3.90 \(\int \frac{e^{\text{sech}^{-1}(c x)} x^3}{1-c^2 x^2} \, dx\)

Optimal. Leaf size=75 \[ -\frac{x \sqrt{1-c x}}{2 c^3 \sqrt{\frac{1}{c x+1}}}-\frac{x}{c^3}+\frac{\sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sin ^{-1}(c x)}{2 c^4}+\frac{\tanh ^{-1}(c x)}{c^4} \]

[Out]

-(x/c^3) - (x*Sqrt[1 - c*x])/(2*c^3*Sqrt[(1 + c*x)^(-1)]) + (Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcSin[c*x])/(
2*c^4) + ArcTanh[c*x]/c^4

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Rubi [A]  time = 0.165123, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318, Rules used = {6341, 1956, 90, 41, 216, 321, 206} \[ -\frac{x \sqrt{1-c x}}{2 c^3 \sqrt{\frac{1}{c x+1}}}-\frac{x}{c^3}+\frac{\sqrt{\frac{1}{c x+1}} \sqrt{c x+1} \sin ^{-1}(c x)}{2 c^4}+\frac{\tanh ^{-1}(c x)}{c^4} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcSech[c*x]*x^3)/(1 - c^2*x^2),x]

[Out]

-(x/c^3) - (x*Sqrt[1 - c*x])/(2*c^3*Sqrt[(1 + c*x)^(-1)]) + (Sqrt[(1 + c*x)^(-1)]*Sqrt[1 + c*x]*ArcSin[c*x])/(
2*c^4) + ArcTanh[c*x]/c^4

Rule 6341

Int[(E^ArcSech[(c_.)*(x_)]*((d_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Dist[d/(a*c), Int[((d*x)^(m
- 1)*Sqrt[1/(1 + c*x)])/Sqrt[1 - c*x], x], x] + Dist[d/c, Int[(d*x)^(m - 1)/(a + b*x^2), x], x] /; FreeQ[{a,
b, c, d, m}, x] && EqQ[b + a*c^2, 0]

Rule 1956

Int[(x_)^(m_.)*((e_.)*((a_) + (b_.)*(x_)^(n_.))^(r_.))^(p_)*((f_.)*((c_) + (d_.)*(x_)^(n_.))^(s_))^(q_), x_Sym
bol] :> Dist[((e*(a + b*x^n)^r)^p*(f*(c + d*x^n)^s)^q)/((a + b*x^n)^(p*r)*(c + d*x^n)^(q*s)), Int[x^m*(a + b*x
^n)^(p*r)*(c + d*x^n)^(q*s), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p, q, r, s}, x]

Rule 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 41

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[(a*c + b*d*x^2)^m, x] /; FreeQ[{a, b
, c, d, m}, x] && EqQ[b*c + a*d, 0] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rule 321

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{e^{\text{sech}^{-1}(c x)} x^3}{1-c^2 x^2} \, dx &=\frac{\int \frac{x^2 \sqrt{\frac{1}{1+c x}}}{\sqrt{1-c x}} \, dx}{c}+\frac{\int \frac{x^2}{1-c^2 x^2} \, dx}{c}\\ &=-\frac{x}{c^3}+\frac{\int \frac{1}{1-c^2 x^2} \, dx}{c^3}+\frac{\left (\sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{x^2}{\sqrt{1-c x} \sqrt{1+c x}} \, dx}{c}\\ &=-\frac{x}{c^3}-\frac{x \sqrt{1-c x}}{2 c^3 \sqrt{\frac{1}{1+c x}}}+\frac{\tanh ^{-1}(c x)}{c^4}+\frac{\left (\sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{1}{\sqrt{1-c x} \sqrt{1+c x}} \, dx}{2 c^3}\\ &=-\frac{x}{c^3}-\frac{x \sqrt{1-c x}}{2 c^3 \sqrt{\frac{1}{1+c x}}}+\frac{\tanh ^{-1}(c x)}{c^4}+\frac{\left (\sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{1}{\sqrt{1-c^2 x^2}} \, dx}{2 c^3}\\ &=-\frac{x}{c^3}-\frac{x \sqrt{1-c x}}{2 c^3 \sqrt{\frac{1}{1+c x}}}+\frac{\sqrt{\frac{1}{1+c x}} \sqrt{1+c x} \sin ^{-1}(c x)}{2 c^4}+\frac{\tanh ^{-1}(c x)}{c^4}\\ \end{align*}

