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

Optimal. Leaf size=349 \[ \frac{b^2 c^3 d \text{PolyLog}\left (2,-\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e \left (c^2 d^2+e^2\right )^{3/2}}-\frac{b^2 c^3 d \text{PolyLog}\left (2,-\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}\right )}{e \left (c^2 d^2+e^2\right )^{3/2}}-\frac{b c \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{\left (c^2 d^2+e^2\right ) (d+e x)}+\frac{b c^3 d \left (a+b \sinh ^{-1}(c x)\right ) \log \left (\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}+1\right )}{e \left (c^2 d^2+e^2\right )^{3/2}}-\frac{b c^3 d \left (a+b \sinh ^{-1}(c x)\right ) \log \left (\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}+1\right )}{e \left (c^2 d^2+e^2\right )^{3/2}}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 e (d+e x)^2}+\frac{b^2 c^2 \log (d+e x)}{e \left (c^2 d^2+e^2\right )} \]

[Out]

-((b*c*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/((c^2*d^2 + e^2)*(d + e*x))) - (a + b*ArcSinh[c*x])^2/(2*e*(d +
e*x)^2) + (b*c^3*d*(a + b*ArcSinh[c*x])*Log[1 + (e*E^ArcSinh[c*x])/(c*d - Sqrt[c^2*d^2 + e^2])])/(e*(c^2*d^2
+ e^2)^(3/2)) - (b*c^3*d*(a + b*ArcSinh[c*x])*Log[1 + (e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2])])/(e*(c^2
*d^2 + e^2)^(3/2)) + (b^2*c^2*Log[d + e*x])/(e*(c^2*d^2 + e^2)) + (b^2*c^3*d*PolyLog[2, -((e*E^ArcSinh[c*x])/(
c*d - Sqrt[c^2*d^2 + e^2]))])/(e*(c^2*d^2 + e^2)^(3/2)) - (b^2*c^3*d*PolyLog[2, -((e*E^ArcSinh[c*x])/(c*d + Sq
rt[c^2*d^2 + e^2]))])/(e*(c^2*d^2 + e^2)^(3/2))

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Rubi [A]  time = 0.60894, antiderivative size = 349, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 10, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.556, Rules used = {5801, 5831, 3324, 3322, 2264, 2190, 2279, 2391, 2668, 31} \[ \frac{b^2 c^3 d \text{PolyLog}\left (2,-\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}\right )}{e \left (c^2 d^2+e^2\right )^{3/2}}-\frac{b^2 c^3 d \text{PolyLog}\left (2,-\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}\right )}{e \left (c^2 d^2+e^2\right )^{3/2}}-\frac{b c \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{\left (c^2 d^2+e^2\right ) (d+e x)}+\frac{b c^3 d \left (a+b \sinh ^{-1}(c x)\right ) \log \left (\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}+1\right )}{e \left (c^2 d^2+e^2\right )^{3/2}}-\frac{b c^3 d \left (a+b \sinh ^{-1}(c x)\right ) \log \left (\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}+1\right )}{e \left (c^2 d^2+e^2\right )^{3/2}}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{2 e (d+e x)^2}+\frac{b^2 c^2 \log (d+e x)}{e \left (c^2 d^2+e^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*c*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/((c^2*d^2 + e^2)*(d + e*x))) - (a + b*ArcSinh[c*x])^2/(2*e*(d +
e*x)^2) + (b*c^3*d*(a + b*ArcSinh[c*x])*Log[1 + (e*E^ArcSinh[c*x])/(c*d - Sqrt[c^2*d^2 + e^2])])/(e*(c^2*d^2
+ e^2)^(3/2)) - (b*c^3*d*(a + b*ArcSinh[c*x])*Log[1 + (e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2])])/(e*(c^2
*d^2 + e^2)^(3/2)) + (b^2*c^2*Log[d + e*x])/(e*(c^2*d^2 + e^2)) + (b^2*c^3*d*PolyLog[2, -((e*E^ArcSinh[c*x])/(
c*d - Sqrt[c^2*d^2 + e^2]))])/(e*(c^2*d^2 + e^2)^(3/2)) - (b^2*c^3*d*PolyLog[2, -((e*E^ArcSinh[c*x])/(c*d + Sq
rt[c^2*d^2 + e^2]))])/(e*(c^2*d^2 + e^2)^(3/2))

Rule 5801

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

Rule 5831

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

Rule 3324

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(b*(c + d*x)^m*Cos[
e + f*x])/(f*(a^2 - b^2)*(a + b*Sin[e + f*x])), x] + (Dist[a/(a^2 - b^2), Int[(c + d*x)^m/(a + b*Sin[e + f*x])
, x], x] - Dist[(b*d*m)/(f*(a^2 - b^2)), Int[((c + d*x)^(m - 1)*Cos[e + f*x])/(a + b*Sin[e + f*x]), x], x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 3322

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]), x_Symbol] :> Dist[2,
Int[((c + d*x)^m*E^(-(I*e) + f*fz*x))/(-(I*b) + 2*a*E^(-(I*e) + f*fz*x) + I*b*E^(2*(-(I*e) + f*fz*x))), x], x]
/; FreeQ[{a, b, c, d, e, f, fz}, x] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]

Rule 2190

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

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 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 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 31

