Integrand size = 35, antiderivative size = 295 \[ \int \frac {x^2 (a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}} \, dx=\frac {x (a+b \arcsin (c x))^2}{c^2 d e \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {i \sqrt {1-c^2 x^2} (a+b \arcsin (c x))^2}{c^3 d e \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {\sqrt {1-c^2 x^2} (a+b \arcsin (c x))^3}{3 b c^3 d e \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {2 b \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) \log \left (1+e^{2 i \arcsin (c x)}\right )}{c^3 d e \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {i b^2 \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )}{c^3 d e \sqrt {d+c d x} \sqrt {e-c e x}} \]
x*(a+b*arcsin(c*x))^2/c^2/d/e/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-I*(a+b*arcs in(c*x))^2*(-c^2*x^2+1)^(1/2)/c^3/d/e/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-1/3 *(a+b*arcsin(c*x))^3*(-c^2*x^2+1)^(1/2)/b/c^3/d/e/(c*d*x+d)^(1/2)/(-c*e*x+ e)^(1/2)+2*b*(a+b*arcsin(c*x))*ln(1+(I*c*x+(-c^2*x^2+1)^(1/2))^2)*(-c^2*x^ 2+1)^(1/2)/c^3/d/e/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-I*b^2*polylog(2,-(I*c* x+(-c^2*x^2+1)^(1/2))^2)*(-c^2*x^2+1)^(1/2)/c^3/d/e/(c*d*x+d)^(1/2)/(-c*e* x+e)^(1/2)
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(636\) vs. \(2(295)=590\).
Time = 3.43 (sec) , antiderivative size = 636, normalized size of antiderivative = 2.16 \[ \int \frac {x^2 (a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}} \, dx=\frac {3 a^2 c \sqrt {d} e x+3 a^2 \sqrt {e} \sqrt {d+c d x} \sqrt {e-c e x} \arctan \left (\frac {c x \sqrt {d+c d x} \sqrt {e-c e x}}{\sqrt {d} \sqrt {e} \left (-1+c^2 x^2\right )}\right )+3 a b \sqrt {d} e \left (2 c x \arcsin (c x)+\sqrt {1-c^2 x^2} \left (-\arcsin (c x)^2+2 \left (\log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )+\log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )\right )\right )\right )+b^2 \sqrt {d} e \left (6 i \pi \sqrt {1-c^2 x^2} \arcsin (c x)+3 c x \arcsin (c x)^2-3 i \sqrt {1-c^2 x^2} \arcsin (c x)^2-\sqrt {1-c^2 x^2} \arcsin (c x)^3+12 \pi \sqrt {1-c^2 x^2} \log \left (1+e^{-i \arcsin (c x)}\right )+3 \pi \sqrt {1-c^2 x^2} \log \left (1-i e^{i \arcsin (c x)}\right )+6 \sqrt {1-c^2 x^2} \arcsin (c x) \log \left (1-i e^{i \arcsin (c x)}\right )-3 \pi \sqrt {1-c^2 x^2} \log \left (1+i e^{i \arcsin (c x)}\right )+6 \sqrt {1-c^2 x^2} \arcsin (c x) \log \left (1+i e^{i \arcsin (c x)}\right )-12 \pi \sqrt {1-c^2 x^2} \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )\right )+3 \pi \sqrt {1-c^2 x^2} \log \left (-\cos \left (\frac {1}{4} (\pi +2 \arcsin (c x))\right )\right )-3 \pi \sqrt {1-c^2 x^2} \log \left (\sin \left (\frac {1}{4} (\pi +2 \arcsin (c x))\right )\right )-6 i \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )-6 i \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )\right )}{3 c^3 d^{3/2} e^2 \sqrt {d+c d x} \sqrt {e-c e x}} \]
