3.92 \(\int \frac{(d+e x^2) (a+b \text{sech}^{-1}(c x))}{x^4} \, dx\)

Optimal. Leaf size=126 \[ -\frac{d \left (a+b \text{sech}^{-1}(c x)\right )}{3 x^3}-\frac{e \left (a+b \text{sech}^{-1}(c x)\right )}{x}+\frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2} \left (2 c^2 d+9 e\right )}{9 x}+\frac{b d \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2}}{9 x^3} \]

[Out]

(b*d*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(9*x^3) + (b*(2*c^2*d + 9*e)*Sqrt[(1 + c*x)^(-1)]*S
qrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(9*x) - (d*(a + b*ArcSech[c*x]))/(3*x^3) - (e*(a + b*ArcSech[c*x]))/x

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Rubi [A]  time = 0.0792224, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {14, 6301, 12, 453, 264} \[ -\frac{d \left (a+b \text{sech}^{-1}(c x)\right )}{3 x^3}-\frac{e \left (a+b \text{sech}^{-1}(c x)\right )}{x}+\frac{b \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2} \left (2 c^2 d+9 e\right )}{9 x}+\frac{b d \sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sqrt{1-c^2 x^2}}{9 x^3} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)*(a + b*ArcSech[c*x]))/x^4,x]

[Out]

(b*d*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(9*x^3) + (b*(2*c^2*d + 9*e)*Sqrt[(1 + c*x)^(-1)]*S
qrt[1 + c*x]*Sqrt[1 - c^2*x^2])/(9*x) - (d*(a + b*ArcSech[c*x]))/(3*x^3) - (e*(a + b*ArcSech[c*x]))/x

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 12

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

Rule 453

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[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 264

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

Rubi steps

\begin{align*} \int \frac{\left (d+e x^2\right ) \left (a+b \text{sech}^{-1}(c x)\right )}{x^4} \, dx &=-\frac{d \left (a+b \text{sech}^{-1}(c x)\right )}{3 x^3}-\frac{e \left (a+b \text{sech}^{-1}(c x)\right )}{x}+\left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{-d-3 e x^2}{3 x^4 \sqrt{1-c^2 x^2}} \, dx\\ &=-\frac{d \left (a+b \text{sech}^{-1}(c x)\right )}{3 x^3}-\frac{e \left (a+b \text{sech}^{-1}(c x)\right )}{x}+\frac{1}{3} \left (b \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{-d-3 e x^2}{x^4 \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}}{9 x^3}-\frac{d \left (a+b \text{sech}^{-1}(c x)\right )}{3 x^3}-\frac{e \left (a+b \text{sech}^{-1}(c x)\right )}{x}+\frac{1}{9} \left (b \left (-2 c^2 d-9 e\right ) \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{1}{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}}{9 x^3}+\frac{b \left (2 c^2 d+9 e\right ) \sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sqrt{1-c^2 x^2}}{9 x}-\frac{d \left (a+b \text{sech}^{-1}(c x)\right )}{3 x^3}-\frac{e \left (a+b \text{sech}^{-1}(c x)\right )}{x}\\ \end{align*}

Mathematica [A]  time = 0.104381, size = 76, normalized size = 0.6 \[ \frac{-3 a \left (d+3 e x^2\right )+b \sqrt{\frac{1-c x}{c x+1}} (c x+1) \left (2 c^2 d x^2+d+9 e x^2\right )-3 b \text{sech}^{-1}(c x) \left (d+3 e x^2\right )}{9 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)*(a + b*ArcSech[c*x]))/x^4,x]

[Out]

(-3*a*(d + 3*e*x^2) + b*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(d + 2*c^2*d*x^2 + 9*e*x^2) - 3*b*(d + 3*e*x^2)*Ar
cSech[c*x])/(9*x^3)

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Maple [A]  time = 0.184, size = 123, normalized size = 1. \begin{align*}{c}^{3} \left ({\frac{a}{{c}^{2}} \left ( -{\frac{e}{cx}}-{\frac{d}{3\,c{x}^{3}}} \right ) }+{\frac{b}{{c}^{2}} \left ( -{\frac{{\rm arcsech} \left (cx\right )e}{cx}}-{\frac{{\rm arcsech} \left (cx\right )d}{3\,c{x}^{3}}}+{\frac{2\,{c}^{4}d{x}^{2}+9\,{c}^{2}{x}^{2}e+{c}^{2}d}{9\,{c}^{2}{x}^{2}}\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((e*x^2+d)*(a+b*arcsech(c*x))/x^4,x)

[Out]

c^3*(a/c^2*(-e/c/x-1/3/c*d/x^3)+b/c^2*(-arcsech(c*x)*e/c/x-1/3*arcsech(c*x)/c*d/x^3+1/9*(-(c*x-1)/c/x)^(1/2)/c
^2/x^2*((c*x+1)/c/x)^(1/2)*(2*c^4*d*x^2+9*c^2*e*x^2+c^2*d)))

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Maxima [A]  time = 0.99585, size = 123, normalized size = 0.98 \begin{align*}{\left (c \sqrt{\frac{1}{c^{2} x^{2}} - 1} - \frac{\operatorname{arsech}\left (c x\right )}{x}\right )} b e + \frac{1}{9} \, b d{\left (\frac{c^{4}{\left (\frac{1}{c^{2} x^{2}} - 1\right )}^{\frac{3}{2}} + 3 \, c^{4} \sqrt{\frac{1}{c^{2} x^{2}} - 1}}{c} - \frac{3 \, \operatorname{arsech}\left (c x\right )}{x^{3}}\right )} - \frac{a e}{x} - \frac{a d}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arcsech(c*x))/x^4,x, algorithm="maxima")

[Out]

(c*sqrt(1/(c^2*x^2) - 1) - arcsech(c*x)/x)*b*e + 1/9*b*d*((c^4*(1/(c^2*x^2) - 1)^(3/2) + 3*c^4*sqrt(1/(c^2*x^2
) - 1))/c - 3*arcsech(c*x)/x^3) - a*e/x - 1/3*a*d/x^3

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Fricas [A]  time = 1.75543, size = 236, normalized size = 1.87 \begin{align*} -\frac{9 \, a e x^{2} + 3 \, a d + 3 \,{\left (3 \, b e x^{2} + b d\right )} \log \left (\frac{c x \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) -{\left (b c d x +{\left (2 \, b c^{3} d + 9 \, b c e\right )} x^{3}\right )} \sqrt{-\frac{c^{2} x^{2} - 1}{c^{2} x^{2}}}}{9 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*(a+b*arcsech(c*x))/x^4,x, algorithm="fricas")

[Out]

-1/9*(9*a*e*x^2 + 3*a*d + 3*(3*b*e*x^2 + b*d)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) - (b*c*d*x +
(2*b*c^3*d + 9*b*c*e)*x^3)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)))/x^3

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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^{4}}\, 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**4,x)

[Out]

Integral((a + b*asech(c*x))*(d + e*x**2)/x**4, 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^{4}}\,{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^4,x, algorithm="giac")

[Out]

integrate((e*x^2 + d)*(b*arcsech(c*x) + a)/x^4, x)