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

Optimal. Leaf size=190 \[ -\frac{c^3 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\text{sech}^{-1}(c x)\right )}{4 b^2}-\frac{3 c^3 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 a}{b}+3 \text{sech}^{-1}(c x)\right )}{4 b^2}+\frac{c^3 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\text{sech}^{-1}(c x)\right )}{4 b^2}+\frac{3 c^3 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \text{sech}^{-1}(c x)\right )}{4 b^2}+\frac{c^2 \sqrt{\frac{1-c x}{c x+1}} (c x+1)}{4 b x \left (a+b \text{sech}^{-1}(c x)\right )}+\frac{c^3 \sinh \left (3 \text{sech}^{-1}(c x)\right )}{4 b \left (a+b \text{sech}^{-1}(c x)\right )} \]

[Out]

(c^2*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x))/(4*b*x*(a + b*ArcSech[c*x])) - (c^3*Cosh[a/b]*CoshIntegral[a/b + Arc
Sech[c*x]])/(4*b^2) - (3*c^3*Cosh[(3*a)/b]*CoshIntegral[(3*a)/b + 3*ArcSech[c*x]])/(4*b^2) + (c^3*Sinh[3*ArcSe
ch[c*x]])/(4*b*(a + b*ArcSech[c*x])) + (c^3*Sinh[a/b]*SinhIntegral[a/b + ArcSech[c*x]])/(4*b^2) + (3*c^3*Sinh[
(3*a)/b]*SinhIntegral[(3*a)/b + 3*ArcSech[c*x]])/(4*b^2)

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Rubi [A]  time = 0.294862, antiderivative size = 190, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {6285, 5448, 3297, 3303, 3298, 3301} \[ -\frac{c^3 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\text{sech}^{-1}(c x)\right )}{4 b^2}-\frac{3 c^3 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 a}{b}+3 \text{sech}^{-1}(c x)\right )}{4 b^2}+\frac{c^3 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\text{sech}^{-1}(c x)\right )}{4 b^2}+\frac{3 c^3 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \text{sech}^{-1}(c x)\right )}{4 b^2}+\frac{c^2 \sqrt{\frac{1-c x}{c x+1}} (c x+1)}{4 b x \left (a+b \text{sech}^{-1}(c x)\right )}+\frac{c^3 \sinh \left (3 \text{sech}^{-1}(c x)\right )}{4 b \left (a+b \text{sech}^{-1}(c x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + b*ArcSech[c*x])^2),x]

[Out]

(c^2*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x))/(4*b*x*(a + b*ArcSech[c*x])) - (c^3*Cosh[a/b]*CoshIntegral[a/b + Arc
Sech[c*x]])/(4*b^2) - (3*c^3*Cosh[(3*a)/b]*CoshIntegral[(3*a)/b + 3*ArcSech[c*x]])/(4*b^2) + (c^3*Sinh[3*ArcSe
ch[c*x]])/(4*b*(a + b*ArcSech[c*x])) + (c^3*Sinh[a/b]*SinhIntegral[a/b + ArcSech[c*x]])/(4*b^2) + (3*c^3*Sinh[
(3*a)/b]*SinhIntegral[(3*a)/b + 3*ArcSech[c*x]])/(4*b^2)

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])

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 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 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 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]

