3.217 \(\int \frac{(c e+d e x)^3}{(a+b \sinh ^{-1}(c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=326 \[ -\frac{2 \sqrt{\pi } e^3 e^{\frac{4 a}{b}} \text{Erf}\left (\frac{2 \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{3 b^{5/2} d}+\frac{\sqrt{2 \pi } e^3 e^{\frac{2 a}{b}} \text{Erf}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{3 b^{5/2} d}+\frac{2 \sqrt{\pi } e^3 e^{-\frac{4 a}{b}} \text{Erfi}\left (\frac{2 \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{3 b^{5/2} d}-\frac{\sqrt{2 \pi } e^3 e^{-\frac{2 a}{b}} \text{Erfi}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{3 b^{5/2} d}-\frac{16 e^3 (c+d x)^4}{3 b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{4 e^3 (c+d x)^2}{b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{2 e^3 \sqrt{(c+d x)^2+1} (c+d x)^3}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}} \]

[Out]

(-2*e^3*(c + d*x)^3*Sqrt[1 + (c + d*x)^2])/(3*b*d*(a + b*ArcSinh[c + d*x])^(3/2)) - (4*e^3*(c + d*x)^2)/(b^2*d
*Sqrt[a + b*ArcSinh[c + d*x]]) - (16*e^3*(c + d*x)^4)/(3*b^2*d*Sqrt[a + b*ArcSinh[c + d*x]]) - (2*e^3*E^((4*a)
/b)*Sqrt[Pi]*Erf[(2*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(3*b^(5/2)*d) + (e^3*E^((2*a)/b)*Sqrt[2*Pi]*Erf[(S
qrt[2]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(3*b^(5/2)*d) + (2*e^3*Sqrt[Pi]*Erfi[(2*Sqrt[a + b*ArcSinh[c +
d*x]])/Sqrt[b]])/(3*b^(5/2)*d*E^((4*a)/b)) - (e^3*Sqrt[2*Pi]*Erfi[(Sqrt[2]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[
b]])/(3*b^(5/2)*d*E^((2*a)/b))

________________________________________________________________________________________

Rubi [A]  time = 1.09579, antiderivative size = 326, normalized size of antiderivative = 1., number of steps used = 26, number of rules used = 10, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {5865, 12, 5667, 5774, 5669, 5448, 3308, 2180, 2204, 2205} \[ -\frac{2 \sqrt{\pi } e^3 e^{\frac{4 a}{b}} \text{Erf}\left (\frac{2 \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{3 b^{5/2} d}+\frac{\sqrt{2 \pi } e^3 e^{\frac{2 a}{b}} \text{Erf}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{3 b^{5/2} d}+\frac{2 \sqrt{\pi } e^3 e^{-\frac{4 a}{b}} \text{Erfi}\left (\frac{2 \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{3 b^{5/2} d}-\frac{\sqrt{2 \pi } e^3 e^{-\frac{2 a}{b}} \text{Erfi}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{3 b^{5/2} d}-\frac{16 e^3 (c+d x)^4}{3 b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{4 e^3 (c+d x)^2}{b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{2 e^3 \sqrt{(c+d x)^2+1} (c+d x)^3}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^3/(a + b*ArcSinh[c + d*x])^(5/2),x]

[Out]

(-2*e^3*(c + d*x)^3*Sqrt[1 + (c + d*x)^2])/(3*b*d*(a + b*ArcSinh[c + d*x])^(3/2)) - (4*e^3*(c + d*x)^2)/(b^2*d
*Sqrt[a + b*ArcSinh[c + d*x]]) - (16*e^3*(c + d*x)^4)/(3*b^2*d*Sqrt[a + b*ArcSinh[c + d*x]]) - (2*e^3*E^((4*a)
/b)*Sqrt[Pi]*Erf[(2*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(3*b^(5/2)*d) + (e^3*E^((2*a)/b)*Sqrt[2*Pi]*Erf[(S
qrt[2]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(3*b^(5/2)*d) + (2*e^3*Sqrt[Pi]*Erfi[(2*Sqrt[a + b*ArcSinh[c +
d*x]])/Sqrt[b]])/(3*b^(5/2)*d*E^((4*a)/b)) - (e^3*Sqrt[2*Pi]*Erfi[(Sqrt[2]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[
b]])/(3*b^(5/2)*d*E^((2*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 3308

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

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]

