3.50 \(\int \frac {(a+b \text {sech}^{-1}(c x))^3}{x^5} \, dx\)

Optimal. Leaf size=242 \[ -\frac {9 b^2 c^2 \left (a+b \text {sech}^{-1}(c x)\right )}{32 x^2}-\frac {3 b^2 \left (a+b \text {sech}^{-1}(c x)\right )}{32 x^4}+\frac {3}{32} c^4 \left (a+b \text {sech}^{-1}(c x)\right )^3+\frac {9 b c^2 \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (a+b \text {sech}^{-1}(c x)\right )^2}{32 x^2}-\frac {\left (a+b \text {sech}^{-1}(c x)\right )^3}{4 x^4}+\frac {3 b \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (a+b \text {sech}^{-1}(c x)\right )^2}{16 x^4}+\frac {45}{256} b^3 c^4 \text {sech}^{-1}(c x)+\frac {45 b^3 c^2 \sqrt {\frac {1-c x}{c x+1}} (c x+1)}{256 x^2}+\frac {3 b^3 \sqrt {\frac {1-c x}{c x+1}} (c x+1)}{128 x^4} \]

[Out]

45/256*b^3*c^4*arcsech(c*x)-3/32*b^2*(a+b*arcsech(c*x))/x^4-9/32*b^2*c^2*(a+b*arcsech(c*x))/x^2+3/32*c^4*(a+b*
arcsech(c*x))^3-1/4*(a+b*arcsech(c*x))^3/x^4+3/128*b^3*(c*x+1)*((-c*x+1)/(c*x+1))^(1/2)/x^4+45/256*b^3*c^2*(c*
x+1)*((-c*x+1)/(c*x+1))^(1/2)/x^2+3/16*b*(c*x+1)*(a+b*arcsech(c*x))^2*((-c*x+1)/(c*x+1))^(1/2)/x^4+9/32*b*c^2*
(c*x+1)*(a+b*arcsech(c*x))^2*((-c*x+1)/(c*x+1))^(1/2)/x^2

________________________________________________________________________________________

Rubi [A]  time = 0.20, antiderivative size = 242, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 6, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {6285, 5447, 3311, 32, 2635, 8} \[ -\frac {9 b^2 c^2 \left (a+b \text {sech}^{-1}(c x)\right )}{32 x^2}-\frac {3 b^2 \left (a+b \text {sech}^{-1}(c x)\right )}{32 x^4}+\frac {9 b c^2 \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (a+b \text {sech}^{-1}(c x)\right )^2}{32 x^2}+\frac {3}{32} c^4 \left (a+b \text {sech}^{-1}(c x)\right )^3-\frac {\left (a+b \text {sech}^{-1}(c x)\right )^3}{4 x^4}+\frac {3 b \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (a+b \text {sech}^{-1}(c x)\right )^2}{16 x^4}+\frac {45 b^3 c^2 \sqrt {\frac {1-c x}{c x+1}} (c x+1)}{256 x^2}+\frac {45}{256} b^3 c^4 \text {sech}^{-1}(c x)+\frac {3 b^3 \sqrt {\frac {1-c x}{c x+1}} (c x+1)}{128 x^4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSech[c*x])^3/x^5,x]

[Out]

(3*b^3*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x))/(128*x^4) + (45*b^3*c^2*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x))/(256*
x^2) + (45*b^3*c^4*ArcSech[c*x])/256 - (3*b^2*(a + b*ArcSech[c*x]))/(32*x^4) - (9*b^2*c^2*(a + b*ArcSech[c*x])
)/(32*x^2) + (3*b*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(a + b*ArcSech[c*x])^2)/(16*x^4) + (9*b*c^2*Sqrt[(1 - c*
x)/(1 + c*x)]*(1 + c*x)*(a + b*ArcSech[c*x])^2)/(32*x^2) + (3*c^4*(a + b*ArcSech[c*x])^3)/32 - (a + b*ArcSech[
c*x])^3/(4*x^4)

