3.218 \(\int \frac{(c e+d e x)^2}{(a+b \sinh ^{-1}(c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=321 \[ -\frac{\sqrt{\pi } e^2 e^{a/b} \text{Erf}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{6 b^{5/2} d}+\frac{\sqrt{3 \pi } e^2 e^{\frac{3 a}{b}} \text{Erf}\left (\frac{\sqrt{3} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{2 b^{5/2} d}-\frac{\sqrt{\pi } e^2 e^{-\frac{a}{b}} \text{Erfi}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{6 b^{5/2} d}+\frac{\sqrt{3 \pi } e^2 e^{-\frac{3 a}{b}} \text{Erfi}\left (\frac{\sqrt{3} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{2 b^{5/2} d}-\frac{4 e^2 (c+d x)^3}{b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{8 e^2 (c+d x)}{3 b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{2 e^2 \sqrt{(c+d x)^2+1} (c+d x)^2}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}} \]

[Out]

(-2*e^2*(c + d*x)^2*Sqrt[1 + (c + d*x)^2])/(3*b*d*(a + b*ArcSinh[c + d*x])^(3/2)) - (8*e^2*(c + d*x))/(3*b^2*d
*Sqrt[a + b*ArcSinh[c + d*x]]) - (4*e^2*(c + d*x)^3)/(b^2*d*Sqrt[a + b*ArcSinh[c + d*x]]) - (e^2*E^(a/b)*Sqrt[
Pi]*Erf[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(6*b^(5/2)*d) + (e^2*E^((3*a)/b)*Sqrt[3*Pi]*Erf[(Sqrt[3]*Sqrt[a
+ b*ArcSinh[c + d*x]])/Sqrt[b]])/(2*b^(5/2)*d) - (e^2*Sqrt[Pi]*Erfi[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(6
*b^(5/2)*d*E^(a/b)) + (e^2*Sqrt[3*Pi]*Erfi[(Sqrt[3]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(2*b^(5/2)*d*E^((3
*a)/b))

________________________________________________________________________________________

Rubi [A]  time = 0.959963, antiderivative size = 321, normalized size of antiderivative = 1., number of steps used = 24, number of rules used = 11, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.44, Rules used = {5865, 12, 5667, 5774, 5669, 5448, 3307, 2180, 2204, 2205, 5657} \[ -\frac{\sqrt{\pi } e^2 e^{a/b} \text{Erf}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{6 b^{5/2} d}+\frac{\sqrt{3 \pi } e^2 e^{\frac{3 a}{b}} \text{Erf}\left (\frac{\sqrt{3} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{2 b^{5/2} d}-\frac{\sqrt{\pi } e^2 e^{-\frac{a}{b}} \text{Erfi}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{6 b^{5/2} d}+\frac{\sqrt{3 \pi } e^2 e^{-\frac{3 a}{b}} \text{Erfi}\left (\frac{\sqrt{3} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{2 b^{5/2} d}-\frac{4 e^2 (c+d x)^3}{b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{8 e^2 (c+d x)}{3 b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{2 e^2 \sqrt{(c+d x)^2+1} (c+d x)^2}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^2/(a + b*ArcSinh[c + d*x])^(5/2),x]

[Out]

(-2*e^2*(c + d*x)^2*Sqrt[1 + (c + d*x)^2])/(3*b*d*(a + b*ArcSinh[c + d*x])^(3/2)) - (8*e^2*(c + d*x))/(3*b^2*d
*Sqrt[a + b*ArcSinh[c + d*x]]) - (4*e^2*(c + d*x)^3)/(b^2*d*Sqrt[a + b*ArcSinh[c + d*x]]) - (e^2*E^(a/b)*Sqrt[
Pi]*Erf[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(6*b^(5/2)*d) + (e^2*E^((3*a)/b)*Sqrt[3*Pi]*Erf[(Sqrt[3]*Sqrt[a
+ b*ArcSinh[c + d*x]])/Sqrt[b]])/(2*b^(5/2)*d) - (e^2*Sqrt[Pi]*Erfi[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(6
*b^(5/2)*d*E^(a/b)) + (e^2*Sqrt[3*Pi]*Erfi[(Sqrt[3]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(2*b^(5/2)*d*E^((3
*a)/b))

