3.1.18 \(\int \frac {(a+b \sinh ^{-1}(c x))^2}{(d+e x)^3} \, dx\) [18]

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

[Out]

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

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Rubi [A]
time = 0.42, antiderivative size = 349, normalized size of antiderivative = 1.00, 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 = {5828, 5843, 3405, 3403, 2296, 2221, 2317, 2438, 2747, 31} \begin {gather*} -\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 )}+\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 {gather*}

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 31

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

Rule 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2296

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 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 2747

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 3403

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 3405

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 5828

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 5843

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)^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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.48, size = 270, normalized size = 0.77 \begin {gather*} \frac {-\frac {2 b c e \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}{(d+e x)^2}+\frac {2 b^2 c^2 \log (d+e x)}{c^2 d^2+e^2}+\frac {2 b c^3 d \left (\left (a+b \sinh ^{-1}(c x)\right ) \left (\log \left (1+\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )-\log \left (1+\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )\right )+b \text {PolyLog}\left (2,\frac {e e^{\sinh ^{-1}(c x)}}{-c d+\sqrt {c^2 d^2+e^2}}\right )-b \text {PolyLog}\left (2,-\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )\right )}{\left (c^2 d^2+e^2\right )^{3/2}}}{2 e} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(1016\) vs. \(2(365)=730\).
time = 12.87, size = 1017, normalized size = 2.91 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c*(-1/2*a^2*c^3/(c*e*x+c*d)^2/e-1/2*b^2*c^5*arcsinh(c*x)^2/e/(c^2*d^2+e^2)/(c*e*x+c*d)^2*d^2-b^2*c^4*arcsinh
(c*x)/(c^2*d^2+e^2)/(c*e*x+c*d)^2*d*(c^2*x^2+1)^(1/2)-b^2*c^4*arcsinh(c*x)*e/(c^2*d^2+e^2)/(c*e*x+c*d)^2*(c^2*
x^2+1)^(1/2)*x+b^2*c^5*arcsinh(c*x)/e/(c^2*d^2+e^2)/(c*e*x+c*d)^2*d^2+2*b^2*c^5*arcsinh(c*x)/(c^2*d^2+e^2)/(c*
e*x+c*d)^2*d*x+b^2*c^5*arcsinh(c*x)*e/(c^2*d^2+e^2)/(c*e*x+c*d)^2*x^2-1/2*b^2*c^3*arcsinh(c*x)^2*e/(c^2*d^2+e^
2)/(c*e*x+c*d)^2+b^2*c^3/e/(c^2*d^2+e^2)*ln(2*(c*x+(c^2*x^2+1)^(1/2))*d*c+e*(c*x+(c^2*x^2+1)^(1/2))^2-e)-2*b^2
*c^3/e/(c^2*d^2+e^2)*ln(c*x+(c^2*x^2+1)^(1/2))+b^2*c^4/e/(c^2*d^2+e^2)^(3/2)*d*arcsinh(c*x)*ln((-c*d-e*(c*x+(c
^2*x^2+1)^(1/2))+(c^2*d^2+e^2)^(1/2))/(-c*d+(c^2*d^2+e^2)^(1/2)))-b^2*c^4/e/(c^2*d^2+e^2)^(3/2)*d*arcsinh(c*x)
*ln((c*d+e*(c*x+(c^2*x^2+1)^(1/2))+(c^2*d^2+e^2)^(1/2))/(c*d+(c^2*d^2+e^2)^(1/2)))+b^2*c^4/e/(c^2*d^2+e^2)^(3/
2)*d*dilog((-c*d-e*(c*x+(c^2*x^2+1)^(1/2))+(c^2*d^2+e^2)^(1/2))/(-c*d+(c^2*d^2+e^2)^(1/2)))-b^2*c^4/e/(c^2*d^2
+e^2)^(3/2)*d*dilog((c*d+e*(c*x+(c^2*x^2+1)^(1/2))+(c^2*d^2+e^2)^(1/2))/(c*d+(c^2*d^2+e^2)^(1/2)))-a*b*c^3/(c*
e*x+c*d)^2/e*arcsinh(c*x)-a*b*c^3/e/(c^2*d^2+e^2)/(c*x+c*d/e)*((c*x+c*d/e)^2-2*d*c/e*(c*x+c*d/e)+(c^2*d^2+e^2)
/e^2)^(1/2)-a*b*c^4/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*d*c/e*(c*x+c*d/e)+
2*((c^2*d^2+e^2)/e^2)^(1/2)*((c*x+c*d/e)^2-2*d*c/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 \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/(e*x+d)^3,x, algorithm="maxima")

[Out]

((c^2*d*arcsinh(c*d*x*e^(-1)/abs(d*e^(-1) + x) - 1/(c*abs(d*e^(-1) + x)))*e^(-4)/(c^2*d^2*e^(-2) + 1)^(3/2) -
sqrt(c^2*x^2 + 1)/(c^2*d^2*x*e + c^2*d^3 + x*e^3 + d*e^2))*c - arcsinh(c*x)/(x^2*e^3 + 2*d*x*e^2 + d^2*e))*a*b
- 1/2*b^2*(log(c*x + sqrt(c^2*x^2 + 1))^2/(x^2*e^3 + 2*d*x*e^2 + d^2*e) - 2*integrate((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*x^5*e^3 + 2*c^3*d*x^4*e^2 + 2*c*d*x^2*e^2 + c*d^2*x*e + (c^
3*d^2*e + c*e^3)*x^3 + (c^2*x^4*e^3 + 2*c^2*d*x^3*e^2 + (c^2*d^2*e + e^3)*x^2 + 2*d*x*e^2 + d^2*e)*sqrt(c^2*x^
2 + 1)), x)) - 1/2*a^2/(x^2*e^3 + 2*d*x*e^2 + d^2*e)

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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/(e*x+d)^3,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}}{\left (d + e x\right )^{3}}\, dx \end {gather*}

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 = 0.00, 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/(e*x+d)^3,x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)^2/(e*x + d)^3, 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}{{\left (d+e\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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