3.21.34 \(\int \frac {e^5 (-x-3 x^3)+3 e^5 x^2 \log (5)+(e^5 (-5+x-45 x^2+9 x^3)+e^5 (30 x-6 x^2) \log (5)) \log (5-x)}{(-15+3 x) \log ^2(5-x)} \, dx\)

Optimal. Leaf size=34 \[ -4+\frac {e^5 x \left (\frac {\frac {x}{3}+x^3}{x}-x \log (5)\right )}{\log (5-x)} \]

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Rubi [C]  time = 0.99, antiderivative size = 214, normalized size of antiderivative = 6.29, number of steps used = 50, number of rules used = 15, integrand size = 80, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {6688, 12, 6742, 2418, 2400, 2399, 2389, 2298, 2390, 2309, 2178, 2297, 2302, 30, 2417} \begin {gather*} -20 e^5 \text {Ei}(2 \log (5-x))+2 e^5 (15-\log (5)) \text {Ei}(2 \log (5-x))-2 e^5 (5-\log (5)) \text {Ei}(2 \log (5-x))+25 e^5 \text {li}(5-x)+5 e^5 (5-\log (5)) \text {li}(5-x)-\frac {2}{3} e^5 (113-15 \log (5)) \text {li}(5-x)+\frac {1}{3} e^5 (76-15 \log (5)) \text {li}(5-x)-\frac {e^5 (5-x) x^2}{\log (5-x)}-\frac {e^5 (5-x) x (5-\log (5))}{\log (5-x)}-\frac {e^5 (5-x) (76-15 \log (5))}{3 \log (5-x)}+\frac {5 e^5 (76-15 \log (5))}{3 \log (5-x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^5*(-x - 3*x^3) + 3*E^5*x^2*Log[5] + (E^5*(-5 + x - 45*x^2 + 9*x^3) + E^5*(30*x - 6*x^2)*Log[5])*Log[5 -
x])/((-15 + 3*x)*Log[5 - x]^2),x]

[Out]

-20*E^5*ExpIntegralEi[2*Log[5 - x]] - 2*E^5*ExpIntegralEi[2*Log[5 - x]]*(5 - Log[5]) + 2*E^5*ExpIntegralEi[2*L
og[5 - x]]*(15 - Log[5]) - (E^5*(5 - x)*x^2)/Log[5 - x] + (5*E^5*(76 - 15*Log[5]))/(3*Log[5 - x]) - (E^5*(5 -
x)*(76 - 15*Log[5]))/(3*Log[5 - x]) - (E^5*(5 - x)*x*(5 - Log[5]))/Log[5 - x] + 25*E^5*LogIntegral[5 - x] + (E
^5*(76 - 15*Log[5])*LogIntegral[5 - x])/3 - (2*E^5*(113 - 15*Log[5])*LogIntegral[5 - x])/3 + 5*E^5*(5 - Log[5]
)*LogIntegral[5 - 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 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rule 2389

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

Rule 2390

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

Rule 2399

Int[((f_.) + (g_.)*(x_))^(q_.)/((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.)), x_Symbol] :> Int[ExpandIn
tegrand[(f + g*x)^q/(a + b*Log[c*(d + e*x)^n]), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g,
0] && IGtQ[q, 0]

Rule 2400

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((
d + e*x)*(f + g*x)^q*(a + b*Log[c*(d + e*x)^n])^(p + 1))/(b*e*n*(p + 1)), x] + (-Dist[(q + 1)/(b*n*(p + 1)), I
nt[(f + g*x)^q*(a + b*Log[c*(d + e*x)^n])^(p + 1), x], x] + Dist[(q*(e*f - d*g))/(b*e*n*(p + 1)), Int[(f + g*x
)^(q - 1)*(a + b*Log[c*(d + e*x)^n])^(p + 1), x], x]) /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g,
0] && LtQ[p, -1] && GtQ[q, 0]

