3.64.37 \(\int \frac {300+600 x+175 x^2-10 x^3+e^x (-60-60 x-18 x^2+x^3)+(-300-85 x+5 x^2+e^x (60+17 x-x^2)) \log (-300-85 x+5 x^2+e^x (60+17 x-x^2))}{300+85 x-5 x^2+e^x (-60-17 x+x^2)} \, dx\)

Optimal. Leaf size=22 \[ x \left (1+x-\log \left (\left (-5+e^x\right ) (20-x) (3+x)\right )\right ) \]

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Rubi [A]  time = 1.63, antiderivative size = 28, normalized size of antiderivative = 1.27, number of steps used = 28, number of rules used = 13, integrand size = 107, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {6741, 6728, 2184, 2190, 2279, 2391, 6688, 6742, 36, 31, 72, 2548, 1612} \begin {gather*} x^2-x \log \left (-\left (\left (5-e^x\right ) \left (-x^2+17 x+60\right )\right )\right )+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(300 + 600*x + 175*x^2 - 10*x^3 + E^x*(-60 - 60*x - 18*x^2 + x^3) + (-300 - 85*x + 5*x^2 + E^x*(60 + 17*x
- x^2))*Log[-300 - 85*x + 5*x^2 + E^x*(60 + 17*x - x^2)])/(300 + 85*x - 5*x^2 + E^x*(-60 - 17*x + x^2)),x]

[Out]

x + x^2 - x*Log[-((5 - E^x)*(60 + 17*x - x^2))]

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 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 1612

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[E
xpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && Poly
Q[Px, x] && IntegersQ[m, n]

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 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 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 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 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 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

