3.301 \(\int \frac{x^4 (a+b \sinh ^{-1}(c x))^2}{(d+c^2 d x^2)^{3/2}} \, dx\)

Optimal. Leaf size=400 \[ -\frac{b^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{c^5 d \sqrt{c^2 d x^2+d}}+\frac{3 x \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^4 d^2}-\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c^2 d \sqrt{c^2 d x^2+d}}-\frac{b x^2 \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^3 d \sqrt{c^2 d x^2+d}}-\frac{\sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^3}{2 b c^5 d \sqrt{c^2 d x^2+d}}+\frac{\sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{c^5 d \sqrt{c^2 d x^2+d}}-\frac{2 b \sqrt{c^2 x^2+1} \log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^5 d \sqrt{c^2 d x^2+d}}+\frac{b^2 x \left (c^2 x^2+1\right )}{4 c^4 d \sqrt{c^2 d x^2+d}}-\frac{b^2 \sqrt{c^2 x^2+1} \sinh ^{-1}(c x)}{4 c^5 d \sqrt{c^2 d x^2+d}} \]

[Out]

(b^2*x*(1 + c^2*x^2))/(4*c^4*d*Sqrt[d + c^2*d*x^2]) - (b^2*Sqrt[1 + c^2*x^2]*ArcSinh[c*x])/(4*c^5*d*Sqrt[d + c
^2*d*x^2]) - (b*x^2*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(2*c^3*d*Sqrt[d + c^2*d*x^2]) - (x^3*(a + b*ArcSin
h[c*x])^2)/(c^2*d*Sqrt[d + c^2*d*x^2]) + (Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^2)/(c^5*d*Sqrt[d + c^2*d*x^2]
) + (3*x*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^2)/(2*c^4*d^2) - (Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^3)/
(2*b*c^5*d*Sqrt[d + c^2*d*x^2]) - (2*b*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])*Log[1 + E^(2*ArcSinh[c*x])])/(c^
5*d*Sqrt[d + c^2*d*x^2]) - (b^2*Sqrt[1 + c^2*x^2]*PolyLog[2, -E^(2*ArcSinh[c*x])])/(c^5*d*Sqrt[d + c^2*d*x^2])

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Rubi [A]  time = 0.66494, antiderivative size = 400, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 13, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.464, Rules used = {5751, 5758, 5677, 5675, 5661, 321, 215, 5767, 5714, 3718, 2190, 2279, 2391} \[ -\frac{b^2 \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{c^5 d \sqrt{c^2 d x^2+d}}+\frac{3 x \sqrt{c^2 d x^2+d} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^4 d^2}-\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c^2 d \sqrt{c^2 d x^2+d}}-\frac{b x^2 \sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^3 d \sqrt{c^2 d x^2+d}}-\frac{\sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^3}{2 b c^5 d \sqrt{c^2 d x^2+d}}+\frac{\sqrt{c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{c^5 d \sqrt{c^2 d x^2+d}}-\frac{2 b \sqrt{c^2 x^2+1} \log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c^5 d \sqrt{c^2 d x^2+d}}+\frac{b^2 x \left (c^2 x^2+1\right )}{4 c^4 d \sqrt{c^2 d x^2+d}}-\frac{b^2 \sqrt{c^2 x^2+1} \sinh ^{-1}(c x)}{4 c^5 d \sqrt{c^2 d x^2+d}} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*ArcSinh[c*x])^2)/(d + c^2*d*x^2)^(3/2),x]

[Out]

