\(\int \frac {-1080 e^4+(25+10 x+x^2) \log (2)}{25+10 x+x^2} \, dx\) [262]

Optimal result

Integrand size = 28, antiderivative size = 20 \[ \int \frac {-1080 e^4+\left (25+10 x+x^2\right ) \log (2)}{25+10 x+x^2} \, dx=\frac {120 e^4 (4-x)}{5+x}+x \log (2) \]

[Out]

120*(-x+4)/(5+x)*exp(4)+x*ln(2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {27, 1864} \[ \int \frac {-1080 e^4+\left (25+10 x+x^2\right ) \log (2)}{25+10 x+x^2} \, dx=\frac {1080 e^4}{x+5}+x \log (2) \]

[In]

Int[(-1080*E^4 + (25 + 10*x + x^2)*Log[2])/(25 + 10*x + x^2),x]

[Out]

(1080*E^4)/(5 + x) + x*Log[2]

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 1864

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps \begin{align*} \text {integral}& = \int \frac {-1080 e^4+\left (25+10 x+x^2\right ) \log (2)}{(5+x)^2} \, dx \\ & = \int \left (-\frac {1080 e^4}{(5+x)^2}+\log (2)\right ) \, dx \\ & = \frac {1080 e^4}{5+x}+x \log (2) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.85 \[ \int \frac {-1080 e^4+\left (25+10 x+x^2\right ) \log (2)}{25+10 x+x^2} \, dx=\frac {1080 e^4}{5+x}+(5+x) \log (2) \]

[In]

Integrate[(-1080*E^4 + (25 + 10*x + x^2)*Log[2])/(25 + 10*x + x^2),x]

[Out]

(1080*E^4)/(5 + x) + (5 + x)*Log[2]

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.75

method	result	size
default	\(x \ln \left (2\right )+\frac {1080 \,{\mathrm e}^{4}}{5+x}\)	\(15\)
risch	\(x \ln \left (2\right )+\frac {1080 \,{\mathrm e}^{4}}{5+x}\)	\(15\)
gosper	\(\frac {x^{2} \ln \left (2\right )-25 \ln \left (2\right )+1080 \,{\mathrm e}^{4}}{5+x}\)	\(22\)
norman	\(\frac {x^{2} \ln \left (2\right )-25 \ln \left (2\right )+1080 \,{\mathrm e}^{4}}{5+x}\)	\(22\)
parallelrisch	\(\frac {x^{2} \ln \left (2\right )-25 \ln \left (2\right )+1080 \,{\mathrm e}^{4}}{5+x}\)	\(22\)
meijerg	\(5 \ln \left (2\right ) \left (\frac {x \left (\frac {3 x}{5}+6\right )}{15+3 x}-2 \ln \left (1+\frac {x}{5}\right )\right )+10 \ln \left (2\right ) \left (-\frac {x}{5 \left (1+\frac {x}{5}\right )}+\ln \left (1+\frac {x}{5}\right )\right )+\frac {\ln \left (2\right ) x}{1+\frac {x}{5}}-\frac {216 \,{\mathrm e}^{4} x}{5 \left (1+\frac {x}{5}\right )}\)	\(74\)

[In]

int(((x^2+10*x+25)*ln(2)-1080*exp(4))/(x^2+10*x+25),x,method=_RETURNVERBOSE)

[Out]

x*ln(2)+1080*exp(4)/(5+x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.05 \[ \int \frac {-1080 e^4+\left (25+10 x+x^2\right ) \log (2)}{25+10 x+x^2} \, dx=\frac {{\left (x^{2} + 5 \, x\right )} \log \left (2\right ) + 1080 \, e^{4}}{x + 5} \]

[In]

integrate(((x^2+10*x+25)*log(2)-1080*exp(4))/(x^2+10*x+25),x, algorithm="fricas")

[Out]

((x^2 + 5*x)*log(2) + 1080*e^4)/(x + 5)

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.60 \[ \int \frac {-1080 e^4+\left (25+10 x+x^2\right ) \log (2)}{25+10 x+x^2} \, dx=x \log {\left (2 \right )} + \frac {1080 e^{4}}{x + 5} \]

[In]

integrate(((x**2+10*x+25)*ln(2)-1080*exp(4))/(x**2+10*x+25),x)

[Out]

x*log(2) + 1080*exp(4)/(x + 5)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {-1080 e^4+\left (25+10 x+x^2\right ) \log (2)}{25+10 x+x^2} \, dx=x \log \left (2\right ) + \frac {1080 \, e^{4}}{x + 5} \]

[In]

integrate(((x^2+10*x+25)*log(2)-1080*exp(4))/(x^2+10*x+25),x, algorithm="maxima")

[Out]

x*log(2) + 1080*e^4/(x + 5)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {-1080 e^4+\left (25+10 x+x^2\right ) \log (2)}{25+10 x+x^2} \, dx=x \log \left (2\right ) + \frac {1080 \, e^{4}}{x + 5} \]

[In]

integrate(((x^2+10*x+25)*log(2)-1080*exp(4))/(x^2+10*x+25),x, algorithm="giac")

[Out]

x*log(2) + 1080*e^4/(x + 5)

Mupad [B] (verification not implemented)

Time = 8.58 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.70 \[ \int \frac {-1080 e^4+\left (25+10 x+x^2\right ) \log (2)}{25+10 x+x^2} \, dx=x\,\ln \left (2\right )+\frac {1080\,{\mathrm {e}}^4}{x+5} \]

[In]

int(-(1080*exp(4) - log(2)*(10*x + x^2 + 25))/(10*x + x^2 + 25),x)

[Out]

x*log(2) + (1080*exp(4))/(x + 5)