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 {300+600 x+175 x^2-10 x^3+e^x \left (-60-60 x-18 x^2+x^3\right )+\left (-300-85 x+5 x^2+e^x \left (60+17 x-x^2\right )\right ) \log \left (-300-85 x+5 x^2+e^x \left (60+17 x-x^2\right )\right )}{\left (5-e^x\right ) \left (60+17 x-x^2\right )} \, dx\\ &=\int \left (-\frac {5 x}{-5+e^x}+\frac {-60-60 x-18 x^2+x^3+60 \log \left (-\left (\left (-5+e^x\right ) \left (-60-17 x+x^2\right )\right )\right )+17 x \log \left (-\left (\left (-5+e^x\right ) \left (-60-17 x+x^2\right )\right )\right )-x^2 \log \left (-\left (\left (-5+e^x\right ) \left (-60-17 x+x^2\right )\right )\right )}{(-20+x) (3+x)}\right ) \, dx\\ &=-\left (5 \int \frac {x}{-5+e^x} \, dx\right )+\int \frac {-60-60 x-18 x^2+x^3+60 \log \left (-\left (\left (-5+e^x\right ) \left (-60-17 x+x^2\right )\right )\right )+17 x \log \left (-\left (\left (-5+e^x\right ) \left (-60-17 x+x^2\right )\right )\right )-x^2 \log \left (-\left (\left (-5+e^x\right ) \left (-60-17 x+x^2\right )\right )\right )}{(-20+x) (3+x)} \, dx\\ &=\frac {x^2}{2}-\int \frac {e^x x}{-5+e^x} \, dx+\int \frac {60+60 x+18 x^2-x^3-\left (60+17 x-x^2\right ) \log \left (-\left (\left (-5+e^x\right ) \left (-60-17 x+x^2\right )\right )\right )}{(20-x) (3+x)} \, dx\\ &=\frac {x^2}{2}-x \log \left (1-\frac {e^x}{5}\right )+\int \log \left (1-\frac {e^x}{5}\right ) \, dx+\int \left (-\frac {60}{(-20+x) (3+x)}-\frac {60 x}{(-20+x) (3+x)}-\frac {18 x^2}{(-20+x) (3+x)}+\frac {x^3}{(-20+x) (3+x)}-\log \left (-\left (\left (-5+e^x\right ) \left (-60-17 x+x^2\right )\right )\right )\right ) \, dx\\ &=\frac {x^2}{2}-x \log \left (1-\frac {e^x}{5}\right )-18 \int \frac {x^2}{(-20+x) (3+x)} \, dx-60 \int \frac {1}{(-20+x) (3+x)} \, dx-60 \int \frac {x}{(-20+x) (3+x)} \, dx+\int \frac {x^3}{(-20+x) (3+x)} \, dx-\int \log \left (-\left (\left (-5+e^x\right ) \left (-60-17 x+x^2\right )\right )\right ) \, dx+\operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{5}\right )}{x} \, dx,x,e^x\right )\\ &=\frac {x^2}{2}-x \log \left (1-\frac {e^x}{5}\right )-x \log \left (-\left (\left (5-e^x\right ) \left (60+17 x-x^2\right )\right )\right )-\text {Li}_2\left (\frac {e^x}{5}\right )-\frac {60}{23} \int \frac {1}{-20+x} \, dx+\frac {60}{23} \int \frac {1}{3+x} \, dx-18 \int \left (1+\frac {400}{23 (-20+x)}-\frac {9}{23 (3+x)}\right ) \, dx-60 \int \left (\frac {20}{23 (-20+x)}+\frac {3}{23 (3+x)}\right ) \, dx+\int \left (17+\frac {8000}{23 (-20+x)}+x+\frac {27}{23 (3+x)}\right ) \, dx+\int \frac {x \left (85-10 x+e^x \left (-77-15 x+x^2\right )\right )}{\left (5-e^x\right ) \left (60+17 x-x^2\right )} \, dx\\ &=-x+x^2-x \log \left (1-\frac {e^x}{5}\right )-20 \log (20-x)+3 \log (3+x)-x \log \left (-\left (\left (5-e^x\right ) \left (60+17 x-x^2\right )\right )\right )-\text {Li}_2\left (\frac {e^x}{5}\right )+\int \left (\frac {5 x}{-5+e^x}+\frac {x \left (-77-15 x+x^2\right )}{(-20+x) (3+x)}\right ) \, dx\\ &=-x+x^2-x \log \left (1-\frac {e^x}{5}\right )-20 \log (20-x)+3 \log (3+x)-x \log \left (-\left (\left (5-e^x\right ) \left (60+17 x-x^2\right )\right )\right )-\text {Li}_2\left (\frac {e^x}{5}\right )+5 \int \frac {x}{-5+e^x} \, dx+\int \frac {x \left (-77-15 x+x^2\right )}{(-20+x) (3+x)} \, dx\\ &=-x+\frac {x^2}{2}-x \log \left (1-\frac {e^x}{5}\right )-20 \log (20-x)+3 \log (3+x)-x \log \left (-\left (\left (5-e^x\right ) \left (60+17 x-x^2\right )\right )\right )-\text {Li}_2\left (\frac {e^x}{5}\right )+\int \frac {e^x x}{-5+e^x} \, dx+\int \left (2+\frac {20}{-20+x}+x-\frac {3}{3+x}\right ) \, dx\\ &=x+x^2-x \log \left (-\left (\left (5-e^x\right ) \left (60+17 x-x^2\right )\right )\right )-\text {Li}_2\left (\frac {e^x}{5}\right )-\int \log \left (1-\frac {e^x}{5}\right ) \, dx\\ &=x+x^2-x \log \left (-\left (\left (5-e^x\right ) \left (60+17 x-x^2\right )\right )\right )-\text {Li}_2\left (\frac {e^x}{5}\right )-\operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{5}\right )}{x} \, dx,x,e^x\right )\\ &=x+x^2-x \log \left (-\left (\left (5-e^x\right ) \left (60+17 x-x^2\right )\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 24, normalized size = 1.09 \begin {gather*} x+x^2-x \log \left (-\left (\left (-5+e^x\right ) \left (-60-17 x+x^2\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(300 + 600*x + 175*x^2 - 10*x^3 + E^x*(-60 - 60*x - 18*x^2 + x^3) + (-300 - 85*x + 5*x^2 + E^x*(60 +
17*x - x^2))*Log[-300 - 85*x + 5*x^2 + E^x*(60 + 17*x - x^2)])/(300 + 85*x - 5*x^2 + E^x*(-60 - 17*x + x^2)),
x]

[Out]

x + x^2 - x*Log[-((-5 + E^x)*(-60 - 17*x + x^2))]

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fricas [A]  time = 0.77, size = 31, normalized size = 1.41 \begin {gather*} x^{2} - x \log \left (5 \, x^{2} - {\left (x^{2} - 17 \, x - 60\right )} e^{x} - 85 \, x - 300\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^2+17*x+60)*exp(x)+5*x^2-85*x-300)*log((-x^2+17*x+60)*exp(x)+5*x^2-85*x-300)+(x^3-18*x^2-60*x-6
0)*exp(x)-10*x^3+175*x^2+600*x+300)/((x^2-17*x-60)*exp(x)-5*x^2+85*x+300),x, algorithm="fricas")

[Out]

x^2 - x*log(5*x^2 - (x^2 - 17*x - 60)*e^x - 85*x - 300) + x

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giac [A]  time = 0.30, size = 35, normalized size = 1.59 \begin {gather*} x^{2} - x \log \left (-x^{2} e^{x} + 5 \, x^{2} + 17 \, x e^{x} - 85 \, x + 60 \, e^{x} - 300\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^2+17*x+60)*exp(x)+5*x^2-85*x-300)*log((-x^2+17*x+60)*exp(x)+5*x^2-85*x-300)+(x^3-18*x^2-60*x-6
0)*exp(x)-10*x^3+175*x^2+600*x+300)/((x^2-17*x-60)*exp(x)-5*x^2+85*x+300),x, algorithm="giac")

