3.1.30 \(\int \frac {a+b \text {sech}^{-1}(c x)}{x^5} \, dx\) [30]

Optimal. Leaf size=126 \[ \frac {b \sqrt {1-c x}}{16 x^4 \sqrt {\frac {1}{1+c x}}}+\frac {3 b c^2 \sqrt {1-c x}}{32 x^2 \sqrt {\frac {1}{1+c x}}}-\frac {a+b \text {sech}^{-1}(c x)}{4 x^4}+\frac {3}{32} b c^4 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \tanh ^{-1}\left (\sqrt {1-c x} \sqrt {1+c x}\right ) \]

[Out]

1/4*(-a-b*arcsech(c*x))/x^4+1/16*b*(-c*x+1)^(1/2)/x^4/(1/(c*x+1))^(1/2)+3/32*b*c^2*(-c*x+1)^(1/2)/x^2/(1/(c*x+
1))^(1/2)+3/32*b*c^4*arctanh((-c*x+1)^(1/2)*(c*x+1)^(1/2))*(1/(c*x+1))^(1/2)*(c*x+1)^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {6418, 105, 12, 94, 214} \begin {gather*} -\frac {a+b \text {sech}^{-1}(c x)}{4 x^4}+\frac {3}{32} b c^4 \sqrt {\frac {1}{c x+1}} \sqrt {c x+1} \tanh ^{-1}\left (\sqrt {1-c x} \sqrt {c x+1}\right )+\frac {3 b c^2 \sqrt {1-c x}}{32 x^2 \sqrt {\frac {1}{c x+1}}}+\frac {b \sqrt {1-c x}}{16 x^4 \sqrt {\frac {1}{c x+1}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSech[c*x])/x^5,x]

[Out]

(b*Sqrt[1 - c*x])/(16*x^4*Sqrt[(1 + c*x)^(-1)]) + (3*b*c^2*Sqrt[1 - c*x])/(32*x^2*Sqrt[(1 + c*x)^(-1)]) - (a +
b*ArcSech[c*x])/(4*x^4) + (3*b*c^4*Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcTanh[Sqrt[1 - c*x]*Sqrt[1 + c*x]])/3
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 94

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))), x_Symbol] :> Dist[b*f, Subst[I
nt[1/(d*(b*e - a*f)^2 + b*f^2*x^2), x], x, Sqrt[a + b*x]*Sqrt[c + d*x]], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
EqQ[2*b*d*e - f*(b*c + a*d), 0]

Rule 105

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && ILtQ[m, -1] &
& (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 6418

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcSech[c*
x])/(d*(m + 1))), x] + Dist[b*(Sqrt[1 + c*x]/(m + 1))*Sqrt[1/(1 + c*x)], Int[(d*x)^m/(Sqrt[1 - c*x]*Sqrt[1 + c
*x]), x], x] /; FreeQ[{a, b, c, d, m}, x] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {a+b \text {sech}^{-1}(c x)}{x^5} \, dx &=-\frac {a+b \text {sech}^{-1}(c x)}{4 x^4}-\frac {1}{4} \left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {1}{x^5 \sqrt {1-c x} \sqrt {1+c x}} \, dx\\ &=\frac {b \sqrt {1-c x}}{16 x^4 \sqrt {\frac {1}{1+c x}}}-\frac {a+b \text {sech}^{-1}(c x)}{4 x^4}+\frac {1}{16} \left (b \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int -\frac {3 c^2}{x^3 \sqrt {1-c x} \sqrt {1+c x}} \, dx\\ &=\frac {b \sqrt {1-c x}}{16 x^4 \sqrt {\frac {1}{1+c x}}}-\frac {a+b \text {sech}^{-1}(c x)}{4 x^4}-\frac {1}{16} \left (3 b c^2 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {1}{x^3 \sqrt {1-c x} \sqrt {1+c x}} \, dx\\ &=\frac {b \sqrt {1-c x}}{16 x^4 \sqrt {\frac {1}{1+c x}}}+\frac {3 b c^2 \sqrt {1-c x}}{32 x^2 \sqrt {\frac {1}{1+c x}}}-\frac {a+b \text {sech}^{-1}(c x)}{4 x^4}-\frac {1}{32} \left (3 b c^2 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {c^2}{x \sqrt {1-c x} \sqrt {1+c x}} \, dx\\ &=\frac {b \sqrt {1-c x}}{16 x^4 \sqrt {\frac {1}{1+c x}}}+\frac {3 b c^2 \sqrt {1-c x}}{32 x^2 \sqrt {\frac {1}{1+c x}}}-\frac {a+b \text {sech}^{-1}(c x)}{4 x^4}-\frac {1}{32} \left (3 b c^4 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \int \frac {1}{x \sqrt {1-c x} \sqrt {1+c x}} \, dx\\ &=\frac {b \sqrt {1-c x}}{16 x^4 \sqrt {\frac {1}{1+c x}}}+\frac {3 b c^2 \sqrt {1-c x}}{32 x^2 \sqrt {\frac {1}{1+c x}}}-\frac {a+b \text {sech}^{-1}(c x)}{4 x^4}+\frac {1}{32} \left (3 b c^5 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x}\right ) \text {Subst}\left (\int \frac {1}{c-c x^2} \, dx,x,\sqrt {1-c x} \sqrt {1+c x}\right )\\ &=\frac {b \sqrt {1-c x}}{16 x^4 \sqrt {\frac {1}{1+c x}}}+\frac {3 b c^2 \sqrt {1-c x}}{32 x^2 \sqrt {\frac {1}{1+c x}}}-\frac {a+b \text {sech}^{-1}(c x)}{4 x^4}+\frac {3}{32} b c^4 \sqrt {\frac {1}{1+c x}} \sqrt {1+c x} \tanh ^{-1}\left (\sqrt {1-c x} \sqrt {1+c x}\right )\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 137, normalized size = 1.09 \begin {gather*} -\frac {a}{4 x^4}+b \left (\frac {1}{16 x^4}+\frac {c}{16 x^3}+\frac {3 c^2}{32 x^2}+\frac {3 c^3}{32 x}\right ) \sqrt {\frac {1-c x}{1+c x}}-\frac {b \text {sech}^{-1}(c x)}{4 x^4}-\frac {3}{32} b c^4 \log (x)+\frac {3}{32} b c^4 \log \left (1+\sqrt {\frac {1-c x}{1+c x}}+c x \sqrt {\frac {1-c x}{1+c x}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSech[c*x])/x^5,x]

