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

Optimal. Leaf size=112 \[ \frac{c^2 \sinh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 a}{b}+2 \text{sech}^{-1}(c x)\right )}{b^3}-\frac{c^2 \cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \text{sech}^{-1}(c x)\right )}{b^3}+\frac{c^2 \cosh \left (2 \text{sech}^{-1}(c x)\right )}{2 b^2 \left (a+b \text{sech}^{-1}(c x)\right )}+\frac{c^2 \sinh \left (2 \text{sech}^{-1}(c x)\right )}{4 b \left (a+b \text{sech}^{-1}(c x)\right )^2} \]

[Out]

(c^2*Cosh[2*ArcSech[c*x]])/(2*b^2*(a + b*ArcSech[c*x])) + (c^2*CoshIntegral[(2*a)/b + 2*ArcSech[c*x]]*Sinh[(2*
a)/b])/b^3 + (c^2*Sinh[2*ArcSech[c*x]])/(4*b*(a + b*ArcSech[c*x])^2) - (c^2*Cosh[(2*a)/b]*SinhIntegral[(2*a)/b
+ 2*ArcSech[c*x]])/b^3

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Rubi [A]  time = 0.206464, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6285, 5448, 12, 3297, 3303, 3298, 3301} \[ \frac{c^2 \sinh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 a}{b}+2 \text{sech}^{-1}(c x)\right )}{b^3}-\frac{c^2 \cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \text{sech}^{-1}(c x)\right )}{b^3}+\frac{c^2 \cosh \left (2 \text{sech}^{-1}(c x)\right )}{2 b^2 \left (a+b \text{sech}^{-1}(c x)\right )}+\frac{c^2 \sinh \left (2 \text{sech}^{-1}(c x)\right )}{4 b \left (a+b \text{sech}^{-1}(c x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*ArcSech[c*x])^3),x]

[Out]

(c^2*Cosh[2*ArcSech[c*x]])/(2*b^2*(a + b*ArcSech[c*x])) + (c^2*CoshIntegral[(2*a)/b + 2*ArcSech[c*x]]*Sinh[(2*
a)/b])/b^3 + (c^2*Sinh[2*ArcSech[c*x]])/(4*b*(a + b*ArcSech[c*x])^2) - (c^2*Cosh[(2*a)/b]*SinhIntegral[(2*a)/b
+ 2*ArcSech[c*x]])/b^3

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])

Rule 5448

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3297

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

Rule 3303

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

Rule 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rule 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rubi steps

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

Mathematica [A]  time = 0.404743, size = 122, normalized size = 1.09 \[ \frac{\frac{b^2 \sqrt{\frac{1-c x}{c x+1}} (c x+1)}{x^2 \left (a+b \text{sech}^{-1}(c x)\right )^2}+2 c^2 \left (\sinh \left (\frac{2 a}{b}\right ) \text{Chi}\left (2 \left (\frac{a}{b}+\text{sech}^{-1}(c x)\right )\right )-\cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (2 \left (\frac{a}{b}+\text{sech}^{-1}(c x)\right )\right )\right )+\frac{b \left (2-c^2 x^2\right )}{x^2 \left (a+b \text{sech}^{-1}(c x)\right )}}{2 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a + b*ArcSech[c*x])^3),x]

[Out]

((b^2*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x))/(x^2*(a + b*ArcSech[c*x])^2) + (b*(2 - c^2*x^2))/(x^2*(a + b*ArcSec
h[c*x])) + 2*c^2*(CoshIntegral[2*(a/b + ArcSech[c*x])]*Sinh[(2*a)/b] - Cosh[(2*a)/b]*SinhIntegral[2*(a/b + Arc
Sech[c*x])]))/(2*b^3)

