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

Optimal. Leaf size=262 \[ -\frac {d^3 \left (a+b \sin ^{-1}(c x)\right )}{2 x^2}+3 d^2 e \log (x) \left (a+b \sin ^{-1}(c x)\right )+\frac {3}{2} d e^2 x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} e^3 x^4 \left (a+b \sin ^{-1}(c x)\right )-\frac {b c d^3 \sqrt {1-c^2 x^2}}{2 x}+\frac {b e^3 x^3 \sqrt {1-c^2 x^2}}{16 c}-\frac {3 b e^2 \left (8 c^2 d+e\right ) \sin ^{-1}(c x)}{32 c^4}+\frac {3 b e^2 x \sqrt {1-c^2 x^2} \left (8 c^2 d+e\right )}{32 c^3}-\frac {3}{2} i b d^2 e \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )-\frac {3}{2} i b d^2 e \sin ^{-1}(c x)^2+3 b d^2 e \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-3 b d^2 e \log (x) \sin ^{-1}(c x) \]

[Out]

-3/32*b*e^2*(8*c^2*d+e)*arcsin(c*x)/c^4-3/2*I*b*d^2*e*arcsin(c*x)^2-1/2*d^3*(a+b*arcsin(c*x))/x^2+3/2*d*e^2*x^
2*(a+b*arcsin(c*x))+1/4*e^3*x^4*(a+b*arcsin(c*x))+3*b*d^2*e*arcsin(c*x)*ln(1-(I*c*x+(-c^2*x^2+1)^(1/2))^2)-3*b
*d^2*e*arcsin(c*x)*ln(x)+3*d^2*e*(a+b*arcsin(c*x))*ln(x)-3/2*I*b*d^2*e*polylog(2,(I*c*x+(-c^2*x^2+1)^(1/2))^2)
-1/2*b*c*d^3*(-c^2*x^2+1)^(1/2)/x+3/32*b*e^2*(8*c^2*d+e)*x*(-c^2*x^2+1)^(1/2)/c^3+1/16*b*e^3*x^3*(-c^2*x^2+1)^
(1/2)/c

________________________________________________________________________________________

Rubi [A]  time = 0.78, antiderivative size = 262, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 16, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.762, Rules used = {266, 43, 4731, 12, 6742, 1807, 1584, 459, 321, 216, 2326, 4625, 3717, 2190, 2279, 2391} \[ -\frac {3}{2} i b d^2 e \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )+3 d^2 e \log (x) \left (a+b \sin ^{-1}(c x)\right )-\frac {d^3 \left (a+b \sin ^{-1}(c x)\right )}{2 x^2}+\frac {3}{2} d e^2 x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} e^3 x^4 \left (a+b \sin ^{-1}(c x)\right )-\frac {b c d^3 \sqrt {1-c^2 x^2}}{2 x}+\frac {3 b e^2 x \sqrt {1-c^2 x^2} \left (8 c^2 d+e\right )}{32 c^3}-\frac {3 b e^2 \left (8 c^2 d+e\right ) \sin ^{-1}(c x)}{32 c^4}+\frac {b e^3 x^3 \sqrt {1-c^2 x^2}}{16 c}-\frac {3}{2} i b d^2 e \sin ^{-1}(c x)^2+3 b d^2 e \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-3 b d^2 e \log (x) \sin ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*c*d^3*Sqrt[1 - c^2*x^2])/(2*x) + (3*b*e^2*(8*c^2*d + e)*x*Sqrt[1 - c^2*x^2])/(32*c^3) + (b*e^3*x^3*Sqrt[1
- c^2*x^2])/(16*c) - (3*b*e^2*(8*c^2*d + e)*ArcSin[c*x])/(32*c^4) - ((3*I)/2)*b*d^2*e*ArcSin[c*x]^2 - (d^3*(a
+ b*ArcSin[c*x]))/(2*x^2) + (3*d*e^2*x^2*(a + b*ArcSin[c*x]))/2 + (e^3*x^4*(a + b*ArcSin[c*x]))/4 + 3*b*d^2*e*
ArcSin[c*x]*Log[1 - E^((2*I)*ArcSin[c*x])] - 3*b*d^2*e*ArcSin[c*x]*Log[x] + 3*d^2*e*(a + b*ArcSin[c*x])*Log[x]
- ((3*I)/2)*b*d^2*e*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 459

