3.2.70 \(\int \cos ^2(a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx\) [170]

Optimal. Leaf size=69 \[ \frac {3 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{10 b}-\frac {\cos (2 a+2 b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{10 b}+\frac {\sin ^{\frac {7}{2}}(2 a+2 b x)}{14 b} \]

[Out]

-3/10*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticE(cos(a+1/4*Pi+b*x),2^(1/2))/b-1/10*cos(2*b*x+2*a)
*sin(2*b*x+2*a)^(3/2)/b+1/14*sin(2*b*x+2*a)^(7/2)/b

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Rubi [A]
time = 0.04, 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 = {4382, 2715, 2719} \begin {gather*} \frac {\sin ^{\frac {7}{2}}(2 a+2 b x)}{14 b}+\frac {3 E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{10 b}-\frac {\sin ^{\frac {3}{2}}(2 a+2 b x) \cos (2 a+2 b x)}{10 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*EllipticE[a - Pi/4 + b*x, 2])/(10*b) - (Cos[2*a + 2*b*x]*Sin[2*a + 2*b*x]^(3/2))/(10*b) + Sin[2*a + 2*b*x]^
(7/2)/(14*b)

Rule 2715

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] && Integ
erQ[2*n]

Rule 2719

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

Rule 4382

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 {5}{2}}(2 a+2 b x) \, dx &=\frac {\sin ^{\frac {7}{2}}(2 a+2 b x)}{14 b}+\frac {1}{2} \int \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx\\ &=-\frac {\cos (2 a+2 b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{10 b}+\frac {\sin ^{\frac {7}{2}}(2 a+2 b x)}{14 b}+\frac {3}{10} \int \sqrt {\sin (2 a+2 b x)} \, dx\\ &=\frac {3 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{10 b}-\frac {\cos (2 a+2 b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{10 b}+\frac {\sin ^{\frac {7}{2}}(2 a+2 b x)}{14 b}\\ \end {align*}

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Mathematica [A]
time = 0.23, size = 66, normalized size = 0.96 \begin {gather*} \frac {84 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )+\sqrt {\sin (2 (a+b x))} (15 \sin (2 (a+b x))-14 \sin (4 (a+b x))-5 \sin (6 (a+b x)))}{280 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(84*EllipticE[a - Pi/4 + b*x, 2] + Sqrt[Sin[2*(a + b*x)]]*(15*Sin[2*(a + b*x)] - 14*Sin[4*(a + b*x)] - 5*Sin[6
*(a + b*x)]))/(280*b)

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Maple [B] result has leaf size over 500,000. Avoiding possible recursion issues.
time = 186.90, size = 358108730, normalized size = 5189981.59

method	result	size
default	\(\text {Expression too large to display}\)	\(358108730\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integral(-(cos(2*b*x + 2*a)^2*cos(b*x + a)^2 - cos(b*x + a)^2)*sqrt(sin(2*b*x + 2*a)), x)

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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