3.62.38 \(\int \frac {3 x^2-5 x^3+e^{3+e^4} (-1+5 e+5 x)+e (2 x-5 x^2)}{-5 e x^2-5 x^3+e^{3+e^4} (5 e+5 x)} \, dx\)

Optimal. Leaf size=24 \[ x-\frac {1}{5} \log \left ((e+x) \left (-e^{3+e^4}+x^2\right )\right ) \]

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Rubi [A]  time = 0.08, antiderivative size = 28, normalized size of antiderivative = 1.17, number of steps used = 3, number of rules used = 2, integrand size = 68, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {2074, 260} \begin {gather*} -\frac {1}{5} \log \left (e^{3+e^4}-x^2\right )+x-\frac {1}{5} \log (x+e) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3*x^2 - 5*x^3 + E^(3 + E^4)*(-1 + 5*E + 5*x) + E*(2*x - 5*x^2))/(-5*E*x^2 - 5*x^3 + E^(3 + E^4)*(5*E + 5*
x)),x]

[Out]

x - Log[E + x]/5 - Log[E^(3 + E^4) - x^2]/5

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (1-\frac {1}{5 (e+x)}-\frac {2 x}{5 \left (-e^{3+e^4}+x^2\right )}\right ) \, dx\\ &=x-\frac {1}{5} \log (e+x)-\frac {2}{5} \int \frac {x}{-e^{3+e^4}+x^2} \, dx\\ &=x-\frac {1}{5} \log (e+x)-\frac {1}{5} \log \left (e^{3+e^4}-x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 32, normalized size = 1.33 \begin {gather*} \frac {1}{5} \left (5 (e+x)-\log (e+x)-\log \left (e^{3+e^4}-x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3*x^2 - 5*x^3 + E^(3 + E^4)*(-1 + 5*E + 5*x) + E*(2*x - 5*x^2))/(-5*E*x^2 - 5*x^3 + E^(3 + E^4)*(5*
E + 5*x)),x]

[Out]

(5*(E + x) - Log[E + x] - Log[E^(3 + E^4) - x^2])/5

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fricas [A]  time = 0.51, size = 26, normalized size = 1.08 \begin {gather*} x - \frac {1}{5} \, \log \left (x^{3} + x^{2} e - {\left (x + e\right )} e^{\left (e^{4} + 3\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(1)+5*x-1)*exp(exp(4)+3)+(-5*x^2+2*x)*exp(1)-5*x^3+3*x^2)/((5*exp(1)+5*x)*exp(exp(4)+3)-5*x^2
*exp(1)-5*x^3),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.13, size = 25, normalized size = 1.04 \begin {gather*} x - \frac {1}{5} \, \log \left ({\left | x^{2} - e^{\left (e^{4} + 3\right )} \right |}\right ) - \frac {1}{5} \, \log \left ({\left | x + e \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(1)+5*x-1)*exp(exp(4)+3)+(-5*x^2+2*x)*exp(1)-5*x^3+3*x^2)/((5*exp(1)+5*x)*exp(exp(4)+3)-5*x^2
*exp(1)-5*x^3),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.13, size = 24, normalized size = 1.00

method	result	size
norman	\(x -\frac {\ln \left (-x^{2}+{\mathrm e}^{{\mathrm e}^{4}+3}\right )}{5}-\frac {\ln \left (x +{\mathrm e}\right )}{5}\)	\(24\)
default	\(x -\frac {\ln \left (x^{2} {\mathrm e}+x^{3}-{\mathrm e}^{{\mathrm e}^{4}+3} x -{\mathrm e}^{4+{\mathrm e}^{4}}\right )}{5}\)	\(31\)
risch	\(x -\frac {\ln \left (x^{2} {\mathrm e}+x^{3}-{\mathrm e}^{{\mathrm e}^{4}+3} x -{\mathrm e}^{4+{\mathrm e}^{4}}\right )}{5}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((5*exp(1)+5*x-1)*exp(exp(4)+3)+(-5*x^2+2*x)*exp(1)-5*x^3+3*x^2)/((5*exp(1)+5*x)*exp(exp(4)+3)-5*x^2*exp(1
)-5*x^3),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.35, size = 23, normalized size = 0.96 \begin {gather*} x - \frac {1}{5} \, \log \left (x^{2} - e^{\left (e^{4} + 3\right )}\right ) - \frac {1}{5} \, \log \left (x + e\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(1)+5*x-1)*exp(exp(4)+3)+(-5*x^2+2*x)*exp(1)-5*x^3+3*x^2)/((5*exp(1)+5*x)*exp(exp(4)+3)-5*x^2
*exp(1)-5*x^3),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 4.57, size = 30, normalized size = 1.25 \begin {gather*} x-\frac {\ln \left (x^3+\mathrm {e}\,x^2-{\mathrm {e}}^{{\mathrm {e}}^4+3}\,x-{\mathrm {e}}^{{\mathrm {e}}^4+4}\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(1)*(2*x - 5*x^2) + exp(exp(4) + 3)*(5*x + 5*exp(1) - 1) + 3*x^2 - 5*x^3)/(5*x^2*exp(1) - exp(exp(4)
+ 3)*(5*x + 5*exp(1)) + 5*x^3),x)

[Out]

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

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sympy [A]  time = 0.38, size = 32, normalized size = 1.33 \begin {gather*} x - \frac {\log {\left (x^{3} + e x^{2} - x e^{3} e^{e^{4}} - e^{4} e^{e^{4}} \right )}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((5*exp(1)+5*x-1)*exp(exp(4)+3)+(-5*x**2+2*x)*exp(1)-5*x**3+3*x**2)/((5*exp(1)+5*x)*exp(exp(4)+3)-5*
x**2*exp(1)-5*x**3),x)

[Out]

x - log(x**3 + E*x**2 - x*exp(3)*exp(exp(4)) - exp(4)*exp(exp(4)))/5

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