3.3.51 \(\int \frac {(a+b \sinh ^{-1}(c x))^2}{x^4 (d+c^2 d x^2)^3} \, dx\) [251]

Optimal. Leaf size=529 \[ -\frac {b^2 c^2}{2 d^3 x}+\frac {b^2 c^2}{6 d^3 x \left (1+c^2 x^2\right )}+\frac {b^2 c^4 x}{12 d^3 \left (1+c^2 x^2\right )}-\frac {b c^3 \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 x^2 \left (1+c^2 x^2\right )^{3/2}}+\frac {29 b c^3 \left (a+b \sinh ^{-1}(c x)\right )}{12 d^3 \sqrt {1+c^2 x^2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^3 x^3 \left (1+c^2 x^2\right )^2}+\frac {7 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^3 x \left (1+c^2 x^2\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )^2}{12 d^3 \left (1+c^2 x^2\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 d^3 \left (1+c^2 x^2\right )}+\frac {35 c^3 \left (a+b \sinh ^{-1}(c x)\right )^2 \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right )}{4 d^3}-\frac {17 b^2 c^3 \text {ArcTan}(c x)}{6 d^3}+\frac {38 b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{3 d^3}+\frac {19 b^2 c^3 \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{3 d^3}-\frac {35 i b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac {35 i b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}-\frac {19 b^2 c^3 \text {PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{3 d^3}+\frac {35 i b^2 c^3 \text {PolyLog}\left (3,-i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}-\frac {35 i b^2 c^3 \text {PolyLog}\left (3,i e^{\sinh ^{-1}(c x)}\right )}{4 d^3} \]

[Out]

