Integrand size = 12, antiderivative size = 63 \[ \int \frac {a+b \arcsin (c x)}{x} \, dx=-\frac {i (a+b \arcsin (c x))^2}{2 b}+(a+b \arcsin (c x)) \log \left (1-e^{2 i \arcsin (c x)}\right )-\frac {1}{2} i b \operatorname {PolyLog}\left (2,e^{2 i \arcsin (c x)}\right ) \] Output:
-1/2*I*(a+b*arcsin(c*x))^2/b+(a+b*arcsin(c*x))*ln(1-(I*c*x+(-c^2*x^2+1)^(1 /2))^2)-1/2*I*b*polylog(2,(I*c*x+(-c^2*x^2+1)^(1/2))^2)
Time = 0.05 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.83 \[ \int \frac {a+b \arcsin (c x)}{x} \, dx=b \arcsin (c x) \log \left (1-e^{2 i \arcsin (c x)}\right )+a \log (x)-\frac {1}{2} i b \left (\arcsin (c x)^2+\operatorname {PolyLog}\left (2,e^{2 i \arcsin (c x)}\right )\right ) \] Input:
Integrate[(a + b*ArcSin[c*x])/x,x]
Output:
b*ArcSin[c*x]*Log[1 - E^((2*I)*ArcSin[c*x])] + a*Log[x] - (I/2)*b*(ArcSin[ c*x]^2 + PolyLog[2, E^((2*I)*ArcSin[c*x])])
Time = 0.36 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.13, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {5136, 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.
\(\displaystyle \int \frac {a+b \arcsin (c x)}{x} \, dx\) |
\(\Big \downarrow \) 5136 |
\(\displaystyle \int \frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{c x}d\arcsin (c x)\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\left (\tan \left (\arcsin (c x)+\frac {\pi }{2}\right ) (a+b \arcsin (c x))\right )d\arcsin (c x)\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int (a+b \arcsin (c x)) \tan \left (\arcsin (c x)+\frac {\pi }{2}\right )d\arcsin (c x)\) |
\(\Big \downarrow \) 4200 |
\(\displaystyle 2 i \int -\frac {e^{2 i \arcsin (c x)} (a+b \arcsin (c x))}{1-e^{2 i \arcsin (c x)}}d\arcsin (c x)-\frac {i (a+b \arcsin (c x))^2}{2 b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -2 i \int \frac {e^{2 i \arcsin (c x)} (a+b \arcsin (c x))}{1-e^{2 i \arcsin (c x)}}d\arcsin (c x)-\frac {i (a+b \arcsin (c x))^2}{2 b}\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle -2 i \left (\frac {1}{2} i \log \left (1-e^{2 i \arcsin (c x)}\right ) (a+b \arcsin (c x))-\frac {1}{2} i b \int \log \left (1-e^{2 i \arcsin (c x)}\right )d\arcsin (c x)\right )-\frac {i (a+b \arcsin (c x))^2}{2 b}\) |
\(\Big \downarrow \) 2715 |
\(\displaystyle -2 i \left (\frac {1}{2} i \log \left (1-e^{2 i \arcsin (c x)}\right ) (a+b \arcsin (c x))-\frac {1}{4} b \int e^{-2 i \arcsin (c x)} \log \left (1-e^{2 i \arcsin (c x)}\right )de^{2 i \arcsin (c x)}\right )-\frac {i (a+b \arcsin (c x))^2}{2 b}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle -2 i \left (\frac {1}{2} i \log \left (1-e^{2 i \arcsin (c x)}\right ) (a+b \arcsin (c x))+\frac {1}{4} b \operatorname {PolyLog}\left (2,e^{2 i \arcsin (c x)}\right )\right )-\frac {i (a+b \arcsin (c x))^2}{2 b}\) |
Input:
Int[(a + b*ArcSin[c*x])/x,x]
Output:
((-1/2*I)*(a + b*ArcSin[c*x])^2)/b - (2*I)*((I/2)*(a + b*ArcSin[c*x])*Log[ 1 - E^((2*I)*ArcSin[c*x])] + (b*PolyLog[2, E^((2*I)*ArcSin[c*x])])/4)
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_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[( a + b*x)^n*Cot[x], x], x, ArcSin[c*x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]
Time = 0.10 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.87
method | result | size |