(b^2*x*(1 + c^2*x^2))/(4*c^4*d*Sqrt[d + c^2*d*x^2]) - (b^2*Sqrt[1 + c^2*x^2]*ArcSinh[c*x])/(4*c^5*d*Sqrt[d + c
^2*d*x^2]) - (b*x^2*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/(2*c^3*d*Sqrt[d + c^2*d*x^2]) - (x^3*(a + b*ArcSin
h[c*x])^2)/(c^2*d*Sqrt[d + c^2*d*x^2]) + (Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^2)/(c^5*d*Sqrt[d + c^2*d*x^2]
) + (3*x*Sqrt[d + c^2*d*x^2]*(a + b*ArcSinh[c*x])^2)/(2*c^4*d^2) - (Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^3)/
(2*b*c^5*d*Sqrt[d + c^2*d*x^2]) - (2*b*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])*Log[1 + E^(2*ArcSinh[c*x])])/(c^
5*d*Sqrt[d + c^2*d*x^2]) - (b^2*Sqrt[1 + c^2*x^2]*PolyLog[2, -E^(2*ArcSinh[c*x])])/(c^5*d*Sqrt[d + c^2*d*x^2])

Rule 5751

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
(f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n)/(2*e*(p + 1)), x] + (-Dist[(f^2*(m - 1))/(2*e*(p
+ 1)), Int[(f*x)^(m - 2)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n, x], x] - Dist[(b*f*n*d^IntPart[p]*(d + e*
x^2)^FracPart[p])/(2*c*(p + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m - 1)*(1 + c^2*x^2)^(p + 1/2)*(a + b*Ar
cSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] && Gt
Q[m, 1]

Rule 5758

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

Rule 5677

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[1 + c^2*x^2]/S
qrt[d + e*x^2], Int[(a + b*ArcSinh[c*x])^n/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e,
c^2*d] &&  !GtQ[d, 0]

Rule 5675

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

Rule 5661

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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 5767

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[
(f*(f*x)^(m - 1)*(d + e*x^2)^(p + 1)*(a + b*ArcSinh[c*x])^n)/(e*(m + 2*p + 1)), x] + (-Dist[(f^2*(m - 1))/(c^2
*(m + 2*p + 1)), Int[(f*x)^(m - 2)*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] - Dist[(b*f*n*d^IntPart[p]*(d
+ e*x^2)^FracPart[p])/(c*(m + 2*p + 1)*(1 + c^2*x^2)^FracPart[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] && GtQ[m
, 1] && NeQ[m + 2*p + 1, 0] && IntegerQ[m]

Rule 5714

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

Rule 3718

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c + d*x)^(m +
1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(1 + E^(2*(-(I*e) + f*fz*x))), x],
x] /; FreeQ[{c, d, e, f, fz}, x] && 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]

