3.1.2 \(\int x^3 \text {sech}^{-1}(a x)^2 \, dx\) [2]

Optimal. Leaf size=104 \[ -\frac {x^2}{12 a^2}-\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)}{3 a^4}-\frac {x^2 \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)}{6 a^2}+\frac {1}{4} x^4 \text {sech}^{-1}(a x)^2-\frac {\log (x)}{3 a^4} \]

[Out]

-1/12*x^2/a^2+1/4*x^4*arcsech(a*x)^2-1/3*ln(x)/a^4-1/3*(a*x+1)*arcsech(a*x)*((-a*x+1)/(a*x+1))^(1/2)/a^4-1/6*x
^2*(a*x+1)*arcsech(a*x)*((-a*x+1)/(a*x+1))^(1/2)/a^2

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Rubi [A]
time = 0.06, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6420, 5526, 4270, 4269, 3556} \begin {gather*} -\frac {\log (x)}{3 a^4}-\frac {\sqrt {\frac {1-a x}{a x+1}} (a x+1) \text {sech}^{-1}(a x)}{3 a^4}-\frac {x^2}{12 a^2}-\frac {x^2 \sqrt {\frac {1-a x}{a x+1}} (a x+1) \text {sech}^{-1}(a x)}{6 a^2}+\frac {1}{4} x^4 \text {sech}^{-1}(a x)^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*ArcSech[a*x]^2,x]

[Out]

-1/12*x^2/a^2 - (Sqrt[(1 - a*x)/(1 + a*x)]*(1 + a*x)*ArcSech[a*x])/(3*a^4) - (x^2*Sqrt[(1 - a*x)/(1 + a*x)]*(1
+ a*x)*ArcSech[a*x])/(6*a^2) + (x^4*ArcSech[a*x]^2)/4 - Log[x]/(3*a^4)

Rule 3556

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

Rule 4269

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(c + d*x)^m)*(Cot[e + f*x]/f), x
] + Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 4270

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(-b^2)*(c + d*x)*Cot[e + f*x]
*((b*Csc[e + f*x])^(n - 2)/(f*(n - 1))), x] + (Dist[b^2*((n - 2)/(n - 1)), Int[(c + d*x)*(b*Csc[e + f*x])^(n -
2), x], x] - Simp[b^2*d*((b*Csc[e + f*x])^(n - 2)/(f^2*(n - 1)*(n - 2))), x]) /; FreeQ[{b, c, d, e, f}, x] &&
GtQ[n, 1] && NeQ[n, 2]

Rule 5526

Int[(x_)^(m_.)*Sech[(a_.) + (b_.)*(x_)^(n_.)]^(p_.)*Tanh[(a_.) + (b_.)*(x_)^(n_.)]^(q_.), x_Symbol] :> Simp[(-
x^(m - n + 1))*(Sech[a + b*x^n]^p/(b*n*p)), x] + Dist[(m - n + 1)/(b*n*p), Int[x^(m - n)*Sech[a + b*x^n]^p, x]
, x] /; FreeQ[{a, b, p}, x] && RationalQ[m] && IntegerQ[n] && GeQ[m - n, 0] && EqQ[q, 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 x^3 \text {sech}^{-1}(a x)^2 \, dx &=-\frac {\text {Subst}\left (\int x^2 \text {sech}^4(x) \tanh (x) \, dx,x,\text {sech}^{-1}(a x)\right )}{a^4}\\ &=\frac {1}{4} x^4 \text {sech}^{-1}(a x)^2-\frac {\text {Subst}\left (\int x \text {sech}^4(x) \, dx,x,\text {sech}^{-1}(a x)\right )}{2 a^4}\\ &=-\frac {x^2}{12 a^2}-\frac {x^2 \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)}{6 a^2}+\frac {1}{4} x^4 \text {sech}^{-1}(a x)^2-\frac {\text {Subst}\left (\int x \text {sech}^2(x) \, dx,x,\text {sech}^{-1}(a x)\right )}{3 a^4}\\ &=-\frac {x^2}{12 a^2}-\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)}{3 a^4}-\frac {x^2 \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)}{6 a^2}+\frac {1}{4} x^4 \text {sech}^{-1}(a x)^2+\frac {\text {Subst}\left (\int \tanh (x) \, dx,x,\text {sech}^{-1}(a x)\right )}{3 a^4}\\ &=-\frac {x^2}{12 a^2}-\frac {\sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)}{3 a^4}-\frac {x^2 \sqrt {\frac {1-a x}{1+a x}} (1+a x) \text {sech}^{-1}(a x)}{6 a^2}+\frac {1}{4} x^4 \text {sech}^{-1}(a x)^2-\frac {\log (x)}{3 a^4}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 77, normalized size = 0.74 \begin {gather*} -\frac {a^2 x^2+2 \sqrt {\frac {1-a x}{1+a x}} \left (2+2 a x+a^2 x^2+a^3 x^3\right ) \text {sech}^{-1}(a x)-3 a^4 x^4 \text {sech}^{-1}(a x)^2+4 \log (x)}{12 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcSech[a*x]^2,x]