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Maple [B]  time = 0.293, size = 277, normalized size = 2.5 \begin{align*}{c}^{2} \left ( -{\frac{2\,b{\rm arcsech} \left (cx\right )+2\,a-b}{8\,{b}^{2}{c}^{2}{x}^{2} \left ( \left ({\rm arcsech} \left (cx\right ) \right ) ^{2}{b}^{2}+2\,{\rm arcsech} \left (cx\right )ab+{a}^{2} \right ) } \left ( 2\,\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}cx+{c}^{2}{x}^{2}-2 \right ) }-{\frac{1}{2\,{b}^{3}}{{\rm e}^{2\,{\frac{a}{b}}}}{\it Ei} \left ( 1,2\,{\frac{a}{b}}+2\,{\rm arcsech} \left (cx\right ) \right ) }-{\frac{1}{8\,{x}^{2}b{c}^{2} \left ( a+b{\rm arcsech} \left (cx\right ) \right ) ^{2}} \left ({c}^{2}{x}^{2}-2-2\,\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}cx \right ) }-{\frac{1}{4\,{b}^{2}{c}^{2}{x}^{2} \left ( a+b{\rm arcsech} \left (cx\right ) \right ) } \left ({c}^{2}{x}^{2}-2-2\,\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}cx \right ) }+{\frac{1}{2\,{b}^{3}}{{\rm e}^{-2\,{\frac{a}{b}}}}{\it Ei} \left ( 1,-2\,{\rm arcsech} \left (cx\right )-2\,{\frac{a}{b}} \right ) } \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(a+b*arcsech(c*x))^3,x)

[Out]

c^2*(-1/8*(2*(-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/2)*c*x+c^2*x^2-2)*(2*b*arcsech(c*x)+2*a-b)/c^2/x^2/b^2/(arc
sech(c*x)^2*b^2+2*arcsech(c*x)*a*b+a^2)-1/2/b^3*exp(2*a/b)*Ei(1,2*a/b+2*arcsech(c*x))-1/8/b*(c^2*x^2-2-2*(-(c*
x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/2)*c*x)/c^2/x^2/(a+b*arcsech(c*x))^2-1/4/b^2*(c^2*x^2-2-2*(-(c*x-1)/c/x)^(1/2
)*((c*x+1)/c/x)^(1/2)*c*x)/c^2/x^2/(a+b*arcsech(c*x))+1/2/b^3*exp(-2*a/b)*Ei(1,-2*arcsech(c*x)-2*a/b))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(a+b*arcsech(c*x))^3,x, algorithm="maxima")

[Out]