Rubi steps

\begin{align*} \int \frac{x^4 \left (a+b \sinh ^{-1}(c x)\right )^2}{\left (d+c^2 d x^2\right )^{3/2}} \, dx &=-\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c^2 d \sqrt{d+c^2 d x^2}}+\frac{3 \int \frac{x^2 \left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt{d+c^2 d x^2}} \, dx}{c^2 d}+\frac{\left (2 b \sqrt{1+c^2 x^2}\right ) \int \frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{c d \sqrt{d+c^2 d x^2}}\\ &=\frac{b x^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{c^3 d \sqrt{d+c^2 d x^2}}-\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c^2 d \sqrt{d+c^2 d x^2}}+\frac{3 x \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^4 d^2}-\frac{3 \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt{d+c^2 d x^2}} \, dx}{2 c^4 d}-\frac{\left (2 b \sqrt{1+c^2 x^2}\right ) \int \frac{x \left (a+b \sinh ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{c^3 d \sqrt{d+c^2 d x^2}}-\frac{\left (3 b \sqrt{1+c^2 x^2}\right ) \int x \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{c^3 d \sqrt{d+c^2 d x^2}}-\frac{\left (b^2 \sqrt{1+c^2 x^2}\right ) \int \frac{x^2}{\sqrt{1+c^2 x^2}} \, dx}{c^2 d \sqrt{d+c^2 d x^2}}\\ &=-\frac{b^2 x \left (1+c^2 x^2\right )}{2 c^4 d \sqrt{d+c^2 d x^2}}-\frac{b x^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^3 d \sqrt{d+c^2 d x^2}}-\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c^2 d \sqrt{d+c^2 d x^2}}+\frac{3 x \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^4 d^2}-\frac{\left (2 b \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int (a+b x) \tanh (x) \, dx,x,\sinh ^{-1}(c x)\right )}{c^5 d \sqrt{d+c^2 d x^2}}-\frac{\left (3 \sqrt{1+c^2 x^2}\right ) \int \frac{\left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt{1+c^2 x^2}} \, dx}{2 c^4 d \sqrt{d+c^2 d x^2}}+\frac{\left (b^2 \sqrt{1+c^2 x^2}\right ) \int \frac{1}{\sqrt{1+c^2 x^2}} \, dx}{2 c^4 d \sqrt{d+c^2 d x^2}}+\frac{\left (3 b^2 \sqrt{1+c^2 x^2}\right ) \int \frac{x^2}{\sqrt{1+c^2 x^2}} \, dx}{2 c^2 d \sqrt{d+c^2 d x^2}}\\ &=\frac{b^2 x \left (1+c^2 x^2\right )}{4 c^4 d \sqrt{d+c^2 d x^2}}+\frac{b^2 \sqrt{1+c^2 x^2} \sinh ^{-1}(c x)}{2 c^5 d \sqrt{d+c^2 d x^2}}-\frac{b x^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^3 d \sqrt{d+c^2 d x^2}}-\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c^2 d \sqrt{d+c^2 d x^2}}+\frac{\sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{c^5 d \sqrt{d+c^2 d x^2}}+\frac{3 x \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^4 d^2}-\frac{\sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{2 b c^5 d \sqrt{d+c^2 d x^2}}-\frac{\left (4 b \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{e^{2 x} (a+b x)}{1+e^{2 x}} \, dx,x,\sinh ^{-1}(c x)\right )}{c^5 d \sqrt{d+c^2 d x^2}}-\frac{\left (3 b^2 \sqrt{1+c^2 x^2}\right ) \int \frac{1}{\sqrt{1+c^2 x^2}} \, dx}{4 c^4 d \sqrt{d+c^2 d x^2}}\\ &=\frac{b^2 x \left (1+c^2 x^2\right )}{4 c^4 d \sqrt{d+c^2 d x^2}}-\frac{b^2 \sqrt{1+c^2 x^2} \sinh ^{-1}(c x)}{4 c^5 d \sqrt{d+c^2 d x^2}}-\frac{b x^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^3 d \sqrt{d+c^2 d x^2}}-\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c^2 d \sqrt{d+c^2 d x^2}}+\frac{\sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{c^5 d \sqrt{d+c^2 d x^2}}+\frac{3 x \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^4 d^2}-\frac{\sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{2 b c^5 d \sqrt{d+c^2 d x^2}}-\frac{2 b \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{c^5 d \sqrt{d+c^2 d x^2}}+\frac{\left (2 b^2 \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c^5 d \sqrt{d+c^2 d x^2}}\\ &=\frac{b^2 x \left (1+c^2 x^2\right )}{4 c^4 d \sqrt{d+c^2 d x^2}}-\frac{b^2 \sqrt{1+c^2 x^2} \sinh ^{-1}(c x)}{4 c^5 d \sqrt{d+c^2 d x^2}}-\frac{b x^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^3 d \sqrt{d+c^2 d x^2}}-\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c^2 d \sqrt{d+c^2 d x^2}}+\frac{\sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{c^5 d \sqrt{d+c^2 d x^2}}+\frac{3 x \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^4 d^2}-\frac{\sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{2 b c^5 d \sqrt{d+c^2 d x^2}}-\frac{2 b \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{c^5 d \sqrt{d+c^2 d x^2}}+\frac{\left (b^2 \sqrt{1+c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{c^5 d \sqrt{d+c^2 d x^2}}\\ &=\frac{b^2 x \left (1+c^2 x^2\right )}{4 c^4 d \sqrt{d+c^2 d x^2}}-\frac{b^2 \sqrt{1+c^2 x^2} \sinh ^{-1}(c x)}{4 c^5 d \sqrt{d+c^2 d x^2}}-\frac{b x^2 \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{2 c^3 d \sqrt{d+c^2 d x^2}}-\frac{x^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c^2 d \sqrt{d+c^2 d x^2}}+\frac{\sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{c^5 d \sqrt{d+c^2 d x^2}}+\frac{3 x \sqrt{d+c^2 d x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 c^4 d^2}-\frac{\sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^3}{2 b c^5 d \sqrt{d+c^2 d x^2}}-\frac{2 b \sqrt{1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{c^5 d \sqrt{d+c^2 d x^2}}-\frac{b^2 \sqrt{1+c^2 x^2} \text{Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{c^5 d \sqrt{d+c^2 d x^2}}\\ \end{align*}

