3.39.87 \(\int \frac {15+5 x+e^{-4+x} (-15+9 x+5 x^2)+(3+x+e^{-4+x} (-3+2 x+x^2)) \log (\frac {1}{15+5 x})+\log (x) (-15-6 x+(-3-x) \log (\frac {1}{15+5 x}))}{3 x^2+x^3} \, dx\)

Optimal. Leaf size=24 \[ \frac {\left (e^{-4+x}+\log (x)\right ) \left (5+\log \left (\frac {1}{5 (3+x)}\right )\right )}{x} \]

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Rubi [B]  time = 2.81, antiderivative size = 136, normalized size of antiderivative = 5.67, number of steps used = 37, number of rules used = 18, integrand size = 81, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {1593, 6742, 6688, 2177, 2178, 2197, 2554, 2357, 2304, 2301, 2317, 2391, 2395, 36, 29, 31, 2376, 2392} \begin {gather*} \frac {5 e^{x-4}}{x}-\frac {1}{3} \log ^2(x)+\frac {1}{3} \log \left (\frac {x}{3}+1\right ) \log (x)-\frac {1}{3} (1-\log (x)) \log (x)+\frac {5 \log (x)}{x}+\frac {1}{3} \log (3) \log (x)+\frac {\log (x)}{3}+\frac {1}{3} (1-\log (x)) \log (x+3)-\frac {1}{3} \log (x+3)-\frac {(1-\log (x)) \log \left (\frac {1}{5 x+15}\right )}{x}+\frac {e^{x-4} \log \left (\frac {1}{5 x+15}\right )}{x}+\frac {\log \left (\frac {1}{5 x+15}\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(15 + 5*x + E^(-4 + x)*(-15 + 9*x + 5*x^2) + (3 + x + E^(-4 + x)*(-3 + 2*x + x^2))*Log[(15 + 5*x)^(-1)] +
Log[x]*(-15 - 6*x + (-3 - x)*Log[(15 + 5*x)^(-1)]))/(3*x^2 + x^3),x]

[Out]

(5*E^(-4 + x))/x + Log[x]/3 + (5*Log[x])/x + (Log[3]*Log[x])/3 + (Log[1 + x/3]*Log[x])/3 - ((1 - Log[x])*Log[x
])/3 - Log[x]^2/3 - Log[3 + x]/3 + ((1 - Log[x])*Log[3 + x])/3 + Log[(15 + 5*x)^(-1)]/x + (E^(-4 + x)*Log[(15
+ 5*x)^(-1)])/x - ((1 - Log[x])*Log[(15 + 5*x)^(-1)])/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 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

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 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rule 2376

Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((g_.)*(x_))^(q_.), x_Sym
bol] :> With[{u = IntHide[(g*x)^q*Log[d*(e + f*x^m)^r], x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[Dist
[1/x, u, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, r, m, n, q}, x] && (IntegerQ[(q + 1)/m] || (RationalQ[m] &
& RationalQ[q])) && NeQ[q, -1]

