3.306 \(\int \frac {(a+b \sinh ^{-1}(c x))^2}{x (d+c^2 d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=412 \[ -\frac {2 b \sqrt {c^2 x^2+1} \text {Li}_2\left (-e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d \sqrt {c^2 d x^2+d}}+\frac {2 b \sqrt {c^2 x^2+1} \text {Li}_2\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d \sqrt {c^2 d x^2+d}}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d \sqrt {c^2 d x^2+d}}-\frac {4 b \sqrt {c^2 x^2+1} \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d \sqrt {c^2 d x^2+d}}-\frac {2 \sqrt {c^2 x^2+1} \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{d \sqrt {c^2 d x^2+d}}+\frac {2 i b^2 \sqrt {c^2 x^2+1} \text {Li}_2\left (-i e^{\sinh ^{-1}(c x)}\right )}{d \sqrt {c^2 d x^2+d}}-\frac {2 i b^2 \sqrt {c^2 x^2+1} \text {Li}_2\left (i e^{\sinh ^{-1}(c x)}\right )}{d \sqrt {c^2 d x^2+d}}+\frac {2 b^2 \sqrt {c^2 x^2+1} \text {Li}_3\left (-e^{\sinh ^{-1}(c x)}\right )}{d \sqrt {c^2 d x^2+d}}-\frac {2 b^2 \sqrt {c^2 x^2+1} \text {Li}_3\left (e^{\sinh ^{-1}(c x)}\right )}{d \sqrt {c^2 d x^2+d}} \]

[Out]

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

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Rubi [A]  time = 0.59, antiderivative size = 412, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 11, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.393, Rules used = {5755, 5764, 5760, 4182, 2531, 2282, 6589, 5693, 4180, 2279, 2391} \[ -\frac {2 b \sqrt {c^2 x^2+1} \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d \sqrt {c^2 d x^2+d}}+\frac {2 b \sqrt {c^2 x^2+1} \text {PolyLog}\left (2,e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d \sqrt {c^2 d x^2+d}}+\frac {2 i b^2 \sqrt {c^2 x^2+1} \text {PolyLog}\left (2,-i e^{\sinh ^{-1}(c x)}\right )}{d \sqrt {c^2 d x^2+d}}-\frac {2 i b^2 \sqrt {c^2 x^2+1} \text {PolyLog}\left (2,i e^{\sinh ^{-1}(c x)}\right )}{d \sqrt {c^2 d x^2+d}}+\frac {2 b^2 \sqrt {c^2 x^2+1} \text {PolyLog}\left (3,-e^{\sinh ^{-1}(c x)}\right )}{d \sqrt {c^2 d x^2+d}}-\frac {2 b^2 \sqrt {c^2 x^2+1} \text {PolyLog}\left (3,e^{\sinh ^{-1}(c x)}\right )}{d \sqrt {c^2 d x^2+d}}+\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d \sqrt {c^2 d x^2+d}}-\frac {4 b \sqrt {c^2 x^2+1} \tan ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d \sqrt {c^2 d x^2+d}}-\frac {2 \sqrt {c^2 x^2+1} \tanh ^{-1}\left (e^{\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{d \sqrt {c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a + b*ArcSinh[c*x])^2/(d*Sqrt[d + c^2*d*x^2]) - (4*b*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])*ArcTan[E^ArcSinh[
c*x]])/(d*Sqrt[d + c^2*d*x^2]) - (2*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^2*ArcTanh[E^ArcSinh[c*x]])/(d*Sqrt[
d + c^2*d*x^2]) - (2*b*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])*PolyLog[2, -E^ArcSinh[c*x]])/(d*Sqrt[d + c^2*d*x
^2]) + ((2*I)*b^2*Sqrt[1 + c^2*x^2]*PolyLog[2, (-I)*E^ArcSinh[c*x]])/(d*Sqrt[d + c^2*d*x^2]) - ((2*I)*b^2*Sqrt
[1 + c^2*x^2]*PolyLog[2, I*E^ArcSinh[c*x]])/(d*Sqrt[d + c^2*d*x^2]) + (2*b*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*
x])*PolyLog[2, E^ArcSinh[c*x]])/(d*Sqrt[d + c^2*d*x^2]) + (2*b^2*Sqrt[1 + c^2*x^2]*PolyLog[3, -E^ArcSinh[c*x]]
)/(d*Sqrt[d + c^2*d*x^2]) - (2*b^2*Sqrt[1 + c^2*x^2]*PolyLog[3, E^ArcSinh[c*x]])/(d*Sqrt[d + c^2*d*x^2])

