\(\int \cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx\) [570]

Optimal result

Integrand size = 20, antiderivative size = 136 \[ \int \cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=-\frac {5 \arcsin (\cos (a+b x)-\sin (a+b x))}{32 b}-\frac {5 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}\right )}{32 b}+\frac {5 \sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{16 b}-\frac {5 \cos (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)}{24 b}+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b} \] Output:

-5/32*arcsin(cos(b*x+a)-sin(b*x+a))/b-5/32*ln(cos(b*x+a)+sin(b*x+a)+sin(2* 
b*x+2*a)^(1/2))/b+5/16*sin(b*x+a)*sin(2*b*x+2*a)^(1/2)/b-5/24*cos(b*x+a)*s 
in(2*b*x+2*a)^(3/2)/b+1/6*sin(b*x+a)*sin(2*b*x+2*a)^(5/2)/b
 

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.72 \[ \int \cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\frac {-5 \left (\arcsin (\cos (a+b x)-\sin (a+b x))+\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right )\right )+\frac {2}{3} \sqrt {\sin (2 (a+b x))} (14 \sin (a+b x)-3 \sin (3 (a+b x))-2 \sin (5 (a+b x)))}{32 b} \] Input:

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

Output:

(-5*(ArcSin[Cos[a + b*x] - Sin[a + b*x]] + Log[Cos[a + b*x] + Sin[a + b*x] 
 + Sqrt[Sin[2*(a + b*x)]]]) + (2*Sqrt[Sin[2*(a + b*x)]]*(14*Sin[a + b*x] - 
 3*Sin[3*(a + b*x)] - 2*Sin[5*(a + b*x)]))/3)/(32*b)
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4789, 3042, 4790, 3042, 4789, 3042, 4794}

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.

\(\displaystyle \int \sin ^{\frac {5}{2}}(2 a+2 b x) \cos (a+b x) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (2 a+2 b x)^{5/2} \cos (a+b x)dx\)

\(\Big \downarrow \) 4789

\(\displaystyle \frac {5}{6} \int \sin (a+b x) \sin ^{\frac {3}{2}}(2 a+2 b x)dx+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5}{6} \int \sin (a+b x) \sin (2 a+2 b x)^{3/2}dx+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b}\)

\(\Big \downarrow \) 4790

\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \int \cos (a+b x) \sqrt {\sin (2 a+2 b x)}dx-\frac {\sin ^{\frac {3}{2}}(2 a+2 b x) \cos (a+b x)}{4 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \int \cos (a+b x) \sqrt {\sin (2 a+2 b x)}dx-\frac {\sin ^{\frac {3}{2}}(2 a+2 b x) \cos (a+b x)}{4 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b}\)

\(\Big \downarrow \) 4789

\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\sin (a+b x)}{\sqrt {\sin (2 a+2 b x)}}dx+\frac {\sqrt {\sin (2 a+2 b x)} \sin (a+b x)}{2 b}\right )-\frac {\sin ^{\frac {3}{2}}(2 a+2 b x) \cos (a+b x)}{4 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \int \frac {\sin (a+b x)}{\sqrt {\sin (2 a+2 b x)}}dx+\frac {\sqrt {\sin (2 a+2 b x)} \sin (a+b x)}{2 b}\right )-\frac {\sin ^{\frac {3}{2}}(2 a+2 b x) \cos (a+b x)}{4 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b}\)

\(\Big \downarrow \) 4794

\(\displaystyle \frac {5}{6} \left (\frac {3}{4} \left (\frac {1}{2} \left (-\frac {\arcsin (\cos (a+b x)-\sin (a+b x))}{2 b}-\frac {\log \left (\sin (a+b x)+\sqrt {\sin (2 a+2 b x)}+\cos (a+b x)\right )}{2 b}\right )+\frac {\sin (a+b x) \sqrt {\sin (2 a+2 b x)}}{2 b}\right )-\frac {\sin ^{\frac {3}{2}}(2 a+2 b x) \cos (a+b x)}{4 b}\right )+\frac {\sin (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x)}{6 b}\)

Input:

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

Output:

(Sin[a + b*x]*Sin[2*a + 2*b*x]^(5/2))/(6*b) + (5*((3*((-1/2*ArcSin[Cos[a + 
 b*x] - Sin[a + b*x]]/b - Log[Cos[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*a + 
 2*b*x]]]/(2*b))/2 + (Sin[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(2*b)))/4 - (Co 
s[a + b*x]*Sin[2*a + 2*b*x]^(3/2))/(4*b)))/6
 

