3.61 \(\int \frac {1}{x^3 (a+b \text {sech}^{-1}(c x))^2} \, dx\)

Optimal. Leaf size=85 \[ -\frac {c^2 \cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 a}{b}+2 \text {sech}^{-1}(c x)\right )}{b^2}+\frac {c^2 \sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 a}{b}+2 \text {sech}^{-1}(c x)\right )}{b^2}+\frac {c^2 \sinh \left (2 \text {sech}^{-1}(c x)\right )}{2 b \left (a+b \text {sech}^{-1}(c x)\right )} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.16, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6285, 5448, 12, 3297, 3303, 3298, 3301} \[ -\frac {c^2 \cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 a}{b}+2 \text {sech}^{-1}(c x)\right )}{b^2}+\frac {c^2 \sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 a}{b}+2 \text {sech}^{-1}(c x)\right )}{b^2}+\frac {c^2 \sinh \left (2 \text {sech}^{-1}(c x)\right )}{2 b \left (a+b \text {sech}^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*ArcSech[c*x])^2),x]

[Out]

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

Rule 12

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

Rule 3297

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 3298

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 3301

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 3303

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 5448

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 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 {1}{x^3 \left (a+b \text {sech}^{-1}(c x)\right )^2} \, dx &=-\left (c^2 \operatorname {Subst}\left (\int \frac {\cosh (x) \sinh (x)}{(a+b x)^2} \, dx,x,\text {sech}^{-1}(c x)\right )\right )\\ &=-\left (c^2 \operatorname {Subst}\left (\int \frac {\sinh (2 x)}{2 (a+b x)^2} \, dx,x,\text {sech}^{-1}(c x)\right )\right )\\ &=-\left (\frac {1}{2} c^2 \operatorname {Subst}\left (\int \frac {\sinh (2 x)}{(a+b x)^2} \, dx,x,\text {sech}^{-1}(c x)\right )\right )\\ &=\frac {c^2 \sinh \left (2 \text {sech}^{-1}(c x)\right )}{2 b \left (a+b \text {sech}^{-1}(c x)\right )}-\frac {c^2 \operatorname {Subst}\left (\int \frac {\cosh (2 x)}{a+b x} \, dx,x,\text {sech}^{-1}(c x)\right )}{b}\\ &=\frac {c^2 \sinh \left (2 \text {sech}^{-1}(c x)\right )}{2 b \left (a+b \text {sech}^{-1}(c x)\right )}-\frac {\left (c^2 \cosh \left (\frac {2 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\text {sech}^{-1}(c x)\right )}{b}+\frac {\left (c^2 \sinh \left (\frac {2 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {2 a}{b}+2 x\right )}{a+b x} \, dx,x,\text {sech}^{-1}(c x)\right )}{b}\\ &=-\frac {c^2 \cosh \left (\frac {2 a}{b}\right ) \text {Chi}\left (\frac {2 a}{b}+2 \text {sech}^{-1}(c x)\right )}{b^2}+\frac {c^2 \sinh \left (2 \text {sech}^{-1}(c x)\right )}{2 b \left (a+b \text {sech}^{-1}(c x)\right )}+\frac {c^2 \sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (\frac {2 a}{b}+2 \text {sech}^{-1}(c x)\right )}{b^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.36, size = 92, normalized size = 1.08 \[ \frac {c^2 \left (-\cosh \left (\frac {2 a}{b}\right )\right ) \text {Chi}\left (2 \left (\frac {a}{b}+\text {sech}^{-1}(c x)\right )\right )+c^2 \sinh \left (\frac {2 a}{b}\right ) \text {Shi}\left (2 \left (\frac {a}{b}+\text {sech}^{-1}(c x)\right )\right )+\frac {b \sqrt {\frac {1-c x}{c x+1}} (c x+1)}{x^2 \left (a+b \text {sech}^{-1}(c x)\right )}}{b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a + b*ArcSech[c*x])^2),x]

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{b^{2} x^{3} \operatorname {arsech}\left (c x\right )^{2} + 2 \, a b x^{3} \operatorname {arsech}\left (c x\right ) + a^{2} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(a+b*arcsech(c*x))^2,x, algorithm="fricas")

[Out]

integral(1/(b^2*x^3*arcsech(c*x)^2 + 2*a*b*x^3*arcsech(c*x) + a^2*x^3), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \operatorname {arsech}\left (c x\right ) + a\right )}^{2} x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(a+b*arcsech(c*x))^2,x, algorithm="giac")

[Out]

integrate(1/((b*arcsech(c*x) + a)^2*x^3), x)

