(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))/(
cos(b*x + a)^2 + 2*cos(b*x + a)*sin(b*x + a) - 1)) - 30*arctan(-(2*sqrt(2)*sqrt(cos(b*x + a)*sin(b*x + a)) - c
os(b*x + a) - sin(b*x + a))/(cos(b*x + a) - sin(b*x + a))) - 15*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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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(sin(b*x+a)*sin(2*b*x+2*a)^(5/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 89.38, size = 183661406, normalized size = 1350451.52 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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