Integrand size = 35, antiderivative size = 287 \[ \int \frac {(a+b \arcsin (c x))^2}{x \sqrt {d+c d x} \sqrt {e-c e x}} \, dx=-\frac {2 \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2 \text {arctanh}\left (e^{i \arcsin (c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}+\frac {2 i b \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) \operatorname {PolyLog}\left (2,-e^{i \arcsin (c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}-\frac {2 i b \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) \operatorname {PolyLog}\left (2,e^{i \arcsin (c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}-\frac {2 b^2 \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (3,-e^{i \arcsin (c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}+\frac {2 b^2 \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (3,e^{i \arcsin (c x)}\right )}{\sqrt {d+c d x} \sqrt {e-c e x}} \]
-2*(a+b*arcsin(c*x))^2*arctanh(I*c*x+(-c^2*x^2+1)^(1/2))*(-c^2*x^2+1)^(1/2 )/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+2*I*b*(a+b*arcsin(c*x))*polylog(2,-I*c* x-(-c^2*x^2+1)^(1/2))*(-c^2*x^2+1)^(1/2)/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)- 2*I*b*(a+b*arcsin(c*x))*polylog(2,I*c*x+(-c^2*x^2+1)^(1/2))*(-c^2*x^2+1)^( 1/2)/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-2*b^2*polylog(3,-I*c*x-(-c^2*x^2+1)^ (1/2))*(-c^2*x^2+1)^(1/2)/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+2*b^2*polylog(3 ,I*c*x+(-c^2*x^2+1)^(1/2))*(-c^2*x^2+1)^(1/2)/(c*d*x+d)^(1/2)/(-c*e*x+e)^( 1/2)
Time = 2.62 (sec) , antiderivative size = 336, normalized size of antiderivative = 1.17 \[ \int \frac {(a+b \arcsin (c x))^2}{x \sqrt {d+c d x} \sqrt {e-c e x}} \, dx=\frac {a^2 \log (c x)}{\sqrt {d} \sqrt {e}}-\frac {a^2 \log \left (d e+\sqrt {d} \sqrt {e} \sqrt {d+c d x} \sqrt {e-c e x}\right )}{\sqrt {d} \sqrt {e}}+\frac {2 a b \sqrt {1-c^2 x^2} \left (\arcsin (c x) \left (\log \left (1-e^{i \arcsin (c x)}\right )-\log \left (1+e^{i \arcsin (c x)}\right )\right )+i \operatorname {PolyLog}\left (2,-e^{i \arcsin (c x)}\right )-i \operatorname {PolyLog}\left (2,e^{i \arcsin (c x)}\right )\right )}{\sqrt {d+c d x} \sqrt {e-c e x}}+\frac {b^2 \sqrt {1-c^2 x^2} \left (\arcsin (c x)^2 \log \left (1-e^{i \arcsin (c x)}\right )-\arcsin (c x)^2 \log \left (1+e^{i \arcsin (c x)}\right )+2 i \arcsin (c x) \operatorname {PolyLog}\left (2,-e^{i \arcsin (c x)}\right )-2 i \arcsin (c x) \operatorname {PolyLog}\left (2,e^{i \arcsin (c x)}\right )-2 \operatorname {PolyLog}\left (3,-e^{i \arcsin (c x)}\right )+2 \operatorname {PolyLog}\left (3,e^{i \arcsin (c x)}\right )\right )}{\sqrt {d+c d x} \sqrt {e-c e x}} \]
(a^2*Log[c*x])/(Sqrt[d]*Sqrt[e]) - (a^2*Log[d*e + Sqrt[d]*Sqrt[e]*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]])/(Sqrt[d]*Sqrt[e]) + (2*a*b*Sqrt[1 - c^2*x^2]*(Ar cSin[c*x]*(Log[1 - E^(I*ArcSin[c*x])] - Log[1 + E^(I*ArcSin[c*x])]) + I*Po lyLog[2, -E^(I*ArcSin[c*x])] - I*PolyLog[2, E^(I*ArcSin[c*x])]))/(Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) + (b^2*Sqrt[1 - c^2*x^2]*(ArcSin[c*x]^2*Log[1 - E ^(I*ArcSin[c*x])] - ArcSin[c*x]^2*Log[1 + E^(I*ArcSin[c*x])] + (2*I)*ArcSi n[c*x]*PolyLog[2, -E^(I*ArcSin[c*x])] - (2*I)*ArcSin[c*x]*PolyLog[2, E^(I* ArcSin[c*x])] - 2*PolyLog[3, -E^(I*ArcSin[c*x])] + 2*PolyLog[3, E^(I*ArcSi n[c*x])]))/(Sqrt[d + c*d*x]*Sqrt[e - c*e*x])
Time = 1.00 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.52, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5238, 5218, 3042, 4671, 3011, 2720, 7143}
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.