Rule 2417

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(Polyx_), x_Symbol] :> Int[ExpandIntegrand[Poly
x*(a + b*Log[c*(d + e*x)^n])^p, x], x] /; FreeQ[{a, b, c, d, e, n, p}, x] && PolynomialQ[Polyx, x]

Rule 2418

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[
(a + b*Log[c*(d + e*x)^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n}, x] && RationalFunct
ionQ[RFx, x] && IntegerQ[p]

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 {e^5 \left (-x \left (-1-3 x^2+3 x \log (5)\right )-(-5+x) \left (1+9 x^2-6 x \log (5)\right ) \log (5-x)\right )}{3 (5-x) \log ^2(5-x)} \, dx\\ &=\frac {1}{3} e^5 \int \frac {-x \left (-1-3 x^2+3 x \log (5)\right )-(-5+x) \left (1+9 x^2-6 x \log (5)\right ) \log (5-x)}{(5-x) \log ^2(5-x)} \, dx\\ &=\frac {1}{3} e^5 \int \left (-\frac {x \left (1+3 x^2-3 x \log (5)\right )}{(-5+x) \log ^2(5-x)}+\frac {1+9 x^2-6 x \log (5)}{\log (5-x)}\right ) \, dx\\ &=-\left (\frac {1}{3} e^5 \int \frac {x \left (1+3 x^2-3 x \log (5)\right )}{(-5+x) \log ^2(5-x)} \, dx\right )+\frac {1}{3} e^5 \int \frac {1+9 x^2-6 x \log (5)}{\log (5-x)} \, dx\\ &=-\left (\frac {1}{3} e^5 \int \left (\frac {3 x^2}{\log ^2(5-x)}+\frac {76 \left (1-\frac {15 \log (5)}{76}\right )}{\log ^2(5-x)}-\frac {3 x (-5+\log (5))}{\log ^2(5-x)}-\frac {5 (-76+15 \log (5))}{(-5+x) \log ^2(5-x)}\right ) \, dx\right )+\frac {1}{3} e^5 \int \left (\frac {9 (5-x)^2}{\log (5-x)}+\frac {6 (5-x) (-15+\log (5))}{\log (5-x)}-\frac {2 (-113+15 \log (5))}{\log (5-x)}\right ) \, dx\\ &=-\left (e^5 \int \frac {x^2}{\log ^2(5-x)} \, dx\right )+\left (3 e^5\right ) \int \frac {(5-x)^2}{\log (5-x)} \, dx-\frac {1}{3} \left (e^5 (76-15 \log (5))\right ) \int \frac {1}{\log ^2(5-x)} \, dx-\frac {1}{3} \left (5 e^5 (76-15 \log (5))\right ) \int \frac {1}{(-5+x) \log ^2(5-x)} \, dx+\frac {1}{3} \left (2 e^5 (113-15 \log (5))\right ) \int \frac {1}{\log (5-x)} \, dx-\left (e^5 (5-\log (5))\right ) \int \frac {x}{\log ^2(5-x)} \, dx-\left (2 e^5 (15-\log (5))\right ) \int \frac {5-x}{\log (5-x)} \, dx\\ &=-\frac {e^5 (5-x) x^2}{\log (5-x)}-\frac {e^5 (5-x) x (5-\log (5))}{\log (5-x)}-\left (3 e^5\right ) \int \frac {x^2}{\log (5-x)} \, dx-\left (3 e^5\right ) \operatorname {Subst}\left (\int \frac {x^2}{\log (x)} \, dx,x,5-x\right )+\left (10 e^5\right ) \int \frac {x}{\log (5-x)} \, dx+\frac {1}{3} \left (e^5 (76-15 \log (5))\right ) \operatorname {Subst}\left (\int \frac {1}{\log ^2(x)} \, dx,x,5-x\right )-\frac {1}{3} \left (5 e^5 (76-15 \log (5))\right ) \operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,5-x\right )-\frac {1}{3} \left (2 e^5 (113-15 \log (5))\right ) \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,5-x\right )-\left (2 e^5 (5-\log (5))\right ) \int \frac {x}{\log (5-x)} \, dx+\left (5 e^5 (5-\log (5))\right ) \int \frac {1}{\log (5-x)} \, dx+\left (2 e^5 (15-\log (5))\right ) \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,5-x\right )\\ &=-\frac {e^5 (5-x) x^2}{\log (5-x)}-\frac {e^5 (5-x) (76-15 \log (5))}{3 \log (5-x)}-\frac {e^5 (5-x) x (5-\log (5))}{\log (5-x)}-\frac {2}{3} e^5 (113-15 \log (5)) \text {li}(5-x)-\left (3 e^5\right ) \int \left (\frac {25}{\log (5-x)}-\frac {10 (5-x)}{\log (5-x)}+\frac {(5-x)^2}{\log (5-x)}\right ) \, dx-\left (3 e^5\right ) \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (5-x)\right )+\left (10 e^5\right ) \int \left (\frac {5}{\log (5-x)}-\frac {5-x}{\log (5-x)}\right ) \, dx+\frac {1}{3} \left (e^5 (76-15 \log (5))\right ) \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,5-x\right )-\frac {1}{3} \left (5 e^5 (76-15 \log (5))\right ) \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (5-x)\right )-\left (2 e^5 (5-\log (5))\right ) \int \left (\frac {5}{\log (5-x)}-\frac {5-x}{\log (5-x)}\right ) \, dx-\left (5 e^5 (5-\log (5))\right ) \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,5-x\right )+\left (2 e^5 (15-\log (5))\right ) \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (5-x)\right )\\ &=-3 e^5 \text {Ei}(3 \log (5-x))+2 e^5 \text {Ei}(2 \log (5-x)) (15-\log (5))-\frac {e^5 (5-x) x^2}{\log (5-x)}+\frac {5 e^5 (76-15 \log (5))}{3 \log (5-x)}-\frac {e^5 (5-x) (76-15 \log (5))}{3 \log (5-x)}-\frac {e^5 (5-x) x (5-\log (5))}{\log (5-x)}+\frac {1}{3} e^5 (76-15 \log (5)) \text {li}(5-x)-\frac {2}{3} e^5 (113-15 \log (5)) \text {li}(5-x)-5 e^5 (5-\log (5)) \text {li}(5-x)-\left (3 e^5\right ) \int \frac {(5-x)^2}{\log (5-x)} \, dx-\left (10 e^5\right ) \int \frac {5-x}{\log (5-x)} \, dx+\left (30 e^5\right ) \int \frac {5-x}{\log (5-x)} \, dx+\left (50 e^5\right ) \int \frac {1}{\log (5-x)} \, dx-\left (75 e^5\right ) \int \frac {1}{\log (5-x)} \, dx+\left (2 e^5 (5-\log (5))\right ) \int \frac {5-x}{\log (5-x)} \, dx-\left (10 e^5 (5-\log (5))\right ) \int \frac {1}{\log (5-x)} \, dx\\ &=-3 e^5 \text {Ei}(3 \log (5-x))+2 e^5 \text {Ei}(2 \log (5-x)) (15-\log (5))-\frac {e^5 (5-x) x^2}{\log (5-x)}+\frac {5 e^5 (76-15 \log (5))}{3 \log (5-x)}-\frac {e^5 (5-x) (76-15 \log (5))}{3 \log (5-x)}-\frac {e^5 (5-x) x (5-\log (5))}{\log (5-x)}+\frac {1}{3} e^5 (76-15 \log (5)) \text {li}(5-x)-\frac {2}{3} e^5 (113-15 \log (5)) \text {li}(5-x)-5 e^5 (5-\log (5)) \text {li}(5-x)+\left (3 e^5\right ) \operatorname {Subst}\left (\int \frac {x^2}{\log (x)} \, dx,x,5-x\right )+\left (10 e^5\right ) \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,5-x\right )-\left (30 e^5\right ) \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,5-x\right )-\left (50 e^5\right ) \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,5-x\right )+\left (75 e^5\right ) \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,5-x\right )-\left (2 e^5 (5-\log (5))\right ) \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,5-x\right )+\left (10 e^5 (5-\log (5))\right ) \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,5-x\right )\\ &=-3 e^5 \text {Ei}(3 \log (5-x))+2 e^5 \text {Ei}(2 \log (5-x)) (15-\log (5))-\frac {e^5 (5-x) x^2}{\log (5-x)}+\frac {5 e^5 (76-15 \log (5))}{3 \log (5-x)}-\frac {e^5 (5-x) (76-15 \log (5))}{3 \log (5-x)}-\frac {e^5 (5-x) x (5-\log (5))}{\log (5-x)}+25 e^5 \text {li}(5-x)+\frac {1}{3} e^5 (76-15 \log (5)) \text {li}(5-x)-\frac {2}{3} e^5 (113-15 \log (5)) \text {li}(5-x)+5 e^5 (5-\log (5)) \text {li}(5-x)+\left (3 e^5\right ) \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (5-x)\right )+\left (10 e^5\right ) \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (5-x)\right )-\left (30 e^5\right ) \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (5-x)\right )-\left (2 e^5 (5-\log (5))\right ) \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (5-x)\right )\\ &=-20 e^5 \text {Ei}(2 \log (5-x))-2 e^5 \text {Ei}(2 \log (5-x)) (5-\log (5))+2 e^5 \text {Ei}(2 \log (5-x)) (15-\log (5))-\frac {e^5 (5-x) x^2}{\log (5-x)}+\frac {5 e^5 (76-15 \log (5))}{3 \log (5-x)}-\frac {e^5 (5-x) (76-15 \log (5))}{3 \log (5-x)}-\frac {e^5 (5-x) x (5-\log (5))}{\log (5-x)}+25 e^5 \text {li}(5-x)+\frac {1}{3} e^5 (76-15 \log (5)) \text {li}(5-x)-\frac {2}{3} e^5 (113-15 \log (5)) \text {li}(5-x)+5 e^5 (5-\log (5)) \text {li}(5-x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 28, normalized size = 0.82 \begin {gather*} \frac {e^5 x \left (1+3 x^2-3 x \log (5)\right )}{3 \log (5-x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^5*(-x - 3*x^3) + 3*E^5*x^2*Log[5] + (E^5*(-5 + x - 45*x^2 + 9*x^3) + E^5*(30*x - 6*x^2)*Log[5])*L
og[5 - x])/((-15 + 3*x)*Log[5 - x]^2),x]