[Out]

-1/12*(a^2*x^2 + 2*Sqrt[(1 - a*x)/(1 + a*x)]*(2 + 2*a*x + a^2*x^2 + a^3*x^3)*ArcSech[a*x] - 3*a^4*x^4*ArcSech[
a*x]^2 + 4*Log[x])/a^4

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Maple [A]
time = 0.37, size = 150, normalized size = 1.44

method	result	size
derivativedivides	\(\frac {-\frac {\mathrm {arcsech}\left (a x \right )}{3}+\frac {a^{4} x^{4} \mathrm {arcsech}\left (a x \right )^{2}}{4}-\frac {\mathrm {arcsech}\left (a x \right ) \sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}\, a^{3} x^{3}}{6}-\frac {\mathrm {arcsech}\left (a x \right ) \sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}\, a x}{3}-\frac {a^{2} x^{2}}{12}+\frac {\ln \left (1+\left (\frac {1}{a x}+\sqrt {\frac {1}{a x}-1}\, \sqrt {1+\frac {1}{a x}}\right )^{2}\right )}{3}}{a^{4}}\)	\(150\)
default	\(\frac {-\frac {\mathrm {arcsech}\left (a x \right )}{3}+\frac {a^{4} x^{4} \mathrm {arcsech}\left (a x \right )^{2}}{4}-\frac {\mathrm {arcsech}\left (a x \right ) \sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}\, a^{3} x^{3}}{6}-\frac {\mathrm {arcsech}\left (a x \right ) \sqrt {-\frac {a x -1}{a x}}\, \sqrt {\frac {a x +1}{a x}}\, a x}{3}-\frac {a^{2} x^{2}}{12}+\frac {\ln \left (1+\left (\frac {1}{a x}+\sqrt {\frac {1}{a x}-1}\, \sqrt {1+\frac {1}{a x}}\right )^{2}\right )}{3}}{a^{4}}\)	\(150\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arcsech(a*x)^2,x,method=_RETURNVERBOSE)

[Out]

1/a^4*(-1/3*arcsech(a*x)+1/4*a^4*x^4*arcsech(a*x)^2-1/6*arcsech(a*x)*(-(a*x-1)/a/x)^(1/2)*((a*x+1)/a/x)^(1/2)*
a^3*x^3-1/3*arcsech(a*x)*(-(a*x-1)/a/x)^(1/2)*((a*x+1)/a/x)^(1/2)*a*x-1/12*a^2*x^2+1/3*ln(1+(1/a/x+(1/a/x-1)^(
1/2)*(1+1/a/x)^(1/2))^2))

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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(x^3*arcsech(a*x)^2,x, algorithm="maxima")

[Out]

integrate(x^3*arcsech(a*x)^2, x)

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Fricas [A]
time = 0.37, size = 125, normalized size = 1.20 \begin {gather*} \frac {3 \, a^{4} x^{4} \log \left (\frac {a x \sqrt {-\frac {a^{2} x^{2} - 1}{a^{2} x^{2}}} + 1}{a x}\right )^{2} - a^{2} x^{2} - 2 \, {\left (a^{3} x^{3} + 2 \, a x\right )} \sqrt {-\frac {a^{2} x^{2} - 1}{a^{2} x^{2}}} \log \left (\frac {a x \sqrt {-\frac {a^{2} x^{2} - 1}{a^{2} x^{2}}} + 1}{a x}\right ) - 4 \, \log \left (x\right )}{12 \, a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arcsech(a*x)^2,x, algorithm="fricas")

[Out]

1/12*(3*a^4*x^4*log((a*x*sqrt(-(a^2*x^2 - 1)/(a^2*x^2)) + 1)/(a*x))^2 - a^2*x^2 - 2*(a^3*x^3 + 2*a*x)*sqrt(-(a
^2*x^2 - 1)/(a^2*x^2))*log((a*x*sqrt(-(a^2*x^2 - 1)/(a^2*x^2)) + 1)/(a*x)) - 4*log(x))/a^4

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \operatorname {asech}^{2}{\left (a x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*asech(a*x)**2,x)

[Out]

Integral(x**3*asech(a*x)**2, 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(x^3*arcsech(a*x)^2,x, algorithm="giac")

[Out]

integrate(x^3*arcsech(a*x)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^3\,{\mathrm {acosh}\left (\frac {1}{a\,x}\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*acosh(1/(a*x))^2,x)

[Out]

int(x^3*acosh(1/(a*x))^2, x)

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