3.619 \(\int \frac {(d+e x^2)^3 (a+b \sin ^{-1}(c x))}{x} \, dx\)

Optimal. Leaf size=357 \[ d^3 \log (x) \left (a+b \sin ^{-1}(c x)\right )+\frac {3}{2} d^2 e x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {3}{4} d e^2 x^4 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{6} e^3 x^6 \left (a+b \sin ^{-1}(c x)\right )-\frac {5 b e^3 \sin ^{-1}(c x)}{96 c^6}-\frac {9 b d e^2 \sin ^{-1}(c x)}{32 c^4}+\frac {3 b d^2 e x \sqrt {1-c^2 x^2}}{4 c}-\frac {3 b d^2 e \sin ^{-1}(c x)}{4 c^2}+\frac {3 b d e^2 x^3 \sqrt {1-c^2 x^2}}{16 c}+\frac {b e^3 x^5 \sqrt {1-c^2 x^2}}{36 c}+\frac {5 b e^3 x \sqrt {1-c^2 x^2}}{96 c^5}+\frac {9 b d e^2 x \sqrt {1-c^2 x^2}}{32 c^3}+\frac {5 b e^3 x^3 \sqrt {1-c^2 x^2}}{144 c^3}-\frac {1}{2} i b d^3 \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )-\frac {1}{2} i b d^3 \sin ^{-1}(c x)^2+b d^3 \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-b d^3 \log (x) \sin ^{-1}(c x) \]

[Out]

-3/4*b*d^2*e*arcsin(c*x)/c^2-9/32*b*d*e^2*arcsin(c*x)/c^4-5/96*b*e^3*arcsin(c*x)/c^6-1/2*I*b*d^3*arcsin(c*x)^2
+3/2*d^2*e*x^2*(a+b*arcsin(c*x))+3/4*d*e^2*x^4*(a+b*arcsin(c*x))+1/6*e^3*x^6*(a+b*arcsin(c*x))+b*d^3*arcsin(c*
x)*ln(1-(I*c*x+(-c^2*x^2+1)^(1/2))^2)-b*d^3*arcsin(c*x)*ln(x)+d^3*(a+b*arcsin(c*x))*ln(x)-1/2*I*b*d^3*polylog(
2,(I*c*x+(-c^2*x^2+1)^(1/2))^2)+3/4*b*d^2*e*x*(-c^2*x^2+1)^(1/2)/c+9/32*b*d*e^2*x*(-c^2*x^2+1)^(1/2)/c^3+5/96*
b*e^3*x*(-c^2*x^2+1)^(1/2)/c^5+3/16*b*d*e^2*x^3*(-c^2*x^2+1)^(1/2)/c+5/144*b*e^3*x^3*(-c^2*x^2+1)^(1/2)/c^3+1/
36*b*e^3*x^5*(-c^2*x^2+1)^(1/2)/c

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Rubi [A]  time = 0.48, antiderivative size = 357, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 13, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.619, Rules used = {266, 43, 4731, 12, 6742, 321, 216, 2326, 4625, 3717, 2190, 2279, 2391} \[ -\frac {1}{2} i b d^3 \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )+\frac {3}{2} d^2 e x^2 \left (a+b \sin ^{-1}(c x)\right )+d^3 \log (x) \left (a+b \sin ^{-1}(c x)\right )+\frac {3}{4} d e^2 x^4 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{6} e^3 x^6 \left (a+b \sin ^{-1}(c x)\right )+\frac {3 b d^2 e x \sqrt {1-c^2 x^2}}{4 c}-\frac {3 b d^2 e \sin ^{-1}(c x)}{4 c^2}+\frac {3 b d e^2 x^3 \sqrt {1-c^2 x^2}}{16 c}+\frac {9 b d e^2 x \sqrt {1-c^2 x^2}}{32 c^3}-\frac {9 b d e^2 \sin ^{-1}(c x)}{32 c^4}+\frac {b e^3 x^5 \sqrt {1-c^2 x^2}}{36 c}+\frac {5 b e^3 x^3 \sqrt {1-c^2 x^2}}{144 c^3}+\frac {5 b e^3 x \sqrt {1-c^2 x^2}}{96 c^5}-\frac {5 b e^3 \sin ^{-1}(c x)}{96 c^6}-\frac {1}{2} i b d^3 \sin ^{-1}(c x)^2+b d^3 \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-b d^3 \log (x) \sin ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)^3*(a + b*ArcSin[c*x]))/x,x]