[Out]

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

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fricas [A]  time = 0.69, size = 31, normalized size = 0.91 \begin {gather*} -\frac {3 \, x^{2} e^{5} \log \relax (5) - {\left (3 \, x^{3} + x\right )} e^{5}}{3 \, \log \left (-x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x^2+30*x)*exp(5)*log(5)+(9*x^3-45*x^2+x-5)*exp(5))*log(5-x)+3*x^2*exp(5)*log(5)+(-3*x^3-x)*exp
(5))/(3*x-15)/log(5-x)^2,x, algorithm="fricas")

[Out]

-1/3*(3*x^2*e^5*log(5) - (3*x^3 + x)*e^5)/log(-x + 5)

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giac [A]  time = 0.26, size = 31, normalized size = 0.91 \begin {gather*} \frac {3 \, x^{3} e^{5} - 3 \, x^{2} e^{5} \log \relax (5) + x e^{5}}{3 \, \log \left (-x + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x^2+30*x)*exp(5)*log(5)+(9*x^3-45*x^2+x-5)*exp(5))*log(5-x)+3*x^2*exp(5)*log(5)+(-3*x^3-x)*exp
(5))/(3*x-15)/log(5-x)^2,x, algorithm="giac")

[Out]

1/3*(3*x^3*e^5 - 3*x^2*e^5*log(5) + x*e^5)/log(-x + 5)

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maple [A]  time = 0.35, size = 26, normalized size = 0.76

method	result	size
risch	\(-\frac {x \,{\mathrm e}^{5} \left (3 x \ln \relax (5)-3 x^{2}-1\right )}{3 \ln \left (5-x \right )}\)	\(26\)
norman	\(\frac {x^{3} {\mathrm e}^{5}+\frac {x \,{\mathrm e}^{5}}{3}-x^{2} {\mathrm e}^{5} \ln \relax (5)}{\ln \left (5-x \right )}\)	\(31\)
derivativedivides	\(2 \,{\mathrm e}^{5} \ln \relax (5) \expIntegralEi \left (1, -2 \ln \left (5-x \right )\right )+3 \,{\mathrm e}^{5} \expIntegralEi \left (1, -3 \ln \left (5-x \right )\right )-10 \,{\mathrm e}^{5} \ln \relax (5) \expIntegralEi \left (1, -\ln \left (5-x \right )\right )+{\mathrm e}^{5} \ln \relax (5) \left (-\frac {\left (5-x \right )^{2}}{\ln \left (5-x \right )}-2 \expIntegralEi \left (1, -2 \ln \left (5-x \right )\right )\right )-30 \,{\mathrm e}^{5} \expIntegralEi \left (1, -2 \ln \left (5-x \right )\right )+{\mathrm e}^{5} \left (-\frac {\left (5-x \right )^{3}}{\ln \left (5-x \right )}-3 \expIntegralEi \left (1, -3 \ln \left (5-x \right )\right )\right )-10 \,{\mathrm e}^{5} \ln \relax (5) \left (-\frac {5-x}{\ln \left (5-x \right )}-\expIntegralEi \left (1, -\ln \left (5-x \right )\right )\right )+\frac {226 \,{\mathrm e}^{5} \expIntegralEi \left (1, -\ln \left (5-x \right )\right )}{3}-15 \,{\mathrm e}^{5} \left (-\frac {\left (5-x \right )^{2}}{\ln \left (5-x \right )}-2 \expIntegralEi \left (1, -2 \ln \left (5-x \right )\right )\right )-\frac {25 \,{\mathrm e}^{5} \ln \relax (5)}{\ln \left (5-x \right )}+\frac {226 \,{\mathrm e}^{5} \left (-\frac {5-x}{\ln \left (5-x \right )}-\expIntegralEi \left (1, -\ln \left (5-x \right )\right )\right )}{3}+\frac {380 \,{\mathrm e}^{5}}{3 \ln \left (5-x \right )}\)	\(270\)
default	\(2 \,{\mathrm e}^{5} \ln \relax (5) \expIntegralEi \left (1, -2 \ln \left (5-x \right )\right )+3 \,{\mathrm e}^{5} \expIntegralEi \left (1, -3 \ln \left (5-x \right )\right )-10 \,{\mathrm e}^{5} \ln \relax (5) \expIntegralEi \left (1, -\ln \left (5-x \right )\right )+{\mathrm e}^{5} \ln \relax (5) \left (-\frac {\left (5-x \right )^{2}}{\ln \left (5-x \right )}-2 \expIntegralEi \left (1, -2 \ln \left (5-x \right )\right )\right )-30 \,{\mathrm e}^{5} \expIntegralEi \left (1, -2 \ln \left (5-x \right )\right )+{\mathrm e}^{5} \left (-\frac {\left (5-x \right )^{3}}{\ln \left (5-x \right )}-3 \expIntegralEi \left (1, -3 \ln \left (5-x \right )\right )\right )-10 \,{\mathrm e}^{5} \ln \relax (5) \left (-\frac {5-x}{\ln \left (5-x \right )}-\expIntegralEi \left (1, -\ln \left (5-x \right )\right )\right )+\frac {226 \,{\mathrm e}^{5} \expIntegralEi \left (1, -\ln \left (5-x \right )\right )}{3}-15 \,{\mathrm e}^{5} \left (-\frac {\left (5-x \right )^{2}}{\ln \left (5-x \right )}-2 \expIntegralEi \left (1, -2 \ln \left (5-x \right )\right )\right )-\frac {25 \,{\mathrm e}^{5} \ln \relax (5)}{\ln \left (5-x \right )}+\frac {226 \,{\mathrm e}^{5} \left (-\frac {5-x}{\ln \left (5-x \right )}-\expIntegralEi \left (1, -\ln \left (5-x \right )\right )\right )}{3}+\frac {380 \,{\mathrm e}^{5}}{3 \ln \left (5-x \right )}\)	\(270\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-6*x^2+30*x)*exp(5)*ln(5)+(9*x^3-45*x^2+x-5)*exp(5))*ln(5-x)+3*x^2*exp(5)*ln(5)+(-3*x^3-x)*exp(5))/(3*x
-15)/ln(5-x)^2,x,method=_RETURNVERBOSE)

