3.1.40 \(\int \frac {(a+b \text {sech}^{-1}(c x))^2}{x^4} \, dx\) [40]

Optimal. Leaf size=122 \[ -\frac {2 b^2}{27 x^3}-\frac {4 b^2 c^2}{9 x}+\frac {2 b \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )}{9 x^3}+\frac {4 b c^2 \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )}{9 x}-\frac {\left (a+b \text {sech}^{-1}(c x)\right )^2}{3 x^3} \]

[Out]

-2/27*b^2/x^3-4/9*b^2*c^2/x-1/3*(a+b*arcsech(c*x))^2/x^3+2/9*b*(c*x+1)*(a+b*arcsech(c*x))*((-c*x+1)/(c*x+1))^(
1/2)/x^3+4/9*b*c^2*(c*x+1)*(a+b*arcsech(c*x))*((-c*x+1)/(c*x+1))^(1/2)/x

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Rubi [A]
time = 0.08, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {6420, 5555, 3391, 3377, 2718} \begin {gather*} \frac {4 b c^2 \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (a+b \text {sech}^{-1}(c x)\right )}{9 x}+\frac {2 b \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (a+b \text {sech}^{-1}(c x)\right )}{9 x^3}-\frac {\left (a+b \text {sech}^{-1}(c x)\right )^2}{3 x^3}-\frac {4 b^2 c^2}{9 x}-\frac {2 b^2}{27 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSech[c*x])^2/x^4,x]

[Out]

(-2*b^2)/(27*x^3) - (4*b^2*c^2)/(9*x) + (2*b*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(a + b*ArcSech[c*x]))/(9*x^3)
+ (4*b*c^2*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(a + b*ArcSech[c*x]))/(9*x) - (a + b*ArcSech[c*x])^2/(3*x^3)

Rule 2718

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

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^4} \, dx &=-\left (c^3 \text {Subst}\left (\int (a+b x)^2 \cosh ^2(x) \sinh (x) \, dx,x,\text {sech}^{-1}(c x)\right )\right )\\ &=-\frac {\left (a+b \text {sech}^{-1}(c x)\right )^2}{3 x^3}+\frac {1}{3} \left (2 b c^3\right ) \text {Subst}\left (\int (a+b x) \cosh ^3(x) \, dx,x,\text {sech}^{-1}(c x)\right )\\ &=-\frac {2 b^2}{27 x^3}+\frac {2 b \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )}{9 x^3}-\frac {\left (a+b \text {sech}^{-1}(c x)\right )^2}{3 x^3}+\frac {1}{9} \left (4 b c^3\right ) \text {Subst}\left (\int (a+b x) \cosh (x) \, dx,x,\text {sech}^{-1}(c x)\right )\\ &=-\frac {2 b^2}{27 x^3}+\frac {2 b \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )}{9 x^3}+\frac {4 b c^2 \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )}{9 x}-\frac {\left (a+b \text {sech}^{-1}(c x)\right )^2}{3 x^3}-\frac {1}{9} \left (4 b^2 c^3\right ) \text {Subst}\left (\int \sinh (x) \, dx,x,\text {sech}^{-1}(c x)\right )\\ &=-\frac {2 b^2}{27 x^3}-\frac {4 b^2 c^2}{9 x}+\frac {2 b \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )}{9 x^3}+\frac {4 b c^2 \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )}{9 x}-\frac {\left (a+b \text {sech}^{-1}(c x)\right )^2}{3 x^3}\\ \end {align*}

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Mathematica [A]
time = 0.16, size = 134, normalized size = 1.10 \begin {gather*} \frac {-9 a^2-2 b^2 \left (1+6 c^2 x^2\right )+6 a b \sqrt {\frac {1-c x}{1+c x}} \left (1+c x+2 c^2 x^2+2 c^3 x^3\right )+6 b \left (-3 a+b \sqrt {\frac {1-c x}{1+c x}} \left (1+c x+2 c^2 x^2+2 c^3 x^3\right )\right ) \text {sech}^{-1}(c x)-9 b^2 \text {sech}^{-1}(c x)^2}{27 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*a^2 - 2*b^2*(1 + 6*c^2*x^2) + 6*a*b*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x + 2*c^2*x^2 + 2*c^3*x^3) + 6*b*(-3*
a + b*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x + 2*c^2*x^2 + 2*c^3*x^3))*ArcSech[c*x] - 9*b^2*ArcSech[c*x]^2)/(27*x^
3)

