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

Optimal. Leaf size=358 \[ \frac{\sqrt{\pi } d^2 e^{\frac{4 a}{b}} \text{Erf}\left (\frac{2 \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{4 b^{3/2} c^2}+\frac{5 \sqrt{\frac{\pi }{2}} d^2 e^{\frac{2 a}{b}} \text{Erf}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c^2}+\frac{\sqrt{\frac{3 \pi }{2}} d^2 e^{\frac{6 a}{b}} \text{Erf}\left (\frac{\sqrt{6} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c^2}+\frac{\sqrt{\pi } d^2 e^{-\frac{4 a}{b}} \text{Erfi}\left (\frac{2 \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{4 b^{3/2} c^2}+\frac{5 \sqrt{\frac{\pi }{2}} d^2 e^{-\frac{2 a}{b}} \text{Erfi}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c^2}+\frac{\sqrt{\frac{3 \pi }{2}} d^2 e^{-\frac{6 a}{b}} \text{Erfi}\left (\frac{\sqrt{6} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c^2}-\frac{2 d^2 x \left (c^2 x^2+1\right )^{5/2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}} \]

[Out]

(-2*d^2*x*(1 + c^2*x^2)^(5/2))/(b*c*Sqrt[a + b*ArcSinh[c*x]]) + (d^2*E^((4*a)/b)*Sqrt[Pi]*Erf[(2*Sqrt[a + b*Ar
cSinh[c*x]])/Sqrt[b]])/(4*b^(3/2)*c^2) + (5*d^2*E^((2*a)/b)*Sqrt[Pi/2]*Erf[(Sqrt[2]*Sqrt[a + b*ArcSinh[c*x]])/
Sqrt[b]])/(16*b^(3/2)*c^2) + (d^2*E^((6*a)/b)*Sqrt[(3*Pi)/2]*Erf[(Sqrt[6]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/
(16*b^(3/2)*c^2) + (d^2*Sqrt[Pi]*Erfi[(2*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(4*b^(3/2)*c^2*E^((4*a)/b)) + (5*
d^2*Sqrt[Pi/2]*Erfi[(Sqrt[2]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(16*b^(3/2)*c^2*E^((2*a)/b)) + (d^2*Sqrt[(3*P
i)/2]*Erfi[(Sqrt[6]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(16*b^(3/2)*c^2*E^((6*a)/b))

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Rubi [A]  time = 1.33021, antiderivative size = 358, normalized size of antiderivative = 1., number of steps used = 32, number of rules used = 9, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.346, Rules used = {5777, 5699, 3312, 3307, 2180, 2204, 2205, 5779, 5448} \[ \frac{\sqrt{\pi } d^2 e^{\frac{4 a}{b}} \text{Erf}\left (\frac{2 \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{4 b^{3/2} c^2}+\frac{5 \sqrt{\frac{\pi }{2}} d^2 e^{\frac{2 a}{b}} \text{Erf}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c^2}+\frac{\sqrt{\frac{3 \pi }{2}} d^2 e^{\frac{6 a}{b}} \text{Erf}\left (\frac{\sqrt{6} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c^2}+\frac{\sqrt{\pi } d^2 e^{-\frac{4 a}{b}} \text{Erfi}\left (\frac{2 \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{4 b^{3/2} c^2}+\frac{5 \sqrt{\frac{\pi }{2}} d^2 e^{-\frac{2 a}{b}} \text{Erfi}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c^2}+\frac{\sqrt{\frac{3 \pi }{2}} d^2 e^{-\frac{6 a}{b}} \text{Erfi}\left (\frac{\sqrt{6} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c^2}-\frac{2 d^2 x \left (c^2 x^2+1\right )^{5/2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d^2*x*(1 + c^2*x^2)^(5/2))/(b*c*Sqrt[a + b*ArcSinh[c*x]]) + (d^2*E^((4*a)/b)*Sqrt[Pi]*Erf[(2*Sqrt[a + b*Ar
cSinh[c*x]])/Sqrt[b]])/(4*b^(3/2)*c^2) + (5*d^2*E^((2*a)/b)*Sqrt[Pi/2]*Erf[(Sqrt[2]*Sqrt[a + b*ArcSinh[c*x]])/
Sqrt[b]])/(16*b^(3/2)*c^2) + (d^2*E^((6*a)/b)*Sqrt[(3*Pi)/2]*Erf[(Sqrt[6]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/
(16*b^(3/2)*c^2) + (d^2*Sqrt[Pi]*Erfi[(2*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(4*b^(3/2)*c^2*E^((4*a)/b)) + (5*
d^2*Sqrt[Pi/2]*Erfi[(Sqrt[2]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(16*b^(3/2)*c^2*E^((2*a)/b)) + (d^2*Sqrt[(3*P
i)/2]*Erfi[(Sqrt[6]*Sqrt[a + b*ArcSinh[c*x]])/Sqrt[b]])/(16*b^(3/2)*c^2*E^((6*a)/b))