-1/2*b^2*c^2/d^3/x+1/6*b^2*c^2/d^3/x/(c^2*x^2+1)+1/12*b^2*c^4*x/d^3/(c^2*x^2+1)-1/6*b*c^3*(a+b*arcsinh(c*x))/d
^3/(c^2*x^2+1)^(3/2)-1/3*b*c*(a+b*arcsinh(c*x))/d^3/x^2/(c^2*x^2+1)^(3/2)-1/3*(a+b*arcsinh(c*x))^2/d^3/x^3/(c^
2*x^2+1)^2+7/3*c^2*(a+b*arcsinh(c*x))^2/d^3/x/(c^2*x^2+1)^2+35/12*c^4*x*(a+b*arcsinh(c*x))^2/d^3/(c^2*x^2+1)^2
+35/8*c^4*x*(a+b*arcsinh(c*x))^2/d^3/(c^2*x^2+1)+35/4*c^3*(a+b*arcsinh(c*x))^2*arctan(c*x+(c^2*x^2+1)^(1/2))/d
^3-17/6*b^2*c^3*arctan(c*x)/d^3+38/3*b*c^3*(a+b*arcsinh(c*x))*arctanh(c*x+(c^2*x^2+1)^(1/2))/d^3+19/3*b^2*c^3*
polylog(2,-c*x-(c^2*x^2+1)^(1/2))/d^3+35/4*I*b*c^3*(a+b*arcsinh(c*x))*polylog(2,I*(c*x+(c^2*x^2+1)^(1/2)))/d^3
-35/4*I*b*c^3*(a+b*arcsinh(c*x))*polylog(2,-I*(c*x+(c^2*x^2+1)^(1/2)))/d^3-19/3*b^2*c^3*polylog(2,c*x+(c^2*x^2
+1)^(1/2))/d^3-35/4*I*b^2*c^3*polylog(3,I*(c*x+(c^2*x^2+1)^(1/2)))/d^3+35/4*I*b^2*c^3*polylog(3,-I*(c*x+(c^2*x
^2+1)^(1/2)))/d^3+29/12*b*c^3*(a+b*arcsinh(c*x))/d^3/(c^2*x^2+1)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.94, antiderivative size = 529, normalized size of antiderivative = 1.00, number of steps used = 43, number of rules used = 17, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.654, Rules used = {5809, 5788, 5789, 4265, 2611, 2320, 6724, 5798, 209, 205, 5811, 5816, 4267, 2317, 2438, 296, 331} \begin {gather*} \frac {35 c^3 \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{4 d^3}-\frac {35 i b c^3 \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3}+\frac {35 i b c^3 \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{4 d^3}+\frac {38 b c^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3}+\frac {7 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^3 x \left (c^2 x^2+1\right )^2}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 x^2 \left (c^2 x^2+1\right )^{3/2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^3 x^3 \left (c^2 x^2+1\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 d^3 \left (c^2 x^2+1\right )}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )^2}{12 d^3 \left (c^2 x^2+1\right )^2}+\frac {29 b c^3 \left (a+b \sinh ^{-1}(c x)\right )}{12 d^3 \sqrt {c^2 x^2+1}}-\frac {b c^3 \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (c^2 x^2+1\right )^{3/2}}-\frac {17 b^2 c^3 \text {ArcTan}(c x)}{6 d^3}+\frac {19 b^2 c^3 \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{3 d^3}-\frac {19 b^2 c^3 \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{3 d^3}+\frac {35 i b^2 c^3 \text {Li}_3\left (-i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}-\frac {35 i b^2 c^3 \text {Li}_3\left (i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac {b^2 c^2}{6 d^3 x \left (c^2 x^2+1\right )}-\frac {b^2 c^2}{2 d^3 x}+\frac {b^2 c^4 x}{12 d^3 \left (c^2 x^2+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(b^2*c^2)/(d^3*x) + (b^2*c^2)/(6*d^3*x*(1 + c^2*x^2)) + (b^2*c^4*x)/(12*d^3*(1 + c^2*x^2)) - (b*c^3*(a +
b*ArcSinh[c*x]))/(6*d^3*(1 + c^2*x^2)^(3/2)) - (b*c*(a + b*ArcSinh[c*x]))/(3*d^3*x^2*(1 + c^2*x^2)^(3/2)) + (2
9*b*c^3*(a + b*ArcSinh[c*x]))/(12*d^3*Sqrt[1 + c^2*x^2]) - (a + b*ArcSinh[c*x])^2/(3*d^3*x^3*(1 + c^2*x^2)^2)
+ (7*c^2*(a + b*ArcSinh[c*x])^2)/(3*d^3*x*(1 + c^2*x^2)^2) + (35*c^4*x*(a + b*ArcSinh[c*x])^2)/(12*d^3*(1 + c^
2*x^2)^2) + (35*c^4*x*(a + b*ArcSinh[c*x])^2)/(8*d^3*(1 + c^2*x^2)) + (35*c^3*(a + b*ArcSinh[c*x])^2*ArcTan[E^
ArcSinh[c*x]])/(4*d^3) - (17*b^2*c^3*ArcTan[c*x])/(6*d^3) + (38*b*c^3*(a + b*ArcSinh[c*x])*ArcTanh[E^ArcSinh[c
*x]])/(3*d^3) + (19*b^2*c^3*PolyLog[2, -E^ArcSinh[c*x]])/(3*d^3) - (((35*I)/4)*b*c^3*(a + b*ArcSinh[c*x])*Poly
Log[2, (-I)*E^ArcSinh[c*x]])/d^3 + (((35*I)/4)*b*c^3*(a + b*ArcSinh[c*x])*PolyLog[2, I*E^ArcSinh[c*x]])/d^3 -
(19*b^2*c^3*PolyLog[2, E^ArcSinh[c*x]])/(3*d^3) + (((35*I)/4)*b^2*c^3*PolyLog[3, (-I)*E^ArcSinh[c*x]])/d^3 - (
((35*I)/4)*b^2*c^3*PolyLog[3, I*E^ArcSinh[c*x]])/d^3

Rule 205

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}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 296

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-(c*x)^(m + 1))*((a + b*x^n)^(p + 1)/
(a*c*n*(p + 1))), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2317

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 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 2438

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 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 4265

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[-2*(c +
d*x)^m*(ArcTanh[E^((-I)*e + f*fz*x)/E^(I*k*Pi)]/(f*fz*I)), x] + (-Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*
Log[1 - E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x] + Dist[d*(m/(f*fz*I)), Int[(c + d*x)^(m - 1)*Log[1 + E^((-I)*e
+ f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[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 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 5789

