3.14.69 \(\int \frac {25000 e^{\frac {25+3 x}{5+x}}-315000 e^{\frac {25+3 x}{5+x}} \log (2+e^{\frac {25+3 x}{5+x}})+1318000 e^{\frac {25+3 x}{5+x}} \log ^2(2+e^{\frac {25+3 x}{5+x}})-1831200 e^{\frac {25+3 x}{5+x}} \log ^3(2+e^{\frac {25+3 x}{5+x}})}{(50+20 x+2 x^2+e^{\frac {25+3 x}{5+x}} (25+10 x+x^2)) \log ^5(2+e^{\frac {25+3 x}{5+x}})} \, dx\)

Optimal. Leaf size=29 \[ \left (5-\left (21-\frac {5}{\log \left (2+e^{5-\frac {2 x}{5+x}}\right )}\right )^2\right )^2 \]

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Rubi [B]  time = 3.49, antiderivative size = 81, normalized size of antiderivative = 2.79, number of steps used = 8, number of rules used = 4, integrand size = 166, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {6688, 12, 6742, 6686} \begin {gather*} \frac {625}{\log ^4\left (e^{\frac {3 x+25}{x+5}}+2\right )}-\frac {10500}{\log ^3\left (e^{\frac {3 x+25}{x+5}}+2\right )}+\frac {65900}{\log ^2\left (e^{\frac {3 x+25}{x+5}}+2\right )}-\frac {183120}{\log \left (e^{\frac {3 x+25}{x+5}}+2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(25000*E^((25 + 3*x)/(5 + x)) - 315000*E^((25 + 3*x)/(5 + x))*Log[2 + E^((25 + 3*x)/(5 + x))] + 1318000*E^
((25 + 3*x)/(5 + x))*Log[2 + E^((25 + 3*x)/(5 + x))]^2 - 1831200*E^((25 + 3*x)/(5 + x))*Log[2 + E^((25 + 3*x)/
(5 + x))]^3)/((50 + 20*x + 2*x^2 + E^((25 + 3*x)/(5 + x))*(25 + 10*x + x^2))*Log[2 + E^((25 + 3*x)/(5 + x))]^5
),x]

[Out]

625/Log[2 + E^((25 + 3*x)/(5 + x))]^4 - 10500/Log[2 + E^((25 + 3*x)/(5 + x))]^3 + 65900/Log[2 + E^((25 + 3*x)/
(5 + x))]^2 - 183120/Log[2 + E^((25 + 3*x)/(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 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

