3.169 \(\int \cos ^2(a+b x) \sin ^{\frac{7}{2}}(2 a+2 b x) \, dx\)

Optimal. Leaf size=98 \[ \frac{5 \text{EllipticF}\left (a+b x-\frac{\pi }{4},2\right )}{42 b}+\frac{\sin ^{\frac{9}{2}}(2 a+2 b x)}{18 b}-\frac{\sin ^{\frac{5}{2}}(2 a+2 b x) \cos (2 a+2 b x)}{14 b}-\frac{5 \sqrt{\sin (2 a+2 b x)} \cos (2 a+2 b x)}{42 b} \]

[Out]

(5*EllipticF[a - Pi/4 + b*x, 2])/(42*b) - (5*Cos[2*a + 2*b*x]*Sqrt[Sin[2*a + 2*b*x]])/(42*b) - (Cos[2*a + 2*b*
x]*Sin[2*a + 2*b*x]^(5/2))/(14*b) + Sin[2*a + 2*b*x]^(9/2)/(18*b)

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Rubi [A]  time = 0.0573313, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {4297, 2635, 2641} \[ \frac{\sin ^{\frac{9}{2}}(2 a+2 b x)}{18 b}+\frac{5 F\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{42 b}-\frac{\sin ^{\frac{5}{2}}(2 a+2 b x) \cos (2 a+2 b x)}{14 b}-\frac{5 \sqrt{\sin (2 a+2 b x)} \cos (2 a+2 b x)}{42 b} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^2*Sin[2*a + 2*b*x]^(7/2),x]

[Out]

(5*EllipticF[a - Pi/4 + b*x, 2])/(42*b) - (5*Cos[2*a + 2*b*x]*Sqrt[Sin[2*a + 2*b*x]])/(42*b) - (Cos[2*a + 2*b*
x]*Sin[2*a + 2*b*x]^(5/2))/(14*b) + Sin[2*a + 2*b*x]^(9/2)/(18*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 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]

Rubi steps

\begin{align*} \int \cos ^2(a+b x) \sin ^{\frac{7}{2}}(2 a+2 b x) \, dx &=\frac{\sin ^{\frac{9}{2}}(2 a+2 b x)}{18 b}+\frac{1}{2} \int \sin ^{\frac{7}{2}}(2 a+2 b x) \, dx\\ &=-\frac{\cos (2 a+2 b x) \sin ^{\frac{5}{2}}(2 a+2 b x)}{14 b}+\frac{\sin ^{\frac{9}{2}}(2 a+2 b x)}{18 b}+\frac{5}{14} \int \sin ^{\frac{3}{2}}(2 a+2 b x) \, dx\\ &=-\frac{5 \cos (2 a+2 b x) \sqrt{\sin (2 a+2 b x)}}{42 b}-\frac{\cos (2 a+2 b x) \sin ^{\frac{5}{2}}(2 a+2 b x)}{14 b}+\frac{\sin ^{\frac{9}{2}}(2 a+2 b x)}{18 b}+\frac{5}{42} \int \frac{1}{\sqrt{\sin (2 a+2 b x)}} \, dx\\ &=\frac{5 F\left (\left .a-\frac{\pi }{4}+b x\right |2\right )}{42 b}-\frac{5 \cos (2 a+2 b x) \sqrt{\sin (2 a+2 b x)}}{42 b}-\frac{\cos (2 a+2 b x) \sin ^{\frac{5}{2}}(2 a+2 b x)}{14 b}+\frac{\sin ^{\frac{9}{2}}(2 a+2 b x)}{18 b}\\ \end{align*}

Mathematica [A]  time = 0.384852, size = 96, normalized size = 0.98 \[ \frac{240 \sqrt{\sin (2 (a+b x))} \text{EllipticF}\left (a+b x-\frac{\pi }{4},2\right )+70 \sin (2 (a+b x))-156 \sin (4 (a+b x))-35 \sin (6 (a+b x))+18 \sin (8 (a+b x))+7 \sin (10 (a+b x))}{2016 b \sqrt{\sin (2 (a+b x))}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]^2*Sin[2*a + 2*b*x]^(7/2),x]

[Out]

(240*EllipticF[a - Pi/4 + b*x, 2]*Sqrt[Sin[2*(a + b*x)]] + 70*Sin[2*(a + b*x)] - 156*Sin[4*(a + b*x)] - 35*Sin
[6*(a + b*x)] + 18*Sin[8*(a + b*x)] + 7*Sin[10*(a + b*x)])/(2016*b*Sqrt[Sin[2*(a + b*x)]])

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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( bx+a \right ) \right ) ^{2} \left ( \sin \left ( 2\,bx+2\,a \right ) \right ) ^{{\frac{7}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)^2*sin(2*b*x+2*a)^(7/2),x)

[Out]

int(cos(b*x+a)^2*sin(2*b*x+2*a)^(7/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \cos \left (b x + a\right )^{2} \sin \left (2 \, b x + 2 \, a\right )^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2*sin(2*b*x+2*a)^(7/2),x, algorithm="maxima")

[Out]

integrate(cos(b*x + a)^2*sin(2*b*x + 2*a)^(7/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (\cos \left (2 \, b x + 2 \, a\right )^{2} \cos \left (b x + a\right )^{2} - \cos \left (b x + a\right )^{2}\right )} \sin \left (2 \, b x + 2 \, a\right )^{\frac{3}{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2*sin(2*b*x+2*a)^(7/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)**2*sin(2*b*x+2*a)**(7/2),x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^2*sin(2*b*x+2*a)^(7/2),x, algorithm="giac")

[Out]

Timed out