3.35.92 \(\int \frac {e^{\frac {15}{-12+5 x}} (-351 x+180 x^2-100 x^3)+(6480-22680 x+27045 x^2-12600 x^3+2000 x^4+e^{\frac {15}{-12+5 x}} (-432+936 x-555 x^2+100 x^3)) \log (\frac {-15+e^{\frac {15}{-12+5 x}}+20 x}{-3+4 x})}{6480-22680 x+27045 x^2-12600 x^3+2000 x^4+e^{\frac {15}{-12+5 x}} (-432+936 x-555 x^2+100 x^3)} \, dx\)

Optimal. Leaf size=24 \[ x \log \left (5+\frac {e^{\frac {3}{-\frac {12}{5}+x}}}{-3+4 x}\right ) \]

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Rubi [A]  time = 5.79, antiderivative size = 29, normalized size of antiderivative = 1.21, number of steps used = 11, number of rules used = 4, integrand size = 150, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {6688, 6742, 2228, 2548} \begin {gather*} x \log \left (\frac {-20 x-e^{-\frac {15}{12-5 x}}+15}{3-4 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(15/(-12 + 5*x))*(-351*x + 180*x^2 - 100*x^3) + (6480 - 22680*x + 27045*x^2 - 12600*x^3 + 2000*x^4 + E^
(15/(-12 + 5*x))*(-432 + 936*x - 555*x^2 + 100*x^3))*Log[(-15 + E^(15/(-12 + 5*x)) + 20*x)/(-3 + 4*x)])/(6480
- 22680*x + 27045*x^2 - 12600*x^3 + 2000*x^4 + E^(15/(-12 + 5*x))*(-432 + 936*x - 555*x^2 + 100*x^3)),x]

[Out]

x*Log[(15 - E^(-15/(12 - 5*x)) - 20*x)/(3 - 4*x)]

Rule 2228

Int[(F_)^((a_.) + (b_.)/((c_.) + (d_.)*(x_)))/(((e_.) + (f_.)*(x_))*((g_.) + (h_.)*(x_))), x_Symbol] :> -Dist[
d/(f*(d*g - c*h)), Subst[Int[F^(a - (b*h)/(d*g - c*h) + (d*b*x)/(d*g - c*h))/x, x], x, (g + h*x)/(c + d*x)], x
] /; FreeQ[{F, a, b, c, d, e, f}, x] && EqQ[d*e - c*f, 0]

Rule 2548

Int[Log[u_], x_Symbol] :> Simp[x*Log[u], x] - Int[SimplifyIntegrand[(x*D[u, x])/u, x], x] /; InverseFunctionFr
eeQ[u, x]