(3*a^2*c*Sqrt[d]*e*x + 3*a^2*Sqrt[e]*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*ArcTa n[(c*x*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])/(Sqrt[d]*Sqrt[e]*(-1 + c^2*x^2))] + 3*a*b*Sqrt[d]*e*(2*c*x*ArcSin[c*x] + Sqrt[1 - c^2*x^2]*(-ArcSin[c*x]^2 + 2*(Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]] + Log[Cos[ArcSin[c*x]/2] + Sin[ArcSin[c*x]/2]]))) + b^2*Sqrt[d]*e*((6*I)*Pi*Sqrt[1 - c^2*x^2]*ArcSi n[c*x] + 3*c*x*ArcSin[c*x]^2 - (3*I)*Sqrt[1 - c^2*x^2]*ArcSin[c*x]^2 - Sqr t[1 - c^2*x^2]*ArcSin[c*x]^3 + 12*Pi*Sqrt[1 - c^2*x^2]*Log[1 + E^((-I)*Arc Sin[c*x])] + 3*Pi*Sqrt[1 - c^2*x^2]*Log[1 - I*E^(I*ArcSin[c*x])] + 6*Sqrt[ 1 - c^2*x^2]*ArcSin[c*x]*Log[1 - I*E^(I*ArcSin[c*x])] - 3*Pi*Sqrt[1 - c^2* x^2]*Log[1 + I*E^(I*ArcSin[c*x])] + 6*Sqrt[1 - c^2*x^2]*ArcSin[c*x]*Log[1 + I*E^(I*ArcSin[c*x])] - 12*Pi*Sqrt[1 - c^2*x^2]*Log[Cos[ArcSin[c*x]/2]] + 3*Pi*Sqrt[1 - c^2*x^2]*Log[-Cos[(Pi + 2*ArcSin[c*x])/4]] - 3*Pi*Sqrt[1 - c^2*x^2]*Log[Sin[(Pi + 2*ArcSin[c*x])/4]] - (6*I)*Sqrt[1 - c^2*x^2]*PolyLo g[2, (-I)*E^(I*ArcSin[c*x])] - (6*I)*Sqrt[1 - c^2*x^2]*PolyLog[2, I*E^(I*A rcSin[c*x])]))/(3*c^3*d^(3/2)*e^2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])
Time = 1.35 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.57, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {5238, 5206, 5152, 5180, 3042, 4202, 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.
| \(\displaystyle \int \frac {x^2 (a+b \arcsin (c x))^2}{(c d x+d)^{3/2} (e-c e x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 5238 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \int \frac {x^2 (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{3/2}}dx}{d e \sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 5206 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (-\frac {2 b \int \frac {x (a+b \arcsin (c x))}{1-c^2 x^2}dx}{c}-\frac {\int \frac {(a+b \arcsin (c x))^2}{\sqrt {1-c^2 x^2}}dx}{c^2}+\frac {x (a+b \arcsin (c x))^2}{c^2 \sqrt {1-c^2 x^2}}\right )}{d e \sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 5152 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (-\frac {2 b \int \frac {x (a+b \arcsin (c x))}{1-c^2 x^2}dx}{c}-\frac {(a+b \arcsin (c x))^3}{3 b c^3}+\frac {x (a+b \arcsin (c x))^2}{c^2 \sqrt {1-c^2 x^2}}\right )}{d e \sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 5180 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (-\frac {2 b \int \frac {c x (a+b \arcsin (c x))}{\sqrt {1-c^2 x^2}}d\arcsin (c x)}{c^3}-\frac {(a+b \arcsin (c x))^3}{3 b c^3}+\frac {x (a+b \arcsin (c x))^2}{c^2 \sqrt {1-c^2 x^2}}\right )}{d e \sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (-\frac {2 b \int (a+b \arcsin (c x)) \tan (\arcsin (c x))d\arcsin (c x)}{c^3}-\frac {(a+b \arcsin (c x))^3}{3 b c^3}+\frac {x (a+b \arcsin (c x))^2}{c^2 \sqrt {1-c^2 x^2}}\right )}{d e \sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 4202 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (-\frac {2 b \left (\frac {i (a+b \arcsin (c x))^2}{2 b}-2 i \int \frac {e^{2 i \arcsin (c x)} (a+b \arcsin (c x))}{1+e^{2 i \arcsin (c x)}}d\arcsin (c x)\right )}{c^3}-\frac {(a+b \arcsin (c x))^3}{3 b c^3}+\frac {x (a+b \arcsin (c x))^2}{c^2 \sqrt {1-c^2 x^2}}\right )}{d e \sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (-\frac {2 b \left (\frac {i (a+b \arcsin (c x))^2}{2 b}-2 i \left (\frac {1}{2} i b \int \log \left (1+e^{2 i \arcsin (c x)}\right )d\arcsin (c x)-\frac {1}{2} i \log \left (1+e^{2 i \arcsin (c x)}\right ) (a+b \arcsin (c x))\right )\right )}{c^3}-\frac {(a+b \arcsin (c x))^3}{3 b c^3}+\frac {x (a+b \arcsin (c x))^2}{c^2 \sqrt {1-c^2 x^2}}\right )}{d e \sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 2715 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (-\frac {2 b \left (\frac {i (a+b \arcsin (c x))^2}{2 b}-2 i \left (\frac {1}{4} b \int