Mathematica [C]  time = 0.153733, size = 110, normalized size = 1.47 \[ -\frac{c^2 x^2 \sqrt{\frac{1-c x}{c x+1}}+2 c x+c x \sqrt{\frac{1-c x}{c x+1}}+\log (1-c x)-\log (c x+1)-i \log \left (2 \sqrt{\frac{1-c x}{c x+1}} (c x+1)-2 i c x\right )}{2 c^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(E^ArcSech[c*x]*x^3)/(1 - c^2*x^2),x]

[Out]

-(2*c*x + c*x*Sqrt[(1 - c*x)/(1 + c*x)] + c^2*x^2*Sqrt[(1 - c*x)/(1 + c*x)] + Log[1 - c*x] - Log[1 + c*x] - I*
Log[(-2*I)*c*x + 2*Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)])/(2*c^4)

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Maple [C]  time = 0.213, size = 117, normalized size = 1.6 \begin{align*} -{\frac{x{\it csgn} \left ( c \right ) }{2\,{c}^{3}}\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}} \left ( x\sqrt{-{c}^{2}{x}^{2}+1}{\it csgn} \left ( c \right ) c-\arctan \left ({x{\it csgn} \left ( c \right ) c{\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}}} \right ) \right ){\frac{1}{\sqrt{-{c}^{2}{x}^{2}+1}}}}-{\frac{x}{{c}^{3}}}-{\frac{\ln \left ( cx-1 \right ) }{2\,{c}^{4}}}+{\frac{\ln \left ( cx+1 \right ) }{2\,{c}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))*x^3/(-c^2*x^2+1),x)

[Out]

-1/2*(-(c*x-1)/c/x)^(1/2)*x*((c*x+1)/c/x)^(1/2)/c^3*(x*(-c^2*x^2+1)^(1/2)*csgn(c)*c-arctan(csgn(c)*c*x/(-c^2*x
^2+1)^(1/2)))/(-c^2*x^2+1)^(1/2)*csgn(c)-x/c^3-1/2/c^4*ln(c*x-1)+1/2/c^4*ln(c*x+1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{x}{c^{3}} + \frac{\log \left (c x + 1\right )}{2 \, c^{4}} - \frac{\log \left (c x - 1\right )}{2 \, c^{4}} - \int \frac{\sqrt{c x + 1} \sqrt{-c x + 1} x^{2}}{c^{3} x^{2} - c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))*x^3/(-c^2*x^2+1),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.0929, size = 213, normalized size = 2.84 \begin{align*} -\frac{c^{2} x^{2} \sqrt{\frac{c x + 1}{c x}} \sqrt{-\frac{c x - 1}{c x}} + 2 \, c x + \arctan \left (\sqrt{\frac{c x + 1}{c x}} \sqrt{-\frac{c x - 1}{c x}}\right ) - \log \left (c x + 1\right ) + \log \left (c x - 1\right )}{2 \, c^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))*x^3/(-c^2*x^2+1),x, algorithm="fricas")

[Out]

-1/2*(c^2*x^2*sqrt((c*x + 1)/(c*x))*sqrt(-(c*x - 1)/(c*x)) + 2*c*x + arctan(sqrt((c*x + 1)/(c*x))*sqrt(-(c*x -
1)/(c*x))) - log(c*x + 1) + log(c*x - 1))/c^4

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(-1+1/c/x)**(1/2)*(1+1/c/x)**(1/2))*x**3/(-c**2*x**2+1),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{x^{3}{\left (\sqrt{\frac{1}{c x} + 1} \sqrt{\frac{1}{c x} - 1} + \frac{1}{c x}\right )}}{c^{2} x^{2} - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((1/c/x+(-1+1/c/x)^(1/2)*(1+1/c/x)^(1/2))*x^3/(-c^2*x^2+1),x, algorithm="giac")

[Out]

integrate(-x^3*(sqrt(1/(c*x) + 1)*sqrt(1/(c*x) - 1) + 1/(c*x))/(c^2*x^2 - 1), x)