Rule 5777

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp
[((f*x)^m*Sqrt[1 + c^2*x^2]*(d + e*x^2)^p*(a + b*ArcSinh[c*x])^(n + 1))/(b*c*(n + 1)), x] + (-Dist[(f*m*d^IntP
art[p]*(d + e*x^2)^FracPart[p])/(b*c*(n + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m - 1)*(1 + c^2*x^2)^(p -
1/2)*(a + b*ArcSinh[c*x])^(n + 1), x], x] - Dist[(c*(m + 2*p + 1)*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(b*f*(
n + 1)*(1 + c^2*x^2)^FracPart[p]), Int[(f*x)^(m + 1)*(1 + c^2*x^2)^(p - 1/2)*(a + b*ArcSinh[c*x])^(n + 1), x],
x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[e, c^2*d] && LtQ[n, -1] && IGtQ[m, -3] && IGtQ[2*p, 0]

Rule 5699

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[d^p/c, Subst[Int[
(a + b*x)^n*Cosh[x]^(2*p + 1), x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && IG
tQ[2*p, 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]

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 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 \frac{x \left (d+c^2 d x^2\right )^2}{\left (a+b \sinh ^{-1}(c x)\right )^{3/2}} \, dx &=-\frac{2 d^2 x \left (1+c^2 x^2\right )^{5/2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}}+\frac{\left (2 d^2\right ) \int \frac{\left (1+c^2 x^2\right )^{3/2}}{\sqrt{a+b \sinh ^{-1}(c x)}} \, dx}{b c}+\frac{\left (12 c d^2\right ) \int \frac{x^2 \left (1+c^2 x^2\right )^{3/2}}{\sqrt{a+b \sinh ^{-1}(c x)}} \, dx}{b}\\ &=-\frac{2 d^2 x \left (1+c^2 x^2\right )^{5/2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}}+\frac{\left (2 d^2\right ) \operatorname{Subst}\left (\int \frac{\cosh ^4(x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{b c^2}+\frac{\left (12 d^2\right ) \operatorname{Subst}\left (\int \frac{\cosh ^4(x) \sinh ^2(x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{b c^2}\\ &=-\frac{2 d^2 x \left (1+c^2 x^2\right )^{5/2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}}+\frac{\left (2 d^2\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 x)\right )}{b c^2}+\frac{\left (12 d^2\right ) \operatorname{Subst}\left (\int \left (-\frac{1}{16 \sqrt{a+b x}}-\frac{\cosh (2 x)}{32 \sqrt{a+b x}}+\frac{\cosh (4 x)}{16 \sqrt{a+b x}}+\frac{\cosh (6 x)}{32 \sqrt{a+b x}}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{b c^2}\\ &=-\frac{2 d^2 x \left (1+c^2 x^2\right )^{5/2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}}+\frac{d^2 \operatorname{Subst}\left (\int \frac{\cosh (4 x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^2}-\frac{\left (3 d^2\right ) \operatorname{Subst}\left (\int \frac{\cosh (2 x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c^2}+\frac{\left (3 d^2\right ) \operatorname{Subst}\left (\int \frac{\cosh (6 x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c^2}+\frac{\left (3 d^2\right ) \operatorname{Subst}\left (\int \frac{\cosh (4 x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{4 b c^2}+\frac{d^2 \operatorname{Subst}\left (\int \frac{\cosh (2 x)}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{b c^2}\\ &=-\frac{2 d^2 x \left (1+c^2 x^2\right )^{5/2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}}+\frac{d^2 \operatorname{Subst}\left (\int \frac{e^{-4 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c^2}+\frac{d^2 \operatorname{Subst}\left (\int \frac{e^{4 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c^2}+\frac{\left (3 d^2\right ) \operatorname{Subst}\left (\int \frac{e^{-6 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^2}-\frac{\left (3 d^2\right ) \operatorname{Subst}\left (\int \frac{e^{-2 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^2}-\frac{\left (3 d^2\right ) \operatorname{Subst}\left (\int \frac{e^{2 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^2}+\frac{\left (3 d^2\right ) \operatorname{Subst}\left (\int \frac{e^{6 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^2}+\frac{\left (3 d^2\right ) \operatorname{Subst}\left (\int \frac{e^{-4 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c^2}+\frac{\left (3 d^2\right ) \operatorname{Subst}\left (\int \frac{e^{4 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c^2}+\frac{d^2 \operatorname{Subst}\left (\int \frac{e^{-2 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{2 b c^2}+\frac{d^2 \operatorname{Subst}\left (\int \frac{e^{2 x}}{\sqrt{a+b x}} \, dx,x,\sinh ^{-1}(c x)\right )}{2 b c^2}\\ &=-\frac{2 d^2 x \left (1+c^2 x^2\right )^{5/2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}}+\frac{d^2 \operatorname{Subst}\left (\int e^{\frac{4 a}{b}-\frac{4 x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c x)}\right )}{4 b^2 c^2}+\frac{d^2 \operatorname{Subst}\left (\int e^{-\frac{4 a}{b}+\frac{4 x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c x)}\right )}{4 b^2 c^2}+\frac{\left (3 d^2\right ) \operatorname{Subst}\left (\int e^{\frac{6 a}{b}-\frac{6 x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c x)}\right )}{8 b^2 c^2}-\frac{\left (3 d^2\right ) \operatorname{Subst}\left (\int e^{\frac{2 a}{b}-\frac{2 x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c x)}\right )}{8 b^2 c^2}-\frac{\left (3 d^2\right ) \operatorname{Subst}\left (\int e^{-\frac{2 a}{b}+\frac{2 x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c x)}\right )}{8 b^2 c^2}+\frac{\left (3 d^2\right ) \operatorname{Subst}\left (\int e^{-\frac{6 a}{b}+\frac{6 x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c x)}\right )}{8 b^2 c^2}+\frac{\left (3 d^2\right ) \operatorname{Subst}\left (\int e^{\frac{4 a}{b}-\frac{4 x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c x)}\right )}{4 b^2 c^2}+\frac{\left (3 d^2\right ) \operatorname{Subst}\left (\int e^{-\frac{4 a}{b}+\frac{4 x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c x)}\right )}{4 b^2 c^2}+\frac{d^2 \operatorname{Subst}\left (\int e^{\frac{2 a}{b}-\frac{2 x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c x)}\right )}{b^2 c^2}+\frac{d^2 \operatorname{Subst}\left (\int e^{-\frac{2 a}{b}+\frac{2 x^2}{b}} \, dx,x,\sqrt{a+b \sinh ^{-1}(c x)}\right )}{b^2 c^2}\\ &=-\frac{2 d^2 x \left (1+c^2 x^2\right )^{5/2}}{b c \sqrt{a+b \sinh ^{-1}(c x)}}+\frac{d^2 e^{\frac{4 a}{b}} \sqrt{\pi } \text{erf}\left (\frac{2 \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{4 b^{3/2} c^2}+\frac{5 d^2 e^{\frac{2 a}{b}} \sqrt{\frac{\pi }{2}} \text{erf}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c^2}+\frac{d^2 e^{\frac{6 a}{b}} \sqrt{\frac{3 \pi }{2}} \text{erf}\left (\frac{\sqrt{6} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c^2}+\frac{d^2 e^{-\frac{4 a}{b}} \sqrt{\pi } \text{erfi}\left (\frac{2 \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{4 b^{3/2} c^2}+\frac{5 d^2 e^{-\frac{2 a}{b}} \sqrt{\frac{\pi }{2}} \text{erfi}\left (\frac{\sqrt{2} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c^2}+\frac{d^2 e^{-\frac{6 a}{b}} \sqrt{\frac{3 \pi }{2}} \text{erfi}\left (\frac{\sqrt{6} \sqrt{a+b \sinh ^{-1}(c x)}}{\sqrt{b}}\right )}{16 b^{3/2} c^2}\\ \end{align*}

