Integrand size = 23, antiderivative size = 167 \[ \int \frac {(a+b \arcsin (c+d x))^3}{(c e+d e x)^3} \, dx=-\frac {3 i b (a+b \arcsin (c+d x))^2}{2 d e^3}-\frac {3 b \sqrt {1-(c+d x)^2} (a+b \arcsin (c+d x))^2}{2 d e^3 (c+d x)}-\frac {(a+b \arcsin (c+d x))^3}{2 d e^3 (c+d x)^2}+\frac {3 b^2 (a+b \arcsin (c+d x)) \log \left (1-e^{2 i \arcsin (c+d x)}\right )}{d e^3}-\frac {3 i b^3 \operatorname {PolyLog}\left (2,e^{2 i \arcsin (c+d x)}\right )}{2 d e^3} \]
-3/2*I*b*(a+b*arcsin(d*x+c))^2/d/e^3-1/2*(a+b*arcsin(d*x+c))^3/d/e^3/(d*x+ c)^2+3*b^2*(a+b*arcsin(d*x+c))*ln(1-(I*(d*x+c)+(1-(d*x+c)^2)^(1/2))^2)/d/e ^3-3/2*I*b^3*polylog(2,(I*(d*x+c)+(1-(d*x+c)^2)^(1/2))^2)/d/e^3-3/2*b*(a+b *arcsin(d*x+c))^2*(1-(d*x+c)^2)^(1/2)/d/e^3/(d*x+c)
Time = 1.15 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.49 \[ \int \frac {(a+b \arcsin (c+d x))^3}{(c e+d e x)^3} \, dx=-\frac {3 b^2 \left (a+b (c+d x) \left (i c+i d x+\sqrt {1-c^2-2 c d x-d^2 x^2}\right )\right ) \arcsin (c+d x)^2+b^3 \arcsin (c+d x)^3+3 b \arcsin (c+d x) \left (a \left (a+2 b (c+d x) \sqrt {1-c^2-2 c d x-d^2 x^2}\right )-2 b^2 (c+d x)^2 \log \left (1-e^{2 i \arcsin (c+d x)}\right )\right )+a \left (a \left (a+3 b (c+d x) \sqrt {1-c^2-2 c d x-d^2 x^2}\right )-6 b^2 (c+d x)^2 \log (c+d x)\right )+3 i b^3 (c+d x)^2 \operatorname {PolyLog}\left (2,e^{2 i \arcsin (c+d x)}\right )}{2 d e^3 (c+d x)^2} \]
-1/2*(3*b^2*(a + b*(c + d*x)*(I*c + I*d*x + Sqrt[1 - c^2 - 2*c*d*x - d^2*x ^2]))*ArcSin[c + d*x]^2 + b^3*ArcSin[c + d*x]^3 + 3*b*ArcSin[c + d*x]*(a*( a + 2*b*(c + d*x)*Sqrt[1 - c^2 - 2*c*d*x - d^2*x^2]) - 2*b^2*(c + d*x)^2*L og[1 - E^((2*I)*ArcSin[c + d*x])]) + a*(a*(a + 3*b*(c + d*x)*Sqrt[1 - c^2 - 2*c*d*x - d^2*x^2]) - 6*b^2*(c + d*x)^2*Log[c + d*x]) + (3*I)*b^3*(c + d *x)^2*PolyLog[2, E^((2*I)*ArcSin[c + d*x])])/(d*e^3*(c + d*x)^2)
Time = 0.71 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.93, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {5304, 27, 5138, 5186, 5136, 3042, 25, 4200, 25, 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.