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 5447

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 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^5} \, dx &=-\left (c^4 \operatorname {Subst}\left (\int (a+b x)^3 \cosh ^3(x) \sinh (x) \, dx,x,\text {sech}^{-1}(c x)\right )\right )\\ &=-\frac {\left (a+b \text {sech}^{-1}(c x)\right )^3}{4 x^4}+\frac {1}{4} \left (3 b c^4\right ) \operatorname {Subst}\left (\int (a+b x)^2 \cosh ^4(x) \, dx,x,\text {sech}^{-1}(c x)\right )\\ &=-\frac {3 b^2 \left (a+b \text {sech}^{-1}(c x)\right )}{32 x^4}+\frac {3 b \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )^2}{16 x^4}-\frac {\left (a+b \text {sech}^{-1}(c x)\right )^3}{4 x^4}+\frac {1}{16} \left (9 b c^4\right ) \operatorname {Subst}\left (\int (a+b x)^2 \cosh ^2(x) \, dx,x,\text {sech}^{-1}(c x)\right )+\frac {1}{32} \left (3 b^3 c^4\right ) \operatorname {Subst}\left (\int \cosh ^4(x) \, dx,x,\text {sech}^{-1}(c x)\right )\\ &=\frac {3 b^3 \sqrt {\frac {1-c x}{1+c x}} (1+c x)}{128 x^4}-\frac {3 b^2 \left (a+b \text {sech}^{-1}(c x)\right )}{32 x^4}-\frac {9 b^2 c^2 \left (a+b \text {sech}^{-1}(c x)\right )}{32 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}{16 x^4}+\frac {9 b c^2 \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )^2}{32 x^2}-\frac {\left (a+b \text {sech}^{-1}(c x)\right )^3}{4 x^4}+\frac {1}{32} \left (9 b c^4\right ) \operatorname {Subst}\left (\int (a+b x)^2 \, dx,x,\text {sech}^{-1}(c x)\right )+\frac {1}{128} \left (9 b^3 c^4\right ) \operatorname {Subst}\left (\int \cosh ^2(x) \, dx,x,\text {sech}^{-1}(c x)\right )+\frac {1}{32} \left (9 b^3 c^4\right ) \operatorname {Subst}\left (\int \cosh ^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)}{128 x^4}+\frac {45 b^3 c^2 \sqrt {\frac {1-c x}{1+c x}} (1+c x)}{256 x^2}-\frac {3 b^2 \left (a+b \text {sech}^{-1}(c x)\right )}{32 x^4}-\frac {9 b^2 c^2 \left (a+b \text {sech}^{-1}(c x)\right )}{32 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}{16 x^4}+\frac {9 b c^2 \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )^2}{32 x^2}+\frac {3}{32} c^4 \left (a+b \text {sech}^{-1}(c x)\right )^3-\frac {\left (a+b \text {sech}^{-1}(c x)\right )^3}{4 x^4}+\frac {1}{256} \left (9 b^3 c^4\right ) \operatorname {Subst}\left (\int 1 \, dx,x,\text {sech}^{-1}(c x)\right )+\frac {1}{64} \left (9 b^3 c^4\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)}{128 x^4}+\frac {45 b^3 c^2 \sqrt {\frac {1-c x}{1+c x}} (1+c x)}{256 x^2}+\frac {45}{256} b^3 c^4 \text {sech}^{-1}(c x)-\frac {3 b^2 \left (a+b \text {sech}^{-1}(c x)\right )}{32 x^4}-\frac {9 b^2 c^2 \left (a+b \text {sech}^{-1}(c x)\right )}{32 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}{16 x^4}+\frac {9 b c^2 \sqrt {\frac {1-c x}{1+c x}} (1+c x) \left (a+b \text {sech}^{-1}(c x)\right )^2}{32 x^2}+\frac {3}{32} c^4 \left (a+b \text {sech}^{-1}(c x)\right )^3-\frac {\left (a+b \text {sech}^{-1}(c x)\right )^3}{4 x^4}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.79, size = 332, normalized size = 1.37 \[ \frac {-9 b c^4 x^4 \left (8 a^2+5 b^2\right ) \log (x)+9 b c^4 x^4 \left (8 a^2+5 b^2\right ) \log \left (c x \sqrt {\frac {1-c x}{c x+1}}+\sqrt {\frac {1-c x}{c x+1}}+1\right )+3 b \sqrt {\frac {1-c x}{c x+1}} (c x+1) \left (8 a^2 \left (3 c^2 x^2+2\right )+b^2 \left (15 c^2 x^2+2\right )\right )-24 b \text {sech}^{-1}(c x) \left (8 a^2-2 a b \sqrt {\frac {1-c x}{c x+1}} \left (3 c^3 x^3+3 c^2 x^2+2 c x+2\right )+b^2 \left (3 c^2 x^2+1\right )\right )-8 a \left (8 a^2+3 b^2\right )-72 a b^2 c^2 x^2+24 b^2 \text {sech}^{-1}(c x)^2 \left (a \left (3 c^4 x^4-8\right )+b \sqrt {\frac {1-c x}{c x+1}} \left (3 c^3 x^3+3 c^2 x^2+2 c x+2\right )\right )+8 b^3 \left (3 c^4 x^4-8\right ) \text {sech}^{-1}(c x)^3}{256 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSech[c*x])^3/x^5,x]

