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

Optimal. Leaf size=328 \[ -\frac{3 \sqrt{\pi } b^{3/2} e^2 e^{a/b} \text{Erf}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{32 d}+\frac{\sqrt{\frac{\pi }{3}} b^{3/2} e^2 e^{\frac{3 a}{b}} \text{Erf}\left (\frac{\sqrt{3} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{96 d}-\frac{3 \sqrt{\pi } b^{3/2} e^2 e^{-\frac{a}{b}} \text{Erfi}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{32 d}+\frac{\sqrt{\frac{\pi }{3}} b^{3/2} e^2 e^{-\frac{3 a}{b}} \text{Erfi}\left (\frac{\sqrt{3} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{96 d}+\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{3 d}-\frac{b e^2 \sqrt{(c+d x)^2+1} (c+d x)^2 \sqrt{a+b \sinh ^{-1}(c+d x)}}{6 d}+\frac{b e^2 \sqrt{(c+d x)^2+1} \sqrt{a+b \sinh ^{-1}(c+d x)}}{3 d} \]

[Out]

(b*e^2*Sqrt[1 + (c + d*x)^2]*Sqrt[a + b*ArcSinh[c + d*x]])/(3*d) - (b*e^2*(c + d*x)^2*Sqrt[1 + (c + d*x)^2]*Sq
rt[a + b*ArcSinh[c + d*x]])/(6*d) + (e^2*(c + d*x)^3*(a + b*ArcSinh[c + d*x])^(3/2))/(3*d) - (3*b^(3/2)*e^2*E^
(a/b)*Sqrt[Pi]*Erf[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(32*d) + (b^(3/2)*e^2*E^((3*a)/b)*Sqrt[Pi/3]*Erf[(Sq
rt[3]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(96*d) - (3*b^(3/2)*e^2*Sqrt[Pi]*Erfi[Sqrt[a + b*ArcSinh[c + d*x
]]/Sqrt[b]])/(32*d*E^(a/b)) + (b^(3/2)*e^2*Sqrt[Pi/3]*Erfi[(Sqrt[3]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(9
6*d*E^((3*a)/b))

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Rubi [A]  time = 0.882012, antiderivative size = 328, normalized size of antiderivative = 1., number of steps used = 24, 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, 5657, 3307, 2180, 2205, 2204, 5669, 5448} \[ -\frac{3 \sqrt{\pi } b^{3/2} e^2 e^{a/b} \text{Erf}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{32 d}+\frac{\sqrt{\frac{\pi }{3}} b^{3/2} e^2 e^{\frac{3 a}{b}} \text{Erf}\left (\frac{\sqrt{3} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{96 d}-\frac{3 \sqrt{\pi } b^{3/2} e^2 e^{-\frac{a}{b}} \text{Erfi}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{32 d}+\frac{\sqrt{\frac{\pi }{3}} b^{3/2} e^2 e^{-\frac{3 a}{b}} \text{Erfi}\left (\frac{\sqrt{3} \sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{96 d}+\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{3 d}-\frac{b e^2 \sqrt{(c+d x)^2+1} (c+d x)^2 \sqrt{a+b \sinh ^{-1}(c+d x)}}{6 d}+\frac{b e^2 \sqrt{(c+d x)^2+1} \sqrt{a+b \sinh ^{-1}(c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*e^2*Sqrt[1 + (c + d*x)^2]*Sqrt[a + b*ArcSinh[c + d*x]])/(3*d) - (b*e^2*(c + d*x)^2*Sqrt[1 + (c + d*x)^2]*Sq
rt[a + b*ArcSinh[c + d*x]])/(6*d) + (e^2*(c + d*x)^3*(a + b*ArcSinh[c + d*x])^(3/2))/(3*d) - (3*b^(3/2)*e^2*E^
(a/b)*Sqrt[Pi]*Erf[Sqrt[a + b*ArcSinh[c + d*x]]/Sqrt[b]])/(32*d) + (b^(3/2)*e^2*E^((3*a)/b)*Sqrt[Pi/3]*Erf[(Sq
rt[3]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(96*d) - (3*b^(3/2)*e^2*Sqrt[Pi]*Erfi[Sqrt[a + b*ArcSinh[c + d*x
]]/Sqrt[b]])/(32*d*E^(a/b)) + (b^(3/2)*e^2*Sqrt[Pi/3]*Erfi[(Sqrt[3]*Sqrt[a + b*ArcSinh[c + d*x]])/Sqrt[b]])/(9
6*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 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]

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 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 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 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]

