Integrand size = 33, antiderivative size = 128 \[ \int \frac {\arcsin (a+b x)^3}{\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}} \, dx=-\frac {i \arcsin (a+b x)^3}{b}+\frac {(a+b x) \arcsin (a+b x)^3}{b \sqrt {1-(a+b x)^2}}+\frac {3 \arcsin (a+b x)^2 \log \left (1+e^{2 i \arcsin (a+b x)}\right )}{b}-\frac {3 i \arcsin (a+b x) \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (a+b x)}\right )}{b}+\frac {3 \operatorname {PolyLog}\left (3,-e^{2 i \arcsin (a+b x)}\right )}{2 b} \]
-I*arcsin(b*x+a)^3/b+3*arcsin(b*x+a)^2*ln(1+(I*(b*x+a)+(1-(b*x+a)^2)^(1/2) )^2)/b-3*I*arcsin(b*x+a)*polylog(2,-(I*(b*x+a)+(1-(b*x+a)^2)^(1/2))^2)/b+3 /2*polylog(3,-(I*(b*x+a)+(1-(b*x+a)^2)^(1/2))^2)/b+(b*x+a)*arcsin(b*x+a)^3 /b/(1-(b*x+a)^2)^(1/2)
Time = 0.44 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.12 \[ \int \frac {\arcsin (a+b x)^3}{\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}} \, dx=\frac {2 \arcsin (a+b x)^2 \left (\frac {\left (a+b x-i \sqrt {1-a^2-2 a b x-b^2 x^2}\right ) \arcsin (a+b x)}{\sqrt {1-a^2-2 a b x-b^2 x^2}}+3 \log \left (1+e^{2 i \arcsin (a+b x)}\right )\right )-6 i \arcsin (a+b x) \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (a+b x)}\right )+3 \operatorname {PolyLog}\left (3,-e^{2 i \arcsin (a+b x)}\right )}{2 b} \]
(2*ArcSin[a + b*x]^2*(((a + b*x - I*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2])*Arc Sin[a + b*x])/Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2] + 3*Log[1 + E^((2*I)*ArcSi n[a + b*x])]) - (6*I)*ArcSin[a + b*x]*PolyLog[2, -E^((2*I)*ArcSin[a + b*x] )] + 3*PolyLog[3, -E^((2*I)*ArcSin[a + b*x])])/(2*b)
Time = 0.65 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {5306, 5160, 5180, 3042, 4202, 2620, 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.
| \(\displaystyle \int \frac {\arcsin (a+b x)^3}{\left (-a^2-2 a b x-b^2 x^2+1\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 5306 |
| \(\displaystyle \frac {\int \frac {\arcsin (a+b x)^3}{\left (1-(a+b x)^2\right )^{3/2}}d(a+b x)}{b}\) |
| \(\Big \downarrow \) 5160 |
| \(\displaystyle \frac {\frac {(a+b x) \arcsin (a+b x)^3}{\sqrt {1-(a+b x)^2}}-3 \int \frac {(a+b x) \arcsin (a+b x)^2}{1-(a+b x)^2}d(a+b x)}{b}\) |
| \(\Big \downarrow \) 5180 |
| \(\displaystyle \frac {\frac {(a+b x) \arcsin (a+b x)^3}{\sqrt {1-(a+b x)^2}}-3 \int \frac {(a+b x) \arcsin (a+b x)^2}{\sqrt {1-(a+b x)^2}}d\arcsin (a+b x)}{b}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {(a+b x) \arcsin (a+b x)^3}{\sqrt {1-(a+b x)^2}}-3 \int \arcsin (a+b x)^2 \tan (\arcsin (a+b x))d\arcsin (a+b x)}{b}\) |
| \(\Big \downarrow \) 4202 |
