3.3.50 \(\int \frac {(a+b \sinh ^{-1}(c x))^2}{x^3 (d+c^2 d x^2)^3} \, dx\) [250]

Optimal. Leaf size=381 \[ \frac {b^2 c^2}{12 d^3 \left (1+c^2 x^2\right )}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{d^3 x \left (1+c^2 x^2\right )^{3/2}}-\frac {5 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {4 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \sqrt {1+c^2 x^2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1+c^2 x^2\right )^2}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 \left (1+c^2 x^2\right )}+\frac {6 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}+\frac {b^2 c^2 \log (x)}{d^3}-\frac {7 b^2 c^2 \log \left (1+c^2 x^2\right )}{6 d^3}+\frac {3 b c^2 \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{d^3}-\frac {3 b c^2 \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )}{d^3}-\frac {3 b^2 c^2 \text {PolyLog}\left (3,-e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}+\frac {3 b^2 c^2 \text {PolyLog}\left (3,e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3} \]

[Out]

1/12*b^2*c^2/d^3/(c^2*x^2+1)-b*c*(a+b*arcsinh(c*x))/d^3/x/(c^2*x^2+1)^(3/2)-5/6*b*c^3*x*(a+b*arcsinh(c*x))/d^3
/(c^2*x^2+1)^(3/2)-3/4*c^2*(a+b*arcsinh(c*x))^2/d^3/(c^2*x^2+1)^2-1/2*(a+b*arcsinh(c*x))^2/d^3/x^2/(c^2*x^2+1)
^2-3/2*c^2*(a+b*arcsinh(c*x))^2/d^3/(c^2*x^2+1)+6*c^2*(a+b*arcsinh(c*x))^2*arctanh((c*x+(c^2*x^2+1)^(1/2))^2)/
d^3+b^2*c^2*ln(x)/d^3-7/6*b^2*c^2*ln(c^2*x^2+1)/d^3+3*b*c^2*(a+b*arcsinh(c*x))*polylog(2,-(c*x+(c^2*x^2+1)^(1/
2))^2)/d^3-3*b*c^2*(a+b*arcsinh(c*x))*polylog(2,(c*x+(c^2*x^2+1)^(1/2))^2)/d^3-3/2*b^2*c^2*polylog(3,-(c*x+(c^
2*x^2+1)^(1/2))^2)/d^3+3/2*b^2*c^2*polylog(3,(c*x+(c^2*x^2+1)^(1/2))^2)/d^3+4/3*b*c^3*x*(a+b*arcsinh(c*x))/d^3
/(c^2*x^2+1)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.55, antiderivative size = 381, normalized size of antiderivative = 1.00, number of steps used = 23, number of rules used = 19, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.731, Rules used = {5809, 5811, 5799, 5569, 4267, 2611, 2320, 6724, 5787, 266, 5788, 267, 277, 198, 197, 5804, 12, 1265, 907} \begin {gather*} \frac {3 b c^2 \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^3}-\frac {3 b c^2 \text {Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^3}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 \left (c^2 x^2+1\right )}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (c^2 x^2+1\right )^2}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{d^3 x \left (c^2 x^2+1\right )^{3/2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (c^2 x^2+1\right )^2}+\frac {6 c^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{d^3}+\frac {4 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \sqrt {c^2 x^2+1}}-\frac {5 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (c^2 x^2+1\right )^{3/2}}-\frac {3 b^2 c^2 \text {Li}_3\left (-e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}+\frac {3 b^2 c^2 \text {Li}_3\left (e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}+\frac {b^2 c^2}{12 d^3 \left (c^2 x^2+1\right )}-\frac {7 b^2 c^2 \log \left (c^2 x^2+1\right )}{6 d^3}+\frac {b^2 c^2 \log (x)}{d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*c^2)/(12*d^3*(1 + c^2*x^2)) - (b*c*(a + b*ArcSinh[c*x]))/(d^3*x*(1 + c^2*x^2)^(3/2)) - (5*b*c^3*x*(a + b*
ArcSinh[c*x]))/(6*d^3*(1 + c^2*x^2)^(3/2)) + (4*b*c^3*x*(a + b*ArcSinh[c*x]))/(3*d^3*Sqrt[1 + c^2*x^2]) - (3*c
^2*(a + b*ArcSinh[c*x])^2)/(4*d^3*(1 + c^2*x^2)^2) - (a + b*ArcSinh[c*x])^2/(2*d^3*x^2*(1 + c^2*x^2)^2) - (3*c
^2*(a + b*ArcSinh[c*x])^2)/(2*d^3*(1 + c^2*x^2)) + (6*c^2*(a + b*ArcSinh[c*x])^2*ArcTanh[E^(2*ArcSinh[c*x])])/
d^3 + (b^2*c^2*Log[x])/d^3 - (7*b^2*c^2*Log[1 + c^2*x^2])/(6*d^3) + (3*b*c^2*(a + b*ArcSinh[c*x])*PolyLog[2, -
E^(2*ArcSinh[c*x])])/d^3 - (3*b*c^2*(a + b*ArcSinh[c*x])*PolyLog[2, E^(2*ArcSinh[c*x])])/d^3 - (3*b^2*c^2*Poly
Log[3, -E^(2*ArcSinh[c*x])])/(2*d^3) + (3*b^2*c^2*PolyLog[3, E^(2*ArcSinh[c*x])])/(2*d^3)