Rubi steps

\begin{align*} \int \frac{(c e+d e x)^3}{\left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{e^3 x^3}{\left (a+b \sinh ^{-1}(x)\right )^{5/2}} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^3 \operatorname{Subst}\left (\int \frac{x^3}{\left (a+b \sinh ^{-1}(x)\right )^{5/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac{2 e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}+\frac{\left (2 e^3\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1+x^2} \left (a+b \sinh ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{b d}+\frac{\left (8 e^3\right ) \operatorname{Subst}\left (\int \frac{x^4}{\sqrt{1+x^2} \left (a+b \sinh ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{3 b d}\\ &=-\frac{2 e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{4 e^3 (c+d x)^2}{b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{16 e^3 (c+d x)^4}{3 b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}+\frac{\left (8 e^3\right ) \operatorname{Subst}\left (\int \frac{x}{\sqrt{a+b \sinh ^{-1}(x)}} \, dx,x,c+d x\right )}{b^2 d}+\frac{\left (64 e^3\right ) \operatorname{Subst}\left (\int \frac{x^3}{\sqrt{a+b \sinh ^{-1}(x)}} \, dx,x,c+d x\right )}{3 b^2 d}\\ &=-\frac{2 e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{4 e^3 (c+d x)^2}{b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{16 e^3 (c+d x)^4}{3 b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}+\frac{\left (8 e^3\right ) \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh (x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}+\frac{\left (64 e^3\right ) \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh ^3(x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 b^2 d}\\ &=-\frac{2 e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{4 e^3 (c+d x)^2}{b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{16 e^3 (c+d x)^4}{3 b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}+\frac{\left (8 e^3\right ) \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{2 \sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}+\frac{\left (64 e^3\right ) \operatorname{Subst}\left (\int \left (-\frac{\sinh (2 x)}{4 \sqrt{a+b x}}+\frac{\sinh (4 x)}{8 \sqrt{a+b x}}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 b^2 d}\\ &=-\frac{2 e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{4 e^3 (c+d x)^2}{b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{16 e^3 (c+d x)^4}{3 b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}+\frac{\left (8 e^3\right ) \operatorname{Subst}\left (\int \frac{\sinh (4 x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 b^2 d}+\frac{\left (4 e^3\right ) \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}-\frac{\left (16 e^3\right ) \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 b^2 d}\\ &=-\frac{2 e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{4 e^3 (c+d x)^2}{b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{16 e^3 (c+d x)^4}{3 b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{\left (4 e^3\right ) \operatorname{Subst}\left (\int \frac{e^{-4 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 b^2 d}+\frac{\left (4 e^3\right ) \operatorname{Subst}\left (\int \frac{e^{4 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 b^2 d}-\frac{\left (2 e^3\right ) \operatorname{Subst}\left (\int \frac{e^{-2 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}+\frac{\left (2 e^3\right ) \operatorname{Subst}\left (\int \frac{e^{2 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{b^2 d}+\frac{\left (8 e^3\right ) \operatorname{Subst}\left (\int \frac{e^{-2 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 b^2 d}-\frac{\left (8 e^3\right ) \operatorname{Subst}\left (\int \frac{e^{2 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{3 b^2 d}\\ &=-\frac{2 e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{4 e^3 (c+d x)^2}{b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{16 e^3 (c+d x)^4}{3 b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{\left (8 e^3\right ) \operatorname{Subst}\left (\int e^{\frac{4 a}{b}-\frac{4 x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c+d x)}\right )}{3 b^3 d}+\frac{\left (8 e^3\right ) \operatorname{Subst}\left (\int e^{-\frac{4 a}{b}+\frac{4 x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c+d x)}\right )}{3 b^3 d}-\frac{\left (4 e^3\right ) \operatorname{Subst}\left (\int e^{\frac{2 a}{b}-\frac{2 x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c+d x)}\right )}{b^3 d}+\frac{\left (4 e^3\right ) \operatorname{Subst}\left (\int e^{-\frac{2 a}{b}+\frac{2 x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c+d x)}\right )}{b^3 d}+\frac{\left (16 e^3\right ) \operatorname{Subst}\left (\int e^{\frac{2 a}{b}-\frac{2 x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c+d x)}\right )}{3 b^3 d}-\frac{\left (16 e^3\right ) \operatorname{Subst}\left (\int e^{-\frac{2 a}{b}+\frac{2 x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c+d x)}\right )}{3 b^3 d}\\ &=-\frac{2 e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{3 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{4 e^3 (c+d x)^2}{b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{16 e^3 (c+d x)^4}{3 b^2 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{2 e^3 e^{\frac{4 a}{b}} \sqrt{\pi } \text{erf}\left (\frac{2 \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{3 b^{5/2} d}+\frac{e^3 e^{\frac{2 a}{b}} \sqrt{2 \pi } \text{erf}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{3 b^{5/2} d}+\frac{2 e^3 e^{-\frac{4 a}{b}} \sqrt{\pi } \text{erfi}\left (\frac{2 \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{3 b^{5/2} d}-\frac{e^3 e^{-\frac{2 a}{b}} \sqrt{2 \pi } \text{erfi}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{3 b^{5/2} d}\\ \end{align*}

