3.622 \(\int \frac{(d+e x^2)^3 (a+b \sin ^{-1}(c x))}{x^4} \, dx\)

Optimal. Leaf size=186 \[ -\frac{3 d^2 e \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac{d^3 \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}+3 d e^2 x \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{3} e^3 x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{6} b c d^2 \left (c^2 d+18 e\right ) \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )-\frac{b c d^3 \sqrt{1-c^2 x^2}}{6 x^2}+\frac{b e^2 \sqrt{1-c^2 x^2} \left (9 c^2 d+e\right )}{3 c^3}-\frac{b e^3 \left (1-c^2 x^2\right )^{3/2}}{9 c^3} \]

[Out]

(b*e^2*(9*c^2*d + e)*Sqrt[1 - c^2*x^2])/(3*c^3) - (b*c*d^3*Sqrt[1 - c^2*x^2])/(6*x^2) - (b*e^3*(1 - c^2*x^2)^(
3/2))/(9*c^3) - (d^3*(a + b*ArcSin[c*x]))/(3*x^3) - (3*d^2*e*(a + b*ArcSin[c*x]))/x + 3*d*e^2*x*(a + b*ArcSin[
c*x]) + (e^3*x^3*(a + b*ArcSin[c*x]))/3 - (b*c*d^2*(c^2*d + 18*e)*ArcTanh[Sqrt[1 - c^2*x^2]])/6

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Rubi [A]  time = 0.315391, antiderivative size = 186, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {270, 4731, 12, 1799, 1621, 897, 1153, 208} \[ -\frac{3 d^2 e \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac{d^3 \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}+3 d e^2 x \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{3} e^3 x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{6} b c d^2 \left (c^2 d+18 e\right ) \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )-\frac{b c d^3 \sqrt{1-c^2 x^2}}{6 x^2}+\frac{b e^2 \sqrt{1-c^2 x^2} \left (9 c^2 d+e\right )}{3 c^3}-\frac{b e^3 \left (1-c^2 x^2\right )^{3/2}}{9 c^3} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)^3*(a + b*ArcSin[c*x]))/x^4,x]

[Out]

(b*e^2*(9*c^2*d + e)*Sqrt[1 - c^2*x^2])/(3*c^3) - (b*c*d^3*Sqrt[1 - c^2*x^2])/(6*x^2) - (b*e^3*(1 - c^2*x^2)^(
3/2))/(9*c^3) - (d^3*(a + b*ArcSin[c*x]))/(3*x^3) - (3*d^2*e*(a + b*ArcSin[c*x]))/x + 3*d*e^2*x*(a + b*ArcSin[
c*x]) + (e^3*x^3*(a + b*ArcSin[c*x]))/3 - (b*c*d^2*(c^2*d + 18*e)*ArcTanh[Sqrt[1 - c^2*x^2]])/6

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 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 12

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

Rule 1799

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*SubstFor[x^2,
Pq, x]*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x^2] && IntegerQ[(m - 1)/2]

Rule 1621

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> With[{Qx = PolynomialQuotient[Px,
a + b*x, x], R = PolynomialRemainder[Px, a + b*x, x]}, Simp[(R*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/((m + 1)*
(b*c - a*d)), x] + Dist[1/((m + 1)*(b*c - a*d)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*ExpandToSum[(m + 1)*(b*c -
a*d)*Qx - d*R*(m + n + 2), x], x], x]] /; FreeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && ILtQ[m, -1] && GtQ[Expo
n[Px, x], 2]

