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

Optimal. Leaf size=297 \[ \frac{\left (d+e x^2\right )^{5/2} \left (a+b \text{sech}^{-1}(c x)\right )}{5 e}-\frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \left (15 c^4 d^2+10 c^2 d e+3 e^2\right ) \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{1-c^2 x^2}}{c \sqrt{d+e x^2}}\right )}{40 c^5 \sqrt{e}}-\frac{b d^{5/2} \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d} \sqrt{1-c^2 x^2}}\right )}{5 e}-\frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c^2}-\frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2} \left (7 c^2 d+3 e\right ) \sqrt{d+e x^2}}{40 c^4} \]

[Out]

-(b*(7*c^2*d + 3*e)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2]*Sqrt[d + e*x^2])/(40*c^4) - (b*Sqrt[(
1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2]*(d + e*x^2)^(3/2))/(20*c^2) + ((d + e*x^2)^(5/2)*(a + b*ArcSech
[c*x]))/(5*e) - (b*(15*c^4*d^2 + 10*c^2*d*e + 3*e^2)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcTan[(Sqrt[e]*Sqrt[1
- c^2*x^2])/(c*Sqrt[d + e*x^2])])/(40*c^5*Sqrt[e]) - (b*d^(5/2)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcTanh[Sq
rt[d + e*x^2]/(Sqrt[d]*Sqrt[1 - c^2*x^2])])/(5*e)

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Rubi [A]  time = 0.43235, antiderivative size = 297, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 11, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.524, Rules used = {6299, 517, 446, 102, 154, 157, 63, 217, 203, 93, 207} \[ \frac{\left (d+e x^2\right )^{5/2} \left (a+b \text{sech}^{-1}(c x)\right )}{5 e}-\frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \left (15 c^4 d^2+10 c^2 d e+3 e^2\right ) \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{1-c^2 x^2}}{c \sqrt{d+e x^2}}\right )}{40 c^5 \sqrt{e}}-\frac{b d^{5/2} \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d} \sqrt{1-c^2 x^2}}\right )}{5 e}-\frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c^2}-\frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2} \left (7 c^2 d+3 e\right ) \sqrt{d+e x^2}}{40 c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(7*c^2*d + 3*e)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2]*Sqrt[d + e*x^2])/(40*c^4) - (b*Sqrt[(
1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2]*(d + e*x^2)^(3/2))/(20*c^2) + ((d + e*x^2)^(5/2)*(a + b*ArcSech
[c*x]))/(5*e) - (b*(15*c^4*d^2 + 10*c^2*d*e + 3*e^2)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcTan[(Sqrt[e]*Sqrt[1
- c^2*x^2])/(c*Sqrt[d + e*x^2])])/(40*c^5*Sqrt[e]) - (b*d^(5/2)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcTanh[Sq
rt[d + e*x^2]/(Sqrt[d]*Sqrt[1 - c^2*x^2])])/(5*e)

Rule 6299

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))*(x_)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)^(p +
1)*(a + b*ArcSech[c*x]))/(2*e*(p + 1)), x] + Dist[(b*Sqrt[1 + c*x]*Sqrt[1/(1 + c*x)])/(2*e*(p + 1)), Int[(d +
e*x^2)^(p + 1)/(x*Sqrt[1 - c*x]*Sqrt[1 + c*x]), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]

Rule 517

Int[(u_.)*((c_) + (d_.)*(x_)^(n_.))^(q_.)*((a1_) + (b1_.)*(x_)^(non2_.))^(p_.)*((a2_) + (b2_.)*(x_)^(non2_.))^
(p_.), x_Symbol] :> Int[u*(a1*a2 + b1*b2*x^n)^p*(c + d*x^n)^q, x] /; FreeQ[{a1, b1, a2, b2, c, d, n, p, q}, x]
&& EqQ[non2, n/2] && EqQ[a2*b1 + a1*b2, 0] && (IntegerQ[p] || (GtQ[a1, 0] && GtQ[a2, 0]))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 102