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 \left (-\frac {e^{\frac {15}{-12+5 x}} x \left (351-180 x+100 x^2\right )}{(12-5 x)^2 (-3+4 x) \left (-15+e^{\frac {15}{-12+5 x}}+20 x\right )}+\log \left (\frac {-15+e^{\frac {15}{-12+5 x}}+20 x}{-3+4 x}\right )\right ) \, dx\\ &=-\int \frac {e^{\frac {15}{-12+5 x}} x \left (351-180 x+100 x^2\right )}{(12-5 x)^2 (-3+4 x) \left (-15+e^{\frac {15}{-12+5 x}}+20 x\right )} \, dx+\int \log \left (\frac {-15+e^{\frac {15}{-12+5 x}}+20 x}{-3+4 x}\right ) \, dx\\ &=x \log \left (\frac {15-e^{-\frac {15}{12-5 x}}-20 x}{3-4 x}\right )-\int \frac {e^{\frac {15}{-12+5 x}} x \left (-351+180 x-100 x^2\right )}{\left (15-e^{\frac {15}{-12+5 x}}-20 x\right ) (12-5 x)^2 (3-4 x)} \, dx-\int \left (\frac {e^{\frac {15}{-12+5 x}}}{-15+e^{\frac {15}{-12+5 x}}+20 x}+\frac {3 e^{\frac {15}{-12+5 x}}}{(-3+4 x) \left (-15+e^{\frac {15}{-12+5 x}}+20 x\right )}+\frac {180 e^{\frac {15}{-12+5 x}}}{(-12+5 x)^2 \left (-15+e^{\frac {15}{-12+5 x}}+20 x\right )}+\frac {15 e^{\frac {15}{-12+5 x}}}{(-12+5 x) \left (-15+e^{\frac {15}{-12+5 x}}+20 x\right )}\right ) \, dx\\ &=x \log \left (\frac {15-e^{-\frac {15}{12-5 x}}-20 x}{3-4 x}\right )-3 \int \frac {e^{\frac {15}{-12+5 x}}}{(-3+4 x) \left (-15+e^{\frac {15}{-12+5 x}}+20 x\right )} \, dx-15 \int \frac {e^{\frac {15}{-12+5 x}}}{(-12+5 x) \left (-15+e^{\frac {15}{-12+5 x}}+20 x\right )} \, dx-180 \int \frac {e^{\frac {15}{-12+5 x}}}{(-12+5 x)^2 \left (-15+e^{\frac {15}{-12+5 x}}+20 x\right )} \, dx-\int \frac {e^{\frac {15}{-12+5 x}}}{-15+e^{\frac {15}{-12+5 x}}+20 x} \, dx-\int \left (-\frac {e^{\frac {15}{-12+5 x}}}{-15+e^{\frac {15}{-12+5 x}}+20 x}-\frac {3 e^{\frac {15}{-12+5 x}}}{(-3+4 x) \left (-15+e^{\frac {15}{-12+5 x}}+20 x\right )}-\frac {180 e^{\frac {15}{-12+5 x}}}{(-12+5 x)^2 \left (-15+e^{\frac {15}{-12+5 x}}+20 x\right )}-\frac {15 e^{\frac {15}{-12+5 x}}}{(-12+5 x) \left (-15+e^{\frac {15}{-12+5 x}}+20 x\right )}\right ) \, dx\\ &=x \log \left (\frac {15-e^{-\frac {15}{12-5 x}}-20 x}{3-4 x}\right )+15 \int \frac {e^{\frac {15}{-12+5 x}}}{(-12+5 x) \left (-15+e^{\frac {15}{-12+5 x}}+20 x\right )} \, dx+\frac {15 \operatorname {Subst}\left (\int \frac {\exp \left (-\frac {300}{240+5 \left (-15+e^{\frac {15}{-12+5 x}}\right )}+\frac {75 x}{240+5 \left (-15+e^{\frac {15}{-12+5 x}}\right )}\right )}{x} \, dx,x,\frac {-15+e^{\frac {15}{-12+5 x}}+20 x}{-12+5 x}\right )}{240+5 \left (-15+e^{\frac {15}{-12+5 x}}\right )}\\ &=x \log \left (\frac {15-e^{-\frac {15}{12-5 x}}-20 x}{3-4 x}\right )+\frac {15 \operatorname {Subst}\left (\int \frac {e^{\frac {15 (-4+x)}{33+e^{\frac {15}{-12+5 x}}}}}{x} \, dx,x,\frac {-15+e^{\frac {15}{-12+5 x}}+20 x}{-12+5 x}\right )}{240+5 \left (-15+e^{\frac {15}{-12+5 x}}\right )}-\frac {15 \operatorname {Subst}\left (\int \frac {\exp \left (-\frac {300}{240+5 \left (-15+e^{\frac {15}{-12+5 x}}\right )}+\frac {75 x}{240+5 \left (-15+e^{\frac {15}{-12+5 x}}\right )}\right )}{x} \, dx,x,\frac {-15+e^{\frac {15}{-12+5 x}}+20 x}{-12+5 x}\right )}{240+5 \left (-15+e^{\frac {15}{-12+5 x}}\right )}\\ &=x \log \left (\frac {15-e^{-\frac {15}{12-5 x}}-20 x}{3-4 x}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.76, size = 27, normalized size = 1.12 \begin {gather*} x \log \left (\frac {-15+e^{\frac {15}{-12+5 x}}+20 x}{-3+4 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(15/(-12 + 5*x))*(-351*x + 180*x^2 - 100*x^3) + (6480 - 22680*x + 27045*x^2 - 12600*x^3 + 2000*x^
4 + E^(15/(-12 + 5*x))*(-432 + 936*x - 555*x^2 + 100*x^3))*Log[(-15 + E^(15/(-12 + 5*x)) + 20*x)/(-3 + 4*x)])/
(6480 - 22680*x + 27045*x^2 - 12600*x^3 + 2000*x^4 + E^(15/(-12 + 5*x))*(-432 + 936*x - 555*x^2 + 100*x^3)),x]

