\(\int (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2} \, dx\) [34]

Optimal result

Integrand size = 23, antiderivative size = 100 \[ \int (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2} \, dx=\frac {6 a e^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d \sqrt {\sin (c+d x)}}-\frac {2 a e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}+\frac {2 b (e \sin (c+d x))^{7/2}}{7 d e} \] Output:

-6/5*a*e^2*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*(e*sin(d*x+c))^(1/ 
2)/d/sin(d*x+c)^(1/2)-2/5*a*e*cos(d*x+c)*(e*sin(d*x+c))^(3/2)/d+2/7*b*(e*s 
in(d*x+c))^(7/2)/d/e
 

Mathematica [A] (verified)

Time = 0.96 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.80 \[ \int (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2} \, dx=\frac {2 (e \sin (c+d x))^{5/2} \left (-21 a E\left (\left .\frac {1}{4} (-2 c+\pi -2 d x)\right |2\right )+\sin ^{\frac {3}{2}}(c+d x) \left (-7 a \cos (c+d x)+5 b \sin ^2(c+d x)\right )\right )}{35 d \sin ^{\frac {5}{2}}(c+d x)} \] Input:

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

Output:

(2*(e*Sin[c + d*x])^(5/2)*(-21*a*EllipticE[(-2*c + Pi - 2*d*x)/4, 2] + Sin 
[c + d*x]^(3/2)*(-7*a*Cos[c + d*x] + 5*b*Sin[c + d*x]^2)))/(35*d*Sin[c + d 
*x]^(5/2))
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.01, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3148, 3042, 3115, 3042, 3121, 3042, 3119}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (e \cos \left (c+d x-\frac {\pi }{2}\right )\right )^{5/2} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )dx\)

\(\Big \downarrow \) 3148

\(\displaystyle a \int (e \sin (c+d x))^{5/2}dx+\frac {2 b (e \sin (c+d x))^{7/2}}{7 d e}\)

\(\Big \downarrow \) 3042

\(\displaystyle a \int (e \sin (c+d x))^{5/2}dx+\frac {2 b (e \sin (c+d x))^{7/2}}{7 d e}\)

\(\Big \downarrow \) 3115

\(\displaystyle a \left (\frac {3}{5} e^2 \int \sqrt {e \sin (c+d x)}dx-\frac {2 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}\right )+\frac {2 b (e \sin (c+d x))^{7/2}}{7 d e}\)

\(\Big \downarrow \) 3042

\(\displaystyle a \left (\frac {3}{5} e^2 \int \sqrt {e \sin (c+d x)}dx-\frac {2 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}\right )+\frac {2 b (e \sin (c+d x))^{7/2}}{7 d e}\)

\(\Big \downarrow \) 3121

\(\displaystyle a \left (\frac {3 e^2 \sqrt {e \sin (c+d x)} \int \sqrt {\sin (c+d x)}dx}{5 \sqrt {\sin (c+d x)}}-\frac {2 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}\right )+\frac {2 b (e \sin (c+d x))^{7/2}}{7 d e}\)

\(\Big \downarrow \) 3042

\(\displaystyle a \left (\frac {3 e^2 \sqrt {e \sin (c+d x)} \int \sqrt {\sin (c+d x)}dx}{5 \sqrt {\sin (c+d x)}}-\frac {2 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}\right )+\frac {2 b (e \sin (c+d x))^{7/2}}{7 d e}\)

\(\Big \downarrow \) 3119

\(\displaystyle a \left (\frac {6 e^2 E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{5 d \sqrt {\sin (c+d x)}}-\frac {2 e \cos (c+d x) (e \sin (c+d x))^{3/2}}{5 d}\right )+\frac {2 b (e \sin (c+d x))^{7/2}}{7 d e}\)

Input:

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

Output:

(2*b*(e*Sin[c + d*x])^(7/2))/(7*d*e) + a*((6*e^2*EllipticE[(c - Pi/2 + d*x 
)/2, 2]*Sqrt[e*Sin[c + d*x]])/(5*d*Sqrt[Sin[c + d*x]]) - (2*e*Cos[c + d*x] 
*(e*Sin[c + d*x])^(3/2))/(5*d))
 

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 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3121

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) 
^n/Sin[c + d*x]^n   Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt 
Q[-1, n, 1] && IntegerQ[2*n]
 

rule 3148

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + 
 Simp[a   Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && 
 (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])
 

Maple [A] (verified)

Time = 3.10 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.71

method	result	size
default	\(\frac {\frac {2 b \left (e \sin \left (d x +c \right )\right )^{\frac {7}{2}}}{7 e}-\frac {e^{3} a \left (6 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c \right )^{4}+2 \sin \left (d x +c \right )^{2}\right )}{5 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\)	\(171\)
parts	\(-\frac {a \,e^{3} \left (6 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticE}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-3 \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c \right )^{4}+2 \sin \left (d x +c \right )^{2}\right )}{5 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {2 b \left (e \sin \left (d x +c \right )\right )^{\frac {7}{2}}}{7 d e}\)	\(173\)

Input:

int((a+cos(d*x+c)*b)*(e*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
 

Output:

(2/7/e*b*(e*sin(d*x+c))^(7/2)-1/5*e^3*a*(6*(1-sin(d*x+c))^(1/2)*(2+2*sin(d 
*x+c))^(1/2)*sin(d*x+c)^(1/2)*EllipticE((1-sin(d*x+c))^(1/2),1/2*2^(1/2))- 
3*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*EllipticF(( 
1-sin(d*x+c))^(1/2),1/2*2^(1/2))-2*sin(d*x+c)^4+2*sin(d*x+c)^2)/cos(d*x+c) 
/(e*sin(d*x+c))^(1/2))/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.10 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.20 \[ \int (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2} \, dx=-\frac {2 \, {\left (-21 i \, a \sqrt {-\frac {1}{2} i \, e} e^{2} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 i \, a \sqrt {\frac {1}{2} i \, e} e^{2} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + {\left (5 \, b e^{2} \cos \left (d x + c\right )^{2} + 7 \, a e^{2} \cos \left (d x + c\right ) - 5 \, b e^{2}\right )} \sqrt {e \sin \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{35 \, d} \] Input:

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

Output:

-2/35*(-21*I*a*sqrt(-1/2*I*e)*e^2*weierstrassZeta(4, 0, weierstrassPInvers 
e(4, 0, cos(d*x + c) + I*sin(d*x + c))) + 21*I*a*sqrt(1/2*I*e)*e^2*weierst 
rassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c) - I*sin(d*x + c))) + 
 (5*b*e^2*cos(d*x + c)^2 + 7*a*e^2*cos(d*x + c) - 5*b*e^2)*sqrt(e*sin(d*x 
+ c))*sin(d*x + c))/d
 

Sympy [F]

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

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

Output:

Integral((e*sin(c + d*x))**(5/2)*(a + b*cos(c + d*x)), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (a+b \cos (c+d x)) (e \sin (c+d x))^{5/2} \, dx=\frac {\sqrt {e}\, e^{2} \left (2 \sqrt {\sin \left (d x +c \right )}\, \sin \left (d x +c \right )^{3} b +7 \left (\int \sqrt {\sin \left (d x +c \right )}\, \sin \left (d x +c \right )^{2}d x \right ) a d \right )}{7 d} \] Input:

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

Output:

(sqrt(e)*e**2*(2*sqrt(sin(c + d*x))*sin(c + d*x)**3*b + 7*int(sqrt(sin(c + 
 d*x))*sin(c + d*x)**2,x)*a*d))/(7*d)