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

Rule 5798

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

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 5816

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

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^4 \left (d+c^2 d x^2\right )^3} \, dx &=-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^3 x^3 \left (1+c^2 x^2\right )^2}-\frac {1}{3} \left (7 c^2\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{x^2 \left (d+c^2 d x^2\right )^3} \, dx+\frac {(2 b c) \int \frac {a+b \sinh ^{-1}(c x)}{x^3 \left (1+c^2 x^2\right )^{5/2}} \, dx}{3 d^3}\\ &=-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 x^2 \left (1+c^2 x^2\right )^{3/2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^3 x^3 \left (1+c^2 x^2\right )^2}+\frac {7 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^3 x \left (1+c^2 x^2\right )^2}+\frac {1}{3} \left (35 c^4\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{\left (d+c^2 d x^2\right )^3} \, dx+\frac {\left (b^2 c^2\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )^2} \, dx}{3 d^3}-\frac {\left (5 b c^3\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \left (1+c^2 x^2\right )^{5/2}} \, dx}{3 d^3}-\frac {\left (14 b c^3\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \left (1+c^2 x^2\right )^{5/2}} \, dx}{3 d^3}\\ &=\frac {b^2 c^2}{6 d^3 x \left (1+c^2 x^2\right )}-\frac {19 b c^3 \left (a+b \sinh ^{-1}(c x)\right )}{9 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 x^2 \left (1+c^2 x^2\right )^{3/2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^3 x^3 \left (1+c^2 x^2\right )^2}+\frac {7 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^3 x \left (1+c^2 x^2\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )^2}{12 d^3 \left (1+c^2 x^2\right )^2}+\frac {\left (b^2 c^2\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx}{2 d^3}-\frac {\left (5 b c^3\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \left (1+c^2 x^2\right )^{3/2}} \, dx}{3 d^3}-\frac {\left (14 b c^3\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \left (1+c^2 x^2\right )^{3/2}} \, dx}{3 d^3}+\frac {\left (5 b^2 c^4\right ) \int \frac {1}{\left (1+c^2 x^2\right )^2} \, dx}{9 d^3}+\frac {\left (14 b^2 c^4\right ) \int \frac {1}{\left (1+c^2 x^2\right )^2} \, dx}{9 d^3}-\frac {\left (35 b c^5\right ) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{6 d^3}+\frac {\left (35 c^4\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{\left (d+c^2 d x^2\right )^2} \, dx}{4 d}\\ &=-\frac {b^2 c^2}{2 d^3 x}+\frac {b^2 c^2}{6 d^3 x \left (1+c^2 x^2\right )}+\frac {19 b^2 c^4 x}{18 d^3 \left (1+c^2 x^2\right )}-\frac {b c^3 \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 x^2 \left (1+c^2 x^2\right )^{3/2}}-\frac {19 b c^3 \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}{3 d^3 x^3 \left (1+c^2 x^2\right )^2}+\frac {7 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^3 x \left (1+c^2 x^2\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )^2}{12 d^3 \left (1+c^2 x^2\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 d^3 \left (1+c^2 x^2\right )}-\frac {\left (5 b c^3\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \sqrt {1+c^2 x^2}} \, dx}{3 d^3}-\frac {\left (14 b c^3\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \sqrt {1+c^2 x^2}} \, dx}{3 d^3}+\frac {\left (5 b^2 c^4\right ) \int \frac {1}{1+c^2 x^2} \, dx}{18 d^3}-\frac {\left (b^2 c^4\right ) \int \frac {1}{1+c^2 x^2} \, dx}{2 d^3}+\frac {\left (7 b^2 c^4\right ) \int \frac {1}{1+c^2 x^2} \, dx}{9 d^3}+\frac {\left (5 b^2 c^4\right ) \int \frac {1}{1+c^2 x^2} \, dx}{3 d^3}-\frac {\left (35 b^2 c^4\right ) \int \frac {1}{\left (1+c^2 