| \(\displaystyle \int \frac {(a+b \arcsin (c x))^2}{x \sqrt {c d x+d} \sqrt {e-c e x}} \, dx\) |
| \(\Big \downarrow \) 5238 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \int \frac {(a+b \arcsin (c x))^2}{x \sqrt {1-c^2 x^2}}dx}{\sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 5218 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \int \frac {(a+b \arcsin (c x))^2}{c x}d\arcsin (c x)}{\sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \int (a+b \arcsin (c x))^2 \csc (\arcsin (c x))d\arcsin (c x)}{\sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 4671 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (-2 b \int (a+b \arcsin (c x)) \log \left (1-e^{i \arcsin (c x)}\right )d\arcsin (c x)+2 b \int (a+b \arcsin (c x)) \log \left (1+e^{i \arcsin (c x)}\right )d\arcsin (c x)-2 \text {arctanh}\left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))^2\right )}{\sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (2 b \left (i \operatorname {PolyLog}\left (2,-e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))-i b \int \operatorname {PolyLog}\left (2,-e^{i \arcsin (c x)}\right )d\arcsin (c x)\right )-2 b \left (i \operatorname {PolyLog}\left (2,e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))-i b \int \operatorname {PolyLog}\left (2,e^{i \arcsin (c x)}\right )d\arcsin (c x)\right )-2 \text {arctanh}\left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))^2\right )}{\sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (2 b \left (i \operatorname {PolyLog}\left (2,-e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))-b \int e^{-i \arcsin (c x)} \operatorname {PolyLog}\left (2,-e^{i \arcsin (c x)}\right )de^{i \arcsin (c x)}\right )-2 b \left (i \operatorname {PolyLog}\left (2,e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))-b \int e^{-i \arcsin (c x)} \operatorname {PolyLog}\left (2,e^{i \arcsin (c x)}\right )de^{i \arcsin (c x)}\right )-2 \text {arctanh}\left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))^2\right )}{\sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (-2 \text {arctanh}\left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))^2+2 b \left (i \operatorname {PolyLog}\left (2,-e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))-b \operatorname {PolyLog}\left (3,-e^{i \arcsin (c x)}\right )\right )-2 b \left (i \operatorname {PolyLog}\left (2,e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))-b \operatorname {PolyLog}\left (3,e^{i \arcsin (c x)}\right )\right )\right )}{\sqrt {c d x+d} \sqrt {e-c e x}}\) |
(Sqrt[1 - c^2*x^2]*(-2*(a + b*ArcSin[c*x])^2*ArcTanh[E^(I*ArcSin[c*x])] + 2*b*(I*(a + b*ArcSin[c*x])*PolyLog[2, -E^(I*ArcSin[c*x])] - b*PolyLog[3, - E^(I*ArcSin[c*x])]) - 2*b*(I*(a + b*ArcSin[c*x])*PolyLog[2, E^(I*ArcSin[c* x])] - b*PolyLog[3, E^(I*ArcSin[c*x])])))/(Sqrt[d + c*d*x]*Sqrt[e - c*e*x] )
3.6.89.3.1 Defintions of rubi rules used
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) *(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F])) Int[(f + g*x)^( m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e , f, g, n}, x] && GtQ[m, 0]
Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[- 2*(c + d*x)^m*(ArcTanh[E^(I*(e + f*x))]/f), x] + (-Simp[d*(m/f) Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Simp[d*(m/f) Int[(c + d*x )^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IG tQ[m, 0]
Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*(x_)^(m_))/Sqrt[(d_) + (e_.)* (x_)^2], x_Symbol] :> Simp[(1/c^(m + 1))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e* x^2]] Subst[Int[(a + b*x)^n*Sin[x]^m, x], x, ArcSin[c*x]], x] /; FreeQ[{a , b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[n, 0] && IntegerQ[m]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((h_.)*(x_))^(m_.)*((d_) + (e_ .)*(x_))^(p_)*((f_) + (g_.)*(x_))^(q_), x_Symbol] :> Simp[((-d^2)*(g/e))^In tPart[q]*(d + e*x)^FracPart[q]*((f + g*x)^FracPart[q]/(1 - c^2*x^2)^FracPar t[q]) Int[(h*x)^m*(d + e*x)^(p - q)*(1 - c^2*x^2)^q*(a + b*ArcSin[c*x])^n , x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && EqQ[e*f + d*g, 0] & & EqQ[c^2*d^2 - e^2, 0] && HalfIntegerQ[p, q] && GeQ[p - q, 0]
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d , e, n, p}, x] && EqQ[b*d, a*e]
\[\int \frac {\left (a +b \arcsin \left (c x \right )\right )^{2}}{x \sqrt {c d x +d}\, \sqrt {-c e x +e}}d x\]
\[ \int \frac {(a+b \arcsin (c x))^2}{x \sqrt {d+c d x} \sqrt {e-c e x}} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{\sqrt {c d x + d} \sqrt {-c e x + e} x} \,d x } \]
integral(-(b^2*arcsin(c*x)^2 + 2*a*b*arcsin(c*x) + a^2)*sqrt(c*d*x + d)*sq rt(-c*e*x + e)/(c^2*d*e*x^3 - d*e*x), x)
\[ \int \frac {(a+b \arcsin (c x))^2}{x \sqrt {d+c d x} \sqrt {e-c e x}} \, dx=\int \frac {\left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{x \sqrt {d \left (c x + 1\right )} \sqrt {- e \left (c x - 1\right )}}\, dx \]
Exception generated. \[ \int \frac {(a+b \arcsin (c x))^2}{x \sqrt {d+c d x} \sqrt {e-c e x}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int \frac {(a+b \arcsin (c x))^2}{x \sqrt {d+c d x} \sqrt {e-c e x}} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{\sqrt {c d x + d} \sqrt {-c e x + e} x} \,d x } \]
Timed out. \[ \int \frac {(a+b \arcsin (c x))^2}{x \sqrt {d+c d x} \sqrt {e-c e x}} \, dx=\int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{x\,\sqrt {d+c\,d\,x}\,\sqrt {e-c\,e\,x}} \,d x \]