\(\int (a+b \cos (c+d x)) (e \sin (c+d x))^{3/2} \, dx\) [35]

Optimal result

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

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

Mathematica [A] (verified)

Time = 0.72 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.80 \[ \int (a+b \cos (c+d x)) (e \sin (c+d x))^{3/2} \, dx=\frac {2 (e \sin (c+d x))^{3/2} \left (-5 a \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),2\right )+\sqrt {\sin (c+d x)} \left (-5 a \cos (c+d x)+3 b \sin ^2(c+d x)\right )\right )}{15 d \sin ^{\frac {3}{2}}(c+d x)} \] Input:

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

Output:

(2*(e*Sin[c + d*x])^(3/2)*(-5*a*EllipticF[(-2*c + Pi - 2*d*x)/4, 2] + Sqrt 
[Sin[c + d*x]]*(-5*a*Cos[c + d*x] + 3*b*Sin[c + d*x]^2)))/(15*d*Sin[c + d* 
x]^(3/2))
 

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, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3148

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

Input:

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

Output:

(2*b*(e*Sin[c + d*x])^(5/2))/(5*d*e) + a*((2*e^2*EllipticF[(c - Pi/2 + d*x 
)/2, 2]*Sqrt[Sin[c + d*x]])/(3*d*Sqrt[e*Sin[c + d*x]]) - (2*e*Cos[c + d*x] 
*Sqrt[e*Sin[c + d*x]])/(3*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 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 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 = 1.85 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.16

method	result	size
default	\(\frac {\frac {2 b \left (e \sin \left (d x +c \right )\right )^{\frac {5}{2}}}{5 e}-\frac {a \,e^{2} \left (\sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c \right )^{3}+2 \sin \left (d x +c \right )\right )}{3 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\)	\(116\)
parts	\(-\frac {a \,e^{2} \left (\sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2+2 \sin \left (d x +c \right )}\, \sqrt {\sin \left (d x +c \right )}\, \operatorname {EllipticF}\left (\sqrt {1-\sin \left (d x +c \right )}, \frac {\sqrt {2}}{2}\right )-2 \sin \left (d x +c \right )^{3}+2 \sin \left (d x +c \right )\right )}{3 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}+\frac {2 b \left (e \sin \left (d x +c \right )\right )^{\frac {5}{2}}}{5 d e}\)	\(118\)

Input:

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

Output:

(2/5/e*b*(e*sin(d*x+c))^(5/2)-1/3*a*e^2*((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)^3+2*sin(d*x+c))/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 = 99, normalized size of antiderivative = 0.99 \[ \int (a+b \cos (c+d x)) (e \sin (c+d x))^{3/2} \, dx=\frac {2 \, {\left (5 \, a \sqrt {-\frac {1}{2} i \, e} e {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 \, a \sqrt {\frac {1}{2} i \, e} e {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - {\left (3 \, b e \cos \left (d x + c\right )^{2} + 5 \, a e \cos \left (d x + c\right ) - 3 \, b e\right )} \sqrt {e \sin \left (d x + c\right )}\right )}}{15 \, d} \] Input:

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

Output:

2/15*(5*a*sqrt(-1/2*I*e)*e*weierstrassPInverse(4, 0, cos(d*x + c) + I*sin( 
d*x + c)) + 5*a*sqrt(1/2*I*e)*e*weierstrassPInverse(4, 0, cos(d*x + c) - I 
*sin(d*x + c)) - (3*b*e*cos(d*x + c)^2 + 5*a*e*cos(d*x + c) - 3*b*e)*sqrt( 
e*sin(d*x + c)))/d
 

Sympy [F]

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

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

Output:

Integral((e*sin(c + d*x))**(3/2)*(a + b*cos(c + d*x)), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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