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

Rubi steps

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

Mathematica [A]  time = 0.874181, size = 270, normalized size = 0.77 \[ \frac{\frac{2 b c^3 d \left (b \text{PolyLog}\left (2,\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}-c d}\right )-b \text{PolyLog}\left (2,-\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}\right )+\left (a+b \sinh ^{-1}(c x)\right ) \left (\log \left (\frac{e e^{\sinh ^{-1}(c x)}}{c d-\sqrt{c^2 d^2+e^2}}+1\right )-\log \left (\frac{e e^{\sinh ^{-1}(c x)}}{\sqrt{c^2 d^2+e^2}+c d}+1\right )\right )\right )}{\left (c^2 d^2+e^2\right )^{3/2}}-\frac{2 b c e \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{\left (c^2 d^2+e^2\right ) (d+e x)}-\frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{(d+e x)^2}+\frac{2 b^2 c^2 \log (d+e x)}{c^2 d^2+e^2}}{2 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*b*c*e*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/((c^2*d^2 + e^2)*(d + e*x)) - (a + b*ArcSinh[c*x])^2/(d + e
*x)^2 + (2*b^2*c^2*Log[d + e*x])/(c^2*d^2 + e^2) + (2*b*c^3*d*((a + b*ArcSinh[c*x])*(Log[1 + (e*E^ArcSinh[c*x]
)/(c*d - Sqrt[c^2*d^2 + e^2])] - Log[1 + (e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2])]) + b*PolyLog[2, (e*E^
ArcSinh[c*x])/(-(c*d) + Sqrt[c^2*d^2 + e^2])] - b*PolyLog[2, -((e*E^ArcSinh[c*x])/(c*d + Sqrt[c^2*d^2 + e^2]))
]))/(c^2*d^2 + e^2)^(3/2))/(2*e)

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Maple [B]  time = 0.262, size = 1013, normalized size = 2.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*c^2*a^2/(c*e*x+c*d)^2/e-1/2*c^4*b^2*arcsinh(c*x)^2/e/(c*e*x+c*d)^2/(c^2*d^2+e^2)*d^2-c^3*b^2*arcsinh(c*x)
*e/(c*e*x+c*d)^2/(c^2*d^2+e^2)*(c^2*x^2+1)^(1/2)*x-c^3*b^2*arcsinh(c*x)/(c*e*x+c*d)^2/(c^2*d^2+e^2)*d*(c^2*x^2
+1)^(1/2)+c^4*b^2*arcsinh(c*x)*e/(c*e*x+c*d)^2/(c^2*d^2+e^2)*x^2+2*c^4*b^2*arcsinh(c*x)/(c*e*x+c*d)^2/(c^2*d^2
+e^2)*d*x+c^4*b^2*arcsinh(c*x)/e/(c*e*x+c*d)^2/(c^2*d^2+e^2)*d^2-1/2*c^2*b^2*arcsinh(c*x)^2*e/(c*e*x+c*d)^2/(c
^2*d^2+e^2)+c^2*b^2/e/(c^2*d^2+e^2)*ln(2*c*d*(c*x+(c^2*x^2+1)^(1/2))+(c*x+(c^2*x^2+1)^(1/2))^2*e-e)-2*c^2*b^2/
e/(c^2*d^2+e^2)*ln(c*x+(c^2*x^2+1)^(1/2))+c^3*b^2/e/(c^2*d^2+e^2)^(3/2)*d*arcsinh(c*x)*ln((-(c*x+(c^2*x^2+1)^(
1/2))*e-c*d+(c^2*d^2+e^2)^(1/2))/(-c*d+(c^2*d^2+e^2)^(1/2)))-c^3*b^2/e/(c^2*d^2+e^2)^(3/2)*d*arcsinh(c*x)*ln((
(c*x+(c^2*x^2+1)^(1/2))*e+c*d+(c^2*d^2+e^2)^(1/2))/(c*d+(c^2*d^2+e^2)^(1/2)))+c^3*b^2/e/(c^2*d^2+e^2)^(3/2)*d*
dilog((-(c*x+(c^2*x^2+1)^(1/2))*e-c*d+(c^2*d^2+e^2)^(1/2))/(-c*d+(c^2*d^2+e^2)^(1/2)))-c^3*b^2/e/(c^2*d^2+e^2)
^(3/2)*d*dilog(((c*x+(c^2*x^2+1)^(1/2))*e+c*d+(c^2*d^2+e^2)^(1/2))/(c*d+(c^2*d^2+e^2)^(1/2)))-c^2*a*b/(c*e*x+c
*d)^2/e*arcsinh(c*x)-c^2*a*b/e/(c^2*d^2+e^2)/(c*x+c*d/e)*((c*x+c*d/e)^2-2*c*d/e*(c*x+c*d/e)+(c^2*d^2+e^2)/e^2)
^(1/2)-c^3*a*b/e^2*d/(c^2*d^2+e^2)/((c^2*d^2+e^2)/e^2)^(1/2)*ln((2*(c^2*d^2+e^2)/e^2-2*c*d/e*(c*x+c*d/e)+2*((c
^2*d^2+e^2)/e^2)^(1/2)*((c*x+c*d/e)^2-2*c*d/e*(c*x+c*d/e)+(c^2*d^2+e^2)/e^2)^(1/2))/(c*x+c*d/e))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} \operatorname{arsinh}\left (c x\right )^{2} + 2 \, a b \operatorname{arsinh}\left (c x\right ) + a^{2}}{e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{asinh}{\left (c x \right )}\right )^{2}}{\left (d + e x\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}^{2}}{{\left (e x + d\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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