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

Optimal. Leaf size=455 \[ -\frac{15 \sqrt{\pi } b^{5/2} e^3 e^{\frac{4 a}{b}} \text{Erf}\left (\frac{2 \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{16384 d}+\frac{15 \sqrt{\frac{\pi }{2}} b^{5/2} e^3 e^{\frac{2 a}{b}} \text{Erf}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{512 d}-\frac{15 \sqrt{\pi } b^{5/2} e^3 e^{-\frac{4 a}{b}} \text{Erfi}\left (\frac{2 \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{16384 d}+\frac{15 \sqrt{\frac{\pi }{2}} b^{5/2} e^3 e^{-\frac{2 a}{b}} \text{Erfi}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{512 d}+\frac{15 b^2 e^3 (c+d x)^4 \sqrt{a+b \sinh ^{-1}(c+d x)}}{256 d}-\frac{45 b^2 e^3 (c+d x)^2 \sqrt{a+b \sinh ^{-1}(c+d x)}}{256 d}-\frac{225 b^2 e^3 \sqrt{a+b \sinh ^{-1}(c+d x)}}{2048 d}+\frac{e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}-\frac{5 b e^3 \sqrt{(c+d x)^2+1} (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{32 d}+\frac{15 b e^3 \sqrt{(c+d x)^2+1} (c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{64 d}-\frac{3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{32 d} \]

[Out]

(-225*b^2*e^3*Sqrt[a + b*ArcSinh[c + d*x]])/(2048*d) - (45*b^2*e^3*(c + d*x)^2*Sqrt[a + b*ArcSinh[c + d*x]])/(
256*d) + (15*b^2*e^3*(c + d*x)^4*Sqrt[a + b*ArcSinh[c + d*x]])/(256*d) + (15*b*e^3*(c + d*x)*Sqrt[1 + (c + d*x
)^2]*(a + b*ArcSinh[c + d*x])^(3/2))/(64*d) - (5*b*e^3*(c + d*x)^3*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*
x])^(3/2))/(32*d) - (3*e^3*(a + b*ArcSinh[c + d*x])^(5/2))/(32*d) + (e^3*(c + d*x)^4*(a + b*ArcSinh[c + d*x])^
(5/2))/(4*d) - (15*b^(5/2)*e^3*E^((4*a)/b)*Sqrt[Pi]*Erf[(2*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(16384*d) +
(15*b^(5/2)*e^3*E^((2*a)/b)*Sqrt[Pi/2]*Erf[(Sqrt[2]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(512*d) - (15*b^(
5/2)*e^3*Sqrt[Pi]*Erfi[(2*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(16384*d*E^((4*a)/b)) + (15*b^(5/2)*e^3*Sqrt
[Pi/2]*Erfi[(Sqrt[2]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(512*d*E^((2*a)/b))