[Out]

(3*b*d^2*e*x*Sqrt[1 - c^2*x^2])/(4*c) + (9*b*d*e^2*x*Sqrt[1 - c^2*x^2])/(32*c^3) + (5*b*e^3*x*Sqrt[1 - c^2*x^2
])/(96*c^5) + (3*b*d*e^2*x^3*Sqrt[1 - c^2*x^2])/(16*c) + (5*b*e^3*x^3*Sqrt[1 - c^2*x^2])/(144*c^3) + (b*e^3*x^
5*Sqrt[1 - c^2*x^2])/(36*c) - (3*b*d^2*e*ArcSin[c*x])/(4*c^2) - (9*b*d*e^2*ArcSin[c*x])/(32*c^4) - (5*b*e^3*Ar
cSin[c*x])/(96*c^6) - (I/2)*b*d^3*ArcSin[c*x]^2 + (3*d^2*e*x^2*(a + b*ArcSin[c*x]))/2 + (3*d*e^2*x^4*(a + b*Ar
cSin[c*x]))/4 + (e^3*x^6*(a + b*ArcSin[c*x]))/6 + b*d^3*ArcSin[c*x]*Log[1 - E^((2*I)*ArcSin[c*x])] - b*d^3*Arc
Sin[c*x]*Log[x] + d^3*(a + b*ArcSin[c*x])*Log[x] - (I/2)*b*d^3*PolyLog[2, E^((2*I)*ArcSin[c*x])]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(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]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[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]

Rule 2326

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(ArcSin[(Rt[-e, 2]*x)/S
qrt[d]]*(a + b*Log[c*x^n]))/Rt[-e, 2], x] - Dist[(b*n)/Rt[-e, 2], Int[ArcSin[(Rt[-e, 2]*x)/Sqrt[d]]/x, x], x]
/; FreeQ[{a, b, c, d, e, n}, x] && GtQ[d, 0] && NegQ[e]

Rule 2391

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]

Rule 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*
(m + 1)), x] - Dist[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]

Rule 4625

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[(a + b*x)^n/Tan[x], x], x, ArcSin[c*
x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 4731

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSin[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/Sqrt[1 -
c^2*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[c^2*d + e, 0] && IntegerQ[p] && (GtQ[p, 0] ||
(IGtQ[(m - 1)/2, 0] && LeQ[m + p, 0]))

