3.80.12 \(\int \frac {-50+45 x+19 e x-12 x^2+x^3}{25 x-10 x^2+x^3} \, dx\)

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

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Rubi [A]  time = 0.04, antiderivative size = 16, normalized size of antiderivative = 0.67, number of steps used = 5, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6, 1594, 27, 1620} \begin {gather*} x+\frac {19 e}{5-x}-2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-50 + 45*x + 19*E*x - 12*x^2 + x^3)/(25*x - 10*x^2 + x^3),x]

[Out]

(19*E)/(5 - x) + x - 2*Log[x]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-50+(45+19 e) x-12 x^2+x^3}{25 x-10 x^2+x^3} \, dx\\ &=\int \frac {-50+(45+19 e) x-12 x^2+x^3}{x \left (25-10 x+x^2\right )} \, dx\\ &=\int \frac {-50+(45+19 e) x-12 x^2+x^3}{(-5+x)^2 x} \, dx\\ &=\int \left (1+\frac {19 e}{(-5+x)^2}-\frac {2}{x}\right ) \, dx\\ &=\frac {19 e}{5-x}+x-2 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 14, normalized size = 0.58 \begin {gather*} -\frac {19 e}{-5+x}+x-2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-50 + 45*x + 19*E*x - 12*x^2 + x^3)/(25*x - 10*x^2 + x^3),x]

[Out]

(-19*E)/(-5 + x) + x - 2*Log[x]

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fricas [A]  time = 0.68, size = 24, normalized size = 1.00 \begin {gather*} \frac {x^{2} - 2 \, {\left (x - 5\right )} \log \relax (x) - 5 \, x - 19 \, e}{x - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((19*x*exp(1)+x^3-12*x^2+45*x-50)/(x^3-10*x^2+25*x),x, algorithm="fricas")

[Out]

(x^2 - 2*(x - 5)*log(x) - 5*x - 19*e)/(x - 5)

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giac [A]  time = 0.18, size = 16, normalized size = 0.67 \begin {gather*} x - \frac {19 \, e}{x - 5} - 2 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((19*x*exp(1)+x^3-12*x^2+45*x-50)/(x^3-10*x^2+25*x),x, algorithm="giac")

[Out]

x - 19*e/(x - 5) - 2*log(abs(x))

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maple [A]  time = 0.04, size = 16, normalized size = 0.67

method	result	size
default	\(x -2 \ln \relax (x )-\frac {19 \,{\mathrm e}}{x -5}\)	\(16\)
risch	\(x -2 \ln \relax (x )-\frac {19 \,{\mathrm e}}{x -5}\)	\(16\)
norman	\(\frac {x^{2}-25-19 \,{\mathrm e}}{x -5}-2 \ln \relax (x )\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((19*x*exp(1)+x^3-12*x^2+45*x-50)/(x^3-10*x^2+25*x),x,method=_RETURNVERBOSE)

[Out]

x-2*ln(x)-19*exp(1)/(x-5)

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maxima [A]  time = 0.36, size = 15, normalized size = 0.62 \begin {gather*} x - \frac {19 \, e}{x - 5} - 2 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((19*x*exp(1)+x^3-12*x^2+45*x-50)/(x^3-10*x^2+25*x),x, algorithm="maxima")

[Out]

x - 19*e/(x - 5) - 2*log(x)

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mupad [B]  time = 0.10, size = 15, normalized size = 0.62 \begin {gather*} x-2\,\ln \relax (x)-\frac {19\,\mathrm {e}}{x-5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((45*x + 19*x*exp(1) - 12*x^2 + x^3 - 50)/(25*x - 10*x^2 + x^3),x)

[Out]

x - 2*log(x) - (19*exp(1))/(x - 5)

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sympy [A]  time = 0.17, size = 14, normalized size = 0.58 \begin {gather*} x - 2 \log {\relax (x )} - \frac {19 e}{x - 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((19*x*exp(1)+x**3-12*x**2+45*x-50)/(x**3-10*x**2+25*x),x)

[Out]

x - 2*log(x) - 19*E/(x - 5)

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