3.2.35 \(\int \frac {(a+b \sinh ^{-1}(c+d x))^2}{(c e+d e x)^4} \, dx\) [135]

Optimal. Leaf size=169 \[ -\frac {b^2}{3 d e^4 (c+d x)}-\frac {b \sqrt {1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}+\frac {2 b \left (a+b \sinh ^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right )}{3 d e^4}+\frac {b^2 \text {PolyLog}\left (2,-e^{\sinh ^{-1}(c+d x)}\right )}{3 d e^4}-\frac {b^2 \text {PolyLog}\left (2,e^{\sinh ^{-1}(c+d x)}\right )}{3 d e^4} \]

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {5859, 12, 5776, 5809, 5816, 4267, 2317, 2438, 30} \begin {gather*} -\frac {b \sqrt {(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d e^4 (c+d x)^2}-\frac {\left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d e^4 (c+d x)^3}+\frac {2 b \tanh ^{-1}\left (e^{\sinh ^{-1}(c+d x)}\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d e^4}+\frac {b^2 \text {Li}_2\left (-e^{\sinh ^{-1}(c+d x)}\right )}{3 d e^4}-\frac {b^2 \text {Li}_2\left (e^{\sinh ^{-1}(c+d x)}\right )}{3 d e^4}-\frac {b^2}{3 d e^4 (c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 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 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 5776

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcS
inh[c*x])^n/(d*(m + 1))), x] - Dist[b*c*(n/(d*(m + 1))), Int[(d*x)^(m + 1)*((a + b*ArcSinh[c*x])^(n - 1)/Sqrt[
1 + c^2*x^2]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -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 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 5859

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((d*e - c*f)/d + f*(x/d))^m*(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
]

Rubi steps

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

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Mathematica [A]
time = 1.13, size = 212, normalized size = 1.25 \begin {gather*} -\frac {4 a^2+a b \left (8 \sinh ^{-1}(c+d x)+2 \sinh \left (2 \sinh ^{-1}(c+d x)\right )+\log \left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c+d x)\right )\right ) \left (-3 (c+d x)+\sinh \left (3 \sinh ^{-1}(c+d x)\right )\right )\right )+b^2 \left (4 (c+d x)^2+4 \sinh ^{-1}(c+d x)^2+4 (c+d x)^3 \text {PolyLog}\left (2,-e^{-\sinh ^{-1}(c+d x)}\right )-4 (c+d x)^3 \text {PolyLog}\left (2,e^{-\sinh ^{-1}(c+d x)}\right )+\sinh ^{-1}(c+d x) \left (2 \sinh \left (2 \sinh ^{-1}(c+d x)\right )+\left (\log \left (1-e^{-\sinh ^{-1}(c+d x)}\right )-\log \left (1+e^{-\sinh ^{-1}(c+d x)}\right )\right ) \left (-3 (c+d x)+\sinh \left (3 \sinh ^{-1}(c+d x)\right )\right )\right )\right )}{12 d e^4 (c+d x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(4*a^2 + a*b*(8*ArcSinh[c + d*x] + 2*Sinh[2*ArcSinh[c + d*x]] + Log[Tanh[ArcSinh[c + d*x]/2]]*(-3*(c + d
*x) + Sinh[3*ArcSinh[c + d*x]])) + b^2*(4*(c + d*x)^2 + 4*ArcSinh[c + d*x]^2 + 4*(c + d*x)^3*PolyLog[2, -E^(-A
rcSinh[c + d*x])] - 4*(c + d*x)^3*PolyLog[2, E^(-ArcSinh[c + d*x])] + ArcSinh[c + d*x]*(2*Sinh[2*ArcSinh[c + d
*x]] + (Log[1 - E^(-ArcSinh[c + d*x])] - Log[1 + E^(-ArcSinh[c + d*x])])*(-3*(c + d*x) + Sinh[3*ArcSinh[c + d*
x]]))))/(d*e^4*(c + d*x)^3)

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Maple [A]
time = 4.79, size = 274, normalized size = 1.62

method	result	size
derivativedivides	\(\frac {-\frac {a^{2}}{3 e^{4} \left (d x +c \right )^{3}}-\frac {b^{2} \arcsinh \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}}{3 e^{4} \left (d x +c \right )^{2}}-\frac {b^{2} \arcsinh \left (d x +c \right )^{2}}{3 e^{4} \left (d x +c \right )^{3}}-\frac {b^{2}}{3 e^{4} \left (d x +c \right )}+\frac {b^{2} \arcsinh \left (d x +c \right ) \ln \left (1+d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{3 e^{4}}+\frac {b^{2} \polylog \left (2, -d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{3 e^{4}}-\frac {b^{2} \arcsinh \left (d x +c \right ) \ln \left (1-d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{3 e^{4}}-\frac {b^{2} \polylog \left (2, d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{3 e^{4}}+\frac {2 a b \left (-\frac {\arcsinh \left (d x +c \right )}{3 \left (d x +c \right )^{3}}-\frac {\sqrt {1+\left (d x +c \right )^{2}}}{6 \left (d x +c \right )^{2}}+\frac {\arctanh \left (\frac {1}{\sqrt {1+\left (d x +c \right )^{2}}}\right )}{6}\right )}{e^{4}}}{d}\)	\(274\)
default	\(\frac {-\frac {a^{2}}{3 e^{4} \left (d x +c \right )^{3}}-\frac {b^{2} \arcsinh \left (d x +c \right ) \sqrt {1+\left (d x +c \right )^{2}}}{3 e^{4} \left (d x +c \right )^{2}}-\frac {b^{2} \arcsinh \left (d x +c \right )^{2}}{3 e^{4} \left (d x +c \right )^{3}}-\frac {b^{2}}{3 e^{4} \left (d x +c \right )}+\frac {b^{2} \arcsinh \left (d x +c \right ) \ln \left (1+d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{3 e^{4}}+\frac {b^{2} \polylog \left (2, -d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{3 e^{4}}-\frac {b^{2} \arcsinh \left (d x +c \right ) \ln \left (1-d x -c -\sqrt {1+\left (d x +c \right )^{2}}\right )}{3 e^{4}}-\frac {b^{2} \polylog \left (2, d x +c +\sqrt {1+\left (d x +c \right )^{2}}\right )}{3 e^{4}}+\frac {2 a b \left (-\frac {\arcsinh \left (d x +c \right )}{3 \left (d x +c \right )^{3}}-\frac {\sqrt {1+\left (d x +c \right )^{2}}}{6 \left (d x +c \right )^{2}}+\frac {\arctanh \left (\frac {1}{\sqrt {1+\left (d x +c \right )^{2}}}\right )}{6}\right )}{e^{4}}}{d}\)	\(274\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(d*x+c))^2/(d*e*x+c*e)^4,x,method=_RETURNVERBOSE)

