3.17.60 $\int \frac{4-x+x^2+e(-1+2x)+(4-x+x^2)\log (x)}{e(4-x+x^2)}dx$

Optimal. Leaf size=20 $$\frac{-4+x\log (x)}{e}+\log (4-x+x^2)$$

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Rubi [A]  time = 0.14, antiderivative size = 17, normalized size of antiderivative = 0.85, number of steps used = 7, number of rules used = 5, integrand size = 40, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.125, Rules used = {12, 6728, 1657, 628, 2295} $$\begin{array}{c}\log (x^2-x+4)+\frac{x\log (x)}{e}\end{array}$$

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Log[x])/E + Log[4 - x + x^2]

Rule 12

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

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1657

Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x + c*x^2)^p, x
], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\frac{\int \frac{4-x+x^2+e(-1+2x)+(4-x+x^2)\log (x)}{4-x+x^2}dx}{e}\\ & =\frac{\int (\frac{4-e-(1-2e)x+x^2}{4-x+x^2}+\log (x))dx}{e}\\ & =\frac{\int \frac{4-e-(1-2e)x+x^2}{4-x+x^2}dx}{e}+\frac{\int \log (x)dx}{e}\\ & =-\frac{x}{e}+\frac{x\log (x)}{e}+\frac{\int (1-\frac{e-2ex}{4-x+x^2})dx}{e}\\ & =\frac{x\log (x)}{e}-\frac{\int \frac{e-2ex}{4-x+x^2}dx}{e}\\ & =\frac{x\log (x)}{e}+\log (4-x+x^2)\end{array}\end{array}$$

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Mathematica [A]  time = 0.03, size = 20, normalized size = 1.00 $$\begin{array}{c}\frac{x\log (x)+e\log (4-x+x^2)}{e}\end{array}$$

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Log[x] + E*Log[4 - x + x^2])/E

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fricas [A]  time = 0.60, size = 20, normalized size = 1.00 $$\begin{array}{c}(e\log (x^2-x+4)+x\log (x))e^{(-1)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 20, normalized size = 1.00 $$\begin{array}{c}(e\log (x^2-x+4)+x\log (x))e^{(-1)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.63, size = 17, normalized size = 0.85

method	result	size
risch	$x{\mathrm{e}}^{-1}\ln (x)+\ln (x^2-x+4)$	$17$
norman	$x{\mathrm{e}}^{-1}\ln (x)+\ln (x^2-x+4)$	$19$
default	${\mathrm{e}}^{-1}(\mathrm{e}\ln (x^2-x+4)+x\ln (x))$	$23$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.90, size = 60, normalized size = 3.00 $$\begin{array}{c}-\frac{1}{15}(2\sqrt{15}\arctan \left(\frac{1}{15}\sqrt{15}(2x-1)\right)e-(2\sqrt{15}\arctan \left(\frac{1}{15}\sqrt{15}(2x-1)\right)+15\log (x^2-x+4))e-15x\log (x))e^{(-1)}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(2*sqrt(15)*arctan(1/15*sqrt(15)*(2*x - 1))*e - (2*sqrt(15)*arctan(1/15*sqrt(15)*(2*x - 1)) + 15*log(x^2
- x + 4))*e - 15*x*log(x))*e^(-1)

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mupad [B]  time = 1.12, size = 16, normalized size = 0.80 $$\begin{array}{c}\ln (x^2-x+4)+x{\mathrm{e}}^{-1}\ln (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x^2 - x + 4) + x*exp(-1)*log(x)

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sympy [A]  time = 0.14, size = 15, normalized size = 0.75 $$\begin{array}{c}\frac{x\log (x)}{e}+\log (x^2-x+4)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*exp(-1)*log(x) + log(x**2 - x + 4)

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