Rule 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1153

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{\left (d+e x^2\right )^3 \left (a+b \sin ^{-1}(c x)\right )}{x^4} \, dx &=-\frac{d^3 \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}-\frac{3 d^2 e \left (a+b \sin ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{3} e^3 x^3 \left (a+b \sin ^{-1}(c x)\right )-(b c) \int \frac{-d^3-9 d^2 e x^2+9 d e^2 x^4+e^3 x^6}{3 x^3 \sqrt{1-c^2 x^2}} \, dx\\ &=-\frac{d^3 \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}-\frac{3 d^2 e \left (a+b \sin ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{3} e^3 x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{3} (b c) \int \frac{-d^3-9 d^2 e x^2+9 d e^2 x^4+e^3 x^6}{x^3 \sqrt{1-c^2 x^2}} \, dx\\ &=-\frac{d^3 \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}-\frac{3 d^2 e \left (a+b \sin ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{3} e^3 x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{6} (b c) \operatorname{Subst}\left (\int \frac{-d^3-9 d^2 e x+9 d e^2 x^2+e^3 x^3}{x^2 \sqrt{1-c^2 x}} \, dx,x,x^2\right )\\ &=-\frac{b c d^3 \sqrt{1-c^2 x^2}}{6 x^2}-\frac{d^3 \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}-\frac{3 d^2 e \left (a+b \sin ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{3} e^3 x^3 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{6} (b c) \operatorname{Subst}\left (\int \frac{\frac{1}{2} d^2 \left (c^2 d+18 e\right )-9 d e^2 x-e^3 x^2}{x \sqrt{1-c^2 x}} \, dx,x,x^2\right )\\ &=-\frac{b c d^3 \sqrt{1-c^2 x^2}}{6 x^2}-\frac{d^3 \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}-\frac{3 d^2 e \left (a+b \sin ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{3} e^3 x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{b \operatorname{Subst}\left (\int \frac{\frac{-9 c^2 d e^2-e^3+\frac{1}{2} c^4 d^2 \left (c^2 d+18 e\right )}{c^4}-\frac{\left (-9 c^2 d e^2-2 e^3\right ) x^2}{c^4}-\frac{e^3 x^4}{c^4}}{\frac{1}{c^2}-\frac{x^2}{c^2}} \, dx,x,\sqrt{1-c^2 x^2}\right )}{3 c}\\ &=-\frac{b c d^3 \sqrt{1-c^2 x^2}}{6 x^2}-\frac{d^3 \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}-\frac{3 d^2 e \left (a+b \sin ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{3} e^3 x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{b \operatorname{Subst}\left (\int \left (-e^2 \left (9 d+\frac{e}{c^2}\right )+\frac{e^3 x^2}{c^2}+\frac{c^2 d^3+18 d^2 e}{2 \left (\frac{1}{c^2}-\frac{x^2}{c^2}\right )}\right ) \, dx,x,\sqrt{1-c^2 x^2}\right )}{3 c}\\ &=\frac{b e^2 \left (9 c^2 d+e\right ) \sqrt{1-c^2 x^2}}{3 c^3}-\frac{b c d^3 \sqrt{1-c^2 x^2}}{6 x^2}-\frac{b e^3 \left (1-c^2 x^2\right )^{3/2}}{9 c^3}-\frac{d^3 \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}-\frac{3 d^2 e \left (a+b \sin ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{3} e^3 x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{\left (b d^2 \left (c^2 d+18 e\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{c^2}-\frac{x^2}{c^2}} \, dx,x,\sqrt{1-c^2 x^2}\right )}{6 c}\\ &=\frac{b e^2 \left (9 c^2 d+e\right ) \sqrt{1-c^2 x^2}}{3 c^3}-\frac{b c d^3 \sqrt{1-c^2 x^2}}{6 x^2}-\frac{b e^3 \left (1-c^2 x^2\right )^{3/2}}{9 c^3}-\frac{d^3 \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}-\frac{3 d^2 e \left (a+b \sin ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{3} e^3 x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{6} b c d^2 \left (c^2 d+18 e\right ) \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.251719, size = 194, normalized size = 1.04 \[ \frac{1}{6} \left (-\frac{18 a d^2 e}{x}-\frac{2 a d^3}{x^3}+18 a d e^2 x+2 a e^3 x^3+\frac{b \sqrt{1-c^2 x^2} \left (-3 c^4 d^3+2 c^2 e^2 x^2 \left (27 d+e x^2\right )+4 e^3 x^2\right )}{3 c^3 x^2}-b c d^2 \left (c^2 d+18 e\right ) \log \left (\sqrt{1-c^2 x^2}+1\right )+b c d^2 \log (x) \left (c^2 d+18 e\right )+\frac{2 b \sin ^{-1}(c x) \left (-9 d^2 e x^2-d^3+9 d e^2 x^4+e^3 x^6\right )}{x^3}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)^3*(a + b*ArcSin[c*x]))/x^4,x]