Rubi steps

\begin{align*} \int \frac{1}{x^4 \left (a+b \text{sech}^{-1}(c x)\right )^2} \, dx &=-\left (c^3 \operatorname{Subst}\left (\int \frac{\cosh ^2(x) \sinh (x)}{(a+b x)^2} \, dx,x,\text{sech}^{-1}(c x)\right )\right )\\ &=-\left (c^3 \operatorname{Subst}\left (\int \left (\frac{\sinh (x)}{4 (a+b x)^2}+\frac{\sinh (3 x)}{4 (a+b x)^2}\right ) \, dx,x,\text{sech}^{-1}(c x)\right )\right )\\ &=-\left (\frac{1}{4} c^3 \operatorname{Subst}\left (\int \frac{\sinh (x)}{(a+b x)^2} \, dx,x,\text{sech}^{-1}(c x)\right )\right )-\frac{1}{4} c^3 \operatorname{Subst}\left (\int \frac{\sinh (3 x)}{(a+b x)^2} \, dx,x,\text{sech}^{-1}(c x)\right )\\ &=\frac{c^2 \sqrt{\frac{1-c x}{1+c x}} (1+c x)}{4 b x \left (a+b \text{sech}^{-1}(c x)\right )}+\frac{c^3 \sinh \left (3 \text{sech}^{-1}(c x)\right )}{4 b \left (a+b \text{sech}^{-1}(c x)\right )}-\frac{c^3 \operatorname{Subst}\left (\int \frac{\cosh (x)}{a+b x} \, dx,x,\text{sech}^{-1}(c x)\right )}{4 b}-\frac{\left (3 c^3\right ) \operatorname{Subst}\left (\int \frac{\cosh (3 x)}{a+b x} \, dx,x,\text{sech}^{-1}(c x)\right )}{4 b}\\ &=\frac{c^2 \sqrt{\frac{1-c x}{1+c x}} (1+c x)}{4 b x \left (a+b \text{sech}^{-1}(c x)\right )}+\frac{c^3 \sinh \left (3 \text{sech}^{-1}(c x)\right )}{4 b \left (a+b \text{sech}^{-1}(c x)\right )}-\frac{\left (c^3 \cosh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\text{sech}^{-1}(c x)\right )}{4 b}-\frac{\left (3 c^3 \cosh \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\text{sech}^{-1}(c x)\right )}{4 b}+\frac{\left (c^3 \sinh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\text{sech}^{-1}(c x)\right )}{4 b}+\frac{\left (3 c^3 \sinh \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\text{sech}^{-1}(c x)\right )}{4 b}\\ &=\frac{c^2 \sqrt{\frac{1-c x}{1+c x}} (1+c x)}{4 b x \left (a+b \text{sech}^{-1}(c x)\right )}-\frac{c^3 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\text{sech}^{-1}(c x)\right )}{4 b^2}-\frac{3 c^3 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 a}{b}+3 \text{sech}^{-1}(c x)\right )}{4 b^2}+\frac{c^3 \sinh \left (3 \text{sech}^{-1}(c x)\right )}{4 b \left (a+b \text{sech}^{-1}(c x)\right )}+\frac{c^3 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\text{sech}^{-1}(c x)\right )}{4 b^2}+\frac{3 c^3 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \text{sech}^{-1}(c x)\right )}{4 b^2}\\ \end{align*}

Mathematica [A]  time = 0.623285, size = 170, normalized size = 0.89 \[ \frac{c^3 \left (-\cosh \left (\frac{a}{b}\right )\right ) \text{Chi}\left (\frac{a}{b}+\text{sech}^{-1}(c x)\right )-3 c^3 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (3 \left (\frac{a}{b}+\text{sech}^{-1}(c x)\right )\right )+c^3 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\text{sech}^{-1}(c x)\right )+3 c^3 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (3 \left (\frac{a}{b}+\text{sech}^{-1}(c x)\right )\right )+\frac{4 b c \sqrt{\frac{1-c x}{c x+1}}}{x^2 \left (a+b \text{sech}^{-1}(c x)\right )}+\frac{4 b \sqrt{\frac{1-c x}{c x+1}}}{x^3 \left (a+b \text{sech}^{-1}(c x)\right )}}{4 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(a + b*ArcSech[c*x])^2),x]

[Out]

((4*b*Sqrt[(1 - c*x)/(1 + c*x)])/(x^3*(a + b*ArcSech[c*x])) + (4*b*c*Sqrt[(1 - c*x)/(1 + c*x)])/(x^2*(a + b*Ar
cSech[c*x])) - c^3*Cosh[a/b]*CoshIntegral[a/b + ArcSech[c*x]] - 3*c^3*Cosh[(3*a)/b]*CoshIntegral[3*(a/b + ArcS
ech[c*x])] + c^3*Sinh[a/b]*SinhIntegral[a/b + ArcSech[c*x]] + 3*c^3*Sinh[(3*a)/b]*SinhIntegral[3*(a/b + ArcSec
h[c*x])])/(4*b^2)

