3.3.42 \(\int \frac {(a+b \sinh ^{-1}(c x))^2}{x^4 (d+c^2 d x^2)^2} \, dx\) [242]

Optimal. Leaf size=401 \[ -\frac {b^2 c^2}{3 d^2 x}+\frac {2 b c^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt {1+c^2 x^2}}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 x^2 \sqrt {1+c^2 x^2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^2 x^3 \left (1+c^2 x^2\right )}+\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^2 x \left (1+c^2 x^2\right )}+\frac {5 c^4 x \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^2 \left (1+c^2 x^2\right )}+\frac {5 c^3 \left (a+b \sinh ^{-1}(c x)\right )^2 \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right )}{d^2}-\frac {b^2 c^3 \text {ArcTan}(c x)}{d^2}+\frac {26 b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{3 d^2}+\frac {13 b^2 c^3 \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right )}{3 d^2}-\frac {5 i b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{d^2}+\frac {5 i b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{d^2}-\frac {13 b^2 c^3 \text {PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right )}{3 d^2}+\frac {5 i b^2 c^3 \text {PolyLog}\left (3,-i e^{\sinh ^{-1}(c x)}\right )}{d^2}-\frac {5 i b^2 c^3 \text {PolyLog}\left (3,i e^{\sinh ^{-1}(c x)}\right )}{d^2} \]

[Out]

-1/3*b^2*c^2/d^2/x-1/3*(a+b*arcsinh(c*x))^2/d^2/x^3/(c^2*x^2+1)+5/3*c^2*(a+b*arcsinh(c*x))^2/d^2/x/(c^2*x^2+1)
+5/2*c^4*x*(a+b*arcsinh(c*x))^2/d^2/(c^2*x^2+1)+5*c^3*(a+b*arcsinh(c*x))^2*arctan(c*x+(c^2*x^2+1)^(1/2))/d^2-b
^2*c^3*arctan(c*x)/d^2+26/3*b*c^3*(a+b*arcsinh(c*x))*arctanh(c*x+(c^2*x^2+1)^(1/2))/d^2+13/3*b^2*c^3*polylog(2
,-c*x-(c^2*x^2+1)^(1/2))/d^2-5*I*b*c^3*(a+b*arcsinh(c*x))*polylog(2,-I*(c*x+(c^2*x^2+1)^(1/2)))/d^2+5*I*b*c^3*
(a+b*arcsinh(c*x))*polylog(2,I*(c*x+(c^2*x^2+1)^(1/2)))/d^2-13/3*b^2*c^3*polylog(2,c*x+(c^2*x^2+1)^(1/2))/d^2+
5*I*b^2*c^3*polylog(3,-I*(c*x+(c^2*x^2+1)^(1/2)))/d^2-5*I*b^2*c^3*polylog(3,I*(c*x+(c^2*x^2+1)^(1/2)))/d^2+2/3
*b*c^3*(a+b*arcsinh(c*x))/d^2/(c^2*x^2+1)^(1/2)-1/3*b*c*(a+b*arcsinh(c*x))/d^2/x^2/(c^2*x^2+1)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.67, antiderivative size = 401, normalized size of antiderivative = 1.00, number of steps used = 32, number of rules used = 15, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.577, Rules used = {5809, 5788, 5789, 4265, 2611, 2320, 6724, 5798, 209, 5811, 5816, 4267, 2317, 2438, 331} \begin {gather*} \frac {5 c^3 \text {ArcTan}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{d^2}-\frac {5 i b c^3 \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^2}+\frac {5 i b c^3 \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d^2}+\frac {26 b c^3 \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2}+\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^2 x \left (c^2 x^2+1\right )}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 x^2 \sqrt {c^2 x^2+1}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^2 x^3 \left (c^2 x^2+1\right )}+\frac {5 c^4 x \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^2 \left (c^2 x^2+1\right )}+\frac {2 b c^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt {c^2 x^2+1}}-\frac {b^2 c^3 \text {ArcTan}(c x)}{d^2}+\frac {13 b^2 c^3 \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{3 d^2}-\frac {13 b^2 c^3 \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{3 d^2}+\frac {5 i b^2 c^3 \text {Li}_3\left (-i e^{\sinh ^{-1}(c x)}\right )}{d^2}-\frac {5 i b^2 c^3 \text {Li}_3\left (i e^{\sinh ^{-1}(c x)}\right )}{d^2}-\frac {b^2 c^2}{3 d^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(b^2*c^2)/(d^2*x) + (2*b*c^3*(a + b*ArcSinh[c*x]))/(3*d^2*Sqrt[1 + c^2*x^2]) - (b*c*(a + b*ArcSinh[c*x]))
/(3*d^2*x^2*Sqrt[1 + c^2*x^2]) - (a + b*ArcSinh[c*x])^2/(3*d^2*x^3*(1 + c^2*x^2)) + (5*c^2*(a + b*ArcSinh[c*x]
)^2)/(3*d^2*x*(1 + c^2*x^2)) + (5*c^4*x*(a + b*ArcSinh[c*x])^2)/(2*d^2*(1 + c^2*x^2)) + (5*c^3*(a + b*ArcSinh[
c*x])^2*ArcTan[E^ArcSinh[c*x]])/d^2 - (b^2*c^3*ArcTan[c*x])/d^2 + (26*b*c^3*(a + b*ArcSinh[c*x])*ArcTanh[E^Arc
Sinh[c*x]])/(3*d^2) + (13*b^2*c^3*PolyLog[2, -E^ArcSinh[c*x]])/(3*d^2) - ((5*I)*b*c^3*(a + b*ArcSinh[c*x])*Pol
yLog[2, (-I)*E^ArcSinh[c*x]])/d^2 + ((5*I)*b*c^3*(a + b*ArcSinh[c*x])*PolyLog[2, I*E^ArcSinh[c*x]])/d^2 - (13*
b^2*c^3*PolyLog[2, E^ArcSinh[c*x]])/(3*d^2) + ((5*I)*b^2*c^3*PolyLog[3, (-I)*E^ArcSinh[c*x]])/d^2 - ((5*I)*b^2
*c^3*PolyLog[3, I*E^ArcSinh[c*x]])/d^2

