3.119 \(\int x^3 \sqrt{d+e x^2} (a+b \text{csch}^{-1}(c x)) \, dx\)

Optimal. Leaf size=302 \[ -\frac{d \left (d+e x^2\right )^{3/2} \left (a+b \text{csch}^{-1}(c x)\right )}{3 e^2}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \text{csch}^{-1}(c x)\right )}{5 e^2}-\frac{2 b c d^{5/2} x \tan ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d} \sqrt{-c^2 x^2-1}}\right )}{15 e^2 \sqrt{-c^2 x^2}}-\frac{b x \left (15 c^4 d^2+10 c^2 d e-9 e^2\right ) \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{-c^2 x^2-1}}{c \sqrt{d+e x^2}}\right )}{120 c^4 e^{3/2} \sqrt{-c^2 x^2}}+\frac{b x \sqrt{-c^2 x^2-1} \left (d+e x^2\right )^{3/2}}{20 c e \sqrt{-c^2 x^2}}+\frac{b x \sqrt{-c^2 x^2-1} \left (c^2 d-9 e\right ) \sqrt{d+e x^2}}{120 c^3 e \sqrt{-c^2 x^2}} \]

[Out]

(b*(c^2*d - 9*e)*x*Sqrt[-1 - c^2*x^2]*Sqrt[d + e*x^2])/(120*c^3*e*Sqrt[-(c^2*x^2)]) + (b*x*Sqrt[-1 - c^2*x^2]*
(d + e*x^2)^(3/2))/(20*c*e*Sqrt[-(c^2*x^2)]) - (d*(d + e*x^2)^(3/2)*(a + b*ArcCsch[c*x]))/(3*e^2) + ((d + e*x^
2)^(5/2)*(a + b*ArcCsch[c*x]))/(5*e^2) - (b*(15*c^4*d^2 + 10*c^2*d*e - 9*e^2)*x*ArcTan[(Sqrt[e]*Sqrt[-1 - c^2*
x^2])/(c*Sqrt[d + e*x^2])])/(120*c^4*e^(3/2)*Sqrt[-(c^2*x^2)]) - (2*b*c*d^(5/2)*x*ArcTan[Sqrt[d + e*x^2]/(Sqrt
[d]*Sqrt[-1 - c^2*x^2])])/(15*e^2*Sqrt[-(c^2*x^2)])

________________________________________________________________________________________

Rubi [A]  time = 0.433657, antiderivative size = 302, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 12, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.522, Rules used = {266, 43, 6302, 12, 573, 154, 157, 63, 217, 203, 93, 204} \[ -\frac{d \left (d+e x^2\right )^{3/2} \left (a+b \text{csch}^{-1}(c x)\right )}{3 e^2}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \text{csch}^{-1}(c x)\right )}{5 e^2}-\frac{2 b c d^{5/2} x \tan ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d} \sqrt{-c^2 x^2-1}}\right )}{15 e^2 \sqrt{-c^2 x^2}}-\frac{b x \left (15 c^4 d^2+10 c^2 d e-9 e^2\right ) \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{-c^2 x^2-1}}{c \sqrt{d+e x^2}}\right )}{120 c^4 e^{3/2} \sqrt{-c^2 x^2}}+\frac{b x \sqrt{-c^2 x^2-1} \left (d+e x^2\right )^{3/2}}{20 c e \sqrt{-c^2 x^2}}+\frac{b x \sqrt{-c^2 x^2-1} \left (c^2 d-9 e\right ) \sqrt{d+e x^2}}{120 c^3 e \sqrt{-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(c^2*d - 9*e)*x*Sqrt[-1 - c^2*x^2]*Sqrt[d + e*x^2])/(120*c^3*e*Sqrt[-(c^2*x^2)]) + (b*x*Sqrt[-1 - c^2*x^2]*
(d + e*x^2)^(3/2))/(20*c*e*Sqrt[-(c^2*x^2)]) - (d*(d + e*x^2)^(3/2)*(a + b*ArcCsch[c*x]))/(3*e^2) + ((d + e*x^
2)^(5/2)*(a + b*ArcCsch[c*x]))/(5*e^2) - (b*(15*c^4*d^2 + 10*c^2*d*e - 9*e^2)*x*ArcTan[(Sqrt[e]*Sqrt[-1 - c^2*
x^2])/(c*Sqrt[d + e*x^2])])/(120*c^4*e^(3/2)*Sqrt[-(c^2*x^2)]) - (2*b*c*d^(5/2)*x*ArcTan[Sqrt[d + e*x^2]/(Sqrt
[d]*Sqrt[-1 - c^2*x^2])])/(15*e^2*Sqrt[-(c^2*x^2)])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 6302

