3.92.3 \(\int \frac {1-3 x^2+e^x (-x+x^3)}{-3 x+3 x^3+e^5 (x-x^3)+e^x (-x+x^3)+(-x+x^3) \log (-\frac {3}{-x+x^3})} \, dx\)

Optimal. Leaf size=26 \[ \log \left (3-e^5+e^x+\log \left (\frac {3}{x \left (1-x^2\right )}\right )\right ) \]

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Rubi [A]  time = 0.49, antiderivative size = 23, normalized size of antiderivative = 0.88, number of steps used = 2, number of rules used = 2, integrand size = 72, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {6741, 6684} \begin {gather*} \log \left (\log \left (\frac {3}{x-x^3}\right )+e^x-e^5+3\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - 3*x^2 + E^x*(-x + x^3))/(-3*x + 3*x^3 + E^5*(x - x^3) + E^x*(-x + x^3) + (-x + x^3)*Log[-3/(-x + x^3)
]),x]

[Out]

Log[3 - E^5 + E^x + Log[3/(x - x^3)]]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-1+3 x^2-e^x \left (-x+x^3\right )}{x \left (1-x^2\right ) \left (e^x+3 \left (1-\frac {e^5}{3}\right )+\log \left (\frac {3}{x-x^3}\right )\right )} \, dx\\ &=\log \left (3-e^5+e^x+\log \left (\frac {3}{x-x^3}\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.89, size = 23, normalized size = 0.88 \begin {gather*} \log \left (3-e^5+e^x+\log \left (\frac {3}{x-x^3}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 - 3*x^2 + E^x*(-x + x^3))/(-3*x + 3*x^3 + E^5*(x - x^3) + E^x*(-x + x^3) + (-x + x^3)*Log[-3/(-x
+ x^3)]),x]

[Out]

Log[3 - E^5 + E^x + Log[3/(x - x^3)]]

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fricas [A]  time = 0.92, size = 21, normalized size = 0.81 \begin {gather*} \log \left (-e^{5} + e^{x} + \log \left (-\frac {3}{x^{3} - x}\right ) + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-x)*exp(x)-3*x^2+1)/((x^3-x)*log(-3/(x^3-x))+(x^3-x)*exp(x)+(-x^3+x)*exp(5)+3*x^3-3*x),x, algor
ithm="fricas")

[Out]

log(-e^5 + e^x + log(-3/(x^3 - x)) + 3)

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giac [A]  time = 0.17, size = 21, normalized size = 0.81 \begin {gather*} \log \left (-e^{5} + e^{x} + \log \left (-\frac {3}{x^{3} - x}\right ) + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-x)*exp(x)-3*x^2+1)/((x^3-x)*log(-3/(x^3-x))+(x^3-x)*exp(x)+(-x^3+x)*exp(5)+3*x^3-3*x),x, algor
ithm="giac")

[Out]

log(-e^5 + e^x + log(-3/(x^3 - x)) + 3)

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maple [A]  time = 1.18, size = 24, normalized size = 0.92

method	result	size
default	\(\ln \left ({\mathrm e}^{5}-{\mathrm e}^{x}-\ln \left (-\frac {3}{x^{3}-x}\right )-3\right )\)	\(24\)
norman	\(\ln \left ({\mathrm e}^{5}-{\mathrm e}^{x}-\ln \left (-\frac {3}{x^{3}-x}\right )-3\right )\)	\(24\)
risch	\(\ln \left (\ln \left (x^{2}-1\right )+\frac {i \left (\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i}{x^{2}-1}\right ) \mathrm {csgn}\left (\frac {i}{x \left (x^{2}-1\right )}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i}{x \left (x^{2}-1\right )}\right )^{2}-\pi \,\mathrm {csgn}\left (\frac {i}{x^{2}-1}\right ) \mathrm {csgn}\left (\frac {i}{x \left (x^{2}-1\right )}\right )^{2}-\pi \mathrm {csgn}\left (\frac {i}{x \left (x^{2}-1\right )}\right )^{3}+2 \pi \mathrm {csgn}\left (\frac {i}{x \left (x^{2}-1\right )}\right )^{2}-2 i {\mathrm e}^{5}+2 i \ln \relax (3)+2 i {\mathrm e}^{x}-2 i \ln \relax (x )-2 \pi +6 i\right )}{2}\right )\)	\(166\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^3-x)*exp(x)-3*x^2+1)/((x^3-x)*ln(-3/(x^3-x))+(x^3-x)*exp(x)+(-x^3+x)*exp(5)+3*x^3-3*x),x,method=_RETUR
NVERBOSE)

[Out]

ln(exp(5)-exp(x)-ln(-3/(x^3-x))-3)

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maxima [A]  time = 0.61, size = 25, normalized size = 0.96 \begin {gather*} \log \left (e^{5} - e^{x} - \log \relax (3) + \log \left (x + 1\right ) + \log \relax (x) + \log \left (-x + 1\right ) - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^3-x)*exp(x)-3*x^2+1)/((x^3-x)*log(-3/(x^3-x))+(x^3-x)*exp(x)+(-x^3+x)*exp(5)+3*x^3-3*x),x, algor
ithm="maxima")

[Out]

log(e^5 - e^x - log(3) + log(x + 1) + log(x) + log(-x + 1) - 3)

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mupad [B]  time = 9.85, size = 21, normalized size = 0.81 \begin {gather*} \ln \left (\ln \left (\frac {3}{x-x^3}\right )-{\mathrm {e}}^5+{\mathrm {e}}^x+3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(x - x^3) + 3*x^2 - 1)/(3*x + exp(x)*(x - x^3) + log(3/(x - x^3))*(x - x^3) - exp(5)*(x - x^3) - 3
*x^3),x)

[Out]

log(log(3/(x - x^3)) - exp(5) + exp(x) + 3)

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sympy [A]  time = 0.56, size = 19, normalized size = 0.73 \begin {gather*} \log {\left (e^{x} + \log {\left (- \frac {3}{x^{3} - x} \right )} - e^{5} + 3 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**3-x)*exp(x)-3*x**2+1)/((x**3-x)*ln(-3/(x**3-x))+(x**3-x)*exp(x)+(-x**3+x)*exp(5)+3*x**3-3*x),x)

[Out]

log(exp(x) + log(-3/(x**3 - x)) - exp(5) + 3)

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