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

Optimal. Leaf size=394 \[ -\frac{15 \sqrt{\pi } b^{5/2} e^2 e^{a/b} \text{Erf}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{64 d}+\frac{5 \sqrt{\frac{\pi }{3}} b^{5/2} e^2 e^{\frac{3 a}{b}} \text{Erf}\left (\frac{\sqrt{3} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{576 d}+\frac{15 \sqrt{\pi } b^{5/2} e^2 e^{-\frac{a}{b}} \text{Erfi}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{64 d}-\frac{5 \sqrt{\frac{\pi }{3}} b^{5/2} e^2 e^{-\frac{3 a}{b}} \text{Erfi}\left (\frac{\sqrt{3} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{576 d}+\frac{5 b^2 e^2 (c+d x)^3 \sqrt{a+b \sinh ^{-1}(c+d x)}}{36 d}-\frac{5 b^2 e^2 (c+d x) \sqrt{a+b \sinh ^{-1}(c+d x)}}{6 d}+\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{3 d}-\frac{5 b e^2 \sqrt{(c+d x)^2+1} (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{18 d}+\frac{5 b e^2 \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{9 d} \]

[Out]

(-5*b^2*e^2*(c + d*x)*Sqrt[a + b*ArcSinh[c + d*x]])/(6*d) + (5*b^2*e^2*(c + d*x)^3*Sqrt[a + b*ArcSinh[c + d*x]
])/(36*d) + (5*b*e^2*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x])^(3/2))/(9*d) - (5*b*e^2*(c + d*x)^2*Sqrt[1
+ (c + d*x)^2]*(a + b*ArcSinh[c + d*x])^(3/2))/(18*d) + (e^2*(c + d*x)^3*(a + b*ArcSinh[c + d*x])^(5/2))/(3*d
) - (15*b^(5/2)*e^2*E^(a/b)*Sqrt[Pi]*Erf[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(64*d) + (5*b^(5/2)*e^2*E^((3*
a)/b)*Sqrt[Pi/3]*Erf[(Sqrt[3]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(576*d) + (15*b^(5/2)*e^2*Sqrt[Pi]*Erfi[
Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(64*d*E^(a/b)) - (5*b^(5/2)*e^2*Sqrt[Pi/3]*Erfi[(Sqrt[3]*Sqrt[a + b*Arc
Sinh[c + d*x]])/Sqrt[b]])/(576*d*E^((3*a)/b))

________________________________________________________________________________________

Rubi [A]  time = 1.24447, antiderivative size = 394, normalized size of antiderivative = 1., number of steps used = 26, number of rules used = 12, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.48, Rules used = {5865, 12, 5663, 5758, 5717, 5653, 5779, 3308, 2180, 2204, 2205, 3312} \[ -\frac{15 \sqrt{\pi } b^{5/2} e^2 e^{a/b} \text{Erf}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{64 d}+\frac{5 \sqrt{\frac{\pi }{3}} b^{5/2} e^2 e^{\frac{3 a}{b}} \text{Erf}\left (\frac{\sqrt{3} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{576 d}+\frac{15 \sqrt{\pi } b^{5/2} e^2 e^{-\frac{a}{b}} \text{Erfi}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{64 d}-\frac{5 \sqrt{\frac{\pi }{3}} b^{5/2} e^2 e^{-\frac{3 a}{b}} \text{Erfi}\left (\frac{\sqrt{3} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{576 d}+\frac{5 b^2 e^2 (c+d x)^3 \sqrt{a+b \sinh ^{-1}(c+d x)}}{36 d}-\frac{5 b^2 e^2 (c+d x) \sqrt{a+b \sinh ^{-1}(c+d x)}}{6 d}+\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{3 d}-\frac{5 b e^2 \sqrt{(c+d x)^2+1} (c+d x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{18 d}+\frac{5 b e^2 \sqrt{(c+d x)^2+1} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*b^2*e^2*(c + d*x)*Sqrt[a + b*ArcSinh[c + d*x]])/(6*d) + (5*b^2*e^2*(c + d*x)^3*Sqrt[a + b*ArcSinh[c + d*x]
])/(36*d) + (5*b*e^2*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*x])^(3/2))/(9*d) - (5*b*e^2*(c + d*x)^2*Sqrt[1
+ (c + d*x)^2]*(a + b*ArcSinh[c + d*x])^(3/2))/(18*d) + (e^2*(c + d*x)^3*(a + b*ArcSinh[c + d*x])^(5/2))/(3*d
) - (15*b^(5/2)*e^2*E^(a/b)*Sqrt[Pi]*Erf[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(64*d) + (5*b^(5/2)*e^2*E^((3*
a)/b)*Sqrt[Pi/3]*Erf[(Sqrt[3]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(576*d) + (15*b^(5/2)*e^2*Sqrt[Pi]*Erfi[
Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(64*d*E^(a/b)) - (5*b^(5/2)*e^2*Sqrt[Pi/3]*Erfi[(Sqrt[3]*Sqrt[a + b*Arc
Sinh[c + d*x]])/Sqrt[b]])/(576*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 5663