Int[((a_.) + ArcCsch[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u
= IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcCsch[c*x], u, x] - Dist[(b*c*x)/Sqrt[-(c^2*x^2)], Int[Simp
lifyIntegrand[u/(x*Sqrt[-1 - c^2*x^2]), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && ((IGtQ[p, 0] &&
!(ILtQ[(m - 1)/2, 0] && GtQ[m + 2*p + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[p, 0] && GtQ[m + 2*p + 3, 0]))
|| (ILtQ[(m + 2*p + 1)/2, 0] &&  !ILtQ[(m - 1)/2, 0]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 573

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^(r_.), x
_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q*(e + f*x)^r, x], x, x^n],
x] /; FreeQ[{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 154

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(h*(a + b*x)^m*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 2)), x] + Dist[1/(d*f*(m + n
+ p + 2)), Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2) - h*(b*c*e*m + a*(d*e*(
n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x]
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2
*n, 2*p]

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int x^3 \sqrt{d+e x^2} \left (a+b \text{csch}^{-1}(c x)\right ) \, dx &=-\frac{d \left (d+e x^2\right )^{3/2} \left (a+b \text{csch}^{-1}(c x)\right )}{3 e^2}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \text{csch}^{-1}(c x)\right )}{5 e^2}-\frac{(b c x) \int \frac{\left (d+e x^2\right )^{3/2} \left (-2 d+3 e x^2\right )}{15 e^2 x \sqrt{-1-c^2 x^2}} \, dx}{\sqrt{-c^2 x^2}}\\ &=-\frac{d \left (d+e x^2\right )^{3/2} \left (a+b \text{csch}^{-1}(c x)\right )}{3 e^2}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \text{csch}^{-1}(c x)\right )}{5 e^2}-\frac{(b c x) \int \frac{\left (d+e x^2\right )^{3/2} \left (-2 d+3 e x^2\right )}{x \sqrt{-1-c^2 x^2}} \, dx}{15 e^2 \sqrt{-c^2 x^2}}\\ &=-\frac{d \left (d+e x^2\right )^{3/2} \left (a+b \text{csch}^{-1}(c x)\right )}{3 e^2}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \text{csch}^{-1}(c x)\right )}{5 e^2}-\frac{(b c x) \operatorname{Subst}\left (\int \frac{(d+e x)^{3/2} (-2 d+3 e x)}{x \sqrt{-1-c^2 x}} \, dx,x,x^2\right )}{30 e^2 \sqrt{-c^2 x^2}}\\ &=\frac{b x \sqrt{-1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c e \sqrt{-c^2 x^2}}-\frac{d \left (d+e x^2\right )^{3/2} \left (a+b \text{csch}^{-1}(c x)\right )}{3 e^2}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \text{csch}^{-1}(c x)\right )}{5 e^2}+\frac{(b x) \operatorname{Subst}\left (\int \frac{\sqrt{d+e x} \left (4 c^2 d^2-\frac{1}{2} \left (c^2 d-9 e\right ) e x\right )}{x \sqrt{-1-c^2 x}} \, dx,x,x^2\right )}{60 c e^2 \sqrt{-c^2 x^2}}\\ &=\frac{b \left (c^2 d-9 e\right ) x \sqrt{-1-c^2 x^2} \sqrt{d+e x^2}}{120 c^3 e \sqrt{-c^2 x^2}}+\frac{b x \sqrt{-1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c e \sqrt{-c^2 x^2}}-\frac{d \left (d+e x^2\right )^{3/2} \left (a+b \text{csch}^{-1}(c x)\right )}{3 e^2}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \text{csch}^{-1}(c x)\right )}{5 e^2}-\frac{(b x) \operatorname{Subst}\left (\int \frac{-4 c^4 d^3-\frac{1}{4} e \left (15 c^4 d^2+10 c^2 d e-9 e^2\right ) x}{x \sqrt{-1-c^2 x} \sqrt{d+e x}} \, dx,x,x^2\right )}{60 c^3 e^2 \sqrt{-c^2 x^2}}\\ &=\frac{b \left (c^2 d-9 e\right ) x \sqrt{-1-c^2 x^2} \sqrt{d+e x^2}}{120 c^3 e \sqrt{-c^2 x^2}}+\frac{b x \sqrt{-1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c e \sqrt{-c^2 x^2}}-\frac{d \left (d+e x^2\right )^{3/2} \left (a+b \text{csch}^{-1}(c x)\right )}{3 e^2}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \text{csch}^{-1}(c x)\right )}{5 e^2}+\frac{\left (b c d^3 x\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{-1-c^2 x} \sqrt{d+e x}} \, dx,x,x^2\right )}{15 e^2 \sqrt{-c^2 x^2}}+\frac{\left (b \left (15 c^4 d^2+10 c^2 d e-9 e^2\right ) x\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-1-c^2 x} \sqrt{d+e x}} \, dx,x,x^2\right )}{240 c^3 e \sqrt{-c^2 x^2}}\\ &=\frac{b \left (c^2 d-9 e\right ) x \sqrt{-1-c^2 x^2} \sqrt{d+e x^2}}{120 c^3 e \sqrt{-c^2 x^2}}+\frac{b x \sqrt{-1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c e \sqrt{-c^2 x^2}}-\frac{d \left (d+e x^2\right )^{3/2} \left (a+b \text{csch}^{-1}(c x)\right )}{3 e^2}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \text{csch}^{-1}(c x)\right )}{5 e^2}+\frac{\left (2 b c d^3 x\right ) \operatorname{Subst}\left (\int \frac{1}{-d-x^2} \, dx,x,\frac{\sqrt{d+e x^2}}{\sqrt{-1-c^2 x^2}}\right )}{15 e^2 \sqrt{-c^2 x^2}}-\frac{\left (b \left (15 c^4 d^2+10 c^2 d e-9 e^2\right ) x\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{d-\frac{e}{c^2}-\frac{e x^2}{c^2}}} \, dx,x,\sqrt{-1-c^2 x^2}\right )}{120 c^5 e \sqrt{-c^2 x^2}}\\ &=\frac{b \left (c^2 d-9 e\right ) x \sqrt{-1-c^2 x^2} \sqrt{d+e x^2}}{120 c^3 e \sqrt{-c^2 x^2}}+\frac{b x \sqrt{-1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c e \sqrt{-c^2 x^2}}-\frac{d \left (d+e x^2\right )^{3/2} \left (a+b \text{csch}^{-1}(c x)\right )}{3 e^2}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \text{csch}^{-1}(c x)\right )}{5 e^2}-\frac{2 b c d^{5/2} x \tan ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d} \sqrt{-1-c^2 x^2}}\right )}{15 e^2 \sqrt{-c^2 x^2}}-\frac{\left (b \left (15 c^4 d^2+10 c^2 d e-9 e^2\right ) x\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{e x^2}{c^2}} \, dx,x,\frac{\sqrt{-1-c^2 x^2}}{\sqrt{d+e x^2}}\right )}{120 c^5 e \sqrt{-c^2 x^2}}\\ &=\frac{b \left (c^2 d-9 e\right ) x \sqrt{-1-c^2 x^2} \sqrt{d+e x^2}}{120 c^3 e \sqrt{-c^2 x^2}}+\frac{b x \sqrt{-1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c e \sqrt{-c^2 x^2}}-\frac{d \left (d+e x^2\right )^{3/2} \left (a+b \text{csch}^{-1}(c x)\right )}{3 e^2}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \text{csch}^{-1}(c x)\right )}{5 e^2}-\frac{b \left (15 c^4 d^2+10 c^2 d e-9 e^2\right ) x \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{-1-c^2 x^2}}{c \sqrt{d+e x^2}}\right )}{120 c^4 e^{3/2} \sqrt{-c^2 x^2}}-\frac{2 b c d^{5/2} x \tan ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d} \sqrt{-1-c^2 x^2}}\right )}{15 e^2 \sqrt{-c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.740532, size = 337, normalized size = 1.12 \[ \frac{\sqrt{d+e x^2} \left (8 a c^3 \left (-2 d^2+d e x^2+3 e^2 x^4\right )+8 b c^3 \text{csch}^{-1}(c x) \left (-2 d^2+d e x^2+3 e^2 x^4\right )+b e x \sqrt{\frac{1}{c^2 x^2}+1} \left (c^2 \left (7 d+6 e x^2\right )-9 e\right )\right )}{120 c^3 e^2}+\frac{b x \sqrt{\frac{1}{c^2 x^2}+1} \left (16 c^7 d^{5/2} \sqrt{-d-e x^2} \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{c^2 x^2+1}}{\sqrt{-d-e x^2}}\right )-\sqrt{c^2} \sqrt{e} \sqrt{c^2 d-e} \left (15 c^4 d^2+10 c^2 d e-9 e^2\right ) \sqrt{\frac{c^2 \left (d+e x^2\right )}{c^2 d-e}} \sinh ^{-1}\left (\frac{c \sqrt{e} \sqrt{c^2 x^2+1}}{\sqrt{c^2} \sqrt{c^2 d-e}}\right )\right )}{120 c^6 e^2 \sqrt{c^2 x^2+1} \sqrt{d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + e*x^2]*(8*a*c^3*(-2*d^2 + d*e*x^2 + 3*e^2*x^4) + b*e*Sqrt[1 + 1/(c^2*x^2)]*x*(-9*e + c^2*(7*d + 6*e*
x^2)) + 8*b*c^3*(-2*d^2 + d*e*x^2 + 3*e^2*x^4)*ArcCsch[c*x]))/(120*c^3*e^2) + (b*Sqrt[1 + 1/(c^2*x^2)]*x*(-(Sq
rt[c^2]*Sqrt[c^2*d - e]*Sqrt[e]*(15*c^4*d^2 + 10*c^2*d*e - 9*e^2)*Sqrt[(c^2*(d + e*x^2))/(c^2*d - e)]*ArcSinh[
(c*Sqrt[e]*Sqrt[1 + c^2*x^2])/(Sqrt[c^2]*Sqrt[c^2*d - e])]) + 16*c^7*d^(5/2)*Sqrt[-d - e*x^2]*ArcTan[(Sqrt[d]*
Sqrt[1 + c^2*x^2])/Sqrt[-d - e*x^2]]))/(120*c^6*e^2*Sqrt[1 + c^2*x^2]*Sqrt[d + e*x^2])

