Integrand size = 33, antiderivative size = 86 \[ \int \frac {\sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{x} \, dx=\cos (e) \operatorname {CosIntegral}(f x) \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}-\sec (e+f x) \sin (e) \sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)} \text {Si}(f x) \]
Ci(f*x)*cos(e)*sec(f*x+e)*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2)-se c(f*x+e)*Si(f*x)*sin(e)*(a-a*sin(f*x+e))^(1/2)*(c+c*sin(f*x+e))^(1/2)
Time = 0.76 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.60 \[ \int \frac {\sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{x} \, dx=\sec (e+f x) \sqrt {c (1+\sin (e+f x))} \sqrt {a-a \sin (e+f x)} (\cos (e) \operatorname {CosIntegral}(f x)-\sin (e) \text {Si}(f x)) \]
Sec[e + f*x]*Sqrt[c*(1 + Sin[e + f*x])]*Sqrt[a - a*Sin[e + f*x]]*(Cos[e]*C osIntegral[f*x] - Sin[e]*SinIntegral[f*x])
Time = 0.49 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.60, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {5115, 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.
\(\displaystyle \int \frac {\sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c}}{x} \, dx\) |
\(\Big \downarrow \) 5115 |
\(\displaystyle \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \int \frac {\cos (e+f x)}{x}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \int \frac {\sin \left (e+f x+\frac {\pi }{2}\right )}{x}dx\) |
\(\Big \downarrow \) 3784 |
\(\displaystyle \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \left (\cos (e) \int \frac {\cos (f x)}{x}dx-\sin (e) \int \frac {\sin (f x)}{x}dx\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \left (\cos (e) \int \frac {\sin \left (f x+\frac {\pi }{2}\right )}{x}dx-\sin (e) \int \frac {\sin (f x)}{x}dx\right )\) |
\(\Big \downarrow \) 3780 |
\(\displaystyle \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} \left (\cos (e) \int \frac {\sin \left (f x+\frac {\pi }{2}\right )}{x}dx-\sin (e) \text {Si}(f x)\right )\) |
\(\Big \downarrow \) 3783 |
\(\displaystyle \sec (e+f x) \sqrt {a-a \sin (e+f x)} \sqrt {c \sin (e+f x)+c} (\cos (e) \operatorname {CosIntegral}(f x)-\sin (e) \text {Si}(f x))\) |
Sec[e + f*x]*Sqrt[a - a*Sin[e + f*x]]*Sqrt[c + c*Sin[e + f*x]]*(Cos[e]*Cos Integral[f*x] - Sin[e]*SinIntegral[f*x])
3.2.68.3.1 Defintions of rubi rules used
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[((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]
\[\int \frac {\sqrt {a -\sin \left (f x +e \right ) a}\, \sqrt {c +c \sin \left (f x +e \right )}}{x}d x\]
Exception generated. \[ \int \frac {\sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{x} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (has polynomial part)
\[ \int \frac {\sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{x} \, dx=\int \frac {\sqrt {c \left (\sin {\left (e + f x \right )} + 1\right )} \sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right )}}{x}\, dx \]
\[ \int \frac {\sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{x} \, dx=\int { \frac {\sqrt {-a \sin \left (f x + e\right ) + a} \sqrt {c \sin \left (f x + e\right ) + c}}{x} \,d x } \]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.32 (sec) , antiderivative size = 274, normalized size of antiderivative = 3.19 \[ \int \frac {\sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{x} \, dx=\frac {{\left (\Re \left ( \operatorname {Ci}\left (f x\right ) \right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \tan \left (\frac {1}{2} \, e\right )^{2} + \Re \left ( \operatorname {Ci}\left (-f x\right ) \right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \tan \left (\frac {1}{2} \, e\right )^{2} + 2 \, \Im \left ( \operatorname {Ci}\left (f x\right ) \right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \tan \left (\frac {1}{2} \, e\right ) - 2 \, \Im \left ( \operatorname {Ci}\left (-f x\right ) \right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \tan \left (\frac {1}{2} \, e\right ) + 4 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \operatorname {Si}\left (f x\right ) \tan \left (\frac {1}{2} \, e\right ) - \Re \left ( \operatorname {Ci}\left (f x\right ) \right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - \Re \left ( \operatorname {Ci}\left (-f x\right ) \right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )\right )} \sqrt {a} \sqrt {c}}{2 \, {\left (\tan \left (\frac {1}{2} \, e\right )^{2} + 1\right )}} \]
1/2*(real_part(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))*tan(1/2*e)^2 + real_part(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)) *tan(1/2*e)^2 + 2*imag_part(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))*tan(1/2*e) - 2*imag_part(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))*tan(1/2*e) + 4*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_integral(f*x)*tan(1/2*e) - real_part(cos_in tegral(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)) - real_part(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)))*sqrt(a)*sqrt(c)/(tan(1/2*e)^2 + 1)
Timed out. \[ \int \frac {\sqrt {a-a \sin (e+f x)} \sqrt {c+c \sin (e+f x)}}{x} \, dx=\int \frac {\sqrt {a-a\,\sin \left (e+f\,x\right )}\,\sqrt {c+c\,\sin \left (e+f\,x\right )}}{x} \,d x \]