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

Optimal. Leaf size=178 \[ -\frac{1}{3} c^6 d^3 x^3 \left (a+b \sin ^{-1}(c x)\right )+3 c^4 d^3 x \left (a+b \sin ^{-1}(c x)\right )+\frac{3 c^2 d^3 \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac{d^3 \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}+\frac{1}{9} b c^3 d^3 \left (1-c^2 x^2\right )^{3/2}+\frac{8}{3} b c^3 d^3 \sqrt{1-c^2 x^2}-\frac{b c d^3 \sqrt{1-c^2 x^2}}{6 x^2}+\frac{17}{6} b c^3 d^3 \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right ) \]

[Out]

(8*b*c^3*d^3*Sqrt[1 - c^2*x^2])/3 - (b*c*d^3*Sqrt[1 - c^2*x^2])/(6*x^2) + (b*c^3*d^3*(1 - c^2*x^2)^(3/2))/9 -
(d^3*(a + b*ArcSin[c*x]))/(3*x^3) + (3*c^2*d^3*(a + b*ArcSin[c*x]))/x + 3*c^4*d^3*x*(a + b*ArcSin[c*x]) - (c^6
*d^3*x^3*(a + b*ArcSin[c*x]))/3 + (17*b*c^3*d^3*ArcTanh[Sqrt[1 - c^2*x^2]])/6

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Rubi [A]  time = 0.251276, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {270, 4687, 12, 1799, 1621, 897, 1153, 208} \[ -\frac{1}{3} c^6 d^3 x^3 \left (a+b \sin ^{-1}(c x)\right )+3 c^4 d^3 x \left (a+b \sin ^{-1}(c x)\right )+\frac{3 c^2 d^3 \left (a+b \sin ^{-1}(c x)\right )}{x}-\frac{d^3 \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}+\frac{1}{9} b c^3 d^3 \left (1-c^2 x^2\right )^{3/2}+\frac{8}{3} b c^3 d^3 \sqrt{1-c^2 x^2}-\frac{b c d^3 \sqrt{1-c^2 x^2}}{6 x^2}+\frac{17}{6} b c^3 d^3 \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*b*c^3*d^3*Sqrt[1 - c^2*x^2])/3 - (b*c*d^3*Sqrt[1 - c^2*x^2])/(6*x^2) + (b*c^3*d^3*(1 - c^2*x^2)^(3/2))/9 -
(d^3*(a + b*ArcSin[c*x]))/(3*x^3) + (3*c^2*d^3*(a + b*ArcSin[c*x]))/x + 3*c^4*d^3*x*(a + b*ArcSin[c*x]) - (c^6
*d^3*x^3*(a + b*ArcSin[c*x]))/3 + (17*b*c^3*d^3*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 4687

