3.3.48 \(\int \frac {(a+b \sinh ^{-1}(c x))^2}{x (d+c^2 d x^2)^3} \, dx\) [248]

Optimal. Leaf size=275 \[ -\frac {b^2}{12 d^3 \left (1+c^2 x^2\right )}-\frac {b c x \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {4 b c x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \sqrt {1+c^2 x^2}}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 \left (1+c^2 x^2\right )}-\frac {2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}+\frac {2 b^2 \log \left (1+c^2 x^2\right )}{3 d^3}-\frac {b \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{d^3}+\frac {b \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )}{d^3}+\frac {b^2 \text {PolyLog}\left (3,-e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}-\frac {b^2 \text {PolyLog}\left (3,e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.36, antiderivative size = 275, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 11, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {5811, 5799, 5569, 4267, 2611, 2320, 6724, 5787, 266, 5788, 267} \begin {gather*} -\frac {4 b c x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \sqrt {c^2 x^2+1}}-\frac {b c x \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (c^2 x^2+1\right )^{3/2}}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^3 \left (c^2 x^2+1\right )}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (c^2 x^2+1\right )^2}-\frac {b \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^3}+\frac {b \text {Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^3}-\frac {2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{d^3}-\frac {b^2}{12 d^3 \left (c^2 x^2+1\right )}+\frac {2 b^2 \log \left (c^2 x^2+1\right )}{3 d^3}+\frac {b^2 \text {Li}_3\left (-e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}-\frac {b^2 \text {Li}_3\left (e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 \left (d+c^2 d x^2\right )^3} \, dx &=\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3 \left (1+c^2 x^2\right )^2}-\frac {(b c) \int \frac {a+b \sinh ^{-1}(c x)}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{2 d^3}+\frac {\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 x \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}+\frac {\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 \left (1+c^2 x^2\right )}-\frac {(b c) \int \frac {a+b \sinh ^{-1}(c x)}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{3 d^3}-\frac {(b c) \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^2\right ) \int \frac {x}{\left (1+c^2 x^2\right )^2} \, dx}{6 d^3}+\frac {\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}{12 d^3 \left (1+c^2 x^2\right )}-\frac {b c 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 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \sqrt {1+c^2 x^2}}+\frac {\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 \left (1+c^2 x^2\right )}+\frac {\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 ) \int \frac {x}{1+c^2 x^2} \, dx}{3 d^3}+\frac {\left (b^2 c^2\right ) \int \frac {x}{1+c^2 x^2} \, dx}{d^3}\\ &=-\frac {b^2}{12 d^3 \left (1+c^2 x^2\right )}-\frac {b c 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 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \sqrt {1+c^2 x^2}}+\frac {\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 \left (1+c^2 x^2\right )}+\frac {2 b^2 \log \left (1+c^2 x^2\right )}{3 d^3}+\frac {2 \text {Subst}\left (\int (a+b x)^2 \text {csch}(2 x) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}\\ &=-\frac {b^2}{12 d^3 \left (1+c^2 x^2\right )}-\frac {b c 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 