3.6.64 \(\int \frac {5+10 x+e^x x}{x} \, dx\)

Optimal. Leaf size=22 \[ 5+e^x+5 x+5 \left (5+\frac {e^5}{3}+x+\log (x)\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 11, normalized size of antiderivative = 0.50, number of steps used = 5, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {14, 2194, 43} \begin {gather*} 10 x+e^x+5 \log (x) \end {gather*}

Antiderivative was successfully verified.

[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 43

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 2194

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

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Mathematica [A]  time = 0.01, size = 11, normalized size = 0.50 \begin {gather*} e^x+10 x+5 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5 + 10*x + E^x*x)/x,x]

[Out]

E^x + 10*x + 5*Log[x]

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fricas [A]  time = 0.62, size = 10, normalized size = 0.45 \begin {gather*} 10 \, x + e^{x} + 5 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

10*x + e^x + 5*log(x)

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giac [A]  time = 0.66, size = 10, normalized size = 0.45 \begin {gather*} 10 \, x + e^{x} + 5 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

10*x + e^x + 5*log(x)

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maple [A]  time = 0.02, size = 11, normalized size = 0.50

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 10, normalized size = 0.45 \begin {gather*} 10 \, x + e^{x} + 5 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

10*x + e^x + 5*log(x)

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mupad [B]  time = 0.44, size = 10, normalized size = 0.45 \begin {gather*} 10\,x+{\mathrm {e}}^x+5\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*x + x*exp(x) + 5)/x,x)

[Out]

10*x + exp(x) + 5*log(x)

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sympy [A]  time = 0.09, size = 10, normalized size = 0.45 \begin {gather*} 10 x + e^{x} + 5 \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

10*x + exp(x) + 5*log(x)

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