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

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 {200 e^{\frac {25+3 x}{5+x}} \left (125-1575 \log \left (2+e^{\frac {25+3 x}{5+x}}\right )+6590 \log ^2\left (2+e^{\frac {25+3 x}{5+x}}\right )-9156 \log ^3\left (2+e^{\frac {25+3 x}{5+x}}\right )\right )}{\left (2+e^{\frac {25+3 x}{5+x}}\right ) (5+x)^2 \log ^5\left (2+e^{\frac {25+3 x}{5+x}}\right )} \, dx\\ &=200 \int \frac {e^{\frac {25+3 x}{5+x}} \left (125-1575 \log \left (2+e^{\frac {25+3 x}{5+x}}\right )+6590 \log ^2\left (2+e^{\frac {25+3 x}{5+x}}\right )-9156 \log ^3\left (2+e^{\frac {25+3 x}{5+x}}\right )\right )}{\left (2+e^{\frac {25+3 x}{5+x}}\right ) (5+x)^2 \log ^5\left (2+e^{\frac {25+3 x}{5+x}}\right )} \, dx\\ &=200 \int \left (\frac {125 e^{\frac {25+3 x}{5+x}}}{\left (2+e^{\frac {25}{5+x}+\frac {3 x}{5+x}}\right ) (5+x)^2 \log ^5\left (2+e^{\frac {25+3 x}{5+x}}\right )}-\frac {1575 e^{\frac {25+3 x}{5+x}}}{\left (2+e^{\frac {25}{5+x}+\frac {3 x}{5+x}}\right ) (5+x)^2 \log ^4\left (2+e^{\frac {25+3 x}{5+x}}\right )}+\frac {6590 e^{\frac {25+3 x}{5+x}}}{\left (2+e^{\frac {25}{5+x}+\frac {3 x}{5+x}}\right ) (5+x)^2 \log ^3\left (2+e^{\frac {25+3 x}{5+x}}\right )}-\frac {9156 e^{\frac {25+3 x}{5+x}}}{\left (2+e^{\frac {25}{5+x}+\frac {3 x}{5+x}}\right ) (5+x)^2 \log ^2\left (2+e^{\frac {25+3 x}{5+x}}\right )}\right ) \, dx\\ &=25000 \int \frac {e^{\frac {25+3 x}{5+x}}}{\left (2+e^{\frac {25}{5+x}+\frac {3 x}{5+x}}\right ) (5+x)^2 \log ^5\left (2+e^{\frac {25+3 x}{5+x}}\right )} \, dx-315000 \int \frac {e^{\frac {25+3 x}{5+x}}}{\left (2+e^{\frac {25}{5+x}+\frac {3 x}{5+x}}\right ) (5+x)^2 \log ^4\left (2+e^{\frac {25+3 x}{5+x}}\right )} \, dx+1318000 \int \frac {e^{\frac {25+3 x}{5+x}}}{\left (2+e^{\frac {25}{5+x}+\frac {3 x}{5+x}}\right ) (5+x)^2 \log ^3\left (2+e^{\frac {25+3 x}{5+x}}\right )} \, dx-1831200 \int \frac {e^{\frac {25+3 x}{5+x}}}{\left (2+e^{\frac {25}{5+x}+\frac {3 x}{5+x}}\right ) (5+x)^2 \log ^2\left (2+e^{\frac {25+3 x}{5+x}}\right )} \, dx\\ &=\frac {625}{\log ^4\left (2+e^{\frac {25+3 x}{5+x}}\right )}-\frac {10500}{\log ^3\left (2+e^{\frac {25+3 x}{5+x}}\right )}+\frac {65900}{\log ^2\left (2+e^{\frac {25+3 x}{5+x}}\right )}-\frac {183120}{\log \left (2+e^{\frac {25+3 x}{5+x}}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.04, size = 91, normalized size = 3.14 \begin {gather*} -200 \left (-\frac {25}{8 \log ^4\left (2+e^{\frac {25+3 x}{5+x}}\right )}+\frac {105}{2 \log ^3\left (2+e^{\frac {25+3 x}{5+x}}\right )}-\frac {659}{2 \log ^2\left (2+e^{\frac {25+3 x}{5+x}}\right )}+\frac {4578}{5 \log \left (2+e^{\frac {25+3 x}{5+x}}\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25000*E^((25 + 3*x)/(5 + x)) - 315000*E^((25 + 3*x)/(5 + x))*Log[2 + E^((25 + 3*x)/(5 + x))] + 1318
000*E^((25 + 3*x)/(5 + x))*Log[2 + E^((25 + 3*x)/(5 + x))]^2 - 1831200*E^((25 + 3*x)/(5 + x))*Log[2 + E^((25 +
3*x)/(5 + x))]^3)/((50 + 20*x + 2*x^2 + E^((25 + 3*x)/(5 + x))*(25 + 10*x + x^2))*Log[2 + E^((25 + 3*x)/(5 +
x))]^5),x]

[Out]

-200*(-25/(8*Log[2 + E^((25 + 3*x)/(5 + x))]^4) + 105/(2*Log[2 + E^((25 + 3*x)/(5 + x))]^3) - 659/(2*Log[2 + E
^((25 + 3*x)/(5 + x))]^2) + 4578/(5*Log[2 + E^((25 + 3*x)/(5 + x))]))

