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

Optimal. Leaf size=18 \[ -\frac {4}{3}+e^{5 (4+x) (13+x)}+\log (2 x) \]

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Rubi [A]  time = 0.03, antiderivative size = 15, normalized size of antiderivative = 0.83, number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {14, 2236} \begin {gather*} e^{5 x^2+85 x+260}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[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 2236

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 15, normalized size = 0.83 \begin {gather*} e^{260+85 x+5 x^2}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(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]

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fricas [A]  time = 0.86, size = 14, normalized size = 0.78 \begin {gather*} e^{\left (5 \, x^{2} + 85 \, x + 260\right )} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.20, size = 14, normalized size = 0.78 \begin {gather*} e^{\left (5 \, x^{2} + 85 \, x + 260\right )} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.03, size = 13, normalized size = 0.72

method	result	size
risch	\(\ln \relax (x )+{\mathrm e}^{5 \left (x +13\right ) \left (4+x \right )}\)	\(13\)
default	\(\ln \relax (x )+{\mathrm e}^{5 x^{2}+85 x +260}\)	\(15\)
norman	\(\ln \relax (x )+{\mathrm e}^{5 x^{2}+85 x +260}\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.49, size = 89, normalized size = 4.94 \begin {gather*} -\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 \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 0.11, size = 16, normalized size = 0.89 \begin {gather*} \ln \relax (x)+{\mathrm {e}}^{85\,x}\,{\mathrm {e}}^{260}\,{\mathrm {e}}^{5\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.10, size = 14, normalized size = 0.78 \begin {gather*} e^{5 x^{2} + 85 x + 260} + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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