________________________________________________________________________________________

Rubi [A]  time = 1.51963, antiderivative size = 455, normalized size of antiderivative = 1., number of steps used = 29, number of rules used = 11, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.44, Rules used = {5865, 12, 5663, 5758, 5675, 5779, 3312, 3307, 2180, 2204, 2205} \[ -\frac{15 \sqrt{\pi } b^{5/2} e^3 e^{\frac{4 a}{b}} \text{Erf}\left (\frac{2 \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{16384 d}+\frac{15 \sqrt{\frac{\pi }{2}} b^{5/2} e^3 e^{\frac{2 a}{b}} \text{Erf}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{512 d}-\frac{15 \sqrt{\pi } b^{5/2} e^3 e^{-\frac{4 a}{b}} \text{Erfi}\left (\frac{2 \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{16384 d}+\frac{15 \sqrt{\frac{\pi }{2}} b^{5/2} e^3 e^{-\frac{2 a}{b}} \text{Erfi}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{512 d}+\frac{15 b^2 e^3 (c+d x)^4 \sqrt{a+b \sinh ^{-1}(c+d x)}}{256 d}-\frac{45 b^2 e^3 (c+d x)^2 \sqrt{a+b \sinh ^{-1}(c+d x)}}{256 d}-\frac{225 b^2 e^3 \sqrt{a+b \sinh ^{-1}(c+d x)}}{2048 d}+\frac{e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}-\frac{5 b e^3 \sqrt{(c+d x)^2+1} (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{32 d}+\frac{15 b e^3 \sqrt{(c+d x)^2+1} (c+d x) \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{64 d}-\frac{3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{32 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-225*b^2*e^3*Sqrt[a + b*ArcSinh[c + d*x]])/(2048*d) - (45*b^2*e^3*(c + d*x)^2*Sqrt[a + b*ArcSinh[c + d*x]])/(
256*d) + (15*b^2*e^3*(c + d*x)^4*Sqrt[a + b*ArcSinh[c + d*x]])/(256*d) + (15*b*e^3*(c + d*x)*Sqrt[1 + (c + d*x
)^2]*(a + b*ArcSinh[c + d*x])^(3/2))/(64*d) - (5*b*e^3*(c + d*x)^3*Sqrt[1 + (c + d*x)^2]*(a + b*ArcSinh[c + d*
x])^(3/2))/(32*d) - (3*e^3*(a + b*ArcSinh[c + d*x])^(5/2))/(32*d) + (e^3*(c + d*x)^4*(a + b*ArcSinh[c + d*x])^
(5/2))/(4*d) - (15*b^(5/2)*e^3*E^((4*a)/b)*Sqrt[Pi]*Erf[(2*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(16384*d) +
(15*b^(5/2)*e^3*E^((2*a)/b)*Sqrt[Pi/2]*Erf[(Sqrt[2]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(512*d) - (15*b^(
5/2)*e^3*Sqrt[Pi]*Erfi[(2*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(16384*d*E^((4*a)/b)) + (15*b^(5/2)*e^3*Sqrt
[Pi/2]*Erfi[(Sqrt[2]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(512*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 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 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 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 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])
)

