Integrand size = 23, antiderivative size = 82 \[ \int \frac {x (a+b \arccos (c x))}{d-c^2 d x^2} \, dx=\frac {i (a+b \arccos (c x))^2}{2 b c^2 d}-\frac {(a+b \arccos (c x)) \log \left (1-e^{2 i \arccos (c x)}\right )}{c^2 d}+\frac {i b \operatorname {PolyLog}\left (2,e^{2 i \arccos (c x)}\right )}{2 c^2 d} \]
1/2*I*(a+b*arccos(c*x))^2/b/c^2/d-(a+b*arccos(c*x))*ln(1-(c*x+I*(-c^2*x^2+ 1)^(1/2))^2)/c^2/d+1/2*I*b*polylog(2,(c*x+I*(-c^2*x^2+1)^(1/2))^2)/c^2/d
Time = 0.21 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.40 \[ \int \frac {x (a+b \arccos (c x))}{d-c^2 d x^2} \, dx=\frac {i \left (b \arccos (c x)^2+2 i b \arccos (c x) \log \left (1-e^{i \arccos (c x)}\right )+2 i b \arccos (c x) \log \left (1+e^{i \arccos (c x)}\right )+i a \log \left (1-c^2 x^2\right )+2 b \operatorname {PolyLog}\left (2,-e^{i \arccos (c x)}\right )+2 b \operatorname {PolyLog}\left (2,e^{i \arccos (c x)}\right )\right )}{2 c^2 d} \]
((I/2)*(b*ArcCos[c*x]^2 + (2*I)*b*ArcCos[c*x]*Log[1 - E^(I*ArcCos[c*x])] + (2*I)*b*ArcCos[c*x]*Log[1 + E^(I*ArcCos[c*x])] + I*a*Log[1 - c^2*x^2] + 2 *b*PolyLog[2, -E^(I*ArcCos[c*x])] + 2*b*PolyLog[2, E^(I*ArcCos[c*x])]))/(c ^2*d)
Time = 0.39 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.96, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {5181, 3042, 25, 4200, 25, 2620, 2715, 2838}
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.
\(\displaystyle \int \frac {x (a+b \arccos (c x))}{d-c^2 d x^2} \, dx\) |
\(\Big \downarrow \) 5181 |
\(\displaystyle -\frac {\int \frac {c x (a+b \arccos (c x))}{\sqrt {1-c^2 x^2}}d\arccos (c x)}{c^2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int -\left ((a+b \arccos (c x)) \tan \left (\arccos (c x)+\frac {\pi }{2}\right )\right )d\arccos (c x)}{c^2 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int (a+b \arccos (c x)) \tan \left (\arccos (c x)+\frac {\pi }{2}\right )d\arccos (c x)}{c^2 d}\) |
\(\Big \downarrow \) 4200 |
\(\displaystyle -\frac {2 i \int -\frac {e^{2 i \arccos (c x)} (a+b \arccos (c x))}{1-e^{2 i \arccos (c x)}}d\arccos (c x)-\frac {i (a+b \arccos (c x))^2}{2 b}}{c^2 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {-2 i \int \frac {e^{2 i \arccos (c x)} (a+b \arccos (c x))}{1-e^{2 i \arccos (c x)}}d\arccos (c x)-\frac {i (a+b \arccos (c x))^2}{2 b}}{c^2 d}\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle -\frac {-2 i \left (\frac {1}{2} i \log \left (1-e^{2 i \arccos (c x)}\right ) (a+b \arccos (c x))-\frac {1}{2} i b \int \log \left (1-e^{2 i \arccos (c x)}\right )d\arccos (c x)\right )-\frac {i (a+b \arccos (c x))^2}{2 b}}{c^2 d}\) |
\(\Big \downarrow \) 2715 |
\(\displaystyle -\frac {-2 i \left (\frac {1}{2} i \log \left (1-e^{2 i \arccos (c x)}\right ) (a+b \arccos (c x))-\frac {1}{4} b \int e^{-2 i \arccos (c x)} \log \left (1-e^{2 i \arccos (c x)}\right )de^{2 i \arccos (c x)}\right )-\frac {i (a+b \arccos (c x))^2}{2 b}}{c^2 d}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle -\frac {-2 i \left (\frac {1}{2} i \log \left (1-e^{2 i \arccos (c x)}\right ) (a+b \arccos (c x))+\frac {1}{4} b \operatorname {PolyLog}\left (2,e^{2 i \arccos (c x)}\right )\right )-\frac {i (a+b \arccos (c x))^2}{2 b}}{c^2 d}\) |
-((((-1/2*I)*(a + b*ArcCos[c*x])^2)/b - (2*I)*((I/2)*(a + b*ArcCos[c*x])*L og[1 - E^((2*I)*ArcCos[c*x])] + (b*PolyLog[2, E^((2*I)*ArcCos[c*x])])/4))/ (c^2*d))
3.1.3.3.1 Defintions of rubi rules used
Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ ((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp [((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si mp[d*(m/(b*f*g*n*Log[F])) Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x )))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Simp[1/(d*e*n*Log[F]) Subst[Int[Log[a + b*x]/x, x], x, (F^(e*(c + d*x) ))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]
Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2 , (-c)*e*x^n]/n, x] /; FreeQ[{c, d, e, n}, x] && EqQ[c*d, 1]
Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol ] :> Simp[I*((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[2*I Int[(c + d*x)^ m*E^(2*I*k*Pi)*(E^(2*I*(e + f*x))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))), x] , x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]
Int[(((a_.) + ArcCos[(c_.)*(x_)]*(b_.))^(n_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[1/e Subst[Int[(a + b*x)^n*Cot[x], x], x, ArcCos[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[n, 0]
Time = 1.22 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.73
method | result | size |
parts | \(-\frac {a \ln \left (c^{2} x^{2}-1\right )}{2 d \,c^{2}}-\frac {b \left (-\frac {i \arccos \left (c x \right )^{2}}{2}+\arccos \left (c x \right ) \ln \left (1-c x -i \sqrt {-c^{2} x^{2}+1}\right )-i \operatorname {polylog}\left (2, c x +i \sqrt {-c^{2} x^{2}+1}\right )+\arccos \left (c x \right ) \ln \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )-i \operatorname {polylog}\left (2, -c x -i \sqrt {-c^{2} x^{2}+1}\right )\right )}{d \,c^{2}}\) | \(142\) |
derivativedivides | \(\frac {-\frac {a \left (\frac {\ln \left (c x -1\right )}{2}+\frac {\ln \left (c x +1\right )}{2}\right )}{d}-\frac {b \left (-\frac {i \arccos \left (c x \right )^{2}}{2}+\arccos \left (c x \right ) \ln \left (1-c x -i \sqrt {-c^{2} x^{2}+1}\right )-i \operatorname {polylog}\left (2, c x +i \sqrt {-c^{2} x^{2}+1}\right )+\arccos \left (c x \right ) \ln \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )-i \operatorname {polylog}\left (2, -c x -i \sqrt {-c^{2} x^{2}+1}\right )\right )}{d}}{c^{2}}\) | \(147\) |
default | \(\frac {-\frac {a \left (\frac {\ln \left (c x -1\right )}{2}+\frac {\ln \left (c x +1\right )}{2}\right )}{d}-\frac {b \left (-\frac {i \arccos \left (c x \right )^{2}}{2}+\arccos \left (c x \right ) \ln \left (1-c x -i \sqrt {-c^{2} x^{2}+1}\right )-i \operatorname {polylog}\left (2, c x +i \sqrt {-c^{2} x^{2}+1}\right )+\arccos \left (c x \right ) \ln \left (1+c x +i \sqrt {-c^{2} x^{2}+1}\right )-i \operatorname {polylog}\left (2, -c x -i \sqrt {-c^{2} x^{2}+1}\right )\right )}{d}}{c^{2}}\) | \(147\) |
-1/2*a/d/c^2*ln(c^2*x^2-1)-b/d/c^2*(-1/2*I*arccos(c*x)^2+arccos(c*x)*ln(1- c*x-I*(-c^2*x^2+1)^(1/2))-I*polylog(2,c*x+I*(-c^2*x^2+1)^(1/2))+arccos(c*x )*ln(1+c*x+I*(-c^2*x^2+1)^(1/2))-I*polylog(2,-c*x-I*(-c^2*x^2+1)^(1/2)))
\[ \int \frac {x (a+b \arccos (c x))}{d-c^2 d x^2} \, dx=\int { -\frac {{\left (b \arccos \left (c x\right ) + a\right )} x}{c^{2} d x^{2} - d} \,d x } \]
\[ \int \frac {x (a+b \arccos (c x))}{d-c^2 d x^2} \, dx=- \frac {\int \frac {a x}{c^{2} x^{2} - 1}\, dx + \int \frac {b x \operatorname {acos}{\left (c x \right )}}{c^{2} x^{2} - 1}\, dx}{d} \]
\[ \int \frac {x (a+b \arccos (c x))}{d-c^2 d x^2} \, dx=\int { -\frac {{\left (b \arccos \left (c x\right ) + a\right )} x}{c^{2} d x^{2} - d} \,d x } \]
1/2*(2*c^2*d*integrate(1/2*(e^(1/2*log(c*x + 1) + 1/2*log(-c*x + 1))*log(c *x + 1) + e^(1/2*log(c*x + 1) + 1/2*log(-c*x + 1))*log(-c*x + 1))/(c^5*d*x ^4 - c^3*d*x^2 + (c^3*d*x^2 - c*d)*e^(log(c*x + 1) + log(-c*x + 1))), x) - (log(c*x + 1) + log(-c*x + 1))*arctan2(sqrt(c*x + 1)*sqrt(-c*x + 1), c*x) )*b/(c^2*d) - 1/2*a*log(c^2*d*x^2 - d)/(c^2*d)
\[ \int \frac {x (a+b \arccos (c x))}{d-c^2 d x^2} \, dx=\int { -\frac {{\left (b \arccos \left (c x\right ) + a\right )} x}{c^{2} d x^{2} - d} \,d x } \]
Timed out. \[ \int \frac {x (a+b \arccos (c x))}{d-c^2 d x^2} \, dx=\int \frac {x\,\left (a+b\,\mathrm {acos}\left (c\,x\right )\right )}{d-c^2\,d\,x^2} \,d x \]