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

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 207

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

Rubi steps

\begin{align*} \int x \left (d+e x^2\right )^{3/2} \left (a+b \text{sech}^{-1}(c x)\right ) \, dx &=\frac{\left (d+e x^2\right )^{5/2} \left (a+b \text{sech}^{-1}(c x)\right )}{5 e}+\frac{\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{\left (d+e x^2\right )^{5/2}}{x \sqrt{1-c x} \sqrt{1+c x}} \, dx}{5 e}\\ &=\frac{\left (d+e x^2\right )^{5/2} \left (a+b \text{sech}^{-1}(c x)\right )}{5 e}+\frac{\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{\left (d+e x^2\right )^{5/2}}{x \sqrt{1-c^2 x^2}} \, dx}{5 e}\\ &=\frac{\left (d+e x^2\right )^{5/2} \left (a+b \text{sech}^{-1}(c x)\right )}{5 e}+\frac{\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \operatorname{Subst}\left (\int \frac{(d+e x)^{5/2}}{x \sqrt{1-c^2 x}} \, dx,x,x^2\right )}{10 e}\\ &=-\frac{b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c^2}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \text{sech}^{-1}(c x)\right )}{5 e}-\frac{\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{d+e x} \left (-2 c^2 d^2-\frac{1}{2} e \left (7 c^2 d+3 e\right ) x\right )}{x \sqrt{1-c^2 x}} \, dx,x,x^2\right )}{20 c^2 e}\\ &=-\frac{b \left (7 c^2 d+3 e\right ) \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2} \sqrt{d+e x^2}}{40 c^4}-\frac{b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c^2}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \text{sech}^{-1}(c x)\right )}{5 e}+\frac{\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \operatorname{Subst}\left (\int \frac{2 c^4 d^3+\frac{1}{4} e \left (15 c^4 d^2+10 c^2 d e+3 e^2\right ) x}{x \sqrt{1-c^2 x} \sqrt{d+e x}} \, dx,x,x^2\right )}{20 c^4 e}\\ &=-\frac{b \left (7 c^2 d+3 e\right ) \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2} \sqrt{d+e x^2}}{40 c^4}-\frac{b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c^2}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \text{sech}^{-1}(c x)\right )}{5 e}+\frac{\left (b d^3 \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-c^2 x} \sqrt{d+e x}} \, dx,x,x^2\right )}{10 e}+\frac{\left (b \left (15 c^4 d^2+10 c^2 d e+3 e^2\right ) \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-c^2 x} \sqrt{d+e x}} \, dx,x,x^2\right )}{80 c^4}\\ &=-\frac{b \left (7 c^2 d+3 e\right ) \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2} \sqrt{d+e x^2}}{40 c^4}-\frac{b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c^2}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \text{sech}^{-1}(c x)\right )}{5 e}+\frac{\left (b d^3 \sqrt{\frac{1}{1+c x}} \sqrt{1+c 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 )}{5 e}-\frac{\left (b \left (15 c^4 d^2+10 c^2 d e+3 e^2\right ) \sqrt{\frac{1}{1+c x}} \sqrt{1+c 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 )}{40 c^6}\\ &=-\frac{b \left (7 c^2 d+3 e\right ) \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2} \sqrt{d+e x^2}}{40 c^4}-\frac{b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c^2}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \text{sech}^{-1}(c x)\right )}{5 e}-\frac{b d^{5/2} \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d} \sqrt{1-c^2 x^2}}\right )}{5 e}-\frac{\left (b \left (15 c^4 d^2+10 c^2 d e+3 e^2\right ) \sqrt{\frac{1}{1+c x}} \sqrt{1+c 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 )}{40 c^6}\\ &=-\frac{b \left (7 c^2 d+3 e\right ) \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2} \sqrt{d+e x^2}}{40 c^4}-\frac{b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c^2}+\frac{\left (d+e x^2\right )^{5/2} \left (a+b \text{sech}^{-1}(c x)\right )}{5 e}-\frac{b \left (15 c^4 d^2+10 c^2 d e+3 e^2\right ) \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \tan ^{-1}\left (\frac{\sqrt{e} \sqrt{1-c^2 x^2}}{c \sqrt{d+e x^2}}\right )}{40 c^5 \sqrt{e}}-\frac{b d^{5/2} \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d} \sqrt{1-c^2 x^2}}\right )}{5 e}\\ \end{align*}