[Out]

1/d*(-1/3*a^2/e^4/(d*x+c)^3-1/3*b^2/e^4/(d*x+c)^2*arcsinh(d*x+c)*(1+(d*x+c)^2)^(1/2)-1/3*b^2/e^4/(d*x+c)^3*arc
sinh(d*x+c)^2-1/3*b^2/e^4/(d*x+c)+1/3*b^2/e^4*arcsinh(d*x+c)*ln(1+d*x+c+(1+(d*x+c)^2)^(1/2))+1/3*b^2/e^4*polyl
og(2,-d*x-c-(1+(d*x+c)^2)^(1/2))-1/3*b^2/e^4*arcsinh(d*x+c)*ln(1-d*x-c-(1+(d*x+c)^2)^(1/2))-1/3*b^2/e^4*polylo
g(2,d*x+c+(1+(d*x+c)^2)^(1/2))+2*a*b/e^4*(-1/3/(d*x+c)^3*arcsinh(d*x+c)-1/6/(d*x+c)^2*(1+(d*x+c)^2)^(1/2)+1/6*
arctanh(1/(1+(d*x+c)^2)^(1/2))))

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

[Out]

-1/3*b^2*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))^2/(d^4*x^3*e^4 + 3*c*d^3*x^2*e^4 + 3*c^2*d^2*x*e^4 +
c^3*d*e^4) - 1/3*a^2/(d^4*x^3*e^4 + 3*c*d^3*x^2*e^4 + 3*c^2*d^2*x*e^4 + c^3*d*e^4) + integrate(2/3*((3*a*b*d^
3 + b^2*d^3)*x^3 + 3*(c^3 + c)*a*b + (c^3 + c)*b^2 + 3*(3*a*b*c*d^2 + b^2*c*d^2)*x^2 + (3*(3*c^2*d + d)*a*b +
(3*c^2*d + d)*b^2)*x + (b^2*c^2 + 3*(c^2 + 1)*a*b + (3*a*b*d^2 + b^2*d^2)*x^2 + 2*(3*a*b*c*d + b^2*c*d)*x)*sqr
t(d^2*x^2 + 2*c*d*x + c^2 + 1))*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))/(d^7*x^7*e^4 + 7*c*d^6*x^6*e^
4 + (21*c^2*d^5 + d^5)*x^5*e^4 + 5*(7*c^3*d^4 + c*d^4)*x^4*e^4 + 5*(7*c^4*d^3 + 2*c^2*d^3)*x^3*e^4 + (21*c^5*d
^2 + 10*c^3*d^2)*x^2*e^4 + (7*c^6*d + 5*c^4*d)*x*e^4 + (c^7 + c^5)*e^4 + (d^6*x^6*e^4 + 6*c*d^5*x^5*e^4 + (15*
c^2*d^4 + d^4)*x^4*e^4 + 4*(5*c^3*d^3 + c*d^3)*x^3*e^4 + 3*(5*c^4*d^2 + 2*c^2*d^2)*x^2*e^4 + 2*(3*c^5*d + 2*c^
3*d)*x*e^4 + (c^6 + c^4)*e^4)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)), 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(d*x+c))^2/(d*e*x+c*e)^4,x, algorithm="fricas")

[Out]

integral((b^2*arcsinh(d*x + c)^2 + 2*a*b*arcsinh(d*x + c) + a^2)*e^(-4)/(d^4*x^4 + 4*c*d^3*x^3 + 6*c^2*d^2*x^2
+ 4*c^3*d*x + c^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} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {b^{2} \operatorname {asinh}^{2}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx + \int \frac {2 a b \operatorname {asinh}{\left (c + d x \right )}}{c^{4} + 4 c^{3} d x + 6 c^{2} d^{2} x^{2} + 4 c d^{3} x^{3} + d^{4} x^{4}}\, dx}{e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(d*x+c))^2/(d*e*x+c*e)^4,x, algorithm="giac")

[Out]

integrate((b*arcsinh(d*x + c) + a)^2/(d*e*x + c*e)^4, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^2}{{\left (c\,e+d\,e\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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