\(\int \cos ^2(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx\) [578]

Optimal result

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

5/42*InverseJacobiAM(a-1/4*Pi+b*x,2^(1/2))/b-5/42*cos(2*b*x+2*a)*sin(2*b*x 
+2*a)^(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
 

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.98 \[ \int \cos ^2(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx=\frac {240 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sqrt {\sin (2 (a+b x))}+70 \sin (2 (a+b x))-156 \sin (4 (a+b x))-35 \sin (6 (a+b x))+18 \sin (8 (a+b x))+7 \sin (10 (a+b x))}{2016 b \sqrt {\sin (2 (a+b x))}} \] Input:

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

Output:

(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)]])
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.10, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {3042, 4785, 3042, 3115, 3042, 3115, 3042, 3120}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (2 a+2 b x)^{7/2} \cos (a+b x)^2dx\)

\(\Big \downarrow \) 4785

\(\displaystyle \frac {1}{2} \int \sin ^{\frac {7}{2}}(2 a+2 b x)dx+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x)}{18 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \int \sin (2 a+2 b x)^{7/2}dx+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x)}{18 b}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {1}{2} \left (\frac {5}{7} \int \sin ^{\frac {3}{2}}(2 a+2 b x)dx-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (2 a+2 b x)}{7 b}\right )+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x)}{18 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\frac {5}{7} \int \sin (2 a+2 b x)^{3/2}dx-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (2 a+2 b x)}{7 b}\right )+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x)}{18 b}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {1}{2} \left (\frac {5}{7} \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {\sqrt {\sin (2 a+2 b x)} \cos (2 a+2 b x)}{3 b}\right )-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (2 a+2 b x)}{7 b}\right )+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x)}{18 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\frac {5}{7} \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {\sqrt {\sin (2 a+2 b x)} \cos (2 a+2 b x)}{3 b}\right )-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (2 a+2 b x)}{7 b}\right )+\frac {\sin ^{\frac {9}{2}}(2 a+2 b x)}{18 b}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\sin ^{\frac {9}{2}}(2 a+2 b x)}{18 b}+\frac {1}{2} \left (\frac {5}{7} \left (\frac {\operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right )}{3 b}-\frac {\sqrt {\sin (2 a+2 b x)} \cos (2 a+2 b x)}{3 b}\right )-\frac {\sin ^{\frac {5}{2}}(2 a+2 b x) \cos (2 a+2 b x)}{7 b}\right )\)

Input:

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

Output:

Sin[2*a + 2*b*x]^(9/2)/(18*b) + ((5*(EllipticF[a - Pi/4 + b*x, 2]/(3*b) - 
(Cos[2*a + 2*b*x]*Sqrt[Sin[2*a + 2*b*x]])/(3*b)))/7 - (Cos[2*a + 2*b*x]*Si 
n[2*a + 2*b*x]^(5/2))/(7*b))/2
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3115

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] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

rule 3120

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 4785

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] + Simp[e^2*((m + p - 1)/(m + 2*p))   Int[(e*Cos[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]
 

Maple [F(-1)]

Timed out.

hanged

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \cos ^2(a+b x) \sin ^{\frac {7}{2}}(2 a+2 b x) \, dx=\int \sqrt {\sin \left (2 b x +2 a \right )}\, \cos \left (b x +a \right )^{2} \sin \left (2 b x +2 a \right )^{3}d x \] Input:

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

Output:

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