[Out]

(-8*a*(8*a^2 + 3*b^2) - 72*a*b^2*c^2*x^2 + 3*b*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)*(8*a^2*(2 + 3*c^2*x^2) + b^
2*(2 + 15*c^2*x^2)) - 24*b*(8*a^2 + b^2*(1 + 3*c^2*x^2) - 2*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] + 24*b^2*(b*Sqrt[(1 - c*x)/(1 + c*x)]*(2 + 2*c*x + 3*c^2*x^2 + 3*c^3*x^3) + a*(-
8 + 3*c^4*x^4))*ArcSech[c*x]^2 + 8*b^3*(-8 + 3*c^4*x^4)*ArcSech[c*x]^3 - 9*b*(8*a^2 + 5*b^2)*c^4*x^4*Log[x] +
9*b*(8*a^2 + 5*b^2)*c^4*x^4*Log[1 + Sqrt[(1 - c*x)/(1 + c*x)] + c*x*Sqrt[(1 - c*x)/(1 + c*x)]])/(256*x^4)

________________________________________________________________________________________

fricas [A]  time = 0.60, size = 351, normalized size = 1.45 \[ -\frac {72 \, a b^{2} c^{2} x^{2} - 8 \, {\left (3 \, b^{3} c^{4} x^{4} - 8 \, b^{3}\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right )^{3} + 64 \, a^{3} + 24 \, a b^{2} - 24 \, {\left (3 \, a b^{2} c^{4} x^{4} - 8 \, a b^{2} + {\left (3 \, b^{3} c^{3} x^{3} + 2 \, b^{3} c x\right )} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}}\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right )^{2} - 3 \, {\left (3 \, {\left (8 \, a^{2} b + 5 \, b^{3}\right )} c^{4} x^{4} - 24 \, b^{3} c^{2} x^{2} - 64 \, a^{2} b - 8 \, b^{3} + 16 \, {\left (3 \, a b^{2} c^{3} x^{3} + 2 \, a b^{2} c x\right )} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}}\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) - 3 \, {\left (3 \, {\left (8 \, a^{2} b + 5 \, b^{3}\right )} c^{3} x^{3} + 2 \, {\left (8 \, a^{2} b + b^{3}\right )} c x\right )} \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}}}{256 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsech(c*x))^3/x^5,x, algorithm="fricas")

[Out]

-1/256*(72*a*b^2*c^2*x^2 - 8*(3*b^3*c^4*x^4 - 8*b^3)*log((c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x))^3 + 6
4*a^3 + 24*a*b^2 - 24*(3*a*b^2*c^4*x^4 - 8*a*b^2 + (3*b^3*c^3*x^3 + 2*b^3*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*(3*(8*a^2*b + 5*b^3)*c^4*x^4 - 24*b^3*c^2*x^2 - 64*
a^2*b - 8*b^3 + 16*(3*a*b^2*c^3*x^3 + 2*a*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)) - 3*(3*(8*a^2*b + 5*b^3)*c^3*x^3 + 2*(8*a^2*b + b^3)*c*x)*sqrt(-(c^2*x^2 - 1)/(c^2*x^
2)))/x^4

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arsech}\left (c x\right ) + a\right )}^{3}}{x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsech(c*x))^3/x^5,x, algorithm="giac")

[Out]

integrate((b*arcsech(c*x) + a)^3/x^5, x)

