Integrand size = 35, antiderivative size = 257 \[ \int \frac {\sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2}{x^2} \, dx=-\frac {\sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2}{x}-\frac {i c \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}-\frac {c \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^3}{3 b \sqrt {1-c^2 x^2}}+\frac {2 b c \sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x)) \log \left (1-e^{2 i \arcsin (c x)}\right )}{\sqrt {1-c^2 x^2}}-\frac {i b^2 c \sqrt {d+c d x} \sqrt {e-c e x} \operatorname {PolyLog}\left (2,e^{2 i \arcsin (c x)}\right )}{\sqrt {1-c^2 x^2}} \]
-(c*d*x+d)^(1/2)*(-c*e*x+e)^(1/2)*(a+b*arcsin(c*x))^2/x-I*c*(a+b*arcsin(c* x))^2*(c*d*x+d)^(1/2)*(-c*e*x+e)^(1/2)/(-c^2*x^2+1)^(1/2)-1/3*c*(a+b*arcsi n(c*x))^3*(c*d*x+d)^(1/2)*(-c*e*x+e)^(1/2)/b/(-c^2*x^2+1)^(1/2)+2*b*c*(a+b *arcsin(c*x))*ln(1-(I*c*x+(-c^2*x^2+1)^(1/2))^2)*(c*d*x+d)^(1/2)*(-c*e*x+e )^(1/2)/(-c^2*x^2+1)^(1/2)-I*b^2*c*polylog(2,(I*c*x+(-c^2*x^2+1)^(1/2))^2) *(c*d*x+d)^(1/2)*(-c*e*x+e)^(1/2)/(-c^2*x^2+1)^(1/2)
Time = 1.35 (sec) , antiderivative size = 374, normalized size of antiderivative = 1.46 \[ \int \frac {\sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2}{x^2} \, dx=\frac {-3 a^2 \sqrt {d+c d x} \sqrt {e-c e x} \sqrt {1-c^2 x^2}-3 i b \sqrt {d+c d x} \sqrt {e-c e x} \left (-i a c x+b c x-i b \sqrt {1-c^2 x^2}\right ) \arcsin (c x)^2-b^2 c x \sqrt {d+c d x} \sqrt {e-c e x} \arcsin (c x)^3+3 a^2 c \sqrt {d} \sqrt {e} x \sqrt {1-c^2 x^2} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {e-c e x}}{\sqrt {d} \sqrt {e} \left (-1+c^2 x^2\right )}\right )+6 b \sqrt {d+c d x} \sqrt {e-c e x} \arcsin (c x) \left (-a \sqrt {1-c^2 x^2}+b c x \log \left (1-e^{2 i \arcsin (c x)}\right )\right )+6 a b c x \sqrt {d+c d x} \sqrt {e-c e x} \log (c x)-3 i b^2 c x \sqrt {d+c d x} \sqrt {e-c e x} \operatorname {PolyLog}\left (2,e^{2 i \arcsin (c x)}\right )}{3 x \sqrt {1-c^2 x^2}} \]
(-3*a^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*Sqrt[1 - c^2*x^2] - (3*I)*b*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*((-I)*a*c*x + b*c*x - I*b*Sqrt[1 - c^2*x^2])*Arc Sin[c*x]^2 - b^2*c*x*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*ArcSin[c*x]^3 + 3*a^2 *c*Sqrt[d]*Sqrt[e]*x*Sqrt[1 - c^2*x^2]*ArcTan[(c*x*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])/(Sqrt[d]*Sqrt[e]*(-1 + c^2*x^2))] + 6*b*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*ArcSin[c*x]*(-(a*Sqrt[1 - c^2*x^2]) + b*c*x*Log[1 - E^((2*I)*ArcSi n[c*x])]) + 6*a*b*c*x*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*Log[c*x] - (3*I)*b^2 *c*x*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*PolyLog[2, E^((2*I)*ArcSin[c*x])])/(3 *x*Sqrt[1 - c^2*x^2])
Time = 1.36 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.62, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.314, Rules used = {5238, 5196, 5136, 3042, 25, 4200, 25, 2620, 2715, 2838, 5152}
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.