x^2\right )^2} \, dx}{18 d^3}+\frac {\left (14 b^2 c^4\right ) \int \frac {1}{1+c^2 x^2} \, dx}{3 d^3}-\frac {\left (35 b c^5\right ) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{4 d^3}+\frac {\left (35 c^4\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d+c^2 d x^2} \, dx}{8 d^2}\\ &=-\frac {b^2 c^2}{2 d^3 x}+\frac {b^2 c^2}{6 d^3 x \left (1+c^2 x^2\right )}+\frac {b^2 c^4 x}{12 d^3 \left (1+c^2 x^2\right )}-\frac {b c^3 \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 x^2 \left (1+c^2 x^2\right )^{3/2}}+\frac {29 b c^3 \left (a+b \sinh ^{-1}(c x)\right )}{12 d^3 \sqrt {1+c^2 x^2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^3 x^3 \left (1+c^2 x^2\right )^2}+\frac {7 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^3 x \left (1+c^2 x^2\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )^2}{12 d^3 \left (1+c^2 x^2\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 d^3 \left (1+c^2 x^2\right )}+\frac {62 b^2 c^3 \tan ^{-1}(c x)}{9 d^3}+\frac {\left (35 c^3\right ) \text {Subst}\left (\int (a+b x)^2 \text {sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{8 d^3}-\frac {\left (5 b c^3\right ) \text {Subst}\left (\int (a+b x) \text {csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{3 d^3}-\frac {\left (14 b c^3\right ) \text {Subst}\left (\int (a+b x) \text {csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{3 d^3}-\frac {\left (35 b^2 c^4\right ) \int \frac {1}{1+c^2 x^2} \, dx}{36 d^3}-\frac {\left (35 b^2 c^4\right ) \int \frac {1}{1+c^2 x^2} \, dx}{4 d^3}\\ &=-\frac {b^2 c^2}{2 d^3 x}+\frac {b^2 c^2}{6 d^3 x \left (1+c^2 x^2\right )}+\frac {b^2 c^4 x}{12 d^3 \left (1+c^2 x^2\right )}-\frac {b c^3 \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 x^2 \left (1+c^2 x^2\right )^{3/2}}+\frac {29 b c^3 \left (a+b \sinh ^{-1}(c x)\right )}{12 d^3 \sqrt {1+c^2 x^2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^3 x^3 \left (1+c^2 x^2\right )^2}+\frac {7 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^3 x \left (1+c^2 x^2\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )^2}{12 d^3 \left (1+c^2 x^2\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 d^3 \left (1+c^2 x^2\right )}+\frac {35 c^3 \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 d^3}-\frac {17 b^2 c^3 \tan ^{-1}(c x)}{6 d^3}+\frac {38 b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{3 d^3}-\frac {\left (35 i b c^3\right ) \text {Subst}\left (\int (a+b x) \log \left (1-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{4 d^3}+\frac {\left (35 i b c^3\right ) \text {Subst}\left (\int (a+b x) \log \left (1+i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{4 d^3}+\frac {\left (5 b^2 c^3\right ) \text {Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{3 d^3}-\frac {\left (5 b^2 c^3\right ) \text {Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{3 d^3}+\frac {\left (14 b^2 c^3\right ) \text {Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{3 d^3}-\frac {\left (14 b^2 c^3\right ) \text {Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{3 d^3}\\ &=-\frac {b^2 c^2}{2 d^3 x}+\frac {b^2 c^2}{6 d^3 x \left (1+c^2 x^2\right )}+\frac {b^2 c^4 x}{12 d^3 \left (1+c^2 x^2\right )}-\frac {b c^3 \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 x^2 \left (1+c^2 x^2\right )^{3/2}}+\frac {29 b c^3 \left (a+b \sinh ^{-1}(c x)\right )}{12 d^3 \sqrt {1+c^2 x^2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^3 x^3 \left (1+c^2 x^2\right )^2}+\frac {7 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^3 x \left (1+c^2 x^2\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )^2}{12 d^3 \left (1+c^2 x^2\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 d^3 \left (1+c^2 x^2\right )}+\frac {35 