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

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

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^3} \, dx &=-\frac {d^3 \left (a+b \sin ^{-1}(c x)\right )}{2 x^2}+\frac {3}{2} d e^2 x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} e^3 x^4 \left (a+b \sin ^{-1}(c x)\right )+3 d^2 e \left (a+b \sin ^{-1}(c x)\right ) \log (x)-(b c) \int \frac {-2 d^3+6 d e^2 x^4+e^3 x^6+12 d^2 e x^2 \log (x)}{4 x^2 \sqrt {1-c^2 x^2}} \, dx\\ &=-\frac {d^3 \left (a+b \sin ^{-1}(c x)\right )}{2 x^2}+\frac {3}{2} d e^2 x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} e^3 x^4 \left (a+b \sin ^{-1}(c x)\right )+3 d^2 e \left (a+b \sin ^{-1}(c x)\right ) \log (x)-\frac {1}{4} (b c) \int \frac {-2 d^3+6 d e^2 x^4+e^3 x^6+12 d^2 e x^2 \log (x)}{x^2 \sqrt {1-c^2 x^2}} \, dx\\ &=-\frac {d^3 \left (a+b \sin ^{-1}(c x)\right )}{2 x^2}+\frac {3}{2} d e^2 x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} e^3 x^4 \left (a+b \sin ^{-1}(c x)\right )+3 d^2 e \left (a+b \sin ^{-1}(c x)\right ) \log (x)-\frac {1}{4} (b c) \int \left (\frac {-2 d^3+6 d e^2 x^4+e^3 x^6}{x^2 \sqrt {1-c^2 x^2}}+\frac {12 d^2 e \log (x)}{\sqrt {1-c^2 x^2}}\right ) \, dx\\ &=-\frac {d^3 \left (a+b \sin ^{-1}(c x)\right )}{2 x^2}+\frac {3}{2} d e^2 x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} e^3 x^4 \left (a+b \sin ^{-1}(c x)\right )+3 d^2 e \left (a+b \sin ^{-1}(c x)\right ) \log (x)-\frac {1}{4} (b c) \int \frac {-2 d^3+6 d e^2 x^4+e^3 x^6}{x^2 \sqrt {1-c^2 x^2}} \, dx-\left (3 b c d^2 e\right ) \int \frac {\log (x)}{\sqrt {1-c^2 x^2}} \, dx\\ &=-\frac {b c d^3 \sqrt {1-c^2 x^2}}{2 x}-\frac {d^3 \left (a+b \sin ^{-1}(c x)\right )}{2 x^2}+\frac {3}{2} d e^2 x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} e^3 x^4 \left (a+b \sin ^{-1}(c x)\right )-3 b d^2 e \sin ^{-1}(c x) \log (x)+3 d^2 e \left (a+b \sin ^{-1}(c x)\right ) \log (x)+\frac {1}{4} (b c) \int \frac {-6 d e^2 x^3-e^3 x^5}{x \sqrt {1-c^2 x^2}} \, dx+\left (3 b d^2 e\right ) \int \frac {\sin ^{-1}(c x)}{x} \, dx\\ &=-\frac {b c d^3 \sqrt {1-c^2 x^2}}{2 x}-\frac {d^3 \left (a+b \sin ^{-1}(c x)\right )}{2 x^2}+\frac {3}{2} d e^2 x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} e^3 x^4 \left (a+b \sin ^{-1}(c x)\right )-3 b d^2 e \sin ^{-1}(c x) \log (x)+3 d^2 e \left (a+b \sin ^{-1}(c x)\right ) \log (x)+\frac {1}{4} (b c) \int \frac {x^2 \left (-6 d e^2-e^3 x^2\right )}{\sqrt {1-c^2 x^2}} \, dx+\left (3 b d^2 e\right ) \operatorname {Subst}\left (\int x \cot (x) \, dx,x,\sin ^{-1}(c x)\right )\\ &=-\frac {b c d^3 \sqrt {1-c^2 x^2}}{2 x}+\frac {b e^3 x^3 \sqrt {1-c^2 x^2}}{16 c}-\frac {3}{2} i b d^2 e \sin ^{-1}(c x)^2-\frac {d^3 \left (a+b \sin ^{-1}(c x)\right )}{2 x^2}+\frac {3}{2} d e^2 x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} e^3 x^4 \left (a+b \sin ^{-1}(c x)\right )-3 b d^2 e \sin ^{-1}(c x) \log (x)+3 d^2 e \left (a+b \sin ^{-1}(c x)\right ) \log (x)-\left (6 i b d^2 e\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 (3 b e^2 \left (8 c^2 d+e\right )\right ) \int \frac {x^2}{\sqrt {1-c^2 x^2}} \, dx}{16 