Rule 12

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

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 198

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p +
1], 0] && NeQ[p, -1]

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 267

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

Rule 277

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

Rule 907

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], 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] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 1265

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
IntegerQ[(m - 1)/2]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 4267

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(Ar
cTanh[E^((-I)*e + f*fz*x)]/(f*fz*I)), x] + (-Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*Log[1 - E^((-I)*e + f*
fz*x)], x], x] + Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*Log[1 + E^((-I)*e + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, x] && IGtQ[m, 0]

Rule 5569

Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dis
t[2^n, Int[(c + d*x)^m*Csch[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d}, x] && RationalQ[m] && IntegerQ[n]

Rule 5787

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[x*((a + b*ArcSinh
[c*x])^n/(d*Sqrt[d + e*x^2])), x] - Dist[b*c*(n/d)*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]], Int[x*((a + b*ArcS
inh[c*x])^(n - 1)/(1 + c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0]

Rule 5788

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*(d + e*x^2)^(
p + 1)*((a + b*ArcSinh[c*x])^n/(2*d*(p + 1))), x] + (Dist[(2*p + 3)/(2*d*(p + 1)), Int[(d + e*x^2)^(p + 1)*(a
+ b*ArcSinh[c*x])^n, x], x] + Dist[b*c*(n/(2*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p], Int[x*(1 + c^2*x^2
)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0] &
& LtQ[p, -1] && NeQ[p, -3/2]

Rule 5799

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Dist[1/d, Subst[Int[(
a + b*x)^n/(Cosh[x]*Sinh[x]), x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[n
, 0]

Rule 5804

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[x
^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSinh[c*x], u, x] - Dist[b*c*Simp[Sqrt[d + e*x^2]/Sqrt[1 + c^2*x^2]], Int[
SimplifyIntegrand[u/Sqrt[d + e*x^2], x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IntegerQ[p -
1/2] && NeQ[p, -2^(-1)] && (IGtQ[(m + 1)/2, 0] || ILtQ[(m + 2*p + 3)/2, 0])

Rule 5809

Int[((a_.) + ArcSinh[(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*ArcSinh[c*x])^n/(d*f*(m + 1))), x] + (-Dist[c^2*((m + 2*p + 3)/(f^2*
(m + 1))), Int[(f*x)^(m + 2)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] - Dist[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*ArcSinh[c*x])^(n - 1), x], x]) /;
FreeQ[{a, b, c, d, e, f, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && ILtQ[m, -1]

Rule 5811

Int[((a_.) + ArcSinh[(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*ArcSinh[c*x])^n/(2*d*f*(p + 1))), x] + (Dist[(m + 2*p + 3)/(2*d*(
p + 1)), Int[(f*x)^m*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n, x], x] + Dist[b*c*(n/(2*f*(p + 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*ArcSinh[c*x])^(n - 1), x], x]) /;
FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] &&  !GtQ[m, 1] && (IntegerQ[m] ||
IntegerQ[p] || EqQ[n, 1])