| \(\displaystyle \int \frac {(a+b \arcsin (c+d x))^3}{(c e+d e x)^3} \, dx\) |
| \(\Big \downarrow \) 5304 |
| \(\displaystyle \frac {\int \frac {(a+b \arcsin (c+d x))^3}{e^3 (c+d x)^3}d(c+d x)}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {(a+b \arcsin (c+d x))^3}{(c+d x)^3}d(c+d x)}{d e^3}\) |
| \(\Big \downarrow \) 5138 |
| \(\displaystyle \frac {\frac {3}{2} b \int \frac {(a+b \arcsin (c+d x))^2}{(c+d x)^2 \sqrt {1-(c+d x)^2}}d(c+d x)-\frac {(a+b \arcsin (c+d x))^3}{2 (c+d x)^2}}{d e^3}\) |
| \(\Big \downarrow \) 5186 |
| \(\displaystyle \frac {\frac {3}{2} b \left (2 b \int \frac {a+b \arcsin (c+d x)}{c+d x}d(c+d x)-\frac {\sqrt {1-(c+d x)^2} (a+b \arcsin (c+d x))^2}{c+d x}\right )-\frac {(a+b \arcsin (c+d x))^3}{2 (c+d x)^2}}{d e^3}\) |
| \(\Big \downarrow \) 5136 |
| \(\displaystyle \frac {\frac {3}{2} b \left (2 b \int \frac {\sqrt {1-(c+d x)^2} (a+b \arcsin (c+d x))}{c+d x}d\arcsin (c+d x)-\frac {\sqrt {1-(c+d x)^2} (a+b \arcsin (c+d x))^2}{c+d x}\right )-\frac {(a+b \arcsin (c+d x))^3}{2 (c+d x)^2}}{d e^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {3}{2} b \left (2 b \int -\left ((a+b \arcsin (c+d x)) \tan \left (\arcsin (c+d x)+\frac {\pi }{2}\right )\right )d\arcsin (c+d x)-\frac {\sqrt {1-(c+d x)^2} (a+b \arcsin (c+d x))^2}{c+d x}\right )-\frac {(a+b \arcsin (c+d x))^3}{2 (c+d x)^2}}{d e^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {3}{2} b \left (-2 b \int (a+b \arcsin (c+d x)) \tan \left (\arcsin (c+d x)+\frac {\pi }{2}\right )d\arcsin (c+d x)-\frac {\sqrt {1-(c+d x)^2} (a+b \arcsin (c+d x))^2}{c+d x}\right )-\frac {(a+b \arcsin (c+d x))^3}{2 (c+d x)^2}}{d e^3}\) |
| \(\Big \downarrow \) 4200 |
| \(\displaystyle \frac {-\frac {(a+b \arcsin (c+d x))^3}{2 (c+d x)^2}+\frac {3}{2} b \left (-\frac {\sqrt {1-(c+d x)^2} (a+b \arcsin (c+d x))^2}{c+d x}+2 b \left (2 i \int -\frac {e^{2 i \arcsin (c+d x)} (a+b \arcsin (c+d x))}{1-e^{2 i \arcsin (c+d x)}}d\arcsin (c+d x)-\frac {i (a+b \arcsin (c+d x))^2}{2 b}\right )\right )}{d e^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {(a+b \arcsin (c+d x))^3}{2 (c+d x)^2}+\frac {3}{2} b \left (-\frac {\sqrt {1-(c+d x)^2} (a+b \arcsin (c+d x))^2}{c+d x}+2 b \left (-2 i \int \frac {e^{2 i \arcsin (c+d x)} (a+b \arcsin (c+d x))}{1-e^{2 i \arcsin (c+d x)}}d\arcsin (c+d x)-\frac {i (a+b \arcsin (c+d x))^2}{2 b}\right )\right )}{d e^3}\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle \frac {-\frac {(a+b \arcsin (c+d x))^3}{2 (c+d x)^2}+\frac {3}{2} b \left (-\frac {\sqrt {1-(c+d x)^2} (a+b \arcsin (c+d x))^2}{c+d x}+2 b \left (-2 i \left (\frac {1}{2} i \log \left (1-e^{2 i \arcsin (c+d x)}\right ) (a+b \arcsin (c+d x))-\frac {1}{2} i b \int \log \left (1-e^{2 i \arcsin (c+d x)}\right )d\arcsin (c+d x)\right )-\frac {i (a+b \arcsin (c+d x))^2}{2 b}\right )\right )}{d e^3}\) |
| \(\Big \downarrow \) 2715 |
| \(\displaystyle \frac {-\frac {(a+b \arcsin (c+d x))^3}{2 (c+d x)^2}+\frac {3}{2} b \left (-\frac {\sqrt {1-(c+d x)^2} (a+b \arcsin (c+d x))^2}{c+d x}+2 b \left (-2 i \left (\frac {1}{2} i \log \left (1-e^{2 i \arcsin (c+d x)}\right ) (a+b \arcsin (c+d x))-\frac {1}{4} b \int e^{-2 i \arcsin (c+d x)} \log (-c-d x+1)de^{2 i \arcsin (c+d x)}\right )-\frac {i (a+b \arcsin (c+d x))^2}{2 b}\right )\right )}{d e^3}\) |
| \(\Big \downarrow \) 2838 |