Rubi steps

\begin{align*} \int (c e+d e x)^2 \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int e^2 x^2 \left (a+b \sinh ^{-1}(x)\right )^{3/2} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^2 \operatorname{Subst}\left (\int x^2 \left (a+b \sinh ^{-1}(x)\right )^{3/2} \, dx,x,c+d x\right )}{d}\\ &=\frac{e^2 (c+d x)^3 \left (a+b \sinh ^{-1}(c+d x)\right )^{3/2}}{3 d}-\frac{\left (b e^2\right ) \operatorname{Subst}\left (\int \frac{x^3 \sqrt{a+b \sinh ^{-1}(x)}}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{2 d}\\ &=-\frac{b e^2 (c+d x)^2 \sqrt{1+(c+d x)^2} \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 )^{3/2}}{3 d}+\frac{\left (b e^2\right ) \operatorname{Subst}\left (\int \frac{x \sqrt{a+b \sinh ^{-1}(x)}}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{3 d}+\frac{\left (b^2 e^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+b \sinh ^{-1}(x)}} \, dx,x,c+d x\right )}{12 d}\\ &=\frac{b e^2 \sqrt{1+(c+d x)^2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{3 d}-\frac{b e^2 (c+d x)^2 \sqrt{1+(c+d x)^2} \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 )^{3/2}}{3 d}+\frac{\left (b^2 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 )}{12 d}-\frac{\left (b^2 e^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b \sinh ^{-1}(x)}} \, dx,x,c+d x\right )}{6 d}\\ &=\frac{b e^2 \sqrt{1+(c+d x)^2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{3 d}-\frac{b e^2 (c+d x)^2 \sqrt{1+(c+d x)^2} \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 )^{3/2}}{3 d}-\frac{\left (b 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 )}{6 d}+\frac{\left (b^2 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 )}{12 d}\\ &=\frac{b e^2 \sqrt{1+(c+d x)^2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{3 d}-\frac{b e^2 (c+d x)^2 \sqrt{1+(c+d x)^2} \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 )^{3/2}}{3 d}-\frac{\left (b 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 )}{12 d}-\frac{\left (b 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 )}{12 d}-\frac{\left (b^2 e^2\right ) \operatorname{Subst}\left (\int \frac{\cosh (x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{48 d}+\frac{\left (b^2 e^2\right ) \operatorname{Subst}\left (\int \frac{\cosh (3 x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{48 d}\\ &=\frac{b e^2 \sqrt{1+(c+d x)^2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{3 d}-\frac{b e^2 (c+d x)^2 \sqrt{1+(c+d x)^2} \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 )^{3/2}}{3 d}-\frac{\left (b 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 )}{6 d}-\frac{\left (b 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 )}{6 d}+\frac{\left (b^2 e^2\right ) \operatorname{Subst}\left (\int \frac{e^{-3 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{96 d}-\frac{\left (b^2 e^2\right ) \operatorname{Subst}\left (\int \frac{e^{-x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{96 d}-\frac{\left (b^2 e^2\right ) \operatorname{Subst}\left (\int \frac{e^x}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{96 d}+\frac{\left (b^2 e^2\right ) \operatorname{Subst}\left (\int \frac{e^{3 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c+d x)\right )}{96 d}\\ &=\frac{b e^2 \sqrt{1+(c+d x)^2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{3 d}-\frac{b e^2 (c+d x)^2 \sqrt{1+(c+d x)^2} \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 )^{3/2}}{3 d}-\frac{b^{3/2} e^2 e^{a/b} \sqrt{\pi } \text{erf}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{12 d}-\frac{b^{3/2} e^2 e^{-\frac{a}{b}} \sqrt{\pi } \text{erfi}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{12 d}+\frac{\left (b 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 )}{48 d}-\frac{\left (b 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 )}{48 d}-\frac{\left (b 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 )}{48 d}+\frac{\left (b 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 )}{48 d}\\ &=\frac{b e^2 \sqrt{1+(c+d x)^2} \sqrt{a+b \sinh ^{-1}(c+d x)}}{3 d}-\frac{b e^2 (c+d x)^2 \sqrt{1+(c+d x)^2} \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 )^{3/2}}{3 d}-\frac{3 b^{3/2} e^2 e^{a/b} \sqrt{\pi } \text{erf}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{32 d}+\frac{b^{3/2} 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 )}{96 d}-\frac{3 b^{3/2} e^2 e^{-\frac{a}{b}} \sqrt{\pi } \text{erfi}\left (\frac{\sqrt{a+b \sinh ^{-1}(c+d x)}}{\sqrt{b}}\right )}{32 d}+\frac{b^{3/2} 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 )}{96 d}\\ \end{align*}

