\(\int \cos ^2(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx\) [169]

Optimal result

Integrand size = 22, antiderivative size = 98 \[ \int \cos ^2(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx=\frac {5 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right )}{42 b}-\frac {5 \cos (2 a+2 b x) \sqrt {\sin (2 a+2 b x)}}{42 b}-\frac {\cos (2 a+2 b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{14 b}+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x)}{18 b} \]

[Out]

-5/42*(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-1/14*cos(2*b*x+2*a)
*sin(2*b*x+2*a)^(5/2)/b+1/18*sin(2*b*x+2*a)^(9/2)/b-5/42*cos(2*b*x+2*a)*sin(2*b*x+2*a)^(1/2)/b

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {4382, 2715, 2720} \[ \int \cos ^2(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx=\frac {\sin ^{\frac {9}{2}}(2 a+2 b x)}{18 b}+\frac {5 \operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right )}{42 b}-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (2 a+2 b x)}{14 b}-\frac {5 \sqrt {\sin (2 a+2 b x)} \cos (2 a+2 b x)}{42 b} \]

[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 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 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(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*} \text {integral}& = \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 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,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] (verified)

Time = 0.72 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.98 \[ \int \cos ^2(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx=\frac {240 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sqrt {\sin (2 (a+b x))}+70 \sin (2 (a+b x))-156 \sin (4 (a+b x))-35 \sin (6 (a+b x))+18 \sin (8 (a+b x))+7 \sin (10 (a+b x))}{2016 b \sqrt {\sin (2 (a+b x))}} \]

[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)]])

Maple [F(-1)]

Timed out.

\[\int \cos \left (x b +a \right )^{2} \sin \left (2 x b +2 a \right )^{\frac {7}{2}}d x\]

[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)

Fricas [F]

\[ \int \cos ^2(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx=\int { \cos \left (b x + a\right )^{2} \sin \left (2 \, b x + 2 \, a\right )^{\frac {7}{2}} \,d x } \]

[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)

Sympy [F(-1)]

Timed out. \[ \int \cos ^2(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \cos ^2(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx=\int { \cos \left (b x + a\right )^{2} \sin \left (2 \, b x + 2 \, a\right )^{\frac {7}{2}} \,d x } \]

[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)

Giac [F]

\[ \int \cos ^2(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx=\int { \cos \left (b x + a\right )^{2} \sin \left (2 \, b x + 2 \, a\right )^{\frac {7}{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \cos ^2(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx=\int {\cos \left (a+b\,x\right )}^2\,{\sin \left (2\,a+2\,b\,x\right )}^{7/2} \,d x \]

[In]

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

[Out]

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