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

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

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 88, normalized size of antiderivative = 1.00, 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 12

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

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])

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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.20, size = 69, normalized size = 0.78 \[ -\frac {3 c^2 x^2+3 \log \left (1-c^2 x^2\right )+2 \sqrt {\frac {1-c x}{c x+1}} \left (c^3 x^3+c^2 x^2+2 c x+2\right )}{6 c^5} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/6*(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])/c^5

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fricas [A]  time = 1.26, size = 69, normalized size = 0.78 \[ -\frac {3 \, c^{2} x^{2} + 2 \, {\left (c^{3} x^{3} + 2 \, c x\right )} \sqrt {\frac {c x + 1}{c x}} \sqrt {-\frac {c x - 1}{c x}} + 3 \, \log \left (c^{2} x^{2} - 1\right )}{6 \, c^{5}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {x^{4} {\left (\sqrt {\frac {1}{c x} + 1} \sqrt {\frac {1}{c x} - 1} + \frac {1}{c x}\right )}}{c^{2} x^{2} - 1}\,{d x} \]

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)

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maple [A]  time = 0.10, size = 69, normalized size = 0.78 \[ -\frac {\sqrt {-\frac {c x -1}{c x}}\, x \sqrt {\frac {c x +1}{c x}}\, \left (c^{2} x^{2}+2\right )}{3 c^{4}}-\frac {x^{2}}{2 c^{3}}-\frac {\ln \left (c^{2} x^{2}-1\right )}{2 c^{5}} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ -\frac {\frac {1}{2} \, x^{2}}{c^{3}} - \frac {\log \left (c x + 1\right )}{2 \, c^{5}} - \frac {\log \left (c x - 1\right )}{2 \, c^{5}} - \int \frac {\sqrt {c x + 1} \sqrt {-c x + 1} x^{3}}{c^{3} x^{2} - c}\,{d x} \]

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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mupad [B]  time = 2.60, size = 76, normalized size = 0.86 \[ -\frac {\ln \left (c^2\,x^2-1\right )+c^2\,x^2}{2\,c^5}-x^3\,\sqrt {\frac {1}{c\,x}-1}\,\left (\frac {\sqrt {\frac {1}{c\,x}+1}}{3\,c^2}+\frac {2\,\sqrt {\frac {1}{c\,x}+1}}{3\,c^4\,x^2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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