Rule 5865

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
]

Rule 12

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

Rule 5667

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

Rule 5774

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

Rule 5669

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[(a + b*x)^n*
Sinh[x]^m*Cosh[x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 5448

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

Rule 3307

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/(E^(
I*k*Pi)*E^(I*(e + f*x))), x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*k*Pi)*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d
, e, f, m}, x] && IntegerQ[2*k]

Rule 2180

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[F^(g*(e - (c*
f)/d) + (f*g*x^2)/d), x], x, Sqrt[c + d*x]], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2205

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erf[(c + d*x)*Rt[-(b*Log[F]),
2]])/(2*d*Rt[-(b*Log[F]), 2]), x] /; FreeQ[{F, a, b, c, d}, x] && NegQ[b]

Rule 5657

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[1/(b*c), Subst[Int[x^n*Cosh[a/b - x/b], x], x,
a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c, n}, x]

Rubi steps

\begin{align*} \int \frac{(c e+d e x)^2}{\left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{e^2 x^2}{\left (a+b \sinh ^{-1}(x)\right )^{5/2}} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^2 \operatorname{Subst}\left (\int \frac{x^2}{\left (a+b \sinh ^{-1}(x)\right )^{5/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac{2 e^2 (c+d x)^2 \sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}+\frac{\left (4 e^2\right ) \operatorname{Subst}\left (\int \frac{x}{\sqrt{1+x^2} \left (a+b \sinh ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{3 b d}+\frac{\left (2 e^2\right ) \operatorname{Subst}\left (\int \frac{x^3}{\sqrt{1+x^2} \left (a+b \sinh ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{b d}\\ &=-\frac{2 e^2 (c+d x)^2 \sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{8 e^2 (c+d x)}{3 b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{4 e^2 (c+d x)^3}{b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}+\frac{\left (8 e^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b \sinh ^{-1}(x)}} \, dx,x,c+d x\right )}{3 b^2 d}+\frac{\left (12 e^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+b \sinh ^{-1}(x)}} \, dx,x,c+d x\right )}{b^2 d}\\ &=-\frac{2 e^2 (c+d x)^2 \sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{8 e^2 (c+d x)}{3 b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{4 e^2 (c+d x)^3}{b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}+\frac{\left (8 e^2\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{a}{b}-\frac{x}{b}\right )}{\sqrt{x}} \, dx,x,a+b \sinh ^{-1}(c+d x)\right )}{3 b^3 d}+\frac{\left (12 e^2\right ) \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh ^2(x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}\\ &=-\frac{2 e^2 (c+d x)^2 \sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{8 e^2 (c+d x)}{3 b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{4 e^2 (c+d x)^3}{b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}+\frac{\left (4 e^2\right ) \operatorname{Subst}\left (\int \frac{e^{-i \left (\frac{i a}{b}-\frac{i x}{b}\right )}}{\sqrt{x}} \, dx,x,a+b \sinh ^{-1}(c+d x)\right )}{3 b^3 d}+\frac{\left (4 e^2\right ) \operatorname{Subst}\left (\int \frac{e^{i \left (\frac{i a}{b}-\frac{i x}{b}\right )}}{\sqrt{x}} \, dx,x,a+b \sinh ^{-1}(c+d x)\right )}{3 b^3 d}+\frac{\left (12 e^2\right ) \operatorname{Subst}\left (\int \left (-\frac{\cosh (x)}{4 \sqrt{a+b