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[(x^(m + 1)*(a + b*ArcSinh[c*x])^n)/
(m + 1), x] - Dist[(b*c*n)/(m + 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] && GtQ[n, 0]

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 5717

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

Rule 5653

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

Rule 5779

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

Rule 3312

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rubi steps

\begin{align*} \int (c e+d e x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2} \, dx &=\frac{\operatorname{Subst}\left (\int 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 x^2 \left (a+b \sinh ^{-1}(x)\right )^{5/2} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{3 d}-\frac{\left (5 b e^2\right ) \operatorname{Subst}\left (\int \frac{x^3 \left (a+b \sinh ^{-1}(x)\right )^{3/2}}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{6 d}\\ &=-\frac{5 b e^2 (c+d x)^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{18 d}+\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{3 d}+\frac{\left (5 b e^2\right ) \operatorname{Subst}\left (\int \frac{x \left (a+b \sinh ^{-1}(x)\right )^{3/2}}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{9 d}+\frac{\left (5 b^2 e^2\right ) \operatorname{Subst}\left (\int x^2 \sqrt{a+b \sinh ^{-1}(x)} \, dx,x,c+d x\right )}{12 d}\\ &=\frac{5 b^2 e^2 (c+d x)^3 \sqrt{a+b \sinh ^{-1}(c+d x)}}{36 d}+\frac{5 b e^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{9 d}-\frac{5 b e^2 (c+d x)^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{18 d}+\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{3 d}-\frac{\left (5 b^2 e^2\right ) \operatorname{Subst}\left (\int \sqrt{a+b \sinh ^{-1}(x)} \, dx,x,c+d x\right )}{6 d}-\frac{\left (5 b^3 e^2\right ) \operatorname{Subst}\left (\int \frac{x^3}{\sqrt{1+x^2} \sqrt{a+b \sinh ^{-1}(x)}} \, dx,x,c+d x\right )}{72 d}\\ &=-\frac{5 b^2 e^2 (c+d x) \sqrt{a+b \sinh ^{-1}(c+d x)}}{6 d}+\frac{5 b^2 e^2 (c+d x)^3 \sqrt{a+b \sinh ^{-1}(c+d x)}}{36 d}+\frac{5 b e^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{9 d}-\frac{5 b e^2 (c+d x)^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{18 d}+\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{3 d}-\frac{\left (5 b^3 e^2\right ) \operatorname{Subst}\left (\int \frac{\sinh ^3(x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{72 d}+\frac{\left (5 b^3 e^2\right ) \operatorname{Subst}\left (\int \frac{x}{\sqrt{1+x^2} \sqrt{a+b \sinh ^{-1}(x)}} \, dx,x,c+d x\right )}{12 d}\\ &=-\frac{5 b^2 e^2 (c+d x) \sqrt{a+b \sinh ^{-1}(c+d x)}}{6 d}+\frac{5 b^2 e^2 (c+d x)^3 \sqrt{a+b \sinh ^{-1}(c+d x)}}{36 d}+\frac{5 b e^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{9 d}-\frac{5 b e^2 (c+d x)^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{18 d}+\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{3 d}-\frac{\left (5 i b^3 e^2\right ) \operatorname{Subst}\left (\int \left (\frac{3 i \sinh (x)}{4 \sqrt{a+b x}}-\frac{i \sinh (3 x)}{4 \sqrt{a+b x}}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{72 d}+\frac{\left (5 b^3 e^2\right ) \operatorname{Subst}\left (\int \frac{\sinh (x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{12 d}\\ &=-\frac{5 b^2 e^2 (c+d x) \sqrt{a+b \sinh ^{-1}(c+d x)}}{6 d}+\frac{5 b^2 e^2 (c+d x)^3 \sqrt{a+b \sinh ^{-1}(c+d x)}}{36 d}+\frac{5 b e^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{9 d}-\frac{5 b e^2 (c+d x)^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{18 d}+\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{3 d}-\frac{\left (5 b^3 e^2\right ) \operatorname{Subst}\left (\int \frac{\sinh (3 x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{288 d}+\frac{\left (5 b^3 e^2\right ) \operatorname{Subst}\left (\int \frac{\sinh (x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{96 d}-\frac{\left (5 b^3 e^2\right ) \operatorname{Subst}\left (\int \frac{e^{-x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{24 d}+\frac{\left (5 b^3 e^2\right ) \operatorname{Subst}\left (\int \frac{e^x}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{24 d}\\ &=-\frac{5 b^2 e^2 (c+d x) \sqrt{a+b \sinh ^{-1}(c+d x)}}{6 d}+\frac{5 b^2 e^2 (c+d x)^3 \sqrt{a+b \sinh ^{-1}(c+d x)}}{36 d}+\frac{5 b e^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{9 d}-\frac{5 b e^2 (c+d x)^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{18 d}+\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{3 d}-\frac{\left (5 b^2 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 )}{12 d}+\frac{\left (5 b^2 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 )}{12 d}+\frac{\left (5 b^3 e^2\right ) \operatorname{Subst}\left (\int \frac{e^{-3 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{576 d}-\frac{\left (5 b^3 e^2\right ) \operatorname{Subst}\left (\int \frac{e^{3 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{576 d}-\frac{\left (5 b^3 e^2\right ) \operatorname{Subst}\left (\int \frac{e^{-x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{192 d}+\frac{\left (5 b^3 e^2\right ) \operatorname{Subst}\left (\int \frac{e^x}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{192 d}\\ &=-\frac{5 b^2 e^2 (c+d x) \sqrt{a+b \sinh ^{-1}(c+d x)}}{6 d}+\frac{5 b^2 e^2 (c+d x)^3 \sqrt{a+b \sinh ^{-1}(c+d x)}}{36 d}+\frac{5 b e^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{9 d}-\frac{5 b e^2 (c+d x)^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{18 d}+\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{3 d}-\frac{5 b^{5/2} e^2 e^{a/b} \sqrt{\pi } \text{erf}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{24 d}+\frac{5 b^{5/2} e^2 e^{-\frac{a}{b}} \sqrt{\pi } \text{erfi}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{24 d}+\frac{\left (5 b^2 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 )}{288 d}-\frac{\left (5 b^2 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 )}{288 d}-\frac{\left (5 b^2 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 )}{96 d}+\frac{\left (5 b^2 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 )}{96 d}\\ &=-\frac{5 b^2 e^2 (c+d x) \sqrt{a+b \sinh ^{-1}(c+d x)}}{6 d}+\frac{5 b^2 e^2 (c+d x)^3 \sqrt{a+b \sinh ^{-1}(c+d x)}}{36 d}+\frac{5 b e^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{9 d}-\frac{5 b e^2 (c+d x)^2 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{18 d}+\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{3 d}-\frac{15 b^{5/2} e^2 e^{a/b} \sqrt{\pi } \text{erf}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{64 d}+\frac{5 b^{5/2} e^2 e^{\frac{3 a}{b}} \sqrt{\frac{\pi }{3}} \text{erf}\left (\frac{\sqrt{3} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{576 d}+\frac{15 b^{5/2} e^2 e^{-\frac{a}{b}} \sqrt{\pi } \text{erfi}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{64 d}-\frac{5 b^{5/2} e^2 e^{-\frac{3 a}{b}} \sqrt{\frac{\pi }{3}} \text{erfi}\left (\frac{\sqrt{3} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{576 d}\\ \end{align*}