Rule 6724

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

Rubi steps

\begin {align*} \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{x^3 \left (d+c^2 d x^2\right )^3} \, dx &=-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1+c^2 x^2\right )^2}-\left (3 c^2\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{x \left (d+c^2 d x^2\right )^3} \, dx+\frac {(b c) \int \frac {a+b \sinh ^{-1}(c x)}{x^2 \left (1+c^2 x^2\right )^{5/2}} \, dx}{d^3}\\ &=-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{d^3 x \left (1+c^2 x^2\right )^{3/2}}-\frac {4 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {8 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \sqrt {1+c^2 x^2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1+c^2 x^2\right )^2}-\frac {\left (b^2 c^2\right ) \int \frac {-3-12 c^2 x^2-8 c^4 x^4}{3 x \left (1+c^2 x^2\right )^2} \, dx}{d^3}+\frac {\left (3 b c^3\right ) \int \frac {a+b \sinh ^{-1}(c x)}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{2 d^3}-\frac {\left (3 c^2\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{x \left (d+c^2 d x^2\right )^2} \, dx}{d}\\ &=-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{d^3 x \left (1+c^2 x^2\right )^{3/2}}-\frac {5 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {8 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \sqrt {1+c^2 x^2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1+c^2 x^2\right )^2}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 \left (1+c^2 x^2\right )}-\frac {\left (b^2 c^2\right ) \int \frac {-3-12 c^2 x^2-8 c^4 x^4}{x \left (1+c^2 x^2\right )^2} \, dx}{3 d^3}+\frac {\left (b c^3\right ) \int \frac {a+b \sinh ^{-1}(c x)}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{d^3}+\frac {\left (3 b c^3\right ) \int \frac {a+b \sinh ^{-1}(c x)}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{d^3}-\frac {\left (b^2 c^4\right ) \int \frac {x}{\left (1+c^2 x^2\right )^2} \, dx}{2 d^3}-\frac {\left (3 c^2\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{x \left (d+c^2 d x^2\right )} \, dx}{d^2}\\ &=\frac {b^2 c^2}{4 d^3 \left (1+c^2 x^2\right )}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{d^3 x \left (1+c^2 x^2\right )^{3/2}}-\frac {5 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {4 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \sqrt {1+c^2 x^2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1+c^2 x^2\right )^2}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 \left (1+c^2 x^2\right )}-\frac {\left (3 c^2\right ) \text {Subst}\left (\int (a+b x)^2 \text {csch}(x) \text {sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}-\frac {\left (b^2 c^2\right ) \text {Subst}\left (\int \frac {-3-12 c^2 x-8 c^4 x^2}{x \left (1+c^2 x\right )^2} \, dx,x,x^2\right )}{6 d^3}-\frac {\left (b^2 c^4\right ) \int \frac {x}{1+c^2 x^2} \, dx}{d^3}-\frac {\left (3 b^2 c^4\right ) \int \frac {x}{1+c^2 x^2} \, dx}{d^3}\\ &=\frac {b^2 c^2}{4 d^3 \left (1+c^2 x^2\right )}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{d^3 x \left (1+c^2 x^2\right )^{3/2}}-\frac {5 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {4 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \sqrt {1+c^2 x^2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1+c^2 x^2\right )^2}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 \left (1+c^2 x^2\right )}-\frac {2 b^2 c^2 \log \left (1+c^2 x^2\right )}{d^3}-\frac {\left (6 c^2\right ) \text {Subst}\left (\int (a+b x)^2 \text {csch}(2 x) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}-\frac {\left (b^2 c^2\right ) \text {Subst}\left (\int \left (-\frac {3}{x}-\frac {c^2}{\left (1+c^2 x\right )^2}-\frac {5 c^2}{1+c^2 x}\right ) \, dx,x,x^2\right )}{6 d^3}\\ &=\frac {b^2 c^2}{12 d^3 \left (1+c^2 x^2\right )}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{d^3 x \left (1+c^2 x^2\right )^{3/2}}-\frac {5 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {4 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \sqrt {1+c^2 x^2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1+c^2 x^2\right )^2}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 \left (1+c^2 x^2\right )}+\frac {6 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}+\frac {b^2 c^2 \log (x)}{d^3}-\frac {7 b^2 c^2 \log \left (1+c^2 x^2\right )}{6 d^3}+\frac {\left (6 b c^2\right ) \text {Subst}\left (\int (a+b x) \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}-\frac {\left (6 b c^2\right ) \text {Subst}\left (\int (a+b x) \log \left (1+e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}\\ &=\frac {b^2 c^2}{12 d^3 \left (1+c^2 x^2\right )}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{d^3 x \left (1+c^2 x^2\right )^{3/2}}-\frac {5 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {4 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \sqrt {1+c^2 x^2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1+c^2 x^2\right )^2}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 \left (1+c^2 x^2\right )}+\frac {6 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}+\frac {b^2 c^2 \log (x)}{d^3}-\frac {7 b^2 c^2 \log \left (1+c^2 x^2\right )}{6 d^3}+\frac {3 b c^2 \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{d^3}-\frac {3 b c^2 \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}-\frac {\left (3 b^2 c^2\right ) \text {Subst}\left (\int \text {Li}_2\left (-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}+\frac {\left (3 b^2 c^2\right ) \text {Subst}\left (\int \text {Li}_2\left (e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}\\ &=\frac {b^2 c^2}{12 d^3 \left (1+c^2 x^2\right )}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{d^3 x \left (1+c^2 x^2\right )^{3/2}}-\frac {5 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {4 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \sqrt {1+c^2 x^2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1+c^2 x^2\right )^2}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 \left (1+c^2 x^2\right )}+\frac {6 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}+\frac {b^2 c^2 \log (x)}{d^3}-\frac {7 b^2 c^2 \log \left (1+c^2 x^2\right )}{6 d^3}+\frac {3 b c^2 \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{d^3}-\frac {3 b c^2 \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}-\frac {\left (3 b^2 c^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2(-x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}+\frac {\left (3 b^2 c^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}\\ &=\frac {b^2 c^2}{12 d^3 \left (1+c^2 x^2\right )}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{d^3 x \left (1+c^2 x^2\right )^{3/2}}-\frac {5 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {4 b c^3 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \sqrt {1+c^2 x^2}}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 x^2 \left (1+c^2 x^2\right )^2}-\frac {3 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 \left (1+c^2 x^2\right )}+\frac {6 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}+\frac {b^2 c^2 \log (x)}{d^3}-\frac {7 b^2 c^2 \log \left (1+c^2 x^2\right )}{6 d^3}+\frac {3 b c^2 \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{d^3}-\frac {3 b c^2 \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}-\frac {3 b^2 c^2 \text {Li}_3\left (-e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}+\frac {3 b^2 c^2 \text {Li}_3\left (e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains complex when optimal does not.