x}}+\frac{\cosh (3 x)}{4 \sqrt{a+b x}}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}\\ &=-\frac{2 e^2 (c+d x)^2 \sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{8 e^2 (c+d x)}{3 b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{4 e^2 (c+d x)^3}{b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}+\frac{\left (8 e^2\right ) \operatorname{Subst}\left (\int e^{\frac{a}{b}-\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c+d x)}\right )}{3 b^3 d}+\frac{\left (8 e^2\right ) \operatorname{Subst}\left (\int e^{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c+d x)}\right )}{3 b^3 d}-\frac{\left (3 e^2\right ) \operatorname{Subst}\left (\int \frac{\cosh (x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}+\frac{\left (3 e^2\right ) \operatorname{Subst}\left (\int \frac{\cosh (3 x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}\\ &=-\frac{2 e^2 (c+d x)^2 \sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{8 e^2 (c+d x)}{3 b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{4 e^2 (c+d x)^3}{b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}+\frac{4 e^2 e^{a/b} \sqrt{\pi } \text{erf}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{3 b^{5/2} d}+\frac{4 e^2 e^{-\frac{a}{b}} \sqrt{\pi } \text{erfi}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{3 b^{5/2} d}+\frac{\left (3 e^2\right ) \operatorname{Subst}\left (\int \frac{e^{-3 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b^2 d}-\frac{\left (3 e^2\right ) \operatorname{Subst}\left (\int \frac{e^{-x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b^2 d}-\frac{\left (3 e^2\right ) \operatorname{Subst}\left (\int \frac{e^x}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b^2 d}+\frac{\left (3 e^2\right ) \operatorname{Subst}\left (\int \frac{e^{3 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2 b^2 d}\\ &=-\frac{2 e^2 (c+d x)^2 \sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{8 e^2 (c+d x)}{3 b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{4 e^2 (c+d x)^3}{b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}+\frac{4 e^2 e^{a/b} \sqrt{\pi } \text{erf}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{3 b^{5/2} d}+\frac{4 e^2 e^{-\frac{a}{b}} \sqrt{\pi } \text{erfi}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{3 b^{5/2} d}+\frac{\left (3 e^2\right ) \operatorname{Subst}\left (\int e^{\frac{3 a}{b}-\frac{3 x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c+d x)}\right )}{b^3 d}-\frac{\left (3 e^2\right ) \operatorname{Subst}\left (\int e^{\frac{a}{b}-\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c+d x)}\right )}{b^3 d}-\frac{\left (3 e^2\right ) \operatorname{Subst}\left (\int e^{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c+d x)}\right )}{b^3 d}+\frac{\left (3 e^2\right ) \operatorname{Subst}\left (\int e^{-\frac{3 a}{b}+\frac{3 x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c+d x)}\right )}{b^3 d}\\ &=-\frac{2 e^2 (c+d x)^2 \sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{8 e^2 (c+d x)}{3 b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{4 e^2 (c+d x)^3}{b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{e^2 e^{a/b} \sqrt{\pi } \text{erf}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{6 b^{5/2} d}+\frac{e^2 e^{\frac{3 a}{b}} \sqrt{3 \pi } \text{erf}\left (\frac{\sqrt{3} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{2 b^{5/2} d}-\frac{e^2 e^{-\frac{a}{b}} \sqrt{\pi } \text{erfi}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{6 b^{5/2} d}+\frac{e^2 e^{-\frac{3 a}{b}} \sqrt{3 \pi } \text{erfi}\left (\frac{\sqrt{3} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{2 b^{5/2} d}\\ \end{align*}

