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

Optimal. Leaf size=163 \[ -\frac {3 b^2 (1-c x) (c x+1) \left (a+b \text {sech}^{-1}(c x)\right )}{4 x^2}-\frac {1}{4} c^2 \left (a+b \text {sech}^{-1}(c x)\right )^3+\frac {3 b \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (a+b \text {sech}^{-1}(c x)\right )^2}{4 x^2}-\frac {(1-c x) (c x+1) \left (a+b \text {sech}^{-1}(c x)\right )^3}{2 x^2}-\frac {3}{8} b^3 c^2 \text {sech}^{-1}(c x)+\frac {3 b^3 \sqrt {\frac {1-c x}{c x+1}} (c x+1)}{8 x^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {6285, 5446, 3311, 32, 2635, 8} \[ -\frac {3 b^2 (1-c x) (c x+1) \left (a+b \text {sech}^{-1}(c x)\right )}{4 x^2}-\frac {1}{4} c^2 \left (a+b \text {sech}^{-1}(c x)\right )^3+\frac {3 b \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (a+b \text {sech}^{-1}(c x)\right )^2}{4 x^2}-\frac {(1-c x) (c x+1) \left (a+b \text {sech}^{-1}(c x)\right )^3}{2 x^2}-\frac {3}{8} b^3 c^2 \text {sech}^{-1}(c x)+\frac {3 b^3 \sqrt {\frac {1-c x}{c x+1}} (c x+1)}{8 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSech[c*x])^3/x^3,x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2635

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

Rule 3311

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

Rule 5446

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

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

________________________________________________________________________________________

Mathematica [A]  time = 0.48, size = 245, normalized size = 1.50 \[ \frac {-4 a^3-3 b c^2 x^2 \left (2 a^2+b^2\right ) \log (x)+3 b c^2 x^2 \left (2 a^2+b^2\right ) \log \left (c x \sqrt {\frac {1-c x}{c x+1}}+\sqrt {\frac {1-c x}{c x+1}}+1\right )+3 b \left (2 a^2+b^2\right ) \sqrt {\frac {1-c x}{c x+1}} (c x+1)-6 b \text {sech}^{-1}(c x) \left (2 a^2-2 a b \sqrt {\frac {1-c x}{c x+1}} (c x+1)+b^2\right )+6 b^2 \text {sech}^{-1}(c x)^2 \left (a \left (c^2 x^2-2\right )+b \sqrt {\frac {1-c x}{c x+1}} (c x+1)\right )-6 a b^2+2 b^3 \left (c^2 x^2-2\right ) \text {sech}^{-1}(c x)^3}{8 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.48, size = 271, normalized size = 1.66 \[ \frac {2 \, {\left (b^{3} c^{2} x^{2} - 2 \, b^{3}\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right )^{3} + 3 \, {\left (2 \, a^{2} b + b^{3}\right )} c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - 4 \, a^{3} - 6 \, a b^{2} + 6 \, {\left (a b^{2} c^{2} x^{2} + b^{3} c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - 2 \, a b^{2}\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right )^{2} + 3 \, {\left (4 \, a b^{2} c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + {\left (2 \, a^{2} b + b^{3}\right )} c^{2} x^{2} - 4 \, a^{2} b - 2 \, b^{3}\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right )}{8 \, x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(2*(b^3*c^2*x^2 - 2*b^3)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x))^3 + 3*(2*a^2*b + b^3)*c*x*sqr
t(-(c^2*x^2 - 1)/(c^2*x^2)) - 4*a^3 - 6*a*b^2 + 6*(a*b^2*c^2*x^2 + b^3*c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) - 2*
a*b^2)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x))^2 + 3*(4*a*b^2*c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) +
(2*a^2*b + b^3)*c^2*x^2 - 4*a^2*b - 2*b^3)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)))/x^2

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.17, size = 321, normalized size = 1.97 \[ c^{2} \left (-\frac {a^{3}}{2 c^{2} x^{2}}+b^{3} \left (-\frac {\mathrm {arcsech}\left (c x \right )^{3}}{2 c^{2} x^{2}}+\frac {3 \mathrm {arcsech}\left (c x \right )^{2} \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}}{4 c x}+\frac {\mathrm {arcsech}\left (c x \right )^{3}}{4}-\frac {3 \,\mathrm {arcsech}\left (c x \right )}{4 c^{2} x^{2}}+\frac {3 \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}}{8 c x}+\frac {3 \,\mathrm {arcsech}\left (c x \right )}{8}\right )+3 a \,b^{2} \left (-\frac {\mathrm {arcsech}\left (c x \right )^{2}}{2 c^{2} x^{2}}+\frac {\mathrm {arcsech}\left (c x \right ) \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}}{2 c x}+\frac {\mathrm {arcsech}\left (c x \right )^{2}}{4}-\frac {1}{4 c^{2} x^{2}}\right )+3 a^{2} b \left (-\frac {\mathrm {arcsech}\left (c x \right )}{2 c^{2} x^{2}}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, \left (\arctanh \left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right ) c^{2} x^{2}+\sqrt {-c^{2} x^{2}+1}\right )}{4 c x \sqrt {-c^{2} x^{2}+1}}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c^2*(-1/2*a^3/c^2/x^2+b^3*(-1/2/c^2/x^2*arcsech(c*x)^3+3/4*arcsech(c*x)^2/c/x*(-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/
x)^(1/2)+1/4*arcsech(c*x)^3-3/4/c^2/x^2*arcsech(c*x)+3/8/c/x*(-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/2)+3/8*arcs
ech(c*x))+3*a*b^2*(-1/2/c^2/x^2*arcsech(c*x)^2+1/2*arcsech(c*x)/c/x*(-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/2)+1
/4*arcsech(c*x)^2-1/4/c^2/x^2)+3*a^2*b*(-1/2/c^2/x^2*arcsech(c*x)+1/4*(-(c*x-1)/c/x)^(1/2)/c/x*((c*x+1)/c/x)^(
1/2)*(arctanh(1/(-c^2*x^2+1)^(1/2))*c^2*x^2+(-c^2*x^2+1)^(1/2))/(-c^2*x^2+1)^(1/2)))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {3}{8} \, a^{2} b {\left (\frac {\frac {2 \, c^{4} x \sqrt {\frac {1}{c^{2} x^{2}} - 1}}{c^{2} x^{2} {\left (\frac {1}{c^{2} x^{2}} - 1\right )} - 1} - c^{3} \log \left (c x \sqrt {\frac {1}{c^{2} x^{2}} - 1} + 1\right ) + c^{3} \log \left (c x \sqrt {\frac {1}{c^{2} x^{2}} - 1} - 1\right )}{c} + \frac {4 \, \operatorname {arsech}\left (c x\right )}{x^{2}}\right )} - \frac {a^{3}}{2 \, x^{2}} + \int \frac {b^{3} \log \left (\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}\right )^{3}}{x^{3}} + \frac {3 \, a b^{2} \log \left (\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}\right )^{2}}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right )}^3}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(1/(c*x)))^3/x^3,x)

[Out]

int((a + b*acosh(1/(c*x)))^3/x^3, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________