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

Optimal. Leaf size=164 \[ \frac{\left (d+e x^2\right )^2 \left (a+b \text{sech}^{-1}(c x)\right )}{4 e}-\frac{b d^2 \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )}{4 e}-\frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2} \left (2 c^2 d+e\right )}{4 c^4}+\frac{b e \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \left (1-c^2 x^2\right )^{3/2}}{12 c^4} \]

[Out]

-(b*(2*c^2*d + e)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(4*c^4) + (b*e*Sqrt[(1 + c*x)^(-1)]*Sq
rt[1 + c*x]*(1 - c^2*x^2)^(3/2))/(12*c^4) + ((d + e*x^2)^2*(a + b*ArcSech[c*x]))/(4*e) - (b*d^2*Sqrt[(1 + c*x)
^(-1)]*Sqrt[1 + c*x]*ArcTanh[Sqrt[1 - c^2*x^2]])/(4*e)

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Rubi [A]  time = 0.193181, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.353, Rules used = {6299, 517, 446, 88, 63, 208} \[ \frac{\left (d+e x^2\right )^2 \left (a+b \text{sech}^{-1}(c x)\right )}{4 e}-\frac{b d^2 \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )}{4 e}-\frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2} \left (2 c^2 d+e\right )}{4 c^4}+\frac{b e \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \left (1-c^2 x^2\right )^{3/2}}{12 c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(2*c^2*d + e)*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(4*c^4) + (b*e*Sqrt[(1 + c*x)^(-1)]*Sq
rt[1 + c*x]*(1 - c^2*x^2)^(3/2))/(12*c^4) + ((d + e*x^2)^2*(a + b*ArcSech[c*x]))/(4*e) - (b*d^2*Sqrt[(1 + c*x)
^(-1)]*Sqrt[1 + c*x]*ArcTanh[Sqrt[1 - c^2*x^2]])/(4*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 88

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

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 208

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

Rubi steps

\begin{align*} \int x \left (d+e x^2\right ) \left (a+b \text{sech}^{-1}(c x)\right ) \, dx &=\frac{\left (d+e x^2\right )^2 \left (a+b \text{sech}^{-1}(c x)\right )}{4 e}+\frac{\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{\left (d+e x^2\right )^2}{x \sqrt{1-c x} \sqrt{1+c x}} \, dx}{4 e}\\ &=\frac{\left (d+e x^2\right )^2 \left (a+b \text{sech}^{-1}(c x)\right )}{4 e}+\frac{\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{\left (d+e x^2\right )^2}{x \sqrt{1-c^2 x^2}} \, dx}{4 e}\\ &=\frac{\left (d+e x^2\right )^2 \left (a+b \text{sech}^{-1}(c x)\right )}{4 e}+\frac{\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \operatorname{Subst}\left (\int \frac{(d+e x)^2}{x \sqrt{1-c^2 x}} \, dx,x,x^2\right )}{8 e}\\ &=\frac{\left (d+e x^2\right )^2 \left (a+b \text{sech}^{-1}(c x)\right )}{4 e}+\frac{\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \operatorname{Subst}\left (\int \left (\frac{e \left (2 c^2 d+e\right )}{c^2 \sqrt{1-c^2 x}}+\frac{d^2}{x \sqrt{1-c^2 x}}-\frac{e^2 \sqrt{1-c^2 x}}{c^2}\right ) \, dx,x,x^2\right )}{8 e}\\ &=-\frac{b \left (2 c^2 d+e\right ) \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{4 c^4}+\frac{b e \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \left (1-c^2 x^2\right )^{3/2}}{12 c^4}+\frac{\left (d+e x^2\right )^2 \left (a+b \text{sech}^{-1}(c x)\right )}{4 e}+\frac{\left (b d^2 \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1-c^2 x}} \, dx,x,x^2\right )}{8 e}\\ &=-\frac{b \left (2 c^2 d+e\right ) \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{4 c^4}+\frac{b e \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \left (1-c^2 x^2\right )^{3/2}}{12 c^4}+\frac{\left (d+e x^2\right )^2 \left (a+b \text{sech}^{-1}(c x)\right )}{4 e}-\frac{\left (b d^2 \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{c^2}-\frac{x^2}{c^2}} \, dx,x,\sqrt{1-c^2 x^2}\right )}{4 c^2 e}\\ &=-\frac{b \left (2 c^2 d+e\right ) \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{4 c^4}+\frac{b e \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \left (1-c^2 x^2\right )^{3/2}}{12 c^4}+\frac{\left (d+e x^2\right )^2 \left (a+b \text{sech}^{-1}(c x)\right )}{4 e}-\frac{b d^2 \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \tanh ^{-1}\left (\sqrt{1-c^2 x^2}\right )}{4 e}\\ \end{align*}