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fricas [B]  time = 0.77, size = 76, normalized size = 2.62 \begin {gather*} -\frac {5 \, {\left (36624 \, \log \left (e^{\left (\frac {3 \, x + 25}{x + 5}\right )} + 2\right )^{3} - 13180 \, \log \left (e^{\left (\frac {3 \, x + 25}{x + 5}\right )} + 2\right )^{2} + 2100 \, \log \left (e^{\left (\frac {3 \, x + 25}{x + 5}\right )} + 2\right ) - 125\right )}}{\log \left (e^{\left (\frac {3 \, x + 25}{x + 5}\right )} + 2\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1831200*exp((25+3*x)/(5+x))*log(exp((25+3*x)/(5+x))+2)^3+1318000*exp((25+3*x)/(5+x))*log(exp((25+3
*x)/(5+x))+2)^2-315000*exp((25+3*x)/(5+x))*log(exp((25+3*x)/(5+x))+2)+25000*exp((25+3*x)/(5+x)))/((x^2+10*x+25
)*exp((25+3*x)/(5+x))+2*x^2+20*x+50)/log(exp((25+3*x)/(5+x))+2)^5,x, algorithm="fricas")

[Out]

-5*(36624*log(e^((3*x + 25)/(x + 5)) + 2)^3 - 13180*log(e^((3*x + 25)/(x + 5)) + 2)^2 + 2100*log(e^((3*x + 25)
/(x + 5)) + 2) - 125)/log(e^((3*x + 25)/(x + 5)) + 2)^4

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giac [B]  time = 3.65, size = 72, normalized size = 2.48 \begin {gather*} -\frac {5 \, {\left (36624 \, \log \left (e^{\left (-\frac {2 \, x}{x + 5} + 5\right )} + 2\right )^{3} - 13180 \, \log \left (e^{\left (-\frac {2 \, x}{x + 5} + 5\right )} + 2\right )^{2} + 2100 \, \log \left (e^{\left (-\frac {2 \, x}{x + 5} + 5\right )} + 2\right ) - 125\right )}}{\log \left (e^{\left (-\frac {2 \, x}{x + 5} + 5\right )} + 2\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1831200*exp((25+3*x)/(5+x))*log(exp((25+3*x)/(5+x))+2)^3+1318000*exp((25+3*x)/(5+x))*log(exp((25+3
*x)/(5+x))+2)^2-315000*exp((25+3*x)/(5+x))*log(exp((25+3*x)/(5+x))+2)+25000*exp((25+3*x)/(5+x)))/((x^2+10*x+25
)*exp((25+3*x)/(5+x))+2*x^2+20*x+50)/log(exp((25+3*x)/(5+x))+2)^5,x, algorithm="giac")

[Out]

-5*(36624*log(e^(-2*x/(x + 5) + 5) + 2)^3 - 13180*log(e^(-2*x/(x + 5) + 5) + 2)^2 + 2100*log(e^(-2*x/(x + 5) +
5) + 2) - 125)/log(e^(-2*x/(x + 5) + 5) + 2)^4

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maple [B]  time = 0.03, size = 77, normalized size = 2.66

method	result	size
risch	\(-\frac {5 \left (36624 \ln \left ({\mathrm e}^{\frac {25+3 x}{5+x}}+2\right )^{3}-13180 \ln \left ({\mathrm e}^{\frac {25+3 x}{5+x}}+2\right )^{2}+2100 \ln \left ({\mathrm e}^{\frac {25+3 x}{5+x}}+2\right )-125\right )}{\ln \left ({\mathrm e}^{\frac {25+3 x}{5+x}}+2\right )^{4}}\)	\(77\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-1831200*exp((25+3*x)/(5+x))*ln(exp((25+3*x)/(5+x))+2)^3+1318000*exp((25+3*x)/(5+x))*ln(exp((25+3*x)/(5+x
))+2)^2-315000*exp((25+3*x)/(5+x))*ln(exp((25+3*x)/(5+x))+2)+25000*exp((25+3*x)/(5+x)))/((x^2+10*x+25)*exp((25
+3*x)/(5+x))+2*x^2+20*x+50)/ln(exp((25+3*x)/(5+x))+2)^5,x,method=_RETURNVERBOSE)

