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

Optimal. Leaf size=263 \[ \frac {2 b c \left (a+b \sinh ^{-1}(c x)\right ) \log \left (\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}+1\right )}{e \sqrt {c^2 d^2+e^2}}-\frac {2 b c \left (a+b \sinh ^{-1}(c x)\right ) \log \left (\frac {e e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 d^2+e^2}+c d}+1\right )}{e \sqrt {c^2 d^2+e^2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{e (d+e x)}+\frac {2 b^2 c \text {Li}_2\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e \sqrt {c^2 d^2+e^2}}-\frac {2 b^2 c \text {Li}_2\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )}{e \sqrt {c^2 d^2+e^2}} \]

[Out]

-(a+b*arcsinh(c*x))^2/e/(e*x+d)+2*b*c*(a+b*arcsinh(c*x))*ln(1+e*(c*x+(c^2*x^2+1)^(1/2))/(c*d-(c^2*d^2+e^2)^(1/
2)))/e/(c^2*d^2+e^2)^(1/2)-2*b*c*(a+b*arcsinh(c*x))*ln(1+e*(c*x+(c^2*x^2+1)^(1/2))/(c*d+(c^2*d^2+e^2)^(1/2)))/
e/(c^2*d^2+e^2)^(1/2)+2*b^2*c*polylog(2,-e*(c*x+(c^2*x^2+1)^(1/2))/(c*d-(c^2*d^2+e^2)^(1/2)))/e/(c^2*d^2+e^2)^
(1/2)-2*b^2*c*polylog(2,-e*(c*x+(c^2*x^2+1)^(1/2))/(c*d+(c^2*d^2+e^2)^(1/2)))/e/(c^2*d^2+e^2)^(1/2)

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Rubi [A]  time = 0.47, antiderivative size = 263, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {5801, 5831, 3322, 2264, 2190, 2279, 2391} \[ \frac {2 b^2 c \text {PolyLog}\left (2,-\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e \sqrt {c^2 d^2+e^2}}-\frac {2 b^2 c \text {PolyLog}\left (2,-\frac {e e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 d^2+e^2}+c d}\right )}{e \sqrt {c^2 d^2+e^2}}+\frac {2 b c \left (a+b \sinh ^{-1}(c x)\right ) \log \left (\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}+1\right )}{e \sqrt {c^2 d^2+e^2}}-\frac {2 b c \left (a+b \sinh ^{-1}(c x)\right ) \log \left (\frac {e e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 d^2+e^2}+c d}+1\right )}{e \sqrt {c^2 d^2+e^2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{e (d+e x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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 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 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])

Rubi steps

\begin {align*} \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{(d+e x)^2} \, dx &=-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{e (d+e x)}+\frac {(2 b c) \int \frac {a+b \sinh ^{-1}(c x)}{(d+e x) \sqrt {1+c^2 x^2}} \, dx}{e}\\ &=-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{e (d+e x)}+\frac {(2 b c) \operatorname {Subst}\left (\int \frac {a+b x}{c d+e \sinh (x)} \, dx,x,\sinh ^{-1}(c x)\right )}{e}\\ &=-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{e (d+e x)}+\frac {(4 b c) \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}\\ &=-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{e (d+e x)}+\frac {(4 b c) \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 )}{\sqrt {c^2 d^2+e^2}}-\frac {(4 b c) \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 )}{\sqrt {c^2 d^2+e^2}}\\ &=-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{e (d+e x)}+\frac {2 b c \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 \sqrt {c^2 d^2+e^2}}-\frac {2 b c \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 \sqrt {c^2 d^2+e^2}}-\frac {\left (2 b^2 c\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 \sqrt {c^2 d^2+e^2}}+\frac {\left (2 b^2 c\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 \sqrt {c^2 d^2+e^2}}\\ &=-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{e (d+e x)}+\frac {2 b c \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 \sqrt {c^2 d^2+e^2}}-\frac {2 b c \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 \sqrt {c^2 d^2+e^2}}-\frac {\left (2 b^2 c\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 \sqrt {c^2 d^2+e^2}}+\frac {\left (2 b^2 c\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 \sqrt {c^2 d^2+e^2}}\\ &=-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{e (d+e x)}+\frac {2 b c \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 \sqrt {c^2 d^2+e^2}}-\frac {2 b c \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 \sqrt {c^2 d^2+e^2}}+\frac {2 b^2 c \text {Li}_2\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e \sqrt {c^2 d^2+e^2}}-\frac {2 b^2 c \text {Li}_2\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )}{e \sqrt {c^2 d^2+e^2}}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 191, normalized size = 0.73 \[ \frac {\frac {2 b c \left (\left (a+b \sinh ^{-1}(c x)\right ) \left (\log \left (\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}+1\right )-\log \left (\frac {e e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 d^2+e^2}+c d}+1\right )\right )+b \text {Li}_2\left (\frac {e e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 d^2+e^2}-c d}\right )-b \text {Li}_2\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )\right )}{\sqrt {c^2 d^2+e^2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{d+e x}}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-((a + b*ArcSinh[c*x])^2/(d + e*x)) + (2*b*c*((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]))]))/Sqrt[c^2*d^2
+ e^2])/e

