\(\int \frac {\sqrt {a+a \sin (c+d x)}}{x^2} \, dx\) [126]

Optimal result

Integrand size = 18, antiderivative size = 130 \[ \int \frac {\sqrt {a+a \sin (c+d x)}}{x^2} \, dx=-\frac {\sqrt {a+a \sin (c+d x)}}{x}-\frac {1}{2} d \operatorname {CosIntegral}\left (\frac {d x}{2}\right ) \csc \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \sin \left (\frac {1}{4} (2 c-\pi )\right ) \sqrt {a+a \sin (c+d x)}-\frac {1}{2} d \csc \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \sin \left (\frac {1}{4} (2 c+\pi )\right ) \sqrt {a+a \sin (c+d x)} \text {Si}\left (\frac {d x}{2}\right ) \] Output:

-(a+a*sin(d*x+c))^(1/2)/x+1/2*d*Ci(1/2*d*x)*csc(1/2*c+1/4*Pi+1/2*d*x)*cos( 
1/2*c+1/4*Pi)*(a+a*sin(d*x+c))^(1/2)-1/2*d*csc(1/2*c+1/4*Pi+1/2*d*x)*sin(1 
/2*c+1/4*Pi)*(a+a*sin(d*x+c))^(1/2)*Si(1/2*d*x)
 

Mathematica [A] (verified)

Time = 0.71 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt {a+a \sin (c+d x)}}{x^2} \, dx=\frac {\sqrt {a (1+\sin (c+d x))} \left (d x \operatorname {CosIntegral}\left (\frac {d x}{2}\right ) \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right )-2 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-d x \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \text {Si}\left (\frac {d x}{2}\right )\right )}{2 x \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \] Input:

Integrate[Sqrt[a + a*Sin[c + d*x]]/x^2,x]
 

Output:

(Sqrt[a*(1 + Sin[c + d*x])]*(d*x*CosIntegral[(d*x)/2]*(Cos[c/2] - Sin[c/2] 
) - 2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) - d*x*(Cos[c/2] + Sin[c/2])*Si 
nIntegral[(d*x)/2]))/(2*x*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.79, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 3800, 3042, 3778, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {a \sin (c+d x)+a}}{x^2}dx\)

\(\Big \downarrow \) 3800

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3778

\(\displaystyle \csc \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (c+d x)+a} \left (\frac {1}{2} d \int \frac {\cos \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{x}dx-\frac {\sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{x}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \csc \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (c+d x)+a} \left (\frac {1}{2} d \int \frac {\sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {3 \pi }{4}\right )}{x}dx-\frac {\sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{x}\right )\)

\(\Big \downarrow \) 3784

\(\displaystyle \csc \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (c+d x)+a} \left (\frac {1}{2} d \left (-\sin \left (\frac {1}{4} (2 c+\pi )\right ) \int \frac {\sin \left (\frac {d x}{2}\right )}{x}dx-\sin \left (\frac {1}{4} (2 c-\pi )\right ) \int \frac {\cos \left (\frac {d x}{2}\right )}{x}dx\right )-\frac {\sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{x}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \csc \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (c+d x)+a} \left (\frac {1}{2} d \left (-\sin \left (\frac {1}{4} (2 c+\pi )\right ) \int \frac {\sin \left (\frac {d x}{2}\right )}{x}dx-\sin \left (\frac {1}{4} (2 c-\pi )\right ) \int \frac {\sin \left (\frac {d x}{2}+\frac {\pi }{2}\right )}{x}dx\right )-\frac {\sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{x}\right )\)

\(\Big \downarrow \) 3780

\(\displaystyle \csc \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (c+d x)+a} \left (\frac {1}{2} d \left (-\sin \left (\frac {1}{4} (2 c-\pi )\right ) \int \frac {\sin \left (\frac {d x}{2}+\frac {\pi }{2}\right )}{x}dx-\sin \left (\frac {1}{4} (2 c+\pi )\right ) \text {Si}\left (\frac {d x}{2}\right )\right )-\frac {\sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{x}\right )\)

\(\Big \downarrow \) 3783

\(\displaystyle \csc \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (c+d x)+a} \left (\frac {1}{2} d \left (\sin \left (\frac {1}{4} (2 c-\pi )\right ) \left (-\operatorname {CosIntegral}\left (\frac {d x}{2}\right )\right )-\sin \left (\frac {1}{4} (2 c+\pi )\right ) \text {Si}\left (\frac {d x}{2}\right )\right )-\frac {\sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{x}\right )\)