Mathematica [A]  time = 1.48017, size = 342, normalized size = 1.15 \[ \frac{\sqrt{d+e x^2} \left (8 a c^4 \left (d+e x^2\right )^2-b e \sqrt{\frac{1-c x}{c x+1}} (c x+1) \left (c^2 \left (9 d+2 e x^2\right )+3 e\right )+8 b c^4 \text{sech}^{-1}(c x) \left (d+e x^2\right )^2\right )}{40 c^4 e}+\frac{b \sqrt{\frac{1-c x}{c x+1}} \sqrt{1-c^2 x^2} \left (\sqrt{-c^2} \sqrt{e} \sqrt{c^2 (-d)-e} \left (15 c^4 d^2+10 c^2 d e+3 e^2\right ) \sqrt{\frac{c^2 \left (d+e x^2\right )}{c^2 d+e}} \sin ^{-1}\left (\frac{c \sqrt{e} \sqrt{1-c^2 x^2}}{\sqrt{-c^2} \sqrt{c^2 (-d)-e}}\right )+8 c^7 d^{5/2} \sqrt{-d-e x^2} \tan ^{-1}\left (\frac{\sqrt{d} \sqrt{1-c^2 x^2}}{\sqrt{-d-e x^2}}\right )\right )}{40 c^7 e (c x-1) \sqrt{d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[d + e*x^2]*(8*a*c^4*(d + e*x^2)^2 - b*e*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(3*e + c^2*(9*d + 2*e*x^2))
+ 8*b*c^4*(d + e*x^2)^2*ArcSech[c*x]))/(40*c^4*e) + (b*Sqrt[(1 - c*x)/(1 + c*x)]*Sqrt[1 - c^2*x^2]*(Sqrt[-c^2]
*Sqrt[-(c^2*d) - e]*Sqrt[e]*(15*c^4*d^2 + 10*c^2*d*e + 3*e^2)*Sqrt[(c^2*(d + e*x^2))/(c^2*d + e)]*ArcSin[(c*Sq
rt[e]*Sqrt[1 - c^2*x^2])/(Sqrt[-c^2]*Sqrt[-(c^2*d) - e])] + 8*c^7*d^(5/2)*Sqrt[-d - e*x^2]*ArcTan[(Sqrt[d]*Sqr
t[1 - c^2*x^2])/Sqrt[-d - e*x^2]]))/(40*c^7*e*(-1 + c*x)*Sqrt[d + e*x^2])

