3.1.66 \(\int \frac {1}{x^2 (a+b \text {sech}^{-1}(c x))^3} \, dx\) [66]

Optimal. Leaf size=114 \[ \frac {\sqrt {\frac {1-c x}{1+c x}} (1+c x)}{2 b x \left (a+b \text {sech}^{-1}(c x)\right )^2}+\frac {1}{2 b^2 x \left (a+b \text {sech}^{-1}(c x)\right )}+\frac {c \text {Chi}\left (\frac {a}{b}+\text {sech}^{-1}(c x)\right ) \sinh \left (\frac {a}{b}\right )}{2 b^3}-\frac {c \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\text {sech}^{-1}(c x)\right )}{2 b^3} \]

[Out]

1/2/b^2/x/(a+b*arcsech(c*x))-1/2*c*cosh(a/b)*Shi(a/b+arcsech(c*x))/b^3+1/2*c*Chi(a/b+arcsech(c*x))*sinh(a/b)/b
^3+1/2*(c*x+1)*((-c*x+1)/(c*x+1))^(1/2)/b/x/(a+b*arcsech(c*x))^2

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Rubi [A]
time = 0.12, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 6, 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 \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\text {sech}^{-1}(c x)\right )}{2 b^3}-\frac {c \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\text {sech}^{-1}(c x)\right )}{2 b^3}+\frac {1}{2 b^2 x \left (a+b \text {sech}^{-1}(c x)\right )}+\frac {\sqrt {\frac {1-c x}{c x+1}} (c x+1)}{2 b x \left (a+b \text {sech}^{-1}(c x)\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a + b*ArcSech[c*x])^3),x]

[Out]

(Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x))/(2*b*x*(a + b*ArcSech[c*x])^2) + 1/(2*b^2*x*(a + b*ArcSech[c*x])) + (c*C
oshIntegral[a/b + ArcSech[c*x]]*Sinh[a/b])/(2*b^3) - (c*Cosh[a/b]*SinhIntegral[a/b + ArcSech[c*x]])/(2*b^3)

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 )^3} \, dx &=-\left (c \text {Subst}\left (\int \frac {\sinh (x)}{(a+b x)^3} \, dx,x,\text {sech}^{-1}(c x)\right )\right )\\ &=\frac {\sqrt {\frac {1-c x}{1+c x}} (1+c x)}{2 b x \left (a+b \text {sech}^{-1}(c x)\right )^2}-\frac {c \text {Subst}\left (\int \frac {\cosh (x)}{(a+b x)^2} \, dx,x,\text {sech}^{-1}(c x)\right )}{2 b}\\ &=\frac {\sqrt {\frac {1-c x}{1+c x}} (1+c x)}{2 b x \left (a+b \text {sech}^{-1}(c x)\right )^2}+\frac {1}{2 b^2 x \left (a+b \text {sech}^{-1}(c x)\right )}-\frac {c \text {Subst}\left (\int \frac {\sinh (x)}{a+b x} \, dx,x,\text {sech}^{-1}(c x)\right )}{2 b^2}\\ &=\frac {\sqrt {\frac {1-c x}{1+c x}} (1+c x)}{2 b x \left (a+b \text {sech}^{-1}(c x)\right )^2}+\frac {1}{2 b^2 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 {\sinh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\text {sech}^{-1}(c x)\right )}{2 b^2}+\frac {\left (c \sinh \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 )}{2 b^2}\\ &=\frac {\sqrt {\frac {1-c x}{1+c x}} (1+c x)}{2 b x \left (a+b \text {sech}^{-1}(c x)\right )^2}+\frac {1}{2 b^2 x \left (a+b \text {sech}^{-1}(c x)\right )}+\frac {c \text {Chi}\left (\frac {a}{b}+\text {sech}^{-1}(c x)\right ) \sinh \left (\frac {a}{b}\right )}{2 b^3}-\frac {c \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\text {sech}^{-1}(c x)\right )}{2 b^3}\\ \end {align*}