[Out]

-1/4*a/x^4 + b*(1/(16*x^4) + c/(16*x^3) + (3*c^2)/(32*x^2) + (3*c^3)/(32*x))*Sqrt[(1 - c*x)/(1 + c*x)] - (b*Ar
cSech[c*x])/(4*x^4) - (3*b*c^4*Log[x])/32 + (3*b*c^4*Log[1 + Sqrt[(1 - c*x)/(1 + c*x)] + c*x*Sqrt[(1 - c*x)/(1
+ c*x)]])/32

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Maple [A]
time = 0.17, size = 135, normalized size = 1.07

method	result	size
derivativedivides	\(c^{4} \left (-\frac {a}{4 c^{4} x^{4}}+b \left (-\frac {\mathrm {arcsech}\left (c x \right )}{4 c^{4} x^{4}}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, \left (3 \arctanh \left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right ) c^{4} x^{4}+3 \sqrt {-c^{2} x^{2}+1}\, c^{2} x^{2}+2 \sqrt {-c^{2} x^{2}+1}\right )}{32 c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}\right )\right )\)	\(135\)
default	\(c^{4} \left (-\frac {a}{4 c^{4} x^{4}}+b \left (-\frac {\mathrm {arcsech}\left (c x \right )}{4 c^{4} x^{4}}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, \left (3 \arctanh \left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right ) c^{4} x^{4}+3 \sqrt {-c^{2} x^{2}+1}\, c^{2} x^{2}+2 \sqrt {-c^{2} x^{2}+1}\right )}{32 c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}\right )\right )\)	\(135\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsech(c*x))/x^5,x,method=_RETURNVERBOSE)

[Out]

c^4*(-1/4*a/c^4/x^4+b*(-1/4/c^4/x^4*arcsech(c*x)+1/32*(-(c*x-1)/c/x)^(1/2)/c^3/x^3*((c*x+1)/c/x)^(1/2)*(3*arct
anh(1/(-c^2*x^2+1)^(1/2))*c^4*x^4+3*(-c^2*x^2+1)^(1/2)*c^2*x^2+2*(-c^2*x^2+1)^(1/2))/(-c^2*x^2+1)^(1/2)))

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Maxima [A]
time = 0.26, size = 147, normalized size = 1.17 \begin {gather*} \frac {1}{64} \, b {\left (\frac {3 \, c^{5} \log \left (c x \sqrt {\frac {1}{c^{2} x^{2}} - 1} + 1\right ) - 3 \, c^{5} \log \left (c x \sqrt {\frac {1}{c^{2} x^{2}} - 1} - 1\right ) - \frac {2 \, {\left (3 \, c^{8} x^{3} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{\frac {3}{2}} - 5 \, c^{6} x \sqrt {\frac {1}{c^{2} x^{2}} - 1}\right )}}{c^{4} x^{4} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{2} - 2 \, c^{2} x^{2} {\left (\frac {1}{c^{2} x^{2}} - 1\right )} + 1}}{c} - \frac {16 \, \operatorname {arsech}\left (c x\right )}{x^{4}}\right )} - \frac {a}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsech(c*x))/x^5,x, algorithm="maxima")

[Out]

1/64*b*((3*c^5*log(c*x*sqrt(1/(c^2*x^2) - 1) + 1) - 3*c^5*log(c*x*sqrt(1/(c^2*x^2) - 1) - 1) - 2*(3*c^8*x^3*(1
/(c^2*x^2) - 1)^(3/2) - 5*c^6*x*sqrt(1/(c^2*x^2) - 1))/(c^4*x^4*(1/(c^2*x^2) - 1)^2 - 2*c^2*x^2*(1/(c^2*x^2) -
1) + 1))/c - 16*arcsech(c*x)/x^4) - 1/4*a/x^4

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Fricas [A]
time = 0.34, size = 90, normalized size = 0.71 \begin {gather*} \frac {{\left (3 \, b c^{4} x^{4} - 8 \, b\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) + {\left (3 \, b c^{3} x^{3} + 2 \, b c x\right )} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - 8 \, a}{32 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsech(c*x))/x^5,x, algorithm="fricas")

[Out]

1/32*((3*b*c^4*x^4 - 8*b)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) + (3*b*c^3*x^3 + 2*b*c*x)*sqrt(-
(c^2*x^2 - 1)/(c^2*x^2)) - 8*a)/x^4

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b \operatorname {asech}{\left (c x \right )}}{x^{5}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asech(c*x))/x**5,x)

[Out]

Integral((a + b*asech(c*x))/x**5, 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((a+b*arcsech(c*x))/x^5,x, algorithm="giac")

[Out]

integrate((b*arcsech(c*x) + a)/x^5, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )}{x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(1/(c*x)))/x^5,x)

[Out]

int((a + b*acosh(1/(c*x)))/x^5, x)

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