3.68 \(\int \frac {1}{x^4 (a+b \text {sech}^{-1}(c x))^3} \, dx\)

Optimal. Leaf size=240 \[ \frac {c^3 \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\text {sech}^{-1}(c x)\right )}{8 b^3}+\frac {9 c^3 \sinh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 a}{b}+3 \text {sech}^{-1}(c x)\right )}{8 b^3}-\frac {c^3 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\text {sech}^{-1}(c x)\right )}{8 b^3}-\frac {9 c^3 \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 a}{b}+3 \text {sech}^{-1}(c x)\right )}{8 b^3}+\frac {3 c^3 \cosh \left (3 \text {sech}^{-1}(c x)\right )}{8 b^2 \left (a+b \text {sech}^{-1}(c x)\right )}+\frac {c^2}{8 b^2 x \left (a+b \text {sech}^{-1}(c x)\right )}+\frac {c^3 \sinh \left (3 \text {sech}^{-1}(c x)\right )}{8 b \left (a+b \text {sech}^{-1}(c x)\right )^2}+\frac {c^2 \sqrt {\frac {1-c x}{c x+1}} (c x+1)}{8 b x \left (a+b \text {sech}^{-1}(c x)\right )^2} \]

[Out]

1/8*c^2/b^2/x/(a+b*arcsech(c*x))+3/8*c^3*cosh(3*arcsech(c*x))/b^2/(a+b*arcsech(c*x))-1/8*c^3*cosh(a/b)*Shi(a/b
+arcsech(c*x))/b^3-9/8*c^3*cosh(3*a/b)*Shi(3*a/b+3*arcsech(c*x))/b^3+1/8*c^3*Chi(a/b+arcsech(c*x))*sinh(a/b)/b
^3+9/8*c^3*Chi(3*a/b+3*arcsech(c*x))*sinh(3*a/b)/b^3+1/8*c^3*sinh(3*arcsech(c*x))/b/(a+b*arcsech(c*x))^2+1/8*c
^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 = 0.37, antiderivative size = 240, normalized size of antiderivative = 1.00, number of steps used = 13, 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} \[ \frac {c^3 \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\text {sech}^{-1}(c x)\right )}{8 b^3}+\frac {9 c^3 \sinh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 a}{b}+3 \text {sech}^{-1}(c x)\right )}{8 b^3}-\frac {c^3 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\text {sech}^{-1}(c x)\right )}{8 b^3}-\frac {9 c^3 \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 a}{b}+3 \text {sech}^{-1}(c x)\right )}{8 b^3}+\frac {c^2}{8 b^2 x \left (a+b \text {sech}^{-1}(c x)\right )}+\frac {3 c^3 \cosh \left (3 \text {sech}^{-1}(c x)\right )}{8 b^2 \left (a+b \text {sech}^{-1}(c x)\right )}+\frac {c^2 \sqrt {\frac {1-c x}{c x+1}} (c x+1)}{8 b x \left (a+b \text {sech}^{-1}(c x)\right )^2}+\frac {c^3 \sinh \left (3 \text {sech}^{-1}(c x)\right )}{8 b \left (a+b \text {sech}^{-1}(c x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + b*ArcSech[c*x])^3),x]

[Out]

(c^2*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x))/(8*b*x*(a + b*ArcSech[c*x])^2) + c^2/(8*b^2*x*(a + b*ArcSech[c*x]))
+ (3*c^3*Cosh[3*ArcSech[c*x]])/(8*b^2*(a + b*ArcSech[c*x])) + (c^3*CoshIntegral[a/b + ArcSech[c*x]]*Sinh[a/b])
/(8*b^3) + (9*c^3*CoshIntegral[(3*a)/b + 3*ArcSech[c*x]]*Sinh[(3*a)/b])/(8*b^3) + (c^3*Sinh[3*ArcSech[c*x]])/(
8*b*(a + b*ArcSech[c*x])^2) - (c^3*Cosh[a/b]*SinhIntegral[a/b + ArcSech[c*x]])/(8*b^3) - (9*c^3*Cosh[(3*a)/b]*
SinhIntegral[(3*a)/b + 3*ArcSech[c*x]])/(8*b^3)

