3.1.60 \(\int \frac {1}{x^2 (a+b \text {sech}^{-1}(c x))^2} \, dx\) [60]

Optimal. Leaf size=86 \[ \frac {\sqrt {\frac {1-c x}{1+c x}} (1+c x)}{b x \left (a+b \text {sech}^{-1}(c x)\right )}-\frac {c \cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\text {sech}^{-1}(c x)\right )}{b^2}+\frac {c \sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\text {sech}^{-1}(c x)\right )}{b^2} \]

[Out]

-c*Chi(a/b+arcsech(c*x))*cosh(a/b)/b^2+c*Shi(a/b+arcsech(c*x))*sinh(a/b)/b^2+(c*x+1)*((-c*x+1)/(c*x+1))^(1/2)/
b/x/(a+b*arcsech(c*x))

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Rubi [A]
time = 0.11, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {6420, 3378, 3384, 3379, 3382} \begin {gather*} -\frac {c \cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\text {sech}^{-1}(c x)\right )}{b^2}+\frac {c \sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\text {sech}^{-1}(c x)\right )}{b^2}+\frac {\sqrt {\frac {1-c x}{c x+1}} (c x+1)}{b x \left (a+b \text {sech}^{-1}(c x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*ArcSech[c*x])^2),x]

[Out]

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

Rule 3378

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

Rule 3379

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[I*(SinhIntegral[c*f*(fz/
d) + f*fz*x]/d), x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rule 3382

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[c*f*(fz/d)
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rule 3384

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

Rule 6420

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

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Mathematica [A]
time = 0.23, size = 82, normalized size = 0.95 \begin {gather*} \frac {\frac {b \sqrt {\frac {1-c x}{1+c x}} (1+c x)}{x \left (a+b \text {sech}^{-1}(c x)\right )}-c \cosh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\text {sech}^{-1}(c x)\right )+c \sinh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\text {sech}^{-1}(c x)\right )}{b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a + b*ArcSech[c*x])^2),x]

[Out]

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

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Maple [A]
time = 0.36, size = 164, normalized size = 1.91

method	result	size
derivativedivides	\(c \left (\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, c x -1}{2 c x b \left (a +b \,\mathrm {arcsech}\left (c x \right )\right )}+\frac {{\mathrm e}^{\frac {a}{b}} \expIntegral \left (1, \frac {a}{b}+\mathrm {arcsech}\left (c x \right )\right )}{2 b^{2}}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, c x +1}{2 b c x \left (a +b \,\mathrm {arcsech}\left (c x \right )\right )}+\frac {{\mathrm e}^{-\frac {a}{b}} \expIntegral \left (1, -\mathrm {arcsech}\left (c x \right )-\frac {a}{b}\right )}{2 b^{2}}\right )\)	\(164\)
default	\(c \left (\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, c x -1}{2 c x b \left (a +b \,\mathrm {arcsech}\left (c x \right )\right )}+\frac {{\mathrm e}^{\frac {a}{b}} \expIntegral \left (1, \frac {a}{b}+\mathrm {arcsech}\left (c x \right )\right )}{2 b^{2}}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, c x +1}{2 b c x \left (a +b \,\mathrm {arcsech}\left (c x \right )\right )}+\frac {{\mathrm e}^{-\frac {a}{b}} \expIntegral \left (1, -\mathrm {arcsech}\left (c x \right )-\frac {a}{b}\right )}{2 b^{2}}\right )\)	\(164\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(a+b*arcsech(c*x))^2,x,method=_RETURNVERBOSE)

[Out]

c*(1/2*((-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/2)*c*x-1)/c/x/b/(a+b*arcsech(c*x))+1/2/b^2*exp(a/b)*Ei(1,a/b+arc
sech(c*x))+1/2/b*((-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/2)*c*x+1)/c/x/(a+b*arcsech(c*x))+1/2/b^2*exp(-a/b)*Ei(
1,-arcsech(c*x)-a/b))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(a+b*arcsech(c*x))^2,x, algorithm="maxima")

[Out]

-(c^2*x^3 + (c^2*x^3 - x)*sqrt(c*x + 1)*sqrt(-c*x + 1) - x)/((b^2*c^2*x^2 - b^2)*x^2*log(x) + ((b^2*c^2*log(c)
- a*b*c^2)*x^2 - b^2*log(c) + a*b)*x^2 - (b^2*x^2*log(x) + (b^2*log(c) - a*b)*x^2)*sqrt(c*x + 1)*sqrt(-c*x +
1) + (sqrt(c*x + 1)*sqrt(-c*x + 1)*b^2*x^2 - (b^2*c^2*x^2 - b^2)*x^2)*log(sqrt(c*x + 1)*sqrt(-c*x + 1) + 1)) +
integrate(-(c^4*x^4 - 2*c^2*x^2 - (c^2*x^2 + 1)*(c*x + 1)*(c*x - 1) - (c^2*x^2 - 2)*sqrt(c*x + 1)*sqrt(-c*x +
1) + 1)/((b^2*c^4*x^4 - 2*b^2*c^2*x^2 + b^2)*x^2*log(x) - (b^2*x^2*log(x) + (b^2*log(c) - a*b)*x^2)*(c*x + 1)
*(c*x - 1) + ((b^2*c^4*log(c) - a*b*c^4)*x^4 - 2*(b^2*c^2*log(c) - a*b*c^2)*x^2 + b^2*log(c) - a*b)*x^2 - 2*((
b^2*c^2*x^2 - b^2)*x^2*log(x) + ((b^2*c^2*log(c) - a*b*c^2)*x^2 - b^2*log(c) + a*b)*x^2)*sqrt(c*x + 1)*sqrt(-c
*x + 1) + ((c*x + 1)*(c*x - 1)*b^2*x^2 + 2*(b^2*c^2*x^2 - b^2)*sqrt(c*x + 1)*sqrt(-c*x + 1)*x^2 - (b^2*c^4*x^4
- 2*b^2*c^2*x^2 + b^2)*x^2)*log(sqrt(c*x + 1)*sqrt(-c*x + 1) + 1)), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(a+b*arcsech(c*x))^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{2} \left (a + b \operatorname {asech}{\left (c x \right )}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(a+b*asech(c*x))**2,x)

[Out]

Integral(1/(x**2*(a + b*asech(c*x))**2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(a+b*arcsech(c*x))^2,x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^2\,{\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a + b*acosh(1/(c*x)))^2),x)

[Out]

int(1/(x^2*(a + b*acosh(1/(c*x)))^2), x)

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