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Maple [F]  time = 0.888, size = 0, normalized size = 0. \begin{align*} \int x \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}} \left ( a+b{\rm arcsech} \left (cx\right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{{\left (e x^{2} + d\right )}^{\frac{5}{2}} a}{5 \, e} + \frac{1}{15} \, b{\left (\frac{{\left ({\left (3 \, e^{2} x^{4} + d e x^{2} - 2 \, d^{2}\right )} x^{3} + 5 \,{\left (d e x^{4} + d^{2} x^{2}\right )} x\right )} \sqrt{e x^{2} + d} \log \left (\sqrt{c x + 1} \sqrt{-c x + 1} + 1\right )}{e x^{3}} - 15 \, \int \frac{{\left (15 \,{\left (c^{2} e^{2} x^{4} \log \left (c\right ) - e^{2} x^{2} \log \left (c\right )\right )} x^{3} + 15 \,{\left (c^{2} d e x^{4} \log \left (c\right ) - d e x^{2} \log \left (c\right )\right )} x +{\left ({\left (3 \,{\left (5 \, e^{2} \log \left (c\right ) + e^{2}\right )} c^{2} x^{4} - 2 \, c^{2} d^{2} +{\left (c^{2} d e - 15 \, e^{2} \log \left (c\right )\right )} x^{2}\right )} x^{3} + 5 \,{\left ({\left (3 \, d e \log \left (c\right ) + d e\right )} c^{2} x^{4} +{\left (c^{2} d^{2} - 3 \, d e \log \left (c\right )\right )} x^{2}\right )} x + 30 \,{\left ({\left (c^{2} e^{2} x^{4} - e^{2} x^{2}\right )} x^{3} +{\left (c^{2} d e x^{4} - d e x^{2}\right )} x\right )} \log \left (\sqrt{x}\right )\right )} e^{\left (\frac{1}{2} \, \log \left (c x + 1\right ) + \frac{1}{2} \, \log \left (-c x + 1\right )\right )} + 30 \,{\left ({\left (c^{2} e^{2} x^{4} - e^{2} x^{2}\right )} x^{3} +{\left (c^{2} d e x^{4} - d e x^{2}\right )} x\right )} \log \left (\sqrt{x}\right )\right )} \sqrt{e x^{2} + d}}{15 \,{\left (c^{2} e x^{4} - e x^{2} +{\left (c^{2} e x^{4} - e x^{2}\right )} e^{\left (\frac{1}{2} \, \log \left (c x + 1\right ) + \frac{1}{2} \, \log \left (-c x + 1\right )\right )}\right )}}\,{d x}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(e*x^2 + d)^(5/2)*a/e + 1/15*b*(((3*e^2*x^4 + d*e*x^2 - 2*d^2)*x^3 + 5*(d*e*x^4 + d^2*x^2)*x)*sqrt(e*x^2 +
d)*log(sqrt(c*x + 1)*sqrt(-c*x + 1) + 1)/(e*x^3) - 15*integrate(1/15*(15*(c^2*e^2*x^4*log(c) - e^2*x^2*log(c)
)*x^3 + 15*(c^2*d*e*x^4*log(c) - d*e*x^2*log(c))*x + ((3*(5*e^2*log(c) + e^2)*c^2*x^4 - 2*c^2*d^2 + (c^2*d*e -
15*e^2*log(c))*x^2)*x^3 + 5*((3*d*e*log(c) + d*e)*c^2*x^4 + (c^2*d^2 - 3*d*e*log(c))*x^2)*x + 30*((c^2*e^2*x^
4 - e^2*x^2)*x^3 + (c^2*d*e*x^4 - d*e*x^2)*x)*log(sqrt(x)))*e^(1/2*log(c*x + 1) + 1/2*log(-c*x + 1)) + 30*((c^
2*e^2*x^4 - e^2*x^2)*x^3 + (c^2*d*e*x^4 - d*e*x^2)*x)*log(sqrt(x)))*sqrt(e*x^2 + d)/(c^2*e*x^4 - e*x^2 + (c^2*
e*x^4 - e*x^2)*e^(1/2*log(c*x + 1) + 1/2*log(-c*x + 1))), x))

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Fricas [B]  time = 8.4908, size = 3606, normalized size = 12.14 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/160*(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 +
3*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*(b*c^5*e^2*x^4 + 2*b*c^5*d*
e*x^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*(8*a*c^5*e^2*x^4 +
16*a*c^5*d*e*x^2 + 8*a*c^5*d^2 - (2*b*c^4*e^2*x^3 + 3*(3*b*c^4*d*e + b*c^2*e^2)*x)*sqrt(-(c^2*x^2 - 1)/(c^2*x^
2)))*sqrt(e*x^2 + d))/(c^5*e), 1/80*(4*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 + 3*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*(b*c^5*e^2*x^4 + 2*b*c^5*d*e*
x^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*(8*a*c^5*e^2*x^4 + 16
*a*c^5*d*e*x^2 + 8*a*c^5*d^2 - (2*b*c^4*e^2*x^3 + 3*(3*b*c^4*d*e + b*c^2*e^2)*x)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)
))*sqrt(e*x^2 + d))/(c^5*e), -1/160*(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 + 3*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*(b*c^5*e^2*x^4
+ 2*b*c^5*d*e*x^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*(8*a*c^
5*e^2*x^4 + 16*a*c^5*d*e*x^2 + 8*a*c^5*d^2 - (2*b*c^4*e^2*x^3 + 3*(3*b*c^4*d*e + b*c^2*e^2)*x)*sqrt(-(c^2*x^2
- 1)/(c^2*x^2)))*sqrt(e*x^2 + d))/(c^5*e), -1/80*(8*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 + 3*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)*sqr
t(-(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*(b*c^5*e^2*x^4 + 2*b*c^5*d*e*x^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*(8*a*c^5*e^2*x^4 + 16*a*c^
5*d*e*x^2 + 8*a*c^5*d^2 - (2*b*c^4*e^2*x^3 + 3*(3*b*c^4*d*e + b*c^2*e^2)*x)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)))*sq
rt(e*x^2 + d))/(c^5*e)]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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