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 2392

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

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -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 {15+5 x+e^{-4+x} \left (-15+9 x+5 x^2\right )+\left (3+x+e^{-4+x} \left (-3+2 x+x^2\right )\right ) \log \left (\frac {1}{15+5 x}\right )+\log (x) \left (-15-6 x+(-3-x) \log \left (\frac {1}{15+5 x}\right )\right )}{x^2 (3+x)} \, dx\\ &=\int \left (\frac {e^{-4+x} \left (-15+9 x+5 x^2-3 \log \left (\frac {1}{15+5 x}\right )+2 x \log \left (\frac {1}{15+5 x}\right )+x^2 \log \left (\frac {1}{15+5 x}\right )\right )}{x^2 (3+x)}+\frac {15+5 x-15 \log (x)-6 x \log (x)+3 \log \left (\frac {1}{15+5 x}\right )+x \log \left (\frac {1}{15+5 x}\right )-3 \log (x) \log \left (\frac {1}{15+5 x}\right )-x \log (x) \log \left (\frac {1}{15+5 x}\right )}{x^2 (3+x)}\right ) \, dx\\ &=\int \frac {e^{-4+x} \left (-15+9 x+5 x^2-3 \log \left (\frac {1}{15+5 x}\right )+2 x \log \left (\frac {1}{15+5 x}\right )+x^2 \log \left (\frac {1}{15+5 x}\right )\right )}{x^2 (3+x)} \, dx+\int \frac {15+5 x-15 \log (x)-6 x \log (x)+3 \log \left (\frac {1}{15+5 x}\right )+x \log \left (\frac {1}{15+5 x}\right )-3 \log (x) \log \left (\frac {1}{15+5 x}\right )-x \log (x) \log \left (\frac {1}{15+5 x}\right )}{x^2 (3+x)} \, dx\\ &=\int \frac {e^{-4+x} \left (-15+9 x+5 x^2+\left (-3+2 x+x^2\right ) \log \left (\frac {1}{15+5 x}\right )\right )}{x^2 (3+x)} \, dx+\int \frac {(3+x) \left (5+\log \left (\frac {1}{15+5 x}\right )\right )-\log (x) \left (15+6 x+(3+x) \log \left (\frac {1}{15+5 x}\right )\right )}{x^2 (3+x)} \, dx\\ &=\int \left (\frac {e^{-4+x} \left (-15+9 x+5 x^2\right )}{x^2 (3+x)}+\frac {e^{-4+x} (-1+x) \log \left (\frac {1}{15+5 x}\right )}{x^2}\right ) \, dx+\int \left (\frac {15+5 x-15 \log (x)-6 x \log (x)}{x^2 (3+x)}-\frac {(-1+\log (x)) \log \left (\frac {1}{15+5 x}\right )}{x^2}\right ) \, dx\\ &=\int \frac {e^{-4+x} \left (-15+9 x+5 x^2\right )}{x^2 (3+x)} \, dx+\int \frac {15+5 x-15 \log (x)-6 x \log (x)}{x^2 (3+x)} \, dx+\int \frac {e^{-4+x} (-1+x) \log \left (\frac {1}{15+5 x}\right )}{x^2} \, dx-\int \frac {(-1+\log (x)) \log \left (\frac {1}{15+5 x}\right )}{x^2} \, dx\\ &=-\frac {1}{3} (1-\log (x)) \log (x)+\frac {1}{3} (1-\log (x)) \log (3+x)+\frac {e^{-4+x} \log \left (\frac {1}{15+5 x}\right )}{x}-\frac {(1-\log (x)) \log \left (\frac {1}{15+5 x}\right )}{x}-\int \frac {e^{-4+x}}{(-3-x) x} \, dx+\int \left (-\frac {5 e^{-4+x}}{x^2}+\frac {14 e^{-4+x}}{3 x}+\frac {e^{-4+x}}{3 (3+x)}\right ) \, dx+\int \left (\frac {5}{x^2}-\frac {3 (5+2 x) \log (x)}{x^2 (3+x)}\right ) \, dx+\int \left (-\frac {\log (x)}{3 x}+\frac {\log (3+x)}{3 x}-\frac {\log \left (\frac {1}{15+5 x}\right )}{x^2}\right ) \, dx\\ &=-\frac {5}{x}-\frac {1}{3} (1-\log (x)) \log (x)+\frac {1}{3} (1-\log (x)) \log (3+x)+\frac {e^{-4+x} \log \left (\frac {1}{15+5 x}\right )}{x}-\frac {(1-\log (x)) \log \left (\frac {1}{15+5 x}\right )}{x}+\frac {1}{3} \int \frac {e^{-4+x}}{3+x} \, dx-\frac {1}{3} \int \frac {\log (x)}{x} \, dx+\frac {1}{3} \int \frac {\log (3+x)}{x} \, dx-3 \int \frac {(5+2 x) \log (x)}{x^2 (3+x)} \, dx+\frac {14}{3} \int \frac {e^{-4+x}}{x} \, dx-5 \int \frac {e^{-4+x}}{x^2} \, dx-\int \left (-\frac {e^{-4+x}}{3 x}+\frac {e^{-4+x}}{3 (3+x)}\right ) \, dx-\int \frac {\log \left (\frac {1}{15+5 x}\right )}{x^2} \, dx\\ &=-\frac {5}{x}+\frac {5 e^{-4+x}}{x}+\frac {14 \text {Ei}(x)}{3 e^4}+\frac {\text {Ei}(3+x)}{3 e^7}+\frac {1}{3} \log (3) \log (x)-\frac {1}{3} (1-\log (x)) \log (x)-\frac {\log ^2(x)}{6}+\frac {1}{3} (1-\log (x)) \log (3+x)+\frac {\log \left (\frac {1}{15+5 x}\right )}{x}+\frac {e^{-4+x} \log \left (\frac {1}{15+5 x}\right )}{x}-\frac {(1-\log (x)) \log \left (\frac {1}{15+5 x}\right )}{x}+\frac {1}{3} \int \frac {e^{-4+x}}{x} \, dx-\frac {1}{3} \int \frac {e^{-4+x}}{3+x} \, dx+\frac {1}{3} \int \frac {\log \left (1+\frac {x}{3}\right )}{x} \, dx-3 \int \left (\frac {5 \log (x)}{3 x^2}+\frac {\log (x)}{9 x}-\frac {\log (x)}{9 (3+x)}\right ) \, dx-5 \int \frac {e^{-4+x}}{x} \, dx+5 \int \frac {1}{x (15+5 x)} \, dx\\ &=-\frac {5}{x}+\frac {5 e^{-4+x}}{x}+\frac {1}{3} \log (3) \log (x)-\frac {1}{3} (1-\log (x)) \log (x)-\frac {\log ^2(x)}{6}+\frac {1}{3} (1-\log (x)) \log (3+x)+\frac {\log \left (\frac {1}{15+5 x}\right )}{x}+\frac {e^{-4+x} \log \left (\frac {1}{15+5 x}\right )}{x}-\frac {(1-\log (x)) \log \left (\frac {1}{15+5 x}\right )}{x}-\frac {\text {Li}_2\left (-\frac {x}{3}\right )}{3}+\frac {1}{3} \int \frac {1}{x} \, dx-\frac {1}{3} \int \frac {\log (x)}{x} \, dx+\frac {1}{3} \int \frac {\log (x)}{3+x} \, dx-\frac {5}{3} \int \frac {1}{15+5 x} \, dx-5 \int \frac {\log (x)}{x^2} \, dx\\ &=\frac {5 e^{-4+x}}{x}+\frac {\log (x)}{3}+\frac {5 \log (x)}{x}+\frac {1}{3} \log (3) \log (x)+\frac {1}{3} \log \left (1+\frac {x}{3}\right ) \log (x)-\frac {1}{3} (1-\log (x)) \log (x)-\frac {\log ^2(x)}{3}-\frac {1}{3} \log (3+x)+\frac {1}{3} (1-\log (x)) \log (3+x)+\frac {\log \left (\frac {1}{15+5 x}\right )}{x}+\frac {e^{-4+x} \log \left (\frac {1}{15+5 x}\right )}{x}-\frac {(1-\log (x)) \log \left (\frac {1}{15+5 x}\right )}{x}-\frac {\text {Li}_2\left (-\frac {x}{3}\right )}{3}-\frac {1}{3} \int \frac {\log \left (1+\frac {x}{3}\right )}{x} \, dx\\ &=\frac {5 e^{-4+x}}{x}+\frac {\log (x)}{3}+\frac {5 \log (x)}{x}+\frac {1}{3} \log (3) \log (x)+\frac {1}{3} \log \left (1+\frac {x}{3}\right ) \log (x)-\frac {1}{3} (1-\log (x)) \log (x)-\frac {\log ^2(x)}{3}-\frac {1}{3} \log (3+x)+\frac {1}{3} (1-\log (x)) \log (3+x)+\frac {\log \left (\frac {1}{15+5 x}\right )}{x}+\frac {e^{-4+x} \log \left (\frac {1}{15+5 x}\right )}{x}-\frac {(1-\log (x)) \log \left (\frac {1}{15+5 x}\right )}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.69, size = 27, normalized size = 1.12 \begin {gather*} \frac {\left (e^x+e^4 \log (x)\right ) \left (5+\log \left (\frac {1}{15+5 x}\right )\right )}{e^4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(15 + 5*x + E^(-4 + x)*(-15 + 9*x + 5*x^2) + (3 + x + E^(-4 + x)*(-3 + 2*x + x^2))*Log[(15 + 5*x)^(-
1)] + Log[x]*(-15 - 6*x + (-3 - x)*Log[(15 + 5*x)^(-1)]))/(3*x^2 + x^3),x]