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {\left (d+e x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{x} \, dx &=\frac {3}{2} d^2 e x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {3}{4} d e^2 x^4 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{6} e^3 x^6 \left (a+b \sin ^{-1}(c x)\right )+d^3 \left (a+b \sin ^{-1}(c x)\right ) \log (x)-(b c) \int \frac {18 d^2 e x^2+9 d e^2 x^4+2 e^3 x^6+12 d^3 \log (x)}{12 \sqrt {1-c^2 x^2}} \, dx\\ &=\frac {3}{2} d^2 e x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {3}{4} d e^2 x^4 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{6} e^3 x^6 \left (a+b \sin ^{-1}(c x)\right )+d^3 \left (a+b \sin ^{-1}(c x)\right ) \log (x)-\frac {1}{12} (b c) \int \frac {18 d^2 e x^2+9 d e^2 x^4+2 e^3 x^6+12 d^3 \log (x)}{\sqrt {1-c^2 x^2}} \, dx\\ &=\frac {3}{2} d^2 e x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {3}{4} d e^2 x^4 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{6} e^3 x^6 \left (a+b \sin ^{-1}(c x)\right )+d^3 \left (a+b \sin ^{-1}(c x)\right ) \log (x)-\frac {1}{12} (b c) \int \left (\frac {18 d^2 e x^2}{\sqrt {1-c^2 x^2}}+\frac {9 d e^2 x^4}{\sqrt {1-c^2 x^2}}+\frac {2 e^3 x^6}{\sqrt {1-c^2 x^2}}+\frac {12 d^3 \log (x)}{\sqrt {1-c^2 x^2}}\right ) \, dx\\ &=\frac {3}{2} d^2 e x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {3}{4} d e^2 x^4 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{6} e^3 x^6 \left (a+b \sin ^{-1}(c x)\right )+d^3 \left (a+b \sin ^{-1}(c x)\right ) \log (x)-\left (b c d^3\right ) \int \frac {\log (x)}{\sqrt {1-c^2 x^2}} \, dx-\frac {1}{2} \left (3 b c d^2 e\right ) \int \frac {x^2}{\sqrt {1-c^2 x^2}} \, dx-\frac {1}{4} \left (3 b c d e^2\right ) \int \frac {x^4}{\sqrt {1-c^2 x^2}} \, dx-\frac {1}{6} \left (b c e^3\right ) \int \frac {x^6}{\sqrt {1-c^2 x^2}} \, dx\\ &=\frac {3 b d^2 e x \sqrt {1-c^2 x^2}}{4 c}+\frac {3 b d e^2 x^3 \sqrt {1-c^2 x^2}}{16 c}+\frac {b e^3 x^5 \sqrt {1-c^2 x^2}}{36 c}+\frac {3}{2} d^2 e x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {3}{4} d e^2 x^4 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{6} e^3 x^6 \left (a+b \sin ^{-1}(c x)\right )-b d^3 \sin ^{-1}(c x) \log (x)+d^3 \left (a+b \sin ^{-1}(c x)\right ) \log (x)+\left (b d^3\right ) \int \frac {\sin ^{-1}(c x)}{x} \, dx-\frac {\left (3 b d^2 e\right ) \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{4 c}-\frac {\left (9 b d e^2\right ) \int \frac {x^2}{\sqrt {1-c^2 x^2}} \, dx}{16 c}-\frac {\left (5 b e^3\right ) \int \frac {x^4}{\sqrt {1-c^2 x^2}} \, dx}{36 c}\\ &=\frac {3 b d^2 e x \sqrt {1-c^2 x^2}}{4 c}+\frac {9 b d e^2 x \sqrt {1-c^2 x^2}}{32 c^3}+\frac {3 b d e^2 x^3 \sqrt {1-c^2 x^2}}{16 c}+\frac {5 b e^3 x^3 \sqrt {1-c^2 x^2}}{144 c^3}+\frac {b e^3 x^5 \sqrt {1-c^2 x^2}}{36 c}-\frac {3 b d^2 e \sin ^{-1}(c x)}{4 c^2}+\frac {3}{2} d^2 e x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {3}{4} d e^2 x^4 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{6} e^3 x^6 \left (a+b \sin ^{-1}(c x)\right )-b d^3 \sin ^{-1}(c x) \log (x)+d^3 \left (a+b \sin ^{-1}(c x)\right ) \log (x)+\left (b d^3\right ) \operatorname {Subst}\left (\int x \cot (x) \, dx,x,\sin ^{-1}(c x)\right )-\frac {\left (9 b d e^2\right ) \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{32 c^3}-\frac {\left (5 b e^3\right ) \int \frac {x^2}{\sqrt {1-c^2 x^2}} \, dx}{48 c^3}\\ &=\frac {3 b d^2 e x \sqrt {1-c^2 x^2}}{4 