3.103.42 \(\int \frac {300+690 x+204 x^2+18 x^3+e^x (-500-2150 x-1215 x^2-240 x^3-15 x^4)+(-300-120 x-12 x^2+e^x (500+700 x+220 x^2+20 x^3)) \log (x)}{100+40 x+4 x^2} \, dx\)

Optimal. Leaf size=26 \[ \left (-3+5 e^x\right ) x \left (-2-\frac {3 x}{4}-\frac {x}{5+x}+\log (x)\right ) \]

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Rubi [B]  time = 1.74, antiderivative size = 71, normalized size of antiderivative = 2.73, number of steps used = 62, number of rules used = 17, integrand size = 84, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.202, Rules used = {27, 12, 6742, 43, 2314, 31, 2351, 2317, 2391, 2295, 6688, 2177, 2178, 2199, 2194, 2176, 2554} \begin {gather*} -\frac {15}{4} e^x x^2+\frac {9 x^2}{4}-15 e^x x+9 x+25 e^x-\frac {125 e^x}{x+5}+\frac {75}{x+5}-3 x \log (x)-5 e^x \log (x)+5 e^x (x+1) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(300 + 690*x + 204*x^2 + 18*x^3 + E^x*(-500 - 2150*x - 1215*x^2 - 240*x^3 - 15*x^4) + (-300 - 120*x - 12*x
^2 + E^x*(500 + 700*x + 220*x^2 + 20*x^3))*Log[x])/(100 + 40*x + 4*x^2),x]

[Out]

25*E^x + 9*x - 15*E^x*x + (9*x^2)/4 - (15*E^x*x^2)/4 + 75/(5 + x) - (125*E^x)/(5 + x) - 5*E^x*Log[x] - 3*x*Log
[x] + 5*E^x*(1 + x)*Log[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 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

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 2317

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

Rule 2351

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = ExpandIntegrand[a + b*Log[c*x^n], (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c,
d, e, f, m, n, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[m] && IntegerQ[r]))

