Integrand size = 26, antiderivative size = 54 \[ \int \frac {x}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx=-\frac {\operatorname {CosIntegral}\left (\frac {a+b \arcsin (c x)}{b}\right ) \sin \left (\frac {a}{b}\right )}{b c^2}+\frac {\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arcsin (c x)}{b}\right )}{b c^2} \]
Time = 0.17 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.83 \[ \int \frac {x}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx=\frac {-\operatorname {CosIntegral}\left (\frac {a}{b}+\arcsin (c x)\right ) \sin \left (\frac {a}{b}\right )+\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\arcsin (c x)\right )}{b c^2} \]
(-(CosIntegral[a/b + ArcSin[c*x]]*Sin[a/b]) + Cos[a/b]*SinIntegral[a/b + A rcSin[c*x]])/(b*c^2)
Time = 0.44 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.91, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {5224, 25, 3042, 3784, 25, 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.
\(\displaystyle \int \frac {x}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx\) |
\(\Big \downarrow \) 5224 |
\(\displaystyle \frac {\int -\frac {\sin \left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right )}{a+b \arcsin (c x)}d(a+b \arcsin (c x))}{b c^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {\sin \left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right )}{a+b \arcsin (c x)}d(a+b \arcsin (c x))}{b c^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \frac {\sin \left (\frac {a}{b}-\frac {a+b \arcsin (c x)}{b}\right )}{a+b \arcsin (c x)}d(a+b \arcsin (c x))}{b c^2}\) |
\(\Big \downarrow \) 3784 |
\(\displaystyle \frac {-\sin \left (\frac {a}{b}\right ) \int \frac {\cos \left (\frac {a+b \arcsin (c x)}{b}\right )}{a+b \arcsin (c x)}d(a+b \arcsin (c x))-\cos \left (\frac {a}{b}\right ) \int -\frac {\sin \left (\frac {a+b \arcsin (c x)}{b}\right )}{a+b \arcsin (c x)}d(a+b \arcsin (c x))}{b c^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\cos \left (\frac {a}{b}\right ) \int \frac {\sin \left (\frac {a+b \arcsin (c x)}{b}\right )}{a+b \arcsin (c x)}d(a+b \arcsin (c x))-\sin \left (\frac {a}{b}\right ) \int \frac {\cos \left (\frac {a+b \arcsin (c x)}{b}\right )}{a+b \arcsin (c x)}d(a+b \arcsin (c x))}{b c^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\cos \left (\frac {a}{b}\right ) \int \frac {\sin \left (\frac {a+b \arcsin (c x)}{b}\right )}{a+b \arcsin (c x)}d(a+b \arcsin (c x))-\sin \left (\frac {a}{b}\right ) \int \frac {\sin \left (\frac {a+b \arcsin (c x)}{b}+\frac {\pi }{2}\right )}{a+b \arcsin (c x)}d(a+b \arcsin (c x))}{b c^2}\) |
\(\Big \downarrow \) 3780 |
\(\displaystyle \frac {\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arcsin (c x)}{b}\right )-\sin \left (\frac {a}{b}\right ) \int \frac {\sin \left (\frac {a+b \arcsin (c x)}{b}+\frac {\pi }{2}\right )}{a+b \arcsin (c x)}d(a+b \arcsin (c x))}{b c^2}\) |
\(\Big \downarrow \) 3783 |
\(\displaystyle \frac {\cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \arcsin (c x)}{b}\right )-\sin \left (\frac {a}{b}\right ) \operatorname {CosIntegral}\left (\frac {a+b \arcsin (c x)}{b}\right )}{b c^2}\) |
(-(CosIntegral[(a + b*ArcSin[c*x])/b]*Sin[a/b]) + Cos[a/b]*SinIntegral[(a + b*ArcSin[c*x])/b])/(b*c^2)
3.4.53.3.1 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[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_.)*((d_) + (e_.)*(x_)^ 2)^(p_.), x_Symbol] :> Simp[(1/(b*c^(m + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x ^2)^p] Subst[Int[x^n*Sin[-a/b + x/b]^m*Cos[-a/b + x/b]^(2*p + 1), x], x, a + b*ArcSin[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && IGtQ[2*p + 2, 0] && IGtQ[m, 0]
Time = 0.12 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85
method | result | size |
default | \(\frac {\operatorname {Si}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right )-\operatorname {Ci}\left (\arcsin \left (c x \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right )}{c^{2} b}\) | \(46\) |
\[ \int \frac {x}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx=\int { \frac {x}{\sqrt {-c^{2} x^{2} + 1} {\left (b \arcsin \left (c x\right ) + a\right )}} \,d x } \]
\[ \int \frac {x}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx=\int \frac {x}{\sqrt {- \left (c x - 1\right ) \left (c x + 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right )}\, dx \]
\[ \int \frac {x}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx=\int { \frac {x}{\sqrt {-c^{2} x^{2} + 1} {\left (b \arcsin \left (c x\right ) + a\right )}} \,d x } \]
Time = 0.32 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.93 \[ \int \frac {x}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx=-\frac {\operatorname {Ci}\left (\frac {a}{b} + \arcsin \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{b c^{2}} + \frac {\cos \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arcsin \left (c x\right )\right )}{b c^{2}} \]
-cos_integral(a/b + arcsin(c*x))*sin(a/b)/(b*c^2) + cos(a/b)*sin_integral( a/b + arcsin(c*x))/(b*c^2)
Timed out. \[ \int \frac {x}{\sqrt {1-c^2 x^2} (a+b \arcsin (c x))} \, dx=\int \frac {x}{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\sqrt {1-c^2\,x^2}} \,d x \]