c}+\frac {9 b d e^2 x \sqrt {1-c^2 x^2}}{32 c^3}+\frac {5 b e^3 x \sqrt {1-c^2 x^2}}{96 c^5}+\frac {3 b d e^2 x^3 \sqrt {1-c^2 x^2}}{16 c}+\frac {5 b e^3 x^3 \sqrt {1-c^2 x^2}}{144 c^3}+\frac {b e^3 x^5 \sqrt {1-c^2 x^2}}{36 c}-\frac {3 b d^2 e \sin ^{-1}(c x)}{4 c^2}-\frac {9 b d e^2 \sin ^{-1}(c x)}{32 c^4}-\frac {1}{2} i b d^3 \sin ^{-1}(c x)^2+\frac {3}{2} d^2 e x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {3}{4} d e^2 x^4 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{6} e^3 x^6 \left (a+b \sin ^{-1}(c x)\right )-b d^3 \sin ^{-1}(c x) \log (x)+d^3 \left (a+b \sin ^{-1}(c x)\right ) \log (x)-\left (2 i b d^3\right ) \operatorname {Subst}\left (\int \frac {e^{2 i x} x}{1-e^{2 i x}} \, dx,x,\sin ^{-1}(c x)\right )-\frac {\left (5 b e^3\right ) \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{96 c^5}\\ &=\frac {3 b d^2 e x \sqrt {1-c^2 x^2}}{4 c}+\frac {9 b d e^2 x \sqrt {1-c^2 x^2}}{32 c^3}+\frac {5 b e^3 x \sqrt {1-c^2 x^2}}{96 c^5}+\frac {3 b d e^2 x^3 \sqrt {1-c^2 x^2}}{16 c}+\frac {5 b e^3 x^3 \sqrt {1-c^2 x^2}}{144 c^3}+\frac {b e^3 x^5 \sqrt {1-c^2 x^2}}{36 c}-\frac {3 b d^2 e \sin ^{-1}(c x)}{4 c^2}-\frac {9 b d e^2 \sin ^{-1}(c x)}{32 c^4}-\frac {5 b e^3 \sin ^{-1}(c x)}{96 c^6}-\frac {1}{2} i b d^3 \sin ^{-1}(c x)^2+\frac {3}{2} d^2 e x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {3}{4} d e^2 x^4 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{6} e^3 x^6 \left (a+b \sin ^{-1}(c x)\right )+b d^3 \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-b d^3 \sin ^{-1}(c x) \log (x)+d^3 \left (a+b \sin ^{-1}(c x)\right ) \log (x)-\left (b d^3\right ) \operatorname {Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )\\ &=\frac {3 b d^2 e x \sqrt {1-c^2 x^2}}{4 c}+\frac {9 b d e^2 x \sqrt {1-c^2 x^2}}{32 c^3}+\frac {5 b e^3 x \sqrt {1-c^2 x^2}}{96 c^5}+\frac {3 b d e^2 x^3 \sqrt {1-c^2 x^2}}{16 c}+\frac {5 b e^3 x^3 \sqrt {1-c^2 x^2}}{144 c^3}+\frac {b e^3 x^5 \sqrt {1-c^2 x^2}}{36 c}-\frac {3 b d^2 e \sin ^{-1}(c x)}{4 c^2}-\frac {9 b d e^2 \sin ^{-1}(c x)}{32 c^4}-\frac {5 b e^3 \sin ^{-1}(c x)}{96 c^6}-\frac {1}{2} i b d^3 \sin ^{-1}(c x)^2+\frac {3}{2} d^2 e x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {3}{4} d e^2 x^4 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{6} e^3 x^6 \left (a+b \sin ^{-1}(c x)\right )+b d^3 \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-b d^3 \sin ^{-1}(c x) \log (x)+d^3 \left (a+b \sin ^{-1}(c x)\right ) \log (x)+\frac {1}{2} \left (i b d^3\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )\\ &=\frac {3 b d^2 e x \sqrt {1-c^2 x^2}}{4 c}+\frac {9 b d e^2 x \sqrt {1-c^2 x^2}}{32 c^3}+\frac {5 b e^3 x \sqrt {1-c^2 x^2}}{96 c^5}+\frac {3 b d e^2 x^3 \sqrt {1-c^2 x^2}}{16 c}+\frac {5 b e^3 x^3 \sqrt {1-c^2 x^2}}{144 c^3}+\frac {b e^3 x^5 \sqrt {1-c^2 x^2}}{36 c}-\frac {3 b d^2 e \sin ^{-1}(c x)}{4 c^2}-\frac {9 b d e^2 \sin ^{-1}(c x)}{32 c^4}-\frac {5 b e^3 \sin ^{-1}(c x)}{96 c^6}-\frac {1}{2} i b d^3 \sin ^{-1}(c x)^2+\frac {3}{2} d^2 e x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {3}{4} d e^2 x^4 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{6} e^3 x^6 \left (a+b \sin ^{-1}(c x)\right )+b d^3 \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-b d^3 \sin ^{-1}(c x) \log (x)+d^3 \left (a+b \sin ^{-1}(c x)\right ) \log (x)-\frac {1}{2} i b d^3 \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )\\ \end {align*}