\(\displaystyle \int \frac {\sqrt {c d x+d} \sqrt {e-c e x} (a+b \arcsin (c x))^2}{x^2} \, dx\) |
\(\Big \downarrow \) 5238 |
\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \int \frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{x^2}dx}{\sqrt {1-c^2 x^2}}\) |
\(\Big \downarrow \) 5196 |
\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (c^2 \left (-\int \frac {(a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}dx\right )+2 b c \int \frac {a+b \arcsin (c x)}{x}dx-\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{x}\right )}{\sqrt {1-c^2 x^2}}\) |
\(\Big \downarrow \) 5136 |
\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (c^2 \left (-\int \frac {(a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}dx\right )+2 b c \int \frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))}{c x}d\arcsin (c x)-\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{x}\right )}{\sqrt {1-c^2 x^2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (c^2 \left (-\int \frac {(a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}dx\right )+2 b c \int -\left ((a+b \arcsin (c x)) \tan \left (\arcsin (c x)+\frac {\pi }{2}\right )\right )d\arcsin (c x)-\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{x}\right )}{\sqrt {1-c^2 x^2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (c^2 \left (-\int \frac {(a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}dx\right )-2 b c \int (a+b \arcsin (c x)) \tan \left (\arcsin (c x)+\frac {\pi }{2}\right )d\arcsin (c x)-\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{x}\right )}{\sqrt {1-c^2 x^2}}\) |
\(\Big \downarrow \) 4200 |
\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (c^2 \left (-\int \frac {(a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}dx\right )+2 b c \left (2 i \int -\frac {e^{2 i \arcsin (c x)} (a+b \arcsin (c x))}{1-e^{2 i \arcsin (c x)}}d\arcsin (c x)-\frac {i (a+b \arcsin (c x))^2}{2 b}\right )-\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{x}\right )}{\sqrt {1-c^2 x^2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (c^2 \left (-\int \frac {(a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}dx\right )+2 b c \left (-2 i \int \frac {e^{2 i \arcsin (c x)} (a+b \arcsin (c x))}{1-e^{2 i \arcsin (c x)}}d\arcsin (c x)-\frac {i (a+b \arcsin (c x))^2}{2 b}\right )-\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{x}\right )}{\sqrt {1-c^2 x^2}}\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (c^2 \left (-\int \frac {(a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}dx\right )+2 b c \left (-2 i \left (\frac {1}{2} i \log \left (1-e^{2 i \arcsin (c x)}\right ) (a+b \arcsin (c x))-\frac {1}{2} i b \int \log \left (1-e^{2 i \arcsin (c x)}\right )d\arcsin (c x)\right )-\frac {i (a+b \arcsin (c x))^2}{2 b}\right )-\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{x}\right )}{\sqrt {1-c^2 x^2}}\) |
\(\Big \downarrow \) 2715 |
\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (c^2 \left (-\int \frac {(a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}dx\right )+2 b c \left (-2 i \left (\frac {1}{2} i \log \left (1-e^{2 i \arcsin (c x)}\right ) (a+b \arcsin (c x))-\frac {1}{4} b \int e^{-2 i \arcsin (c x)} \log \left (1-e^{2 i \arcsin (c x)}\right )de^{2 i \arcsin (c x)}\right )-\frac {i (a+b \arcsin (c x))^2}{2 b}\right )-\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{x}\right )}{\sqrt {1-c^2 x^2}}\) |
\(\Big \downarrow \) 2838 |
\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (c^2 \left (-\int \frac {(a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}dx\right )-\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{x}+2 b c \left (-2 i \left (\frac {1}{2} i \log \left (1-e^{2 i \arcsin (c x)}\right ) (a+b \arcsin (c x))+\frac {1}{4} b \operatorname {PolyLog}\left (2,e^{2 i \arcsin (c x)}\right )\right )-\frac {i (a+b \arcsin (c x))^2}{2 b}\right )\right )}{\sqrt {1-c^2 x^2}}\) |
\(\Big \downarrow \) 5152 |
\(\displaystyle \frac {\sqrt {c d x+d} \sqrt {e-c e x} \left (-\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{x}+2 b c \left (-2 i \left (\frac {1}{2} i \log \left (1-e^{2 i \arcsin (c x)}\right ) (a+b \arcsin (c x))+\frac {1}{4} b \operatorname {PolyLog}\left (2,e^{2 i \arcsin (c x)}\right )\right )-\frac {i (a+b \arcsin (c x))^2}{2 b}\right )-\frac {c (a+b \arcsin (c x))^3}{3 b}\right )}{\sqrt {1-c^2 x^2}}\) |
(Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*(-((Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]) ^2)/x) - (c*(a + b*ArcSin[c*x])^3)/(3*b) + 2*b*c*(((-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))))/Sqrt[1 - c^2*x^2]
3.6.80.3.1 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_.) + 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]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_S ymbol] :> Simp[(1/(b*c*(n + 1)))*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSin[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[c^2*d + e, 0] && NeQ[n, -1]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(f*x)^(m + 1)*Sqrt[d + e*x^2]*((a + b*ArcS in[c*x])^n/(f*(m + 1))), x] + (-Simp[b*c*(n/(f*(m + 1)))*Simp[Sqrt[d + e*x^ 2]/Sqrt[1 - c^2*x^2]] Int[(f*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1), x], x] + Simp[(c^2/(f^2*(m + 1)))*Simp[Sqrt[d + e*x^2]/Sqrt[1 - c^2*x^2]] Int [(f*x)^(m + 2)*((a + b*ArcSin[c*x])^n/Sqrt[1 - c^2*x^2]), x], x]) /; FreeQ[ {a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[m, -1]
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]
\[\int \frac {\sqrt {c d x +d}\, \sqrt {-c e x +e}\, \left (a +b \arcsin \left (c x \right )\right )^{2}}{x^{2}}d x\]
\[ \int \frac {\sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2}{x^2} \, dx=\int { \frac {\sqrt {c d x + d} \sqrt {-c e x + e} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{x^{2}} \,d x } \]
integral((b^2*arcsin(c*x)^2 + 2*a*b*arcsin(c*x) + a^2)*sqrt(c*d*x + d)*sqr t(-c*e*x + e)/x^2, x)
\[ \int \frac {\sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2}{x^2} \, dx=\int \frac {\sqrt {d \left (c x + 1\right )} \sqrt {- e \left (c x - 1\right )} \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{x^{2}}\, dx \]
Exception generated. \[ \int \frac {\sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2}{x^2} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {\sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2}{x^2} \, dx=\int { \frac {\sqrt {c d x + d} \sqrt {-c e x + e} {\left (b \arcsin \left (c x\right ) + a\right )}^{2}}{x^{2}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {d+c d x} \sqrt {e-c e x} (a+b \arcsin (c x))^2}{x^2} \, dx=\int \frac {{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,\sqrt {d+c\,d\,x}\,\sqrt {e-c\,e\,x}}{x^2} \,d x \]