| \(\displaystyle \frac {-\frac {(a+b \arcsin (c+d x))^3}{2 (c+d x)^2}+\frac {3}{2} b \left (-\frac {\sqrt {1-(c+d x)^2} (a+b \arcsin (c+d x))^2}{c+d x}+2 b \left (-2 i \left (\frac {1}{2} i \log \left (1-e^{2 i \arcsin (c+d x)}\right ) (a+b \arcsin (c+d x))+\frac {1}{4} b \operatorname {PolyLog}\left (2,e^{2 i \arcsin (c+d x)}\right )\right )-\frac {i (a+b \arcsin (c+d x))^2}{2 b}\right )\right )}{d e^3}\) |
(-1/2*(a + b*ArcSin[c + d*x])^3/(c + d*x)^2 + (3*b*(-((Sqrt[1 - (c + d*x)^ 2]*(a + b*ArcSin[c + d*x])^2)/(c + d*x)) + 2*b*(((-1/2*I)*(a + b*ArcSin[c + d*x])^2)/b - (2*I)*((I/2)*(a + b*ArcSin[c + d*x])*Log[1 - E^((2*I)*ArcSi n[c + d*x])] + (b*PolyLog[2, E^((2*I)*ArcSin[c + d*x])])/4))))/2)/(d*e^3)
3.3.4.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcSin[c*x])^n/(d*(m + 1))), x] - Simp[b*c*(n /(d*(m + 1))) Int[(d*x)^(m + 1)*((a + b*ArcSin[c*x])^(n - 1)/Sqrt[1 - c^2 *x^2]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]
Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_. )*(x_)^2)^(p_), x_Symbol] :> Simp[(f*x)^(m + 1)*(d + e*x^2)^(p + 1)*((a + b *ArcSin[c*x])^n/(d*f*(m + 1))), x] - Simp[b*c*(n/(f*(m + 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*A rcSin[c*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[c^ 2*d + e, 0] && GtQ[n, 0] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]
Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m _.), x_Symbol] :> Simp[1/d Subst[Int[((d*e - c*f)/d + f*(x/d))^m*(a + b*A rcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]
Time = 0.91 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.94
| method | result | size |
| derivativedivides | \(\frac {-\frac {a^{3}}{2 e^{3} \left (d x +c \right )^{2}}+\frac {b^{3} \left (-\frac {\arcsin \left (d x +c \right )^{2} \left (-3 i \left (d x +c \right )^{2}+3 \left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}+\arcsin \left (d x +c \right )\right )}{2 \left (d x +c \right )^{2}}+3 \arcsin \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )+3 \arcsin \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )-3 i \arcsin \left (d x +c \right )^{2}-3 i \operatorname {polylog}\left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )-3 i \operatorname {polylog}\left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )\right )}{e^{3}}+\frac {3 a \,b^{2} \left (-\frac {\arcsin \left (d x +c \right )^{2}}{2 \left (d x +c \right )^{2}}-\frac {\arcsin \left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}}{d x +c}+\ln \left (d x +c \right )\right )}{e^{3}}+\frac {3 a^{2} b \left (-\frac {\arcsin \left (d x +c \right )}{2 \left (d x +c \right )^{2}}-\frac {\sqrt {1-\left (d x +c \right )^{2}}}{2 \left (d x +c \right )}\right )}{e^{3}}}{d}\) | \(324\) |
| default | \(\frac {-\frac {a^{3}}{2 e^{3} \left (d x +c \right )^{2}}+\frac {b^{3} \left (-\frac {\arcsin \left (d x +c \right )^{2} \left (-3 i \left (d x +c \right )^{2}+3 \left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}+\arcsin \left (d x +c \right )\right )}{2 \left (d x +c \right )^{2}}+3 \arcsin \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )+3 \arcsin \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )-3 i \arcsin \left (d x +c \right )^{2}-3 i \operatorname {polylog}\left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )-3 i \operatorname {polylog}\left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )\right )}{e^{3}}+\frac {3 a \,b^{2} \left (-\frac {\arcsin \left (d x +c \right )^{2}}{2 \left (d x +c \right )^{2}}-\frac {\arcsin \left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}}{d x +c}+\ln \left (d x +c \right )\right )}{e^{3}}+\frac {3 a^{2} b \left (-\frac {\arcsin \left (d x +c \right )}{2 \left (d x +c \right )^{2}}-\frac {\sqrt {1-\left (d x +c \right )^{2}}}{2 \left (d x +c \right )}\right )}{e^{3}}}{d}\) | \(324\) |