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \sqrt {1+c^2 x^2}}+\frac {\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 \left (1+c^2 x^2\right )}-\frac {2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}+\frac {2 b^2 \log \left (1+c^2 x^2\right )}{3 d^3}-\frac {(2 b) \text {Subst}\left (\int (a+b x) \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}+\frac {(2 b) \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}{12 d^3 \left (1+c^2 x^2\right )}-\frac {b c 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 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \sqrt {1+c^2 x^2}}+\frac {\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 \left (1+c^2 x^2\right )}-\frac {2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}+\frac {2 b^2 \log \left (1+c^2 x^2\right )}{3 d^3}-\frac {b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{d^3}+\frac {b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}+\frac {b^2 \text {Subst}\left (\int \text {Li}_2\left (-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}-\frac {b^2 \text {Subst}\left (\int \text {Li}_2\left (e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^3}\\ &=-\frac {b^2}{12 d^3 \left (1+c^2 x^2\right )}-\frac {b c 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 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \sqrt {1+c^2 x^2}}+\frac {\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 \left (1+c^2 x^2\right )}-\frac {2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}+\frac {2 b^2 \log \left (1+c^2 x^2\right )}{3 d^3}-\frac {b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{d^3}+\frac {b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}+\frac {b^2 \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 \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}{12 d^3 \left (1+c^2 x^2\right )}-\frac {b c 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 x \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 \sqrt {1+c^2 x^2}}+\frac {\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 \left (1+c^2 x^2\right )}-\frac {2 \left (a+b \sinh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}+\frac {2 b^2 \log \left (1+c^2 x^2\right )}{3 d^3}-\frac {b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{d^3}+\frac {b \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )}{d^3}+\frac {b^2 \text {Li}_3\left (-e^{2 \sinh ^{-1}(c x)}\right )}{2 d^3}-\frac {b^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 = 2.62, size = 560, normalized size = 2.04 \begin {gather*} \frac {\frac {6 a^2}{\left (1+c^2 x^2\right )^2}+\frac {12 a^2}{1+c^2 x^2}+24 a^2 \log (c x)-12 a^2 \log \left (1+c^2 x^2\right )+a b \left (-\frac {15 \left (\sqrt {1+c^2 x^2}-i \sinh ^{-1}(c x)\right )}{i+c x}-\frac {15 \left (\sqrt {1+c^2 x^2}+i \sinh ^{-1}(c x)\right )}{-i+c x}-24 \sinh ^{-1}(c x)^2-\frac {(-2 i+c x) \sqrt {1+c^2 x^2}+3 \sinh ^{-1}(c x)}{(-i+c x)^2}-\frac {(2 i+c x) \sqrt {1+c^2 x^2}+3 \sinh ^{-1}(c x)}{(i+c x)^2}+48 \sinh ^{-1}(c x) \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )+12 \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 )+12 \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 )+24 \text {PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )\right )+b^2 \left (i \pi ^3-\frac {2}{1+c^2 x^2}-\frac {4 c x \sinh ^{-1}(c x)}{\left (1+c^2 x^2\right )^{3/2}}-\frac {32 c x \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}}+\frac {6 \sinh ^{-1}(c x)^2}{\left (1+c^2 x^2\right )^2}+\frac {12 \sinh ^{-1}(c x)^2}{1+c^2 x^2}-16 \sinh ^{-1}(c x)^3-24 \sinh ^{-1}(c x)^2 \log \left (1+e^{-2 \sinh ^{-1}(c x)}\right )+24 \sinh ^{-1}(c x)^2 \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )+16 \log \left (1+c^2 x^2\right )+24 \sinh ^{-1}(c x) \text {PolyLog}\left (2,-e^{-2 \sinh ^{-1}(c x)}\right )+24 \sinh ^{-1}(c x) \text {PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )+12 \text {PolyLog}\left (3,-e^{-2 \sinh ^{-1}(c x)}\right )-12 \text {PolyLog}\left (3,e^{2 \sinh ^{-1}(c x)}\right )\right )}{24 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((6*a^2)/(1 + c^2*x^2)^2 + (12*a^2)/(1 + c^2*x^2) + 24*a^2*Log[c*x] - 12*a^2*Log[1 + c^2*x^2] + a*b*((-15*(Sqr
t[1 + c^2*x^2] - I*ArcSinh[c*x]))/(I + c*x) - (15*(Sqrt[1 + c^2*x^2] + I*ArcSinh[c*x]))/(-I + c*x) - 24*ArcSin
h[c*x]^2 - ((-2*I + c*x)*Sqrt[1 + c^2*x^2] + 3*ArcSinh[c*x])/(-I + c*x)^2 - ((2*I + c*x)*Sqrt[1 + c^2*x^2] + 3
*ArcSinh[c*x])/(I + c*x)^2 + 48*ArcSinh[c*x]*Log[1 - E^(2*ArcSinh[c*x])] + 12*(ArcSinh[c*x]*(ArcSinh[c*x] - 4*
Log[1 + I*E^ArcSinh[c*x]]) - 4*PolyLog[2, (-I)*E^ArcSinh[c*x]]) + 12*(ArcSinh[c*x]*(ArcSinh[c*x] - 4*Log[1 - I
*E^ArcSinh[c*x]]) - 4*PolyLog[2, I*E^ArcSinh[c*x]]) + 24*PolyLog[2, E^(2*ArcSinh[c*x])]) + b^2*(I*Pi^3 - 2/(1
+ c^2*x^2) - (4*c*x*ArcSinh[c*x])/(1 + c^2*x^2)^(3/2) - (32*c*x*ArcSinh[c*x])/Sqrt[1 + c^2*x^2] + (6*ArcSinh[c
*x]^2)/(1 + c^2*x^2)^2 + (12*ArcSinh[c*x]^2)/(1 + c^2*x^2) - 16*ArcSinh[c*x]^3 - 24*ArcSinh[c*x]^2*Log[1 + E^(
-2*ArcSinh[c*x])] + 24*ArcSinh[c*x]^2*Log[1 - E^(2*ArcSinh[c*x])] + 16*Log[1 + c^2*x^2] + 24*ArcSinh[c*x]*Poly
Log[2, -E^(-2*ArcSinh[c*x])] + 24*ArcSinh[c*x]*PolyLog[2, E^(2*ArcSinh[c*x])] + 12*PolyLog[3, -E^(-2*ArcSinh[c
*x])] - 12*PolyLog[3, E^(2*ArcSinh[c*x])]))/(24*d^3)

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1128\) vs. \(2(300)=600\).
time = 5.57, size = 1129, normalized size = 4.11

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1129\)
default	\(\text {Expression too large to display}\)	\(1129\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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+1/4*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+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+1/2*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^2-2*b^2/d^3*polylog(3,-c*x-(c^2*x^2+1)^(1/2))-2*b^2/d^3*polylog(3,c*x+(c^2*x^2+1)^(1/2))-3/2*a*b/d^3/(c^4
*x^4+2*c^2*x^2+1)*c*x*(c^2*x^2+1)^(1/2)+4/3*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)-1/2*a^2/d^3*ln(c^2*x^2+1)+1/2*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^3-3/2*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+1/2*b^2*polylog(3,-(c*x+(c^2*x^2+1)^(1/2))^2)/d^3+3/2*a*b/d^3/(c^4*x^4+2*c^2*x^2+1)*arcsinh
(c*x)-2*a*b/d^3*arcsinh(c*x)*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)+2*a*b/d^3*arcsinh(c*x)*ln(1+c*x+(c^2*x^2+1)^(1/2)
)+2*a*b/d^3*arcsinh(c*x)*ln(1-c*x-(c^2*x^2+1)^(1/2))+3/4*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)+b^2/d^3*arcsinh(c*x)^2*ln(1+c*x+(c^2*x^2+1)^(1/2))+2*b^2/d^3*arcsinh(c*
x)*polylog(2,-c*x-(c^2*x^2+1)^(1/2))+b^2/d^3*arcsinh(c*x)^2*ln(1-c*x-(c^2*x^2+1)^(1/2))+2*b^2/d^3*arcsinh(c*x)
*polylog(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)-a*b/d^3*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))
^2)+2*a*b/d^3*polylog(2,-c*x-(c^2*x^2+1)^(1/2))+2*a*b/d^3*polylog(2,c*x+(c^2*x^2+1)^(1/2))-b^2/d^3*arcsinh(c*x
)^2*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)-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/(c^2*d*x^2+d)^3,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

________________________________________________________________________________________