Integrand size = 12, antiderivative size = 63 \[ \int \frac {x}{a+b \arccos (c x)} \, dx=\frac {\operatorname {CosIntegral}\left (\frac {2 (a+b \arccos (c x))}{b}\right ) \sin \left (\frac {2 a}{b}\right )}{2 b c^2}-\frac {\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arccos (c x))}{b}\right )}{2 b c^2} \]
-1/2*cos(2*a/b)*Si(2*(a+b*arccos(c*x))/b)/b/c^2+1/2*Ci(2*(a+b*arccos(c*x)) /b)*sin(2*a/b)/b/c^2
Time = 0.06 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.89 \[ \int \frac {x}{a+b \arccos (c x)} \, dx=-\frac {-\operatorname {CosIntegral}\left (\frac {2 a}{b}+2 \arccos (c x)\right ) \sin \left (\frac {2 a}{b}\right )+\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 a}{b}+2 \arccos (c x)\right )}{2 b c^2} \]
-1/2*(-(CosIntegral[(2*a)/b + 2*ArcCos[c*x]]*Sin[(2*a)/b]) + Cos[(2*a)/b]* SinIntegral[(2*a)/b + 2*ArcCos[c*x]])/(b*c^2)
Time = 0.44 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.89, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.833, Rules used = {5147, 25, 4906, 27, 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.
\(\displaystyle \int \frac {x}{a+b \arccos (c x)} \, dx\) |
\(\Big \downarrow \) 5147 |
\(\displaystyle -\frac {\int -\frac {\cos \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right ) \sin \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b c^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {\cos \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right ) \sin \left (\frac {a}{b}-\frac {a+b \arccos (c x)}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{b c^2}\) |
\(\Big \downarrow \) 4906 |
\(\displaystyle \frac {\int \frac {\sin \left (\frac {2 a}{b}-\frac {2 (a+b \arccos (c x))}{b}\right )}{2 (a+b \arccos (c x))}d(a+b \arccos (c x))}{b c^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\sin \left (\frac {2 a}{b}-\frac {2 (a+b \arccos (c x))}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{2 b c^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\sin \left (\frac {2 a}{b}-\frac {2 (a+b \arccos (c x))}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{2 b c^2}\) |
\(\Big \downarrow \) 3784 |
\(\displaystyle -\frac {-\sin \left (\frac {2 a}{b}\right ) \int \frac {\cos \left (\frac {2 (a+b \arccos (c x))}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))-\cos \left (\frac {2 a}{b}\right ) \int -\frac {\sin \left (\frac {2 (a+b \arccos (c x))}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{2 b c^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\cos \left (\frac {2 a}{b}\right ) \int \frac {\sin \left (\frac {2 (a+b \arccos (c x))}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))-\sin \left (\frac {2 a}{b}\right ) \int \frac {\cos \left (\frac {2 (a+b \arccos (c x))}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{2 b c^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\cos \left (\frac {2 a}{b}\right ) \int \frac {\sin \left (\frac {2 (a+b \arccos (c x))}{b}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))-\sin \left (\frac {2 a}{b}\right ) \int \frac {\sin \left (\frac {2 (a+b \arccos (c x))}{b}+\frac {\pi }{2}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{2 b c^2}\) |
\(\Big \downarrow \) 3780 |
\(\displaystyle -\frac {\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arccos (c x))}{b}\right )-\sin \left (\frac {2 a}{b}\right ) \int \frac {\sin \left (\frac {2 (a+b \arccos (c x))}{b}+\frac {\pi }{2}\right )}{a+b \arccos (c x)}d(a+b \arccos (c x))}{2 b c^2}\) |
\(\Big \downarrow \) 3783 |
\(\displaystyle -\frac {\cos \left (\frac {2 a}{b}\right ) \text {Si}\left (\frac {2 (a+b \arccos (c x))}{b}\right )-\sin \left (\frac {2 a}{b}\right ) \operatorname {CosIntegral}\left (\frac {2 (a+b \arccos (c x))}{b}\right )}{2 b c^2}\) |
-1/2*(-(CosIntegral[(2*(a + b*ArcCos[c*x]))/b]*Sin[(2*a)/b]) + Cos[(2*a)/b ]*SinIntegral[(2*(a + b*ArcCos[c*x]))/b])/(b*c^2)
3.2.59.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b _.)*(x_)]^(n_.), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin[a + b*x ]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IG tQ[p, 0]
Int[((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[- (b*c^(m + 1))^(-1) Subst[Int[x^n*Cos[-a/b + x/b]^m*Sin[-a/b + x/b], x], x , a + b*ArcCos[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]
Time = 0.41 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.92
method | result | size |
derivativedivides | \(\frac {-\frac {\operatorname {Si}\left (2 \arccos \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right )}{2 b}+\frac {\operatorname {Ci}\left (2 \arccos \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right )}{2 b}}{c^{2}}\) | \(58\) |
default | \(\frac {-\frac {\operatorname {Si}\left (2 \arccos \left (c x \right )+\frac {2 a}{b}\right ) \cos \left (\frac {2 a}{b}\right )}{2 b}+\frac {\operatorname {Ci}\left (2 \arccos \left (c x \right )+\frac {2 a}{b}\right ) \sin \left (\frac {2 a}{b}\right )}{2 b}}{c^{2}}\) | \(58\) |
\[ \int \frac {x}{a+b \arccos (c x)} \, dx=\int { \frac {x}{b \arccos \left (c x\right ) + a} \,d x } \]
\[ \int \frac {x}{a+b \arccos (c x)} \, dx=\int \frac {x}{a + b \operatorname {acos}{\left (c x \right )}}\, dx \]
\[ \int \frac {x}{a+b \arccos (c x)} \, dx=\int { \frac {x}{b \arccos \left (c x\right ) + a} \,d x } \]
Time = 0.29 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.37 \[ \int \frac {x}{a+b \arccos (c x)} \, dx=\frac {\cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (c x\right )\right ) \sin \left (\frac {a}{b}\right )}{b c^{2}} - \frac {\cos \left (\frac {a}{b}\right )^{2} \operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (c x\right )\right )}{b c^{2}} + \frac {\operatorname {Si}\left (\frac {2 \, a}{b} + 2 \, \arccos \left (c x\right )\right )}{2 \, b c^{2}} \]
cos(a/b)*cos_integral(2*a/b + 2*arccos(c*x))*sin(a/b)/(b*c^2) - cos(a/b)^2 *sin_integral(2*a/b + 2*arccos(c*x))/(b*c^2) + 1/2*sin_integral(2*a/b + 2* arccos(c*x))/(b*c^2)
Timed out. \[ \int \frac {x}{a+b \arccos (c x)} \, dx=\int \frac {x}{a+b\,\mathrm {acos}\left (c\,x\right )} \,d x \]