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Mathematica [A]
time = 0.19, size = 103, normalized size = 0.90 \begin {gather*} \frac {\frac {b^2 \sqrt {\frac {1-c x}{1+c x}} (1+c x)}{x \left (a+b \text {sech}^{-1}(c x)\right )^2}+\frac {b}{a x+b x \text {sech}^{-1}(c x)}+c \left (\text {Chi}\left (\frac {a}{b}+\text {sech}^{-1}(c x)\right ) \sinh \left (\frac {a}{b}\right )-\cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\text {sech}^{-1}(c x)\right )\right )}{2 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b^2*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x))/(x*(a + b*ArcSech[c*x])^2) + b/(a*x + b*x*ArcSech[c*x]) + c*(CoshIn
tegral[a/b + ArcSech[c*x]]*Sinh[a/b] - Cosh[a/b]*SinhIntegral[a/b + ArcSech[c*x]]))/(2*b^3)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(243\) vs. \(2(104)=208\).
time = 0.35, size = 244, normalized size = 2.14

method	result	size
derivativedivides	\(c \left (-\frac {\left (\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, c x -1\right ) \left (b \,\mathrm {arcsech}\left (c x \right )+a -b \right )}{4 c x \,b^{2} \left (b^{2} \mathrm {arcsech}\left (c x \right )^{2}+2 a b \,\mathrm {arcsech}\left (c x \right )+a^{2}\right )}-\frac {{\mathrm e}^{\frac {a}{b}} \expIntegral \left (1, \frac {a}{b}+\mathrm {arcsech}\left (c x \right )\right )}{4 b^{3}}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, c x +1}{4 b c x \left (a +b \,\mathrm {arcsech}\left (c x \right )\right )^{2}}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, c x +1}{4 b^{2} c x \left (a +b \,\mathrm {arcsech}\left (c x \right )\right )}+\frac {{\mathrm e}^{-\frac {a}{b}} \expIntegral \left (1, -\mathrm {arcsech}\left (c x \right )-\frac {a}{b}\right )}{4 b^{3}}\right )\)	\(244\)
default	\(c \left (-\frac {\left (\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, c x -1\right ) \left (b \,\mathrm {arcsech}\left (c x \right )+a -b \right )}{4 c x \,b^{2} \left (b^{2} \mathrm {arcsech}\left (c x \right )^{2}+2 a b \,\mathrm {arcsech}\left (c x \right )+a^{2}\right )}-\frac {{\mathrm e}^{\frac {a}{b}} \expIntegral \left (1, \frac {a}{b}+\mathrm {arcsech}\left (c x \right )\right )}{4 b^{3}}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, c x +1}{4 b c x \left (a +b \,\mathrm {arcsech}\left (c x \right )\right )^{2}}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, c x +1}{4 b^{2} c x \left (a +b \,\mathrm {arcsech}\left (c x \right )\right )}+\frac {{\mathrm e}^{-\frac {a}{b}} \expIntegral \left (1, -\mathrm {arcsech}\left (c x \right )-\frac {a}{b}\right )}{4 b^{3}}\right )\)	\(244\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*(-1/4*((-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/2)*c*x-1)*(b*arcsech(c*x)+a-b)/c/x/b^2/(b^2*arcsech(c*x)^2+2*a*
b*arcsech(c*x)+a^2)-1/4/b^3*exp(a/b)*Ei(1,a/b+arcsech(c*x))+1/4/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))^2+1/4/b^2*((-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/2)*c*x+1)/c/x/(a+b*arcsech(c*x))+
1/4/b^3*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))^3,x, algorithm="maxima")

[Out]