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 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[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 \frac {300+690 x+204 x^2+18 x^3+e^x \left (-500-2150 x-1215 x^2-240 x^3-15 x^4\right )+\left (-300-120 x-12 x^2+e^x \left (500+700 x+220 x^2+20 x^3\right )\right ) \log (x)}{4 (5+x)^2} \, dx\\ &=\frac {1}{4} \int \frac {300+690 x+204 x^2+18 x^3+e^x \left (-500-2150 x-1215 x^2-240 x^3-15 x^4\right )+\left (-300-120 x-12 x^2+e^x \left (500+700 x+220 x^2+20 x^3\right )\right ) \log (x)}{(5+x)^2} \, dx\\ &=\frac {1}{4} \int \left (\frac {300}{(5+x)^2}+\frac {690 x}{(5+x)^2}+\frac {204 x^2}{(5+x)^2}+\frac {18 x^3}{(5+x)^2}-\frac {300 \log (x)}{(5+x)^2}-\frac {120 x \log (x)}{(5+x)^2}-\frac {12 x^2 \log (x)}{(5+x)^2}-\frac {5 e^x \left (100+430 x+243 x^2+48 x^3+3 x^4-100 \log (x)-140 x \log (x)-44 x^2 \log (x)-4 x^3 \log (x)\right )}{(5+x)^2}\right ) \, dx\\ &=-\frac {75}{5+x}-\frac {5}{4} \int \frac {e^x \left (100+430 x+243 x^2+48 x^3+3 x^4-100 \log (x)-140 x \log (x)-44 x^2 \log (x)-4 x^3 \log (x)\right )}{(5+x)^2} \, dx-3 \int \frac {x^2 \log (x)}{(5+x)^2} \, dx+\frac {9}{2} \int \frac {x^3}{(5+x)^2} \, dx-30 \int \frac {x \log (x)}{(5+x)^2} \, dx+51 \int \frac {x^2}{(5+x)^2} \, dx-75 \int \frac {\log (x)}{(5+x)^2} \, dx+\frac {345}{2} \int \frac {x}{(5+x)^2} \, dx\\ &=-\frac {75}{5+x}-\frac {15 x \log (x)}{5+x}-\frac {5}{4} \int \frac {e^x \left (100+430 x+243 x^2+48 x^3+3 x^4-4 (1+x) (5+x)^2 \log (x)\right )}{(5+x)^2} \, dx-3 \int \left (\log (x)+\frac {25 \log (x)}{(5+x)^2}-\frac {10 \log (x)}{5+x}\right ) \, dx+\frac {9}{2} \int \left (-10+x-\frac {125}{(5+x)^2}+\frac {75}{5+x}\right ) \, dx+15 \int \frac {1}{5+x} \, dx-30 \int \left (-\frac {5 \log (x)}{(5+x)^2}+\frac {\log (x)}{5+x}\right ) \, dx+51 \int \left (1+\frac {25}{(5+x)^2}-\frac {10}{5+x}\right ) \, dx+\frac {345}{2} \int \left (-\frac {5}{(5+x)^2}+\frac {1}{5+x}\right ) \, dx\\ &=6 x+\frac {9 x^2}{4}+\frac {75}{5+x}-\frac {15 x \log (x)}{5+x}+15 \log (5+x)-\frac {5}{4} \int \left (\frac {100 e^x}{(5+x)^2}+\frac {430 e^x x}{(5+x)^2}+\frac {243 e^x x^2}{(5+x)^2}+\frac {48 e^x x^3}{(5+x)^2}+\frac {3 e^x x^4}{(5+x)^2}-4 e^x (1+x) \log (x)\right ) \, dx-3 \int \log (x) \, dx-75 \int \frac {\log (x)}{(5+x)^2} \, dx+150 \int \frac {\log (x)}{(5+x)^2} \, dx\\ &=9 x+\frac {9 x^2}{4}+\frac {75}{5+x}-3 x \log (x)+15 \log (5+x)-\frac {15}{4} \int \frac {e^x x^4}{(5+x)^2} \, dx+5 \int e^x (1+x) \log (x) \, dx+15 \int \frac {1}{5+x} \, dx-30 \int \frac {1}{5+x} \, dx-60 \int \frac {e^x x^3}{(5+x)^2} \, dx-125 \int \frac {e^x}{(5+x)^2} \, dx-\frac {1215}{4} \int \frac {e^x x^2}{(5+x)^2} \, dx-\frac {1075}{2} \int \frac {e^x x}{(5+x)^2} \, dx\\ &=9 x+\frac {9 x^2}{4}+\frac {75}{5+x}+\frac {125 e^x}{5+x}-5 e^x \log (x)-3 x \log (x)+5 e^x (1+x) \log (x)-\frac {15}{4} \int \left (75 e^x-10 e^x x+e^x x^2+\frac {625 e^x}{(5+x)^2}-\frac {500 e^x}{5+x}\right ) \, dx-5 \int e^x \, dx-60 \int \left (-10 e^x+e^x x-\frac {125 e^x}{(5+x)^2}+\frac {75 e^x}{5+x}\right ) \, dx-125 \int \frac {e^x}{5+x} \, dx-\frac {1215}{4} \int \left (e^x+\frac {25 e^x}{(5+x)^2}-\frac {10 e^x}{5+x}\right ) \, dx-\frac {1075}{2} \int \left (-\frac {5 e^x}{(5+x)^2}+\frac {e^x}{5+x}\right ) \, dx\\ &=-5 e^x+9 x+\frac {9 x^2}{4}+\frac {75}{5+x}+\frac {125 e^x}{5+x}-\frac {125 \text {Ei}(5+x)}{e^5}-5 e^x \log (x)-3 x \log (x)+5 e^x (1+x) \log (x)-\frac {15}{4} \int e^x x^2 \, dx+\frac {75}{2} \int e^x x \, dx-60 \int e^x x \, dx-\frac {1125 \int e^x \, dx}{4}-\frac {1215 \int e^x \, dx}{4}-\frac {1075}{2} \int \frac {e^x}{5+x} \, dx+600 \int e^x \, dx+1875 \int \frac {e^x}{5+x} \, dx-\frac {9375}{4} \int \frac {e^x}{(5+x)^2} \, dx+\frac {5375}{2} \int \frac {e^x}{(5+x)^2} \, dx+\frac {6075}{2} \int \frac {e^x}{5+x} \, dx-4500 \int \frac {e^x}{5+x} \, dx+7500 \int \frac {e^x}{(5+x)^2} \, dx-\frac {30375}{4} \int \frac {e^x}{(5+x)^2} \, dx\\ &=10 e^x+9 x-\frac {45 e^x x}{2}+\frac {9 x^2}{4}-\frac {15 e^x x^2}{4}+\frac {75}{5+x}-\frac {125 e^x}{5+x}-\frac {250 \text {Ei}(5+x)}{e^5}-5 e^x \log (x)-3 x \log (x)+5 e^x (1+x) \log (x)+\frac {15}{2} \int e^x x \, dx-\frac {75 \int e^x \, dx}{2}+60 \int e^x \, dx-\frac {9375}{4} \int \frac {e^x}{5+x} \, dx+\frac {5375}{2} \int \frac {e^x}{5+x} \, dx+7500 \int \frac {e^x}{5+x} \, dx-\frac {30375}{4} \int \frac {e^x}{5+x} \, dx\\ &=\frac {65 e^x}{2}+9 x-15 e^x x+\frac {9 x^2}{4}-\frac {15 e^x x^2}{4}+\frac {75}{5+x}-\frac {125 e^x}{5+x}-5 e^x \log (x)-3 x \log (x)+5 e^x (1+x) \log (x)-\frac {15 \int e^x \, dx}{2}\\ &=25 e^x+9 x-15 e^x x+\frac {9 x^2}{4}-\frac {15 e^x x^2}{4}+\frac {75}{5+x}-\frac {125 e^x}{5+x}-5 e^x \log (x)-3 x \log (x)+5 e^x (1+x) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.78, size = 53, normalized size = 2.04 \begin {gather*} \frac {1}{4} \left (36 x+9 x^2+\frac {300}{5+x}+e^x \left (100-60 x-15 x^2-\frac {500}{5+x}\right )+4 \left (-3+5 e^x\right ) x \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(300 + 690*x + 204*x^2 + 18*x^3 + E^x*(-500 - 2150*x - 1215*x^2 - 240*x^3 - 15*x^4) + (-300 - 120*x
- 12*x^2 + E^x*(500 + 700*x + 220*x^2 + 20*x^3))*Log[x])/(100 + 40*x + 4*x^2),x]

