\(\int \cos ^2(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx\) [580]

Optimal result

Integrand size = 22, antiderivative size = 69 \[ \int \cos ^2(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\frac {\operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right )}{6 b}-\frac {\cos (2 a+2 b x) \sqrt {\sin (2 a+2 b x)}}{6 b}+\frac {\sin ^{\frac {5}{2}}(2 a+2 b x)}{10 b} \] Output:

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

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.10 \[ \int \cos ^2(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\frac {20 \operatorname {EllipticF}\left (a-\frac {\pi }{4}+b x,2\right ) \sqrt {\sin (2 (a+b x))}+9 \sin (2 (a+b x))-10 \sin (4 (a+b x))-3 \sin (6 (a+b x))}{120 b \sqrt {\sin (2 (a+b x))}} \] Input:

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

Output:

(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) 
]])
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {3042, 4785, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (2 a+2 b x)^{3/2} \cos (a+b x)^2dx\)

\(\Big \downarrow \) 4785

\(\displaystyle \frac {1}{2} \int \sin ^{\frac {3}{2}}(2 a+2 b x)dx+\frac {\sin ^{\frac {5}{2}}(2 a+2 b x)}{10 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \int \sin (2 a+2 b x)^{3/2}dx+\frac {\sin ^{\frac {5}{2}}(2 a+2 b x)}{10 b}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {\sqrt {\sin (2 a+2 b x)} \cos (2 a+2 b x)}{3 b}\right )+\frac {\sin ^{\frac {5}{2}}(2 a+2 b x)}{10 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin (2 a+2 b x)}}dx-\frac {\sqrt {\sin (2 a+2 b x)} \cos (2 a+2 b x)}{3 b}\right )+\frac {\sin ^{\frac {5}{2}}(2 a+2 b x)}{10 b}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\sin ^{\frac {5}{2}}(2 a+2 b x)}{10 b}+\frac {1}{2} \left (\frac {\operatorname {EllipticF}\left (a+b x-\frac {\pi }{4},2\right )}{3 b}-\frac {\sqrt {\sin (2 a+2 b x)} \cos (2 a+2 b x)}{3 b}\right )\)

Input:

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

Output:

(EllipticF[a - Pi/4 + b*x, 2]/(3*b) - (Cos[2*a + 2*b*x]*Sqrt[Sin[2*a + 2*b 
*x]])/(3*b))/2 + Sin[2*a + 2*b*x]^(5/2)/(10*b)
 

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 [B] (warning: unable to verify)

result has leaf size over 500,000. Avoiding possible recursion issues.

Time = 22.22 (sec) , antiderivative size = 91497799, normalized size of antiderivative = 1326055.06

method	result	size
default	\(\text {Expression too large to display}\)	\(91497799\)

Input:

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

Output:

result too large to display
 

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \cos ^2(a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x) \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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