-1/2*((b*c^6*(log(c) - 1) - a*c^6)*x^7 - 3*(b*c^4*(log(c) - 1) - a*c^4)*x^5 - (b*c^2*x^3 - (b*c^4*log(c) - a*c
^4)*x^5 + (b*(log(c) - 1) - a)*x - (b*c^4*x^5 - b*x)*log(x))*(c*x + 1)^(3/2)*(-c*x + 1)^(3/2) + 3*(b*c^2*(log(
c) - 1) - a*c^2)*x^3 - (2*b*c^4*x^5 + (b*c^2*(3*log(c) - 5) - 3*a*c^2)*x^3 - 3*(b*(log(c) - 1) - a)*x + 3*(b*c
^2*x^3 - b*x)*log(x))*(c*x + 1)*(c*x - 1) + ((b*c^6*(log(c) - 1) - a*c^6)*x^7 - (b*c^4*(4*log(c) - 5) - 4*a*c^
4)*x^5 + (b*c^2*(6*log(c) - 7) - 6*a*c^2)*x^3 - 3*(b*(log(c) - 1) - a)*x + (b*c^6*x^7 - 4*b*c^4*x^5 + 6*b*c^2*
x^3 - 3*b*x)*log(x))*sqrt(c*x + 1)*sqrt(-c*x + 1) - (b*(log(c) - 1) - a)*x - (b*c^6*x^7 - 3*b*c^4*x^5 + 3*b*c^
2*x^3 + (b*c^4*x^5 - b*x)*(c*x + 1)^(3/2)*(-c*x + 1)^(3/2) - 3*(b*c^2*x^3 - b*x)*(c*x + 1)*(c*x - 1) + (b*c^6*
x^7 - 4*b*c^4*x^5 + 6*b*c^2*x^3 - 3*b*x)*sqrt(c*x + 1)*sqrt(-c*x + 1) - b*x)*log(sqrt(c*x + 1)*sqrt(-c*x + 1)
+ 1) + (b*c^6*x^7 - 3*b*c^4*x^5 + 3*b*c^2*x^3 - b*x)*log(x))/((b^4*c^6*x^6 - 3*b^4*c^4*x^4 + 3*b^4*c^2*x^2 - b
^4)*x^2*log(x)^2 - (b^4*x^2*log(x)^2 + 2*(b^4*log(c) - a*b^3)*x^2*log(x) + (b^4*log(c)^2 - 2*a*b^3*log(c) + a^
2*b^2)*x^2)*(c*x + 1)^(3/2)*(-c*x + 1)^(3/2) + 2*((b^4*c^6*log(c) - a*b^3*c^6)*x^6 - 3*(b^4*c^4*log(c) - a*b^3
*c^4)*x^4 - b^4*log(c) + a*b^3 + 3*(b^4*c^2*log(c) - a*b^3*c^2)*x^2)*x^2*log(x) - 3*((b^4*c^2*x^2 - b^4)*x^2*l
og(x)^2 - 2*(b^4*log(c) - a*b^3 - (b^4*c^2*log(c) - a*b^3*c^2)*x^2)*x^2*log(x) - (b^4*log(c)^2 - 2*a*b^3*log(c
) + a^2*b^2 - (b^4*c^2*log(c)^2 - 2*a*b^3*c^2*log(c) + a^2*b^2*c^2)*x^2)*x^2)*(c*x + 1)*(c*x - 1) + ((b^4*c^6*
log(c)^2 - 2*a*b^3*c^6*log(c) + a^2*b^2*c^6)*x^6 - b^4*log(c)^2 - 3*(b^4*c^4*log(c)^2 - 2*a*b^3*c^4*log(c) + a
^2*b^2*c^4)*x^4 + 2*a*b^3*log(c) - a^2*b^2 + 3*(b^4*c^2*log(c)^2 - 2*a*b^3*c^2*log(c) + a^2*b^2*c^2)*x^2)*x^2
- ((c*x + 1)^(3/2)*(-c*x + 1)^(3/2)*b^4*x^2 + 3*(b^4*c^2*x^2 - b^4)*(c*x + 1)*(c*x - 1)*x^2 + 3*(b^4*c^4*x^4 -
2*b^4*c^2*x^2 + b^4)*sqrt(c*x + 1)*sqrt(-c*x + 1)*x^2 - (b^4*c^6*x^6 - 3*b^4*c^4*x^4 + 3*b^4*c^2*x^2 - b^4)*x
^2)*log(sqrt(c*x + 1)*sqrt(-c*x + 1) + 1)^2 - 3*((b^4*c^4*x^4 - 2*b^4*c^2*x^2 + b^4)*x^2*log(x)^2 + 2*((b^4*c^
4*log(c) - a*b^3*c^4)*x^4 + b^4*log(c) - a*b^3 - 2*(b^4*c^2*log(c) - a*b^3*c^2)*x^2)*x^2*log(x) + (b^4*log(c)^
2 + (b^4*c^4*log(c)^2 - 2*a*b^3*c^4*log(c) + a^2*b^2*c^4)*x^4 - 2*a*b^3*log(c) + a^2*b^2 - 2*(b^4*c^2*log(c)^2
- 2*a*b^3*c^2*log(c) + a^2*b^2*c^2)*x^2)*x^2)*sqrt(c*x + 1)*sqrt(-c*x + 1) + 2*((b^4*x^2*log(x) + (b^4*log(c)
