\(\int \frac {\sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{x^2} \, dx\) [91]

Optimal result

Integrand size = 33, antiderivative size = 123 \[ \int \frac {\sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{x^2} \, dx=-\frac {\sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{x}-f \operatorname {CosIntegral}(f x) \sec (e+f x) \sin (e) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}-f \cos (e) \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \text {Si}(f x) \] Output:

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

Mathematica [A] (verified)

Time = 0.85 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.53 \[ \int \frac {\sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{x^2} \, dx=-\frac {\sec (e+f x) \sqrt {c (1+\sin (e+f x))} \sqrt {a-a \sin (e+f x)} (\cos (e+f x)+f x \operatorname {CosIntegral}(f x) \sin (e)+f x \cos (e) \text {Si}(f x))}{x} \] Input:

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

Output:

-((Sec[e + f*x]*Sqrt[c*(1 + Sin[e + f*x])]*Sqrt[a - a*Sin[e + f*x]]*(Cos[e 
 + f*x] + f*x*CosIntegral[f*x]*Sin[e] + f*x*Cos[e]*SinIntegral[f*x]))/x)
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.54, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {5115, 3042, 3778, 25, 3042, 3784, 3042, 3780, 3783}

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.

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

\(\Big \downarrow \) 5115

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3778

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3784

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3780

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

\(\Big \downarrow \) 3783

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042

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

rule 3778

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(c 
 + d*x)^(m + 1)*(Sin[e + f*x]/(d*(m + 1))), x] - Simp[f/(d*(m + 1))   Int[( 
c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[m, - 
1]
 

rule 3780

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinInte 
gral[e + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*e - c*f, 0]
 

rule 3783

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosInte 
gral[e - Pi/2 + f*x]/d, x] /; FreeQ[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - 
c*f, 0]
 

rule 3784

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[Cos[(d* 
e - c*f)/d]   Int[Sin[c*(f/d) + f*x]/(c + d*x), x], x] + Simp[Sin[(d*e - c* 
f)/d]   Int[Cos[c*(f/d) + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f}, x] 
&& NeQ[d*e - c*f, 0]
 

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F(-2)]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.28 (sec) , antiderivative size = 886, normalized size of antiderivative = 7.20 \[ \int \frac {\sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{x^2} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/2*(f*x*imag_part(cos_integral(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))*tan(1/2*f*x)^2*tan(1/2*e)^2 - f*x*ima 
g_part(cos_integral(-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))*tan(1/2*f*x)^2*tan(1/2*e)^2 + 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_integral(f* 
x)*tan(1/2*f*x)^2*tan(1/2*e)^2 - 2*f*x*real_part(cos_integral(f*x))*sgn(co 
s(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))*tan(1/2* 
f*x)^2*tan(1/2*e) - 2*f*x*real_part(cos_integral(-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))*tan(1/2*f*x)^2*tan(1 
/2*e) - f*x*imag_part(cos_integral(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))*tan(1/2*f*x)^2 + f*x*imag_part(cos_ 
integral(-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))*tan(1/2*f*x)^2 - 2*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))*sin_integral(f*x)*tan(1/2*f*x)^2 + f*x* 
imag_part(cos_integral(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))*tan(1/2*e)^2 - f*x*imag_part(cos_integral(-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))* 
tan(1/2*e)^2 + 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_integral(f*x)*tan(1/2*e)^2 - 2*f*x*real_part(cos_in 
tegral(f*x))*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sgn(sin(-1/4*pi + 1/2*...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(c)*sqrt(a)*( - sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1) + in 
t((sqrt(sin(e + f*x) + 1)*sqrt( - sin(e + f*x) + 1)*cos(e + f*x)*sin(e + f 
*x))/(sin(e + f*x)**2*x - x),x)*f*x))/x