\(\int \frac {4 e^{10}+3 x^2+e^5 (-1+7 x)}{4 e^{10}+7 e^5 x+3 x^2} \, dx\) [6889]

Optimal result

Integrand size = 40, antiderivative size = 21 \[ \int \frac {4 e^{10}+3 x^2+e^5 (-1+7 x)}{4 e^{10}+7 e^5 x+3 x^2} \, dx=e^4+x+\log \left (3+\frac {x}{x+\frac {x^2}{e^5}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {1671, 630, 31} \[ \int \frac {4 e^{10}+3 x^2+e^5 (-1+7 x)}{4 e^{10}+7 e^5 x+3 x^2} \, dx=x-\log \left (x+e^5\right )+\log \left (3 x+4 e^5\right ) \]

[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 630

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 1671

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{align*} \text {integral}& = \int \left (1-\frac {e^5}{4 e^{10}+7 e^5 x+3 x^2}\right ) \, dx \\ & = x-e^5 \int \frac {1}{4 e^{10}+7 e^5 x+3 x^2} \, dx \\ & = x-3 \int \frac {1}{3 e^5+3 x} \, dx+3 \int \frac {1}{4 e^5+3 x} \, dx \\ & = x-\log \left (e^5+x\right )+\log \left (4 e^5+3 x\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {4 e^{10}+3 x^2+e^5 (-1+7 x)}{4 e^{10}+7 e^5 x+3 x^2} \, dx=x-\log \left (e^5+x\right )+\log \left (4 e^5+3 x\right ) \]

[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]

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81

method	result	size
parallelrisch	\(x -\ln \left ({\mathrm e}^{5}+x \right )+\ln \left (x +\frac {4 \,{\mathrm e}^{5}}{3}\right )\)	\(17\)
norman	\(x -\ln \left ({\mathrm e}^{5}+x \right )+\ln \left (4 \,{\mathrm e}^{5}+3 x \right )\)	\(19\)
risch	\(x -\ln \left ({\mathrm e}^{5}+x \right )+\ln \left (4 \,{\mathrm e}^{5}+3 x \right )\)	\(19\)
default	\(x +\frac {2 \,{\mathrm e}^{5} \operatorname {arctanh}\left (\frac {7 \,{\mathrm e}^{5}+6 x}{\sqrt {{\mathrm e}^{10}}}\right )}{\sqrt {{\mathrm e}^{10}}}\)	\(43\)

[In]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {4 e^{10}+3 x^2+e^5 (-1+7 x)}{4 e^{10}+7 e^5 x+3 x^2} \, dx=x + \log \left (3 \, x + 4 \, e^{5}\right ) - \log \left (x + e^{5}\right ) \]

[In]

integrate((4*exp(5)^2+(-1+7*x)*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)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {4 e^{10}+3 x^2+e^5 (-1+7 x)}{4 e^{10}+7 e^5 x+3 x^2} \, dx=x - \log {\left (x + e^{5} \right )} + \log {\left (x + \frac {4 e^{5}}{3} \right )} \]

[In]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {4 e^{10}+3 x^2+e^5 (-1+7 x)}{4 e^{10}+7 e^5 x+3 x^2} \, dx=x + \log \left (3 \, x + 4 \, e^{5}\right ) - \log \left (x + e^{5}\right ) \]

[In]

integrate((4*exp(5)^2+(-1+7*x)*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)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {4 e^{10}+3 x^2+e^5 (-1+7 x)}{4 e^{10}+7 e^5 x+3 x^2} \, dx=x + \log \left ({\left | 3 \, x + 4 \, e^{5} \right |}\right ) - \log \left ({\left | x + e^{5} \right |}\right ) \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 12.13 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.57 \[ \int \frac {4 e^{10}+3 x^2+e^5 (-1+7 x)}{4 e^{10}+7 e^5 x+3 x^2} \, dx=x+2\,\mathrm {atanh}\left (6\,x\,{\mathrm {e}}^{-5}+7\right ) \]

[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)