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Maple [F]  time = 0.414, size = 0, normalized size = 0. \begin{align*} \int{x}^{3} \left ( a+b{\rm arccsch} \left (cx\right ) \right ) \sqrt{e{x}^{2}+d}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arccsch(c*x))*(e*x^2+d)^(1/2),x)

[Out]

int(x^3*(a+b*arccsch(c*x))*(e*x^2+d)^(1/2),x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arccsch(c*x))*(e*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 9.54428, size = 3614, normalized size = 11.97 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arccsch(c*x))*(e*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

[1/480*(16*b*c^5*d^(5/2)*log(((c^4*d^2 + 6*c^2*d*e + e^2)*x^4 + 8*(c^2*d^2 + d*e)*x^2 + 4*((c^3*d + c*e)*x^3 +
2*c*d*x)*sqrt(e*x^2 + d)*sqrt(d)*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 8*d^2)/x^4) - (15*b*c^4*d^2 + 10*b*c^2*d*e -
9*b*e^2)*sqrt(e)*log(8*c^4*e^2*x^4 + c^4*d^2 + 6*c^2*d*e + 8*(c^4*d*e + c^2*e^2)*x^2 + 4*(2*c^4*e*x^3 + (c^4*
d + c^2*e)*x)*sqrt(e*x^2 + d)*sqrt(e)*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + e^2) + 32*(3*b*c^5*e^2*x^4 + b*c^5*d*e*x
^2 - 2*b*c^5*d^2)*sqrt(e*x^2 + d)*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x)) + 4*(24*a*c^5*e^2*x^4 + 8
*a*c^5*d*e*x^2 - 16*a*c^5*d^2 + (6*b*c^4*e^2*x^3 + (7*b*c^4*d*e - 9*b*c^2*e^2)*x)*sqrt((c^2*x^2 + 1)/(c^2*x^2)
))*sqrt(e*x^2 + d))/(c^5*e^2), 1/240*(8*b*c^5*d^(5/2)*log(((c^4*d^2 + 6*c^2*d*e + e^2)*x^4 + 8*(c^2*d^2 + d*e)
*x^2 + 4*((c^3*d + c*e)*x^3 + 2*c*d*x)*sqrt(e*x^2 + d)*sqrt(d)*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 8*d^2)/x^4) + (
15*b*c^4*d^2 + 10*b*c^2*d*e - 9*b*e^2)*sqrt(-e)*arctan(1/2*(2*c^2*e*x^3 + (c^2*d + e)*x)*sqrt(e*x^2 + d)*sqrt(
-e)*sqrt((c^2*x^2 + 1)/(c^2*x^2))/(c^2*e^2*x^4 + (c^2*d*e + e^2)*x^2 + d*e)) + 16*(3*b*c^5*e^2*x^4 + b*c^5*d*e
