3.66 \(\int \frac{1}{x^2 (a+b \text{sech}^{-1}(c x))^3} \, dx\)
Optimal. Leaf size=114 \[ \frac{c \sinh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\text{sech}^{-1}(c x)\right )}{2 b^3}-\frac{c \cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\text{sech}^{-1}(c x)\right )}{2 b^3}+\frac{1}{2 b^2 x \left (a+b \text{sech}^{-1}(c x)\right )}+\frac{\sqrt{\frac{1-c x}{c x+1}} (c x+1)}{2 b x \left (a+b \text{sech}^{-1}(c x)\right )^2} \]
[Out]
(Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x))/(2*b*x*(a + b*ArcSech[c*x])^2) + 1/(2*b^2*x*(a + b*ArcSech[c*x])) + (c*C
oshIntegral[a/b + ArcSech[c*x]]*Sinh[a/b])/(2*b^3) - (c*Cosh[a/b]*SinhIntegral[a/b + ArcSech[c*x]])/(2*b^3)
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Rubi [A] time = 0.175679, antiderivative size = 114, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used =
{6285, 3297, 3303, 3298, 3301} \[ \frac{c \sinh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\text{sech}^{-1}(c x)\right )}{2 b^3}-\frac{c \cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\text{sech}^{-1}(c x)\right )}{2 b^3}+\frac{1}{2 b^2 x \left (a+b \text{sech}^{-1}(c x)\right )}+\frac{\sqrt{\frac{1-c x}{c x+1}} (c x+1)}{2 b x \left (a+b \text{sech}^{-1}(c x)\right )^2} \]
Antiderivative was successfully verified.
[In]
Int[1/(x^2*(a + b*ArcSech[c*x])^3),x]
[Out]
(Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x))/(2*b*x*(a + b*ArcSech[c*x])^2) + 1/(2*b^2*x*(a + b*ArcSech[c*x])) + (c*C
oshIntegral[a/b + ArcSech[c*x]]*Sinh[a/b])/(2*b^3) - (c*Cosh[a/b]*SinhIntegral[a/b + ArcSech[c*x]])/(2*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 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^2 \left (a+b \text{sech}^{-1}(c x)\right )^3} \, dx &=-\left (c \operatorname{Subst}\left (\int \frac{\sinh (x)}{(a+b x)^3} \, dx,x,\text{sech}^{-1}(c x)\right )\right )\\ &=\frac{\sqrt{\frac{1-c x}{1+c x}} (1+c x)}{2 b x \left (a+b \text{sech}^{-1}(c x)\right )^2}-\frac{c \operatorname{Subst}\left (\int \frac{\cosh (x)}{(a+b x)^2} \, dx,x,\text{sech}^{-1}(c x)\right )}{2 b}\\ &=\frac{\sqrt{\frac{1-c x}{1+c x}} (1+c x)}{2 b x \left (a+b \text{sech}^{-1}(c x)\right )^2}+\frac{1}{2 b^2 x \left (a+b \text{sech}^{-1}(c x)\right )}-\frac{c \operatorname{Subst}\left (\int \frac{\sinh (x)}{a+b x} \, dx,x,\text{sech}^{-1}(c x)\right )}{2 b^2}\\ &=\frac{\sqrt{\frac{1-c x}{1+c x}} (1+c x)}{2 b x \left (a+b \text{sech}^{-1}(c x)\right )^2}+\frac{1}{2 b^2 x \left (a+b \text{sech}^{-1}(c x)\right )}-\frac{\left (c \cosh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\text{sech}^{-1}(c x)\right )}{2 b^2}+\frac{\left (c \sinh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\text{sech}^{-1}(c x)\right )}{2 b^2}\\ &=\frac{\sqrt{\frac{1-c x}{1+c x}} (1+c x)}{2 b x \left (a+b \text{sech}^{-1}(c x)\right )^2}+\frac{1}{2 b^2 x \left (a+b \text{sech}^{-1}(c x)\right )}+\frac{c \text{Chi}\left (\frac{a}{b}+\text{sech}^{-1}(c x)\right ) \sinh \left (\frac{a}{b}\right )}{2 b^3}-\frac{c \cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\text{sech}^{-1}(c x)\right )}{2 b^3}\\ \end{align*}
Mathematica [A] time = 0.287881, size = 103, normalized size = 0.9 \[ \frac{\frac{b^2 \sqrt{\frac{1-c x}{c x+1}} (c x+1)}{x \left (a+b \text{sech}^{-1}(c x)\right )^2}+c \left (\sinh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\text{sech}^{-1}(c x)\right )-\cosh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\text{sech}^{-1}(c x)\right )\right )+\frac{b}{a x+b x \text{sech}^{-1}(c x)}}{2 b^3} \]
Antiderivative was successfully verified.
