3.223 \(\int \frac{(c e+d e x)^3}{(a+b \sinh ^{-1}(c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=420 \[ \frac{16 \sqrt{\pi } e^3 e^{\frac{4 a}{b}} \text{Erf}\left (\frac{2 \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{15 b^{7/2} d}-\frac{4 \sqrt{2 \pi } e^3 e^{\frac{2 a}{b}} \text{Erf}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{15 b^{7/2} d}+\frac{16 \sqrt{\pi } e^3 e^{-\frac{4 a}{b}} \text{Erfi}\left (\frac{2 \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{15 b^{7/2} d}-\frac{4 \sqrt{2 \pi } e^3 e^{-\frac{2 a}{b}} \text{Erfi}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{15 b^{7/2} d}-\frac{16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{128 e^3 \sqrt{(c+d x)^2+1} (c+d x)^3}{15 b^3 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{16 e^3 \sqrt{(c+d x)^2+1} (c+d x)}{5 b^3 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{2 e^3 \sqrt{(c+d x)^2+1} (c+d x)^3}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}} \]

[Out]

(-2*e^3*(c + d*x)^3*Sqrt[1 + (c + d*x)^2])/(5*b*d*(a + b*ArcSinh[c + d*x])^(5/2)) - (4*e^3*(c + d*x)^2)/(5*b^2
*d*(a + b*ArcSinh[c + d*x])^(3/2)) - (16*e^3*(c + d*x)^4)/(15*b^2*d*(a + b*ArcSinh[c + d*x])^(3/2)) - (16*e^3*
(c + d*x)*Sqrt[1 + (c + d*x)^2])/(5*b^3*d*Sqrt[a + b*ArcSinh[c + d*x]]) - (128*e^3*(c + d*x)^3*Sqrt[1 + (c + d
*x)^2])/(15*b^3*d*Sqrt[a + b*ArcSinh[c + d*x]]) + (16*e^3*E^((4*a)/b)*Sqrt[Pi]*Erf[(2*Sqrt[a + b*ArcSinh[c + d
*x]])/Sqrt[b]])/(15*b^(7/2)*d) - (4*e^3*E^((2*a)/b)*Sqrt[2*Pi]*Erf[(Sqrt[2]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt
[b]])/(15*b^(7/2)*d) + (16*e^3*Sqrt[Pi]*Erfi[(2*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(15*b^(7/2)*d*E^((4*a)
/b)) - (4*e^3*Sqrt[2*Pi]*Erfi[(Sqrt[2]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(15*b^(7/2)*d*E^((2*a)/b))

________________________________________________________________________________________

Rubi [A]  time = 1.09819, antiderivative size = 420, normalized size of antiderivative = 1., number of steps used = 23, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {5865, 12, 5667, 5774, 5665, 3307, 2180, 2204, 2205} \[ \frac{16 \sqrt{\pi } e^3 e^{\frac{4 a}{b}} \text{Erf}\left (\frac{2 \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{15 b^{7/2} d}-\frac{4 \sqrt{2 \pi } e^3 e^{\frac{2 a}{b}} \text{Erf}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{15 b^{7/2} d}+\frac{16 \sqrt{\pi } e^3 e^{-\frac{4 a}{b}} \text{Erfi}\left (\frac{2 \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{15 b^{7/2} d}-\frac{4 \sqrt{2 \pi } e^3 e^{-\frac{2 a}{b}} \text{Erfi}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{15 b^{7/2} d}-\frac{16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{128 e^3 \sqrt{(c+d x)^2+1} (c+d x)^3}{15 b^3 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{16 e^3 \sqrt{(c+d x)^2+1} (c+d x)}{5 b^3 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{2 e^3 \sqrt{(c+d x)^2+1} (c+d x)^3}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^3/(a + b*ArcSinh[c + d*x])^(7/2),x]

[Out]

(-2*e^3*(c + d*x)^3*Sqrt[1 + (c + d*x)^2])/(5*b*d*(a + b*ArcSinh[c + d*x])^(5/2)) - (4*e^3*(c + d*x)^2)/(5*b^2
*d*(a + b*ArcSinh[c + d*x])^(3/2)) - (16*e^3*(c + d*x)^4)/(15*b^2*d*(a + b*ArcSinh[c + d*x])^(3/2)) - (16*e^3*
(c + d*x)*Sqrt[1 + (c + d*x)^2])/(5*b^3*d*Sqrt[a + b*ArcSinh[c + d*x]]) - (128*e^3*(c + d*x)^3*Sqrt[1 + (c + d
*x)^2])/(15*b^3*d*Sqrt[a + b*ArcSinh[c + d*x]]) + (16*e^3*E^((4*a)/b)*Sqrt[Pi]*Erf[(2*Sqrt[a + b*ArcSinh[c + d
*x]])/Sqrt[b]])/(15*b^(7/2)*d) - (4*e^3*E^((2*a)/b)*Sqrt[2*Pi]*Erf[(Sqrt[2]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt
[b]])/(15*b^(7/2)*d) + (16*e^3*Sqrt[Pi]*Erfi[(2*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(15*b^(7/2)*d*E^((4*a)
/b)) - (4*e^3*Sqrt[2*Pi]*Erfi[(Sqrt[2]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(15*b^(7/2)*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 5665