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^4 \left (a+b \text {sech}^{-1}(c x)\right )^3} \, dx &=-\left (c^3 \operatorname {Subst}\left (\int \frac {\cosh ^2(x) \sinh (x)}{(a+b x)^3} \, dx,x,\text {sech}^{-1}(c x)\right )\right )\\ &=-\left (c^3 \operatorname {Subst}\left (\int \left (\frac {\sinh (x)}{4 (a+b x)^3}+\frac {\sinh (3 x)}{4 (a+b x)^3}\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)^3} \, 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)^3} \, dx,x,\text {sech}^{-1}(c x)\right )\\ &=\frac {c^2 \sqrt {\frac {1-c x}{1+c x}} (1+c x)}{8 b x \left (a+b \text {sech}^{-1}(c x)\right )^2}+\frac {c^3 \sinh \left (3 \text {sech}^{-1}(c x)\right )}{8 b \left (a+b \text {sech}^{-1}(c x)\right )^2}-\frac {c^3 \operatorname {Subst}\left (\int \frac {\cosh (x)}{(a+b x)^2} \, dx,x,\text {sech}^{-1}(c x)\right )}{8 b}-\frac {\left (3 c^3\right ) \operatorname {Subst}\left (\int \frac {\cosh (3 x)}{(a+b x)^2} \, dx,x,\text {sech}^{-1}(c x)\right )}{8 b}\\ &=\frac {c^2 \sqrt {\frac {1-c x}{1+c x}} (1+c x)}{8 b x \left (a+b \text {sech}^{-1}(c x)\right )^2}+\frac {c^2}{8 b^2 x \left (a+b \text {sech}^{-1}(c x)\right )}+\frac {3 c^3 \cosh \left (3 \text {sech}^{-1}(c x)\right )}{8 b^2 \left (a+b \text {sech}^{-1}(c x)\right )}+\frac {c^3 \sinh \left (3 \text {sech}^{-1}(c x)\right )}{8 b \left (a+b \text {sech}^{-1}(c x)\right )^2}-\frac {c^3 \operatorname {Subst}\left (\int \frac {\sinh (x)}{a+b x} \, dx,x,\text {sech}^{-1}(c x)\right )}{8 b^2}-\frac {\left (9 c^3\right ) \operatorname {Subst}\left (\int \frac {\sinh (3 x)}{a+b x} \, dx,x,\text {sech}^{-1}(c x)\right )}{8 b^2}\\ &=\frac {c^2 \sqrt {\frac {1-c x}{1+c x}} (1+c x)}{8 b x \left (a+b \text {sech}^{-1}(c x)\right )^2}+\frac {c^2}{8 b^2 x \left (a+b \text {sech}^{-1}(c x)\right )}+\frac {3 c^3 \cosh \left (3 \text {sech}^{-1}(c x)\right )}{8 b^2 \left (a+b \text {sech}^{-1}(c x)\right )}+\frac {c^3 \sinh \left (3 \text {sech}^{-1}(c x)\right )}{8 b \left (a+b \text {sech}^{-1}(c x)\right )^2}-\frac {\left (c^3 \cosh \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 )}{8 b^2}-\frac {\left (9 c^3 \cosh \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 )}{8 b^2}+\frac {\left (c^3 \sinh \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 )}{8 b^2}+\frac {\left (9 c^3 \sinh \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 )}{8 b^2}\\ &=\frac {c^2 \sqrt {\frac {1-c x}{1+c x}} (1+c x)}{8 b x \left (a+b \text {sech}^{-1}(c x)\right )^2}+\frac {c^2}{8 b^2 x \left (a+b \text {sech}^{-1}(c x)\right )}+\frac {3 c^3 \cosh \left (3 \text {sech}^{-1}(c x)\right )}{8 b^2 \left (a+b \text {sech}^{-1}(c x)\right )}+\frac {c^3 \text {Chi}\left (\frac {a}{b}+\text {sech}^{-1}(c x)\right ) \sinh \left (\frac {a}{b}\right )}{8 b^3}+\frac {9 c^3 \text {Chi}\left (\frac {3 a}{b}+3 \text {sech}^{-1}(c x)\right ) \sinh \left (\frac {3 a}{b}\right )}{8 b^3}+\frac {c^3 \sinh \left (3 \text {sech}^{-1}(c x)\right )}{8 b \left (a+b \text {sech}^{-1}(c x)\right )^2}-\frac {c^3 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\text {sech}^{-1}(c x)\right )}{8 b^3}-\frac {9 c^3 \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 a}{b}+3 \text {sech}^{-1}(c x)\right )}{8 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.63, size = 204, normalized size = 0.85 \[ \frac {\frac {4 b^2 \sqrt {\frac {1-c x}{c x+1}} (c x+1)}{x^3 \left (a+b \text {sech}^{-1}(c x)\right )^2}-8 c^3 \left (\sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\text {sech}^{-1}(c x)\right )-\cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\text {sech}^{-1}(c x)\right )\right )+9 c^3 \left (\sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\text {sech}^{-1}(c x)\right )+\sinh \left (\frac {3 a}{b}\right ) \text {Chi}\left (3 \left (\frac {a}{b}+\text {sech}^{-1}(c x)\right )\right )-\cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\text {sech}^{-1}(c x)\right )-\cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (3 \left (\frac {a}{b}+\text {sech}^{-1}(c x)\right )\right )\right )+\frac {4 b \left (3-2 c^2 x^2\right )}{x^3 \left (a+b \text {sech}^{-1}(c x)\right )}}{8 b^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(x^4*(a + b*ArcSech[c*x])^3),x]