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Mathematica [A]  time = 0.38, size = 278, normalized size = 0.78 \[ a d^3 \log (x)+\frac {3}{2} a d^2 e x^2+\frac {3}{4} a d e^2 x^4+\frac {1}{6} a e^3 x^6+\frac {3 b d^2 e \left (c x \sqrt {1-c^2 x^2}-\sin ^{-1}(c x)\right )}{4 c^2}+\frac {3 b d e^2 \left (c x \sqrt {1-c^2 x^2} \left (2 c^2 x^2+3\right )-3 \sin ^{-1}(c x)\right )}{32 c^4}+\frac {b e^3 \left (c x \sqrt {1-c^2 x^2} \left (8 c^4 x^4+10 c^2 x^2+15\right )-15 \sin ^{-1}(c x)\right )}{288 c^6}-\frac {1}{2} i b d^3 \left (\sin ^{-1}(c x)^2+\text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )\right )+b d^3 \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )+\frac {3}{2} b d^2 e x^2 \sin ^{-1}(c x)+\frac {3}{4} b d e^2 x^4 \sin ^{-1}(c x)+\frac {1}{6} b e^3 x^6 \sin ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)^3*(a + b*ArcSin[c*x]))/x,x]

[Out]

(3*a*d^2*e*x^2)/2 + (3*a*d*e^2*x^4)/4 + (a*e^3*x^6)/6 + (b*e^3*(c*x*Sqrt[1 - c^2*x^2]*(15 + 10*c^2*x^2 + 8*c^4
*x^4) - 15*ArcSin[c*x]))/(288*c^6) + (3*b*d*e^2*(c*x*Sqrt[1 - c^2*x^2]*(3 + 2*c^2*x^2) - 3*ArcSin[c*x]))/(32*c
^4) + (3*b*d^2*e*(c*x*Sqrt[1 - c^2*x^2] - ArcSin[c*x]))/(4*c^2) + (3*b*d^2*e*x^2*ArcSin[c*x])/2 + (3*b*d*e^2*x
^4*ArcSin[c*x])/4 + (b*e^3*x^6*ArcSin[c*x])/6 + b*d^3*ArcSin[c*x]*Log[1 - E^((2*I)*ArcSin[c*x])] + a*d^3*Log[x
] - (I/2)*b*d^3*(ArcSin[c*x]^2 + PolyLog[2, E^((2*I)*ArcSin[c*x])])

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a e^{3} x^{6} + 3 \, a d e^{2} x^{4} + 3 \, a d^{2} e x^{2} + a d^{3} + {\left (b e^{3} x^{6} + 3 \, b d e^{2} x^{4} + 3 \, b d^{2} e x^{2} + b d^{3}\right )} \arcsin \left (c x\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^3*(a+b*arcsin(c*x))/x,x, algorithm="fricas")

[Out]

integral((a*e^3*x^6 + 3*a*d*e^2*x^4 + 3*a*d^2*e*x^2 + a*d^3 + (b*e^3*x^6 + 3*b*d*e^2*x^4 + 3*b*d^2*e*x^2 + b*d
^3)*arcsin(c*x))/x, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x^{2} + d\right )}^{3} {\left (b \arcsin \left (c x\right ) + a\right )}}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^3*(a+b*arcsin(c*x))/x,x, algorithm="giac")

[Out]

integrate((e*x^2 + d)^3*(b*arcsin(c*x) + a)/x, x)

