3.18.75 \(\int \frac {-306000+e^{2 x/25} (3600-288 x)}{180625-4250 e^{2 x/25}+25 e^{4 x/25}} \, dx\)

Optimal. Leaf size=20 \[ \frac {9 x}{-5+\frac {1}{16} \left (-5+e^{2 x/25}\right )} \]

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Rubi [B]  time = 0.43, antiderivative size = 98, normalized size of antiderivative = 4.90, number of steps used = 18, number of rules used = 13, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.342, Rules used = {6741, 12, 6742, 2185, 2184, 2190, 2279, 2391, 2191, 2282, 36, 31, 29} \begin {gather*} -\frac {144 x^2}{2125}+\frac {36 (25-2 x)^2}{2125}-\frac {144 x}{85-e^{2 x/25}}+\frac {144 x}{85}+\frac {72}{85} (25-2 x) \log \left (1-\frac {1}{85} e^{2 x/25}\right )-\frac {360}{17} \log \left (85-e^{2 x/25}\right )+\frac {144}{85} x \log \left (1-\frac {1}{85} e^{2 x/25}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-306000 + E^((2*x)/25)*(3600 - 288*x))/(180625 - 4250*E^((2*x)/25) + 25*E^((4*x)/25)),x]

[Out]

(36*(25 - 2*x)^2)/2125 + (144*x)/85 - (144*x)/(85 - E^((2*x)/25)) - (144*x^2)/2125 - (360*Log[85 - E^((2*x)/25
)])/17 + (72*(25 - 2*x)*Log[1 - E^((2*x)/25)/85])/85 + (144*x*Log[1 - E^((2*x)/25)/85])/85

Rule 12

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 2184

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[(c
+ d*x)^(m + 1)/(a*d*(m + 1)), x] - Dist[b/a, Int[((c + d*x)^m*(F^(g*(e + f*x)))^n)/(a + b*(F^(g*(e + f*x)))^n)
, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2185

Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Dis
t[1/a, Int[(c + d*x)^m*(a + b*(F^(g*(e + f*x)))^n)^(p + 1), x], x] - Dist[b/a, Int[(c + d*x)^m*(F^(g*(e + f*x)
))^n*(a + b*(F^(g*(e + f*x)))^n)^p, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && ILtQ[p, 0] && IGtQ[m, 0
]

Rule 2190

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

Rule 2191

Int[((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((a_.) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_.)*
((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m*(a + b*(F^(g*(e + f*x)))^n)^(p + 1))/(b*f*g*n*(p +
1)*Log[F]), x] - Dist[(d*m)/(b*f*g*n*(p + 1)*Log[F]), Int[(c + d*x)^(m - 1)*(a + b*(F^(g*(e + f*x)))^n)^(p + 1
), x], x] /; FreeQ[{F, a, b, c, d, e, f, g, m, n, p}, x] && NeQ[p, -1]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 6741

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

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 {-306000+e^{2 x/25} (3600-288 x)}{25 \left (85-e^{2 x/25}\right )^2} \, dx\\ &=\frac {1}{25} \int \frac {-306000+e^{2 x/25} (3600-288 x)}{\left (85-e^{2 x/25}\right )^2} \, dx\\ &=\frac {1}{25} \int \left (-\frac {24480 x}{\left (-85+e^{2 x/25}\right )^2}-\frac {144 (-25+2 x)}{-85+e^{2 x/25}}\right ) \, dx\\ &=-\left (\frac {144}{25} \int \frac {-25+2 x}{-85+e^{2 x/25}} \, dx\right )-\frac {4896}{5} \int \frac {x}{\left (-85+e^{2 x/25}\right )^2} \, dx\\ &=\frac {36 (25-2 x)^2}{2125}-\frac {144 \int \frac {e^{2 x/25} (-25+2 x)}{-85+e^{2 x/25}} \, dx}{2125}-\frac {288}{25} \int \frac {e^{2 x/25} x}{\left (-85+e^{2 x/25}\right )^2} \, dx+\frac {288}{25} \int \frac {x}{-85+e^{2 x/25}} \, dx\\ &=\frac {36 (25-2 x)^2}{2125}-\frac {144 x}{85-e^{2 x/25}}-\frac {144 x^2}{2125}+\frac {72}{85} (25-2 x) \log \left (1-\frac {1}{85} e^{2 x/25}\right )+\frac {288 \int \frac {e^{2 x/25} x}{-85+e^{2 x/25}} \, dx}{2125}+\frac {144}{85} \int \log \left (1-\frac {1}{85} e^{2 x/25}\right ) \, dx-144 \int \frac {1}{-85+e^{2 x/25}} \, dx\\ &=\frac {36 (25-2 x)^2}{2125}-\frac {144 x}{85-e^{2 x/25}}-\frac {144 x^2}{2125}+\frac {72}{85} (25-2 x) \log \left (1-\frac {1}{85} e^{2 x/25}\right )+\frac {144}{85} x \log \left (1-\frac {1}{85} e^{2 x/25}\right )-\frac {144}{85} \int \log \left (1-\frac {1}{85} e^{2 x/25}\right ) \, dx+\frac {360}{17} \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{85}\right )}{x} \, dx,x,e^{2 x/25}\right )-1800 \operatorname {Subst}\left (\int \frac {1}{(-85+x) x} \, dx,x,e^{2 x/25}\right )\\ &=\frac {36 (25-2 x)^2}{2125}-\frac {144 x}{85-e^{2 x/25}}-\frac {144 x^2}{2125}+\frac {72}{85} (25-2 x) \log \left (1-\frac {1}{85} e^{2 x/25}\right )+\frac {144}{85} x \log \left (1-\frac {1}{85} e^{2 x/25}\right )-\frac {360}{17} \text {Li}_2\left (\frac {1}{85} e^{2 x/25}\right )-\frac {360}{17} \operatorname {Subst}\left (\int \frac {1}{-85+x} \, dx,x,e^{2 x/25}\right )+\frac {360}{17} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^{2 x/25}\right )-\frac {360}{17} \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{85}\right )}{x} \, dx,x,e^{2 x/25}\right )\\ &=\frac {36 (25-2 x)^2}{2125}+\frac {144 x}{85}-\frac {144 x}{85-e^{2 x/25}}-\frac {144 x^2}{2125}-\frac {360}{17} \log \left (85-e^{2 x/25}\right )+\frac {72}{85} (25-2 x) \log \left (1-\frac {1}{85} e^{2 x/25}\right )+\frac {144}{85} x \log \left (1-\frac {1}{85} e^{2 x/25}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 14, normalized size = 0.70 \begin {gather*} \frac {144 x}{-85+e^{2 x/25}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-306000 + E^((2*x)/25)*(3600 - 288*x))/(180625 - 4250*E^((2*x)/25) + 25*E^((4*x)/25)),x]