c^3 \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 d^3}-\frac {17 b^2 c^3 \tan ^{-1}(c x)}{6 d^3}+\frac {38 b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{3 d^3}-\frac {35 i b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac {35 i b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac {\left (35 i b^2 c^3\right ) \text {Subst}\left (\int \text {Li}_2\left (-i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{4 d^3}-\frac {\left (35 i b^2 c^3\right ) \text {Subst}\left (\int \text {Li}_2\left (i e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{4 d^3}+\frac {\left (5 b^2 c^3\right ) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{3 d^3}-\frac {\left (5 b^2 c^3\right ) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{3 d^3}+\frac {\left (14 b^2 c^3\right ) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{3 d^3}-\frac {\left (14 b^2 c^3\right ) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{3 d^3}\\ &=-\frac {b^2 c^2}{2 d^3 x}+\frac {b^2 c^2}{6 d^3 x \left (1+c^2 x^2\right )}+\frac {b^2 c^4 x}{12 d^3 \left (1+c^2 x^2\right )}-\frac {b c^3 \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 x^2 \left (1+c^2 x^2\right )^{3/2}}+\frac {29 b c^3 \left (a+b \sinh ^{-1}(c x)\right )}{12 d^3 \sqrt {1+c^2 x^2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^3 x^3 \left (1+c^2 x^2\right )^2}+\frac {7 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^3 x \left (1+c^2 x^2\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )^2}{12 d^3 \left (1+c^2 x^2\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 d^3 \left (1+c^2 x^2\right )}+\frac {35 c^3 \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 d^3}-\frac {17 b^2 c^3 \tan ^{-1}(c x)}{6 d^3}+\frac {38 b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{3 d^3}+\frac {19 b^2 c^3 \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{3 d^3}-\frac {35 i b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac {35 i b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}-\frac {19 b^2 c^3 \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{3 d^3}+\frac {\left (35 i b^2 c^3\right ) \text {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{4 d^3}-\frac {\left (35 i b^2 c^3\right ) \text {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{4 d^3}\\ &=-\frac {b^2 c^2}{2 d^3 x}+\frac {b^2 c^2}{6 d^3 x \left (1+c^2 x^2\right )}+\frac {b^2 c^4 x}{12 d^3 \left (1+c^2 x^2\right )}-\frac {b c^3 \left (a+b \sinh ^{-1}(c x)\right )}{6 d^3 \left (1+c^2 x^2\right )^{3/2}}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{3 d^3 x^2 \left (1+c^2 x^2\right )^{3/2}}+\frac {29 b c^3 \left (a+b \sinh ^{-1}(c x)\right )}{12 d^3 \sqrt {1+c^2 x^2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^3 x^3 \left (1+c^2 x^2\right )^2}+\frac {7 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^3 x \left (1+c^2 x^2\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )^2}{12 d^3 \left (1+c^2 x^2\right )^2}+\frac {35 c^4 x \left (a+b \sinh ^{-1}(c x)\right )^2}{8 d^3 \left (1+c^2 x^2\right )}+\frac {35 c^3 \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{4 d^3}-\frac {17 b^2 c^3 \tan ^{-1}(c x)}{6 d^3}+\frac {38 b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{3 d^3}+\frac {19 b^2 c^3 \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{3 d^3}-\frac {35 i b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}+\frac {35 i b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}-\frac {19 b^2 c^3 \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{3 d^3}+\frac {35 i b^2 c^3 \text {Li}_3\left (-i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}-\frac {35 i b^2 c^3 \text {Li}_3\left (i e^{\sinh ^{-1}(c x)}\right )}{4 d^3}\\ \end {align*}