[Out]

(36*x + 9*x^2 + 300/(5 + x) + E^x*(100 - 60*x - 15*x^2 - 500/(5 + x)) + 4*(-3 + 5*E^x)*x*Log[x])/4

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fricas [B]  time = 0.74, size = 64, normalized size = 2.46 \begin {gather*} \frac {9 \, x^{3} + 81 \, x^{2} - 5 \, {\left (3 \, x^{3} + 27 \, x^{2} + 40 \, x\right )} e^{x} - 4 \, {\left (3 \, x^{2} - 5 \, {\left (x^{2} + 5 \, x\right )} e^{x} + 15 \, x\right )} \log \relax (x) + 180 \, x + 300}{4 \, {\left (x + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((20*x^3+220*x^2+700*x+500)*exp(x)-12*x^2-120*x-300)*log(x)+(-15*x^4-240*x^3-1215*x^2-2150*x-500)*e
xp(x)+18*x^3+204*x^2+690*x+300)/(4*x^2+40*x+100),x, algorithm="fricas")

[Out]

1/4*(9*x^3 + 81*x^2 - 5*(3*x^3 + 27*x^2 + 40*x)*e^x - 4*(3*x^2 - 5*(x^2 + 5*x)*e^x + 15*x)*log(x) + 180*x + 30
0)/(x + 5)

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giac [B]  time = 0.18, size = 69, normalized size = 2.65 \begin {gather*} -\frac {15 \, x^{3} e^{x} - 20 \, x^{2} e^{x} \log \relax (x) - 9 \, x^{3} + 135 \, x^{2} e^{x} + 12 \, x^{2} \log \relax (x) - 100 \, x e^{x} \log \relax (x) - 81 \, x^{2} + 200 \, x e^{x} + 60 \, x \log \relax (x) - 180 \, x - 300}{4 \, {\left (x + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((20*x^3+220*x^2+700*x+500)*exp(x)-12*x^2-120*x-300)*log(x)+(-15*x^4-240*x^3-1215*x^2-2150*x-500)*e
xp(x)+18*x^3+204*x^2+690*x+300)/(4*x^2+40*x+100),x, algorithm="giac")

[Out]

-1/4*(15*x^3*e^x - 20*x^2*e^x*log(x) - 9*x^3 + 135*x^2*e^x + 12*x^2*log(x) - 100*x*e^x*log(x) - 81*x^2 + 200*x
*e^x + 60*x*log(x) - 180*x - 300)/(x + 5)