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[1/(b*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[(a + b*x)^(n +
1), Sinh[x]^(m - 1)*(m + (m + 1)*Sinh[x]^2), x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0]
&& GeQ[n, -2] && LtQ[n, -1]

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]

Rubi steps

\begin{align*} \int \frac{(c e+d e x)^3}{\left (a+b \sinh ^{-1}(c+d x)\right )^{7/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{e^3 x^3}{\left (a+b \sinh ^{-1}(x)\right )^{7/2}} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^3 \operatorname{Subst}\left (\int \frac{x^3}{\left (a+b \sinh ^{-1}(x)\right )^{7/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac{2 e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}+\frac{\left (6 e^3\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1+x^2} \left (a+b \sinh ^{-1}(x)\right )^{5/2}} \, dx,x,c+d x\right )}{5 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 )^{5/2}} \, dx,x,c+d x\right )}{5 b d}\\ &=-\frac{2 e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}-\frac{4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}+\frac{\left (8 e^3\right ) \operatorname{Subst}\left (\int \frac{x}{\left (a+b \sinh ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{5 b^2 d}+\frac{\left (64 e^3\right ) \operatorname{Subst}\left (\int \frac{x^3}{\left (a+b \sinh ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{15 b^2 d}\\ &=-\frac{2 e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}-\frac{4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{16 e^3 (c+d x) \sqrt{1+(c+d x)^2}}{5 b^3 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{128 e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{15 b^3 d \sqrt{a+b \sinh ^{-1}(c+d x)}}+\frac{\left (16 e^3\right ) \operatorname{Subst}\left (\int \frac{\cosh (2 x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{5 b^3 d}+\frac{\left (128 e^3\right ) \operatorname{Subst}\left (\int \left (-\frac{\cosh (2 x)}{2 \sqrt{a+b x}}+\frac{\cosh (4 x)}{2 \sqrt{a+b x}}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{15 b^3 d}\\ &=-\frac{2 e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}-\frac{4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{16 e^3 (c+d x) \sqrt{1+(c+d x)^2}}{5 b^3 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{128 e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{15 b^3 d \sqrt{a+b \sinh ^{-1}(c+d x)}}+\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 )}{5 b^3 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 )}{5 b^3 d}-\frac{\left (64 e^3\right ) \operatorname{Subst}\left (\int \frac{\cosh (2 x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{15 b^3 d}+\frac{\left (64 e^3\right ) \operatorname{Subst}\left (\int \frac{\cosh (4 x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{15 b^3 d}\\ &=-\frac{2 e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}-\frac{4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{16 e^3 (c+d x) \sqrt{1+(c+d x)^2}}{5 b^3 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{128 e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{15 b^3 d \sqrt{a+b \sinh ^{-1}(c+d x)}}+\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 )}{5 b^4 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 )}{5 b^4 d}+\frac{\left (32 e^3\right ) \operatorname{Subst}\left (\int \frac{e^{-4 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{15 b^3 d}-\frac{\left (32 e^3\right ) \operatorname{Subst}\left (\int \frac{e^{-2 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{15 b^3 d}-\frac{\left (32 e^3\right ) \operatorname{Subst}\left (\int \frac{e^{2 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{15 b^3 d}+\frac{\left (32 e^3\right ) \operatorname{Subst}\left (\int \frac{e^{4 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{15 b^3 d}\\ &=-\frac{2 e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}-\frac{4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{16 e^3 (c+d x) \sqrt{1+(c+d x)^2}}{5 b^3 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{128 e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{15 b^3 d \sqrt{a+b \sinh ^{-1}(c+d x)}}+\frac{4 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 )}{5 b^{7/2} d}+\frac{4 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 )}{5 b^{7/2} d}+\frac{\left (64 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 )}{15 b^4 d}-\frac{\left (64 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 )}{15 b^4 d}-\frac{\left (64 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 )}{15 b^4 d}+\frac{\left (64 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 )}{15 b^4 d}\\ &=-\frac{2 e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{5 b d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}-\frac{4 e^3 (c+d x)^2}{5 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{16 e^3 (c+d x)^4}{15 b^2 d \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}-\frac{16 e^3 (c+d x) \sqrt{1+(c+d x)^2}}{5 b^3 d \sqrt{a+b \sinh ^{-1}(c+d x)}}-\frac{128 e^3 (c+d x)^3 \sqrt{1+(c+d x)^2}}{15 b^3 d \sqrt{a+b \sinh ^{-1}(c+d x)}}+\frac{16 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 )}{15 b^{7/2} d}-\frac{4 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 )}{15 b^{7/2} d}+\frac{16 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 )}{15 b^{7/2} d}-\frac{4 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 )}{15 b^{7/2} d}\\ \end{align*}

