\(\int \frac {(a+a \sin (e+f x))^{3/2}}{x^2} \, dx\) [132]

Optimal result

Integrand size = 18, antiderivative size = 263 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{x^2} \, dx=-\frac {3}{4} a f \operatorname {CosIntegral}\left (\frac {f x}{2}\right ) \csc \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \sin \left (\frac {1}{4} (2 e-\pi )\right ) \sqrt {a+a \sin (e+f x)}+\frac {3}{4} a f \operatorname {CosIntegral}\left (\frac {3 f x}{2}\right ) \csc \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \sin \left (\frac {1}{4} (6 e+\pi )\right ) \sqrt {a+a \sin (e+f x)}-\frac {2 a \sin ^2\left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \sqrt {a+a \sin (e+f x)}}{x}-\frac {3}{4} a f \csc \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \sin \left (\frac {1}{4} (2 e+\pi )\right ) \sqrt {a+a \sin (e+f x)} \text {Si}\left (\frac {f x}{2}\right )+\frac {3}{4} a f \cos \left (\frac {1}{4} (6 e+\pi )\right ) \csc \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right ) \sqrt {a+a \sin (e+f x)} \text {Si}\left (\frac {3 f x}{2}\right ) \] Output:

3/4*a*f*Ci(1/2*f*x)*csc(1/2*e+1/4*Pi+1/2*f*x)*cos(1/2*e+1/4*Pi)*(a+a*sin(f 
*x+e))^(1/2)+3/4*a*f*Ci(3/2*f*x)*csc(1/2*e+1/4*Pi+1/2*f*x)*sin(3/2*e+1/4*P 
i)*(a+a*sin(f*x+e))^(1/2)-2*a*sin(1/2*e+1/4*Pi+1/2*f*x)^2*(a+a*sin(f*x+e)) 
^(1/2)/x-3/4*a*f*csc(1/2*e+1/4*Pi+1/2*f*x)*sin(1/2*e+1/4*Pi)*(a+a*sin(f*x+ 
e))^(1/2)*Si(1/2*f*x)+3/4*a*f*cos(3/2*e+1/4*Pi)*csc(1/2*e+1/4*Pi+1/2*f*x)* 
(a+a*sin(f*x+e))^(1/2)*Si(3/2*f*x)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.46 (sec) , antiderivative size = 226, normalized size of antiderivative = 0.86 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{x^2} \, dx=\frac {i \left (-i a e^{-i (e+f x)} \left (i+e^{i (e+f x)}\right )^2\right )^{3/2} \left (2-6 i e^{i (e+f x)}-6 e^{2 i (e+f x)}+2 i e^{3 i (e+f x)}+3 e^{i e+\frac {3 i f x}{2}} f x \operatorname {ExpIntegralEi}\left (-\frac {1}{2} i f x\right )+3 i e^{2 i e+\frac {3 i f x}{2}} f x \operatorname {ExpIntegralEi}\left (\frac {i f x}{2}\right )+3 i e^{\frac {3 i f x}{2}} f x \operatorname {ExpIntegralEi}\left (-\frac {3}{2} i f x\right )+3 e^{\frac {3}{2} i (2 e+f x)} f x \operatorname {ExpIntegralEi}\left (\frac {3 i f x}{2}\right )\right )}{4 \sqrt {2} \left (i+e^{i (e+f x)}\right )^3 x} \] Input:

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

Output:

((I/4)*(((-I)*a*(I + E^(I*(e + f*x)))^2)/E^(I*(e + f*x)))^(3/2)*(2 - (6*I) 
*E^(I*(e + f*x)) - 6*E^((2*I)*(e + f*x)) + (2*I)*E^((3*I)*(e + f*x)) + 3*E 
^(I*e + ((3*I)/2)*f*x)*f*x*ExpIntegralEi[(-1/2*I)*f*x] + (3*I)*E^((2*I)*e 
+ ((3*I)/2)*f*x)*f*x*ExpIntegralEi[(I/2)*f*x] + (3*I)*E^(((3*I)/2)*f*x)*f* 
x*ExpIntegralEi[((-3*I)/2)*f*x] + 3*E^(((3*I)/2)*(2*e + f*x))*f*x*ExpInteg 
ralEi[((3*I)/2)*f*x]))/(Sqrt[2]*(I + E^(I*(e + f*x)))^3*x)
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.58, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {3042, 3800, 3042, 3794, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3800

\(\displaystyle 2 a \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (e+f x)+a} \int \frac {\sin ^3\left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{x^2}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle 2 a \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (e+f x)+a} \int \frac {\sin \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )^3}{x^2}dx\)

\(\Big \downarrow \) 3794

\(\displaystyle 2 a \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (e+f x)+a} \left (\frac {3}{2} f \int \left (\frac {\cos \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{4 x}+\frac {\sin \left (\frac {3 e}{2}+\frac {3 f x}{2}+\frac {\pi }{4}\right )}{4 x}\right )dx-\frac {\sin ^3\left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{x}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 a \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (e+f x)+a} \left (\frac {3}{2} f \left (-\frac {1}{4} \sin \left (\frac {1}{4} (2 e-\pi )\right ) \operatorname {CosIntegral}\left (\frac {f x}{2}\right )+\frac {1}{4} \sin \left (\frac {1}{4} (6 e+\pi )\right ) \operatorname {CosIntegral}\left (\frac {3 f x}{2}\right )-\frac {1}{4} \sin \left (\frac {1}{4} (2 e+\pi )\right ) \text {Si}\left (\frac {f x}{2}\right )+\frac {1}{4} \cos \left (\frac {1}{4} (6 e+\pi )\right ) \text {Si}\left (\frac {3 f x}{2}\right )\right )-\frac {\sin ^3\left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{x}\right )\)