[Out]

x^2 - x*log(-x^2*e^x + 5*x^2 + 17*x*e^x - 85*x + 60*e^x - 300) + x

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maple [A]  time = 0.22, size = 33, normalized size = 1.50

method	result	size
norman	\(x +x^{2}-x \ln \left (\left (-x^{2}+17 x +60\right ) {\mathrm e}^{x}+5 x^{2}-85 x -300\right )\)	\(33\)
risch	\(-x \ln \left (x^{2}-17 x -60\right )-x \ln \left ({\mathrm e}^{x}-5\right )+\frac {i \pi x \,\mathrm {csgn}\left (i \left ({\mathrm e}^{x}-5\right )\right ) \mathrm {csgn}\left (i \left (x^{2}-17 x -60\right )\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{x}-5\right ) \left (x^{2}-17 x -60\right )\right )}{2}-\frac {i \pi x \,\mathrm {csgn}\left (i \left ({\mathrm e}^{x}-5\right )\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{x}-5\right ) \left (x^{2}-17 x -60\right )\right )^{2}}{2}-\frac {i \pi x \,\mathrm {csgn}\left (i \left (x^{2}-17 x -60\right )\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{x}-5\right ) \left (x^{2}-17 x -60\right )\right )^{2}}{2}-\frac {i \pi x \mathrm {csgn}\left (i \left ({\mathrm e}^{x}-5\right ) \left (x^{2}-17 x -60\right )\right )^{3}}{2}+i \pi x \mathrm {csgn}\left (i \left ({\mathrm e}^{x}-5\right ) \left (x^{2}-17 x -60\right )\right )^{2}-i x \pi +x^{2}+x\)	\(184\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-x^2+17*x+60)*exp(x)+5*x^2-85*x-300)*ln((-x^2+17*x+60)*exp(x)+5*x^2-85*x-300)+(x^3-18*x^2-60*x-60)*exp(
x)-10*x^3+175*x^2+600*x+300)/((x^2-17*x-60)*exp(x)-5*x^2+85*x+300),x,method=_RETURNVERBOSE)

[Out]

x+x^2-x*ln((-x^2+17*x+60)*exp(x)+5*x^2-85*x-300)

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maxima [A]  time = 0.45, size = 29, normalized size = 1.32 \begin {gather*} x^{2} - x \log \left (x + 3\right ) - x \log \left (x - 20\right ) - x \log \left (-e^{x} + 5\right ) + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x^2+17*x+60)*exp(x)+5*x^2-85*x-300)*log((-x^2+17*x+60)*exp(x)+5*x^2-85*x-300)+(x^3-18*x^2-60*x-6
0)*exp(x)-10*x^3+175*x^2+600*x+300)/((x^2-17*x-60)*exp(x)-5*x^2+85*x+300),x, algorithm="maxima")

[Out]

x^2 - x*log(x + 3) - x*log(x - 20) - x*log(-e^x + 5) + x

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mupad [B]  time = 4.58, size = 31, normalized size = 1.41 \begin {gather*} x\,\left (x-\ln \left ({\mathrm {e}}^x\,\left (-x^2+17\,x+60\right )-85\,x+5\,x^2-300\right )+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((600*x + 175*x^2 - 10*x^3 - exp(x)*(60*x + 18*x^2 - x^3 + 60) - log(exp(x)*(17*x - x^2 + 60) - 85*x + 5*x^
2 - 300)*(85*x - exp(x)*(17*x - x^2 + 60) - 5*x^2 + 300) + 300)/(85*x - exp(x)*(17*x - x^2 + 60) - 5*x^2 + 300
),x)

[Out]

x*(x - log(exp(x)*(17*x - x^2 + 60) - 85*x + 5*x^2 - 300) + 1)

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sympy [B]  time = 0.77, size = 56, normalized size = 2.55 \begin {gather*} x^{2} + x + \left (\frac {17}{6} - x\right ) \log {\left (5 x^{2} - 85 x + \left (- x^{2} + 17 x + 60\right ) e^{x} - 300 \right )} - \frac {17 \log {\left (e^{x} - 5 \right )}}{6} - \frac {17 \log {\left (x^{2} - 17 x - 60 \right )}}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-x**2+17*x+60)*exp(x)+5*x**2-85*x-300)*ln((-x**2+17*x+60)*exp(x)+5*x**2-85*x-300)+(x**3-18*x**2-6
0*x-60)*exp(x)-10*x**3+175*x**2+600*x+300)/((x**2-17*x-60)*exp(x)-5*x**2+85*x+300),x)

[Out]

x**2 + x + (17/6 - x)*log(5*x**2 - 85*x + (-x**2 + 17*x + 60)*exp(x) - 300) - 17*log(exp(x) - 5)/6 - 17*log(x*
*2 - 17*x - 60)/6

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