Defintions of rubi rules used

rule 3042

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

rule 4789

Int[cos[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] 
 :> Simp[2*Sin[a + b*x]*((g*Sin[c + d*x])^p/(d*(2*p + 1))), x] + Simp[2*p*( 
g/(2*p + 1))   Int[Sin[a + b*x]*(g*Sin[c + d*x])^(p - 1), x], x] /; FreeQ[{ 
a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ[p] && 
GtQ[p, 0] && IntegerQ[2*p]
 

rule 4790

Int[sin[(a_.) + (b_.)*(x_)]*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] 
 :> Simp[-2*Cos[a + b*x]*((g*Sin[c + d*x])^p/(d*(2*p + 1))), x] + Simp[2*p* 
(g/(2*p + 1))   Int[Cos[a + b*x]*(g*Sin[c + d*x])^(p - 1), x], x] /; FreeQ[ 
{a, b, c, d, g}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] &&  !IntegerQ[p] && 
 GtQ[p, 0] && IntegerQ[2*p]
 

rule 4794

Int[sin[(a_.) + (b_.)*(x_)]/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Sim 
p[-ArcSin[Cos[a + b*x] - Sin[a + b*x]]/d, x] - Simp[Log[Cos[a + b*x] + Sin[ 
a + b*x] + Sqrt[Sin[c + d*x]]]/d, x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - 
 a*d, 0] && EqQ[d/b, 2]
 

Maple [F(-1)]

Timed out.

hanged

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 290 vs. \(2 (118) = 236\).

Time = 0.10 (sec) , antiderivative size = 290, normalized size of antiderivative = 2.13 \[ \int \cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=-\frac {8 \, \sqrt {2} {\left (32 \, \cos \left (b x + a\right )^{4} - 12 \, \cos \left (b x + a\right )^{2} - 15\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} \sin \left (b x + a\right ) - 30 \, \arctan \left (-\frac {\sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} {\left (\cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} + \cos \left (b x + a\right ) \sin \left (b x + a\right )}{\cos \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) - 1}\right ) + 30 \, \arctan \left (-\frac {2 \, \sqrt {2} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} - \cos \left (b x + a\right ) - \sin \left (b x + a\right )}{\cos \left (b x + a\right ) - \sin \left (b x + a\right )}\right ) - 15 \, \log \left (-32 \, \cos \left (b x + a\right )^{4} + 4 \, \sqrt {2} {\left (4 \, \cos \left (b x + a\right )^{3} - {\left (4 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right ) - 5 \, \cos \left (b x + a\right )\right )} \sqrt {\cos \left (b x + a\right ) \sin \left (b x + a\right )} + 32 \, \cos \left (b x + a\right )^{2} + 16 \, \cos \left (b x + a\right ) \sin \left (b x + a\right ) + 1\right )}{384 \, b} \] Input:

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

Output:

-1/384*(8*sqrt(2)*(32*cos(b*x + a)^4 - 12*cos(b*x + a)^2 - 15)*sqrt(cos(b* 
x + a)*sin(b*x + a))*sin(b*x + a) - 30*arctan(-(sqrt(2)*sqrt(cos(b*x + a)* 
sin(b*x + a))*(cos(b*x + a) - sin(b*x + a)) + cos(b*x + a)*sin(b*x + a))/( 
cos(b*x + a)^2 + 2*cos(b*x + a)*sin(b*x + a) - 1)) + 30*arctan(-(2*sqrt(2) 
*sqrt(cos(b*x + a)*sin(b*x + a)) - cos(b*x + a) - sin(b*x + a))/(cos(b*x + 
 a) - sin(b*x + a))) - 15*log(-32*cos(b*x + a)^4 + 4*sqrt(2)*(4*cos(b*x + 
a)^3 - (4*cos(b*x + a)^2 + 1)*sin(b*x + a) - 5*cos(b*x + a))*sqrt(cos(b*x 
+ a)*sin(b*x + a)) + 32*cos(b*x + a)^2 + 16*cos(b*x + a)*sin(b*x + a) + 1) 
)/b
 

Sympy [F(-1)]

Timed out. \[ \int \cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

\[ \int \cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int { \cos \left (b x + a\right ) \sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int { \cos \left (b x + a\right ) \sin \left (2 \, b x + 2 \, a\right )^{\frac {5}{2}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int \cos \left (a+b\,x\right )\,{\sin \left (2\,a+2\,b\,x\right )}^{5/2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \cos (a+b x) \sin ^{\frac {5}{2}}(2 a+2 b x) \, dx=\int \sqrt {\sin \left (2 b x +2 a \right )}\, \cos \left (b x +a \right ) \sin \left (2 b x +2 a \right )^{2}d x \] Input:

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

Output:

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