Input:

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

Output:

2*a*Csc[e/2 + Pi/4 + (f*x)/2]*Sqrt[a + a*Sin[e + f*x]]*(-(Sin[e/2 + Pi/4 + 
 (f*x)/2]^3/x) + (3*f*(-1/4*(CosIntegral[(f*x)/2]*Sin[(2*e - Pi)/4]) + (Co 
sIntegral[(3*f*x)/2]*Sin[(6*e + Pi)/4])/4 - (Sin[(2*e + Pi)/4]*SinIntegral 
[(f*x)/2])/4 + (Cos[(6*e + Pi)/4]*SinIntegral[(3*f*x)/2])/4))/2)
 

Defintions of rubi rules used

rule 2009

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

rule 3042

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

rule 3794

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Si 
mp[(c + d*x)^(m + 1)*(Sin[e + f*x]^n/(d*(m + 1))), x] - Simp[f*(n/(d*(m + 1 
)))   Int[ExpandTrigReduce[(c + d*x)^(m + 1), Cos[e + f*x]*Sin[e + f*x]^(n 
- 1), x], x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && GeQ[m, -2] & 
& LtQ[m, -1]
 

rule 3800

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), 
 x_Symbol] :> Simp[(2*a)^IntPart[n]*((a + b*Sin[e + f*x])^FracPart[n]/Sin[e 
/2 + a*(Pi/(4*b)) + f*(x/2)]^(2*FracPart[n]))   Int[(c + d*x)^m*Sin[e/2 + a 
*(Pi/(4*b)) + f*(x/2)]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && 
EqQ[a^2 - b^2, 0] && IntegerQ[n + 1/2] && (GtQ[n, 0] || IGtQ[m, 0])
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F(-2)]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 505 vs. \(2 (197) = 394\).

Time = 0.39 (sec) , antiderivative size = 505, normalized size of antiderivative = 1.92 \[ \int \frac {(a+a \sin (e+f x))^{3/2}}{x^2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/8*sqrt(2)*(3*pi*a*f^2*cos_integral(3/2*f*x)*sgn(cos(-1/4*pi + 1/2*f*x + 
1/2*e))*sin(3/4*pi - 3/2*e) - 3*(pi - 2*f*x - 2*e)*a*f^2*cos_integral(3/2* 
f*x)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(3/4*pi - 3/2*e) - 6*a*e*f^2*c 
os_integral(3/2*f*x)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(3/4*pi - 3/2* 
e) + 3*pi*a*f^2*cos_integral(1/2*f*x)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))* 
sin(1/4*pi - 1/2*e) - 3*(pi - 2*f*x - 2*e)*a*f^2*cos_integral(1/2*f*x)*sgn 
(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(1/4*pi - 1/2*e) - 6*a*e*f^2*cos_integ 
ral(1/2*f*x)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin(1/4*pi - 1/2*e) - 3*p 
i*a*f^2*cos(3/4*pi - 3/2*e)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin_integr 
al(3/2*f*x) + 3*(pi - 2*f*x - 2*e)*a*f^2*cos(3/4*pi - 3/2*e)*sgn(cos(-1/4* 
pi + 1/2*f*x + 1/2*e))*sin_integral(3/2*f*x) + 6*a*e*f^2*cos(3/4*pi - 3/2* 
e)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin_integral(3/2*f*x) - 3*pi*a*f^2* 
cos(1/4*pi - 1/2*e)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e))*sin_integral(1/2*f 
*x) + 3*(pi - 2*f*x - 2*e)*a*f^2*cos(1/4*pi - 1/2*e)*sgn(cos(-1/4*pi + 1/2 
*f*x + 1/2*e))*sin_integral(1/2*f*x) + 6*a*e*f^2*cos(1/4*pi - 1/2*e)*sgn(c 
os(-1/4*pi + 1/2*f*x + 1/2*e))*sin_integral(1/2*f*x) - 12*a*f^2*cos(-1/4*p 
i + 1/2*f*x + 1/2*e)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)) - 4*a*f^2*cos(-3/ 
4*pi + 3/2*f*x + 3/2*e)*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))*sqrt(a)/(f^2* 
x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(a+a \sin (e+f x))^{3/2}}{x^2} \, dx=\frac {\sqrt {a}\, a \left (-2 \sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )+4 \sqrt {\sin \left (f x +e \right )+1}+6 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}}{\sin \left (f x +e \right ) x^{2}+x^{2}}d x \right ) x +3 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \cos \left (f x +e \right ) \sin \left (f x +e \right )}{\sin \left (f x +e \right ) x +x}d x \right ) f x +6 \left (\int \frac {\sqrt {\sin \left (f x +e \right )+1}\, \sin \left (f x +e \right )}{\sin \left (f x +e \right ) x^{2}+x^{2}}d x \right ) x \right )}{2 x} \] Input:

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

Output:

(sqrt(a)*a*( - 2*sqrt(sin(e + f*x) + 1)*sin(e + f*x) + 4*sqrt(sin(e + f*x) 
 + 1) + 6*int(sqrt(sin(e + f*x) + 1)/(sin(e + f*x)*x**2 + x**2),x)*x + 3*i 
nt((sqrt(sin(e + f*x) + 1)*cos(e + f*x)*sin(e + f*x))/(sin(e + f*x)*x + x) 
,x)*f*x + 6*int((sqrt(sin(e + f*x) + 1)*sin(e + f*x))/(sin(e + f*x)*x**2 + 
 x**2),x)*x))/(2*x)