[Out]

-5*(36624*ln(exp((25+3*x)/(5+x))+2)^3-13180*ln(exp((25+3*x)/(5+x))+2)^2+2100*ln(exp((25+3*x)/(5+x))+2)-125)/ln
(exp((25+3*x)/(5+x))+2)^4

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maxima [B]  time = 0.54, size = 297, normalized size = 10.24 \begin {gather*} -\frac {45780 \, \log \left (e^{\left (\frac {3 \, x}{x + 5} + \frac {25}{x + 5}\right )} + 2\right )^{3}}{\log \left (e^{\left (\frac {10}{x + 5} + 3\right )} + 2\right )^{4}} - \frac {45780 \, \log \left (e^{\left (\frac {3 \, x}{x + 5} + \frac {25}{x + 5}\right )} + 2\right )^{2}}{\log \left (e^{\left (\frac {10}{x + 5} + 3\right )} + 2\right )^{3}} - \frac {45780 \, \log \left (e^{\left (\frac {3 \, x}{x + 5} + \frac {25}{x + 5}\right )} + 2\right )}{\log \left (e^{\left (\frac {10}{x + 5} + 3\right )} + 2\right )^{2}} - \frac {45780}{\log \left (e^{\left (\frac {10}{x + 5} + 3\right )} + 2\right )} + \frac {32950 \, \log \left (e^{\left (\frac {3 \, x}{x + 5} + \frac {25}{x + 5}\right )} + 2\right )^{2}}{\log \left (e^{\left (\frac {10}{x + 5} + 3\right )} + 2\right )^{4}} + \frac {65900 \, \log \left (e^{\left (\frac {3 \, x}{x + 5} + \frac {25}{x + 5}\right )} + 2\right )}{3 \, \log \left (e^{\left (\frac {10}{x + 5} + 3\right )} + 2\right )^{3}} + \frac {32950}{3 \, \log \left (e^{\left (\frac {10}{x + 5} + 3\right )} + 2\right )^{2}} - \frac {7875 \, \log \left (e^{\left (\frac {3 \, x}{x + 5} + \frac {25}{x + 5}\right )} + 2\right )}{\log \left (e^{\left (\frac {10}{x + 5} + 3\right )} + 2\right )^{4}} - \frac {2625}{\log \left (e^{\left (\frac {10}{x + 5} + 3\right )} + 2\right )^{3}} + \frac {625}{\log \left (e^{\left (\frac {10}{x + 5} + 3\right )} + 2\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1831200*exp((25+3*x)/(5+x))*log(exp((25+3*x)/(5+x))+2)^3+1318000*exp((25+3*x)/(5+x))*log(exp((25+3
*x)/(5+x))+2)^2-315000*exp((25+3*x)/(5+x))*log(exp((25+3*x)/(5+x))+2)+25000*exp((25+3*x)/(5+x)))/((x^2+10*x+25
)*exp((25+3*x)/(5+x))+2*x^2+20*x+50)/log(exp((25+3*x)/(5+x))+2)^5,x, algorithm="maxima")

[Out]

-45780*log(e^(3*x/(x + 5) + 25/(x + 5)) + 2)^3/log(e^(10/(x + 5) + 3) + 2)^4 - 45780*log(e^(3*x/(x + 5) + 25/(
x + 5)) + 2)^2/log(e^(10/(x + 5) + 3) + 2)^3 - 45780*log(e^(3*x/(x + 5) + 25/(x + 5)) + 2)/log(e^(10/(x + 5) +
3) + 2)^2 - 45780/log(e^(10/(x + 5) + 3) + 2) + 32950*log(e^(3*x/(x + 5) + 25/(x + 5)) + 2)^2/log(e^(10/(x +
5) + 3) + 2)^4 + 65900/3*log(e^(3*x/(x + 5) + 25/(x + 5)) + 2)/log(e^(10/(x + 5) + 3) + 2)^3 + 32950/3/log(e^(
10/(x + 5) + 3) + 2)^2 - 7875*log(e^(3*x/(x + 5) + 25/(x + 5)) + 2)/log(e^(10/(x + 5) + 3) + 2)^4 - 2625/log(e
^(10/(x + 5) + 3) + 2)^3 + 625/log(e^(10/(x + 5) + 3) + 2)^4

