3.97.13 \(\int \frac {1+e^x x-4 x \log ^2(5)}{x} \, dx\)

Optimal. Leaf size=24 \[ 6+e^x-\frac {4 \left (-2 x+x^2 \log ^2(5)\right )}{x}+\log (x) \]

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Rubi [A]  time = 0.01, antiderivative size = 13, normalized size of antiderivative = 0.54, number of steps used = 5, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {14, 2194, 43} \begin {gather*} e^x-4 x \log ^2(5)+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + E^x*x - 4*x*Log[5]^2)/x,x]

[Out]

E^x - 4*x*Log[5]^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 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 {1-4 x \log ^2(5)}{x}\right ) \, dx\\ &=\int e^x \, dx+\int \frac {1-4 x \log ^2(5)}{x} \, dx\\ &=e^x+\int \left (\frac {1}{x}-4 \log ^2(5)\right ) \, dx\\ &=e^x-4 x \log ^2(5)+\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 13, normalized size = 0.54 \begin {gather*} e^x-4 x \log ^2(5)+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + E^x*x - 4*x*Log[5]^2)/x,x]

[Out]

E^x - 4*x*Log[5]^2 + Log[x]

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fricas [A]  time = 0.74, size = 12, normalized size = 0.50 \begin {gather*} -4 \, x \log \relax (5)^{2} + e^{x} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)*x-4*x*log(5)^2+1)/x,x, algorithm="fricas")

[Out]

-4*x*log(5)^2 + e^x + log(x)

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giac [A]  time = 0.21, size = 12, normalized size = 0.50 \begin {gather*} -4 \, x \log \relax (5)^{2} + e^{x} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)*x-4*x*log(5)^2+1)/x,x, algorithm="giac")

[Out]

-4*x*log(5)^2 + e^x + log(x)

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maple [A]  time = 0.03, size = 13, normalized size = 0.54

method	result	size
default	\(\ln \relax (x )-4 x \ln \relax (5)^{2}+{\mathrm e}^{x}\)	\(13\)
norman	\(\ln \relax (x )-4 x \ln \relax (5)^{2}+{\mathrm e}^{x}\)	\(13\)
risch	\(\ln \relax (x )-4 x \ln \relax (5)^{2}+{\mathrm e}^{x}\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*x-4*x*ln(5)^2+1)/x,x,method=_RETURNVERBOSE)

[Out]

ln(x)-4*x*ln(5)^2+exp(x)

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maxima [A]  time = 0.38, size = 12, normalized size = 0.50 \begin {gather*} -4 \, x \log \relax (5)^{2} + e^{x} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)*x-4*x*log(5)^2+1)/x,x, algorithm="maxima")

[Out]

-4*x*log(5)^2 + e^x + log(x)

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mupad [B]  time = 0.05, size = 12, normalized size = 0.50 \begin {gather*} {\mathrm {e}}^x+\ln \relax (x)-4\,x\,{\ln \relax (5)}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*exp(x) - 4*x*log(5)^2 + 1)/x,x)

[Out]

exp(x) + log(x) - 4*x*log(5)^2

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sympy [A]  time = 0.10, size = 14, normalized size = 0.58 \begin {gather*} - 4 x \log {\relax (5 )}^{2} + e^{x} + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)*x-4*x*ln(5)**2+1)/x,x)

[Out]

-4*x*log(5)**2 + exp(x) + log(x)

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