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maple [B]  time = 0.12, size = 55, normalized size = 2.12

method	result	size
risch	\(\left (5 \,{\mathrm e}^{x} x -3 x \right ) \ln \relax (x )-\frac {15 \,{\mathrm e}^{x} x^{3}-9 x^{3}+135 \,{\mathrm e}^{x} x^{2}-81 x^{2}+200 \,{\mathrm e}^{x} x -180 x -300}{4 \left (5+x \right )}\)	\(55\)
default	\(\frac {-135 \,{\mathrm e}^{x} x^{2}-200 \,{\mathrm e}^{x} x -15 \,{\mathrm e}^{x} x^{3}+20 x^{2} {\mathrm e}^{x} \ln \relax (x )+100 x \,{\mathrm e}^{x} \ln \relax (x )}{20+4 x}+\frac {9 x^{2}}{4}+9 x +\frac {75}{5+x}-3 x \ln \relax (x )\)	\(65\)
norman	\(\frac {\frac {81 x^{2}}{4}+\frac {9 x^{3}}{4}-15 x \ln \relax (x )-3 x^{2} \ln \relax (x )-50 \,{\mathrm e}^{x} x -\frac {135 \,{\mathrm e}^{x} x^{2}}{4}-\frac {15 \,{\mathrm e}^{x} x^{3}}{4}+25 x \,{\mathrm e}^{x} \ln \relax (x )+5 x^{2} {\mathrm e}^{x} \ln \relax (x )-150}{5+x}\)	\(66\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((20*x^3+220*x^2+700*x+500)*exp(x)-12*x^2-120*x-300)*ln(x)+(-15*x^4-240*x^3-1215*x^2-2150*x-500)*exp(x)+1
8*x^3+204*x^2+690*x+300)/(4*x^2+40*x+100),x,method=_RETURNVERBOSE)

[Out]

(5*exp(x)*x-3*x)*ln(x)-1/4*(15*exp(x)*x^3-9*x^3+135*exp(x)*x^2-81*x^2+200*exp(x)*x-180*x-300)/(5+x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 5 \, x e^{x} \log \relax (x) + \frac {9}{4} \, x^{2} - 3 \, x \log \relax (x) + 9 \, x + \frac {125 \, e^{\left (-5\right )} E_{2}\left (-x - 5\right )}{x + 5} + \frac {75}{x + 5} - \frac {1}{4} \, \int \frac {5 \, {\left (3 \, x^{4} + 48 \, x^{3} + 247 \, x^{2} + 470 \, x + 100\right )} e^{x}}{x^{2} + 10 \, x + 25}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((20*x^3+220*x^2+700*x+500)*exp(x)-12*x^2-120*x-300)*log(x)+(-15*x^4-240*x^3-1215*x^2-2150*x-500)*e
xp(x)+18*x^3+204*x^2+690*x+300)/(4*x^2+40*x+100),x, algorithm="maxima")

[Out]

5*x*e^x*log(x) + 9/4*x^2 - 3*x*log(x) + 9*x + 125*e^(-5)*exp_integral_e(2, -x - 5)/(x + 5) + 75/(x + 5) - 1/4*
integrate(5*(3*x^4 + 48*x^3 + 247*x^2 + 470*x + 100)*e^x/(x^2 + 10*x + 25), x)

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mupad [B]  time = 6.34, size = 52, normalized size = 2.00 \begin {gather*} 9\,x-\ln \relax (x)\,\left (3\,x-5\,x\,{\mathrm {e}}^x\right )+\frac {75}{x+5}+\frac {9\,x^2}{4}-\frac {{\mathrm {e}}^x\,\left (\frac {15\,x^3}{4}+\frac {135\,x^2}{4}+50\,x\right )}{x+5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((690*x - exp(x)*(2150*x + 1215*x^2 + 240*x^3 + 15*x^4 + 500) - log(x)*(120*x + 12*x^2 - exp(x)*(700*x + 22
0*x^2 + 20*x^3 + 500) + 300) + 204*x^2 + 18*x^3 + 300)/(40*x + 4*x^2 + 100),x)

[Out]

9*x - log(x)*(3*x - 5*x*exp(x)) + 75/(x + 5) + (9*x^2)/4 - (exp(x)*(50*x + (135*x^2)/4 + (15*x^3)/4))/(x + 5)

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sympy [B]  time = 0.46, size = 58, normalized size = 2.23 \begin {gather*} \frac {9 x^{2}}{4} - 3 x \log {\relax (x )} + 9 x + \frac {\left (- 15 x^{3} + 20 x^{2} \log {\relax (x )} - 135 x^{2} + 100 x \log {\relax (x )} - 200 x\right ) e^{x}}{4 x + 20} + \frac {75}{x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((20*x**3+220*x**2+700*x+500)*exp(x)-12*x**2-120*x-300)*ln(x)+(-15*x**4-240*x**3-1215*x**2-2150*x-5
00)*exp(x)+18*x**3+204*x**2+690*x+300)/(4*x**2+40*x+100),x)

[Out]

9*x**2/4 - 3*x*log(x) + 9*x + (-15*x**3 + 20*x**2*log(x) - 135*x**2 + 100*x*log(x) - 200*x)*exp(x)/(4*x + 20)
+ 75/(x + 5)

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