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 (c e+d e x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2} \, dx &=\frac{\operatorname{Subst}\left (\int 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 x^3 \left (a+b \sinh ^{-1}(x)\right )^{5/2} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}-\frac{\left (5 b e^3\right ) \operatorname{Subst}\left (\int \frac{x^4 \left (a+b \sinh ^{-1}(x)\right )^{3/2}}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{8 d}\\ &=-\frac{5 b e^3 (c+d x)^3 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{32 d}+\frac{e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}+\frac{\left (15 b e^3\right ) \operatorname{Subst}\left (\int \frac{x^2 \left (a+b \sinh ^{-1}(x)\right )^{3/2}}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{32 d}+\frac{\left (15 b^2 e^3\right ) \operatorname{Subst}\left (\int x^3 \sqrt{a+b \sinh ^{-1}(x)} \, dx,x,c+d x\right )}{64 d}\\ &=\frac{15 b^2 e^3 (c+d x)^4 \sqrt{a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac{15 b e^3 (c+d x) \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{64 d}-\frac{5 b e^3 (c+d x)^3 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{32 d}+\frac{e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}-\frac{\left (15 b e^3\right ) \operatorname{Subst}\left (\int \frac{\left (a+b \sinh ^{-1}(x)\right )^{3/2}}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{64 d}-\frac{\left (45 b^2 e^3\right ) \operatorname{Subst}\left (\int x \sqrt{a+b \sinh ^{-1}(x)} \, dx,x,c+d x\right )}{128 d}-\frac{\left (15 b^3 e^3\right ) \operatorname{Subst}\left (\int \frac{x^4}{\sqrt{1+x^2} \sqrt{a+b \sinh ^{-1}(x)}} \, dx,x,c+d x\right )}{512 d}\\ &=-\frac{45 b^2 e^3 (c+d x)^2 \sqrt{a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac{15 b^2 e^3 (c+d x)^4 \sqrt{a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac{15 b e^3 (c+d x) \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{64 d}-\frac{5 b e^3 (c+d x)^3 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{32 d}-\frac{3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{32 d}+\frac{e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}-\frac{\left (15 b^3 e^3\right ) \operatorname{Subst}\left (\int \frac{\sinh ^4(x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{512 d}+\frac{\left (45 b^3 e^3\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{1+x^2} \sqrt{a+b \sinh ^{-1}(x)}} \, dx,x,c+d x\right )}{512 d}\\ &=-\frac{45 b^2 e^3 (c+d x)^2 \sqrt{a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac{15 b^2 e^3 (c+d x)^4 \sqrt{a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac{15 b e^3 (c+d x) \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{64 d}-\frac{5 b e^3 (c+d x)^3 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{32 d}-\frac{3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{32 d}+\frac{e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}-\frac{\left (15 b^3 e^3\right ) \operatorname{Subst}\left (\int \left (\frac{3}{8 \sqrt{a+b x}}-\frac{\cosh (2 x)}{2 \sqrt{a+b x}}+\frac{\cosh (4 x)}{8 \sqrt{a+b x}}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{512 d}+\frac{\left (45 b^3 e^3\right ) \operatorname{Subst}\left (\int \frac{\sinh ^2(x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{512 d}\\ &=-\frac{45 b^2 e^3 \sqrt{a+b \sinh ^{-1}(c+d x)}}{2048 d}-\frac{45 b^2 e^3 (c+d x)^2 \sqrt{a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac{15 b^2 e^3 (c+d x)^4 \sqrt{a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac{15 b e^3 (c+d x) \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{64 d}-\frac{5 b e^3 (c+d x)^3 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{32 d}-\frac{3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{32 d}+\frac{e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}-\frac{\left (15 b^3 e^3\right ) \operatorname{Subst}\left (\int \frac{\cosh (4 x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{4096 d}+\frac{\left (15 b^3 e^3\right ) \operatorname{Subst}\left (\int \frac{\cosh (2 x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{1024 d}-\frac{\left (45 b^3 e^3\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2 \sqrt{a+b x}}-\frac{\cosh (2 x)}{2 \sqrt{a+b x}}\right ) \, dx,x,\sinh ^{-1}(c+d x)\right )}{512 d}\\ &=-\frac{225 b^2 e^3 \sqrt{a+b \sinh ^{-1}(c+d x)}}{2048 d}-\frac{45 b^2 e^3 (c+d x)^2 \sqrt{a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac{15 b^2 e^3 (c+d x)^4 \sqrt{a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac{15 b e^3 (c+d x) \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{64 d}-\frac{5 b e^3 (c+d x)^3 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{32 d}-\frac{3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{32 d}+\frac{e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}-\frac{\left (15 b^3 e^3\right ) \operatorname{Subst}\left (\int \frac{e^{-4 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8192 d}-\frac{\left (15 b^3 e^3\right ) \operatorname{Subst}\left (\int \frac{e^{4 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{8192 d}+\frac{\left (15 b^3 e^3\right ) \operatorname{Subst}\left (\int \frac{e^{-2 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2048 d}+\frac{\left (15 b^3 e^3\right ) \operatorname{Subst}\left (\int \frac{e^{2 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2048 d}+\frac{\left (45 b^3 e^3\right ) \operatorname{Subst}\left (\int \frac{\cosh (2 x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{1024 d}\\ &=-\frac{225 b^2 e^3 \sqrt{a+b \sinh ^{-1}(c+d x)}}{2048 d}-\frac{45 b^2 e^3 (c+d x)^2 \sqrt{a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac{15 b^2 e^3 (c+d x)^4 \sqrt{a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac{15 b e^3 (c+d x) \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{64 d}-\frac{5 b e^3 (c+d x)^3 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{32 d}-\frac{3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{32 d}+\frac{e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}-\frac{\left (15 b^2 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 )}{4096 d}-\frac{\left (15 b^2 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 )}{4096 d}+\frac{\left (15 b^2 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 )}{1024 d}+\frac{\left (15 b^2 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 )}{1024 d}+\frac{\left (45 b^3 e^3\right ) \operatorname{Subst}\left (\int \frac{e^{-2 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2048 d}+\frac{\left (45 b^3 e^3\right ) \operatorname{Subst}\left (\int \frac{e^{2 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{2048 d}\\ &=-\frac{225 b^2 e^3 \sqrt{a+b \sinh ^{-1}(c+d x)}}{2048 d}-\frac{45 b^2 e^3 (c+d x)^2 \sqrt{a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac{15 b^2 e^3 (c+d x)^4 \sqrt{a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac{15 b e^3 (c+d x) \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{64 d}-\frac{5 b e^3 (c+d x)^3 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{32 d}-\frac{3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{32 d}+\frac{e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}-\frac{15 b^{5/2} 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 )}{16384 d}+\frac{15 b^{5/2} e^3 e^{\frac{2 a}{b}} \sqrt{\frac{\pi }{2}} \text{erf}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{2048 d}-\frac{15 b^{5/2} 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 )}{16384 d}+\frac{15 b^{5/2} e^3 e^{-\frac{2 a}{b}} \sqrt{\frac{\pi }{2}} \text{erfi}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{2048 d}+\frac{\left (45 b^2 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 )}{1024 d}+\frac{\left (45 b^2 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 )}{1024 d}\\ &=-\frac{225 b^2 e^3 \sqrt{a+b \sinh ^{-1}(c+d x)}}{2048 d}-\frac{45 b^2 e^3 (c+d x)^2 \sqrt{a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac{15 b^2 e^3 (c+d x)^4 \sqrt{a+b \sinh ^{-1}(c+d x)}}{256 d}+\frac{15 b e^3 (c+d x) \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{64 d}-\frac{5 b e^3 (c+d x)^3 \sqrt{1+(c+d x)^2} \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{32 d}-\frac{3 e^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{32 d}+\frac{e^3 (c+d x)^4 \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2}}{4 d}-\frac{15 b^{5/2} 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 )}{16384 d}+\frac{15 b^{5/2} e^3 e^{\frac{2 a}{b}} \sqrt{\frac{\pi }{2}} \text{erf}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{512 d}-\frac{15 b^{5/2} 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 )}{16384 d}+\frac{15 b^{5/2} e^3 e^{-\frac{2 a}{b}} \sqrt{\frac{\pi }{2}} \text{erfi}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{512 d}\\ \end{align*}

