3.1.41 \(\int \frac {(a+b \text {sech}^{-1}(c x))^2}{x^5} \, dx\) [41]

Optimal. Leaf size=151 \[ -\frac {b^2}{32 x^4}-\frac {3 b^2 c^2}{32 x^2}+\frac {3}{16} a b c^4 \text {sech}^{-1}(c x)+\frac {3}{32} b^2 c^4 \text {sech}^{-1}(c x)^2+\frac {b \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )}{8 x^4}+\frac {3 b c^2 \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )}{16 x^2}-\frac {\left (a+b \text {sech}^{-1}(c x)\right )^2}{4 x^4} \]

[Out]

-1/32*b^2/x^4-3/32*b^2*c^2/x^2+3/16*a*b*c^4*arcsech(c*x)+3/32*b^2*c^4*arcsech(c*x)^2-1/4*(a+b*arcsech(c*x))^2/
x^4+1/8*b*(c*x+1)*(a+b*arcsech(c*x))*((-c*x+1)/(c*x+1))^(1/2)/x^4+3/16*b*c^2*(c*x+1)*(a+b*arcsech(c*x))*((-c*x
+1)/(c*x+1))^(1/2)/x^2

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Rubi [A]
time = 0.08, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {6420, 5555, 3391} \begin {gather*} \frac {3}{16} a b c^4 \text {sech}^{-1}(c x)+\frac {3 b c^2 \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (a+b \text {sech}^{-1}(c x)\right )}{16 x^2}-\frac {\left (a+b \text {sech}^{-1}(c x)\right )^2}{4 x^4}+\frac {b \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (a+b \text {sech}^{-1}(c x)\right )}{8 x^4}+\frac {3}{32} b^2 c^4 \text {sech}^{-1}(c x)^2-\frac {3 b^2 c^2}{32 x^2}-\frac {b^2}{32 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSech[c*x])^2/x^5,x]

[Out]

-1/32*b^2/x^4 - (3*b^2*c^2)/(32*x^2) + (3*a*b*c^4*ArcSech[c*x])/16 + (3*b^2*c^4*ArcSech[c*x]^2)/32 + (b*Sqrt[(
1 - c*x)/(1 + c*x)]*(1 + c*x)*(a + b*ArcSech[c*x]))/(8*x^4) + (3*b*c^2*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(a
+ b*ArcSech[c*x]))/(16*x^2) - (a + b*ArcSech[c*x])^2/(4*x^4)

Rule 3391

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*((b*Sin[e + f*x])^n/(f^2*n^
2)), x] + (Dist[b^2*((n - 1)/n), Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[b*(c + d*x)*Cos[e + f*x
]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

Rule 5555

Int[Cosh[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)], x_Symbol] :> Simp[(c +
d*x)^m*(Cosh[a + b*x]^(n + 1)/(b*(n + 1))), x] - Dist[d*(m/(b*(n + 1))), Int[(c + d*x)^(m - 1)*Cosh[a + b*x]^
(n + 1), x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && NeQ[n, -1]

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 {\left (a+b \text {sech}^{-1}(c x)\right )^2}{x^5} \, dx &=-\left (c^4 \text {Subst}\left (\int (a+b x)^2 \cosh ^3(x) \sinh (x) \, dx,x,\text {sech}^{-1}(c x)\right )\right )\\ &=-\frac {\left (a+b \text {sech}^{-1}(c x)\right )^2}{4 x^4}+\frac {1}{2} \left (b c^4\right ) \text {Subst}\left (\int (a+b x) \cosh ^4(x) \, dx,x,\text {sech}^{-1}(c x)\right )\\ &=-\frac {b^2}{32 x^4}+\frac {b \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )}{8 x^4}-\frac {\left (a+b \text {sech}^{-1}(c x)\right )^2}{4 x^4}+\frac {1}{8} \left (3 b c^4\right ) \text {Subst}\left (\int (a+b x) \cosh ^2(x) \, dx,x,\text {sech}^{-1}(c x)\right )\\ &=-\frac {b^2}{32 x^4}-\frac {3 b^2 c^2}{32 x^2}+\frac {b \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )}{8 x^4}+\frac {3 b c^2 \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )}{16 x^2}-\frac {\left (a+b \text {sech}^{-1}(c x)\right )^2}{4 x^4}+\frac {1}{16} \left (3 b c^4\right ) \text {Subst}\left (\int (a+b x) \, dx,x,\text {sech}^{-1}(c x)\right )\\ &=-\frac {b^2}{32 x^4}-\frac {3 b^2 c^2}{32 x^2}+\frac {3}{16} a b c^4 \text {sech}^{-1}(c x)+\frac {3}{32} b^2 c^4 \text {sech}^{-1}(c x)^2+\frac {b \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )}{8 x^4}+\frac {3 b c^2 \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )}{16 x^2}-\frac {\left (a+b \text {sech}^{-1}(c x)\right )^2}{4 x^4}\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 268, normalized size = 1.77 \begin {gather*} \frac {-8 a^2-b^2-3 b^2 c^2 x^2+4 a b \sqrt {\frac {1-c x}{1+c x}}+4 a b c x \sqrt {\frac {1-c x}{1+c x}}+6 a b c^2 x^2 \sqrt {\frac {1-c x}{1+c x}}+6 a b c^3 x^3 \sqrt {\frac {1-c x}{1+c x}}+2 b \left (-8 a+b \sqrt {\frac {1-c x}{1+c x}} \left (2+2 c x+3 c^2 x^2+3 c^3 x^3\right )\right ) \text {sech}^{-1}(c x)+b^2 \left (-8+3 c^4 x^4\right ) \text {sech}^{-1}(c x)^2-6 a b c^4 x^4 \log (x)+6 a b c^4 x^4 \log \left (1+\sqrt {\frac {1-c x}{1+c x}}+c x \sqrt {\frac {1-c x}{1+c x}}\right )}{32 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSech[c*x])^2/x^5,x]