[Out]

-1/3*x*exp(5)*(3*x*ln(5)-3*x^2-1)/ln(5-x)

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maxima [A]  time = 0.65, size = 36, normalized size = 1.06 \begin {gather*} \frac {3 \, x^{3} e^{5} - 3 \, x^{2} e^{5} \log \relax (5) + x e^{5}}{3 \, \log \left (-x + 5\right )} + \frac {5}{3} \, e^{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x^2+30*x)*exp(5)*log(5)+(9*x^3-45*x^2+x-5)*exp(5))*log(5-x)+3*x^2*exp(5)*log(5)+(-3*x^3-x)*exp
(5))/(3*x-15)/log(5-x)^2,x, algorithm="maxima")

[Out]

1/3*(3*x^3*e^5 - 3*x^2*e^5*log(5) + x*e^5)/log(-x + 5) + 5/3*e^5

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mupad [B]  time = 1.30, size = 82, normalized size = 2.41 \begin {gather*} \frac {3\,{\mathrm {e}}^5\,x^5-3\,{\mathrm {e}}^5\,\left (\ln \relax (5)+10\right )\,x^4+2\,{\mathrm {e}}^5\,\left (15\,\ln \relax (5)+38\right )\,x^3-5\,{\mathrm {e}}^5\,\left (15\,\ln \relax (5)+2\right )\,x^2+25\,{\mathrm {e}}^5\,x}{75\,\ln \left (5-x\right )-30\,x\,\ln \left (5-x\right )+3\,x^2\,\ln \left (5-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(5 - x)*(exp(5)*(x - 45*x^2 + 9*x^3 - 5) + exp(5)*log(5)*(30*x - 6*x^2)) - exp(5)*(x + 3*x^3) + 3*x^2*
exp(5)*log(5))/(log(5 - x)^2*(3*x - 15)),x)

[Out]

(25*x*exp(5) + 3*x^5*exp(5) - 3*x^4*exp(5)*(log(5) + 10) - 5*x^2*exp(5)*(15*log(5) + 2) + 2*x^3*exp(5)*(15*log
(5) + 38))/(75*log(5 - x) - 30*x*log(5 - x) + 3*x^2*log(5 - x))

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sympy [A]  time = 0.13, size = 31, normalized size = 0.91 \begin {gather*} \frac {3 x^{3} e^{5} - 3 x^{2} e^{5} \log {\relax (5 )} + x e^{5}}{3 \log {\left (5 - x \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x**2+30*x)*exp(5)*ln(5)+(9*x**3-45*x**2+x-5)*exp(5))*ln(5-x)+3*x**2*exp(5)*ln(5)+(-3*x**3-x)*e
xp(5))/(3*x-15)/ln(5-x)**2,x)

[Out]

(3*x**3*exp(5) - 3*x**2*exp(5)*log(5) + x*exp(5))/(3*log(5 - x))

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