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fricas [F]  time = 0.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \operatorname {arsinh}\left (c x\right )^{2} + 2 \, a b \operatorname {arsinh}\left (c x\right ) + a^{2}}{e^{2} x^{2} + 2 \, d e x + d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.34, size = 529, normalized size = 2.01 \[ -\frac {c \,a^{2}}{\left (c e x +c d \right ) e}-\frac {c \,b^{2} \arcsinh \left (c x \right )^{2}}{e \left (c e x +c d \right )}+\frac {2 c \,b^{2} \arcsinh \left (c x \right ) \ln \left (\frac {-\left (c x +\sqrt {c^{2} x^{2}+1}\right ) e -c d +\sqrt {c^{2} d^{2}+e^{2}}}{-c d +\sqrt {c^{2} d^{2}+e^{2}}}\right )}{e \sqrt {c^{2} d^{2}+e^{2}}}-\frac {2 c \,b^{2} \arcsinh \left (c x \right ) \ln \left (\frac {\left (c x +\sqrt {c^{2} x^{2}+1}\right ) e +c d +\sqrt {c^{2} d^{2}+e^{2}}}{c d +\sqrt {c^{2} d^{2}+e^{2}}}\right )}{e \sqrt {c^{2} d^{2}+e^{2}}}+\frac {2 c \,b^{2} \dilog \left (\frac {-\left (c x +\sqrt {c^{2} x^{2}+1}\right ) e -c d +\sqrt {c^{2} d^{2}+e^{2}}}{-c d +\sqrt {c^{2} d^{2}+e^{2}}}\right )}{e \sqrt {c^{2} d^{2}+e^{2}}}-\frac {2 c \,b^{2} \dilog \left (\frac {\left (c x +\sqrt {c^{2} x^{2}+1}\right ) e +c d +\sqrt {c^{2} d^{2}+e^{2}}}{c d +\sqrt {c^{2} d^{2}+e^{2}}}\right )}{e \sqrt {c^{2} d^{2}+e^{2}}}-\frac {2 c a b \arcsinh \left (c x \right )}{\left (c e x +c d \right ) e}-\frac {2 c a b \ln \left (\frac {\frac {2 c^{2} d^{2}+2 e^{2}}{e^{2}}-\frac {2 c d \left (c x +\frac {c d}{e}\right )}{e}+2 \sqrt {\frac {c^{2} d^{2}+e^{2}}{e^{2}}}\, \sqrt {\left (c x +\frac {c d}{e}\right )^{2}-\frac {2 c d \left (c x +\frac {c d}{e}\right )}{e}+\frac {c^{2} d^{2}+e^{2}}{e^{2}}}}{c x +\frac {c d}{e}}\right )}{e^{2} \sqrt {\frac {c^{2} d^{2}+e^{2}}{e^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-c*a^2/(c*e*x+c*d)/e-c*b^2*arcsinh(c*x)^2/e/(c*e*x+c*d)+2*c*b^2/e*arcsinh(c*x)/(c^2*d^2+e^2)^(1/2)*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)))-2*c*b^2/e*arcsinh(c*x)/(c^2*d^2+e^2)^
(1/2)*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)))+2*c*b^2/e/(c^2*d^2+e^2
)^(1/2)*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)))-2*c*b^2/e/(c^2*
d^2+e^2)^(1/2)*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)))-2*c*a*b/(c
*e*x+c*d)/e*arcsinh(c*x)-2*c*a*b/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]  time = 0.00, size = 0, normalized size = 0.00 \[ -b^{2} {\left (\frac {\log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2}}{e^{2} x + d e} - \int \frac {2 \, {\left (c^{3} x^{2} + \sqrt {c^{2} x^{2} + 1} c^{2} x + c\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{c^{3} e^{2} x^{4} + c^{3} d e x^{3} + c e^{2} x^{2} + c d e x + {\left (c^{2} e^{2} x^{3} + c^{2} d e x^{2} + e^{2} x + d e\right )} \sqrt {c^{2} x^{2} + 1}}\,{d x}\right )} - 2 \, a b {\left (\frac {\operatorname {arsinh}\left (c x\right )}{e^{2} x + d e} - \frac {c \operatorname {arsinh}\left (\frac {c d e x}{{\left | e^{2} x + d e \right |}} - \frac {e^{2}}{c {\left | e^{2} x + d e \right |}}\right )}{\sqrt {\frac {c^{2} d^{2}}{e^{2}} + 1} e^{2}}\right )} - \frac {a^{2}}{e^{2} x + d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b^2*(log(c*x + sqrt(c^2*x^2 + 1))^2/(e^2*x + d*e) - integrate(2*(c^3*x^2 + sqrt(c^2*x^2 + 1)*c^2*x + c)*log(c
*x + sqrt(c^2*x^2 + 1))/(c^3*e^2*x^4 + c^3*d*e*x^3 + c*e^2*x^2 + c*d*e*x + (c^2*e^2*x^3 + c^2*d*e*x^2 + e^2*x
+ d*e)*sqrt(c^2*x^2 + 1)), x)) - 2*a*b*(arcsinh(c*x)/(e^2*x + d*e) - c*arcsinh(c*d*e*x/abs(e^2*x + d*e) - e^2/
(c*abs(e^2*x + d*e)))/(sqrt(c^2*d^2/e^2 + 1)*e^2)) - a^2/(e^2*x + d*e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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