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Maple [A]
time = 0.28, size = 192, normalized size = 1.57

method	result	size
derivativedivides	\(c^{3} \left (-\frac {a^{2}}{3 c^{3} x^{3}}+b^{2} \left (-\frac {\mathrm {arcsech}\left (c x \right )^{2}}{3 c^{3} x^{3}}+\frac {4 \,\mathrm {arcsech}\left (c x \right ) \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}}{9}+\frac {2 \,\mathrm {arcsech}\left (c x \right ) \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}}{9 c^{2} x^{2}}-\frac {4}{9 c x}-\frac {2}{27 c^{3} x^{3}}\right )+2 a b \left (-\frac {\mathrm {arcsech}\left (c x \right )}{3 c^{3} x^{3}}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, \left (2 c^{2} x^{2}+1\right )}{9 c^{2} x^{2}}\right )\right )\)	\(192\)
default	\(c^{3} \left (-\frac {a^{2}}{3 c^{3} x^{3}}+b^{2} \left (-\frac {\mathrm {arcsech}\left (c x \right )^{2}}{3 c^{3} x^{3}}+\frac {4 \,\mathrm {arcsech}\left (c x \right ) \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}}{9}+\frac {2 \,\mathrm {arcsech}\left (c x \right ) \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}}{9 c^{2} x^{2}}-\frac {4}{9 c x}-\frac {2}{27 c^{3} x^{3}}\right )+2 a b \left (-\frac {\mathrm {arcsech}\left (c x \right )}{3 c^{3} x^{3}}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, \left (2 c^{2} x^{2}+1\right )}{9 c^{2} x^{2}}\right )\right )\)	\(192\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsech(c*x))^2/x^4,x,method=_RETURNVERBOSE)

[Out]

c^3*(-1/3*a^2/c^3/x^3+b^2*(-1/3*arcsech(c*x)^2/c^3/x^3+4/9*arcsech(c*x)*(-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/
2)+2/9*arcsech(c*x)/c^2/x^2*(-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/2)-4/9/c/x-2/27/c^3/x^3)+2*a*b*(-1/3/c^3/x^3
*arcsech(c*x)+1/9*(-(c*x-1)/c/x)^(1/2)/c^2/x^2*((c*x+1)/c/x)^(1/2)*(2*c^2*x^2+1)))

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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^4,x, algorithm="maxima")

[Out]

2/9*a*b*((c^4*(1/(c^2*x^2) - 1)^(3/2) + 3*c^4*sqrt(1/(c^2*x^2) - 1))/c - 3*arcsech(c*x)/x^3) + b^2*integrate(l
og(sqrt(1/(c*x) + 1)*sqrt(1/(c*x) - 1) + 1/(c*x))^2/x^4, x) - 1/3*a^2/x^3

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Fricas [A]
time = 0.35, size = 181, normalized size = 1.48 \begin {gather*} -\frac {12 \, b^{2} c^{2} x^{2} + 9 \, b^{2} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right )^{2} + 9 \, a^{2} + 2 \, b^{2} + 6 \, {\left (3 \, a b - {\left (2 \, b^{2} c^{3} x^{3} + 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 ) - 6 \, {\left (2 \, a b c^{3} x^{3} + a b c x\right )} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}}}{27 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/27*(12*b^2*c^2*x^2 + 9*b^2*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x))^2 + 9*a^2 + 2*b^2 + 6*(3*a*b
- (2*b^2*c^3*x^3 + 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
)) - 6*(2*a*b*c^3*x^3 + a*b*c*x)*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)))/x^3

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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^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asech(c*x))**2/x**4, 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^4,x, algorithm="giac")

[Out]

integrate((b*arcsech(c*x) + a)^2/x^4, 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^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(1/(c*x)))^2/x^4,x)

[Out]

int((a + b*acosh(1/(c*x)))^2/x^4, x)

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