time = 7.33, size = 688, normalized size = 1.81 \begin {gather*} -\frac {\frac {2 a^2}{x^2}+\frac {a^2 c^2}{\left (1+c^2 x^2\right )^2}+\frac {4 a^2 c^2}{1+c^2 x^2}+12 a^2 c^2 \log (x)-6 a^2 c^2 \log \left (1+c^2 x^2\right )-\frac {1}{6} a b \left (\frac {27 c^2 \left (\sqrt {1+c^2 x^2}-i \sinh ^{-1}(c x)\right )}{i+c x}+\frac {27 c^2 \left (\sqrt {1+c^2 x^2}+i \sinh ^{-1}(c x)\right )}{-i+c x}-\frac {24 \left (c x \sqrt {1+c^2 x^2}+\sinh ^{-1}(c x)\right )}{x^2}+\frac {c^2 \left ((-2 i+c x) \sqrt {1+c^2 x^2}+3 \sinh ^{-1}(c x)\right )}{(-i+c x)^2}+\frac {c^2 \left ((2 i+c x) \sqrt {1+c^2 x^2}+3 \sinh ^{-1}(c x)\right )}{(i+c x)^2}-36 c^2 \left (\sinh ^{-1}(c x) \left (\sinh ^{-1}(c x)-4 \log \left (1+i e^{\sinh ^{-1}(c x)}\right )\right )-4 \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )\right )-36 c^2 \left (\sinh ^{-1}(c x) \left (\sinh ^{-1}(c x)-4 \log \left (1-i e^{\sinh ^{-1}(c x)}\right )\right )-4 \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )\right )+72 c^2 \left (\sinh ^{-1}(c x) \left (\sinh ^{-1}(c x)-2 \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )\right )-\text {PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )\right )\right )-4 b^2 c^2 \left (-\frac {i \pi ^3}{8}+\frac {1}{12+12 c^2 x^2}+\frac {c x \sinh ^{-1}(c x)}{6 \left (1+c^2 x^2\right )^{3/2}}+\frac {7 c x \sinh ^{-1}(c x)}{3 \sqrt {1+c^2 x^2}}-\frac {\sqrt {1+c^2 x^2} \sinh ^{-1}(c x)}{c x}-\frac {\sinh ^{-1}(c x)^2}{2 c^2 x^2}-\frac {\sinh ^{-1}(c x)^2}{4 \left (1+c^2 x^2\right )^2}-\frac {\sinh ^{-1}(c x)^2}{1+c^2 x^2}+2 \sinh ^{-1}(c x)^3+3 \sinh ^{-1}(c x)^2 \log \left (1+e^{-2 \sinh ^{-1}(c x)}\right )-3 \sinh ^{-1}(c x)^2 \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )+\log (c x)-\frac {7}{6} \log \left (1+c^2 x^2\right )-3 \sinh ^{-1}(c x) \text {PolyLog}\left (2,-e^{-2 \sinh ^{-1}(c x)}\right )-3 \sinh ^{-1}(c x) \text {PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )-\frac {3}{2} \text {PolyLog}\left (3,-e^{-2 \sinh ^{-1}(c x)}\right )+\frac {3}{2} \text {PolyLog}\left (3,e^{2 \sinh ^{-1}(c x)}\right )\right )}{4 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*((2*a^2)/x^2 + (a^2*c^2)/(1 + c^2*x^2)^2 + (4*a^2*c^2)/(1 + c^2*x^2) + 12*a^2*c^2*Log[x] - 6*a^2*c^2*Log[
1 + c^2*x^2] - (a*b*((27*c^2*(Sqrt[1 + c^2*x^2] - I*ArcSinh[c*x]))/(I + c*x) + (27*c^2*(Sqrt[1 + c^2*x^2] + I*
ArcSinh[c*x]))/(-I + c*x) - (24*(c*x*Sqrt[1 + c^2*x^2] + ArcSinh[c*x]))/x^2 + (c^2*((-2*I + c*x)*Sqrt[1 + c^2*
x^2] + 3*ArcSinh[c*x]))/(-I + c*x)^2 + (c^2*((2*I + c*x)*Sqrt[1 + c^2*x^2] + 3*ArcSinh[c*x]))/(I + c*x)^2 - 36
*c^2*(ArcSinh[c*x]*(ArcSinh[c*x] - 4*Log[1 + I*E^ArcSinh[c*x]]) - 4*PolyLog[2, (-I)*E^ArcSinh[c*x]]) - 36*c^2*
(ArcSinh[c*x]*(ArcSinh[c*x] - 4*Log[1 - I*E^ArcSinh[c*x]]) - 4*PolyLog[2, I*E^ArcSinh[c*x]]) + 72*c^2*(ArcSinh
[c*x]*(ArcSinh[c*x] - 2*Log[1 - E^(2*ArcSinh[c*x])]) - PolyLog[2, E^(2*ArcSinh[c*x])])))/6 - 4*b^2*c^2*((-1/8*
I)*Pi^3 + (12 + 12*c^2*x^2)^(-1) + (c*x*ArcSinh[c*x])/(6*(1 + c^2*x^2)^(3/2)) + (7*c*x*ArcSinh[c*x])/(3*Sqrt[1
+ c^2*x^2]) - (Sqrt[1 + c^2*x^2]*ArcSinh[c*x])/(c*x) - ArcSinh[c*x]^2/(2*c^2*x^2) - ArcSinh[c*x]^2/(4*(1 + c^
2*x^2)^2) - ArcSinh[c*x]^2/(1 + c^2*x^2) + 2*ArcSinh[c*x]^3 + 3*ArcSinh[c*x]^2*Log[1 + E^(-2*ArcSinh[c*x])] -
3*ArcSinh[c*x]^2*Log[1 - E^(2*ArcSinh[c*x])] + Log[c*x] - (7*Log[1 + c^2*x^2])/6 - 3*ArcSinh[c*x]*PolyLog[2, -
E^(-2*ArcSinh[c*x])] - 3*ArcSinh[c*x]*PolyLog[2, E^(2*ArcSinh[c*x])] - (3*PolyLog[3, -E^(-2*ArcSinh[c*x])])/2
+ (3*PolyLog[3, E^(2*ArcSinh[c*x])])/2))/d^3