Rule 2279

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 2282

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 2391

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 2531

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 4180

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 4182

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 5693

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 5755

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*d^IntPart[p]*(d + e*x^2)^F
racPart[p])/(2*f*(p + 1)*(1 + c^2*x^2)^FracPart[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 5760

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

Rule 5764

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist
[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2], Int[((f*x)^m*(a + b*ArcSinh[c*x])^n)/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a
, b, c, d, e, f, m}, x] && EqQ[e, c^2*d] && GtQ[n, 0] &&  !GtQ[d, 0] && (IntegerQ[m] || EqQ[n, 1])

Rule 6589

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

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Mathematica [A]  time = 1.54, size = 568, normalized size = 1.38 \[ \frac {a^2 \sqrt {d} \sqrt {c^2 d x^2+d} \log (c x)-a^2 \sqrt {d} \sqrt {c^2 d x^2+d} \log \left (\sqrt {d} \sqrt {c^2 d x^2+d}+d\right )+a^2 d+2 a b d \left (\sqrt {c^2 x^2+1} \text {Li}_2\left (-e^{-\sinh ^{-1}(c x)}\right )-\sqrt {c^2 x^2+1} \text {Li}_2\left (e^{-\sinh ^{-1}(c x)}\right )+\sqrt {c^2 x^2+1} \sinh ^{-1}(c x) \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-\sqrt {c^2 x^2+1} \sinh ^{-1}(c x) \log \left (e^{-\sinh ^{-1}(c x)}+1\right )-2 \sqrt {c^2 x^2+1} \tan ^{-1}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+\sinh ^{-1}(c x)\right )+b^2 d \left (2 \sqrt {c^2 x^2+1} \sinh ^{-1}(c x) \text {Li}_2\left (-e^{-\sinh ^{-1}(c x)}\right )-2 \sqrt {c^2 x^2+1} \sinh ^{-1}(c x) \text {Li}_2\left (e^{-\sinh ^{-1}(c x)}\right )+2 i \sqrt {c^2 x^2+1} \text {Li}_2\left (-i e^{-\sinh ^{-1}(c x)}\right )-2 i \sqrt {c^2 x^2+1} \text {Li}_2\left (i e^{-\sinh ^{-1}(c x)}\right )+2 \sqrt {c^2 x^2+1} \text {Li}_3\left (-e^{-\sinh ^{-1}(c x)}\right )-2 \sqrt {c^2 x^2+1} \text {Li}_3\left (e^{-\sinh ^{-1}(c x)}\right )+\sqrt {c^2 x^2+1} \sinh ^{-1}(c x)^2 \log \left (1-e^{-\sinh ^{-1}(c x)}\right )-\sqrt {c^2 x^2+1} \sinh ^{-1}(c x)^2 \log \left (e^{-\sinh ^{-1}(c x)}+1\right )+2 i \sqrt {c^2 x^2+1} \sinh ^{-1}(c x) \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )-2 i \sqrt {c^2 x^2+1} \sinh ^{-1}(c x) \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )+\sinh ^{-1}(c x)^2\right )}{d^2 \sqrt {c^2 d x^2+d}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^2*d + a^2*Sqrt[d]*Sqrt[d + c^2*d*x^2]*Log[c*x] - a^2*Sqrt[d]*Sqrt[d + c^2*d*x^2]*Log[d + Sqrt[d]*Sqrt[d + c
^2*d*x^2]] + 2*a*b*d*(ArcSinh[c*x] - 2*Sqrt[1 + c^2*x^2]*ArcTan[Tanh[ArcSinh[c*x]/2]] + Sqrt[1 + c^2*x^2]*ArcS
inh[c*x]*Log[1 - E^(-ArcSinh[c*x])] - Sqrt[1 + c^2*x^2]*ArcSinh[c*x]*Log[1 + E^(-ArcSinh[c*x])] + Sqrt[1 + c^2
*x^2]*PolyLog[2, -E^(-ArcSinh[c*x])] - Sqrt[1 + c^2*x^2]*PolyLog[2, E^(-ArcSinh[c*x])]) + b^2*d*(ArcSinh[c*x]^
2 + Sqrt[1 + c^2*x^2]*ArcSinh[c*x]^2*Log[1 - E^(-ArcSinh[c*x])] + (2*I)*Sqrt[1 + c^2*x^2]*ArcSinh[c*x]*Log[1 -
I/E^ArcSinh[c*x]] - (2*I)*Sqrt[1 + c^2*x^2]*ArcSinh[c*x]*Log[1 + I/E^ArcSinh[c*x]] - Sqrt[1 + c^2*x^2]*ArcSin
h[c*x]^2*Log[1 + E^(-ArcSinh[c*x])] + 2*Sqrt[1 + c^2*x^2]*ArcSinh[c*x]*PolyLog[2, -E^(-ArcSinh[c*x])] + (2*I)*
Sqrt[1 + c^2*x^2]*PolyLog[2, (-I)/E^ArcSinh[c*x]] - (2*I)*Sqrt[1 + c^2*x^2]*PolyLog[2, I/E^ArcSinh[c*x]] - 2*S
qrt[1 + c^2*x^2]*ArcSinh[c*x]*PolyLog[2, E^(-ArcSinh[c*x])] + 2*Sqrt[1 + c^2*x^2]*PolyLog[3, -E^(-ArcSinh[c*x]
)] - 2*Sqrt[1 + c^2*x^2]*PolyLog[3, E^(-ArcSinh[c*x])]))/(d^2*Sqrt[d + c^2*d*x^2])