-1/2*((b*c^6*(2*log(c) - 1) - 2*a*c^6)*x^7 - 3*(b*c^4*(2*log(c) - 1) - 2*a*c^4)*x^5 + ((b*c^2*(2*log(c) - 1) -
2*a*c^2)*x^3 - (b*(2*log(c) - 1) - 2*a)*x + 2*(b*c^2*x^3 - b*x)*log(x))*(c*x + 1)^(3/2)*(-c*x + 1)^(3/2) + 3*
(b*c^2*(2*log(c) - 1) - 2*a*c^2)*x^3 - ((b*c^6*log(c) - a*c^6)*x^7 - (b*c^4*(5*log(c) - 2) - 5*a*c^4)*x^5 + 5*
(b*c^2*(2*log(c) - 1) - 2*a*c^2)*x^3 - 3*(b*(2*log(c) - 1) - 2*a)*x + (b*c^6*x^7 - 5*b*c^4*x^5 + 10*b*c^2*x^3
- 6*b*x)*log(x))*(c*x + 1)*(c*x - 1) + ((b*c^6*(3*log(c) - 1) - 3*a*c^6)*x^7 - (b*c^4*(11*log(c) - 5) - 11*a*c
^4)*x^5 + 7*(b*c^2*(2*log(c) - 1) - 2*a*c^2)*x^3 - 3*(b*(2*log(c) - 1) - 2*a)*x + (3*b*c^6*x^7 - 11*b*c^4*x^5
+ 14*b*c^2*x^3 - 6*b*x)*log(x))*sqrt(c*x + 1)*sqrt(-c*x + 1) - (b*(2*log(c) - 1) - 2*a)*x - (2*b*c^6*x^7 - 6*b
*c^4*x^5 + 6*b*c^2*x^3 + 2*(b*c^2*x^3 - b*x)*(c*x + 1)^(3/2)*(-c*x + 1)^(3/2) - (b*c^6*x^7 - 5*b*c^4*x^5 + 10*
b*c^2*x^3 - 6*b*x)*(c*x + 1)*(c*x - 1) + (3*b*c^6*x^7 - 11*b*c^4*x^5 + 14*b*c^2*x^3 - 6*b*x)*sqrt(c*x + 1)*sqr
t(-c*x + 1) - 2*b*x)*log(sqrt(c*x + 1)*sqrt(-c*x + 1) + 1) + 2*(b*c^6*x^7 - 3*b*c^4*x^5 + 3*b*c^2*x^3 - b*x)*l
og(x))/((b^4*c^6*x^6 - 3*b^4*c^4*x^4 + 3*b^4*c^2*x^2 - b^4)*x^3*log(x)^2 + 2*((b^4*c^6*log(c) - a*b^3*c^6)*x^6
- 3*(b^4*c^4*log(c) - a*b^3*c^4)*x^4 - b^4*log(c) + a*b^3 + 3*(b^4*c^2*log(c) - a*b^3*c^2)*x^2)*x^3*log(x) -
(b^4*x^3*log(x)^2 + 2*(b^4*log(c) - a*b^3)*x^3*log(x) + (b^4*log(c)^2 - 2*a*b^3*log(c) + a^2*b^2)*x^3)*(c*x +
1)^(3/2)*(-c*x + 1)^(3/2) + ((b^4*c^6*log(c)^2 - 2*a*b^3*c^6*log(c) + a^2*b^2*c^6)*x^6 - b^4*log(c)^2 - 3*(b^4
*c^4*log(c)^2 - 2*a*b^3*c^4*log(c) + a^2*b^2*c^4)*x^4 + 2*a*b^3*log(c) - a^2*b^2 + 3*(b^4*c^2*log(c)^2 - 2*a*b
^3*c^2*log(c) + a^2*b^2*c^2)*x^2)*x^3 - 3*((b^4*c^2*x^2 - b^4)*x^3*log(x)^2 - 2*(b^4*log(c) - a*b^3 - (b^4*c^2
*log(c) - a*b^3*c^2)*x^2)*x^3*log(x) - (b^4*log(c)^2 - 2*a*b^3*log(c) + a^2*b^2 - (b^4*c^2*log(c)^2 - 2*a*b^3*
c^2*log(c) + a^2*b^2*c^2)*x^2)*x^3)*(c*x + 1)*(c*x - 1) - ((c*x + 1)^(3/2)*(-c*x + 1)^(3/2)*b^4*x^3 + 3*(b^4*c
^2*x^2 - b^4)*(c*x + 1)*(c*x - 1)*x^3 + 3*(b^4*c^4*x^4 - 2*b^4*c^2*x^2 + b^4)*sqrt(c*x + 1)*sqrt(-c*x + 1)*x^3
- (b^4*c^6*x^6 - 3*b^4*c^4*x^4 + 3*b^4*c^2*x^2 - b^4)*x^3)*log(sqrt(c*x + 1)*sqrt(-c*x + 1) + 1)^2 - 3*((b^4*
c^4*x^4 - 2*b^4*c^2*x^2 + b^4)*x^3*log(x)^2 + 2*((b^4*c^4*log(c) - a*b^3*c^4)*x^4 + b^4*log(c) - a*b^3 - 2*(b^
4*c^2*log(c) - a*b^3*c^2)*x^2)*x^3*log(x) + (b^4*log(c)^2 + (b^4*c^4*log(c)^2 - 2*a*b^3*c^4*log(c) + a^2*b^2*c
^4)*x^4 - 2*a*b^3*log(c) + a^2*b^2 - 2*(b^4*c^2*log(c)^2 - 2*a*b^3*c^2*log(c) + a^2*b^2*c^2)*x^2)*x^3)*sqrt(c*
x + 1)*sqrt(-c*x + 1) - 2*((b^4*c^6*x^6 - 3*b^4*c^4*x^4 + 3*b^4*c^2*x^2 - b^4)*x^3*log(x) - (b^4*x^3*log(x) +
(b^4*log(c) - a*b^3)*x^3)*(c*x + 1)^(3/2)*(-c*x + 1)^(3/2) + ((b^4*c^6*log(c) - a*b^3*c^6)*x^6 - 3*(b^4*c^4*lo
g(c) - a*b^3*c^4)*x^4 - b^4*log(c) + a*b^3 + 3*(b^4*c^2*log(c) - a*b^3*c^2)*x^2)*x^3 - 3*((b^4*c^2*x^2 - b^4)*