[Out]

x*Log[(-15 + E^(15/(-12 + 5*x)) + 20*x)/(-3 + 4*x)]

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fricas [A]  time = 0.54, size = 26, normalized size = 1.08 \begin {gather*} x \log \left (\frac {20 \, x + e^{\left (\frac {15}{5 \, x - 12}\right )} - 15}{4 \, x - 3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((100*x^3-555*x^2+936*x-432)*exp(15/(5*x-12))+2000*x^4-12600*x^3+27045*x^2-22680*x+6480)*log((exp(1
5/(5*x-12))+20*x-15)/(4*x-3))+(-100*x^3+180*x^2-351*x)*exp(15/(5*x-12)))/((100*x^3-555*x^2+936*x-432)*exp(15/(
5*x-12))+2000*x^4-12600*x^3+27045*x^2-22680*x+6480),x, algorithm="fricas")

[Out]

x*log((20*x + e^(15/(5*x - 12)) - 15)/(4*x - 3))

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giac [A]  time = 0.39, size = 29, normalized size = 1.21 \begin {gather*} x \log \left (\frac {20 \, x + e^{\left (\frac {25 \, x}{4 \, {\left (5 \, x - 12\right )}} - \frac {5}{4}\right )} - 15}{4 \, x - 3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((100*x^3-555*x^2+936*x-432)*exp(15/(5*x-12))+2000*x^4-12600*x^3+27045*x^2-22680*x+6480)*log((exp(1
5/(5*x-12))+20*x-15)/(4*x-3))+(-100*x^3+180*x^2-351*x)*exp(15/(5*x-12)))/((100*x^3-555*x^2+936*x-432)*exp(15/(
5*x-12))+2000*x^4-12600*x^3+27045*x^2-22680*x+6480),x, algorithm="giac")

[Out]

x*log((20*x + e^(25/4*x/(5*x - 12) - 5/4) - 15)/(4*x - 3))

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maple [B]  time = 0.18, size = 66, normalized size = 2.75

method	result	size
norman	\(\frac {-12 \ln \left (\frac {{\mathrm e}^{\frac {15}{5 x -12}}+20 x -15}{4 x -3}\right ) x +5 \ln \left (\frac {{\mathrm e}^{\frac {15}{5 x -12}}+20 x -15}{4 x -3}\right ) x^{2}}{5 x -12}\)	\(66\)
risch	\(x \ln \left (\frac {{\mathrm e}^{\frac {15}{5 x -12}}}{20}+x -\frac {3}{4}\right )-x \ln \left (x -\frac {3}{4}\right )-\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{x -\frac {3}{4}}\right ) \mathrm {csgn}\left (i \left (\frac {{\mathrm e}^{\frac {15}{5 x -12}}}{20}+x -\frac {3}{4}\right )\right ) \mathrm {csgn}\left (\frac {i \left (\frac {{\mathrm e}^{\frac {15}{5 x -12}}}{20}+x -\frac {3}{4}\right )}{x -\frac {3}{4}}\right )}{2}+\frac {i \pi x \,\mathrm {csgn}\left (\frac {i}{x -\frac {3}{4}}\right ) \mathrm {csgn}\left (\frac {i \left (\frac {{\mathrm e}^{\frac {15}{5 x -12}}}{20}+x -\frac {3}{4}\right )}{x -\frac {3}{4}}\right )^{2}}{2}+\frac {i \pi x \,\mathrm {csgn}\left (i \left (\frac {{\mathrm e}^{\frac {15}{5 x -12}}}{20}+x -\frac {3}{4}\right )\right ) \mathrm {csgn}\left (\frac {i \left (\frac {{\mathrm e}^{\frac {15}{5 x -12}}}{20}+x -\frac {3}{4}\right )}{x -\frac {3}{4}}\right )^{2}}{2}-\frac {i \pi x \mathrm {csgn}\left (\frac {i \left (\frac {{\mathrm e}^{\frac {15}{5 x -12}}}{20}+x -\frac {3}{4}\right )}{x -\frac {3}{4}}\right )^{3}}{2}+x \ln \relax (5)\)	\(209\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((100*x^3-555*x^2+936*x-432)*exp(15/(5*x-12))+2000*x^4-12600*x^3+27045*x^2-22680*x+6480)*ln((exp(15/(5*x-
12))+20*x-15)/(4*x-3))+(-100*x^3+180*x^2-351*x)*exp(15/(5*x-12)))/((100*x^3-555*x^2+936*x-432)*exp(15/(5*x-12)
)+2000*x^4-12600*x^3+27045*x^2-22680*x+6480),x,method=_RETURNVERBOSE)