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1364\) vs. \(2(402)=804\).
time = 6.08, size = 1365, normalized size = 3.58

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1365\)
default	\(\text {Expression too large to display}\)	\(1365\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))^2/x^3/(c^2*d*x^2+d)^3,x,method=_RETURNVERBOSE)

[Out]

c^2*(-3*a*b/d^3/(c^4*x^4+2*c^2*x^2+1)*arcsinh(c*x)*c^2*x^2-8/3*a*b/d^3/(c^4*x^4+2*c^2*x^2+1)*c^2*x^2-1/2*b^2/d
^3/(c^4*x^4+2*c^2*x^2+1)/c^2/x^2*arcsinh(c*x)^2-1/4*a^2/d^3/(c^2*x^2+1)^2-4/3*a*b/d^3/(c^4*x^4+2*c^2*x^2+1)*c^
4*x^4-3*a^2/d^3*ln(c*x)-4/3*b^2/d^3/(c^4*x^4+2*c^2*x^2+1)*arcsinh(c*x)*c^4*x^4-3/2*b^2/d^3/(c^4*x^4+2*c^2*x^2+
1)*arcsinh(c*x)^2*c^2*x^2-8/3*b^2/d^3/(c^4*x^4+2*c^2*x^2+1)*arcsinh(c*x)*c^2*x^2+4/3*a*b/d^3/(c^4*x^4+2*c^2*x^
2+1)*(c^2*x^2+1)^(1/2)*c^3*x^3+1/12*b^2/d^3/(c^4*x^4+2*c^2*x^2+1)*c^2*x^2+6*b^2/d^3*polylog(3,-c*x-(c^2*x^2+1)
^(1/2))+6*b^2/d^3*polylog(3,c*x+(c^2*x^2+1)^(1/2))-a*b/d^3/(c^4*x^4+2*c^2*x^2+1)/c/x*(c^2*x^2+1)^(1/2)-a*b/d^3
/(c^4*x^4+2*c^2*x^2+1)/c^2/x^2*arcsinh(c*x)-b^2/d^3/(c^4*x^4+2*c^2*x^2+1)/c/x*arcsinh(c*x)*(c^2*x^2+1)^(1/2)+1
/2*a*b/d^3/(c^4*x^4+2*c^2*x^2+1)*c*x*(c^2*x^2+1)^(1/2)+b^2/d^3*ln(1+c*x+(c^2*x^2+1)^(1/2))+b^2/d^3*ln(c*x+(c^2
*x^2+1)^(1/2)-1)-1/2*a^2/d^3/c^2/x^2-7/3*b^2/d^3*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)+1/12*b^2/d^3/(c^4*x^4+2*c^2*x
^2+1)+3/2*a^2/d^3*ln(c^2*x^2+1)-a^2/d^3/(c^2*x^2+1)+8/3*b^2/d^3*ln(c*x+(c^2*x^2+1)^(1/2))+4/3*b^2/d^3/(c^4*x^4
+2*c^2*x^2+1)*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*c^3*x^3+1/2*b^2/d^3/(c^4*x^4+2*c^2*x^2+1)*arcsinh(c*x)*(c^2*x^2+1
)^(1/2)*c*x-3/2*b^2*polylog(3,-(c*x+(c^2*x^2+1)^(1/2))^2)/d^3-9/2*a*b/d^3/(c^4*x^4+2*c^2*x^2+1)*arcsinh(c*x)+6
*a*b/d^3*arcsinh(c*x)*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)-6*a*b/d^3*arcsinh(c*x)*ln(1+c*x+(c^2*x^2+1)^(1/2))-6*a*b
/d^3*arcsinh(c*x)*ln(1-c*x-(c^2*x^2+1)^(1/2))-9/4*b^2/d^3/(c^4*x^4+2*c^2*x^2+1)*arcsinh(c*x)^2-4/3*b^2/d^3/(c^
4*x^4+2*c^2*x^2+1)*arcsinh(c*x)-3*b^2/d^3*arcsinh(c*x)^2*ln(1+c*x+(c^2*x^2+1)^(1/2))-6*b^2/d^3*arcsinh(c*x)*po
lylog(2,-c*x-(c^2*x^2+1)^(1/2))-3*b^2/d^3*arcsinh(c*x)^2*ln(1-c*x-(c^2*x^2+1)^(1/2))-6*b^2/d^3*arcsinh(c*x)*po
lylog(2,c*x+(c^2*x^2+1)^(1/2))-4/3*a*b/d^3/(c^4*x^4+2*c^2*x^2+1)+3*a*b/d^3*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))^
2)-6*a*b/d^3*polylog(2,-c*x-(c^2*x^2+1)^(1/2))-6*a*b/d^3*polylog(2,c*x+(c^2*x^2+1)^(1/2))+3*b^2/d^3*arcsinh(c*
x)^2*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)+3*b^2/d^3*arcsinh(c*x)*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))^2))