- a*b^3)*x^2)*(c*x + 1)^(3/2)*(-c*x + 1)^(3/2) - (b^4*c^6*x^6 - 3*b^4*c^4*x^4 + 3*b^4*c^2*x^2 - b^4)*x^2*log(
x) + 3*((b^4*c^2*x^2 - b^4)*x^2*log(x) - (b^4*log(c) - a*b^3 - (b^4*c^2*log(c) - a*b^3*c^2)*x^2)*x^2)*(c*x + 1
)*(c*x - 1) - ((b^4*c^6*log(c) - a*b^3*c^6)*x^6 - 3*(b^4*c^4*log(c) - a*b^3*c^4)*x^4 - b^4*log(c) + a*b^3 + 3*
(b^4*c^2*log(c) - a*b^3*c^2)*x^2)*x^2 + 3*((b^4*c^4*x^4 - 2*b^4*c^2*x^2 + b^4)*x^2*log(x) + ((b^4*c^4*log(c) -
a*b^3*c^4)*x^4 + b^4*log(c) - a*b^3 - 2*(b^4*c^2*log(c) - a*b^3*c^2)*x^2)*x^2)*sqrt(c*x + 1)*sqrt(-c*x + 1))*
log(sqrt(c*x + 1)*sqrt(-c*x + 1) + 1)) + integrate(-1/2*(c^8*x^8 - 4*c^6*x^6 + 6*c^4*x^4 + (3*c^4*x^4 + 1)*(c*
x + 1)^2*(c*x - 1)^2 + (3*c^4*x^4 - 4*c^2*x^2 + 4)*(c*x + 1)^(3/2)*(-c*x + 1)^(3/2) - 4*c^2*x^2 - 3*(c^6*x^6 +
c^4*x^4 - 4*c^2*x^2 + 2)*(c*x + 1)*(c*x - 1) - (c^6*x^6 - 9*c^4*x^4 + 12*c^2*x^2 - 4)*sqrt(c*x + 1)*sqrt(-c*x
+ 1) + 1)/((b^3*x^2*log(x) + (b^3*log(c) - a*b^2)*x^2)*(c*x + 1)^2*(c*x - 1)^2 - 4*((b^3*c^2*x^2 - b^3)*x^2*l
og(x) - (b^3*log(c) - a*b^2 - (b^3*c^2*log(c) - a*b^2*c^2)*x^2)*x^2)*(c*x + 1)^(3/2)*(-c*x + 1)^(3/2) + (b^3*c
^8*x^8 - 4*b^3*c^6*x^6 + 6*b^3*c^4*x^4 - 4*b^3*c^2*x^2 + b^3)*x^2*log(x) - 6*((b^3*c^4*x^4 - 2*b^3*c^2*x^2 + b
^3)*x^2*log(x) + ((b^3*c^4*log(c) - a*b^2*c^4)*x^4 + b^3*log(c) - a*b^2 - 2*(b^3*c^2*log(c) - a*b^2*c^2)*x^2)*
x^2)*(c*x + 1)*(c*x - 1) + ((b^3*c^8*log(c) - a*b^2*c^8)*x^8 - 4*(b^3*c^6*log(c) - a*b^2*c^6)*x^6 + 6*(b^3*c^4
*log(c) - a*b^2*c^4)*x^4 + b^3*log(c) - a*b^2 - 4*(b^3*c^2*log(c) - a*b^2*c^2)*x^2)*x^2 - 4*((b^3*c^6*x^6 - 3*
b^3*c^4*x^4 + 3*b^3*c^2*x^2 - b^3)*x^2*log(x) + ((b^3*c^6*log(c) - a*b^2*c^6)*x^6 - 3*(b^3*c^4*log(c) - a*b^2*
c^4)*x^4 - b^3*log(c) + a*b^2 + 3*(b^3*c^2*log(c) - a*b^2*c^2)*x^2)*x^2)*sqrt(c*x + 1)*sqrt(-c*x + 1) - ((c*x
+ 1)^2*(c*x - 1)^2*b^3*x^2 - 4*(b^3*c^2*x^2 - b^3)*(c*x + 1)^(3/2)*(-c*x + 1)^(3/2)*x^2 - 6*(b^3*c^4*x^4 - 2*b
^3*c^2*x^2 + b^3)*(c*x + 1)*(c*x - 1)*x^2 - 4*(b^3*c^6*x^6 - 3*b^3*c^4*x^4 + 3*b^3*c^2*x^2 - b^3)*sqrt(c*x + 1
)*sqrt(-c*x + 1)*x^2 + (b^3*c^8*x^8 - 4*b^3*c^6*x^6 + 6*b^3*c^4*x^4 - 4*b^3*c^2*x^2 + b^3)*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))^3,x, algorithm="fricas")

[Out]

integral(1/(b^3*x^2*arcsech(c*x)^3 + 3*a*b^2*x^2*arcsech(c*x)^2 + 3*a^2*b*x^2*arcsech(c*x) + a^3*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 )^{3}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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