[Out]

((4*b^2*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x))/(x^3*(a + b*ArcSech[c*x])^2) + (4*b*(3 - 2*c^2*x^2))/(x^3*(a + b*
ArcSech[c*x])) - 8*c^3*(CoshIntegral[a/b + ArcSech[c*x]]*Sinh[a/b] - Cosh[a/b]*SinhIntegral[a/b + ArcSech[c*x]
]) + 9*c^3*(CoshIntegral[a/b + ArcSech[c*x]]*Sinh[a/b] + CoshIntegral[3*(a/b + ArcSech[c*x])]*Sinh[(3*a)/b] -
Cosh[a/b]*SinhIntegral[a/b + ArcSech[c*x]] - Cosh[(3*a)/b]*SinhIntegral[3*(a/b + ArcSech[c*x])]))/(8*b^3)

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{b^{3} x^{4} \operatorname {arsech}\left (c x\right )^{3} + 3 \, a b^{2} x^{4} \operatorname {arsech}\left (c x\right )^{2} + 3 \, a^{2} b x^{4} \operatorname {arsech}\left (c x\right ) + a^{3} x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(a+b*arcsech(c*x))^3,x, algorithm="fricas")

[Out]

integral(1/(b^3*x^4*arcsech(c*x)^3 + 3*a*b^2*x^4*arcsech(c*x)^2 + 3*a^2*b*x^4*arcsech(c*x) + a^3*x^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \operatorname {arsech}\left (c x\right ) + a\right )}^{3} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(a+b*arcsech(c*x))^3,x, algorithm="giac")

[Out]

integrate(1/((b*arcsech(c*x) + a)^3*x^4), x)

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maple [B]  time = 0.47, size = 628, normalized size = 2.62 \[ c^{3} \left (\frac {\left (\sqrt {\frac {c x +1}{c x}}\, \sqrt {-\frac {c x -1}{c x}}\, c^{3} x^{3}-4 \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, c x -3 c^{2} x^{2}+4\right ) \left (3 b \,\mathrm {arcsech}\left (c x \right )+3 a -b \right )}{16 c^{3} x^{3} b^{2} \left (\mathrm {arcsech}\left (c x \right )^{2} b^{2}+2 \,\mathrm {arcsech}\left (c x \right ) a b +a^{2}\right )}-\frac {9 \,{\mathrm e}^{\frac {3 a}{b}} \Ei \left (1, \frac {3 a}{b}+3 \,\mathrm {arcsech}\left (c x \right )\right )}{16 b^{3}}-\frac {\left (\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, c x -1\right ) \left (b \,\mathrm {arcsech}\left (c x \right )+a -b \right )}{16 c x \,b^{2} \left (\mathrm {arcsech}\left (c x \right )^{2} b^{2}+2 \,\mathrm {arcsech}\left (c x \right ) a b +a^{2}\right )}-\frac {{\mathrm e}^{\frac {a}{b}} \Ei \left (1, \frac {a}{b}+\mathrm {arcsech}\left (c x \right )\right )}{16 b^{3}}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, c x +1}{16 b c x \left (a +b \,\mathrm {arcsech}\left (c x \right )\right )^{2}}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, c x +1}{16 b^{2} c x \left (a +b \,\mathrm {arcsech}\left (c x \right )\right )}+\frac {{\mathrm e}^{-\frac {a}{b}} \Ei \left (1, -\mathrm {arcsech}\left (c x \right )-\frac {a}{b}\right )}{16 b^{3}}-\frac {\sqrt {\frac {c x +1}{c x}}\, \sqrt {-\frac {c x -1}{c x}}\, c^{3} x^{3}-4 \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, c x +3 c^{2} x^{2}-4}{16 b \,c^{3} x^{3} \left (a +b \,\mathrm {arcsech}\left (c x \right )\right )^{2}}-\frac {3 \left (\sqrt {\frac {c x +1}{c x}}\, \sqrt {-\frac {c x -1}{c x}}\, c^{3} x^{3}-4 \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, c x +3 c^{2} x^{2}-4\right )}{16 b^{2} c^{3} x^{3} \left (a +b \,\mathrm {arcsech}\left (c x \right )\right )}+\frac {9 \,{\mathrm e}^{-\frac {3 a}{b}} \Ei \left (1, -3 \,\mathrm {arcsech}\left (c x \right )-\frac {3 a}{b}\right )}{16 b^{3}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(a+b*arcsech(c*x))^3,x)

