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

Optimal. Leaf size=349 \[ -\frac {b c \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{\left (c^2 d^2+e^2\right ) (d+e x)}+\frac {b c^3 d \left (a+b \sinh ^{-1}(c x)\right ) \log \left (\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}+1\right )}{e \left (c^2 d^2+e^2\right )^{3/2}}-\frac {b c^3 d \left (a+b \sinh ^{-1}(c x)\right ) \log \left (\frac {e e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 d^2+e^2}+c d}+1\right )}{e \left (c^2 d^2+e^2\right )^{3/2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{2 e (d+e x)^2}+\frac {b^2 c^2 \log (d+e x)}{e \left (c^2 d^2+e^2\right )}+\frac {b^2 c^3 d \text {Li}_2\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}\right )}{e \left (c^2 d^2+e^2\right )^{3/2}}-\frac {b^2 c^3 d \text {Li}_2\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )}{e \left (c^2 d^2+e^2\right )^{3/2}} \]

[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 = 0.61, 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 = {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 31

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

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 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 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 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 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)^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*}

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Mathematica [A]  time = 0.71, size = 270, normalized size = 0.77 \[ \frac {-\frac {2 b c e \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{\left (c^2 d^2+e^2\right ) (d+e x)}+\frac {2 b c^3 d \left (\left (a+b \sinh ^{-1}(c x)\right ) \left (\log \left (\frac {e e^{\sinh ^{-1}(c x)}}{c d-\sqrt {c^2 d^2+e^2}}+1\right )-\log \left (\frac {e e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 d^2+e^2}+c d}+1\right )\right )+b \text {Li}_2\left (\frac {e e^{\sinh ^{-1}(c x)}}{\sqrt {c^2 d^2+e^2}-c d}\right )-b \text {Li}_2\left (-\frac {e e^{\sinh ^{-1}(c x)}}{c d+\sqrt {c^2 d^2+e^2}}\right )\right )}{\left (c^2 d^2+e^2\right )^{3/2}}-\frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{(d+e x)^2}+\frac {2 b^2 c^2 \log (d+e x)}{c^2 d^2+e^2}}{2 e} \]

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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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} \operatorname {arsinh}\left (c x\right )^{2} + 2 \, a b \operatorname {arsinh}\left (c x\right ) + a^{2}}{e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}}, x\right ) \]

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

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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maple [B]  time = 0.67, size = 1013, normalized size = 2.90 \[ -\frac {c^{2} a^{2}}{2 \left (c e x +c d \right )^{2} e}-\frac {c^{4} b^{2} \arcsinh \left (c x \right )^{2} d^{2}}{2 e \left (c e x +c d \right )^{2} \left (c^{2} d^{2}+e^{2}\right )}-\frac {c^{3} b^{2} \arcsinh \left (c x \right ) e \sqrt {c^{2} x^{2}+1}\, x}{\left (c e x +c d \right )^{2} \left (c^{2} d^{2}+e^{2}\right )}-\frac {c^{3} b^{2} \arcsinh \left (c x \right ) d \sqrt {c^{2} x^{2}+1}}{\left (c e x +c d \right )^{2} \left (c^{2} d^{2}+e^{2}\right )}+\frac {c^{4} b^{2} \arcsinh \left (c x \right ) e \,x^{2}}{\left (c e x +c d \right )^{2} \left (c^{2} d^{2}+e^{2}\right )}+\frac {2 c^{4} b^{2} \arcsinh \left (c x \right ) d x}{\left (c e x +c d \right )^{2} \left (c^{2} d^{2}+e^{2}\right )}+\frac {c^{4} b^{2} \arcsinh \left (c x \right ) d^{2}}{e \left (c e x +c d \right )^{2} \left (c^{2} d^{2}+e^{2}\right )}-\frac {c^{2} b^{2} \arcsinh \left (c x \right )^{2} e}{2 \left (c e x +c d \right )^{2} \left (c^{2} d^{2}+e^{2}\right )}-\frac {2 c^{2} b^{2} \ln \left (c x +\sqrt {c^{2} x^{2}+1}\right )}{e \left (c^{2} d^{2}+e^{2}\right )}+\frac {c^{2} b^{2} \ln \left (2 c d \left (c x +\sqrt {c^{2} x^{2}+1}\right )+\left (c x +\sqrt {c^{2} x^{2}+1}\right )^{2} e -e \right )}{e \left (c^{2} d^{2}+e^{2}\right )}+\frac {c^{3} b^{2} d \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 \left (c^{2} d^{2}+e^{2}\right )^{\frac {3}{2}}}-\frac {c^{3} b^{2} d \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 \left (c^{2} d^{2}+e^{2}\right )^{\frac {3}{2}}}+\frac {c^{3} b^{2} d \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 \left (c^{2} d^{2}+e^{2}\right )^{\frac {3}{2}}}-\frac {c^{3} b^{2} d \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 \left (c^{2} d^{2}+e^{2}\right )^{\frac {3}{2}}}-\frac {c^{2} a b \arcsinh \left (c x \right )}{\left (c e x +c d \right )^{2} e}-\frac {c^{2} a b \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}}}}{e \left (c^{2} d^{2}+e^{2}\right ) \left (c x +\frac {c d}{e}\right )}-\frac {c^{3} a b d \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} \left (c^{2} d^{2}+e^{2}\right ) \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)^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)-2*c^2*b^2/e/(c^2*d^2+e^2)*ln(c*x+(c^2*x^2+1)^(1/2))+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)+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]  time = 0.00, size = 0, normalized size = 0.00 \[ -{\left (c {\left (\frac {\sqrt {c^{2} x^{2} + 1}}{c^{2} d^{2} e x + c^{2} d^{3} + e^{3} x + d e^{2}} - \frac {c^{2} d \operatorname {arsinh}\left (\frac {c d x}{e {\left | x + \frac {d}{e} \right |}} - \frac {1}{c {\left | x + \frac {d}{e} \right |}}\right )}{{\left (\frac {c^{2} d^{2}}{e^{2}} + 1\right )}^{\frac {3}{2}} e^{4}}\right )} + \frac {\operatorname {arsinh}\left (c x\right )}{e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e}\right )} a b - \frac {1}{2} \, b^{2} {\left (\frac {\log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2}}{e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e} - 2 \, \int \frac {{\left (c^{3} x^{2} + \sqrt {c^{2} x^{2} + 1} c^{2} x + c\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{c^{3} e^{3} x^{5} + 2 \, c^{3} d e^{2} x^{4} + 2 \, c d e^{2} x^{2} + c d^{2} e x + {\left (c^{3} d^{2} e + c e^{3}\right )} x^{3} + {\left (c^{2} e^{3} x^{4} + 2 \, c^{2} d e^{2} x^{3} + 2 \, d e^{2} x + d^{2} e + {\left (c^{2} d^{2} e + e^{3}\right )} x^{2}\right )} \sqrt {c^{2} x^{2} + 1}}\,{d x}\right )} - \frac {a^{2}}{2 \, {\left (e^{3} x^{2} + 2 \, d e^{2} x + d^{2} e\right )}} \]

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

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

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

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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