| parts | \(-\frac {a^{3}}{2 e^{3} \left (d x +c \right )^{2} d}+\frac {b^{3} \left (-\frac {\arcsin \left (d x +c \right )^{2} \left (-3 i \left (d x +c \right )^{2}+3 \left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}+\arcsin \left (d x +c \right )\right )}{2 \left (d x +c \right )^{2}}+3 \arcsin \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )+3 \arcsin \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )-3 i \arcsin \left (d x +c \right )^{2}-3 i \operatorname {polylog}\left (2, i \left (d x +c \right )+\sqrt {1-\left (d x +c \right )^{2}}\right )-3 i \operatorname {polylog}\left (2, -i \left (d x +c \right )-\sqrt {1-\left (d x +c \right )^{2}}\right )\right )}{e^{3} d}+\frac {3 a \,b^{2} \left (-\frac {\arcsin \left (d x +c \right )^{2}}{2 \left (d x +c \right )^{2}}-\frac {\arcsin \left (d x +c \right ) \sqrt {1-\left (d x +c \right )^{2}}}{d x +c}+\ln \left (d x +c \right )\right )}{e^{3} d}+\frac {3 a^{2} b \left (-\frac {\arcsin \left (d x +c \right )}{2 \left (d x +c \right )^{2}}-\frac {\sqrt {1-\left (d x +c \right )^{2}}}{2 \left (d x +c \right )}\right )}{e^{3} d}\) | \(332\) |
1/d*(-1/2*a^3/e^3/(d*x+c)^2+b^3/e^3*(-1/2*arcsin(d*x+c)^2*(-3*I*(d*x+c)^2+ 3*(d*x+c)*(1-(d*x+c)^2)^(1/2)+arcsin(d*x+c))/(d*x+c)^2+3*arcsin(d*x+c)*ln( 1-I*(d*x+c)-(1-(d*x+c)^2)^(1/2))+3*arcsin(d*x+c)*ln(1+I*(d*x+c)+(1-(d*x+c) ^2)^(1/2))-3*I*arcsin(d*x+c)^2-3*I*polylog(2,I*(d*x+c)+(1-(d*x+c)^2)^(1/2) )-3*I*polylog(2,-I*(d*x+c)-(1-(d*x+c)^2)^(1/2)))+3*a*b^2/e^3*(-1/2*arcsin( d*x+c)^2/(d*x+c)^2-arcsin(d*x+c)*(1-(d*x+c)^2)^(1/2)/(d*x+c)+ln(d*x+c))+3* a^2*b/e^3*(-1/2/(d*x+c)^2*arcsin(d*x+c)-1/2/(d*x+c)*(1-(d*x+c)^2)^(1/2)))
\[ \int \frac {(a+b \arcsin (c+d x))^3}{(c e+d e x)^3} \, dx=\int { \frac {{\left (b \arcsin \left (d x + c\right ) + a\right )}^{3}}{{\left (d e x + c e\right )}^{3}} \,d x } \]
integral((b^3*arcsin(d*x + c)^3 + 3*a*b^2*arcsin(d*x + c)^2 + 3*a^2*b*arcs in(d*x + c) + a^3)/(d^3*e^3*x^3 + 3*c*d^2*e^3*x^2 + 3*c^2*d*e^3*x + c^3*e^ 3), x)
\[ \int \frac {(a+b \arcsin (c+d x))^3}{(c e+d e x)^3} \, dx=\frac {\int \frac {a^{3}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {b^{3} \operatorname {asin}^{3}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {3 a b^{2} \operatorname {asin}^{2}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx + \int \frac {3 a^{2} b \operatorname {asin}{\left (c + d x \right )}}{c^{3} + 3 c^{2} d x + 3 c d^{2} x^{2} + d^{3} x^{3}}\, dx}{e^{3}} \]
(Integral(a**3/(c**3 + 3*c**2*d*x + 3*c*d**2*x**2 + d**3*x**3), x) + Integ ral(b**3*asin(c + d*x)**3/(c**3 + 3*c**2*d*x + 3*c*d**2*x**2 + d**3*x**3), x) + Integral(3*a*b**2*asin(c + d*x)**2/(c**3 + 3*c**2*d*x + 3*c*d**2*x** 2 + d**3*x**3), x) + Integral(3*a**2*b*asin(c + d*x)/(c**3 + 3*c**2*d*x + 3*c*d**2*x**2 + d**3*x**3), x))/e**3
Timed out. \[ \int \frac {(a+b \arcsin (c+d x))^3}{(c e+d e x)^3} \, dx=\text {Timed out} \]
\[ \int \frac {(a+b \arcsin (c+d x))^3}{(c e+d e x)^3} \, dx=\int { \frac {{\left (b \arcsin \left (d x + c\right ) + a\right )}^{3}}{{\left (d e x + c e\right )}^{3}} \,d x } \]
Timed out. \[ \int \frac {(a+b \arcsin (c+d x))^3}{(c e+d e x)^3} \, dx=\int \frac {{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^3}{{\left (c\,e+d\,e\,x\right )}^3} \,d x \]