Integrand size = 18, antiderivative size = 174 \[ \int \frac {\sqrt {a+a \sin (c+d x)}}{x^3} \, dx=-\frac {\sqrt {a+a \sin (c+d x)}}{2 x^2}-\frac {d \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \sqrt {a+a \sin (c+d x)}}{4 x}-\frac {1}{8} d^2 \operatorname {CosIntegral}\left (\frac {d x}{2}\right ) \csc \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \sin \left (\frac {1}{4} (2 c+\pi )\right ) \sqrt {a+a \sin (c+d x)}-\frac {1}{8} d^2 \cos \left (\frac {1}{4} (2 c+\pi )\right ) \csc \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right ) \sqrt {a+a \sin (c+d x)} \text {Si}\left (\frac {d x}{2}\right ) \]
-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*cos(1/2*c+1/4*Pi)*csc(1/2*c+1/4*Pi+1/2*d*x)*Si(1/2* d*x)*(a+a*sin(d*x+c))^(1/2)-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)
Time = 0.51 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.88 \[ \int \frac {\sqrt {a+a \sin (c+d x)}}{x^3} \, dx=-\frac {\sqrt {a (1+\sin (c+d x))} \left (4 \cos \left (\frac {1}{2} (c+d x)\right )+2 d x \cos \left (\frac {1}{2} (c+d x)\right )+d^2 x^2 \operatorname {CosIntegral}\left (\frac {d x}{2}\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right )+4 \sin \left (\frac {1}{2} (c+d x)\right )-2 d x \sin \left (\frac {1}{2} (c+d x)\right )+d^2 x^2 \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \text {Si}\left (\frac {d x}{2}\right )\right )}{8 x^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )} \]
-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]))
Time = 0.66 (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 |
\(\displaystyle \csc \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right ) \sqrt {a \sin (c+d x)+a} \left (\frac {1}{4} d \left (-\frac {1}{2} d \left (\sin \left (\frac {1}{4} (2 c+\pi )\right ) \operatorname {CosIntegral}\left (\frac {d x}{2}\right )+\cos \left (\frac {1}{4} (2 c+\pi )\right ) \text {Si}\left (\frac {d x}{2}\right )\right )-\frac {\cos \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{x}\right )-\frac {\sin \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{2 x^2}\right )\) |
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)
3.2.27.3.1 Defintions of rubi rules used
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]
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]
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]
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]
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])
\[\int \frac {\sqrt {a +a \sin \left (d x +c \right )}}{x^{3}}d x\]
Exception generated. \[ \int \frac {\sqrt {a+a \sin (c+d x)}}{x^3} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (has polynomial part)
\[ \int \frac {\sqrt {a+a \sin (c+d x)}}{x^3} \, dx=\int \frac {\sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )}}{x^{3}}\, dx \]
\[ \int \frac {\sqrt {a+a \sin (c+d x)}}{x^3} \, dx=\int { \frac {\sqrt {a \sin \left (d x + c\right ) + a}}{x^{3}} \,d x } \]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.48 (sec) , antiderivative size = 1487, normalized size of antiderivative = 8.55 \[ \int \frac {\sqrt {a+a \sin (c+d x)}}{x^3} \, dx=\text {Too large to display} \]
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...
Timed out. \[ \int \frac {\sqrt {a+a \sin (c+d x)}}{x^3} \, dx=\int \frac {\sqrt {a+a\,\sin \left (c+d\,x\right )}}{x^3} \,d x \]