\(\int \frac {\sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2}}{x^3} \, dx\) [98]

Optimal result

Integrand size = 33, antiderivative size = 385 \[ \int \frac {\sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2}}{x^3} \, dx=-\frac {c \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{2 x^2}-\frac {c f \cos (2 e+2 f x) \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{2 x}-\frac {1}{2} c f^2 \cos (e) \operatorname {CosIntegral}(f x) \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}-c f^2 \operatorname {CosIntegral}(2 f x) \sec (e+f x) \sin (2 e) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}-\frac {c \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \sin (2 e+2 f x)}{4 x^2}+\frac {1}{2} c f^2 \sec (e+f x) \sin (e) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \text {Si}(f x)-c f^2 \cos (2 e) \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \text {Si}(2 f x)+\frac {c f \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \tan (e+f x)}{2 x} \] Output:

-1/2*c*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2)/x^2-1/2*c*f*cos(2*f*x 
+2*e)*sec(f*x+e)*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2)/x-1/2*c*f^2 
*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)-c 
*f^2*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)-1/4*c*sec(f*x+e)*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2)*sin( 
2*f*x+2*e)/x^2+1/2*c*f^2*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)-c*f^2*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)+1/2*c*f*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x 
+e))^(1/2)*tan(f*x+e)/x
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 2.88 (sec) , antiderivative size = 317, normalized size of antiderivative = 0.82 \[ \int \frac {\sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2}}{x^3} \, dx=\frac {c^2 e^{-2 i (e+f x)} \left (i+e^{i (e+f x)}\right ) \left (-1+2 i e^{i (e+f x)}+2 i e^{3 i (e+f x)}+e^{4 i (e+f x)}+2 i f x+2 e^{i (e+f x)} f x-2 e^{3 i (e+f x)} f x+2 i e^{4 i (e+f x)} f x+2 i e^{i (e+2 f x)} f^2 x^2 \operatorname {ExpIntegralEi}(-i f x)+2 i e^{3 i e+2 i f x} f^2 x^2 \operatorname {ExpIntegralEi}(i f x)-4 e^{2 i f x} f^2 x^2 \operatorname {ExpIntegralEi}(-2 i f x)+4 e^{2 i (2 e+f x)} f^2 x^2 \operatorname {ExpIntegralEi}(2 i f x)\right ) \sqrt {a-a \sin (e+f x)}}{4 \sqrt {2} \left (-i+e^{i (e+f x)}\right ) \sqrt {-i c e^{-i (e+f x)} \left (i+e^{i (e+f x)}\right )^2} x^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.94 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.39, 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.

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

\(\Big \downarrow \) 5115

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

\(\Big \downarrow \) 7292

\(\displaystyle \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \int \frac {c \cos (e+f x) (\sin (e+f x)+1)}{x^3}dx\)

\(\Big \downarrow \) 27

\(\displaystyle c \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \int \frac {\cos (e+f x) (\sin (e+f x)+1)}{x^3}dx\)

\(\Big \downarrow \) 7293

\(\displaystyle c \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \int \left (\frac {\cos (e+f x)}{x^3}+\frac {\sin (2 e+2 f x)}{2 x^3}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle c \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \left (-f^2 \sin (2 e) \operatorname {CosIntegral}(2 f x)-\frac {1}{2} f^2 \cos (e) \operatorname {CosIntegral}(f x)+\frac {1}{2} f^2 \sin (e) \text {Si}(f x)-f^2 \cos (2 e) \text {Si}(2 f x)-\frac {\sin (2 e+2 f x)}{4 x^2}-\frac {\cos (e+f x)}{2 x^2}+\frac {f \sin (e+f x)}{2 x}-\frac {f \cos (2 e+2 f x)}{2 x}\right )\)

Input:

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

Output:

c*Sec[e + f*x]*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]]*(-1/2*Cos 
[e + f*x]/x^2 - (f*Cos[2*e + 2*f*x])/(2*x) - (f^2*Cos[e]*CosIntegral[f*x]) 
/2 - f^2*CosIntegral[2*f*x]*Sin[2*e] + (f*Sin[e + f*x])/(2*x) - Sin[2*e + 
2*f*x]/(4*x^2) + (f^2*Sin[e]*SinIntegral[f*x])/2 - f^2*Cos[2*e]*SinIntegra 
l[2*f*x])
 

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]

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

Input:

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

Output:

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

Fricas [F(-2)]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1502 vs. \(2 (341) = 682\).

Time = 0.33 (sec) , antiderivative size = 1502, normalized size of antiderivative = 3.90 \[ \int \frac {\sqrt {a-a \sin (e+f x)} (c+c \sin (e+f x))^{3/2}}{x^3} \, dx=\text {Too large to display} \] Input:

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

Output:

1/2*(pi^2*c*f^3*cos(e)*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)) - 2*pi*(pi - 2*f*x - 2*e)*c*f^3*cos 
(e)*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)) + (pi - 2*f*x - 2*e)^2*c*f^3*cos(e)*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)) - 
4*pi*c*e*f^3*cos(e)*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)) + 4*(pi - 2*f*x - 2*e)*c*e*f^3*cos(e)* 
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)) + 4*c*e^2*f^3*cos(e)*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)) + 2*pi^2*c*f^3*cos_in 
tegral(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) - 4*pi*(pi - 2*f*x - 2*e)*c*f^3*cos_integral(2*f*x)*s 
gn(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) + 2*(pi - 2*f*x - 2*e)^2*c*f^3*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) - 8*pi*c*e 
*f^3*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) + 8*(pi - 2*f*x - 2*e)*c*e*f^3*cos_integra 
l(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) + 8*c*e^2*f^3*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) + 2*pi^2*c*f^3...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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