3.86.13 \(\int \frac {3 e^{10}+x+3 x^2+e^5 (7+6 x)+e^5 \log (x)}{10 e^{10}+20 e^5 x+10 x^2} \, dx\)

Optimal. Leaf size=30 \[ \frac {2}{5} \left (x+\frac {1}{4} \left (-x+\frac {6+\log (x)}{1+\frac {e^5}{x}}\right )\right ) \]

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Rubi [B]  time = 0.15, antiderivative size = 69, normalized size of antiderivative = 2.30, number of steps used = 12, number of rules used = 6, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.128, Rules used = {27, 12, 6742, 43, 2314, 31} \begin {gather*} \frac {3 x}{10}-\frac {e^5 \left (7-6 e^5\right )}{10 \left (x+e^5\right )}-\frac {3 e^{10}}{5 \left (x+e^5\right )}+\frac {e^5}{10 \left (x+e^5\right )}+\frac {x \log (x)}{10 \left (x+e^5\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3*E^10 + x + 3*x^2 + E^5*(7 + 6*x) + E^5*Log[x])/(10*E^10 + 20*E^5*x + 10*x^2),x]

[Out]

(3*x)/10 + E^5/(10*(E^5 + x)) - (3*E^10)/(5*(E^5 + x)) - (E^5*(7 - 6*E^5))/(10*(E^5 + x)) + (x*Log[x])/(10*(E^
5 + 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 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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 2314

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp[(x*(d + e*x^r)^(q
+ 1)*(a + b*Log[c*x^n]))/d, x] - Dist[(b*n)/d, Int[(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, n, q,
r}, x] && EqQ[r*(q + 1) + 1, 0]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 e^{10}+x+3 x^2+e^5 (7+6 x)+e^5 \log (x)}{10 \left (e^5+x\right )^2} \, dx\\ &=\frac {1}{10} \int \frac {3 e^{10}+x+3 x^2+e^5 (7+6 x)+e^5 \log (x)}{\left (e^5+x\right )^2} \, dx\\ &=\frac {1}{10} \int \left (\frac {3 e^{10}}{\left (e^5+x\right )^2}+\frac {x}{\left (e^5+x\right )^2}+\frac {3 x^2}{\left (e^5+x\right )^2}+\frac {e^5 (7+6 x)}{\left (e^5+x\right )^2}+\frac {e^5 \log (x)}{\left (e^5+x\right )^2}\right ) \, dx\\ &=-\frac {3 e^{10}}{10 \left (e^5+x\right )}+\frac {1}{10} \int \frac {x}{\left (e^5+x\right )^2} \, dx+\frac {3}{10} \int \frac {x^2}{\left (e^5+x\right )^2} \, dx+\frac {1}{10} e^5 \int \frac {7+6 x}{\left (e^5+x\right )^2} \, dx+\frac {1}{10} e^5 \int \frac {\log (x)}{\left (e^5+x\right )^2} \, dx\\ &=-\frac {3 e^{10}}{10 \left (e^5+x\right )}+\frac {x \log (x)}{10 \left (e^5+x\right )}-\frac {1}{10} \int \frac {1}{e^5+x} \, dx+\frac {1}{10} \int \left (-\frac {e^5}{\left (e^5+x\right )^2}+\frac {1}{e^5+x}\right ) \, dx+\frac {3}{10} \int \left (1+\frac {e^{10}}{\left (e^5+x\right )^2}-\frac {2 e^5}{e^5+x}\right ) \, dx+\frac {1}{10} e^5 \int \left (\frac {7-6 e^5}{\left (e^5+x\right )^2}+\frac {6}{e^5+x}\right ) \, dx\\ &=\frac {3 x}{10}+\frac {e^5}{10 \left (e^5+x\right )}-\frac {3 e^{10}}{5 \left (e^5+x\right )}-\frac {e^5 \left (7-6 e^5\right )}{10 \left (e^5+x\right )}+\frac {x \log (x)}{10 \left (e^5+x\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 26, normalized size = 0.87 \begin {gather*} \frac {1}{10} \left (3 x+\log (x)-\frac {e^5 (6+\log (x))}{e^5+x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3*E^10 + x + 3*x^2 + E^5*(7 + 6*x) + E^5*Log[x])/(10*E^10 + 20*E^5*x + 10*x^2),x]

[Out]

(3*x + Log[x] - (E^5*(6 + Log[x]))/(E^5 + x))/10

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fricas [A]  time = 1.07, size = 25, normalized size = 0.83 \begin {gather*} \frac {3 \, x^{2} + 3 \, {\left (x - 2\right )} e^{5} + x \log \relax (x)}{10 \, {\left (x + e^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(5)*log(x)+3*exp(5)^2+(6*x+7)*exp(5)+3*x^2+x)/(10*exp(5)^2+20*x*exp(5)+10*x^2),x, algorithm="fri
cas")

[Out]

1/10*(3*x^2 + 3*(x - 2)*e^5 + x*log(x))/(x + e^5)

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giac [A]  time = 0.22, size = 27, normalized size = 0.90 \begin {gather*} \frac {3 \, x^{2} + 3 \, x e^{5} + x \log \relax (x) - 6 \, e^{5}}{10 \, {\left (x + e^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(5)*log(x)+3*exp(5)^2+(6*x+7)*exp(5)+3*x^2+x)/(10*exp(5)^2+20*x*exp(5)+10*x^2),x, algorithm="gia
c")