[Out]

(-12*ln((exp(15/(5*x-12))+20*x-15)/(4*x-3))*x+5*ln((exp(15/(5*x-12))+20*x-15)/(4*x-3))*x^2)/(5*x-12)

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maxima [A]  time = 0.53, size = 28, normalized size = 1.17 \begin {gather*} x \log \left (20 \, x + e^{\left (\frac {15}{5 \, x - 12}\right )} - 15\right ) - x \log \left (4 \, x - 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((100*x^3-555*x^2+936*x-432)*exp(15/(5*x-12))+2000*x^4-12600*x^3+27045*x^2-22680*x+6480)*log((exp(1
5/(5*x-12))+20*x-15)/(4*x-3))+(-100*x^3+180*x^2-351*x)*exp(15/(5*x-12)))/((100*x^3-555*x^2+936*x-432)*exp(15/(
5*x-12))+2000*x^4-12600*x^3+27045*x^2-22680*x+6480),x, algorithm="maxima")

[Out]

x*log(20*x + e^(15/(5*x - 12)) - 15) - x*log(4*x - 3)

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mupad [B]  time = 0.42, size = 26, normalized size = 1.08 \begin {gather*} x\,\ln \left (\frac {20\,x+{\mathrm {e}}^{\frac {15}{5\,x-12}}-15}{4\,x-3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(15/(5*x - 12))*(351*x - 180*x^2 + 100*x^3) - log((20*x + exp(15/(5*x - 12)) - 15)/(4*x - 3))*(exp(15
/(5*x - 12))*(936*x - 555*x^2 + 100*x^3 - 432) - 22680*x + 27045*x^2 - 12600*x^3 + 2000*x^4 + 6480))/(exp(15/(
5*x - 12))*(936*x - 555*x^2 + 100*x^3 - 432) - 22680*x + 27045*x^2 - 12600*x^3 + 2000*x^4 + 6480),x)

[Out]

x*log((20*x + exp(15/(5*x - 12)) - 15)/(4*x - 3))

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sympy [A]  time = 0.79, size = 20, normalized size = 0.83 \begin {gather*} x \log {\left (\frac {20 x + e^{\frac {15}{5 x - 12}} - 15}{4 x - 3} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((100*x**3-555*x**2+936*x-432)*exp(15/(5*x-12))+2000*x**4-12600*x**3+27045*x**2-22680*x+6480)*ln((e
xp(15/(5*x-12))+20*x-15)/(4*x-3))+(-100*x**3+180*x**2-351*x)*exp(15/(5*x-12)))/((100*x**3-555*x**2+936*x-432)*
exp(15/(5*x-12))+2000*x**4-12600*x**3+27045*x**2-22680*x+6480),x)

[Out]

x*log((20*x + exp(15/(5*x - 12)) - 15)/(4*x - 3))

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