3.177 \(\int \cos ^3(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx\)

Optimal. Leaf size=136 \[ \frac {7 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{48 b}-\frac {7 \sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{64 b}+\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (a+b x)}{12 b}-\frac {7 \sqrt {\sin (2 a+2 b x)} \cos (a+b x)}{32 b}+\frac {7 \log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{64 b} \]

[Out]

-7/64*arcsin(cos(b*x+a)-sin(b*x+a))/b+7/64*ln(cos(b*x+a)+sin(b*x+a)+sin(2*b*x+2*a)^(1/2))/b+7/48*sin(b*x+a)*si
n(2*b*x+2*a)^(3/2)/b+1/12*cos(b*x+a)*sin(2*b*x+2*a)^(5/2)/b-7/32*cos(b*x+a)*sin(2*b*x+2*a)^(1/2)/b

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Rubi [A]  time = 0.10, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4297, 4301, 4302, 4305} \[ \frac {7 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{48 b}+\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (a+b x)}{12 b}-\frac {7 \sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{64 b}-\frac {7 \sqrt {\sin (2 a+2 b x)} \cos (a+b x)}{32 b}+\frac {7 \log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{64 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*ArcSin[Cos[a + b*x] - Sin[a + b*x]])/(64*b) + (7*Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*a + 2*b*x]]]
)/(64*b) - (7*Cos[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(32*b) + (7*Sin[a + b*x]*Sin[2*a + 2*b*x]^(3/2))/(48*b) + (
Cos[a + b*x]*Sin[2*a + 2*b*x]^(5/2))/(12*b)

Rule 4297

Int[(cos[(a_.) + (b_.)*(x_)]*(e_.))^(m_)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Simp[(e^2*(e*Cos[a
+ b*x])^(m - 2)*(g*Sin[c + d*x])^(p + 1))/(2*b*g*(m + 2*p)), x] + Dist[(e^2*(m + p - 1))/(m + 2*p), Int[(e*Co
s[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] && GtQ[m, 1] && NeQ[m + 2*p, 0] && IntegersQ[2*m, 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 \cos ^3(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx &=\frac {\cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{12 b}+\frac {7}{12} \int \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx\\ &=\frac {7 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{48 b}+\frac {\cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{12 b}+\frac {7}{16} \int \sin (a+b x) \sqrt {\sin (2 a+2 b x)} \, dx\\ &=-\frac {7 \cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{32 b}+\frac {7 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{48 b}+\frac {\cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{12 b}+\frac {7}{32} \int \frac {\cos (a+b x)}{\sqrt {\sin (2 a+2 b x)}} \, dx\\ &=-\frac {7 \sin ^{-1}(\cos (a+b x)-\sin (a+b x))}{64 b}+\frac {7 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{64 b}-\frac {7 \cos (a+b x) \sqrt {\sin (2 a+2 b x)}}{32 b}+\frac {7 \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{48 b}+\frac {\cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{12 b}\\ \end {align*}

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Mathematica [A]  time = 0.34, size = 99, normalized size = 0.73 \[ \frac {-7 \sin ^{-1}(\cos (a+b x)-\sin (a+b x))-\frac {2}{3} \sqrt {\sin (2 (a+b x))} (10 \cos (a+b x)+9 \cos (3 (a+b x))+2 \cos (5 (a+b x)))+7 \log \left (\sin (a+b x)+\sqrt {\sin (2 (a+b x))}+\cos (a+b x)\right )}{64 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*ArcSin[Cos[a + b*x] - Sin[a + b*x]] + 7*Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*(a + b*x)]]] - (2*(10
*Cos[a + b*x] + 9*Cos[3*(a + b*x)] + 2*Cos[5*(a + b*x)])*Sqrt[Sin[2*(a + b*x)]])/3)/(64*b)

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fricas [B]  time = 0.46, size = 291, normalized size = 2.14 \[ -\frac {8 \, \sqrt {2} {\left (32 \, \cos \left (b x + a\right )^{5} - 4 \, \cos \left (b x + a\right )^{3} - 7 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - 42 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) + 42 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) + 21 \, \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{768 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/768*(8*sqrt(2)*(32*cos(b*x + a)^5 - 4*cos(b*x + a)^3 - 7*cos(b*x + a))*sqrt(cos(b*x + a)*sin(b*x + a)) - 42
*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)) + 42*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))) + 21*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(cos(b*x+a)^3*sin(2*b*x+2*a)^(3/2),x, algorithm="giac")

[Out]

Timed out

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maple [F(-1)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{3}\left (b x +a \right )\right ) \left (\sin ^{\frac {3}{2}}\left (2 b x +2 a \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \cos \left (b x + a\right )^{3} \sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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