3.20.1 \(\int \frac {(2+2 x) \log (5)+e^x (x+x^2+x \log ^2(5))}{2 x \log (5)} \, dx\)

Optimal. Leaf size=27 \[ 3 e+x-\log (5)+\frac {1}{2} e^x \left (\frac {x}{\log (5)}+\log (5)\right )+\log (x) \]

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Rubi [A]  time = 0.04, antiderivative size = 33, normalized size of antiderivative = 1.22, number of steps used = 7, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {12, 14, 43, 2176, 2194} \begin {gather*} x+\frac {e^x \left (x+1+\log ^2(5)\right )}{2 \log (5)}+\log (x)-\frac {e^x}{2 \log (5)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((2 + 2*x)*Log[5] + E^x*(x + x^2 + x*Log[5]^2))/(2*x*Log[5]),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

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

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Mathematica [A]  time = 0.04, size = 18, normalized size = 0.67 \begin {gather*} x+\frac {e^x \left (x+\log ^2(5)\right )}{\log (25)}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x + (E^x*(x + Log[5]^2))/Log[25] + Log[x]

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fricas [A]  time = 0.65, size = 27, normalized size = 1.00 \begin {gather*} \frac {{\left (\log \relax (5)^{2} + x\right )} e^{x} + 2 \, x \log \relax (5) + 2 \, \log \relax (5) \log \relax (x)}{2 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((log(5)^2 + x)*e^x + 2*x*log(5) + 2*log(5)*log(x))/log(5)

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giac [A]  time = 0.25, size = 29, normalized size = 1.07 \begin {gather*} \frac {e^{x} \log \relax (5)^{2} + x e^{x} + 2 \, x \log \relax (5) + 2 \, \log \relax (5) \log \relax (x)}{2 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(e^x*log(5)^2 + x*e^x + 2*x*log(5) + 2*log(5)*log(x))/log(5)

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maple [A]  time = 0.04, size = 19, normalized size = 0.70

method	result	size
risch	\(x +\ln \relax (x )+\frac {\left (\ln \relax (5)^{2}+x \right ) {\mathrm e}^{x}}{2 \ln \relax (5)}\)	\(19\)
norman	\(x +\frac {{\mathrm e}^{x} \ln \relax (5)}{2}+\frac {{\mathrm e}^{x} x}{2 \ln \relax (5)}+\ln \relax (x )\)	\(20\)
default	\(\frac {{\mathrm e}^{x} x +\ln \relax (5)^{2} {\mathrm e}^{x}+2 \ln \relax (5) \ln \relax (x )+2 x \ln \relax (5)}{2 \ln \relax (5)}\)	\(30\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x+ln(x)+1/2/ln(5)*(ln(5)^2+x)*exp(x)

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maxima [A]  time = 0.69, size = 33, normalized size = 1.22 \begin {gather*} \frac {e^{x} \log \relax (5)^{2} + {\left (x - 1\right )} e^{x} + 2 \, x \log \relax (5) + 2 \, \log \relax (5) \log \relax (x) + e^{x}}{2 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(e^x*log(5)^2 + (x - 1)*e^x + 2*x*log(5) + 2*log(5)*log(x) + e^x)/log(5)

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mupad [B]  time = 1.12, size = 19, normalized size = 0.70 \begin {gather*} x+\ln \relax (x)+\frac {{\mathrm {e}}^x\,\ln \relax (5)}{2}+\frac {x\,{\mathrm {e}}^x}{2\,\ln \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + log(x) + (exp(x)*log(5))/2 + (x*exp(x))/(2*log(5))

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sympy [A]  time = 0.13, size = 19, normalized size = 0.70 \begin {gather*} x + \frac {\left (x + \log {\relax (5 )}^{2}\right ) e^{x}}{2 \log {\relax (5 )}} + \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + (x + log(5)**2)*exp(x)/(2*log(5)) + log(x)

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