| \(\displaystyle \frac {\frac {(a+b x) \arcsin (a+b x)^3}{\sqrt {1-(a+b x)^2}}-3 \left (\frac {1}{3} i \arcsin (a+b x)^3-2 i \int \frac {e^{2 i \arcsin (a+b x)} \arcsin (a+b x)^2}{1+e^{2 i \arcsin (a+b x)}}d\arcsin (a+b x)\right )}{b}\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle \frac {\frac {(a+b x) \arcsin (a+b x)^3}{\sqrt {1-(a+b x)^2}}-3 \left (\frac {1}{3} i \arcsin (a+b x)^3-2 i \left (i \int \arcsin (a+b x) \log \left (1+e^{2 i \arcsin (a+b x)}\right )d\arcsin (a+b x)-\frac {1}{2} i \arcsin (a+b x)^2 \log \left (1+e^{2 i \arcsin (a+b x)}\right )\right )\right )}{b}\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle \frac {\frac {(a+b x) \arcsin (a+b x)^3}{\sqrt {1-(a+b x)^2}}-3 \left (\frac {1}{3} i \arcsin (a+b x)^3-2 i \left (i \left (\frac {1}{2} i \arcsin (a+b x) \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (a+b x)}\right )-\frac {1}{2} i \int \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (a+b x)}\right )d\arcsin (a+b x)\right )-\frac {1}{2} i \arcsin (a+b x)^2 \log \left (1+e^{2 i \arcsin (a+b x)}\right )\right )\right )}{b}\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {\frac {(a+b x) \arcsin (a+b x)^3}{\sqrt {1-(a+b x)^2}}-3 \left (\frac {1}{3} i \arcsin (a+b x)^3-2 i \left (i \left (\frac {1}{2} i \arcsin (a+b x) \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (a+b x)}\right )-\frac {1}{4} \int e^{-2 i \arcsin (a+b x)} \operatorname {PolyLog}(2,-a-b x)de^{2 i \arcsin (a+b x)}\right )-\frac {1}{2} i \arcsin (a+b x)^2 \log \left (1+e^{2 i \arcsin (a+b x)}\right )\right )\right )}{b}\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle \frac {\frac {(a+b x) \arcsin (a+b x)^3}{\sqrt {1-(a+b x)^2}}-3 \left (\frac {1}{3} i \arcsin (a+b x)^3-2 i \left (i \left (-\frac {1}{4} \operatorname {PolyLog}(3,-a-b x)+\frac {1}{2} i \arcsin (a+b x) \operatorname {PolyLog}\left (2,-e^{2 i \arcsin (a+b x)}\right )\right )-\frac {1}{2} i \arcsin (a+b x)^2 \log \left (1+e^{2 i \arcsin (a+b x)}\right )\right )\right )}{b}\) |
(((a + b*x)*ArcSin[a + b*x]^3)/Sqrt[1 - (a + b*x)^2] - 3*((I/3)*ArcSin[a + b*x]^3 - (2*I)*((-1/2*I)*ArcSin[a + b*x]^2*Log[1 + E^((2*I)*ArcSin[a + b* x])] + I*((I/2)*ArcSin[a + b*x]*PolyLog[2, -E^((2*I)*ArcSin[a + b*x])] - P olyLog[3, -a - b*x]/4))))/b
3.4.33.3.1 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[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[((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_.)/((d_) + (e_.)*(x_)^2)^(3/2), x _Symbol] :> Simp[x*((a + b*ArcSin[c*x])^n/(d*Sqrt[d + e*x^2])), x] - Simp[b *c*(n/d)*Simp[Sqrt[1 - c^2*x^2]/Sqrt[d + e*x^2]] Int[x*((a + b*ArcSin[c*x ])^(n - 1)/(1 - c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[n, 0]
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_) + (d_.)*(x_)]*(b_.))^(n_.)*((A_.) + (B_.)*(x_) + ( C_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/d Subst[Int[(-C/d^2 + (C/d^2)*x^2 )^p*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, A, B, C, n, p}, x] && EqQ[B*(1 - c^2) + 2*A*c*d, 0] && EqQ[2*c*C - B*d, 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]
Time = 3.27 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.79
| method | result | size |