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 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 )^2} \, dx &=-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^2 x^3 \left (1+c^2 x^2\right )}-\frac {1}{3} \left (5 c^2\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{x^2 \left (d+c^2 d x^2\right )^2} \, dx+\frac {(2 b c) \int \frac {a+b \sinh ^{-1}(c x)}{x^3 \left (1+c^2 x^2\right )^{3/2}} \, dx}{3 d^2}\\ &=-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 x^2 \sqrt {1+c^2 x^2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^2 x^3 \left (1+c^2 x^2\right )}+\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^2 x \left (1+c^2 x^2\right )}+\left (5 c^4\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{\left (d+c^2 d x^2\right )^2} \, dx+\frac {\left (b^2 c^2\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx}{3 d^2}-\frac {\left (b c^3\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \left (1+c^2 x^2\right )^{3/2}} \, dx}{d^2}-\frac {\left (10 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^2}\\ &=-\frac {b^2 c^2}{3 d^2 x}-\frac {13 b c^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt {1+c^2 x^2}}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 x^2 \sqrt {1+c^2 x^2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^2 x^3 \left (1+c^2 x^2\right )}+\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^2 x \left (1+c^2 x^2\right )}+\frac {5 c^4 x \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^2 \left (1+c^2 x^2\right )}-\frac {\left (b c^3\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \sqrt {1+c^2 x^2}} \, dx}{d^2}-\frac {\left (10 b c^3\right ) \int \frac {a+b \sinh ^{-1}(c x)}{x \sqrt {1+c^2 x^2}} \, dx}{3 d^2}-\frac {\left (b^2 c^4\right ) \int \frac {1}{1+c^2 x^2} \, dx}{3 d^2}+\frac {\left (b^2 c^4\right ) \int \frac {1}{1+c^2 x^2} \, dx}{d^2}+\frac {\left (10 b^2 c^4\right ) \int \frac {1}{1+c^2 x^2} \, dx}{3 d^2}-\frac {\left (5 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}{d^2}+\frac {\left (5 c^4\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d+c^2 d x^2} \, dx}{2 d}\\ &=-\frac {b^2 c^2}{3 d^2 x}+\frac {2 b c^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt {1+c^2 x^2}}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 x^2 \sqrt {1+c^2 x^2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^2 x^3 \left (1+c^2 x^2\right )}+\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^2 x \left (1+c^2 x^2\right )}+\frac {5 c^4 x \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^2 \left (1+c^2 x^2\right )}+\frac {4 b^2 c^3 \tan ^{-1}(c x)}{d^2}+\frac {\left (5 c^3\right ) \text {Subst}\left (\int (a+b x)^2 \text {sech}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{2 d^2}-\frac {\left (b c^3\right ) \text {Subst}\left (\int (a+b x) \text {csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{d^2}-\frac {\left (10 b c^3\right ) \text {Subst}\left (\int (a+b x) \text {csch}(x) \, dx,x,\sinh ^{-1}(c x)\right )}{3 d^2}-\frac {\left (5 b^2 c^4\right ) \int \frac {1}{1+c^2 x^2} \, dx}{d^2}\\ &=-\frac {b^2 c^2}{3 d^2 x}+\frac {2 b c^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt {1+c^2 x^2}}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 x^2 \sqrt {1+c^2 x^2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^2 x^3 \left (1+c^2 x^2\right )}+\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^2 x \left (1+c^2 x^2\right )}+\frac {5 c^4 x \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^2 \left (1+c^2 x^2\right )}+\frac {5 c^3 \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d^2}-\frac {b^2 c^3 \tan ^{-1}(c x)}{d^2}+\frac {26 b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{3 d^2}-\frac {\left (5 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 )}{d^2}+\frac {\left (5 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 )}{d^2}+\frac {\left (b^2 c^3\right ) \text {Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^2}-\frac {\left (b^2 c^3\right ) \text {Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d^2}+\frac {\left (10 b^2 c^3\right ) \text {Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{3 d^2}-\frac {\left (10 b^2 c^3\right ) \text {Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{3 d^2}\\ &=-\frac {b^2 c^2}{3 d^2 x}+\frac {2 b c^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt {1+c^2 x^2}}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 x^2 \sqrt {1+c^2 x^2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^2 x^3 \left (1+c^2 x^2\right )}+\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^2 x \left (1+c^2 x^2\right )}+\frac {5 c^4 x \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^2 \left (1+c^2 x^2\right )}+\frac {5 c^3 \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d^2}-\frac {b^2 c^3 \tan ^{-1}(c x)}{d^2}+\frac {26 b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{3 d^2}-\frac {5 i b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{d^2}+\frac {5 i b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{d^2}+\frac {\left (5 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 )}{d^2}-\frac {\left (5 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 )}{d^2}+\frac {\left (b^2 c^3\right ) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{d^2}-\frac {\left (b^2 c^3\right ) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{d^2}+\frac {\left (10 b^2 c^3\right ) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{3 d^2}-\frac {\left (10 b^2 c^3\right ) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{\sinh ^{-1}(c x)}\right )}{3 d^2}\\ &=-\frac {b^2 c^2}{3 d^2 x}+\frac {2 b c^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt {1+c^2 x^2}}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 x^2 \sqrt {1+c^2 x^2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^2 x^3 \left (1+c^2 x^2\right )}+\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^2 x \left (1+c^2 x^2\right )}+\frac {5 c^4 x \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^2 \left (1+c^2 x^2\right )}+\frac {5 c^3 \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d^2}-\frac {b^2 c^3 \tan ^{-1}(c x)}{d^2}+\frac {26 b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{3 d^2}+\frac {13 b^2 c^3 \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{3 d^2}-\frac {5 i b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{d^2}+\frac {5 i b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{d^2}-\frac {13 b^2 c^3 \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{3 d^2}+\frac {\left (5 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 )}{d^2}-\frac {\left (5 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 )}{d^2}\\ &=-\frac {b^2 c^2}{3 d^2 x}+\frac {2 b c^3 \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 \sqrt {1+c^2 x^2}}-\frac {b c \left (a+b \sinh ^{-1}(c x)\right )}{3 d^2 x^2 \sqrt {1+c^2 x^2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^2 x^3 \left (1+c^2 x^2\right )}+\frac {5 c^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{3 d^2 x \left (1+c^2 x^2\right )}+\frac {5 c^4 x \left (a+b \sinh ^{-1}(c x)\right )^2}{2 d^2 \left (1+c^2 x^2\right )}+\frac {5 c^3 \left (a+b \sinh ^{-1}(c x)\right )^2 \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{d^2}-\frac {b^2 c^3 \tan ^{-1}(c x)}{d^2}+\frac {26 b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right )}{3 d^2}+\frac {13 b^2 c^3 \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right )}{3 d^2}-\frac {5 i b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{d^2}+\frac {5 i b c^3 \left (a+b \sinh ^{-1}(c x)\right ) \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{d^2}-\frac {13 b^2 c^3 \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right )}{3 d^2}+\frac {5 i b^2 c^3 \text {Li}_3\left (-i e^{\sinh ^{-1}(c x)}\right )}{d^2}-\frac {5 i b^2 c^3 \text {Li}_3\left (i e^{\sinh ^{-1}(c x)}\right )}{d^2}\\ \end {align*}