[Out]

((-2*a*d^3)/x^3 - (18*a*d^2*e)/x + 18*a*d*e^2*x + 2*a*e^3*x^3 + (b*Sqrt[1 - c^2*x^2]*(-3*c^4*d^3 + 4*e^3*x^2 +
2*c^2*e^2*x^2*(27*d + e*x^2)))/(3*c^3*x^2) + (2*b*(-d^3 - 9*d^2*e*x^2 + 9*d*e^2*x^4 + e^3*x^6)*ArcSin[c*x])/x
^3 + b*c*d^2*(c^2*d + 18*e)*Log[x] - b*c*d^2*(c^2*d + 18*e)*Log[1 + Sqrt[1 - c^2*x^2]])/6

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Maple [A]  time = 0.01, size = 249, normalized size = 1.3 \begin{align*}{c}^{3} \left ({\frac{a}{{c}^{6}} \left ({\frac{{e}^{3}{c}^{3}{x}^{3}}{3}}+3\,{c}^{3}xd{e}^{2}-3\,{\frac{{c}^{3}{d}^{2}e}{x}}-{\frac{{c}^{3}{d}^{3}}{3\,{x}^{3}}} \right ) }+{\frac{b}{{c}^{6}} \left ({\frac{\arcsin \left ( cx \right ){e}^{3}{c}^{3}{x}^{3}}{3}}+3\,\arcsin \left ( cx \right ){c}^{3}xd{e}^{2}-3\,{\frac{\arcsin \left ( cx \right ){c}^{3}{d}^{2}e}{x}}-{\frac{\arcsin \left ( cx \right ){d}^{3}{c}^{3}}{3\,{x}^{3}}}-{\frac{{e}^{3}}{3} \left ( -{\frac{{c}^{2}{x}^{2}}{3}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{2}{3}\sqrt{-{c}^{2}{x}^{2}+1}} \right ) }+3\,{c}^{2}d{e}^{2}\sqrt{-{c}^{2}{x}^{2}+1}-3\,{c}^{4}{d}^{2}e{\it Artanh} \left ({\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}} \right ) +{\frac{{d}^{3}{c}^{6}}{3} \left ( -{\frac{1}{2\,{c}^{2}{x}^{2}}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{1}{2}{\it Artanh} \left ({\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}} \right ) } \right ) } \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^3*(a/c^6*(1/3*e^3*c^3*x^3+3*c^3*x*d*e^2-3*c^3*d^2*e/x-1/3*d^3*c^3/x^3)+b/c^6*(1/3*arcsin(c*x)*e^3*c^3*x^3+3*
arcsin(c*x)*c^3*x*d*e^2-3*arcsin(c*x)*c^3*d^2*e/x-1/3*arcsin(c*x)*d^3*c^3/x^3-1/3*e^3*(-1/3*c^2*x^2*(-c^2*x^2+
1)^(1/2)-2/3*(-c^2*x^2+1)^(1/2))+3*c^2*d*e^2*(-c^2*x^2+1)^(1/2)-3*c^4*d^2*e*arctanh(1/(-c^2*x^2+1)^(1/2))+1/3*
d^3*c^6*(-1/2/c^2/x^2*(-c^2*x^2+1)^(1/2)-1/2*arctanh(1/(-c^2*x^2+1)^(1/2)))))