Mathematica [A]  time = 1.9296, size = 288, normalized size = 0.72 \[ \frac{b^2 \sqrt{d} \left (8 \sqrt{c^2 x^2+1} \text{PolyLog}\left (2,-e^{-2 \sinh ^{-1}(c x)}\right )+\sqrt{c^2 x^2+1} \left (-4 \sinh ^{-1}(c x)^3+2 \left (\sinh \left (2 \sinh ^{-1}(c x)\right )-4\right ) \sinh ^{-1}(c x)^2+\sinh \left (2 \sinh ^{-1}(c x)\right )-2 \sinh ^{-1}(c x) \left (8 \log \left (e^{-2 \sinh ^{-1}(c x)}+1\right )+\cosh \left (2 \sinh ^{-1}(c x)\right )\right )\right )+8 c x \sinh ^{-1}(c x)^2\right )+4 a^2 c \sqrt{d} x \left (c^2 x^2+3\right )-12 a^2 \sqrt{c^2 d x^2+d} \log \left (\sqrt{d} \sqrt{c^2 d x^2+d}+c d x\right )+2 a b \sqrt{d} \left (8 c x \sinh ^{-1}(c x)-\sqrt{c^2 x^2+1} \left (4 \log \left (c^2 x^2+1\right )+6 \sinh ^{-1}(c x)^2-2 \sinh \left (2 \sinh ^{-1}(c x)\right ) \sinh ^{-1}(c x)+\cosh \left (2 \sinh ^{-1}(c x)\right )\right )\right )}{8 c^5 d^{3/2} \sqrt{c^2 d x^2+d}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^4*(a + b*ArcSinh[c*x])^2)/(d + c^2*d*x^2)^(3/2),x]

[Out]

(4*a^2*c*Sqrt[d]*x*(3 + c^2*x^2) - 12*a^2*Sqrt[d + c^2*d*x^2]*Log[c*d*x + Sqrt[d]*Sqrt[d + c^2*d*x^2]] + b^2*S
qrt[d]*(8*c*x*ArcSinh[c*x]^2 + 8*Sqrt[1 + c^2*x^2]*PolyLog[2, -E^(-2*ArcSinh[c*x])] + Sqrt[1 + c^2*x^2]*(-4*Ar
cSinh[c*x]^3 - 2*ArcSinh[c*x]*(Cosh[2*ArcSinh[c*x]] + 8*Log[1 + E^(-2*ArcSinh[c*x])]) + 2*ArcSinh[c*x]^2*(-4 +
Sinh[2*ArcSinh[c*x]]) + Sinh[2*ArcSinh[c*x]])) + 2*a*b*Sqrt[d]*(8*c*x*ArcSinh[c*x] - Sqrt[1 + c^2*x^2]*(6*Arc
Sinh[c*x]^2 + Cosh[2*ArcSinh[c*x]] + 4*Log[1 + c^2*x^2] - 2*ArcSinh[c*x]*Sinh[2*ArcSinh[c*x]])))/(8*c^5*d^(3/2
)*Sqrt[d + c^2*d*x^2])

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Maple [B]  time = 0.386, size = 816, normalized size = 2. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d)^(3/2),x)

[Out]