________________________________________________________________________________________

maple [B]  time = 0.26, size = 186, normalized size = 2.19 \[ c^{2} \left (\frac {2 \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, c x +c^{2} x^{2}-2}{4 c^{2} x^{2} \left (a +b \,\mathrm {arcsech}\left (c x \right )\right ) b}+\frac {{\mathrm e}^{\frac {2 a}{b}} \Ei \left (1, \frac {2 a}{b}+2 \,\mathrm {arcsech}\left (c x \right )\right )}{2 b^{2}}-\frac {c^{2} x^{2}-2-2 \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, c x}{4 b \,c^{2} x^{2} \left (a +b \,\mathrm {arcsech}\left (c x \right )\right )}+\frac {{\mathrm e}^{-\frac {2 a}{b}} \Ei \left (1, -2 \,\mathrm {arcsech}\left (c x \right )-\frac {2 a}{b}\right )}{2 b^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(a+b*arcsech(c*x))^2,x)

[Out]

c^2*(1/4*(2*(-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/2)*c*x+c^2*x^2-2)/c^2/x^2/(a+b*arcsech(c*x))/b+1/2/b^2*exp(2
*a/b)*Ei(1,2*a/b+2*arcsech(c*x))-1/4/b*(c^2*x^2-2-2*(-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/2)*c*x)/c^2/x^2/(a+b
*arcsech(c*x))+1/2/b^2*exp(-2*a/b)*Ei(1,-2*arcsech(c*x)-2*a/b))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {c^{2} x^{3} + {\left (c^{2} x^{3} - x\right )} \sqrt {c x + 1} \sqrt {-c x + 1} - x}{{\left (b^{2} c^{2} x^{2} - b^{2}\right )} x^{3} \log \relax (x) + {\left ({\left (b^{2} c^{2} \log \relax (c) - a b c^{2}\right )} x^{2} - b^{2} \log \relax (c) + a b\right )} x^{3} - {\left (b^{2} x^{3} \log \relax (x) + {\left (b^{2} \log \relax (c) - a b\right )} x^{3}\right )} \sqrt {c x + 1} \sqrt {-c x + 1} + {\left (\sqrt {c x + 1} \sqrt {-c x + 1} b^{2} x^{3} - {\left (b^{2} c^{2} x^{2} - b^{2}\right )} x^{3}\right )} \log \left (\sqrt {c x + 1} \sqrt {-c x + 1} + 1\right )} + \int -\frac {2 \, c^{4} x^{4} - 4 \, c^{2} x^{2} - 2 \, {\left (c x + 1\right )} {\left (c x - 1\right )} + {\left (c^{4} x^{4} - 4 \, c^{2} x^{2} + 4\right )} \sqrt {c x + 1} \sqrt {-c x + 1} + 2}{{\left (b^{2} c^{4} x^{4} - 2 \, b^{2} c^{2} x^{2} + b^{2}\right )} x^{3} \log \relax (x) + {\left ({\left (b^{2} c^{4} \log \relax (c) - a b c^{4}\right )} x^{4} - 2 \, {\left (b^{2} c^{2} \log \relax (c) - a b c^{2}\right )} x^{2} + b^{2} \log \relax (c) - a b\right )} x^{3} - {\left (b^{2} x^{3} \log \relax (x) + {\left (b^{2} \log \relax (c) - a b\right )} x^{3}\right )} {\left (c x + 1\right )} {\left (c x - 1\right )} - 2 \, {\left ({\left (b^{2} c^{2} x^{2} - b^{2}\right )} x^{3} \log \relax (x) + {\left ({\left (b^{2} c^{2} \log \relax (c) - a b c^{2}\right )} x^{2} - b^{2} \log \relax (c) + a b\right )} x^{3}\right )} \sqrt {c x + 1} \sqrt {-c x + 1} + {\left ({\left (c x + 1\right )} {\left (c x - 1\right )} b^{2} x^{3} + 2 \, {\left (b^{2} c^{2} x^{2} - b^{2}\right )} \sqrt {c x + 1} \sqrt {-c x + 1} x^{3} - {\left (b^{2} c^{4} x^{4} - 2 \, b^{2} c^{2} x^{2} + b^{2}\right )} x^{3}\right )} \log \left (\sqrt {c x + 1} \sqrt {-c x + 1} + 1\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^3\,{\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(a + b*acosh(1/(c*x)))^2),x)

[Out]

int(1/(x^3*(a + b*acosh(1/(c*x)))^2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{3} \left (a + b \operatorname {asech}{\left (c x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(a+b*asech(c*x))**2,x)

[Out]

Integral(1/(x**3*(a + b*asech(c*x))**2), x)

________________________________________________________________________________________