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Mathematica [A]
time = 8.80, size = 937, normalized size = 1.77 \begin {gather*} -\frac {a^2}{3 d^3 x^3}+\frac {3 a^2 c^2}{d^3 x}+\frac {a^2 c^4 x}{4 d^3 \left (1+c^2 x^2\right )^2}+\frac {11 a^2 c^4 x}{8 d^3 \left (1+c^2 x^2\right )}+\frac {35 a^2 c^3 \text {ArcTan}(c x)}{8 d^3}+\frac {2 a b \left (-\frac {c \sqrt {1+c^2 x^2}}{6 x^2}+\frac {i c^3 \left ((2 i-c x) \sqrt {1+c^2 x^2}-3 \sinh ^{-1}(c x)\right )}{48 (-i+c x)^2}-\frac {11 c^3 \left (\sqrt {1+c^2 x^2}+i \sinh ^{-1}(c x)\right )}{16 (-1-i c x)}-\frac {\sinh ^{-1}(c x)}{3 x^3}+\frac {11 c^4 \left (i \sqrt {1+c^2 x^2}+\sinh ^{-1}(c x)\right )}{16 \left (i c+c^2 x\right )}+\frac {i c^3 \left ((2 i+c x) \sqrt {1+c^2 x^2}+3 \sinh ^{-1}(c x)\right )}{48 (i+c x)^2}+\frac {1}{6} c^3 \tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right )-3 c^2 \left (-\frac {\sinh ^{-1}(c x)}{x}-c \tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right )\right )-\frac {35}{16} i c^4 \left (-\frac {\sinh ^{-1}(c x)^2}{2 c}+\frac {2 \sinh ^{-1}(c x) \log \left (1+i e^{\sinh ^{-1}(c x)}\right )}{c}+\frac {2 \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{c}\right )+\frac {35}{16} i c^4 \left (-\frac {\sinh ^{-1}(c x)^2}{2 c}+\frac {2 \sinh ^{-1}(c x) \log \left (1-i e^{\sinh ^{-1}(c x)}\right )}{c}+\frac {2 \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{c}\right )\right )}{d^3}+\frac {b^2 c^3 \left (-\frac {2 c x}{1+c^2 x^2}+\frac {4 \sinh ^{-1}(c x)}{\left (1+c^2 x^2\right )^{3/2}}+\frac {66 \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}}+\frac {6 c x \sinh ^{-1}(c x)^2}{\left (1+c^2 x^2\right )^2}+\frac {33 c x \sinh ^{-1}(c x)^2}{1+c^2 x^2}-136 \text {ArcTan}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )-4 \coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+38 \sinh ^{-1}(c x)^2 \coth \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-2 \sinh ^{-1}(c x) \text {csch}^2\left (\frac {1}{2} \sinh ^{-1}(c x)\right )-\frac {1}{2} c x \sinh ^{-1}(c x)^2 \text {csch}^4\left (\frac {1}{2} \sinh ^{-1}(c x)\right )-152 \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-105 i \sinh ^{-1}(c x)^2 \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )+105 i \sinh ^{-1}(c x)^2 \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )+152 \sinh ^{-1}(c x) \log \left (1+e^{-\sinh ^{-1}(c x)}\right )-152 \text {PolyLog}\left (2,-e^{-\sinh ^{-1}(c x)}\right )-210 i \sinh ^{-1}(c x) \text {PolyLog}\left (2,-i e^{-\sinh ^{-1}(c x)}\right )+210 i \sinh ^{-1}(c x) \text {PolyLog}\left (2,i e^{-\sinh ^{-1}(c x)}\right )+152 \text {PolyLog}\left (2,e^{-\sinh ^{-1}(c x)}\right )-210 i \text {PolyLog}\left (3,-i e^{-\sinh ^{-1}(c x)}\right )+210 i \text {PolyLog}\left (3,i e^{-\sinh ^{-1}(c x)}\right )-2 \sinh ^{-1}(c x) \text {sech}^2\left (\frac {1}{2} \sinh ^{-1}(c x)\right )-\frac {8 \sinh ^{-1}(c x)^2 \sinh ^4\left (\frac {1}{2} \sinh ^{-1}(c x)\right )}{c^3 x^3}+4 \tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-38 \sinh ^{-1}(c x)^2 \tanh \left (\frac {1}{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^4*(d + c^2*d*x^2)^3),x]