Mathematica [A]  time = 0.413291, size = 238, normalized size = 0.6 \[ -\frac{e^2 e^{-\frac{3 a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2} \left (81 e^{\frac{4 a}{b}} \sqrt{-\frac{a+b \sinh ^{-1}(c+d x)}{b}} \text{Gamma}\left (\frac{7}{2},\frac{a}{b}+\sinh ^{-1}(c+d x)\right )+\sqrt{3} \sqrt{\frac{a}{b}+\sinh ^{-1}(c+d x)} \text{Gamma}\left (\frac{7}{2},-\frac{3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )-81 e^{\frac{2 a}{b}} \sqrt{\frac{a}{b}+\sinh ^{-1}(c+d x)} \text{Gamma}\left (\frac{7}{2},-\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )-\sqrt{3} e^{\frac{6 a}{b}} \sqrt{-\frac{a+b \sinh ^{-1}(c+d x)}{b}} \text{Gamma}\left (\frac{7}{2},\frac{3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )\right )}{648 d \left (-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^2}{b^2}\right )^{3/2}} \]

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*(a + b*ArcSinh[c + d*x])^(5/2)*(81*E^((4*a)/b)*Sqrt[-((a + b*ArcSinh[c + d*x])/b)]*Gamma[7/2, a/b + ArcS
inh[c + d*x]] + Sqrt[3]*Sqrt[a/b + ArcSinh[c + d*x]]*Gamma[7/2, (-3*(a + b*ArcSinh[c + d*x]))/b] - 81*E^((2*a)
/b)*Sqrt[a/b + ArcSinh[c + d*x]]*Gamma[7/2, -((a + b*ArcSinh[c + d*x])/b)] - Sqrt[3]*E^((6*a)/b)*Sqrt[-((a + b
*ArcSinh[c + d*x])/b)]*Gamma[7/2, (3*(a + b*ArcSinh[c + d*x]))/b]))/(648*d*E^((3*a)/b)*(-((a + b*ArcSinh[c + d
*x])^2/b^2))^(3/2))

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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{\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(-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)**2*(a+b*asinh(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\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)