\(\int \frac {\sqrt {a+a \sin (c+d x)}}{x^3} \, dx\) [127]

Optimal result

Integrand size = 18, antiderivative size = 174 \[ \int \frac {\sqrt {a+a \sin (c+d x)}}{x^3} \, dx=-\frac {\sqrt {a+a \sin (c+d x)}}{2 x^2}-\frac {d \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \sqrt {a+a \sin (c+d x)}}{4 x}-\frac {1}{8} d^2 \operatorname {CosIntegral}\left (\frac {d x}{2}\right ) \csc \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \sin \left (\frac {1}{4} (2 c+\pi )\right ) \sqrt {a+a \sin (c+d x)}-\frac {1}{8} d^2 \cos \left (\frac {1}{4} (2 c+\pi )\right ) \csc \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \sqrt {a+a \sin (c+d x)} \text {Si}\left (\frac {d x}{2}\right ) \] Output:

-1/2*(a+a*sin(d*x+c))^(1/2)/x^2-1/4*d*cot(1/2*c+1/4*Pi+1/2*d*x)*(a+a*sin(d 
*x+c))^(1/2)/x-1/8*d^2*Ci(1/2*d*x)*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)-1/8*d^2*cos(1/2*c+1/4*Pi)*csc(1/2*c+1/4*Pi+1/2 
*d*x)*(a+a*sin(d*x+c))^(1/2)*Si(1/2*d*x)
 

Mathematica [A] (verified)

Time = 0.81 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.88 \[ \int \frac {\sqrt {a+a \sin (c+d x)}}{x^3} \, dx=-\frac {\sqrt {a (1+\sin (c+d x))} \left (4 \cos \left (\frac {1}{2} (c+d x)\right )+2 d x \cos \left (\frac {1}{2} (c+d x)\right )+d^2 x^2 \operatorname {CosIntegral}\left (\frac {d x}{2}\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right )+4 \sin \left (\frac {1}{2} (c+d x)\right )-2 d x \sin \left (\frac {1}{2} (c+d x)\right )+d^2 x^2 \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \text {Si}\left (\frac {d x}{2}\right )\right )}{8 x^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \] Input:

Integrate[Sqrt[a + a*Sin[c + d*x]]/x^3,x]
 

Output:

-1/8*(Sqrt[a*(1 + Sin[c + d*x])]*(4*Cos[(c + d*x)/2] + 2*d*x*Cos[(c + d*x) 
/2] + d^2*x^2*CosIntegral[(d*x)/2]*(Cos[c/2] + Sin[c/2]) + 4*Sin[(c + d*x) 
/2] - 2*d*x*Sin[(c + d*x)/2] + d^2*x^2*(Cos[c/2] - Sin[c/2])*SinIntegral[( 
d*x)/2]))/(x^2*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.75, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 3800, 3042, 3778, 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.

\(\displaystyle \int \frac {\sqrt {a \sin (c+d x)+a}}{x^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sqrt {a \sin (c+d x)+a}}{x^3}dx\)

\(\Big \downarrow \) 3800

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3778

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3778

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3784

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3780

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

\(\Big \downarrow \) 3783

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

Input:

Int[Sqrt[a + a*Sin[c + d*x]]/x^3,x]
 

Output:

Csc[c/2 + Pi/4 + (d*x)/2]*Sqrt[a + a*Sin[c + d*x]]*(-1/2*Sin[c/2 + Pi/4 + 
(d*x)/2]/x^2 + (d*(-(Cos[c/2 + Pi/4 + (d*x)/2]/x) - (d*(CosIntegral[(d*x)/ 
2]*Sin[(2*c + Pi)/4] + Cos[(2*c + Pi)/4]*SinIntegral[(d*x)/2]))/2))/4)
 

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 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]

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

Input:

int((a+a*sin(d*x+c))^(1/2)/x^3,x)
 

Output:

int((a+a*sin(d*x+c))^(1/2)/x^3,x)
 

Fricas [F(-2)]

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

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

Output:

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

Sympy [F]

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

integrate((a+a*sin(d*x+c))**(1/2)/x**3,x)
 

Output:

Integral(sqrt(a*(sin(c + d*x) + 1))/x**3, x)
 

Maxima [F]

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

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

Output:

integrate(sqrt(a*sin(d*x + c) + a)/x^3, x)
 

Giac [C] (verification not implemented)

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

Time = 0.51 (sec) , antiderivative size = 1487, normalized size of antiderivative = 8.55 \[ \int \frac {\sqrt {a+a \sin (c+d x)}}{x^3} \, dx=\text {Too large to display} \] Input:

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

Output:

1/16*sqrt(2)*(d^2*x^2*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^2*x^2*imag_part(cos_integ 
ral(-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^2*x^2*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^2*x^2*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^2*x^2*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^2*x^2*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^2*x^2*imag_pa 
rt(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^2*x^2*real_part(cos_integral(1/2*d*x))*sgn(cos(-1/4*p 
i + 1/2*d*x + 1/2*c))*tan(1/4*d*x)^2*tan(1/4*c) - 2*d^2*x^2*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^2*x^2*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^2*x^2*imag_part(cos_integral(1/2*d*x))*s 
gn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*tan(1/4*d*x)^2 + d^2*x^2*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 
^2*x^2*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^2*x^2*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^2*x^2*sgn(cos(-1/4*pi + 1/2...
 

Mupad [F(-1)]

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

int((a + a*sin(c + d*x))^(1/2)/x^3,x)
 

Output:

int((a + a*sin(c + d*x))^(1/2)/x^3, x)
 

Reduce [F]

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

int((a+a*sin(d*x+c))^(1/2)/x^3,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**2 + x**2),x)*d*x**2))/(4*x**2)