Mathematica [A]  time = 1.74964, size = 390, normalized size = 1.2 \[ \frac{e^3 e^{-4 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )} \left (-8 b e^{4 \sinh ^{-1}(c+d x)} \left (-\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )^{3/2} \text{Gamma}\left (\frac{1}{2},-\frac{4 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )+4 \sqrt{2} b e^{\frac{2 a}{b}+4 \sinh ^{-1}(c+d x)} \left (-\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )^{3/2} \text{Gamma}\left (\frac{1}{2},-\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )+\frac{1}{2} e^{\frac{4 a}{b}} \left (-8 \sqrt{2} e^{\frac{2 a}{b}+4 \sinh ^{-1}(c+d x)} \sqrt{\frac{a}{b}+\sinh ^{-1}(c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Gamma}\left (\frac{1}{2},\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )+16 e^{4 \left (\frac{a}{b}+\sinh ^{-1}(c+d x)\right )} \sqrt{\frac{a}{b}+\sinh ^{-1}(c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Gamma}\left (\frac{1}{2},\frac{4 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )+\left (e^{2 \sinh ^{-1}(c+d x)}-1\right )^2 \left (-\left (8 a \left (e^{2 \sinh ^{-1}(c+d x)}+e^{4 \sinh ^{-1}(c+d x)}+1\right )+b \left (e^{4 \sinh ^{-1}(c+d x)}-1\right )+8 b \left (e^{2 \sinh ^{-1}(c+d x)}+e^{4 \sinh ^{-1}(c+d x)}+1\right ) \sinh ^{-1}(c+d x)\right )\right )\right )\right )}{12 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c*e + d*e*x)^3/(a + b*ArcSinh[c + d*x])^(5/2),x]

[Out]

(e^3*(-8*b*E^(4*ArcSinh[c + d*x])*(-((a + b*ArcSinh[c + d*x])/b))^(3/2)*Gamma[1/2, (-4*(a + b*ArcSinh[c + d*x]
))/b] + 4*Sqrt[2]*b*E^((2*a)/b + 4*ArcSinh[c + d*x])*(-((a + b*ArcSinh[c + d*x])/b))^(3/2)*Gamma[1/2, (-2*(a +
b*ArcSinh[c + d*x]))/b] + (E^((4*a)/b)*(-((-1 + E^(2*ArcSinh[c + d*x]))^2*(b*(-1 + E^(4*ArcSinh[c + d*x])) +
8*a*(1 + E^(2*ArcSinh[c + d*x]) + E^(4*ArcSinh[c + d*x])) + 8*b*(1 + E^(2*ArcSinh[c + d*x]) + E^(4*ArcSinh[c +
d*x]))*ArcSinh[c + d*x])) - 8*Sqrt[2]*E^((2*a)/b + 4*ArcSinh[c + d*x])*Sqrt[a/b + ArcSinh[c + d*x]]*(a + b*Ar
cSinh[c + d*x])*Gamma[1/2, (2*(a + b*ArcSinh[c + d*x]))/b] + 16*E^(4*(a/b + ArcSinh[c + d*x]))*Sqrt[a/b + ArcS
inh[c + d*x]]*(a + b*ArcSinh[c + d*x])*Gamma[1/2, (4*(a + b*ArcSinh[c + d*x]))/b]))/2))/(12*b^2*d*E^(4*(a/b +
ArcSinh[c + d*x]))*(a + b*ArcSinh[c + d*x])^(3/2))

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Maple [F]  time = 0.197, size = 0, normalized size = 0. \begin{align*} \int{ \left ( dex+ce \right ) ^{3} \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)^3/(a+b*arcsinh(d*x+c))^(5/2),x)

[Out]

int((d*e*x+c*e)^3/(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 )}^{3}}{{\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)^3/(a+b*arcsinh(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((d*e*x + c*e)^3/(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)^3/(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^{3} \left (\int \frac{c^{3}}{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^{3} x^{3}}{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{3 c 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{3 c^{2} 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)**3/(a+b*asinh(d*x+c))**(5/2),x)

[Out]

e**3*(Integral(c**3/(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**3*x**3/(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(3*c*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) + Integral(3*c**2*d*x/(a**2*sqrt(a + b*asinh(c + d*x)) + 2*a*b*s
qrt(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 )}^{3}}{{\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)^3/(a+b*arcsinh(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((d*e*x + c*e)^3/(b*arcsinh(d*x + c) + a)^(5/2), x)