3.1.74 \(\int \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx\) [74]

Optimal. Leaf size=110 \[ -\frac {3 \text {ArcSin}(\cos (a+b x)-\sin (a+b x))}{16 b}-\frac {3 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{16 b}+\frac {3 \sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{8 b}-\frac {\cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b} \]

[Out]

-3/16*arcsin(cos(b*x+a)-sin(b*x+a))/b-3/16*ln(cos(b*x+a)+sin(b*x+a)+sin(2*b*x+2*a)^(1/2))/b-1/4*cos(b*x+a)*sin
(2*b*x+2*a)^(3/2)/b+3/8*sin(b*x+a)*sin(2*b*x+2*a)^(1/2)/b

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Rubi [A]
time = 0.05, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4387, 4386, 4391} \begin {gather*} -\frac {3 \text {ArcSin}(\cos (a+b x)-\sin (a+b x))}{16 b}+\frac {3 \sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{8 b}-\frac {\sin ^{\frac {3}{2}}(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 )}{16 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]*Sin[2*a + 2*b*x]^(3/2),x]

[Out]

(-3*ArcSin[Cos[a + b*x] - Sin[a + b*x]])/(16*b) - (3*Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*a + 2*b*x]]]
)/(16*b) + (3*Sin[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(8*b) - (Cos[a + b*x]*Sin[2*a + 2*b*x]^(3/2))/(4*b)

Rule 4386

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 4387

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 4391

Int[sin[(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 {3}{2}}(2 a+2 b x) \, dx &=-\frac {\cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}+\frac {3}{4} \int \cos (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx\\ &=\frac {3 \sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{8 b}-\frac {\cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}+\frac {3}{8} \int \frac {\sin (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx\\ &=-\frac {3 \sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{16 b}-\frac {3 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{16 b}+\frac {3 \sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{8 b}-\frac {\cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{4 b}\\ \end {align*}

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Mathematica [A]
time = 0.22, size = 86, normalized size = 0.78 \begin {gather*} \frac {-3 \left (\text {ArcSin}(\cos (a+b x)-\sin (a+b x))+\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right )\right )+2 \sqrt {\sin (2 (a+b x))} (2 \sin (a+b x)-\sin (3 (a+b x)))}{16 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]*Sin[2*a + 2*b*x]^(3/2),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*Sqrt
[Sin[2*(a + b*x)]]*(2*Sin[a + b*x] - Sin[3*(a + b*x)]))/(16*b)

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Maple [B] result has leaf size over 500,000. Avoiding possible recursion issues.
time = 28.84, size = 73720488, normalized size = 670186.25

method	result	size
default	\(\text {Expression too large to display}\)	\(73720488\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)*sin(2*b*x+2*a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

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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(sin(b*x+a)*sin(2*b*x+2*a)^(3/2),x, algorithm="maxima")

[Out]

integrate(sin(2*b*x + 2*a)^(3/2)*sin(b*x + a), x)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 280 vs. \(2 (96) = 192\).
time = 6.04, size = 280, normalized size = 2.55 \begin {gather*} -\frac {8 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{2} - 3\right )} \sqrt {\cos \left (b x + a\right ) \sin \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 )}{64 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)*sin(2*b*x+2*a)^(3/2),x, algorithm="fricas")

[Out]

-1/64*(8*sqrt(2)*(4*cos(b*x + a)^2 - 3)*sqrt(cos(b*x + a)*sin(b*x + a))*sin(b*x + a) - 6*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)) + 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)*s
in(b*x + a) + 1))/b

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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(sin(b*x+a)*sin(2*b*x+2*a)**(3/2),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(sin(b*x+a)*sin(2*b*x+2*a)^(3/2),x, algorithm="giac")

[Out]

integrate(sin(2*b*x + 2*a)^(3/2)*sin(b*x + a), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \sin \left (a+b\,x\right )\,{\sin \left (2\,a+2\,b\,x\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)*sin(2*a + 2*b*x)^(3/2),x)

[Out]

int(sin(a + b*x)*sin(2*a + 2*b*x)^(3/2), x)

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