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Mathematica [A]
time = 7.94, size = 764, normalized size = 1.91 \begin {gather*} -\frac {a^2}{3 d^2 x^3}+\frac {2 a^2 c^2}{d^2 x}+\frac {a^2 c^4 x}{2 d^2 \left (1+c^2 x^2\right )}+\frac {5 a^2 c^3 \text {ArcTan}(c x)}{2 d^2}+\frac {2 a b \left (-\frac {c \sqrt {1+c^2 x^2}}{6 x^2}-\frac {c^3 \left (\sqrt {1+c^2 x^2}+i \sinh ^{-1}(c x)\right )}{4 (-1-i c x)}-\frac {\sinh ^{-1}(c x)}{3 x^3}+\frac {c^4 \left (i \sqrt {1+c^2 x^2}+\sinh ^{-1}(c x)\right )}{4 \left (i c+c^2 x\right )}+\frac {1}{6} c^3 \tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right )-2 c^2 \left (-\frac {\sinh ^{-1}(c x)}{x}-c \tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right )\right )-\frac {5}{4} 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 {5}{4} 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^2}+\frac {b^2 c^3 \left (\frac {24 \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}}+\frac {12 c x \sinh ^{-1}(c x)^2}{1+c^2 x^2}-48 \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 )+26 \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 )-104 \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-60 i \sinh ^{-1}(c x)^2 \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )+60 i \sinh ^{-1}(c x)^2 \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )+104 \sinh ^{-1}(c x) \log \left (1+e^{-\sinh ^{-1}(c x)}\right )-104 \text {PolyLog}\left (2,-e^{-\sinh ^{-1}(c x)}\right )-120 i \sinh ^{-1}(c x) \text {PolyLog}\left (2,-i e^{-\sinh ^{-1}(c x)}\right )+120 i \sinh ^{-1}(c x) \text {PolyLog}\left (2,i e^{-\sinh ^{-1}(c x)}\right )+104 \text {PolyLog}\left (2,e^{-\sinh ^{-1}(c x)}\right )-120 i \text {PolyLog}\left (3,-i e^{-\sinh ^{-1}(c x)}\right )+120 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 )-26 \sinh ^{-1}(c x)^2 \tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )}{24 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*a^2/(d^2*x^3) + (2*a^2*c^2)/(d^2*x) + (a^2*c^4*x)/(2*d^2*(1 + c^2*x^2)) + (5*a^2*c^3*ArcTan[c*x])/(2*d^2)
+ (2*a*b*(-1/6*(c*Sqrt[1 + c^2*x^2])/x^2 - (c^3*(Sqrt[1 + c^2*x^2] + I*ArcSinh[c*x]))/(4*(-1 - I*c*x)) - ArcS
inh[c*x]/(3*x^3) + (c^4*(I*Sqrt[1 + c^2*x^2] + ArcSinh[c*x]))/(4*(I*c + c^2*x)) + (c^3*ArcTanh[Sqrt[1 + c^2*x^
2]])/6 - 2*c^2*(-(ArcSinh[c*x]/x) - c*ArcTanh[Sqrt[1 + c^2*x^2]]) - ((5*I)/4)*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) + ((5*I)/4)*c^4*(-1/2*ArcSi
nh[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^2 + (b^2*
c^3*((24*ArcSinh[c*x])/Sqrt[1 + c^2*x^2] + (12*c*x*ArcSinh[c*x]^2)/(1 + c^2*x^2) - 48*ArcTan[Tanh[ArcSinh[c*x]
/2]] - 4*Coth[ArcSinh[c*x]/2] + 26*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 - 104*ArcSinh[c*x]*Log[1 - E^(-ArcSinh[c*x])] - (60*I)*ArcSin
h[c*x]^2*Log[1 - I/E^ArcSinh[c*x]] + (60*I)*ArcSinh[c*x]^2*Log[1 + I/E^ArcSinh[c*x]] + 104*ArcSinh[c*x]*Log[1
+ E^(-ArcSinh[c*x])] - 104*PolyLog[2, -E^(-ArcSinh[c*x])] - (120*I)*ArcSinh[c*x]*PolyLog[2, (-I)/E^ArcSinh[c*x
]] + (120*I)*ArcSinh[c*x]*PolyLog[2, I/E^ArcSinh[c*x]] + 104*PolyLog[2, E^(-ArcSinh[c*x])] - (120*I)*PolyLog[3
, (-I)/E^ArcSinh[c*x]] + (120*I)*PolyLog[3, I/E^ArcSinh[c*x]] - 2*ArcSinh[c*x]*Sech[ArcSinh[c*x]/2]^2 - (8*Arc
Sinh[c*x]^2*Sinh[ArcSinh[c*x]/2]^4)/(c^3*x^3) + 4*Tanh[ArcSinh[c*x]/2] - 26*ArcSinh[c*x]^2*Tanh[ArcSinh[c*x]/2
]))/(24*d^2)