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mupad [B]  time = 1.17, size = 101, normalized size = 3.48 \begin {gather*} \frac {65900}{{\ln \left ({\mathrm {e}}^{\frac {3\,x}{x+5}}\,{\mathrm {e}}^{\frac {25}{x+5}}+2\right )}^2}-\frac {183120}{\ln \left ({\mathrm {e}}^{\frac {3\,x}{x+5}}\,{\mathrm {e}}^{\frac {25}{x+5}}+2\right )}-\frac {10500}{{\ln \left ({\mathrm {e}}^{\frac {3\,x}{x+5}}\,{\mathrm {e}}^{\frac {25}{x+5}}+2\right )}^3}+\frac {625}{{\ln \left ({\mathrm {e}}^{\frac {3\,x}{x+5}}\,{\mathrm {e}}^{\frac {25}{x+5}}+2\right )}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((25000*exp((3*x + 25)/(x + 5)) - 315000*exp((3*x + 25)/(x + 5))*log(exp((3*x + 25)/(x + 5)) + 2) + 1318000
*exp((3*x + 25)/(x + 5))*log(exp((3*x + 25)/(x + 5)) + 2)^2 - 1831200*exp((3*x + 25)/(x + 5))*log(exp((3*x + 2
5)/(x + 5)) + 2)^3)/(log(exp((3*x + 25)/(x + 5)) + 2)^5*(20*x + exp((3*x + 25)/(x + 5))*(10*x + x^2 + 25) + 2*
x^2 + 50)),x)

[Out]

65900/log(exp((3*x)/(x + 5))*exp(25/(x + 5)) + 2)^2 - 183120/log(exp((3*x)/(x + 5))*exp(25/(x + 5)) + 2) - 105
00/log(exp((3*x)/(x + 5))*exp(25/(x + 5)) + 2)^3 + 625/log(exp((3*x)/(x + 5))*exp(25/(x + 5)) + 2)^4

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sympy [B]  time = 0.38, size = 65, normalized size = 2.24 \begin {gather*} \frac {- 183120 \log {\left (e^{\frac {3 x + 25}{x + 5}} + 2 \right )}^{3} + 65900 \log {\left (e^{\frac {3 x + 25}{x + 5}} + 2 \right )}^{2} - 10500 \log {\left (e^{\frac {3 x + 25}{x + 5}} + 2 \right )} + 625}{\log {\left (e^{\frac {3 x + 25}{x + 5}} + 2 \right )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-1831200*exp((25+3*x)/(5+x))*ln(exp((25+3*x)/(5+x))+2)**3+1318000*exp((25+3*x)/(5+x))*ln(exp((25+3*
x)/(5+x))+2)**2-315000*exp((25+3*x)/(5+x))*ln(exp((25+3*x)/(5+x))+2)+25000*exp((25+3*x)/(5+x)))/((x**2+10*x+25
)*exp((25+3*x)/(5+x))+2*x**2+20*x+50)/ln(exp((25+3*x)/(5+x))+2)**5,x)

[Out]

(-183120*log(exp((3*x + 25)/(x + 5)) + 2)**3 + 65900*log(exp((3*x + 25)/(x + 5)) + 2)**2 - 10500*log(exp((3*x
+ 25)/(x + 5)) + 2) + 625)/log(exp((3*x + 25)/(x + 5)) + 2)**4

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