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maple [A]  time = 0.44, size = 391, normalized size = 1.10 \[ \frac {a \,e^{3} x^{6}}{6}+\frac {3 a d \,e^{2} x^{4}}{4}+\frac {3 a \,d^{2} e \,x^{2}}{2}+d^{3} a \ln \left (c x \right )-\frac {i b \,d^{3} \arcsin \left (c x \right )^{2}}{2}+d^{3} b \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )+d^{3} b \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )-i d^{3} b \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )-i d^{3} b \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )-\frac {b \arcsin \left (c x \right ) e^{3} \cos \left (6 \arcsin \left (c x \right )\right )}{192 c^{6}}+\frac {b \,e^{3} \sin \left (6 \arcsin \left (c x \right )\right )}{1152 c^{6}}+\frac {3 b \cos \left (4 \arcsin \left (c x \right )\right ) \arcsin \left (c x \right ) d \,e^{2}}{32 c^{4}}+\frac {b \cos \left (4 \arcsin \left (c x \right )\right ) \arcsin \left (c x \right ) e^{3}}{32 c^{6}}-\frac {3 b \sin \left (4 \arcsin \left (c x \right )\right ) d \,e^{2}}{128 c^{4}}-\frac {b \sin \left (4 \arcsin \left (c x \right )\right ) e^{3}}{128 c^{6}}-\frac {3 b \cos \left (2 \arcsin \left (c x \right )\right ) \arcsin \left (c x \right ) d^{2} e}{4 c^{2}}-\frac {3 b \cos \left (2 \arcsin \left (c x \right )\right ) \arcsin \left (c x \right ) d \,e^{2}}{8 c^{4}}-\frac {5 b \cos \left (2 \arcsin \left (c x \right )\right ) \arcsin \left (c x \right ) e^{3}}{64 c^{6}}+\frac {3 b \sin \left (2 \arcsin \left (c x \right )\right ) d^{2} e}{8 c^{2}}+\frac {3 b \sin \left (2 \arcsin \left (c x \right )\right ) d \,e^{2}}{16 c^{4}}+\frac {5 b \sin \left (2 \arcsin \left (c x \right )\right ) e^{3}}{128 c^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)^3*(a+b*arcsin(c*x))/x,x)

[Out]

1/6*a*e^3*x^6+3/4*a*d*e^2*x^4+3/2*a*d^2*e*x^2+d^3*a*ln(c*x)-I*d^3*b*polylog(2,-I*c*x-(-c^2*x^2+1)^(1/2))+d^3*b
*arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1)^(1/2))+d^3*b*arcsin(c*x)*ln(1-I*c*x-(-c^2*x^2+1)^(1/2))-1/2*I*b*d^3*arcsi
n(c*x)^2-I*d^3*b*polylog(2,I*c*x+(-c^2*x^2+1)^(1/2))-1/192*b/c^6*arcsin(c*x)*e^3*cos(6*arcsin(c*x))+1/1152*b/c
^6*e^3*sin(6*arcsin(c*x))+3/32*b/c^4*cos(4*arcsin(c*x))*arcsin(c*x)*d*e^2+1/32*b/c^6*cos(4*arcsin(c*x))*arcsin
(c*x)*e^3-3/128*b/c^4*sin(4*arcsin(c*x))*d*e^2-1/128*b/c^6*sin(4*arcsin(c*x))*e^3-3/4*b/c^2*cos(2*arcsin(c*x))
*arcsin(c*x)*d^2*e-3/8*b/c^4*cos(2*arcsin(c*x))*arcsin(c*x)*d*e^2-5/64*b/c^6*cos(2*arcsin(c*x))*arcsin(c*x)*e^
3+3/8*b/c^2*sin(2*arcsin(c*x))*d^2*e+3/16*b/c^4*sin(2*arcsin(c*x))*d*e^2+5/128*b/c^6*sin(2*arcsin(c*x))*e^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{6} \, a e^{3} x^{6} + \frac {3}{4} \, a d e^{2} x^{4} + \frac {3}{2} \, a d^{2} e x^{2} + a d^{3} \log \relax (x) + \int \frac {{\left (b e^{3} x^{6} + 3 \, b d e^{2} x^{4} + 3 \, b d^{2} e x^{2} + b d^{3}\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^3*(a+b*arcsin(c*x))/x,x, algorithm="maxima")

[Out]

1/6*a*e^3*x^6 + 3/4*a*d*e^2*x^4 + 3/2*a*d^2*e*x^2 + a*d^3*log(x) + integrate((b*e^3*x^6 + 3*b*d*e^2*x^4 + 3*b*
d^2*e*x^2 + b*d^3)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))/x, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,{\left (e\,x^2+d\right )}^3}{x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*asin(c*x))*(d + e*x^2)^3)/x,x)

[Out]

int(((a + b*asin(c*x))*(d + e*x^2)^3)/x, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {asin}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{3}}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)**3*(a+b*asin(c*x))/x,x)

[Out]

Integral((a + b*asin(c*x))*(d + e*x**2)**3/x, x)

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