[In]
Integrate[1/(x^2*(a + b*ArcSech[c*x])^3),x]
[Out]
((b^2*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x))/(x*(a + b*ArcSech[c*x])^2) + b/(a*x + b*x*ArcSech[c*x]) + c*(CoshIn
tegral[a/b + ArcSech[c*x]]*Sinh[a/b] - Cosh[a/b]*SinhIntegral[a/b + ArcSech[c*x]]))/(2*b^3)
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Maple [B] time = 0.231, size = 244, normalized size = 2.1 \begin{align*} c \left ( -{\frac{b{\rm arcsech} \left (cx\right )+a-b}{4\,cx{b}^{2} \left ( \left ({\rm arcsech} \left (cx\right ) \right ) ^{2}{b}^{2}+2\,{\rm arcsech} \left (cx\right )ab+{a}^{2} \right ) } \left ( \sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}cx-1 \right ) }-{\frac{1}{4\,{b}^{3}}{{\rm e}^{{\frac{a}{b}}}}{\it Ei} \left ( 1,{\frac{a}{b}}+{\rm arcsech} \left (cx\right ) \right ) }+{\frac{1}{4\,xbc \left ( a+b{\rm arcsech} \left (cx\right ) \right ) ^{2}} \left ( \sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}cx+1 \right ) }+{\frac{1}{4\,cx{b}^{2} \left ( a+b{\rm arcsech} \left (cx\right ) \right ) } \left ( \sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}cx+1 \right ) }+{\frac{1}{4\,{b}^{3}}{{\rm e}^{-{\frac{a}{b}}}}{\it Ei} \left ( 1,-{\rm arcsech} \left (cx\right )-{\frac{a}{b}} \right ) } \right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/x^2/(a+b*arcsech(c*x))^3,x)
[Out]
c*(-1/4*((-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/2)*c*x-1)*(b*arcsech(c*x)+a-b)/c/x/b^2/(arcsech(c*x)^2*b^2+2*ar
csech(c*x)*a*b+a^2)-1/4/b^3*exp(a/b)*Ei(1,a/b+arcsech(c*x))+1/4/b*((-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/2)*c*
x+1)/c/x/(a+b*arcsech(c*x))^2+1/4/b^2*((-(c*x-1)/c/x)^(1/2)*((c*x+1)/c/x)^(1/2)*c*x+1)/c/x/(a+b*arcsech(c*x))+
1/4/b^3*exp(-a/b)*Ei(1,-arcsech(c*x)-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^2/(a+b*arcsech(c*x))^3,x, algorithm="maxima")
[Out]
-1/2*((b*c^6*(log(c) - 1) - a*c^6)*x^7 - 3*(b*c^4*(log(c) - 1) - a*c^4)*x^5 - (b*c^2*x^3 - (b*c^4*log(c) - a*c
^4)*x^5 + (b*(log(c) - 1) - a)*x - (b*c^4*x^5 - b*x)*log(x))*(c*x + 1)^(3/2)*(-c*x + 1)^(3/2) + 3*(b*c^2*(log(
c) - 1) - a*c^2)*x^3 - (2*b*c^4*x^5 + (b*c^2*(3*log(c) - 5) - 3*a*c^2)*x^3 - 3*(b*(log(c) - 1) - a)*x + 3*(b*c
^2*x^3 - b*x)*log(x))*(c*x + 1)*(c*x - 1) + ((b*c^6*(log(c) - 1) - a*c^6)*x^7 - (b*c^4*(4*log(c) - 5) - 4*a*c^
4)*x^5 + (b*c^2*(6*log(c) - 7) - 6*a*c^2)*x^3 - 3*(b*(log(c) - 1) - a)*x + (b*c^6*x^7 - 4*b*c^4*x^5 + 6*b*c^2*
x^3 - 3*b*x)*log(x))*sqrt(c*x + 1)*sqrt(-c*x + 1) - (b*(log(c) - 1) - a)*x - (b*c^6*x^7 - 3*b*c^4*x^5 + 3*b*c^
2*x^3 + (b*c^4*x^5 - b*x)*(c*x + 1)^(3/2)*(-c*x + 1)^(3/2) - 3*(b*c^2*x^3 - b*x)*(c*x + 1)*(c*x - 1) + (b*c^6*
x^7 - 4*b*c^4*x^5 + 6*b*c^2*x^3 - 3*b*x)*sqrt(c*x + 1)*sqrt(-c*x + 1) - b*x)*log(sqrt(c*x + 1)*sqrt(-c*x + 1)
+ 1) + (b*c^6*x^7 - 3*b*c^4*x^5 + 3*b*c^2*x^3 - b*x)*log(x))/((b^4*c^6*x^6 - 3*b^4*c^4*x^4 + 3*b^4*c^2*x^2 - b
^4)*x^2*log(x)^2 - (b^4*x^2*log(x)^2 + 2*(b^4*log(c) - a*b^3)*x^2*log(x) + (b^4*log(c)^2 - 2*a*b^3*log(c) + a^
2*b^2)*x^2)*(c*x + 1)^(3/2)*(-c*x + 1)^(3/2) + 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^2*log(x) - 3*((b^4*c^2*x^2 - b^4)*x^2*l