Mathematica [A]  time = 0.125561, size = 85, normalized size = 0.52 \[ \frac{1}{12} \left (3 a x^2 \left (2 d+e x^2\right )-\frac{b \sqrt{\frac{1-c x}{c x+1}} (c x+1) \left (c^2 \left (6 d+e x^2\right )+2 e\right )}{c^4}+3 b x^2 \text{sech}^{-1}(c x) \left (2 d+e x^2\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*a*x^2*(2*d + e*x^2) - (b*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(2*e + c^2*(6*d + e*x^2)))/c^4 + 3*b*x^2*(2*d
+ e*x^2)*ArcSech[c*x])/12

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Maple [A]  time = 0.178, size = 113, normalized size = 0.7 \begin{align*}{\frac{1}{{c}^{2}} \left ({\frac{a}{{c}^{2}} \left ({\frac{{c}^{4}{x}^{4}e}{4}}+{\frac{{x}^{2}{c}^{4}d}{2}} \right ) }+{\frac{b}{{c}^{2}} \left ({\frac{{\rm arcsech} \left (cx\right ){c}^{4}{x}^{4}e}{4}}+{\frac{{\rm arcsech} \left (cx\right ){c}^{4}{x}^{2}d}{2}}-{\frac{cx \left ({c}^{2}{x}^{2}e+6\,{c}^{2}d+2\,e \right ) }{12}\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^2*(a/c^2*(1/4*c^4*x^4*e+1/2*x^2*c^4*d)+b/c^2*(1/4*arcsech(c*x)*c^4*x^4*e+1/2*arcsech(c*x)*c^4*x^2*d-1/12*(
-(c*x-1)/c/x)^(1/2)*c*x*((c*x+1)/c/x)^(1/2)*(c^2*e*x^2+6*c^2*d+2*e)))

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Maxima [A]  time = 0.998229, size = 130, normalized size = 0.79 \begin{align*} \frac{1}{4} \, a e x^{4} + \frac{1}{2} \, a d x^{2} + \frac{1}{2} \,{\left (x^{2} \operatorname{arsech}\left (c x\right ) - \frac{x \sqrt{\frac{1}{c^{2} x^{2}} - 1}}{c}\right )} b d + \frac{1}{12} \,{\left (3 \, x^{4} \operatorname{arsech}\left (c x\right ) + \frac{c^{2} x^{3}{\left (\frac{1}{c^{2} x^{2}} - 1\right )}^{\frac{3}{2}} - 3 \, x \sqrt{\frac{1}{c^{2} x^{2}} - 1}}{c^{3}}\right )} b e \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*a*e*x^4 + 1/2*a*d*x^2 + 1/2*(x^2*arcsech(c*x) - x*sqrt(1/(c^2*x^2) - 1)/c)*b*d + 1/12*(3*x^4*arcsech(c*x)
+ (c^2*x^3*(1/(c^2*x^2) - 1)^(3/2) - 3*x*sqrt(1/(c^2*x^2) - 1))/c^3)*b*e

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Fricas [A]  time = 2.0441, size = 269, normalized size = 1.64 \begin{align*} \frac{3 \, a c^{3} e x^{4} + 6 \, a c^{3} d x^{2} + 3 \,{\left (b c^{3} e x^{4} + 2 \, b c^{3} d x^{2}\right )} \log \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) -{\left (b c^{2} e x^{3} + 2 \,{\left (3 \, b c^{2} d + b e\right )} x\right )} \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}}}{12 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*a*c^3*e*x^4 + 6*a*c^3*d*x^2 + 3*(b*c^3*e*x^4 + 2*b*c^3*d*x^2)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2))
+ 1)/(c*x)) - (b*c^2*e*x^3 + 2*(3*b*c^2*d + b*e)*x)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)))/c^3

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Sympy [A]  time = 4.15568, size = 126, normalized size = 0.77 \begin{align*} \begin{cases} \frac{a d x^{2}}{2} + \frac{a e x^{4}}{4} + \frac{b d x^{2} \operatorname{asech}{\left (c x \right )}}{2} + \frac{b e x^{4} \operatorname{asech}{\left (c x \right )}}{4} - \frac{b d \sqrt{- c^{2} x^{2} + 1}}{2 c^{2}} - \frac{b e x^{2} \sqrt{- c^{2} x^{2} + 1}}{12 c^{2}} - \frac{b e \sqrt{- c^{2} x^{2} + 1}}{6 c^{4}} & \text{for}\: c \neq 0 \\\left (a + \infty b\right ) \left (\frac{d x^{2}}{2} + \frac{e x^{4}}{4}\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*d*x**2/2 + a*e*x**4/4 + b*d*x**2*asech(c*x)/2 + b*e*x**4*asech(c*x)/4 - b*d*sqrt(-c**2*x**2 + 1)/
(2*c**2) - b*e*x**2*sqrt(-c**2*x**2 + 1)/(12*c**2) - b*e*sqrt(-c**2*x**2 + 1)/(6*c**4), Ne(c, 0)), ((a + oo*b)
*(d*x**2/2 + e*x**4/4), True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x^{2} + d\right )}{\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)*(a+b*arcsech(c*x)),x, algorithm="giac")

[Out]

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