[Out]

-1/3*a^2/(d^3*x^3) + (3*a^2*c^2)/(d^3*x) + (a^2*c^4*x)/(4*d^3*(1 + c^2*x^2)^2) + (11*a^2*c^4*x)/(8*d^3*(1 + c^
2*x^2)) + (35*a^2*c^3*ArcTan[c*x])/(8*d^3) + (2*a*b*(-1/6*(c*Sqrt[1 + c^2*x^2])/x^2 + ((I/48)*c^3*((2*I - c*x)
*Sqrt[1 + c^2*x^2] - 3*ArcSinh[c*x]))/(-I + c*x)^2 - (11*c^3*(Sqrt[1 + c^2*x^2] + I*ArcSinh[c*x]))/(16*(-1 - I
*c*x)) - ArcSinh[c*x]/(3*x^3) + (11*c^4*(I*Sqrt[1 + c^2*x^2] + ArcSinh[c*x]))/(16*(I*c + c^2*x)) + ((I/48)*c^3
*((2*I + c*x)*Sqrt[1 + c^2*x^2] + 3*ArcSinh[c*x]))/(I + c*x)^2 + (c^3*ArcTanh[Sqrt[1 + c^2*x^2]])/6 - 3*c^2*(-
(ArcSinh[c*x]/x) - c*ArcTanh[Sqrt[1 + c^2*x^2]]) - ((35*I)/16)*c^4*(-1/2*ArcSinh[c*x]^2/c + (2*ArcSinh[c*x]*Lo
g[1 + I*E^ArcSinh[c*x]])/c + (2*PolyLog[2, (-I)*E^ArcSinh[c*x]])/c) + ((35*I)/16)*c^4*(-1/2*ArcSinh[c*x]^2/c +
(2*ArcSinh[c*x]*Log[1 - I*E^ArcSinh[c*x]])/c + (2*PolyLog[2, I*E^ArcSinh[c*x]])/c)))/d^3 + (b^2*c^3*((-2*c*x)
/(1 + c^2*x^2) + (4*ArcSinh[c*x])/(1 + c^2*x^2)^(3/2) + (66*ArcSinh[c*x])/Sqrt[1 + c^2*x^2] + (6*c*x*ArcSinh[c
*x]^2)/(1 + c^2*x^2)^2 + (33*c*x*ArcSinh[c*x]^2)/(1 + c^2*x^2) - 136*ArcTan[Tanh[ArcSinh[c*x]/2]] - 4*Coth[Arc
Sinh[c*x]/2] + 38*ArcSinh[c*x]^2*Coth[ArcSinh[c*x]/2] - 2*ArcSinh[c*x]*Csch[ArcSinh[c*x]/2]^2 - (c*x*ArcSinh[c
*x]^2*Csch[ArcSinh[c*x]/2]^4)/2 - 152*ArcSinh[c*x]*Log[1 - E^(-ArcSinh[c*x])] - (105*I)*ArcSinh[c*x]^2*Log[1 -
I/E^ArcSinh[c*x]] + (105*I)*ArcSinh[c*x]^2*Log[1 + I/E^ArcSinh[c*x]] + 152*ArcSinh[c*x]*Log[1 + E^(-ArcSinh[c
*x])] - 152*PolyLog[2, -E^(-ArcSinh[c*x])] - (210*I)*ArcSinh[c*x]*PolyLog[2, (-I)/E^ArcSinh[c*x]] + (210*I)*Ar
cSinh[c*x]*PolyLog[2, I/E^ArcSinh[c*x]] + 152*PolyLog[2, E^(-ArcSinh[c*x])] - (210*I)*PolyLog[3, (-I)/E^ArcSin
h[c*x]] + (210*I)*PolyLog[3, I/E^ArcSinh[c*x]] - 2*ArcSinh[c*x]*Sech[ArcSinh[c*x]/2]^2 - (8*ArcSinh[c*x]^2*Sin
h[ArcSinh[c*x]/2]^4)/(c^3*x^3) + 4*Tanh[ArcSinh[c*x]/2] - 38*ArcSinh[c*x]^2*Tanh[ArcSinh[c*x]/2]))/(24*d^3)

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Maple [F]
time = 0.10, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \arcsinh \left (c x \right )\right )^{2}}{x^{4} \left (c^{2} d \,x^{2}+d \right )^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^4/(c^2*d*x^2+d)^3,x, algorithm="maxima")

[Out]

1/24*a^2*(105*c^3*arctan(c*x)/d^3 + (105*c^6*x^6 + 175*c^4*x^4 + 56*c^2*x^2 - 8)/(c^4*d^3*x^7 + 2*c^2*d^3*x^5
+ d^3*x^3)) + integrate(b^2*log(c*x + sqrt(c^2*x^2 + 1))^2/(c^6*d^3*x^10 + 3*c^4*d^3*x^8 + 3*c^2*d^3*x^6 + d^3
*x^4) + 2*a*b*log(c*x + sqrt(c^2*x^2 + 1))/(c^6*d^3*x^10 + 3*c^4*d^3*x^8 + 3*c^2*d^3*x^6 + d^3*x^4), x)

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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^4/(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^10 + 3*c^4*d^3*x^8 + 3*c^2*d^3*x^6 + d^3*x
^4), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a^{2}}{c^{6} x^{10} + 3 c^{4} x^{8} + 3 c^{2} x^{6} + x^{4}}\, dx + \int \frac {b^{2} \operatorname {asinh}^{2}{\left (c x \right )}}{c^{6} x^{10} + 3 c^{4} x^{8} + 3 c^{2} x^{6} + x^{4}}\, dx + \int \frac {2 a b \operatorname {asinh}{\left (c x \right )}}{c^{6} x^{10} + 3 c^{4} x^{8} + 3 c^{2} x^{6} + x^{4}}\, dx}{d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a**2/(c**6*x**10 + 3*c**4*x**8 + 3*c**2*x**6 + x**4), x) + Integral(b**2*asinh(c*x)**2/(c**6*x**10 +
3*c**4*x**8 + 3*c**2*x**6 + x**4), x) + Integral(2*a*b*asinh(c*x)/(c**6*x**10 + 3*c**4*x**8 + 3*c**2*x**6 + x
**4), x))/d**3

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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^4/(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^4), x)

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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^4\,{\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^4*(d + c^2*d*x^2)^3),x)

[Out]

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

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