Mathematica [A]  time = 1.44945, size = 389, normalized size = 1.21 \[ \frac{e^2 e^{-3 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )} \left (-6 \sqrt{3} b e^{3 \sinh ^{-1}(c+d x)} \left (-\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )^{3/2} \text{Gamma}\left (\frac{1}{2},-\frac{3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )+2 b e^{\frac{2 a}{b}+3 \sinh ^{-1}(c+d x)} \left (-\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )^{3/2} \text{Gamma}\left (\frac{1}{2},-\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )+2 e^{\frac{4 a}{b}+3 \sinh ^{-1}(c+d x)} \sqrt{\frac{a}{b}+\sinh ^{-1}(c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Gamma}\left (\frac{1}{2},\frac{a}{b}+\sinh ^{-1}(c+d x)\right )-e^{\frac{3 a}{b}} \left (6 \sqrt{3} e^{3 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )} \sqrt{\frac{a}{b}+\sinh ^{-1}(c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Gamma}\left (\frac{1}{2},\frac{3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )+\left (e^{2 \sinh ^{-1}(c+d x)}-1\right ) \left (a \left (4 e^{2 \sinh ^{-1}(c+d x)}+6 e^{4 \sinh ^{-1}(c+d x)}+6\right )+b \left (e^{4 \sinh ^{-1}(c+d x)}-1\right )+2 b \left (2 e^{2 \sinh ^{-1}(c+d x)}+3 e^{4 \sinh ^{-1}(c+d x)}+3\right ) \sinh ^{-1}(c+d x)\right )\right )\right )}{12 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c*e + d*e*x)^2/(a + b*ArcSinh[c + d*x])^(5/2),x]

[Out]

(e^2*(2*E^((4*a)/b + 3*ArcSinh[c + d*x])*Sqrt[a/b + ArcSinh[c + d*x]]*(a + b*ArcSinh[c + d*x])*Gamma[1/2, a/b
+ ArcSinh[c + d*x]] - 6*Sqrt[3]*b*E^(3*ArcSinh[c + d*x])*(-((a + b*ArcSinh[c + d*x])/b))^(3/2)*Gamma[1/2, (-3*
(a + b*ArcSinh[c + d*x]))/b] + 2*b*E^((2*a)/b + 3*ArcSinh[c + d*x])*(-((a + b*ArcSinh[c + d*x])/b))^(3/2)*Gamm
a[1/2, -((a + b*ArcSinh[c + d*x])/b)] - E^((3*a)/b)*((-1 + E^(2*ArcSinh[c + d*x]))*(b*(-1 + E^(4*ArcSinh[c + d
*x])) + a*(6 + 4*E^(2*ArcSinh[c + d*x]) + 6*E^(4*ArcSinh[c + d*x])) + 2*b*(3 + 2*E^(2*ArcSinh[c + d*x]) + 3*E^
(4*ArcSinh[c + d*x]))*ArcSinh[c + d*x]) + 6*Sqrt[3]*E^(3*(a/b + ArcSinh[c + d*x]))*Sqrt[a/b + ArcSinh[c + d*x]
]*(a + b*ArcSinh[c + d*x])*Gamma[1/2, (3*(a + b*ArcSinh[c + d*x]))/b])))/(12*b^2*d*E^(3*(a/b + ArcSinh[c + d*x
]))*(a + b*ArcSinh[c + d*x])^(3/2))

________________________________________________________________________________________

Maple [F]  time = 0.203, size = 0, normalized size = 0. \begin{align*} \int{ \left ( dex+ce \right ) ^{2} \left ( a+b{\it Arcsinh} \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^2/(a+b*arcsinh(d*x+c))^(5/2),x)

[Out]

int((d*e*x+c*e)^2/(a+b*arcsinh(d*x+c))^(5/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d e x + c e\right )}^{2}}{{\left (b \operatorname{arsinh}\left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^2/(a+b*arcsinh(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((d*e*x + c*e)^2/(b*arcsinh(d*x + c) + a)^(5/2), x)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^2/(a+b*arcsinh(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**2/(a+b*asinh(d*x+c))**(5/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d e x + c e\right )}^{2}}{{\left (b \operatorname{arsinh}\left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^2/(a+b*arcsinh(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((d*e*x + c*e)^2/(b*arcsinh(d*x + c) + a)^(5/2), x)