[Out]

((E^x + E^4*Log[x])*(5 + Log[(15 + 5*x)^(-1)]))/(E^4*x)

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fricas [A]  time = 0.51, size = 37, normalized size = 1.54 \begin {gather*} \frac {{\left (\log \left (\frac {1}{5 \, {\left (x + 3\right )}}\right ) + 5\right )} \log \relax (x) + e^{\left (x - 4\right )} \log \left (\frac {1}{5 \, {\left (x + 3\right )}}\right ) + 5 \, e^{\left (x - 4\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3-x)*log(1/(5*x+15))-6*x-15)*log(x)+((x^2+2*x-3)*exp(x-4)+3+x)*log(1/(5*x+15))+(5*x^2+9*x-15)*ex
p(x-4)+5*x+15)/(x^3+3*x^2),x, algorithm="fricas")

[Out]

((log(1/5/(x + 3)) + 5)*log(x) + e^(x - 4)*log(1/5/(x + 3)) + 5*e^(x - 4))/x

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giac [A]  time = 0.18, size = 38, normalized size = 1.58 \begin {gather*} -\frac {{\left (e^{4} \log \left (5 \, x + 15\right ) \log \relax (x) + e^{x} \log \left (5 \, x + 15\right ) - 5 \, e^{4} \log \relax (x) - 5 \, e^{x}\right )} e^{\left (-4\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3-x)*log(1/(5*x+15))-6*x-15)*log(x)+((x^2+2*x-3)*exp(x-4)+3+x)*log(1/(5*x+15))+(5*x^2+9*x-15)*ex
p(x-4)+5*x+15)/(x^3+3*x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.23, size = 36, normalized size = 1.50

method	result	size
risch	\(-\frac {\left (\ln \relax (x )+{\mathrm e}^{x -4}\right ) \ln \left (3+x \right )}{x}-\frac {\left (-10+2 \ln \relax (5)\right ) \left (\ln \relax (x )+{\mathrm e}^{x -4}\right )}{2 x}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-3-x)*ln(1/(5*x+15))-6*x-15)*ln(x)+((x^2+2*x-3)*exp(x-4)+3+x)*ln(1/(5*x+15))+(5*x^2+9*x-15)*exp(x-4)+5*
x+15)/(x^3+3*x^2),x,method=_RETURNVERBOSE)

[Out]

-(ln(x)+exp(x-4))/x*ln(3+x)-1/2*(-10+2*ln(5))*(ln(x)+exp(x-4))/x

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maxima [B]  time = 0.50, size = 47, normalized size = 1.96 \begin {gather*} -\frac {{\left ({\left (\log \relax (5) - 5\right )} e^{4} \log \relax (x) + {\left (\log \relax (5) - 5\right )} e^{x} + {\left (e^{4} \log \relax (x) + e^{x}\right )} \log \left (x + 3\right ) - 5 \, e^{4}\right )} e^{\left (-4\right )}}{x} - \frac {5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3-x)*log(1/(5*x+15))-6*x-15)*log(x)+((x^2+2*x-3)*exp(x-4)+3+x)*log(1/(5*x+15))+(5*x^2+9*x-15)*ex
p(x-4)+5*x+15)/(x^3+3*x^2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {5\,x+{\mathrm {e}}^{x-4}\,\left (5\,x^2+9\,x-15\right )-\ln \relax (x)\,\left (6\,x+\ln \left (\frac {1}{5\,x+15}\right )\,\left (x+3\right )+15\right )+\ln \left (\frac {1}{5\,x+15}\right )\,\left (x+{\mathrm {e}}^{x-4}\,\left (x^2+2\,x-3\right )+3\right )+15}{x^3+3\,x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x + exp(x - 4)*(9*x + 5*x^2 - 15) - log(x)*(6*x + log(1/(5*x + 15))*(x + 3) + 15) + log(1/(5*x + 15))*(
x + exp(x - 4)*(2*x + x^2 - 3) + 3) + 15)/(3*x^2 + x^3),x)

[Out]

int((5*x + exp(x - 4)*(9*x + 5*x^2 - 15) - log(x)*(6*x + log(1/(5*x + 15))*(x + 3) + 15) + log(1/(5*x + 15))*(
x + exp(x - 4)*(2*x + x^2 - 3) + 3) + 15)/(3*x^2 + x^3), x)

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sympy [A]  time = 0.89, size = 36, normalized size = 1.50 \begin {gather*} \frac {\left (\log {\left (\frac {1}{5 x + 15} \right )} + 5\right ) e^{x - 4}}{x} + \frac {\log {\relax (x )} \log {\left (\frac {1}{5 x + 15} \right )}}{x} + \frac {5 \log {\relax (x )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3-x)*ln(1/(5*x+15))-6*x-15)*ln(x)+((x**2+2*x-3)*exp(x-4)+3+x)*ln(1/(5*x+15))+(5*x**2+9*x-15)*exp
(x-4)+5*x+15)/(x**3+3*x**2),x)

[Out]

(log(1/(5*x + 15)) + 5)*exp(x - 4)/x + log(x)*log(1/(5*x + 15))/x + 5*log(x)/x

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