3.89 \(\int \frac{e^{\text{sech}^{-1}(c x)} x^4}{1-c^2 x^2} \, dx\)

Optimal. Leaf size=88 \[ -\frac{x^2 \sqrt{1-c x}}{3 c^3 \sqrt{\frac{1}{c x+1}}}-\frac{x^2}{2 c^3}-\frac{\log \left (1-c^2 x^2\right )}{2 c^5}-\frac{2 \sqrt{1-c x}}{3 c^5 \sqrt{\frac{1}{c x+1}}} \]

[Out]

-x^2/(2*c^3) - (2*Sqrt[1 - c*x])/(3*c^5*Sqrt[(1 + c*x)^(-1)]) - (x^2*Sqrt[1 - c*x])/(3*c^3*Sqrt[(1 + c*x)^(-1)
]) - Log[1 - c^2*x^2]/(2*c^5)

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Rubi [A]  time = 0.180897, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.318, Rules used = {6341, 1956, 100, 12, 74, 266, 43} \[ -\frac{x^2 \sqrt{1-c x}}{3 c^3 \sqrt{\frac{1}{c x+1}}}-\frac{x^2}{2 c^3}-\frac{\log \left (1-c^2 x^2\right )}{2 c^5}-\frac{2 \sqrt{1-c x}}{3 c^5 \sqrt{\frac{1}{c x+1}}} \]

Antiderivative was successfully verified.

[In]

Int[(E^ArcSech[c*x]*x^4)/(1 - c^2*x^2),x]

[Out]

-x^2/(2*c^3) - (2*Sqrt[1 - c*x])/(3*c^5*Sqrt[(1 + c*x)^(-1)]) - (x^2*Sqrt[1 - c*x])/(3*c^3*Sqrt[(1 + c*x)^(-1)
]) - Log[1 - c^2*x^2]/(2*c^5)

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 100

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

Rule 12

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

Rule 74

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{e^{\text{sech}^{-1}(c x)} x^4}{1-c^2 x^2} \, dx &=\frac{\int \frac{x^3 \sqrt{\frac{1}{1+c x}}}{\sqrt{1-c x}} \, dx}{c}+\frac{\int \frac{x^3}{1-c^2 x^2} \, dx}{c}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x}{1-c^2 x} \, dx,x,x^2\right )}{2 c}+\frac{\left (\sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{x^3}{\sqrt{1-c x} \sqrt{1+c x}} \, dx}{c}\\ &=-\frac{x^2 \sqrt{1-c x}}{3 c^3 \sqrt{\frac{1}{1+c x}}}+\frac{\operatorname{Subst}\left (\int \left (-\frac{1}{c^2}-\frac{1}{c^2 \left (-1+c^2 x\right )}\right ) \, dx,x,x^2\right )}{2 c}-\frac{\left (\sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int -\frac{2 x}{\sqrt{1-c x} \sqrt{1+c x}} \, dx}{3 c^3}\\ &=-\frac{x^2}{2 c^3}-\frac{x^2 \sqrt{1-c x}}{3 c^3 \sqrt{\frac{1}{1+c x}}}-\frac{\log \left (1-c^2 x^2\right )}{2 c^5}+\frac{\left (2 \sqrt{\frac{1}{1+c x}} \sqrt{1+c x}\right ) \int \frac{x}{\sqrt{1-c x} \sqrt{1+c x}} \, dx}{3 c^3}\\ &=-\frac{x^2}{2 c^3}-\frac{2 \sqrt{1-c x}}{3 c^5 \sqrt{\frac{1}{1+c x}}}-\frac{x^2 \sqrt{1-c x}}{3 c^3 \sqrt{\frac{1}{1+c x}}}-\frac{\log \left (1-c^2 x^2\right )}{2 c^5}\\ \end{align*}

Mathematica [A]  time = 0.18891, size = 69, normalized size = 0.78 \[ -\frac{3 c^2 x^2+2 \sqrt{\frac{1-c x}{c x+1}} \left (c^3 x^3+c^2 x^2+2 c x+2\right )+3 \log \left (1-c^2 x^2\right )}{6 c^5} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(E^ArcSech[c*x]*x^4)/(1 - c^2*x^2),x]

[Out]

-(3*c^2*x^2 + 2*Sqrt[(1 - c*x)/(1 + c*x)]*(2 + 2*c*x + c^2*x^2 + c^3*x^3) + 3*Log[1 - c^2*x^2])/(6*c^5)

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Maple [A]  time = 0.211, size = 69, normalized size = 0.8 \begin{align*} -{\frac{x \left ({c}^{2}{x}^{2}+2 \right ) }{3\,{c}^{4}}\sqrt{-{\frac{cx-1}{cx}}}\sqrt{{\frac{cx+1}{cx}}}}-{\frac{{x}^{2}}{2\,{c}^{3}}}-{\frac{\ln \left ({c}^{2}{x}^{2}-1 \right ) }{2\,{c}^{5}}} \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^4/(-c^2*x^2+1),x)

[Out]

-1/3*(-(c*x-1)/c/x)^(1/2)*x*((c*x+1)/c/x)^(1/2)*(c^2*x^2+2)/c^4-1/2*x^2/c^3-1/2/c^5*ln(c^2*x^2-1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{\frac{1}{2} \, x^{2}}{c^{3}} - \frac{\log \left (c x + 1\right )}{2 \, c^{5}} - \frac{\log \left (c x - 1\right )}{2 \, c^{5}} - \int \frac{\sqrt{c x + 1} \sqrt{-c x + 1} x^{3}}{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^4/(-c^2*x^2+1),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.06668, size = 149, normalized size = 1.69 \begin{align*} -\frac{3 \, c^{2} x^{2} + 2 \,{\left (c^{3} x^{3} + 2 \, c x\right )} \sqrt{\frac{c x + 1}{c x}} \sqrt{-\frac{c x - 1}{c x}} + 3 \, \log \left (c^{2} x^{2} - 1\right )}{6 \, c^{5}} \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^4/(-c^2*x^2+1),x, algorithm="fricas")

[Out]

-1/6*(3*c^2*x^2 + 2*(c^3*x^3 + 2*c*x)*sqrt((c*x + 1)/(c*x))*sqrt(-(c*x - 1)/(c*x)) + 3*log(c^2*x^2 - 1))/c^5

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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**4/(-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^{4}{\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^4/(-c^2*x^2+1),x, algorithm="giac")

[Out]

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