Integrand size = 33, antiderivative size = 244 \[ \int \frac {x (a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}} \, dx=\frac {(a+b \arcsin (c x))^2}{c^2 d e \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {4 i b \sqrt {1-c^2 x^2} (a+b \arcsin (c x)) \arctan \left (e^{i \arcsin (c x)}\right )}{c^2 d e \sqrt {d+c d x} \sqrt {e-c e x}}-\frac {2 i b^2 \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )}{c^2 d e \sqrt {d+c d x} \sqrt {e-c e x}}+\frac {2 i b^2 \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{c^2 d e \sqrt {d+c d x} \sqrt {e-c e x}} \]
(a+b*arcsin(c*x))^2/c^2/d/e/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+4*I*b*(a+b*ar csin(c*x))*arctan(I*c*x+(-c^2*x^2+1)^(1/2))*(-c^2*x^2+1)^(1/2)/c^2/d/e/(c* d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-2*I*b^2*polylog(2,-I*(I*c*x+(-c^2*x^2+1)^(1/ 2)))*(-c^2*x^2+1)^(1/2)/c^2/d/e/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+2*I*b^2*p olylog(2,I*(I*c*x+(-c^2*x^2+1)^(1/2)))*(-c^2*x^2+1)^(1/2)/c^2/d/e/(c*d*x+d )^(1/2)/(-c*e*x+e)^(1/2)
Time = 3.78 (sec) , antiderivative size = 453, normalized size of antiderivative = 1.86 \[ \int \frac {x (a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}} \, dx=\frac {a^2+2 a b \arcsin (c x)+i b^2 \pi \sqrt {1-c^2 x^2} \arcsin (c x)+b^2 \arcsin (c x)^2-b^2 \pi \sqrt {1-c^2 x^2} \log \left (1-i e^{i \arcsin (c x)}\right )-2 b^2 \sqrt {1-c^2 x^2} \arcsin (c x) \log \left (1-i e^{i \arcsin (c x)}\right )-b^2 \pi \sqrt {1-c^2 x^2} \log \left (1+i e^{i \arcsin (c x)}\right )+2 b^2 \sqrt {1-c^2 x^2} \arcsin (c x) \log \left (1+i e^{i \arcsin (c x)}\right )+b^2 \pi \sqrt {1-c^2 x^2} \log \left (-\cos \left (\frac {1}{4} (\pi +2 \arcsin (c x))\right )\right )+2 a b \sqrt {1-c^2 x^2} \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )-\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )-2 a b \sqrt {1-c^2 x^2} \log \left (\cos \left (\frac {1}{2} \arcsin (c x)\right )+\sin \left (\frac {1}{2} \arcsin (c x)\right )\right )+b^2 \pi \sqrt {1-c^2 x^2} \log \left (\sin \left (\frac {1}{4} (\pi +2 \arcsin (c x))\right )\right )-2 i b^2 \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )+2 i b^2 \sqrt {1-c^2 x^2} \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )}{c^2 d e \sqrt {d+c d x} \sqrt {e-c e x}} \]
(a^2 + 2*a*b*ArcSin[c*x] + I*b^2*Pi*Sqrt[1 - c^2*x^2]*ArcSin[c*x] + b^2*Ar cSin[c*x]^2 - b^2*Pi*Sqrt[1 - c^2*x^2]*Log[1 - I*E^(I*ArcSin[c*x])] - 2*b^ 2*Sqrt[1 - c^2*x^2]*ArcSin[c*x]*Log[1 - I*E^(I*ArcSin[c*x])] - b^2*Pi*Sqrt [1 - c^2*x^2]*Log[1 + I*E^(I*ArcSin[c*x])] + 2*b^2*Sqrt[1 - c^2*x^2]*ArcSi n[c*x]*Log[1 + I*E^(I*ArcSin[c*x])] + b^2*Pi*Sqrt[1 - c^2*x^2]*Log[-Cos[(P i + 2*ArcSin[c*x])/4]] + 2*a*b*Sqrt[1 - c^2*x^2]*Log[Cos[ArcSin[c*x]/2] - Sin[ArcSin[c*x]/2]] - 2*a*b*Sqrt[1 - c^2*x^2]*Log[Cos[ArcSin[c*x]/2] + Sin [ArcSin[c*x]/2]] + b^2*Pi*Sqrt[1 - c^2*x^2]*Log[Sin[(Pi + 2*ArcSin[c*x])/4 ]] - (2*I)*b^2*Sqrt[1 - c^2*x^2]*PolyLog[2, (-I)*E^(I*ArcSin[c*x])] + (2*I )*b^2*Sqrt[1 - c^2*x^2]*PolyLog[2, I*E^(I*ArcSin[c*x])])/(c^2*d*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])
Time = 0.92 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.59, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {5238, 5182, 5164, 3042, 4669, 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.