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*a^2*((6*c^4*x^4 + 9*c^2*x^2 + 2)/(c^4*d^3*x^6 + 2*c^2*d^3*x^4 + d^3*x^2) - 6*c^2*log(c^2*x^2 + 1)/d^3 + 1
2*c^2*log(x)/d^3) + integrate(b^2*log(c*x + sqrt(c^2*x^2 + 1))^2/(c^6*d^3*x^9 + 3*c^4*d^3*x^7 + 3*c^2*d^3*x^5
+ d^3*x^3) + 2*a*b*log(c*x + sqrt(c^2*x^2 + 1))/(c^6*d^3*x^9 + 3*c^4*d^3*x^7 + 3*c^2*d^3*x^5 + d^3*x^3), x)

________________________________________________________________________________________

Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*arcsinh(c*x)^2 + 2*a*b*arcsinh(c*x) + a^2)/(c^6*d^3*x^9 + 3*c^4*d^3*x^7 + 3*c^2*d^3*x^5 + d^3*x^
3), x)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a^{2}}{c^{6} x^{9} + 3 c^{4} x^{7} + 3 c^{2} x^{5} + x^{3}}\, dx + \int \frac {b^{2} \operatorname {asinh}^{2}{\left (c x \right )}}{c^{6} x^{9} + 3 c^{4} x^{7} + 3 c^{2} x^{5} + x^{3}}\, dx + \int \frac {2 a b \operatorname {asinh}{\left (c x \right )}}{c^{6} x^{9} + 3 c^{4} x^{7} + 3 c^{2} x^{5} + x^{3}}\, dx}{d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a**2/(c**6*x**9 + 3*c**4*x**7 + 3*c**2*x**5 + x**3), x) + Integral(b**2*asinh(c*x)**2/(c**6*x**9 + 3
*c**4*x**7 + 3*c**2*x**5 + x**3), x) + Integral(2*a*b*asinh(c*x)/(c**6*x**9 + 3*c**4*x**7 + 3*c**2*x**5 + x**3
), x))/d**3

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2}{x^3\,{\left (d\,c^2\,x^2+d\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________