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

Optimal. Leaf size=69 \[ \frac {\sin ^{\frac {5}{2}}(2 a+2 b x)}{10 b}+\frac {F\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{6 b}-\frac {\sqrt {\sin (2 a+2 b x)} \cos (2 a+2 b x)}{6 b} \]

[Out]

-1/6*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticF(cos(a+1/4*Pi+b*x),2^(1/2))/b+1/10*sin(2*b*x+2*a)^
(5/2)/b-1/6*cos(2*b*x+2*a)*sin(2*b*x+2*a)^(1/2)/b

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Rubi [A]  time = 0.05, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {4297, 2635, 2641} \[ \frac {\sin ^{\frac {5}{2}}(2 a+2 b x)}{10 b}+\frac {F\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{6 b}-\frac {\sqrt {\sin (2 a+2 b x)} \cos (2 a+2 b x)}{6 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

EllipticF[a - Pi/4 + b*x, 2]/(6*b) - (Cos[2*a + 2*b*x]*Sqrt[Sin[2*a + 2*b*x]])/(6*b) + Sin[2*a + 2*b*x]^(5/2)/
(10*b)

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

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]

Rubi steps

\begin {align*} \int \cos ^2(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx &=\frac {\sin ^{\frac {5}{2}}(2 a+2 b x)}{10 b}+\frac {1}{2} \int \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx\\ &=-\frac {\cos (2 a+2 b x) \sqrt {\sin (2 a+2 b x)}}{6 b}+\frac {\sin ^{\frac {5}{2}}(2 a+2 b x)}{10 b}+\frac {1}{6} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}} \, dx\\ &=\frac {F\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{6 b}-\frac {\cos (2 a+2 b x) \sqrt {\sin (2 a+2 b x)}}{6 b}+\frac {\sin ^{\frac {5}{2}}(2 a+2 b x)}{10 b}\\ \end {align*}

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Mathematica [A]  time = 0.35, size = 76, normalized size = 1.10 \[ \frac {9 \sin (2 (a+b x))-10 \sin (4 (a+b x))-3 \sin (6 (a+b x))+20 \sqrt {\sin (2 (a+b x))} F\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{120 b \sqrt {\sin (2 (a+b x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(20*EllipticF[a - Pi/4 + b*x, 2]*Sqrt[Sin[2*(a + b*x)]] + 9*Sin[2*(a + b*x)] - 10*Sin[4*(a + b*x)] - 3*Sin[6*(
a + b*x)])/(120*b*Sqrt[Sin[2*(a + b*x)]])

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\cos \left (b x + a\right )^{2} \sin \left (2 \, b x + 2 \, a\right )^{\frac {3}{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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maple [B]  time = 61.35, size = 174364216, normalized size = 2527017.62 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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