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

Optimal. Leaf size=61 \[ \frac {2 b \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (a+b \text {sech}^{-1}(c x)\right )}{x}-\frac {\left (a+b \text {sech}^{-1}(c x)\right )^2}{x}-\frac {2 b^2}{x} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {6285, 3296, 2638} \[ \frac {2 b \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (a+b \text {sech}^{-1}(c x)\right )}{x}-\frac {\left (a+b \text {sech}^{-1}(c x)\right )^2}{x}-\frac {2 b^2}{x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSech[c*x])^2/x^2,x]

[Out]

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

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 6285

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> -Dist[(c^(m + 1))^(-1), Subst[Int[(a + b
*x)^n*Sech[x]^(m + 1)*Tanh[x], x], x, ArcSech[c*x]], x] /; FreeQ[{a, b, c}, x] && IntegerQ[n] && IntegerQ[m] &
& (GtQ[n, 0] || LtQ[m, -1])

Rubi steps

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

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Mathematica [A]  time = 0.24, size = 87, normalized size = 1.43 \[ -\frac {a^2-2 a b \sqrt {\frac {1-c x}{c x+1}} (c x+1)-2 b \text {sech}^{-1}(c x) \left (b \sqrt {\frac {1-c x}{c x+1}} (c x+1)-a\right )+b^2 \text {sech}^{-1}(c x)^2+2 b^2}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.59, size = 143, normalized size = 2.34 \[ \frac {2 \, a b c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - b^{2} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right )^{2} - a^{2} - 2 \, b^{2} + 2 \, {\left (b^{2} c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - a b\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arsech}\left (c x\right ) + a\right )}^{2}}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.14, size = 124, normalized size = 2.03 \[ c \left (-\frac {a^{2}}{c x}+b^{2} \left (-\frac {\mathrm {arcsech}\left (c x \right )^{2}}{c x}+2 \,\mathrm {arcsech}\left (c x \right ) \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}-\frac {2}{c x}\right )+2 a b \left (-\frac {\mathrm {arcsech}\left (c x \right )}{c x}+\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 78, normalized size = 1.28 \[ 2 \, {\left (c \sqrt {\frac {1}{c^{2} x^{2}} - 1} - \frac {\operatorname {arsech}\left (c x\right )}{x}\right )} a b + 2 \, {\left (c \sqrt {\frac {1}{c^{2} x^{2}} - 1} \operatorname {arsech}\left (c x\right ) - \frac {1}{x}\right )} b^{2} - \frac {b^{2} \operatorname {arsech}\left (c x\right )^{2}}{x} - \frac {a^{2}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(c*sqrt(1/(c^2*x^2) - 1) - arcsech(c*x)/x)*a*b + 2*(c*sqrt(1/(c^2*x^2) - 1)*arcsech(c*x) - 1/x)*b^2 - b^2*ar
csech(c*x)^2/x - a^2/x

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right )}^2}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(1/(c*x)))^2/x^2,x)

[Out]

int((a + b*acosh(1/(c*x)))^2/x^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {asech}{\left (c x \right )}\right )^{2}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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