parts | \(a \ln \left (x \right )+b \left (-\frac {i \arcsin \left (c x \right )^{2}}{2}+\arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )-i \operatorname {polylog}\left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )+\arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )-i \operatorname {polylog}\left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )\right )\) | \(118\) |
derivativedivides | \(a \ln \left (c x \right )+b \left (-\frac {i \arcsin \left (c x \right )^{2}}{2}+\arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )-i \operatorname {polylog}\left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )+\arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )-i \operatorname {polylog}\left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )\right )\) | \(120\) |
default | \(a \ln \left (c x \right )+b \left (-\frac {i \arcsin \left (c x \right )^{2}}{2}+\arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )-i \operatorname {polylog}\left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )+\arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )-i \operatorname {polylog}\left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )\right )\) | \(120\) |
Input:
int((a+b*arcsin(c*x))/x,x,method=_RETURNVERBOSE)
Output:
a*ln(x)+b*(-1/2*I*arcsin(c*x)^2+arcsin(c*x)*ln(1-I*c*x-(-c^2*x^2+1)^(1/2)) -I*polylog(2,I*c*x+(-c^2*x^2+1)^(1/2))+arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1) ^(1/2))-I*polylog(2,-I*c*x-(-c^2*x^2+1)^(1/2)))
\[ \int \frac {a+b \arcsin (c x)}{x} \, dx=\int { \frac {b \arcsin \left (c x\right ) + a}{x} \,d x } \] Input:
integrate((a+b*arcsin(c*x))/x,x, algorithm="fricas")
Output:
integral((b*arcsin(c*x) + a)/x, x)
\[ \int \frac {a+b \arcsin (c x)}{x} \, dx=\int \frac {a + b \operatorname {asin}{\left (c x \right )}}{x}\, dx \] Input:
integrate((a+b*asin(c*x))/x,x)
Output:
Integral((a + b*asin(c*x))/x, x)
\[ \int \frac {a+b \arcsin (c x)}{x} \, dx=\int { \frac {b \arcsin \left (c x\right ) + a}{x} \,d x } \] Input:
integrate((a+b*arcsin(c*x))/x,x, algorithm="maxima")
Output:
b*integrate(arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/x, x) + a*log(x)
Exception generated. \[ \int \frac {a+b \arcsin (c x)}{x} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate((a+b*arcsin(c*x))/x,x, algorithm="giac")
Output:
Exception raised: RuntimeError >> an error occurred running a Giac command :INPUT:sage2OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const ve cteur & l) Error: Bad Argument Value
Time = 0.17 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.76 \[ \int \frac {a+b \arcsin (c x)}{x} \, dx=a\,\ln \left (x\right )-\frac {b\,\mathrm {polylog}\left (2,{\mathrm {e}}^{\mathrm {asin}\left (c\,x\right )\,2{}\mathrm {i}}\right )\,1{}\mathrm {i}}{2}-\frac {b\,{\mathrm {asin}\left (c\,x\right )}^2\,1{}\mathrm {i}}{2}+b\,\ln \left (1-{\mathrm {e}}^{\mathrm {asin}\left (c\,x\right )\,2{}\mathrm {i}}\right )\,\mathrm {asin}\left (c\,x\right ) \] Input:
int((a + b*asin(c*x))/x,x)
Output:
a*log(x) - (b*polylog(2, exp(asin(c*x)*2i))*1i)/2 - (b*asin(c*x)^2*1i)/2 + b*log(1 - exp(asin(c*x)*2i))*asin(c*x)
\[ \int \frac {a+b \arcsin (c x)}{x} \, dx=\left (\int \frac {\mathit {asin} \left (c x \right )}{x}d x \right ) b +\mathrm {log}\left (x \right ) a \] Input:
int((a+b*asin(c*x))/x,x)
Output:
int(asin(c*x)/x,x)*b + log(x)*a