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Maxima [A]  time = 1.47079, size = 312, normalized size = 1.68 \begin{align*} \frac{1}{3} \, a e^{3} x^{3} - \frac{1}{6} \,{\left ({\left (c^{2} \log \left (\frac{2 \, \sqrt{-c^{2} x^{2} + 1}}{{\left | x \right |}} + \frac{2}{{\left | x \right |}}\right ) + \frac{\sqrt{-c^{2} x^{2} + 1}}{x^{2}}\right )} c + \frac{2 \, \arcsin \left (c x\right )}{x^{3}}\right )} b d^{3} - 3 \,{\left (c \log \left (\frac{2 \, \sqrt{-c^{2} x^{2} + 1}}{{\left | x \right |}} + \frac{2}{{\left | x \right |}}\right ) + \frac{\arcsin \left (c x\right )}{x}\right )} b d^{2} e + \frac{1}{9} \,{\left (3 \, x^{3} \arcsin \left (c x\right ) + c{\left (\frac{\sqrt{-c^{2} x^{2} + 1} x^{2}}{c^{2}} + \frac{2 \, \sqrt{-c^{2} x^{2} + 1}}{c^{4}}\right )}\right )} b e^{3} + 3 \, a d e^{2} x + \frac{3 \,{\left (c x \arcsin \left (c x\right ) + \sqrt{-c^{2} x^{2} + 1}\right )} b d e^{2}}{c} - \frac{3 \, a d^{2} e}{x} - \frac{a d^{3}}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^3*(a+b*arcsin(c*x))/x^4,x, algorithm="maxima")

[Out]

1/3*a*e^3*x^3 - 1/6*((c^2*log(2*sqrt(-c^2*x^2 + 1)/abs(x) + 2/abs(x)) + sqrt(-c^2*x^2 + 1)/x^2)*c + 2*arcsin(c
*x)/x^3)*b*d^3 - 3*(c*log(2*sqrt(-c^2*x^2 + 1)/abs(x) + 2/abs(x)) + arcsin(c*x)/x)*b*d^2*e + 1/9*(3*x^3*arcsin
(c*x) + c*(sqrt(-c^2*x^2 + 1)*x^2/c^2 + 2*sqrt(-c^2*x^2 + 1)/c^4))*b*e^3 + 3*a*d*e^2*x + 3*(c*x*arcsin(c*x) +
sqrt(-c^2*x^2 + 1))*b*d*e^2/c - 3*a*d^2*e/x - 1/3*a*d^3/x^3

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Fricas [A]  time = 4.12765, size = 541, normalized size = 2.91 \begin{align*} \frac{12 \, a c^{3} e^{3} x^{6} + 108 \, a c^{3} d e^{2} x^{4} - 108 \, a c^{3} d^{2} e x^{2} - 12 \, a c^{3} d^{3} - 3 \,{\left (b c^{6} d^{3} + 18 \, b c^{4} d^{2} e\right )} x^{3} \log \left (\sqrt{-c^{2} x^{2} + 1} + 1\right ) + 3 \,{\left (b c^{6} d^{3} + 18 \, b c^{4} d^{2} e\right )} x^{3} \log \left (\sqrt{-c^{2} x^{2} + 1} - 1\right ) + 12 \,{\left (b c^{3} e^{3} x^{6} + 9 \, b c^{3} d e^{2} x^{4} - 9 \, b c^{3} d^{2} e x^{2} - b c^{3} d^{3}\right )} \arcsin \left (c x\right ) + 2 \,{\left (2 \, b c^{2} e^{3} x^{5} - 3 \, b c^{4} d^{3} x + 2 \,{\left (27 \, b c^{2} d e^{2} + 2 \, b e^{3}\right )} x^{3}\right )} \sqrt{-c^{2} x^{2} + 1}}{36 \, c^{3} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^3*(a+b*arcsin(c*x))/x^4,x, algorithm="fricas")

[Out]

1/36*(12*a*c^3*e^3*x^6 + 108*a*c^3*d*e^2*x^4 - 108*a*c^3*d^2*e*x^2 - 12*a*c^3*d^3 - 3*(b*c^6*d^3 + 18*b*c^4*d^
2*e)*x^3*log(sqrt(-c^2*x^2 + 1) + 1) + 3*(b*c^6*d^3 + 18*b*c^4*d^2*e)*x^3*log(sqrt(-c^2*x^2 + 1) - 1) + 12*(b*
c^3*e^3*x^6 + 9*b*c^3*d*e^2*x^4 - 9*b*c^3*d^2*e*x^2 - b*c^3*d^3)*arcsin(c*x) + 2*(2*b*c^2*e^3*x^5 - 3*b*c^4*d^
3*x + 2*(27*b*c^2*d*e^2 + 2*b*e^3)*x^3)*sqrt(-c^2*x^2 + 1))/(c^3*x^3)