[Out]

c^3*(1/16*(((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)*(3*b*arcsech(c*x)+3*a-b)/c^3/x^3/b^2/(arcsech(c*x)^2*b^2+2*arcsech(c*x)*a*b+a^2)-9/16/b^3*exp(3*a/b)
*Ei(1,3*a/b+3*arcsech(c*x))-1/16*((-(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
/(arcsech(c*x)^2*b^2+2*arcsech(c*x)*a*b+a^2)-1/16/b^3*exp(a/b)*Ei(1,a/b+arcsech(c*x))+1/16/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+1/16/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))+1/16/b^3*exp(-a/b)*Ei(1,-arcsech(c*x)-a/b)-1/16/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))^2-3/1
6/b^2*(((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))+9/16/b^3*exp(-3*a/b)*Ei(1,-3*arcsech(c*x)-3*a/b))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(a+b*arcsech(c*x))^3,x, algorithm="maxima")

[Out]

-1/2*((b*c^6*(3*log(c) - 1) - 3*a*c^6)*x^7 - 3*(b*c^4*(3*log(c) - 1) - 3*a*c^4)*x^5 - ((b*c^4*log(c) - a*c^4)*
x^5 - (b*c^2*(4*log(c) - 1) - 4*a*c^2)*x^3 + (b*(3*log(c) - 1) - 3*a)*x + (b*c^4*x^5 - 4*b*c^2*x^3 + 3*b*x)*lo
g(x))*(c*x + 1)^(3/2)*(-c*x + 1)^(3/2) + 3*(b*c^2*(3*log(c) - 1) - 3*a*c^2)*x^3 - (2*(b*c^6*log(c) - a*c^6)*x^
7 - 2*(b*c^4*(5*log(c) - 1) - 5*a*c^4)*x^5 + (b*c^2*(17*log(c) - 5) - 17*a*c^2)*x^3 - 3*(b*(3*log(c) - 1) - 3*
a)*x + (2*b*c^6*x^7 - 10*b*c^4*x^5 + 17*b*c^2*x^3 - 9*b*x)*log(x))*(c*x + 1)*(c*x - 1) + ((b*c^6*(5*log(c) - 1
) - 5*a*c^6)*x^7 - (b*c^4*(18*log(c) - 5) - 18*a*c^4)*x^5 + (b*c^2*(22*log(c) - 7) - 22*a*c^2)*x^3 - 3*(b*(3*l
og(c) - 1) - 3*a)*x + (5*b*c^6*x^7 - 18*b*c^4*x^5 + 22*b*c^2*x^3 - 9*b*x)*log(x))*sqrt(c*x + 1)*sqrt(-c*x + 1)
- (b*(3*log(c) - 1) - 3*a)*x - (3*b*c^6*x^7 - 9*b*c^4*x^5 + 9*b*c^2*x^3 - (b*c^4*x^5 - 4*b*c^2*x^3 + 3*b*x)*(
c*x + 1)^(3/2)*(-c*x + 1)^(3/2) - (2*b*c^6*x^7 - 10*b*c^4*x^5 + 17*b*c^2*x^3 - 9*b*x)*(c*x + 1)*(c*x - 1) + (5
*b*c^6*x^7 - 18*b*c^4*x^5 + 22*b*c^2*x^3 - 9*b*x)*sqrt(c*x + 1)*sqrt(-c*x + 1) - 3*b*x)*log(sqrt(c*x + 1)*sqrt
(-c*x + 1) + 1) + 3*(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^4*log(x)^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^4*log(x) + ((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^4 - (b^4*x^4*log(x)^2 + 2*(b^4*log(
c) - a*b^3)*x^4*log(x) + (b^4*log(c)^2 - 2*a*b^3*log(c) + a^2*b^2)*x^4)*(c*x + 1)^(3/2)*(-c*x + 1)^(3/2) - 3*(
(b^4*c^2*x^2 - b^4)*x^4*log(x)^2 - 2*(b^4*log(c) - a*b^3 - (b^4*c^2*log(c) - a*b^3*c^2)*x^2)*x^4*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^4)*(c*x +
1)*(c*x - 1) - ((c*x + 1)^(3/2)*(-c*x + 1)^(3/2)*b^4*x^4 + 3*(b^4*c^2*x^2 - b^4)*(c*x + 1)*(c*x - 1)*x^4 + 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^4 - (b^4*c^6*x^6 - 3*b^4*c^4*x^4 + 3*b^4*c^2
