3.69.89 $\int \frac{4e^{10}+3x^2+e^5(-1+7x)}{4e^{10}+7e^{5}x+3x^2}dx$

Optimal. Leaf size=21 $$e^4+x+\log (3+\frac{x}{x+\frac{x^2}{e^5}})$$

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Rubi [A]  time = 0.05, antiderivative size = 20, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 3, integrand size = 40, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.075, Rules used = {1657, 616, 31} $$\begin{array}{c}x-\log (x+e^5)+\log (3x+4e^5)\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(4*E^10 + 3*x^2 + E^5*(-1 + 7*x))/(4*E^10 + 7*E^5*x + 3*x^2),x]

[Out]

x - Log[E^5 + x] + Log[4*E^5 + 3*x]

Rule 31

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

Rule 616

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[c/q, Int[1/Simp
[b/2 - q/2 + c*x, x], x], x] - Dist[c/q, Int[1/Simp[b/2 + q/2 + c*x, x], x], x]] /; FreeQ[{a, b, c}, x] && NeQ
[b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c] && PerfectSquareQ[b^2 - 4*a*c]

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]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (1-\frac{e^5}{4e^{10}+7e^{5}x+3x^2})dx\\ & =x-e^5\int \frac{1}{4e^{10}+7e^{5}x+3x^2}dx\\ & =x-3\int \frac{1}{3e^5+3x}dx+3\int \frac{1}{4e^5+3x}dx\\ & =x-\log (e^5+x)+\log (4e^5+3x)\end{array}\end{array}$$

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Mathematica [A]  time = 0.01, size = 20, normalized size = 0.95 $$\begin{array}{c}x-\log (e^5+x)+\log (4e^5+3x)\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(4*E^10 + 3*x^2 + E^5*(-1 + 7*x))/(4*E^10 + 7*E^5*x + 3*x^2),x]

[Out]

x - Log[E^5 + x] + Log[4*E^5 + 3*x]

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fricas [A]  time = 0.88, size = 18, normalized size = 0.86 $$\begin{array}{c}x+\log (3x+4e^5)-\log (x+e^5)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(5)^2+(7*x-1)*exp(5)+3*x^2)/(4*exp(5)^2+7*x*exp(5)+3*x^2),x, algorithm="fricas")

[Out]

x + log(3*x + 4*e^5) - log(x + e^5)

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giac [A]  time = 0.15, size = 26, normalized size = 1.24 $$\begin{array}{c}x-\log \left(\frac{|6x+6e^5|}{|6x+8e^5|}\right)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(5)^2+(7*x-1)*exp(5)+3*x^2)/(4*exp(5)^2+7*x*exp(5)+3*x^2),x, algorithm="giac")

[Out]

x - log(abs(6*x + 6*e^5)/abs(6*x + 8*e^5))

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maple [A]  time = 0.10, size = 19, normalized size = 0.90

method	result	size
norman	$x-\ln ({\mathrm{e}}^5+x)+\ln (4{\mathrm{e}}^5+3x)$	$19$
risch	$x-\ln ({\mathrm{e}}^5+x)+\ln (4{\mathrm{e}}^5+3x)$	$19$
default	$x+\frac{2{\mathrm{e}}^5\mathrm{arctanh}\left(\frac{7{\mathrm{e}}^5+6x}{\sqrt{{\mathrm{e}}^{10}}}\right)}{\sqrt{{\mathrm{e}}^{10}}}$	$43$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*exp(5)^2+(7*x-1)*exp(5)+3*x^2)/(4*exp(5)^2+7*x*exp(5)+3*x^2),x,method=_RETURNVERBOSE)

[Out]

x-ln(exp(5)+x)+ln(4*exp(5)+3*x)

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maxima [A]  time = 0.36, size = 18, normalized size = 0.86 $$\begin{array}{c}x+\log (3x+4e^5)-\log (x+e^5)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(5)^2+(7*x-1)*exp(5)+3*x^2)/(4*exp(5)^2+7*x*exp(5)+3*x^2),x, algorithm="maxima")

[Out]

x + log(3*x + 4*e^5) - log(x + e^5)

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mupad [B]  time = 4.09, size = 12, normalized size = 0.57 $$\begin{array}{c}x+2\mathrm{atanh}(6x{\mathrm{e}}^{-5}+7)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*exp(10) + 3*x^2 + exp(5)*(7*x - 1))/(4*exp(10) + 7*x*exp(5) + 3*x^2),x)

[Out]

x + 2*atanh(6*x*exp(-5) + 7)

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sympy [A]  time = 0.19, size = 17, normalized size = 0.81 $$\begin{array}{c}x-\log (x+e^5)+\log (x+\frac{4e^5}{3})\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*exp(5)**2+(7*x-1)*exp(5)+3*x**2)/(4*exp(5)**2+7*x*exp(5)+3*x**2),x)

[Out]

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

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