________________________________________________________________________________________

maple [B]  time = 0.48, size = 485, normalized size = 2.00 \[ c^{4} \left (-\frac {a^{3}}{4 c^{4} x^{4}}+b^{3} \left (-\frac {\mathrm {arcsech}\left (c x \right )^{3}}{4 c^{4} x^{4}}+\frac {3 \mathrm {arcsech}\left (c x \right )^{2} \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}}{16 c^{3} x^{3}}+\frac {9 \mathrm {arcsech}\left (c x \right )^{2} \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}}{32 c x}+\frac {3 \mathrm {arcsech}\left (c x \right )^{3}}{32}-\frac {3 \,\mathrm {arcsech}\left (c x \right )}{32 c^{4} x^{4}}+\frac {3 \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}}{128 c^{3} x^{3}}+\frac {45 \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}}{256 c x}+\frac {45 \,\mathrm {arcsech}\left (c x \right )}{256}-\frac {9 \,\mathrm {arcsech}\left (c x \right )}{32 c^{2} x^{2}}\right )+3 a \,b^{2} \left (-\frac {\mathrm {arcsech}\left (c x \right )^{2}}{4 c^{4} x^{4}}+\frac {\mathrm {arcsech}\left (c x \right ) \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}}{8 c^{3} x^{3}}+\frac {3 \,\mathrm {arcsech}\left (c x \right ) \sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}}{16 c x}+\frac {3 \mathrm {arcsech}\left (c x \right )^{2}}{32}-\frac {1}{32 c^{4} x^{4}}-\frac {3}{32 c^{2} x^{2}}\right )+3 a^{2} b \left (-\frac {\mathrm {arcsech}\left (c x \right )}{4 c^{4} x^{4}}+\frac {\sqrt {-\frac {c x -1}{c x}}\, \sqrt {\frac {c x +1}{c x}}\, \left (3 \arctanh \left (\frac {1}{\sqrt {-c^{2} x^{2}+1}}\right ) c^{4} x^{4}+3 c^{2} x^{2} \sqrt {-c^{2} x^{2}+1}+2 \sqrt {-c^{2} x^{2}+1}\right )}{32 c^{3} x^{3} \sqrt {-c^{2} x^{2}+1}}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsech(c*x))^3/x^5,x)

[Out]

c^4*(-1/4*a^3/c^4/x^4+b^3*(-1/4/c^4/x^4*arcsech(c*x)^3+3/16*arcsech(c*x)^2/c^3/x^3*(-(c*x-1)/c/x)^(1/2)*((c*x+
1)/c/x)^(1/2)+9/32*arcsech(c*x)^2/c/x*(-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/2)+3/32*arcsech(c*x)^3-3/32/c^4/x^
4*arcsech(c*x)+3/128/c^3/x^3*(-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/2)+45/256/c/x*(-(c*x-1)/c/x)^(1/2)*((c*x+1)
/c/x)^(1/2)+45/256*arcsech(c*x)-9/32/c^2/x^2*arcsech(c*x))+3*a*b^2*(-1/4/c^4/x^4*arcsech(c*x)^2+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^4-3/32/c^2/x^2)+3*a^2*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*(-c^2*x^2+1)^(1/2)+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 \[ \frac {3}{64} \, a^{2} b {\left (\frac {3 \, c^{5} \log \left (c x \sqrt {\frac {1}{c^{2} x^{2}} - 1} + 1\right ) - 3 \, c^{5} \log \left (c x \sqrt {\frac {1}{c^{2} x^{2}} - 1} - 1\right ) - \frac {2 \, {\left (3 \, c^{8} x^{3} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{\frac {3}{2}} - 5 \, c^{6} x \sqrt {\frac {1}{c^{2} x^{2}} - 1}\right )}}{c^{4} x^{4} {\left (\frac {1}{c^{2} x^{2}} - 1\right )}^{2} - 2 \, c^{2} x^{2} {\left (\frac {1}{c^{2} x^{2}} - 1\right )} + 1}}{c} - \frac {16 \, \operatorname {arsech}\left (c x\right )}{x^{4}}\right )} - \frac {a^{3}}{4 \, x^{4}} + \int \frac {b^{3} \log \left (\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}\right )^{3}}{x^{5}} + \frac {3 \, a b^{2} \log \left (\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}\right )^{2}}{x^{5}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsech(c*x))^3/x^5,x, algorithm="maxima")

[Out]

3/64*a^2*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^3/x^4 + integrate(b^3*log(sqrt(1/(c*x) + 1)*sqrt(1/(c*x) - 1) +
1/(c*x))^3/x^5 + 3*a*b^2*log(sqrt(1/(c*x) + 1)*sqrt(1/(c*x) - 1) + 1/(c*x))^2/x^5, x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right )}^3}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*acosh(1/(c*x)))^3/x^5,x)

[Out]

int((a + b*acosh(1/(c*x)))^3/x^5, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {asech}{\left (c x \right )}\right )^{3}}{x^{5}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asech(c*x))**3/x**5,x)

[Out]

Integral((a + b*asech(c*x))**3/x**5, x)

________________________________________________________________________________________