\(\int \frac {1+e^{260+85 x+5 x^2} (85 x+10 x^2)}{x} \, dx\) [8182]

Optimal result

Integrand size = 28, antiderivative size = 18 \[ \int \frac {1+e^{260+85 x+5 x^2} \left (85 x+10 x^2\right )}{x} \, dx=-\frac {4}{3}+e^{5 (4+x) (13+x)}+\log (2 x) \]

[Out]

ln(2*x)-4/3+exp((13+x)*(20+5*x))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {14, 2268} \[ \int \frac {1+e^{260+85 x+5 x^2} \left (85 x+10 x^2\right )}{x} \, dx=e^{5 x^2+85 x+260}+\log (x) \]

[In]

Int[(1 + E^(260 + 85*x + 5*x^2)*(85*x + 10*x^2))/x,x]

[Out]

E^(260 + 85*x + 5*x^2) + Log[x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2268

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

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{x}+5 e^{260+85 x+5 x^2} (17+2 x)\right ) \, dx \\ & = \log (x)+5 \int e^{260+85 x+5 x^2} (17+2 x) \, dx \\ & = e^{260+85 x+5 x^2}+\log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.72 \[ \int \frac {1+e^{260+85 x+5 x^2} \left (85 x+10 x^2\right )}{x} \, dx=e^{5 (4+x) (13+x)}+\log (x) \]

[In]

Integrate[(1 + E^(260 + 85*x + 5*x^2)*(85*x + 10*x^2))/x,x]

[Out]

E^(5*(4 + x)*(13 + x)) + Log[x]

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.72

method	result	size
risch	\({\mathrm e}^{5 \left (x +13\right ) \left (4+x \right )}+\ln \left (x \right )\)	\(13\)
norman	\({\mathrm e}^{5 x^{2}+85 x +260}+\ln \left (x \right )\)	\(15\)
parallelrisch	\({\mathrm e}^{5 x^{2}+85 x +260}+\ln \left (x \right )\)	\(15\)
default	\(\ln \left (x \right )-\frac {17 i {\mathrm e}^{260} \sqrt {\pi }\, {\mathrm e}^{-\frac {1445}{4}} \sqrt {5}\, \operatorname {erf}\left (i \sqrt {5}\, x +\frac {17 i \sqrt {5}}{2}\right )}{2}+10 \,{\mathrm e}^{260} \left (\frac {{\mathrm e}^{5 x^{2}+85 x}}{10}+\frac {17 i \sqrt {\pi }\, {\mathrm e}^{-\frac {1445}{4}} \sqrt {5}\, \operatorname {erf}\left (i \sqrt {5}\, x +\frac {17 i \sqrt {5}}{2}\right )}{20}\right )\)	\(75\)
parts	\(\ln \left (x \right )-\frac {17 i {\mathrm e}^{260} \sqrt {\pi }\, {\mathrm e}^{-\frac {1445}{4}} \sqrt {5}\, \operatorname {erf}\left (i \sqrt {5}\, x +\frac {17 i \sqrt {5}}{2}\right )}{2}+10 \,{\mathrm e}^{260} \left (\frac {{\mathrm e}^{5 x^{2}+85 x}}{10}+\frac {17 i \sqrt {\pi }\, {\mathrm e}^{-\frac {1445}{4}} \sqrt {5}\, \operatorname {erf}\left (i \sqrt {5}\, x +\frac {17 i \sqrt {5}}{2}\right )}{20}\right )\)	\(75\)

[In]

int(((10*x^2+85*x)*exp(5*x^2+85*x+260)+1)/x,x,method=_RETURNVERBOSE)

[Out]

exp(5*(x+13)*(4+x))+ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {1+e^{260+85 x+5 x^2} \left (85 x+10 x^2\right )}{x} \, dx=e^{\left (5 \, x^{2} + 85 \, x + 260\right )} + \log \left (x\right ) \]

[In]

integrate(((10*x^2+85*x)*exp(5*x^2+85*x+260)+1)/x,x, algorithm="fricas")

[Out]

e^(5*x^2 + 85*x + 260) + log(x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {1+e^{260+85 x+5 x^2} \left (85 x+10 x^2\right )}{x} \, dx=e^{5 x^{2} + 85 x + 260} + \log {\left (x \right )} \]

[In]

integrate(((10*x**2+85*x)*exp(5*x**2+85*x+260)+1)/x,x)

[Out]

exp(5*x**2 + 85*x + 260) + log(x)

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.31 (sec) , antiderivative size = 89, normalized size of antiderivative = 4.94 \[ \int \frac {1+e^{260+85 x+5 x^2} \left (85 x+10 x^2\right )}{x} \, dx=-\frac {17}{2} i \, \sqrt {5} \sqrt {\pi } \operatorname {erf}\left (i \, \sqrt {5} x + \frac {17}{2} i \, \sqrt {5}\right ) e^{\left (-\frac {405}{4}\right )} - \frac {1}{10} \, \sqrt {5} {\left (\frac {85 \, \sqrt {\pi } {\left (2 \, x + 17\right )} {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {5} \sqrt {-{\left (2 \, x + 17\right )}^{2}}\right ) - 1\right )}}{\sqrt {-{\left (2 \, x + 17\right )}^{2}}} - 2 \, \sqrt {5} e^{\left (\frac {5}{4} \, {\left (2 \, x + 17\right )}^{2}\right )}\right )} e^{\left (-\frac {405}{4}\right )} + \log \left (x\right ) \]

[In]

integrate(((10*x^2+85*x)*exp(5*x^2+85*x+260)+1)/x,x, algorithm="maxima")

[Out]

-17/2*I*sqrt(5)*sqrt(pi)*erf(I*sqrt(5)*x + 17/2*I*sqrt(5))*e^(-405/4) - 1/10*sqrt(5)*(85*sqrt(pi)*(2*x + 17)*(
erf(1/2*sqrt(5)*sqrt(-(2*x + 17)^2)) - 1)/sqrt(-(2*x + 17)^2) - 2*sqrt(5)*e^(5/4*(2*x + 17)^2))*e^(-405/4) + l
og(x)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int \frac {1+e^{260+85 x+5 x^2} \left (85 x+10 x^2\right )}{x} \, dx=e^{\left (5 \, x^{2} + 85 \, x + 260\right )} + \log \left (x\right ) \]

[In]

integrate(((10*x^2+85*x)*exp(5*x^2+85*x+260)+1)/x,x, algorithm="giac")

[Out]

e^(5*x^2 + 85*x + 260) + log(x)

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {1+e^{260+85 x+5 x^2} \left (85 x+10 x^2\right )}{x} \, dx=\ln \left (x\right )+{\mathrm {e}}^{85\,x}\,{\mathrm {e}}^{260}\,{\mathrm {e}}^{5\,x^2} \]

[In]

int((exp(85*x + 5*x^2 + 260)*(85*x + 10*x^2) + 1)/x,x)

[Out]

log(x) + exp(85*x)*exp(260)*exp(5*x^2)