c}\\ &=-\frac {b c d^3 \sqrt {1-c^2 x^2}}{2 x}+\frac {3 b e^2 \left (8 c^2 d+e\right ) x \sqrt {1-c^2 x^2}}{32 c^3}+\frac {b e^3 x^3 \sqrt {1-c^2 x^2}}{16 c}-\frac {3}{2} i b d^2 e \sin ^{-1}(c x)^2-\frac {d^3 \left (a+b \sin ^{-1}(c x)\right )}{2 x^2}+\frac {3}{2} d e^2 x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} e^3 x^4 \left (a+b \sin ^{-1}(c x)\right )+3 b d^2 e \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-3 b d^2 e \sin ^{-1}(c x) \log (x)+3 d^2 e \left (a+b \sin ^{-1}(c x)\right ) \log (x)-\left (3 b d^2 e\right ) \operatorname {Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )-\frac {\left (3 b e^2 \left (8 c^2 d+e\right )\right ) \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx}{32 c^3}\\ &=-\frac {b c d^3 \sqrt {1-c^2 x^2}}{2 x}+\frac {3 b e^2 \left (8 c^2 d+e\right ) x \sqrt {1-c^2 x^2}}{32 c^3}+\frac {b e^3 x^3 \sqrt {1-c^2 x^2}}{16 c}-\frac {3 b e^2 \left (8 c^2 d+e\right ) \sin ^{-1}(c x)}{32 c^4}-\frac {3}{2} i b d^2 e \sin ^{-1}(c x)^2-\frac {d^3 \left (a+b \sin ^{-1}(c x)\right )}{2 x^2}+\frac {3}{2} d e^2 x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} e^3 x^4 \left (a+b \sin ^{-1}(c x)\right )+3 b d^2 e \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-3 b d^2 e \sin ^{-1}(c x) \log (x)+3 d^2 e \left (a+b \sin ^{-1}(c x)\right ) \log (x)+\frac {1}{2} \left (3 i b d^2 e\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )\\ &=-\frac {b c d^3 \sqrt {1-c^2 x^2}}{2 x}+\frac {3 b e^2 \left (8 c^2 d+e\right ) x \sqrt {1-c^2 x^2}}{32 c^3}+\frac {b e^3 x^3 \sqrt {1-c^2 x^2}}{16 c}-\frac {3 b e^2 \left (8 c^2 d+e\right ) \sin ^{-1}(c x)}{32 c^4}-\frac {3}{2} i b d^2 e \sin ^{-1}(c x)^2-\frac {d^3 \left (a+b \sin ^{-1}(c x)\right )}{2 x^2}+\frac {3}{2} d e^2 x^2 \left (a+b \sin ^{-1}(c x)\right )+\frac {1}{4} e^3 x^4 \left (a+b \sin ^{-1}(c x)\right )+3 b d^2 e \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-3 b d^2 e \sin ^{-1}(c x) \log (x)+3 d^2 e \left (a+b \sin ^{-1}(c x)\right ) \log (x)-\frac {3}{2} i b d^2 e \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.47, size = 220, normalized size = 0.84 \[ \frac {1}{32} \left (-\frac {16 a d^3}{x^2}+96 a d^2 e \log (x)+48 a d e^2 x^2+8 a e^3 x^4-\frac {16 b d^3 \left (c x \sqrt {1-c^2 x^2}+\sin ^{-1}(c x)\right )}{x^2}+\frac {24 b d e^2 \left (c x \sqrt {1-c^2 x^2}+\left (2 c^2 x^2-1\right ) \sin ^{-1}(c x)\right )}{c^2}+\frac {b e^3 \left (\left (8 c^4 x^4-3\right ) \sin ^{-1}(c x)+c x \sqrt {1-c^2 x^2} \left (2 c^2 x^2+3\right )\right )}{c^4}+96 b d^2 e \left (\sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-\frac {1}{2} i \left (\sin ^{-1}(c x)^2+\text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-16*a*d^3)/x^2 + 48*a*d*e^2*x^2 + 8*a*e^3*x^4 - (16*b*d^3*(c*x*Sqrt[1 - c^2*x^2] + ArcSin[c*x]))/x^2 + (24*b
*d*e^2*(c*x*Sqrt[1 - c^2*x^2] + (-1 + 2*c^2*x^2)*ArcSin[c*x]))/c^2 + (b*e^3*(c*x*Sqrt[1 - c^2*x^2]*(3 + 2*c^2*
x^2) + (-3 + 8*c^4*x^4)*ArcSin[c*x]))/c^4 + 96*a*d^2*e*Log[x] + 96*b*d^2*e*(ArcSin[c*x]*Log[1 - E^((2*I)*ArcSi
n[c*x])] - (I/2)*(ArcSin[c*x]^2 + PolyLog[2, E^((2*I)*ArcSin[c*x])])))/32