| default | \(\frac {\left (-x b \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}+i b^{2} x^{2}-a \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}+2 i a b x +i a^{2}-i\right ) \arcsin \left (b x +a \right )^{3}}{b \left (b^{2} x^{2}+2 a b x +a^{2}-1\right )}+\frac {-4 i \arcsin \left (b x +a \right )^{3}+6 \arcsin \left (b x +a \right )^{2} \ln \left (1+\left (i \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}\right )^{2}\right )-6 i \arcsin \left (b x +a \right ) \operatorname {polylog}\left (2, -\left (i \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}\right )^{2}\right )+3 \operatorname {polylog}\left (3, -\left (i \left (b x +a \right )+\sqrt {1-\left (b x +a \right )^{2}}\right )^{2}\right )}{2 b}\) | \(229\) |
(-x*b*(-b^2*x^2-2*a*b*x-a^2+1)^(1/2)+I*b^2*x^2-a*(-b^2*x^2-2*a*b*x-a^2+1)^ (1/2)+2*I*a*b*x+I*a^2-I)/b/(b^2*x^2+2*a*b*x+a^2-1)*arcsin(b*x+a)^3+1/2*(-4 *I*arcsin(b*x+a)^3+6*arcsin(b*x+a)^2*ln(1+(I*(b*x+a)+(1-(b*x+a)^2)^(1/2))^ 2)-6*I*arcsin(b*x+a)*polylog(2,-(I*(b*x+a)+(1-(b*x+a)^2)^(1/2))^2)+3*polyl og(3,-(I*(b*x+a)+(1-(b*x+a)^2)^(1/2))^2))/b
\[ \int \frac {\arcsin (a+b x)^3}{\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}} \, dx=\int { \frac {\arcsin \left (b x + a\right )^{3}}{{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]
integral(sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)*arcsin(b*x + a)^3/(b^4*x^4 + 4 *a*b^3*x^3 + 2*(3*a^2 - 1)*b^2*x^2 + a^4 + 4*(a^3 - a)*b*x - 2*a^2 + 1), x )
\[ \int \frac {\arcsin (a+b x)^3}{\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}} \, dx=\int \frac {\operatorname {asin}^{3}{\left (a + b x \right )}}{\left (- \left (a + b x - 1\right ) \left (a + b x + 1\right )\right )^{\frac {3}{2}}}\, dx \]
Time = 0.48 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.07 \[ \int \frac {\arcsin (a+b x)^3}{\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}} \, dx=\frac {3}{2} \, b {\left (\frac {\log \left (b x + a + 1\right )}{b^{2}} + \frac {\log \left (b x + a - 1\right )}{b^{2}}\right )} \arcsin \left (b x + a\right )^{2} + {\left (\frac {b^{2} x}{{\left (a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}} + \frac {a b}{{\left (a^{2} b^{2} - {\left (a^{2} - 1\right )} b^{2}\right )} \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}\right )} \arcsin \left (b x + a\right )^{3} \]
3/2*b*(log(b*x + a + 1)/b^2 + log(b*x + a - 1)/b^2)*arcsin(b*x + a)^2 + (b ^2*x/((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)) + a*b/ ((a^2*b^2 - (a^2 - 1)*b^2)*sqrt(-b^2*x^2 - 2*a*b*x - a^2 + 1)))*arcsin(b*x + a)^3
\[ \int \frac {\arcsin (a+b x)^3}{\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}} \, dx=\int { \frac {\arcsin \left (b x + a\right )^{3}}{{\left (-b^{2} x^{2} - 2 \, a b x - a^{2} + 1\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\arcsin (a+b x)^3}{\left (1-a^2-2 a b x-b^2 x^2\right )^{3/2}} \, dx=\int \frac {{\mathrm {asin}\left (a+b\,x\right )}^3}{{\left (-a^2-2\,a\,b\,x-b^2\,x^2+1\right )}^{3/2}} \,d x \]