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Maple [F]
time = 0.11, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \arcsinh \left (c x \right )\right )^{2}}{x^{4} \left (c^{2} d \,x^{2}+d \right )^{2}}\, dx\]

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)^2,x)

[Out]

int((a+b*arcsinh(c*x))^2/x^4/(c^2*d*x^2+d)^2,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)^2,x, algorithm="maxima")

[Out]

1/6*(15*c^3*arctan(c*x)/d^2 + (15*c^4*x^4 + 10*c^2*x^2 - 2)/(c^2*d^2*x^5 + d^2*x^3))*a^2 + integrate(b^2*log(c
*x + sqrt(c^2*x^2 + 1))^2/(c^4*d^2*x^8 + 2*c^2*d^2*x^6 + d^2*x^4) + 2*a*b*log(c*x + sqrt(c^2*x^2 + 1))/(c^4*d^
2*x^8 + 2*c^2*d^2*x^6 + d^2*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)^2,x, algorithm="fricas")

[Out]

integral((b^2*arcsinh(c*x)^2 + 2*a*b*arcsinh(c*x) + a^2)/(c^4*d^2*x^8 + 2*c^2*d^2*x^6 + d^2*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^{4} x^{8} + 2 c^{2} x^{6} + x^{4}}\, dx + \int \frac {b^{2} \operatorname {asinh}^{2}{\left (c x \right )}}{c^{4} x^{8} + 2 c^{2} x^{6} + x^{4}}\, dx + \int \frac {2 a b \operatorname {asinh}{\left (c x \right )}}{c^{4} x^{8} + 2 c^{2} x^{6} + x^{4}}\, dx}{d^{2}} \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)**2,x)

[Out]

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

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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)^2,x, algorithm="giac")

[Out]

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

[Out]

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

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