________________________________________________________________________________________

fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a e^{3} x^{6} + 3 \, a d e^{2} x^{4} + 3 \, a d^{2} e x^{2} + a d^{3} + {\left (b e^{3} x^{6} + 3 \, b d e^{2} x^{4} + 3 \, b d^{2} e x^{2} + b d^{3}\right )} \arcsin \left (c x\right )}{x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.73, size = 360, normalized size = 1.37 \[ \frac {a \,e^{3} x^{4}}{4}+\frac {3 a \,x^{2} d \,e^{2}}{2}+3 a \,d^{2} e \ln \left (c x \right )-\frac {a \,d^{3}}{2 x^{2}}+3 b \,d^{2} e \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )-\frac {b \sin \left (4 \arcsin \left (c x \right )\right ) e^{3}}{128 c^{4}}+\frac {b \cos \left (4 \arcsin \left (c x \right )\right ) \arcsin \left (c x \right ) e^{3}}{32 c^{4}}-3 i b \,d^{2} e \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )-3 i b \,d^{2} e \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )+\frac {3 b \sqrt {-c^{2} x^{2}+1}\, x d \,e^{2}}{4 c}+\frac {i c^{2} b \,d^{3}}{2}+3 b \,d^{2} e \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )-\frac {3 i b \,d^{2} e \arcsin \left (c x \right )^{2}}{2}-\frac {b c \,d^{3} \sqrt {-c^{2} x^{2}+1}}{2 x}+\frac {b \sqrt {-c^{2} x^{2}+1}\, x \,e^{3}}{8 c^{3}}-\frac {3 b \arcsin \left (c x \right ) d \,e^{2}}{4 c^{2}}-\frac {b \arcsin \left (c x \right ) e^{3}}{8 c^{4}}+\frac {b \arcsin \left (c x \right ) x^{2} e^{3}}{4 c^{2}}+\frac {3 b \arcsin \left (c x \right ) x^{2} d \,e^{2}}{2}-\frac {b \arcsin \left (c x \right ) d^{3}}{2 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*e^3*x^4+3/2*a*x^2*d*e^2+3*a*d^2*e*ln(c*x)-1/2*a*d^3/x^2+3*b*d^2*e*arcsin(c*x)*ln(1+I*c*x+(-c^2*x^2+1)^(1
/2))-1/128/c^4*b*sin(4*arcsin(c*x))*e^3+1/32/c^4*b*cos(4*arcsin(c*x))*arcsin(c*x)*e^3-3*I*b*d^2*e*polylog(2,-I
*c*x-(-c^2*x^2+1)^(1/2))+1/2*I*c^2*b*d^3+3/4/c*b*(-c^2*x^2+1)^(1/2)*x*d*e^2-3*I*b*d^2*e*polylog(2,I*c*x+(-c^2*
x^2+1)^(1/2))+3*b*d^2*e*arcsin(c*x)*ln(1-I*c*x-(-c^2*x^2+1)^(1/2))-3/2*I*b*d^2*e*arcsin(c*x)^2-1/2*b*c*d^3*(-c
^2*x^2+1)^(1/2)/x+1/8/c^3*b*(-c^2*x^2+1)^(1/2)*x*e^3-3/4/c^2*b*arcsin(c*x)*d*e^2-1/8/c^4*b*arcsin(c*x)*e^3+1/4
/c^2*b*arcsin(c*x)*x^2*e^3+3/2*b*arcsin(c*x)*x^2*d*e^2-1/2*b*arcsin(c*x)*d^3/x^2

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________