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

Optimal. Leaf size=96 \[ -\frac{d \left (a+b \text{sech}^{-1}(c x)\right )}{x}+e x \left (a+b \text{sech}^{-1}(c x)\right )+\frac{b d \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2}}{x}+\frac{b e \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sin ^{-1}(c x)}{c} \]

[Out]

(b*d*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/x - (d*(a + b*ArcSech[c*x]))/x + e*x*(a + b*ArcSech
[c*x]) + (b*e*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcSin[c*x])/c

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Rubi [A]  time = 0.0658553, antiderivative size = 96, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {14, 6301, 451, 216} \[ -\frac{d \left (a+b \text{sech}^{-1}(c x)\right )}{x}+e x \left (a+b \text{sech}^{-1}(c x)\right )+\frac{b d \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2}}{x}+\frac{b e \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sin ^{-1}(c x)}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/x - (d*(a + b*ArcSech[c*x]))/x + e*x*(a + b*ArcSech
[c*x]) + (b*e*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcSin[c*x])/c

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 6301

Int[((a_.) + ArcSech[(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*ArcSech[c*x], u, x] + Dist[b*Sqrt[1 + c*x]*Sqrt[1/(1 + c*x)],
Int[SimplifyIntegrand[u/(x*Sqrt[1 - c*x]*Sqrt[1 + c*x]), 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 451

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

Rule 216

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

Rubi steps

\begin{align*} \int \frac{\left (d+e x^2\right ) \left (a+b \text{sech}^{-1}(c x)\right )}{x^2} \, dx &=-\frac{d \left (a+b \text{sech}^{-1}(c x)\right )}{x}+e x \left (a+b \text{sech}^{-1}(c x)\right )+\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{-d+e x^2}{x^2 \sqrt{1-c^2 x^2}} \, dx\\ &=\frac{b d \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{x}-\frac{d \left (a+b \text{sech}^{-1}(c x)\right )}{x}+e x \left (a+b \text{sech}^{-1}(c x)\right )+\left (b e \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{1}{\sqrt{1-c^2 x^2}} \, dx\\ &=\frac{b d \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{x}-\frac{d \left (a+b \text{sech}^{-1}(c x)\right )}{x}+e x \left (a+b \text{sech}^{-1}(c x)\right )+\frac{b e \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sin ^{-1}(c x)}{c}\\ \end{align*}

Mathematica [A]  time = 0.238545, size = 107, normalized size = 1.11 \[ -\frac{a d}{x}+a e x-\frac{b e \sqrt{\frac{1-c x}{c x+1}} \sqrt{1-c^2 x^2} \sin ^{-1}(c x)}{c (c x-1)}+b d \left (c+\frac{1}{x}\right ) \sqrt{\frac{1-c x}{c x+1}}-\frac{b d \text{sech}^{-1}(c x)}{x}+b e x \text{sech}^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*d)/x) + a*e*x + b*d*(c + x^(-1))*Sqrt[(1 - c*x)/(1 + c*x)] - (b*d*ArcSech[c*x])/x + b*e*x*ArcSech[c*x] -
(b*e*Sqrt[(1 - c*x)/(1 + c*x)]*Sqrt[1 - c^2*x^2]*ArcSin[c*x])/(c*(-1 + c*x))

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Maple [A]  time = 0.21, size = 114, normalized size = 1.2 \begin{align*} c \left ({\frac{a}{{c}^{2}} \left ( cxe-{\frac{cd}{x}} \right ) }+{\frac{b}{{c}^{2}} \left ({\rm arcsech} \left (cx\right )cxe-{\frac{{\rm arcsech} \left (cx\right )cd}{x}}+{\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}} \left ({c}^{2}d\sqrt{-{c}^{2}{x}^{2}+1}+\arcsin \left ( cx \right ) cxe \right ){\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.00703, size = 89, normalized size = 0.93 \begin{align*}{\left (c \sqrt{\frac{1}{c^{2} x^{2}} - 1} - \frac{\operatorname{arsech}\left (c x\right )}{x}\right )} b d + a e x + \frac{{\left (c x \operatorname{arsech}\left (c x\right ) - \arctan \left (\sqrt{\frac{1}{c^{2} x^{2}} - 1}\right )\right )} b e}{c} - \frac{a d}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.08469, size = 401, normalized size = 4.18 \begin{align*} \frac{b c^{2} d x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} + a c e x^{2} - 2 \, b e x \arctan \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{c x}\right ) - a c d +{\left (b c d - b c e\right )} x \log \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{x}\right ) +{\left (b c e x^{2} - b c d +{\left (b c d - b c e\right )} x\right )} \log \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right )}{c x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*c^2*d*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + a*c*e*x^2 - 2*b*e*x*arctan((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) - 1
)/(c*x)) - a*c*d + (b*c*d - b*c*e)*x*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) - 1)/x) + (b*c*e*x^2 - b*c*d + (b
*c*d - b*c*e)*x)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)))/(c*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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