Mathematica [A]  time = 0.663044, size = 351, normalized size = 0.98 \[ -\frac{d^2 e^{-\frac{6 a}{b}} \left (-\sqrt{6} \sqrt{-\frac{a+b \sinh ^{-1}(c x)}{b}} \text{Gamma}\left (\frac{1}{2},-\frac{6 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )-8 e^{\frac{2 a}{b}} \sqrt{-\frac{a+b \sinh ^{-1}(c x)}{b}} \text{Gamma}\left (\frac{1}{2},-\frac{4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )-5 \sqrt{2} e^{\frac{4 a}{b}} \sqrt{-\frac{a+b \sinh ^{-1}(c x)}{b}} \text{Gamma}\left (\frac{1}{2},-\frac{2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )+5 \sqrt{2} e^{\frac{8 a}{b}} \sqrt{\frac{a}{b}+\sinh ^{-1}(c x)} \text{Gamma}\left (\frac{1}{2},\frac{2 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )+8 e^{\frac{10 a}{b}} \sqrt{\frac{a}{b}+\sinh ^{-1}(c x)} \text{Gamma}\left (\frac{1}{2},\frac{4 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )+\sqrt{6} e^{\frac{12 a}{b}} \sqrt{\frac{a}{b}+\sinh ^{-1}(c x)} \text{Gamma}\left (\frac{1}{2},\frac{6 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )+10 e^{\frac{6 a}{b}} \sinh \left (2 \sinh ^{-1}(c x)\right )+8 e^{\frac{6 a}{b}} \sinh \left (4 \sinh ^{-1}(c x)\right )+2 e^{\frac{6 a}{b}} \sinh \left (6 \sinh ^{-1}(c x)\right )\right )}{32 b c^2 \sqrt{a+b \sinh ^{-1}(c x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(d^2*(-(Sqrt[6]*Sqrt[-((a + b*ArcSinh[c*x])/b)]*Gamma[1/2, (-6*(a + b*ArcSinh[c*x]))/b]) - 8*E^((2*a)/b)*Sqrt
[-((a + b*ArcSinh[c*x])/b)]*Gamma[1/2, (-4*(a + b*ArcSinh[c*x]))/b] - 5*Sqrt[2]*E^((4*a)/b)*Sqrt[-((a + b*ArcS
inh[c*x])/b)]*Gamma[1/2, (-2*(a + b*ArcSinh[c*x]))/b] + 5*Sqrt[2]*E^((8*a)/b)*Sqrt[a/b + ArcSinh[c*x]]*Gamma[1
/2, (2*(a + b*ArcSinh[c*x]))/b] + 8*E^((10*a)/b)*Sqrt[a/b + ArcSinh[c*x]]*Gamma[1/2, (4*(a + b*ArcSinh[c*x]))/
b] + Sqrt[6]*E^((12*a)/b)*Sqrt[a/b + ArcSinh[c*x]]*Gamma[1/2, (6*(a + b*ArcSinh[c*x]))/b] + 10*E^((6*a)/b)*Sin
h[2*ArcSinh[c*x]] + 8*E^((6*a)/b)*Sinh[4*ArcSinh[c*x]] + 2*E^((6*a)/b)*Sinh[6*ArcSinh[c*x]]))/(32*b*c^2*E^((6*
a)/b)*Sqrt[a + b*ArcSinh[c*x]])

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Maple [F]  time = 0.231, size = 0, normalized size = 0. \begin{align*} \int{x \left ({c}^{2}d{x}^{2}+d \right ) ^{2} \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c^{2} d x^{2} + d\right )}^{2} x}{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} d^{2} \left (\int \frac{x}{a \sqrt{a + b \operatorname{asinh}{\left (c x \right )}} + b \sqrt{a + b \operatorname{asinh}{\left (c x \right )}} \operatorname{asinh}{\left (c x \right )}}\, dx + \int \frac{2 c^{2} x^{3}}{a \sqrt{a + b \operatorname{asinh}{\left (c x \right )}} + b \sqrt{a + b \operatorname{asinh}{\left (c x \right )}} \operatorname{asinh}{\left (c x \right )}}\, dx + \int \frac{c^{4} x^{5}}{a \sqrt{a + b \operatorname{asinh}{\left (c x \right )}} + b \sqrt{a + b \operatorname{asinh}{\left (c x \right )}} \operatorname{asinh}{\left (c x \right )}}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c**2*d*x**2+d)**2/(a+b*asinh(c*x))**(3/2),x)

[Out]

d**2*(Integral(x/(a*sqrt(a + b*asinh(c*x)) + b*sqrt(a + b*asinh(c*x))*asinh(c*x)), x) + Integral(2*c**2*x**3/(
a*sqrt(a + b*asinh(c*x)) + b*sqrt(a + b*asinh(c*x))*asinh(c*x)), x) + Integral(c**4*x**5/(a*sqrt(a + b*asinh(c
*x)) + b*sqrt(a + b*asinh(c*x))*asinh(c*x)), x))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (c^{2} d x^{2} + d\right )}^{2} x}{{\left (b \operatorname{arsinh}\left (c x\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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