e^{-2 i \arcsin (c x)} \log \left (1+e^{2 i \arcsin (c x)}\right )de^{2 i \arcsin (c x)}-\frac {1}{2} i \log \left (1+e^{2 i \arcsin (c x)}\right ) (a+b \arcsin (c x))\right )\right )}{c^3}-\frac {(a+b \arcsin (c x))^3}{3 b c^3}+\frac {x (a+b \arcsin (c x))^2}{c^2 \sqrt {1-c^2 x^2}}\right )}{d e \sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 2838 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (-\frac {2 b \left (\frac {i (a+b \arcsin (c x))^2}{2 b}-2 i \left (-\frac {1}{2} i \log \left (1+e^{2 i \arcsin (c x)}\right ) (a+b \arcsin (c x))-\frac {1}{4} b \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (c x)}\right )\right )\right )}{c^3}-\frac {(a+b \arcsin (c x))^3}{3 b c^3}+\frac {x (a+b \arcsin (c x))^2}{c^2 \sqrt {1-c^2 x^2}}\right )}{d e \sqrt {c d x+d} \sqrt {e-c e x}}\) |
(Sqrt[1 - c^2*x^2]*((x*(a + b*ArcSin[c*x])^2)/(c^2*Sqrt[1 - c^2*x^2]) - (a + b*ArcSin[c*x])^3/(3*b*c^3) - (2*b*(((I/2)*(a + b*ArcSin[c*x])^2)/b - (2 *I)*((-1/2*I)*(a + b*ArcSin[c*x])*Log[1 + E^((2*I)*ArcSin[c*x])] - (b*Poly Log[2, -E^((2*I)*ArcSin[c*x])])/4)))/c^3))/(d*e*Sqrt[d + c*d*x]*Sqrt[e - c *e*x])
3.6.91.3.1 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_.) + (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*( e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] && IGt Q[m, 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_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[-e^(-1) Subst[Int[(a + b*x)^n*Tan[x], x], x, ArcSin[c*x ]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && IGtQ[n, 0]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_. )*(x_)^2)^(p_), x_Symbol] :> Simp[f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*((a + b*ArcSin[c*x])^n/(2*e*(p + 1))), x] + (-Simp[f^2*((m - 1)/(2*e*(p + 1))) Int[(f*x)^(m - 2)*(d + e*x^2)^(p + 1)*(a + b*ArcSin[c*x])^n, x], x] + Simp [b*f*(n/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p] Int[(f*x)^(m - 1)*(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x]) /; FreeQ[{ a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && LtQ[p, -1] && IG tQ[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]
\[\int \frac {x^{2} \left (a +b \arcsin \left (c x \right )\right )^{2}}{\left (c d x +d \right )^{\frac {3}{2}} \left (-c e x +e \right )^{\frac {3}{2}}}d x\]
\[ \int \frac {x^2 (a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2} x^{2}}{{\left (c d x + d\right )}^{\frac {3}{2}} {\left (-c e x + e\right )}^{\frac {3}{2}}} \,d x } \]
integral((b^2*x^2*arcsin(c*x)^2 + 2*a*b*x^2*arcsin(c*x) + a^2*x^2)*sqrt(c* d*x + d)*sqrt(-c*e*x + e)/(c^4*d^2*e^2*x^4 - 2*c^2*d^2*e^2*x^2 + d^2*e^2), x)
\[ \int \frac {x^2 (a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}} \, dx=\int \frac {x^{2} \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{\left (d \left (c x + 1\right )\right )^{\frac {3}{2}} \left (- e \left (c x - 1\right )\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {x^2 (a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {x^2 (a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2} x^{2}}{{\left (c d x + d\right )}^{\frac {3}{2}} {\left (-c e x + e\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {x^2 (a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}} \, dx=\int \frac {x^2\,{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{{\left (d+c\,d\,x\right )}^{3/2}\,{\left (e-c\,e\,x\right )}^{3/2}} \,d x \]