Input:

Int[Sqrt[a + a*Sin[c + d*x]]/x^2,x]
 

Output:

Csc[c/2 + Pi/4 + (d*x)/2]*Sqrt[a + a*Sin[c + d*x]]*(-(Sin[c/2 + Pi/4 + (d* 
x)/2]/x) + (d*(-(CosIntegral[(d*x)/2]*Sin[(2*c - Pi)/4]) - Sin[(2*c + Pi)/ 
4]*SinIntegral[(d*x)/2]))/2)
 

Defintions of rubi rules used

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 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]

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

Input:

int((a+a*sin(d*x+c))^(1/2)/x^2,x)
 

Output:

int((a+a*sin(d*x+c))^(1/2)/x^2,x)
 

Fricas [F(-2)]

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

integrate((a+a*sin(d*x+c))^(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]

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

integrate((a+a*sin(d*x+c))**(1/2)/x**2,x)
 

Output:

Integral(sqrt(a*(sin(c + d*x) + 1))/x**2, x)
 

Maxima [F]

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

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

Output:

integrate(sqrt(a*sin(d*x + c) + a)/x^2, x)
 

Giac [C] (verification not implemented)

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

Time = 0.47 (sec) , antiderivative size = 1140, normalized size of antiderivative = 8.77 \[ \int \frac {\sqrt {a+a \sin (c+d x)}}{x^2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*sqrt(2)*(d*x*imag_part(cos_integral(1/2*d*x))*sgn(cos(-1/4*pi + 1/2*d* 
x + 1/2*c))*tan(1/4*d*x)^2*tan(1/4*c)^2 - d*x*imag_part(cos_integral(-1/2* 
d*x))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*d*x)^2*tan(1/4*c)^2 - d* 
x*real_part(cos_integral(1/2*d*x))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan 
(1/4*d*x)^2*tan(1/4*c)^2 - d*x*real_part(cos_integral(-1/2*d*x))*sgn(cos(- 
1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*d*x)^2*tan(1/4*c)^2 + 2*d*x*sgn(cos(-1/ 
4*pi + 1/2*d*x + 1/2*c))*sin_integral(1/2*d*x)*tan(1/4*d*x)^2*tan(1/4*c)^2 
 - 2*d*x*imag_part(cos_integral(1/2*d*x))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2* 
c))*tan(1/4*d*x)^2*tan(1/4*c) + 2*d*x*imag_part(cos_integral(-1/2*d*x))*sg 
n(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*d*x)^2*tan(1/4*c) - 2*d*x*real_p 
art(cos_integral(1/2*d*x))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*d*x 
)^2*tan(1/4*c) - 2*d*x*real_part(cos_integral(-1/2*d*x))*sgn(cos(-1/4*pi + 
 1/2*d*x + 1/2*c))*tan(1/4*d*x)^2*tan(1/4*c) - 4*d*x*sgn(cos(-1/4*pi + 1/2 
*d*x + 1/2*c))*sin_integral(1/2*d*x)*tan(1/4*d*x)^2*tan(1/4*c) - d*x*imag_ 
part(cos_integral(1/2*d*x))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*d* 
x)^2 + d*x*imag_part(cos_integral(-1/2*d*x))*sgn(cos(-1/4*pi + 1/2*d*x + 1 
/2*c))*tan(1/4*d*x)^2 + d*x*real_part(cos_integral(1/2*d*x))*sgn(cos(-1/4* 
pi + 1/2*d*x + 1/2*c))*tan(1/4*d*x)^2 + d*x*real_part(cos_integral(-1/2*d* 
x))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*d*x)^2 - 2*d*x*sgn(cos(-1/ 
4*pi + 1/2*d*x + 1/2*c))*sin_integral(1/2*d*x)*tan(1/4*d*x)^2 + d*x*ima...
 

Mupad [F(-1)]

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

int((a + a*sin(c + d*x))^(1/2)/x^2,x)
 

Output:

int((a + a*sin(c + d*x))^(1/2)/x^2, x)
 

Reduce [F]

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

int((a+a*sin(d*x+c))^(1/2)/x^2,x)
 

Output:

(sqrt(a)*( - 2*sqrt(sin(c + d*x) + 1) + int((sqrt(sin(c + d*x) + 1)*cos(c 
+ d*x))/(sin(c + d*x)*x + x),x)*d*x))/(2*x)