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u = I
ntHide[(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] && EqQ[c^2*d + e, 0] && IGtQ[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-c^2 d 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 c^2 d^3 \left (a+b \sin ^{-1}(c x)\right )}{x}+3 c^4 d^3 x \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{3} c^6 d^3 x^3 \left (a+b \sin ^{-1}(c x)\right )-(b c) \int \frac{d^3 \left (-1+9 c^2 x^2+9 c^4 x^4-c^6 x^6\right )}{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 c^2 d^3 \left (a+b \sin ^{-1}(c x)\right )}{x}+3 c^4 d^3 x \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{3} c^6 d^3 x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{3} \left (b c d^3\right ) \int \frac{-1+9 c^2 x^2+9 c^4 x^4-c^6 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 c^2 d^3 \left (a+b \sin ^{-1}(c x)\right )}{x}+3 c^4 d^3 x \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{3} c^6 d^3 x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{6} \left (b c d^3\right ) \operatorname{Subst}\left (\int \frac{-1+9 c^2 x+9 c^4 x^2-c^6 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 c^2 d^3 \left (a+b \sin ^{-1}(c x)\right )}{x}+3 c^4 d^3 x \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{3} c^6 d^3 x^3 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{6} \left (b c d^3\right ) \operatorname{Subst}\left (\int \frac{-\frac{17 c^2}{2}-9 c^4 x+c^6 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 c^2 d^3 \left (a+b \sin ^{-1}(c x)\right )}{x}+3 c^4 d^3 x \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{3} c^6 d^3 x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{\left (b d^3\right ) \operatorname{Subst}\left (\int \frac{-\frac{33 c^2}{2}+7 c^2 x^2+c^2 x^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 c^2 d^3 \left (a+b \sin ^{-1}(c x)\right )}{x}+3 c^4 d^3 x \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{3} c^6 d^3 x^3 \left (a+b \sin ^{-1}(c x)\right )-\frac{\left (b d^3\right ) \operatorname{Subst}\left (\int \left (-8 c^4-c^4 x^2-\frac{17 c^2}{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{8}{3} b c^3 d^3 \sqrt{1-c^2 x^2}-\frac{b c d^3 \sqrt{1-c^2 x^2}}{6 x^2}+\frac{1}{9} b c^3 d^3 \left (1-c^2 x^2\right )^{3/2}-\frac{d^3 \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}+\frac{3 c^2 d^3 \left (a+b \sin ^{-1}(c x)\right )}{x}+3 c^4 d^3 x \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{3} c^6 d^3 x^3 \left (a+b \sin ^{-1}(c x)\right )+\frac{1}{6} \left (17 b c d^3\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 )\\ &=\frac{8}{3} b c^3 d^3 \sqrt{1-c^2 x^2}-\frac{b c d^3 \sqrt{1-c^2 x^2}}{6 x^2}+\frac{1}{9} b c^3 d^3 \left (1-c^2 x^2\right )^{3/2}-\frac{d^3 \left (a+b \sin ^{-1}(c x)\right )}{3 x^3}+\frac{3 c^2 d^3 \left (a+b \sin ^{-1}(c x)\right )}{x}+3 c^4 d^3 x \left (a+b \sin ^{-1}(c x)\right )-\frac{1}{3} c^6 d^3 x^3 \left (a+b \sin ^{-1}(c x)\right )+\frac{17}{6} b c^3 d^3 \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )\\ \end{align*}

Mathematica [A]  time = 0.15279, size = 175, normalized size = 0.98 \[ -\frac{d^3 \left (6 a c^6 x^6-54 a c^4 x^4-54 a c^2 x^2+6 a+2 b c^5 x^5 \sqrt{1-c^2 x^2}-50 b c^3 x^3 \sqrt{1-c^2 x^2}+3 b c x \sqrt{1-c^2 x^2}+51 b c^3 x^3 \log (x)-51 b c^3 x^3 \log \left (\sqrt{1-c^2 x^2}+1\right )+6 b \left (c^6 x^6-9 c^4 x^4-9 c^2 x^2+1\right ) \sin ^{-1}(c x)\right )}{18 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d^3*(6*a - 54*a*c^2*x^2 - 54*a*c^4*x^4 + 6*a*c^6*x^6 + 3*b*c*x*Sqrt[1 - c^2*x^2] - 50*b*c^3*x^3*Sqrt[1 - c^2
*x^2] + 2*b*c^5*x^5*Sqrt[1 - c^2*x^2] + 6*b*(1 - 9*c^2*x^2 - 9*c^4*x^4 + c^6*x^6)*ArcSin[c*x] + 51*b*c^3*x^3*L
og[x] - 51*b*c^3*x^3*Log[1 + Sqrt[1 - c^2*x^2]]))/(18*x^3)

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Maple [A]  time = 0.01, size = 161, normalized size = 0.9 \begin{align*}{c}^{3} \left ( -{d}^{3}a \left ({\frac{{c}^{3}{x}^{3}}{3}}-3\,cx-3\,{\frac{1}{cx}}+{\frac{1}{3\,{c}^{3}{x}^{3}}} \right ) -{d}^{3}b \left ({\frac{{c}^{3}{x}^{3}\arcsin \left ( cx \right ) }{3}}-3\,cx\arcsin \left ( cx \right ) -3\,{\frac{\arcsin \left ( cx \right ) }{cx}}+{\frac{\arcsin \left ( cx \right ) }{3\,{c}^{3}{x}^{3}}}+{\frac{{c}^{2}{x}^{2}}{9}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{25}{9}\sqrt{-{c}^{2}{x}^{2}+1}}-{\frac{17}{6}{\it Artanh} \left ({\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}} \right ) }+{\frac{1}{6\,{c}^{2}{x}^{2}}\sqrt{-{c}^{2}{x}^{2}+1}} \right ) \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^3*(-d^3*a*(1/3*c^3*x^3-3*c*x-3/c/x+1/3/c^3/x^3)-d^3*b*(1/3*c^3*x^3*arcsin(c*x)-3*c*x*arcsin(c*x)-3/c/x*arcsi
n(c*x)+1/3/c^3/x^3*arcsin(c*x)+1/9*c^2*x^2*(-c^2*x^2+1)^(1/2)-25/9*(-c^2*x^2+1)^(1/2)-17/6*arctanh(1/(-c^2*x^2
+1)^(1/2))+1/6/c^2/x^2*(-c^2*x^2+1)^(1/2)))