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Maple [B]  time = 0.273, size = 420, normalized size = 2.2 \begin{align*}{c}^{3} \left ( -{\frac{1}{8\,{x}^{3}b{c}^{3} \left ( a+b{\rm arcsech} \left (cx\right ) \right ) } \left ( \sqrt{{\frac{cx+1}{cx}}}\sqrt{-{\frac{cx-1}{cx}}}{c}^{3}{x}^{3}-4\,\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}cx-3\,{c}^{2}{x}^{2}+4 \right ) }+{\frac{3}{8\,{b}^{2}}{{\rm e}^{3\,{\frac{a}{b}}}}{\it Ei} \left ( 1,3\,{\frac{a}{b}}+3\,{\rm arcsech} \left (cx\right ) \right ) }+{\frac{1}{8\,xbc \left ( a+b{\rm arcsech} \left (cx\right ) \right ) } \left ( \sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}cx-1 \right ) }+{\frac{1}{8\,{b}^{2}}{{\rm e}^{{\frac{a}{b}}}}{\it Ei} \left ( 1,{\frac{a}{b}}+{\rm arcsech} \left (cx\right ) \right ) }+{\frac{1}{8\,xbc \left ( a+b{\rm arcsech} \left (cx\right ) \right ) } \left ( \sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}cx+1 \right ) }+{\frac{1}{8\,{b}^{2}}{{\rm e}^{-{\frac{a}{b}}}}{\it Ei} \left ( 1,-{\rm arcsech} \left (cx\right )-{\frac{a}{b}} \right ) }-{\frac{1}{8\,{x}^{3}b{c}^{3} \left ( a+b{\rm arcsech} \left (cx\right ) \right ) } \left ( \sqrt{{\frac{cx+1}{cx}}}\sqrt{-{\frac{cx-1}{cx}}}{c}^{3}{x}^{3}-4\,\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}cx+3\,{c}^{2}{x}^{2}-4 \right ) }+{\frac{3}{8\,{b}^{2}}{{\rm e}^{-3\,{\frac{a}{b}}}}{\it Ei} \left ( 1,-3\,{\rm arcsech} \left (cx\right )-3\,{\frac{a}{b}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(a+b*arcsech(c*x))^2,x)

[Out]