og(x)^2 - 2*(b^4*log(c) - a*b^3 - (b^4*c^2*log(c) - a*b^3*c^2)*x^2)*x^2*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^2)*(c*x + 1)*(c*x - 1) + ((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^2
- ((c*x + 1)^(3/2)*(-c*x + 1)^(3/2)*b^4*x^2 + 3*(b^4*c^2*x^2 - b^4)*(c*x + 1)*(c*x - 1)*x^2 + 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^2 - (b^4*c^6*x^6 - 3*b^4*c^4*x^4 + 3*b^4*c^2*x^2 - b^4)*x
^2)*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^2*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^2*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^2)*sqrt(c*x + 1)*sqrt(-c*x + 1) + 2*((b^4*x^2*log(x) + (b^4*log(c)
- a*b^3)*x^2)*(c*x + 1)^(3/2)*(-c*x + 1)^(3/2) - (b^4*c^6*x^6 - 3*b^4*c^4*x^4 + 3*b^4*c^2*x^2 - b^4)*x^2*log(
x) + 3*((b^4*c^2*x^2 - b^4)*x^2*log(x) - (b^4*log(c) - a*b^3 - (b^4*c^2*log(c) - a*b^3*c^2)*x^2)*x^2)*(c*x + 1
)*(c*x - 1) - ((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^2 + 3*((b^4*c^4*x^4 - 2*b^4*c^2*x^2 + b^4)*x^2*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^2)*sqrt(c*x + 1)*sqrt(-c*x + 1))*
log(sqrt(c*x + 1)*sqrt(-c*x + 1) + 1)) + integrate(-1/2*(c^8*x^8 - 4*c^6*x^6 + 6*c^4*x^4 + (3*c^4*x^4 + 1)*(c*
x + 1)^2*(c*x - 1)^2 + (3*c^4*x^4 - 4*c^2*x^2 + 4)*(c*x + 1)^(3/2)*(-c*x + 1)^(3/2) - 4*c^2*x^2 - 3*(c^6*x^6 +
c^4*x^4 - 4*c^2*x^2 + 2)*(c*x + 1)*(c*x - 1) - (c^6*x^6 - 9*c^4*x^4 + 12*c^2*x^2 - 4)*sqrt(c*x + 1)*sqrt(-c*x
+ 1) + 1)/((b^3*x^2*log(x) + (b^3*log(c) - a*b^2)*x^2)*(c*x + 1)^2*(c*x - 1)^2 - 4*((b^3*c^2*x^2 - b^3)*x^2*l
og(x) - (b^3*log(c) - a*b^2 - (b^3*c^2*log(c) - a*b^2*c^2)*x^2)*x^2)*(c*x + 1)^(3/2)*(-c*x + 1)^(3/2) + (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^2*log(x) - 6*((b^3*c^4*x^4 - 2*b^3*c^2*x^2 + b
^3)*x^2*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^2)*(c*x + 1)*(c*x - 1) + ((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^2 - 4*((b^3*c^6*x^6 - 3*
b^3*c^4*x^4 + 3*b^3*c^2*x^2 - b^3)*x^2*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^2)*sqrt(c*x + 1)*sqrt(-c*x + 1) - ((c*x
+ 1)^2*(c*x - 1)^2*b^3*x^2 - 4*(b^3*c^2*x^2 - b^3)*(c*x + 1)^(3/2)*(-c*x + 1)^(3/2)*x^2 - 6*(b^3*c^4*x^4 - 2*b
^3*c^2*x^2 + b^3)*(c*x + 1)*(c*x - 1)*x^2 - 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^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^2)*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^{2} \operatorname{arsech}\left (c x\right )^{3} + 3 \, a b^{2} x^{2} \operatorname{arsech}\left (c x\right )^{2} + 3 \, a^{2} b x^{2} \operatorname{arsech}\left (c x\right ) + a^{3} x^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x^2/(a+b*arcsech(c*x))^3,x, algorithm="fricas")
[Out]
integral(1/(b^3*x^2*arcsech(c*x)^3 + 3*a*b^2*x^2*arcsech(c*x)^2 + 3*a^2*b*x^2*arcsech(c*x) + a^3*x^2), x)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{2} \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**2/(a+b*asech(c*x))**3,x)
[Out]
Integral(1/(x**2*(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^{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/x^2/(a+b*arcsech(c*x))^3,x, algorithm="giac")
[Out]
integrate(1/((b*arcsech(c*x) + a)^3*x^2), x)