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Sympy [A]  time = 10.3274, size = 311, normalized size = 1.67 \begin{align*} - \frac{a d^{3}}{3 x^{3}} - \frac{3 a d^{2} e}{x} + 3 a d e^{2} x + \frac{a e^{3} x^{3}}{3} + \frac{b c d^{3} \left (\begin{cases} - \frac{c^{2} \operatorname{acosh}{\left (\frac{1}{c x} \right )}}{2} - \frac{c \sqrt{-1 + \frac{1}{c^{2} x^{2}}}}{2 x} & \text{for}\: \frac{1}{\left |{c^{2} x^{2}}\right |} > 1 \\\frac{i c^{2} \operatorname{asin}{\left (\frac{1}{c x} \right )}}{2} - \frac{i c}{2 x \sqrt{1 - \frac{1}{c^{2} x^{2}}}} + \frac{i}{2 c x^{3} \sqrt{1 - \frac{1}{c^{2} x^{2}}}} & \text{otherwise} \end{cases}\right )}{3} + 3 b c d^{2} e \left (\begin{cases} - \operatorname{acosh}{\left (\frac{1}{c x} \right )} & \text{for}\: \frac{1}{\left |{c^{2} x^{2}}\right |} > 1 \\i \operatorname{asin}{\left (\frac{1}{c x} \right )} & \text{otherwise} \end{cases}\right ) - \frac{b c e^{3} \left (\begin{cases} - \frac{x^{2} \sqrt{- c^{2} x^{2} + 1}}{3 c^{2}} - \frac{2 \sqrt{- c^{2} x^{2} + 1}}{3 c^{4}} & \text{for}\: c \neq 0 \\\frac{x^{4}}{4} & \text{otherwise} \end{cases}\right )}{3} - \frac{b d^{3} \operatorname{asin}{\left (c x \right )}}{3 x^{3}} - \frac{3 b d^{2} e \operatorname{asin}{\left (c x \right )}}{x} + 3 b d e^{2} \left (\begin{cases} 0 & \text{for}\: c = 0 \\x \operatorname{asin}{\left (c x \right )} + \frac{\sqrt{- c^{2} x^{2} + 1}}{c} & \text{otherwise} \end{cases}\right ) + \frac{b e^{3} x^{3} \operatorname{asin}{\left (c x \right )}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)**3*(a+b*asin(c*x))/x**4,x)

[Out]

-a*d**3/(3*x**3) - 3*a*d**2*e/x + 3*a*d*e**2*x + a*e**3*x**3/3 + b*c*d**3*Piecewise((-c**2*acosh(1/(c*x))/2 -
c*sqrt(-1 + 1/(c**2*x**2))/(2*x), 1/Abs(c**2*x**2) > 1), (I*c**2*asin(1/(c*x))/2 - I*c/(2*x*sqrt(1 - 1/(c**2*x
**2))) + I/(2*c*x**3*sqrt(1 - 1/(c**2*x**2))), True))/3 + 3*b*c*d**2*e*Piecewise((-acosh(1/(c*x)), 1/Abs(c**2*
x**2) > 1), (I*asin(1/(c*x)), True)) - b*c*e**3*Piecewise((-x**2*sqrt(-c**2*x**2 + 1)/(3*c**2) - 2*sqrt(-c**2*
x**2 + 1)/(3*c**4), Ne(c, 0)), (x**4/4, True))/3 - b*d**3*asin(c*x)/(3*x**3) - 3*b*d**2*e*asin(c*x)/x + 3*b*d*
e**2*Piecewise((0, Eq(c, 0)), (x*asin(c*x) + sqrt(-c**2*x**2 + 1)/c, True)) + b*e**3*x**3*asin(c*x)/3

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^3*(a+b*arcsin(c*x))/x^4,x, algorithm="giac")

[Out]

sage0*x