*x^2 - 2*b*c^5*d^2)*sqrt(e*x^2 + d)*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x)) + 2*(24*a*c^5*e^2*x^4 +
8*a*c^5*d*e*x^2 - 16*a*c^5*d^2 + (6*b*c^4*e^2*x^3 + (7*b*c^4*d*e - 9*b*c^2*e^2)*x)*sqrt((c^2*x^2 + 1)/(c^2*x^
2)))*sqrt(e*x^2 + d))/(c^5*e^2), -1/480*(32*b*c^5*sqrt(-d)*d^2*arctan(1/2*((c^3*d + c*e)*x^3 + 2*c*d*x)*sqrt(e
*x^2 + d)*sqrt(-d)*sqrt((c^2*x^2 + 1)/(c^2*x^2))/(c^2*d*e*x^4 + (c^2*d^2 + d*e)*x^2 + d^2)) + (15*b*c^4*d^2 +
10*b*c^2*d*e - 9*b*e^2)*sqrt(e)*log(8*c^4*e^2*x^4 + c^4*d^2 + 6*c^2*d*e + 8*(c^4*d*e + c^2*e^2)*x^2 + 4*(2*c^4
*e*x^3 + (c^4*d + c^2*e)*x)*sqrt(e*x^2 + d)*sqrt(e)*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + e^2) - 32*(3*b*c^5*e^2*x^4
+ b*c^5*d*e*x^2 - 2*b*c^5*d^2)*sqrt(e*x^2 + d)*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x)) - 4*(24*a*c
^5*e^2*x^4 + 8*a*c^5*d*e*x^2 - 16*a*c^5*d^2 + (6*b*c^4*e^2*x^3 + (7*b*c^4*d*e - 9*b*c^2*e^2)*x)*sqrt((c^2*x^2
+ 1)/(c^2*x^2)))*sqrt(e*x^2 + d))/(c^5*e^2), -1/240*(16*b*c^5*sqrt(-d)*d^2*arctan(1/2*((c^3*d + c*e)*x^3 + 2*c
*d*x)*sqrt(e*x^2 + d)*sqrt(-d)*sqrt((c^2*x^2 + 1)/(c^2*x^2))/(c^2*d*e*x^4 + (c^2*d^2 + d*e)*x^2 + d^2)) - (15*
b*c^4*d^2 + 10*b*c^2*d*e - 9*b*e^2)*sqrt(-e)*arctan(1/2*(2*c^2*e*x^3 + (c^2*d + e)*x)*sqrt(e*x^2 + d)*sqrt(-e)
*sqrt((c^2*x^2 + 1)/(c^2*x^2))/(c^2*e^2*x^4 + (c^2*d*e + e^2)*x^2 + d*e)) - 16*(3*b*c^5*e^2*x^4 + b*c^5*d*e*x^
2 - 2*b*c^5*d^2)*sqrt(e*x^2 + d)*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x)) - 2*(24*a*c^5*e^2*x^4 + 8*
a*c^5*d*e*x^2 - 16*a*c^5*d^2 + (6*b*c^4*e^2*x^3 + (7*b*c^4*d*e - 9*b*c^2*e^2)*x)*sqrt((c^2*x^2 + 1)/(c^2*x^2))
)*sqrt(e*x^2 + d))/(c^5*e^2)]

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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(x**3*(a+b*acsch(c*x))*(e*x**2+d)**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arccsch(c*x))*(e*x^2+d)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(e*x^2 + d)*(b*arccsch(c*x) + a)*x^3, x)