[Out]

1/10*(3*x^2 + 3*x*e^5 + x*log(x) - 6*e^5)/(x + e^5)

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maple [A]  time = 0.42, size = 29, normalized size = 0.97

method	result	size
norman	\(\frac {\frac {3 x^{2}}{10}+\frac {x \ln \relax (x )}{10}-\frac {3 \,{\mathrm e}^{10}}{10}-\frac {3 \,{\mathrm e}^{5}}{5}}{{\mathrm e}^{5}+x}\)	\(29\)
risch	\(-\frac {{\mathrm e}^{5} \ln \relax (x )}{10 \left ({\mathrm e}^{5}+x \right )}+\frac {{\mathrm e}^{5} \ln \relax (x )+3 x \,{\mathrm e}^{5}+x \ln \relax (x )+3 x^{2}-6 \,{\mathrm e}^{5}}{10 \,{\mathrm e}^{5}+10 x}\)	\(46\)
default	\(\frac {3 x^{2}-3 \left ({\mathrm e}^{5}\right )^{2}-6 \,{\mathrm e}^{5}}{10 \,{\mathrm e}^{5}+10 x}+\frac {\ln \left ({\mathrm e}^{5}+x \right )}{10}+\frac {{\mathrm e}^{5} \ln \relax (x ) \ln \left (\frac {{\mathrm e}^{5}-\sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}+x}{{\mathrm e}^{5}-\sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}}\right )}{20 \sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}}-\frac {{\mathrm e}^{5} \ln \relax (x ) \ln \left (\frac {{\mathrm e}^{5}+\sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}+x}{{\mathrm e}^{5}+\sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}}\right )}{20 \sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}}+\frac {{\mathrm e}^{5} \dilog \left (\frac {{\mathrm e}^{5}-\sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}+x}{{\mathrm e}^{5}-\sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}}\right )}{20 \sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}}-\frac {{\mathrm e}^{5} \dilog \left (\frac {{\mathrm e}^{5}+\sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}+x}{{\mathrm e}^{5}+\sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}}\right )}{20 \sqrt {\left ({\mathrm e}^{5}\right )^{2}-{\mathrm e}^{10}}}\)	\(237\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(5)*ln(x)+3*exp(5)^2+(6*x+7)*exp(5)+3*x^2+x)/(10*exp(5)^2+20*x*exp(5)+10*x^2),x,method=_RETURNVERBOSE)

[Out]

(3/10*x^2+1/10*x*ln(x)-3/10*exp(5)^2-3/5*exp(5))/(exp(5)+x)

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maxima [B]  time = 0.37, size = 87, normalized size = 2.90 \begin {gather*} -\frac {1}{10} \, {\left (e^{\left (-5\right )} \log \left (x + e^{5}\right ) - e^{\left (-5\right )} \log \relax (x) + \frac {\log \relax (x)}{x + e^{5}}\right )} e^{5} + \frac {3}{5} \, {\left (\frac {e^{5}}{x + e^{5}} + \log \left (x + e^{5}\right )\right )} e^{5} - \frac {3}{5} \, e^{5} \log \left (x + e^{5}\right ) + \frac {3}{10} \, x - \frac {3 \, e^{10}}{5 \, {\left (x + e^{5}\right )}} - \frac {3 \, e^{5}}{5 \, {\left (x + e^{5}\right )}} + \frac {1}{10} \, \log \left (x + e^{5}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(5)*log(x)+3*exp(5)^2+(6*x+7)*exp(5)+3*x^2+x)/(10*exp(5)^2+20*x*exp(5)+10*x^2),x, algorithm="max
ima")

[Out]

-1/10*(e^(-5)*log(x + e^5) - e^(-5)*log(x) + log(x)/(x + e^5))*e^5 + 3/5*(e^5/(x + e^5) + log(x + e^5))*e^5 -
3/5*e^5*log(x + e^5) + 3/10*x - 3/5*e^10/(x + e^5) - 3/5*e^5/(x + e^5) + 1/10*log(x + e^5)

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mupad [B]  time = 5.72, size = 23, normalized size = 0.77 \begin {gather*} \frac {x\,\left (3\,x+3\,{\mathrm {e}}^5+\ln \relax (x)+6\right )}{10\,\left (x+{\mathrm {e}}^5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + 3*exp(10) + exp(5)*log(x) + 3*x^2 + exp(5)*(6*x + 7))/(10*exp(10) + 20*x*exp(5) + 10*x^2),x)

[Out]

(x*(3*x + 3*exp(5) + log(x) + 6))/(10*(x + exp(5)))

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sympy [A]  time = 0.27, size = 37, normalized size = 1.23 \begin {gather*} \frac {3 x}{10} + \frac {\log {\relax (x )}}{10} - \frac {e^{5} \log {\relax (x )}}{10 x + 10 e^{5}} - \frac {3 e^{5}}{5 x + 5 e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(5)*ln(x)+3*exp(5)**2+(6*x+7)*exp(5)+3*x**2+x)/(10*exp(5)**2+20*x*exp(5)+10*x**2),x)

[Out]

3*x/10 + log(x)/10 - exp(5)*log(x)/(10*x + 10*exp(5)) - 3*exp(5)/(5*x + 5*exp(5))

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