3.1.56 \(\int \frac {1}{x^4 (a+b \text {sech}^{-1}(c x))} \, dx\) [56]

Optimal. Leaf size=117 \[ \frac {c^3 \text {Chi}\left (\frac {a}{b}+\text {sech}^{-1}(c x)\right ) \sinh \left (\frac {a}{b}\right )}{4 b}+\frac {c^3 \text {Chi}\left (\frac {3 a}{b}+3 \text {sech}^{-1}(c x)\right ) \sinh \left (\frac {3 a}{b}\right )}{4 b}-\frac {c^3 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\text {sech}^{-1}(c x)\right )}{4 b}-\frac {c^3 \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 a}{b}+3 \text {sech}^{-1}(c x)\right )}{4 b} \]

[Out]

-1/4*c^3*cosh(a/b)*Shi(a/b+arcsech(c*x))/b-1/4*c^3*cosh(3*a/b)*Shi(3*a/b+3*arcsech(c*x))/b+1/4*c^3*Chi(a/b+arc
sech(c*x))*sinh(a/b)/b+1/4*c^3*Chi(3*a/b+3*arcsech(c*x))*sinh(3*a/b)/b

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Rubi [A]
time = 0.18, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {6420, 5556, 3384, 3379, 3382} \begin {gather*} \frac {c^3 \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\text {sech}^{-1}(c x)\right )}{4 b}+\frac {c^3 \sinh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 a}{b}+3 \text {sech}^{-1}(c x)\right )}{4 b}-\frac {c^3 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\text {sech}^{-1}(c x)\right )}{4 b}-\frac {c^3 \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 a}{b}+3 \text {sech}^{-1}(c x)\right )}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + b*ArcSech[c*x])),x]

[Out]

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

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 5556

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

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Mathematica [A]
time = 0.12, size = 91, normalized size = 0.78 \begin {gather*} -\frac {c^3 \left (-\text {Chi}\left (\frac {a}{b}+\text {sech}^{-1}(c x)\right ) \sinh \left (\frac {a}{b}\right )-\text {Chi}\left (3 \left (\frac {a}{b}+\text {sech}^{-1}(c x)\right )\right ) \sinh \left (\frac {3 a}{b}\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 )}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(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])]))/b

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Maple [A]
time = 0.48, size = 110, normalized size = 0.94

method	result	size
derivativedivides	\(c^{3} \left (-\frac {{\mathrm e}^{\frac {3 a}{b}} \expIntegral \left (1, \frac {3 a}{b}+3 \,\mathrm {arcsech}\left (c x \right )\right )}{8 b}-\frac {{\mathrm e}^{\frac {a}{b}} \expIntegral \left (1, \frac {a}{b}+\mathrm {arcsech}\left (c x \right )\right )}{8 b}+\frac {{\mathrm e}^{-\frac {a}{b}} \expIntegral \left (1, -\mathrm {arcsech}\left (c x \right )-\frac {a}{b}\right )}{8 b}+\frac {{\mathrm e}^{-\frac {3 a}{b}} \expIntegral \left (1, -3 \,\mathrm {arcsech}\left (c x \right )-\frac {3 a}{b}\right )}{8 b}\right )\)	\(110\)
default	\(c^{3} \left (-\frac {{\mathrm e}^{\frac {3 a}{b}} \expIntegral \left (1, \frac {3 a}{b}+3 \,\mathrm {arcsech}\left (c x \right )\right )}{8 b}-\frac {{\mathrm e}^{\frac {a}{b}} \expIntegral \left (1, \frac {a}{b}+\mathrm {arcsech}\left (c x \right )\right )}{8 b}+\frac {{\mathrm e}^{-\frac {a}{b}} \expIntegral \left (1, -\mathrm {arcsech}\left (c x \right )-\frac {a}{b}\right )}{8 b}+\frac {{\mathrm e}^{-\frac {3 a}{b}} \expIntegral \left (1, -3 \,\mathrm {arcsech}\left (c x \right )-\frac {3 a}{b}\right )}{8 b}\right )\)	\(110\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(a+b*arcsech(c*x)),x,method=_RETURNVERBOSE)

[Out]

c^3*(-1/8/b*exp(3*a/b)*Ei(1,3*a/b+3*arcsech(c*x))-1/8/b*exp(a/b)*Ei(1,a/b+arcsech(c*x))+1/8/b*exp(-a/b)*Ei(1,-
arcsech(c*x)-a/b)+1/8/b*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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^4\,\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(a + b*acosh(1/(c*x)))),x)

[Out]

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

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