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Maxima [A]  time = 1.56809, size = 327, normalized size = 1.84 \begin{align*} -\frac{1}{3} \, a c^{6} d^{3} x^{3} - \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 c^{6} d^{3} + 3 \, a c^{4} d^{3} x + 3 \,{\left (c x \arcsin \left (c x\right ) + \sqrt{-c^{2} x^{2} + 1}\right )} b c^{3} 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 c^{2} d^{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} + \frac{3 \, a c^{2} d^{3}}{x} - \frac{a d^{3}}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.80443, size = 440, normalized size = 2.47 \begin{align*} -\frac{12 \, a c^{6} d^{3} x^{6} - 108 \, a c^{4} d^{3} x^{4} - 51 \, b c^{3} d^{3} x^{3} \log \left (\sqrt{-c^{2} x^{2} + 1} + 1\right ) + 51 \, b c^{3} d^{3} x^{3} \log \left (\sqrt{-c^{2} x^{2} + 1} - 1\right ) - 108 \, a c^{2} d^{3} x^{2} + 12 \, a d^{3} + 12 \,{\left (b c^{6} d^{3} x^{6} - 9 \, b c^{4} d^{3} x^{4} - 9 \, b c^{2} d^{3} x^{2} + b d^{3}\right )} \arcsin \left (c x\right ) + 2 \,{\left (2 \, b c^{5} d^{3} x^{5} - 50 \, b c^{3} d^{3} x^{3} + 3 \, b c d^{3} x\right )} \sqrt{-c^{2} x^{2} + 1}}{36 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/36*(12*a*c^6*d^3*x^6 - 108*a*c^4*d^3*x^4 - 51*b*c^3*d^3*x^3*log(sqrt(-c^2*x^2 + 1) + 1) + 51*b*c^3*d^3*x^3*
log(sqrt(-c^2*x^2 + 1) - 1) - 108*a*c^2*d^3*x^2 + 12*a*d^3 + 12*(b*c^6*d^3*x^6 - 9*b*c^4*d^3*x^4 - 9*b*c^2*d^3
*x^2 + b*d^3)*arcsin(c*x) + 2*(2*b*c^5*d^3*x^5 - 50*b*c^3*d^3*x^3 + 3*b*c*d^3*x)*sqrt(-c^2*x^2 + 1))/x^3

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Sympy [A]  time = 17.5578, size = 326, normalized size = 1.83 \begin{align*} - \frac{a c^{6} d^{3} x^{3}}{3} + 3 a c^{4} d^{3} x + \frac{3 a c^{2} d^{3}}{x} - \frac{a d^{3}}{3 x^{3}} + \frac{b c^{7} d^{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 c^{6} d^{3} x^{3} \operatorname{asin}{\left (c x \right )}}{3} + 3 b c^{4} d^{3} \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 ) - 3 b c^{3} d^{3} \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{3 b c^{2} d^{3} \operatorname{asin}{\left (c x \right )}}{x} + \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} - \frac{b d^{3} \operatorname{asin}{\left (c x \right )}}{3 x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*c**6*d**3*x**3/3 + 3*a*c**4*d**3*x + 3*a*c**2*d**3/x - a*d**3/(3*x**3) + b*c**7*d**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*c**6*d**3*x**3*as
in(c*x)/3 + 3*b*c**4*d**3*Piecewise((0, Eq(c, 0)), (x*asin(c*x) + sqrt(-c**2*x**2 + 1)/c, True)) - 3*b*c**3*d*
*3*Piecewise((-acosh(1/(c*x)), 1/Abs(c**2*x**2) > 1), (I*asin(1/(c*x)), True)) + 3*b*c**2*d**3*asin(c*x)/x + 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*a
sin(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 - b*d**3*a
sin(c*x)/(3*x**3)

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out