Mathematica [A]  time = 0.314397, size = 223, normalized size = 0.49 \[ -\frac{e^3 e^{-\frac{4 a}{b}} \left (a+b \sinh ^{-1}(c+d x)\right )^{5/2} \left (\sqrt{\frac{a}{b}+\sinh ^{-1}(c+d x)} \text{Gamma}\left (\frac{7}{2},-\frac{4 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )-16 \sqrt{2} e^{\frac{2 a}{b}} \sqrt{\frac{a}{b}+\sinh ^{-1}(c+d x)} \text{Gamma}\left (\frac{7}{2},-\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )+e^{\frac{6 a}{b}} \sqrt{-\frac{a+b \sinh ^{-1}(c+d x)}{b}} \left (e^{\frac{2 a}{b}} \text{Gamma}\left (\frac{7}{2},\frac{4 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )-16 \sqrt{2} \text{Gamma}\left (\frac{7}{2},\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )\right )\right )}{2048 d \left (-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^2}{b^2}\right )^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(e^3*(a + b*ArcSinh[c + d*x])^(5/2)*(Sqrt[a/b + ArcSinh[c + d*x]]*Gamma[7/2, (-4*(a + b*ArcSinh[c + d*x]))/b]
- 16*Sqrt[2]*E^((2*a)/b)*Sqrt[a/b + ArcSinh[c + d*x]]*Gamma[7/2, (-2*(a + b*ArcSinh[c + d*x]))/b] + E^((6*a)/
b)*Sqrt[-((a + b*ArcSinh[c + d*x])/b)]*(-16*Sqrt[2]*Gamma[7/2, (2*(a + b*ArcSinh[c + d*x]))/b] + E^((2*a)/b)*G
amma[7/2, (4*(a + b*ArcSinh[c + d*x]))/b])))/(2048*d*E^((4*a)/b)*(-((a + b*ArcSinh[c + d*x])^2/b^2))^(3/2))

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Maple [F]  time = 0.194, 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{\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(-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))**(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 )}^{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)