Mathematica [A]  time = 2.06066, size = 429, normalized size = 1.02 \[ \frac{e^3 \left (4 \left (a+b \sinh ^{-1}(c+d x)\right ) \left (4 \sqrt{2} b e^{-\frac{2 a}{b}} \left (-\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )^{3/2} \text{Gamma}\left (\frac{1}{2},-\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )+4 \sqrt{2} e^{\frac{2 a}{b}} \sqrt{\frac{a}{b}+\sinh ^{-1}(c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Gamma}\left (\frac{1}{2},\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )+e^{2 \sinh ^{-1}(c+d x)} \left (4 a+4 b \sinh ^{-1}(c+d x)+b\right )-4 a e^{-2 \sinh ^{-1}(c+d x)}+b e^{-2 \sinh ^{-1}(c+d x)}-4 b e^{-2 \sinh ^{-1}(c+d x)} \sinh ^{-1}(c+d x)\right )-4 \left (a+b \sinh ^{-1}(c+d x)\right ) \left (16 b e^{-\frac{4 a}{b}} \left (-\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )^{3/2} \text{Gamma}\left (\frac{1}{2},-\frac{4 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )+16 e^{\frac{4 a}{b}} \sqrt{\frac{a}{b}+\sinh ^{-1}(c+d x)} \left (a+b \sinh ^{-1}(c+d x)\right ) \text{Gamma}\left (\frac{1}{2},\frac{4 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )+e^{4 \sinh ^{-1}(c+d x)} \left (8 a+8 b \sinh ^{-1}(c+d x)+b\right )-8 a e^{-4 \sinh ^{-1}(c+d x)}+b e^{-4 \sinh ^{-1}(c+d x)} \left (1-8 \sinh ^{-1}(c+d x)\right )\right )+6 b^2 \sinh \left (2 \sinh ^{-1}(c+d x)\right )-3 b^2 \sinh \left (4 \sinh ^{-1}(c+d x)\right )\right )}{60 b^3 d \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c*e + d*e*x)^3/(a + b*ArcSinh[c + d*x])^(7/2),x]

[Out]

(e^3*(4*(a + b*ArcSinh[c + d*x])*((-4*a)/E^(2*ArcSinh[c + d*x]) + b/E^(2*ArcSinh[c + d*x]) - (4*b*ArcSinh[c +
d*x])/E^(2*ArcSinh[c + d*x]) + E^(2*ArcSinh[c + d*x])*(4*a + b + 4*b*ArcSinh[c + d*x]) + (4*Sqrt[2]*b*(-((a +
b*ArcSinh[c + d*x])/b))^(3/2)*Gamma[1/2, (-2*(a + b*ArcSinh[c + d*x]))/b])/E^((2*a)/b) + 4*Sqrt[2]*E^((2*a)/b)
*Sqrt[a/b + ArcSinh[c + d*x]]*(a + b*ArcSinh[c + d*x])*Gamma[1/2, (2*(a + b*ArcSinh[c + d*x]))/b]) - 4*(a + b*
ArcSinh[c + d*x])*((-8*a)/E^(4*ArcSinh[c + d*x]) + (b*(1 - 8*ArcSinh[c + d*x]))/E^(4*ArcSinh[c + d*x]) + E^(4*
ArcSinh[c + d*x])*(8*a + b + 8*b*ArcSinh[c + d*x]) + (16*b*(-((a + b*ArcSinh[c + d*x])/b))^(3/2)*Gamma[1/2, (-
4*(a + b*ArcSinh[c + d*x]))/b])/E^((4*a)/b) + 16*E^((4*a)/b)*Sqrt[a/b + ArcSinh[c + d*x]]*(a + b*ArcSinh[c + d
*x])*Gamma[1/2, (4*(a + b*ArcSinh[c + d*x]))/b]) + 6*b^2*Sinh[2*ArcSinh[c + d*x]] - 3*b^2*Sinh[4*ArcSinh[c + d
*x]]))/(60*b^3*d*(a + b*ArcSinh[c + d*x])^(5/2))

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Maple [F]  time = 0.192, size = 0, normalized size = 0. \begin{align*} \int{ \left ( dex+ce \right ) ^{3} \left ( a+b{\it Arcsinh} \left ( dx+c \right ) \right ) ^{-{\frac{7}{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))^(7/2),x)

[Out]

int((d*e*x+c*e)^3/(a+b*arcsinh(d*x+c))^(7/2),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{7}{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))^(7/2),x, algorithm="maxima")

[Out]

integrate((d*e*x + c*e)^3/(b*arcsinh(d*x + c) + a)^(7/2), 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))^(7/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))**(7/2),x)

[Out]

Timed out

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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{7}{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))^(7/2),x, algorithm="giac")

[Out]

integrate((d*e*x + c*e)^3/(b*arcsinh(d*x + c) + a)^(7/2), x)