| \(\displaystyle \int \frac {x (a+b \arcsin (c x))^2}{(c d x+d)^{3/2} (e-c e x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 5238 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \int \frac {x (a+b \arcsin (c x))^2}{\left (1-c^2 x^2\right )^{3/2}}dx}{d e \sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 5182 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (\frac {(a+b \arcsin (c x))^2}{c^2 \sqrt {1-c^2 x^2}}-\frac {2 b \int \frac {a+b \arcsin (c x)}{1-c^2 x^2}dx}{c}\right )}{d e \sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 5164 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (\frac {(a+b \arcsin (c x))^2}{c^2 \sqrt {1-c^2 x^2}}-\frac {2 b \int \frac {a+b \arcsin (c x)}{\sqrt {1-c^2 x^2}}d\arcsin (c x)}{c^2}\right )}{d e \sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (\frac {(a+b \arcsin (c x))^2}{c^2 \sqrt {1-c^2 x^2}}-\frac {2 b \int (a+b \arcsin (c x)) \csc \left (\arcsin (c x)+\frac {\pi }{2}\right )d\arcsin (c x)}{c^2}\right )}{d e \sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 4669 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (\frac {(a+b \arcsin (c x))^2}{c^2 \sqrt {1-c^2 x^2}}-\frac {2 b \left (-b \int \log \left (1-i e^{i \arcsin (c x)}\right )d\arcsin (c x)+b \int \log \left (1+i e^{i \arcsin (c x)}\right )d\arcsin (c x)-2 i \arctan \left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))\right )}{c^2}\right )}{d e \sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 2715 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (\frac {(a+b \arcsin (c x))^2}{c^2 \sqrt {1-c^2 x^2}}-\frac {2 b \left (i b \int e^{-i \arcsin (c x)} \log \left (1-i e^{i \arcsin (c x)}\right )de^{i \arcsin (c x)}-i b \int e^{-i \arcsin (c x)} \log \left (1+i e^{i \arcsin (c x)}\right )de^{i \arcsin (c x)}-2 i \arctan \left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))\right )}{c^2}\right )}{d e \sqrt {c d x+d} \sqrt {e-c e x}}\) |
| \(\Big \downarrow \) 2838 |
| \(\displaystyle \frac {\sqrt {1-c^2 x^2} \left (\frac {(a+b \arcsin (c x))^2}{c^2 \sqrt {1-c^2 x^2}}-\frac {2 b \left (-2 i \arctan \left (e^{i \arcsin (c x)}\right ) (a+b \arcsin (c x))+i b \operatorname {PolyLog}\left (2,-i e^{i \arcsin (c x)}\right )-i b \operatorname {PolyLog}\left (2,i e^{i \arcsin (c x)}\right )\right )}{c^2}\right )}{d e \sqrt {c d x+d} \sqrt {e-c e x}}\) |
(Sqrt[1 - c^2*x^2]*((a + b*ArcSin[c*x])^2/(c^2*Sqrt[1 - c^2*x^2]) - (2*b*( (-2*I)*(a + b*ArcSin[c*x])*ArcTan[E^(I*ArcSin[c*x])] + I*b*PolyLog[2, (-I) *E^(I*ArcSin[c*x])] - I*b*PolyLog[2, I*E^(I*ArcSin[c*x])]))/c^2))/(d*e*Sqr t[d + c*d*x]*Sqrt[e - c*e*x])
3.6.92.3.1 Defintions of rubi rules used
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[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol ] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^(I*k*Pi)*E^(I*(e + f*x))]/f), x] + (-Si mp[d*(m/f) Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))], x], x] + Simp[d*(m/f) Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x ))], x], x]) /; FreeQ[{c, d, e, f}, x] && IntegerQ[2*k] && IGtQ[m, 0]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2), x_Symbo l] :> Simp[1/(c*d) Subst[Int[(a + b*x)^n*Sec[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_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_ .), x_Symbol] :> Simp[(d + e*x^2)^(p + 1)*((a + b*ArcSin[c*x])^n/(2*e*(p + 1))), x] + Simp[b*(n/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 - c^2*x^2)^p] I nt[(1 - c^2*x^2)^(p + 1/2)*(a + b*ArcSin[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0] && NeQ[p, -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]
\[\int \frac {x \left (a +b \arcsin \left (c x \right )\right )^{2}}{\left (c d x +d \right )^{\frac {3}{2}} \left (-c e x +e \right )^{\frac {3}{2}}}d x\]
\[ \int \frac {x (a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2} x}{{\left (c d x + d\right )}^{\frac {3}{2}} {\left (-c e x + e\right )}^{\frac {3}{2}}} \,d x } \]
integral((b^2*x*arcsin(c*x)^2 + 2*a*b*x*arcsin(c*x) + a^2*x)*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)
\[ \int \frac {x (a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}} \, dx=\int \frac {x \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{\left (d \left (c x + 1\right )\right )^{\frac {3}{2}} \left (- e \left (c x - 1\right )\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {x (a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2} x}{{\left (c d x + d\right )}^{\frac {3}{2}} {\left (-c e x + e\right )}^{\frac {3}{2}}} \,d x } \]
sqrt(d)*sqrt(e)*integrate((b^2*x*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1) )^2 + 2*a*b*x*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)))*sqrt(c*x + 1)*sq rt(-c*x + 1)/(c^4*d^2*e^2*x^4 - 2*c^2*d^2*e^2*x^2 + d^2*e^2), x) + a^2/(sq rt(-c^2*d*e*x^2 + d*e)*c^2*d*e)
\[ \int \frac {x (a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}} \, dx=\int { \frac {{\left (b \arcsin \left (c x\right ) + a\right )}^{2} x}{{\left (c d x + d\right )}^{\frac {3}{2}} {\left (-c e x + e\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {x (a+b \arcsin (c x))^2}{(d+c d x)^{3/2} (e-c e x)^{3/2}} \, dx=\int \frac {x\,{\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2}{{\left (d+c\,d\,x\right )}^{3/2}\,{\left (e-c\,e\,x\right )}^{3/2}} \,d x \]