\(\int \frac {\sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2}}{x} \, dx\) [96]

Optimal result

Integrand size = 33, antiderivative size = 186 \[ \int \frac {\sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2}}{x} \, dx=c \cos (e) \operatorname {CosIntegral}(f x) \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}+\frac {1}{2} c \operatorname {CosIntegral}(2 f x) \sec (e+f x) \sin (2 e) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}-c \sec (e+f x) \sin (e) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \text {Si}(f x)+\frac {1}{2} c \cos (2 e) \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \text {Si}(2 f x) \] Output:

c*cos(e)*Ci(f*x)*sec(f*x+e)*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2)+ 
1/2*c*Ci(2*f*x)*sec(f*x+e)*sin(2*e)*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e) 
)^(1/2)-c*sec(f*x+e)*sin(e)*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2)* 
Si(f*x)+1/2*c*cos(2*e)*sec(f*x+e)*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^ 
(1/2)*Si(2*f*x)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.40 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.81 \[ \int \frac {\sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2}}{x} \, dx=\frac {c e^{-i (e-f x)} \sqrt {-i c e^{-i (e+f x)} \left (i+e^{i (e+f x)}\right )^2} \left (2 e^{i e} \operatorname {ExpIntegralEi}(-i f x)+2 e^{3 i e} \operatorname {ExpIntegralEi}(i f x)+i \left (\operatorname {ExpIntegralEi}(-2 i f x)-e^{4 i e} \operatorname {ExpIntegralEi}(2 i f x)\right )\right ) \sqrt {a-a \sin (e+f x)}}{2 \sqrt {2} \left (1+e^{2 i (e+f x)}\right )} \] Input:

Integrate[(Sqrt[a - a*Sin[e + f*x]]*(c + c*Sin[e + f*x])^(3/2))/x,x]
 

Output:

(c*Sqrt[((-I)*c*(I + E^(I*(e + f*x)))^2)/E^(I*(e + f*x))]*(2*E^(I*e)*ExpIn 
tegralEi[(-I)*f*x] + 2*E^((3*I)*e)*ExpIntegralEi[I*f*x] + I*(ExpIntegralEi 
[(-2*I)*f*x] - E^((4*I)*e)*ExpIntegralEi[(2*I)*f*x]))*Sqrt[a - a*Sin[e + f 
*x]])/(2*Sqrt[2]*E^(I*(e - f*x))*(1 + E^((2*I)*(e + f*x))))
 

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.42, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {5115, 7292, 27, 7293, 2009}

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.

\(\displaystyle \int \frac {\sqrt {a-a \sin (e+f x)} (c \sin (e+f x)+c)^{3/2}}{x} \, dx\)

\(\Big \downarrow \) 5115

\(\displaystyle \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \int \frac {\cos (e+f x) (\sin (e+f x) c+c)}{x}dx\)

\(\Big \downarrow \) 7292

\(\displaystyle \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \int \frac {c \cos (e+f x) (\sin (e+f x)+1)}{x}dx\)

\(\Big \downarrow \) 27

\(\displaystyle c \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \int \frac {\cos (e+f x) (\sin (e+f x)+1)}{x}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle c \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \int \left (\frac {\cos (e+f x)}{x}+\frac {\sin (2 e+2 f x)}{2 x}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle c \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \left (\frac {1}{2} \sin (2 e) \operatorname {CosIntegral}(2 f x)+\cos (e) \operatorname {CosIntegral}(f x)-\sin (e) \text {Si}(f x)+\frac {1}{2} \cos (2 e) \text {Si}(2 f x)\right )\)

Input:

Int[(Sqrt[a - a*Sin[e + f*x]]*(c + c*Sin[e + f*x])^(3/2))/x,x]
 

Output:

c*Sec[e + f*x]*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]]*(Cos[e]*C 
osIntegral[f*x] + (CosIntegral[2*f*x]*Sin[2*e])/2 - Sin[e]*SinIntegral[f*x 
] + (Cos[2*e]*SinIntegral[2*f*x])/2)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 5115

Int[((g_.) + (h_.)*(x_))^(p_.)*((a_) + (b_.)*Sin[(e_.) + (f_.)*(x_)])^(m_)* 
((c_) + (d_.)*Sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a^IntPart[m] 
*c^IntPart[m]*(a + b*Sin[e + f*x])^FracPart[m]*((c + d*Sin[e + f*x])^FracPa 
rt[m]/Cos[e + f*x]^(2*FracPart[m]))   Int[(g + h*x)^p*Cos[e + f*x]^(2*m)*(c 
 + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x] && 
 EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[p] && IntegerQ[2*m] && 
IGeQ[n - m, 0]
 

rule 7292

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =! 
= u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

Maple [F]

\[\int \frac {\sqrt {a -a \sin \left (f x +e \right )}\, \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}}}{x}d x\]

Input:

int((a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(3/2)/x,x)
 

Output:

int((a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(3/2)/x,x)
 

Fricas [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2}}{x} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(3/2)/x,x, algorithm="fr 
icas")
 

Output:

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (has polynomial part)
 

Sympy [F]

\[ \int \frac {\sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2}}{x} \, dx=\int \frac {\left (c \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}} \sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right )}}{x}\, dx \] Input:

integrate((a-a*sin(f*x+e))**(1/2)*(c+c*sin(f*x+e))**(3/2)/x,x)
 

Output:

Integral((c*(sin(e + f*x) + 1))**(3/2)*sqrt(-a*(sin(e + f*x) - 1))/x, x)
 

Maxima [F]

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

integrate((a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(3/2)/x,x, algorithm="ma 
xima")
 

Output:

integrate(sqrt(-a*sin(f*x + e) + a)*(c*sin(f*x + e) + c)^(3/2)/x, x)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2}}{x} \, dx=-\frac {{\left (2 \, c f \cos \left (e\right ) \operatorname {Ci}\left (f x\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + c f \operatorname {Ci}\left (2 \, f x\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (2 \, e\right ) + c f \cos \left (2 \, e\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \operatorname {Si}\left (2 \, f x\right ) - 2 \, c f \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \sin \left (e\right ) \operatorname {Si}\left (f x\right )\right )} \sqrt {a} \sqrt {c}}{2 \, f} \] Input:

integrate((a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(3/2)/x,x, algorithm="gi 
ac")
 

Output:

-1/2*(2*c*f*cos(e)*cos_integral(f*x)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*s 
gn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) + c*f*cos_integral(2*f*x)*sgn(cos(-1/4* 
pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(2*e) + c*f* 
cos(2*e)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1 
/2*e))*sin_integral(2*f*x) - 2*c*f*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn 
(sin(-1/4*pi + 1/2*f*x + 1/2*e))*sin(e)*sin_integral(f*x))*sqrt(a)*sqrt(c) 
/f
 

Mupad [F(-1)]

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

int(((a - a*sin(e + f*x))^(1/2)*(c + c*sin(e + f*x))^(3/2))/x,x)
 

Output:

int(((a - a*sin(e + f*x))^(1/2)*(c + c*sin(e + f*x))^(3/2))/x, x)
 

Reduce [F]

\[ \int \frac {\sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2}}{x} \, dx=\sqrt {c}\, \sqrt {a}\, c \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{x}d x +\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sqrt {-\sin \left (f x +e \right )+1}}{x}d x \right ) \] Input:

int((a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(3/2)/x,x)
 

Output:

sqrt(c)*sqrt(a)*c*(int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*s 
in(e + f*x))/x,x) + int((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)) 
/x,x))