[Out]

(144*x)/(-85 + E^((2*x)/25))

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fricas [A]  time = 0.82, size = 11, normalized size = 0.55 \begin {gather*} \frac {144 \, x}{e^{\left (\frac {2}{25} \, x\right )} - 85} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-288*x+3600)*exp(1/25*x)^2-306000)/(25*exp(1/25*x)^4-4250*exp(1/25*x)^2+180625),x, algorithm="fric
as")

[Out]

144*x/(e^(2/25*x) - 85)

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giac [A]  time = 0.51, size = 11, normalized size = 0.55 \begin {gather*} \frac {144 \, x}{e^{\left (\frac {2}{25} \, x\right )} - 85} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-288*x+3600)*exp(1/25*x)^2-306000)/(25*exp(1/25*x)^4-4250*exp(1/25*x)^2+180625),x, algorithm="giac
")

[Out]

144*x/(e^(2/25*x) - 85)

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maple [A]  time = 0.10, size = 12, normalized size = 0.60

method	result	size
risch	\(\frac {144 x}{{\mathrm e}^{\frac {2 x}{25}}-85}\)	\(12\)
norman	\(\frac {144 x}{{\mathrm e}^{\frac {2 x}{25}}-85}\)	\(14\)
derivativedivides	\(-\frac {720 \ln \left ({\mathrm e}^{\frac {x}{25}}\right )}{17}+\frac {144 x \,{\mathrm e}^{\frac {2 x}{25}}}{85 \left ({\mathrm e}^{\frac {2 x}{25}}-85\right )}\)	\(28\)
default	\(-\frac {720 \ln \left ({\mathrm e}^{\frac {x}{25}}\right )}{17}+\frac {144 x \,{\mathrm e}^{\frac {2 x}{25}}}{85 \left ({\mathrm e}^{\frac {2 x}{25}}-85\right )}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-288*x+3600)*exp(1/25*x)^2-306000)/(25*exp(1/25*x)^4-4250*exp(1/25*x)^2+180625),x,method=_RETURNVERBOSE)

[Out]

144*x/(exp(2/25*x)-85)

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maxima [A]  time = 0.44, size = 19, normalized size = 0.95 \begin {gather*} -\frac {144}{85} \, x + \frac {144 \, x e^{\left (\frac {2}{25} \, x\right )}}{85 \, {\left (e^{\left (\frac {2}{25} \, x\right )} - 85\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-288*x+3600)*exp(1/25*x)^2-306000)/(25*exp(1/25*x)^4-4250*exp(1/25*x)^2+180625),x, algorithm="maxi
ma")

[Out]

-144/85*x + 144/85*x*e^(2/25*x)/(e^(2/25*x) - 85)

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mupad [B]  time = 1.16, size = 11, normalized size = 0.55 \begin {gather*} \frac {144\,x}{{\mathrm {e}}^{\frac {2\,x}{25}}-85} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((2*x)/25)*(288*x - 3600) + 306000)/(25*exp((4*x)/25) - 4250*exp((2*x)/25) + 180625),x)

[Out]

(144*x)/(exp((2*x)/25) - 85)

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sympy [A]  time = 0.10, size = 10, normalized size = 0.50 \begin {gather*} \frac {144 x}{e^{\frac {2 x}{25}} - 85} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-288*x+3600)*exp(1/25*x)**2-306000)/(25*exp(1/25*x)**4-4250*exp(1/25*x)**2+180625),x)

[Out]

144*x/(exp(2*x/25) - 85)

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