x^3*log(x) - (b^4*log(c) - a*b^3 - (b^4*c^2*log(c) - a*b^3*c^2)*x^2)*x^3)*(c*x + 1)*(c*x - 1) - 3*((b^4*c^4*x^
4 - 2*b^4*c^2*x^2 + b^4)*x^3*log(x) + ((b^4*c^4*log(c) - a*b^3*c^4)*x^4 + b^4*log(c) - a*b^3 - 2*(b^4*c^2*log(
c) - a*b^3*c^2)*x^2)*x^3)*sqrt(c*x + 1)*sqrt(-c*x + 1))*log(sqrt(c*x + 1)*sqrt(-c*x + 1) + 1)) + integrate(-1/
2*(4*c^8*x^8 - 16*c^6*x^6 + 24*c^4*x^4 + 4*(c*x + 1)^2*(c*x - 1)^2 + (3*c^6*x^6 - 16*c^2*x^2 + 16)*(c*x + 1)^(
3/2)*(-c*x + 1)^(3/2) - 16*c^2*x^2 - 24*(c^4*x^4 - 2*c^2*x^2 + 1)*(c*x + 1)*(c*x - 1) + (3*c^8*x^8 - 19*c^6*x^
6 + 48*c^4*x^4 - 48*c^2*x^2 + 16)*sqrt(c*x + 1)*sqrt(-c*x + 1) + 4)/((b^3*x^3*log(x) + (b^3*log(c) - a*b^2)*x^
3)*(c*x + 1)^2*(c*x - 1)^2 + (b^3*c^8*x^8 - 4*b^3*c^6*x^6 + 6*b^3*c^4*x^4 - 4*b^3*c^2*x^2 + b^3)*x^3*log(x) -
4*((b^3*c^2*x^2 - b^3)*x^3*log(x) - (b^3*log(c) - a*b^2 - (b^3*c^2*log(c) - a*b^2*c^2)*x^2)*x^3)*(c*x + 1)^(3/
2)*(-c*x + 1)^(3/2) + ((b^3*c^8*log(c) - a*b^2*c^8)*x^8 - 4*(b^3*c^6*log(c) - a*b^2*c^6)*x^6 + 6*(b^3*c^4*log(
c) - a*b^2*c^4)*x^4 + b^3*log(c) - a*b^2 - 4*(b^3*c^2*log(c) - a*b^2*c^2)*x^2)*x^3 - 6*((b^3*c^4*x^4 - 2*b^3*c
^2*x^2 + b^3)*x^3*log(x) + ((b^3*c^4*log(c) - a*b^2*c^4)*x^4 + b^3*log(c) - a*b^2 - 2*(b^3*c^2*log(c) - a*b^2*
c^2)*x^2)*x^3)*(c*x + 1)*(c*x - 1) - 4*((b^3*c^6*x^6 - 3*b^3*c^4*x^4 + 3*b^3*c^2*x^2 - b^3)*x^3*log(x) + ((b^3
*c^6*log(c) - a*b^2*c^6)*x^6 - 3*(b^3*c^4*log(c) - a*b^2*c^4)*x^4 - b^3*log(c) + a*b^2 + 3*(b^3*c^2*log(c) - a
*b^2*c^2)*x^2)*x^3)*sqrt(c*x + 1)*sqrt(-c*x + 1) - ((c*x + 1)^2*(c*x - 1)^2*b^3*x^3 - 4*(b^3*c^2*x^2 - b^3)*(c
*x + 1)^(3/2)*(-c*x + 1)^(3/2)*x^3 - 6*(b^3*c^4*x^4 - 2*b^3*c^2*x^2 + b^3)*(c*x + 1)*(c*x - 1)*x^3 - 4*(b^3*c^
6*x^6 - 3*b^3*c^4*x^4 + 3*b^3*c^2*x^2 - b^3)*sqrt(c*x + 1)*sqrt(-c*x + 1)*x^3 + (b^3*c^8*x^8 - 4*b^3*c^6*x^6 +
6*b^3*c^4*x^4 - 4*b^3*c^2*x^2 + b^3)*x^3)*log(sqrt(c*x + 1)*sqrt(-c*x + 1) + 1)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{b^{3} x^{3} \operatorname{arsech}\left (c x\right )^{3} + 3 \, a b^{2} x^{3} \operatorname{arsech}\left (c x\right )^{2} + 3 \, a^{2} b x^{3} \operatorname{arsech}\left (c x\right ) + a^{3} x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(a+b*arcsech(c*x))^3,x, algorithm="fricas")

[Out]

integral(1/(b^3*x^3*arcsech(c*x)^3 + 3*a*b^2*x^3*arcsech(c*x)^2 + 3*a^2*b*x^3*arcsech(c*x) + a^3*x^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{3} \left (a + b \operatorname{asech}{\left (c x \right )}\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(a+b*asech(c*x))**3,x)

[Out]

Integral(1/(x**3*(a + b*asech(c*x))**3), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \operatorname{arsech}\left (c x\right ) + a\right )}^{3} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(a+b*arcsech(c*x))^3,x, algorithm="giac")

[Out]

integrate(1/((b*arcsech(c*x) + a)^3*x^3), x)