*x^2 - b^4)*x^4)*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^4*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^4*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^4)*sqrt(c*x + 1)*sqrt(-c*x + 1) - 2*((b^4*c^6*x^6 - 3
*b^4*c^4*x^4 + 3*b^4*c^2*x^2 - b^4)*x^4*log(x) + ((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^4 - (b^4*x^4*log(x) + (b^4*log(c) - a*b
^3)*x^4)*(c*x + 1)^(3/2)*(-c*x + 1)^(3/2) - 3*((b^4*c^2*x^2 - b^4)*x^4*log(x) - (b^4*log(c) - a*b^3 - (b^4*c^2
*log(c) - a*b^3*c^2)*x^2)*x^4)*(c*x + 1)*(c*x - 1) - 3*((b^4*c^4*x^4 - 2*b^4*c^2*x^2 + b^4)*x^4*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^4)*sqrt(c*x + 1)*sqr
t(-c*x + 1))*log(sqrt(c*x + 1)*sqrt(-c*x + 1) + 1)) - integrate(1/2*(9*c^8*x^8 - 36*c^6*x^6 + 54*c^4*x^4 - (c^
4*x^4 + 4*c^2*x^2 - 9)*(c*x + 1)^2*(c*x - 1)^2 + (2*c^6*x^6 + 13*c^4*x^4 - 48*c^2*x^2 + 36)*(c*x + 1)^(3/2)*(-
c*x + 1)^(3/2) - 36*c^2*x^2 - (2*c^8*x^8 - 19*c^6*x^6 + 83*c^4*x^4 - 120*c^2*x^2 + 54)*(c*x + 1)*(c*x - 1) + (
10*c^8*x^8 - 57*c^6*x^6 + 123*c^4*x^4 - 112*c^2*x^2 + 36)*sqrt(c*x + 1)*sqrt(-c*x + 1) + 9)/((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^4*log(x) + (b^3*x^4*log(x) + (b^3*log(c) - a*b^2)*x^4)*(c
*x + 1)^2*(c*x - 1)^2 + ((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*lo
g(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^4 - 4*((b^3*c^2*x^2 - b^3)*
x^4*log(x) - (b^3*log(c) - a*b^2 - (b^3*c^2*log(c) - a*b^2*c^2)*x^2)*x^4)*(c*x + 1)^(3/2)*(-c*x + 1)^(3/2) - 6
*((b^3*c^4*x^4 - 2*b^3*c^2*x^2 + b^3)*x^4*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^4)*(c*x + 1)*(c*x - 1) - 4*((b^3*c^6*x^6 - 3*b^3*c^4*x^4 + 3*b^3*c^2*x^2 -
b^3)*x^4*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^4)*sqrt(c*x + 1)*sqrt(-c*x + 1) - ((c*x + 1)^2*(c*x - 1)^2*b^3*x^4 -
4*(b^3*c^2*x^2 - b^3)*(c*x + 1)^(3/2)*(-c*x + 1)^(3/2)*x^4 - 6*(b^3*c^4*x^4 - 2*b^3*c^2*x^2 + b^3)*(c*x + 1)*(
c*x - 1)*x^4 - 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^4 + (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^4)*log(sqrt(c*x + 1)*sqrt(-c*x + 1) + 1)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{x^4\,{\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(a + b*acosh(1/(c*x)))^3),x)

[Out]

int(1/(x^4*(a + b*acosh(1/(c*x)))^3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{4} \left (a + b \operatorname {asech}{\left (c x \right )}\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(a+b*asech(c*x))**3,x)

[Out]

Integral(1/(x**4*(a + b*asech(c*x))**3), x)

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