1/2*a^2*x^3/c^2/d/(c^2*d*x^2+d)^(1/2)+3/2*a^2/c^4*x/d/(c^2*d*x^2+d)^(1/2)-3/2*a^2/c^4/d*ln(x*c^2*d/(c^2*d)^(1/
2)+(c^2*d*x^2+d)^(1/2))/(c^2*d)^(1/2)-2*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^5/d^2*arcsinh(c*x)*ln(1+
(c*x+(c^2*x^2+1)^(1/2))^2)+1/4*b^2*(d*(c^2*x^2+1))^(1/2)/c^2/d^2/(c^2*x^2+1)*x^3+1/4*b^2*(d*(c^2*x^2+1))^(1/2)
/c^4/d^2/(c^2*x^2+1)*x-b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^5/d^2*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))^
2)+b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^5/d^2*arcsinh(c*x)^2-1/2*b^2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1
)^(1/2)/c^5/d^2*arcsinh(c*x)^3+1/2*b^2*(d*(c^2*x^2+1))^(1/2)/c^2/d^2/(c^2*x^2+1)*arcsinh(c*x)^2*x^3+3/2*b^2*(d
*(c^2*x^2+1))^(1/2)/c^4/d^2/(c^2*x^2+1)*arcsinh(c*x)^2*x-1/4*b^2*(d*(c^2*x^2+1))^(1/2)/c^5/d^2/(c^2*x^2+1)^(1/
2)*arcsinh(c*x)-1/2*b^2*(d*(c^2*x^2+1))^(1/2)/c^3/d^2/(c^2*x^2+1)^(1/2)*arcsinh(c*x)*x^2-3/2*a*b*(d*(c^2*x^2+1
))^(1/2)/(c^2*x^2+1)^(1/2)/c^5/d^2*arcsinh(c*x)^2+a*b*(d*(c^2*x^2+1))^(1/2)/c^2/d^2/(c^2*x^2+1)*arcsinh(c*x)*x
^3-1/2*a*b*(d*(c^2*x^2+1))^(1/2)/c^3/d^2/(c^2*x^2+1)^(1/2)*x^2+3*a*b*(d*(c^2*x^2+1))^(1/2)/c^4/d^2/(c^2*x^2+1)
*arcsinh(c*x)*x+2*a*b*(d*(c^2*x^2+1))^(1/2)/c^5/d^2/(c^2*x^2+1)^(1/2)*arcsinh(c*x)-1/4*a*b*(d*(c^2*x^2+1))^(1/
2)/c^5/d^2/(c^2*x^2+1)^(1/2)-2*a*b*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c^5/d^2*ln(1+(c*x+(c^2*x^2+1)^(1/2)
)^2)

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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(x^4*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d)^(3/2),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{{\left (b^{2} x^{4} \operatorname{arsinh}\left (c x\right )^{2} + 2 \, a b x^{4} \operatorname{arsinh}\left (c x\right ) + a^{2} x^{4}\right )} \sqrt{c^{2} d x^{2} + d}}{c^{4} d^{2} x^{4} + 2 \, c^{2} d^{2} x^{2} + d^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d)^(3/2),x, algorithm="fricas")

[Out]

integral((b^2*x^4*arcsinh(c*x)^2 + 2*a*b*x^4*arcsinh(c*x) + a^2*x^4)*sqrt(c^2*d*x^2 + d)/(c^4*d^2*x^4 + 2*c^2*
d^2*x^2 + d^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4} \left (a + b \operatorname{asinh}{\left (c x \right )}\right )^{2}}{\left (d \left (c^{2} x^{2} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(a+b*asinh(c*x))**2/(c**2*d*x**2+d)**(3/2),x)

[Out]

Integral(x**4*(a + b*asinh(c*x))**2/(d*(c**2*x**2 + 1))**(3/2), 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} x^{4}}{{\left (c^{2} d x^{2} + d\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arcsinh(c*x))^2/(c^2*d*x^2+d)^(3/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)^2*x^4/(c^2*d*x^2 + d)^(3/2), x)