Mathematica [A]  time = 0.289547, size = 238, normalized size = 0.73 \[ -\frac{b e^2 e^{-\frac{3 a}{b}} \sqrt{a+b \sinh ^{-1}(c+d x)} \left (-27 e^{\frac{4 a}{b}} \sqrt{-\frac{a+b \sinh ^{-1}(c+d x)}{b}} \text{Gamma}\left (\frac{5}{2},\frac{a}{b}+\sinh ^{-1}(c+d x)\right )+\sqrt{3} \sqrt{\frac{a}{b}+\sinh ^{-1}(c+d x)} \text{Gamma}\left (\frac{5}{2},-\frac{3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )-27 e^{\frac{2 a}{b}} \sqrt{\frac{a}{b}+\sinh ^{-1}(c+d x)} \text{Gamma}\left (\frac{5}{2},-\frac{a+b \sinh ^{-1}(c+d x)}{b}\right )+\sqrt{3} e^{\frac{6 a}{b}} \sqrt{-\frac{a+b \sinh ^{-1}(c+d x)}{b}} \text{Gamma}\left (\frac{5}{2},\frac{3 \left (a+b \sinh ^{-1}(c+d x)\right )}{b}\right )\right )}{216 d \sqrt{-\frac{\left (a+b \sinh ^{-1}(c+d x)\right )^2}{b^2}}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(b*e^2*Sqrt[a + b*ArcSinh[c + d*x]]*(-27*E^((4*a)/b)*Sqrt[-((a + b*ArcSinh[c + d*x])/b)]*Gamma[5/2, a/b + Arc
Sinh[c + d*x]] + Sqrt[3]*Sqrt[a/b + ArcSinh[c + d*x]]*Gamma[5/2, (-3*(a + b*ArcSinh[c + d*x]))/b] - 27*E^((2*a
)/b)*Sqrt[a/b + ArcSinh[c + d*x]]*Gamma[5/2, -((a + b*ArcSinh[c + d*x])/b)] + Sqrt[3]*E^((6*a)/b)*Sqrt[-((a +
b*ArcSinh[c + d*x])/b)]*Gamma[5/2, (3*(a + b*ArcSinh[c + d*x]))/b]))/(216*d*E^((3*a)/b)*Sqrt[-((a + b*ArcSinh[
c + d*x])^2/b^2)])

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Maple [F]  time = 0.218, size = 0, normalized size = 0. \begin{align*} \int \left ( dex+ce \right ) ^{2} \left ( a+b{\it Arcsinh} \left ( dx+c \right ) \right ) ^{{\frac{3}{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))^(3/2),x)

[Out]

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

[Out]

integrate((d*e*x + c*e)^2*(b*arcsinh(d*x + c) + a)^(3/2), 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))^(3/2),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 a c^{2} \sqrt{a + b \operatorname{asinh}{\left (c + d x \right )}}\, dx + \int a d^{2} x^{2} \sqrt{a + b \operatorname{asinh}{\left (c + d x \right )}}\, dx + \int b c^{2} \sqrt{a + b \operatorname{asinh}{\left (c + d x \right )}} \operatorname{asinh}{\left (c + d x \right )}\, dx + \int 2 a c d x \sqrt{a + b \operatorname{asinh}{\left (c + d x \right )}}\, dx + \int b d^{2} x^{2} \sqrt{a + b \operatorname{asinh}{\left (c + d x \right )}} \operatorname{asinh}{\left (c + d x \right )}\, dx + \int 2 b c d x \sqrt{a + b \operatorname{asinh}{\left (c + d x \right )}} \operatorname{asinh}{\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))**(3/2),x)

[Out]

e**2*(Integral(a*c**2*sqrt(a + b*asinh(c + d*x)), x) + Integral(a*d**2*x**2*sqrt(a + b*asinh(c + d*x)), x) + I
ntegral(b*c**2*sqrt(a + b*asinh(c + d*x))*asinh(c + d*x), x) + Integral(2*a*c*d*x*sqrt(a + b*asinh(c + d*x)),
x) + Integral(b*d**2*x**2*sqrt(a + b*asinh(c + d*x))*asinh(c + d*x), x) + Integral(2*b*c*d*x*sqrt(a + b*asinh(
c + d*x))*asinh(c + d*x), x))

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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{3}{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))^(3/2),x, algorithm="giac")

[Out]

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