\(\int \frac {5+10 x+e^x x}{x} \, dx\) [564]

Optimal result

Integrand size = 14, antiderivative size = 22 \[ \int \frac {5+10 x+e^x x}{x} \, dx=5+e^x+5 x+5 \left (5+\frac {e^5}{3}+x+\log (x)\right ) \]

[Out]

10*x+5*ln(x)+5/3*exp(5)+30+exp(x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.50, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {14, 2225, 45} \[ \int \frac {5+10 x+e^x x}{x} \, dx=10 x+e^x+5 \log (x) \]

[In]

Int[(5 + 10*x + E^x*x)/x,x]

[Out]

E^x + 10*x + 5*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 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (e^x+\frac {5 (1+2 x)}{x}\right ) \, dx \\ & = 5 \int \frac {1+2 x}{x} \, dx+\int e^x \, dx \\ & = e^x+5 \int \left (2+\frac {1}{x}\right ) \, dx \\ & = e^x+10 x+5 \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.50 \[ \int \frac {5+10 x+e^x x}{x} \, dx=e^x+10 x+5 \log (x) \]

[In]

Integrate[(5 + 10*x + E^x*x)/x,x]

[Out]

E^x + 10*x + 5*Log[x]

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.50

method	result	size
default	\(10 x +5 \ln \left (x \right )+{\mathrm e}^{x}\)	\(11\)
norman	\(10 x +5 \ln \left (x \right )+{\mathrm e}^{x}\)	\(11\)
risch	\(10 x +5 \ln \left (x \right )+{\mathrm e}^{x}\)	\(11\)
parallelrisch	\(10 x +5 \ln \left (x \right )+{\mathrm e}^{x}\)	\(11\)
parts	\(10 x +5 \ln \left (x \right )+{\mathrm e}^{x}\)	\(11\)

[In]

int((exp(x)*x+10*x+5)/x,x,method=_RETURNVERBOSE)

[Out]

10*x+5*ln(x)+exp(x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.45 \[ \int \frac {5+10 x+e^x x}{x} \, dx=10 \, x + e^{x} + 5 \, \log \left (x\right ) \]

[In]

integrate((exp(x)*x+10*x+5)/x,x, algorithm="fricas")

[Out]

10*x + e^x + 5*log(x)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.45 \[ \int \frac {5+10 x+e^x x}{x} \, dx=10 x + e^{x} + 5 \log {\left (x \right )} \]

[In]

integrate((exp(x)*x+10*x+5)/x,x)

[Out]

10*x + exp(x) + 5*log(x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.45 \[ \int \frac {5+10 x+e^x x}{x} \, dx=10 \, x + e^{x} + 5 \, \log \left (x\right ) \]

[In]

integrate((exp(x)*x+10*x+5)/x,x, algorithm="maxima")

[Out]

10*x + e^x + 5*log(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.45 \[ \int \frac {5+10 x+e^x x}{x} \, dx=10 \, x + e^{x} + 5 \, \log \left (x\right ) \]

[In]

integrate((exp(x)*x+10*x+5)/x,x, algorithm="giac")

[Out]

10*x + e^x + 5*log(x)

Mupad [B] (verification not implemented)

Time = 7.33 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.45 \[ \int \frac {5+10 x+e^x x}{x} \, dx=10\,x+{\mathrm {e}}^x+5\,\ln \left (x\right ) \]

[In]

int((10*x + x*exp(x) + 5)/x,x)

[Out]

10*x + exp(x) + 5*log(x)