c^3*(-1/8*(((c*x+1)/c/x)^(1/2)*(-(c*x-1)/c/x)^(1/2)*c^3*x^3-4*(-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/2)*c*x-3*c
^2*x^2+4)/c^3/x^3/b/(a+b*arcsech(c*x))+3/8/b^2*exp(3*a/b)*Ei(1,3*a/b+3*arcsech(c*x))+1/8*((-(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))+1/8/b^2*exp(a/b)*Ei(1,a/b+arcsech(c*x))+1/8/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))+1/8/b^2*exp(-a/b)*Ei(1,-arcsech(c*x)-a/b)-1/8/b*(
((c*x+1)/c/x)^(1/2)*(-(c*x-1)/c/x)^(1/2)*c^3*x^3-4*(-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/2)*c*x+3*c^2*x^2-4)/c
^3/x^3/(a+b*arcsech(c*x))+3/8/b^2*exp(-3*a/b)*Ei(1,-3*arcsech(c*x)-3*a/b))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\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^{4} \log \left (x\right ) +{\left ({\left (b^{2} c^{2} \log \left (c\right ) - a b c^{2}\right )} x^{2} - b^{2} \log \left (c\right ) + a b\right )} x^{4} -{\left (b^{2} x^{4} \log \left (x\right ) +{\left (b^{2} \log \left (c\right ) - a b\right )} x^{4}\right )} \sqrt{c x + 1} \sqrt{-c x + 1} +{\left (\sqrt{c x + 1} \sqrt{-c x + 1} b^{2} x^{4} -{\left (b^{2} c^{2} x^{2} - b^{2}\right )} x^{4}\right )} \log \left (\sqrt{c x + 1} \sqrt{-c x + 1} + 1\right )} - \int \frac{3 \, c^{4} x^{4} - 6 \, c^{2} x^{2} +{\left (c^{2} x^{2} - 3\right )}{\left (c x + 1\right )}{\left (c x - 1\right )} +{\left (2 \, c^{4} x^{4} - 7 \, c^{2} x^{2} + 6\right )} \sqrt{c x + 1} \sqrt{-c x + 1} + 3}{{\left (b^{2} c^{4} x^{4} - 2 \, b^{2} c^{2} x^{2} + b^{2}\right )} x^{4} \log \left (x\right ) +{\left ({\left (b^{2} c^{4} \log \left (c\right ) - a b c^{4}\right )} x^{4} - 2 \,{\left (b^{2} c^{2} \log \left (c\right ) - a b c^{2}\right )} x^{2} + b^{2} \log \left (c\right ) - a b\right )} x^{4} -{\left (b^{2} x^{4} \log \left (x\right ) +{\left (b^{2} \log \left (c\right ) - a b\right )} x^{4}\right )}{\left (c x + 1\right )}{\left (c x - 1\right )} - 2 \,{\left ({\left (b^{2} c^{2} x^{2} - b^{2}\right )} x^{4} \log \left (x\right ) +{\left ({\left (b^{2} c^{2} \log \left (c\right ) - a b c^{2}\right )} x^{2} - b^{2} \log \left (c\right ) + a b\right )} x^{4}\right )} \sqrt{c x + 1} \sqrt{-c x + 1} +{\left ({\left (c x + 1\right )}{\left (c x - 1\right )} b^{2} x^{4} + 2 \,{\left (b^{2} c^{2} x^{2} - b^{2}\right )} \sqrt{c x + 1} \sqrt{-c x + 1} x^{4} -{\left (b^{2} c^{4} x^{4} - 2 \, b^{2} c^{2} x^{2} + b^{2}\right )} x^{4}\right )} \log \left (\sqrt{c x + 1} \sqrt{-c x + 1} + 1\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(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^4*log(x) + ((b^2*c^2*log(c)
- a*b*c^2)*x^2 - b^2*log(c) + a*b)*x^4 - (b^2*x^4*log(x) + (b^2*log(c) - a*b)*x^4)*sqrt(c*x + 1)*sqrt(-c*x +
1) + (sqrt(c*x + 1)*sqrt(-c*x + 1)*b^2*x^4 - (b^2*c^2*x^2 - b^2)*x^4)*log(sqrt(c*x + 1)*sqrt(-c*x + 1) + 1)) -
integrate((3*c^4*x^4 - 6*c^2*x^2 + (c^2*x^2 - 3)*(c*x + 1)*(c*x - 1) + (2*c^4*x^4 - 7*c^2*x^2 + 6)*sqrt(c*x +
1)*sqrt(-c*x + 1) + 3)/((b^2*c^4*x^4 - 2*b^2*c^2*x^2 + b^2)*x^4*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^4 - (b^2*x^4*log(x) + (b^2*log(c) - a*b)*x^4)*(c*x + 1)*(
c*x - 1) - 2*((b^2*c^2*x^2 - b^2)*x^4*log(x) + ((b^2*c^2*log(c) - a*b*c^2)*x^2 - b^2*log(c) + a*b)*x^4)*sqrt(c
*x + 1)*sqrt(-c*x + 1) + ((c*x + 1)*(c*x - 1)*b^2*x^4 + 2*(b^2*c^2*x^2 - b^2)*sqrt(c*x + 1)*sqrt(-c*x + 1)*x^4
- (b^2*c^4*x^4 - 2*b^2*c^2*x^2 + b^2)*x^4)*log(sqrt(c*x + 1)*sqrt(-c*x + 1) + 1)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{b^{2} x^{4} \operatorname{arsech}\left (c x\right )^{2} + 2 \, a b x^{4} \operatorname{arsech}\left (c x\right ) + a^{2} x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(a+b*arcsech(c*x))^2,x, algorithm="fricas")

[Out]

integral(1/(b^2*x^4*arcsech(c*x)^2 + 2*a*b*x^4*arcsech(c*x) + a^2*x^4), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{4} \left (a + b \operatorname{asech}{\left (c x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(a+b*asech(c*x))**2,x)

[Out]

Integral(1/(x**4*(a + b*asech(c*x))**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \operatorname{arsech}\left (c x\right ) + a\right )}^{2} x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(a+b*arcsech(c*x))^2,x, algorithm="giac")

[Out]

integrate(1/((b*arcsech(c*x) + a)^2*x^4), x)