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {c^{2} d x^{2} + d} {\left (b^{2} \operatorname {arsinh}\left (c x\right )^{2} + 2 \, a b \operatorname {arsinh}\left (c x\right ) + a^{2}\right )}}{c^{4} d^{2} x^{5} + 2 \, c^{2} d^{2} x^{3} + d^{2} x}, x\right ) \]

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)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} x}\,{d x} \]

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)^(3/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.40, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arcsinh \left (c x \right )\right )^{2}}{x \left (c^{2} d \,x^{2}+d \right )^{\frac {3}{2}}}\, dx \]

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)^(3/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -a^{2} {\left (\frac {\operatorname {arsinh}\left (\frac {1}{c {\left | x \right |}}\right )}{d^{\frac {3}{2}}} - \frac {1}{\sqrt {c^{2} d x^{2} + d} d}\right )} + \int \frac {b^{2} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2}}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} x} + \frac {2 \, a b \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{{\left (c^{2} d x^{2} + d\right )}^{\frac {3}{2}} x}\,{d x} \]

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)^(3/2),x, algorithm="maxima")

[Out]

-a^2*(arcsinh(1/(c*abs(x)))/d^(3/2) - 1/(sqrt(c^2*d*x^2 + d)*d)) + integrate(b^2*log(c*x + sqrt(c^2*x^2 + 1))^
2/((c^2*d*x^2 + d)^(3/2)*x) + 2*a*b*log(c*x + sqrt(c^2*x^2 + 1))/((c^2*d*x^2 + d)^(3/2)*x), x)

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

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)^(3/2)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}}{x \left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

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)**(3/2),x)

[Out]

Integral((a + b*asinh(c*x))**2/(x*(d*(c**2*x**2 + 1))**(3/2)), x)

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