[Out]

(-8*a^2 - b^2 - 3*b^2*c^2*x^2 + 4*a*b*Sqrt[(1 - c*x)/(1 + c*x)] + 4*a*b*c*x*Sqrt[(1 - c*x)/(1 + c*x)] + 6*a*b*
c^2*x^2*Sqrt[(1 - c*x)/(1 + c*x)] + 6*a*b*c^3*x^3*Sqrt[(1 - c*x)/(1 + c*x)] + 2*b*(-8*a + b*Sqrt[(1 - c*x)/(1
+ c*x)]*(2 + 2*c*x + 3*c^2*x^2 + 3*c^3*x^3))*ArcSech[c*x] + b^2*(-8 + 3*c^4*x^4)*ArcSech[c*x]^2 - 6*a*b*c^4*x^
4*Log[x] + 6*a*b*c^4*x^4*Log[1 + Sqrt[(1 - c*x)/(1 + c*x)] + c*x*Sqrt[(1 - c*x)/(1 + c*x)]])/(32*x^4)

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Maple [A]
time = 0.33, size = 264, normalized size = 1.75

method	result	size
derivativedivides	\(c^{4} \left (-\frac {a^{2}}{4 c^{4} x^{4}}+b^{2} \left (-\frac {\mathrm {arcsech}\left (c x \right )^{2}}{4 c^{4} x^{4}}+\frac {\mathrm {arcsech}\left (c x \right ) \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}}{8 c^{3} x^{3}}+\frac {3 \,\mathrm {arcsech}\left (c x \right ) \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}}{16 c x}+\frac {3 \mathrm {arcsech}\left (c x \right )^{2}}{32}-\frac {1}{32 c^{4} x^{4}}-\frac {3}{32 c^{2} x^{2}}\right )+2 a b \left (-\frac {\mathrm {arcsech}\left (c x \right )}{4 c^{4} x^{4}}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, \left (3 \arctanh \left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right ) c^{4} x^{4}+3 \sqrt {-c^{2} x^{2}+1}\, c^{2} x^{2}+2 \sqrt {-c^{2} x^{2}+1}\right )}{32 c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}\right )\right )\)	\(264\)
default	\(c^{4} \left (-\frac {a^{2}}{4 c^{4} x^{4}}+b^{2} \left (-\frac {\mathrm {arcsech}\left (c x \right )^{2}}{4 c^{4} x^{4}}+\frac {\mathrm {arcsech}\left (c x \right ) \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}}{8 c^{3} x^{3}}+\frac {3 \,\mathrm {arcsech}\left (c x \right ) \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}}{16 c x}+\frac {3 \mathrm {arcsech}\left (c x \right )^{2}}{32}-\frac {1}{32 c^{4} x^{4}}-\frac {3}{32 c^{2} x^{2}}\right )+2 a b \left (-\frac {\mathrm {arcsech}\left (c x \right )}{4 c^{4} x^{4}}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, \left (3 \arctanh \left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right ) c^{4} x^{4}+3 \sqrt {-c^{2} x^{2}+1}\, c^{2} x^{2}+2 \sqrt {-c^{2} x^{2}+1}\right )}{32 c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}\right )\right )\)	\(264\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsech(c*x))^2/x^5,x,method=_RETURNVERBOSE)

[Out]

c^4*(-1/4*a^2/c^4/x^4+b^2*(-1/4*arcsech(c*x)^2/c^4/x^4+1/8*arcsech(c*x)/c^3/x^3*(-(c*x-1)/c/x)^(1/2)*((c*x+1)/
c/x)^(1/2)+3/16*arcsech(c*x)/c/x*(-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/2)+3/32*arcsech(c*x)^2-1/32/c^4/x^4-3/3
2/c^2/x^2)+2*a*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*arctanh(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 [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((a+b*arcsech(c*x))^2/x^5,x, algorithm="maxima")

[Out]

1/32*a*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) + b^2*integrate(log(sqrt(1/(c*x) + 1)*sqrt(1/(c*x) - 1) + 1/(c*x))^2/x^5,
x) - 1/4*a^2/x^4

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Fricas [A]
time = 0.35, size = 204, normalized size = 1.35 \begin {gather*} -\frac {3 \, b^{2} c^{2} x^{2} - {\left (3 \, b^{2} c^{4} x^{4} - 8 \, b^{2}\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right )^{2} + 8 \, a^{2} + b^{2} - 2 \, {\left (3 \, a b c^{4} x^{4} - 8 \, a b + {\left (3 \, b^{2} c^{3} x^{3} + 2 \, b^{2} c x\right )} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}}\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) - 2 \, {\left (3 \, a b c^{3} x^{3} + 2 \, a b c x\right )} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}}}{32 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsech(c*x))^2/x^5,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {asech}{\left (c x \right )}\right )^{2}}{x^{5}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asech(c*x))**2/x**5,x)

[Out]

Integral((a + b*asech(c*x))**2/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))^2/x^5,x, algorithm="giac")

[Out]

integrate((b*arcsech(c*x) + a)^2/x^5, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right )}^2}{x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(1/(c*x)))^2/x^5,x)

[Out]

int((a + b*acosh(1/(c*x)))^2/x^5, x)

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