3.5.77 \(\int \frac {e^5 (5184-122688 x)+e^5 (-9120+431840 x) \log (2 x)+e^5 (-67800+684600 x) \log ^2(2 x)+e^5 (-31500+220500 x) \log ^3(2 x)}{8+60 \log (2 x)+150 \log ^2(2 x)+125 \log ^3(2 x)} \, dx\)

Optimal. Leaf size=28 \[ 2 e^5 \left (3-x-20 \left (x+\frac {x}{\frac {2}{5}+\log (2 x)}\right )\right )^2 \]

________________________________________________________________________________________

Rubi [B]  time = 0.45, antiderivative size = 70, normalized size of antiderivative = 2.50, number of steps used = 23, number of rules used = 9, integrand size = 80, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.112, Rules used = {6688, 12, 6742, 2306, 2309, 2178, 2320, 2330, 2299} \begin {gather*} \frac {8000 e^5 x^2}{5 \log (2 x)+2}+\frac {20000 e^5 x^2}{(5 \log (2 x)+2)^2}+18 e^5 (1-7 x)^2-\frac {400 e^5 (3-x) x}{5 \log (2 x)+2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^5*(5184 - 122688*x) + E^5*(-9120 + 431840*x)*Log[2*x] + E^5*(-67800 + 684600*x)*Log[2*x]^2 + E^5*(-3150
0 + 220500*x)*Log[2*x]^3)/(8 + 60*Log[2*x] + 150*Log[2*x]^2 + 125*Log[2*x]^3),x]

[Out]

18*E^5*(1 - 7*x)^2 + (20000*E^5*x^2)/(2 + 5*Log[2*x])^2 - (400*E^5*(3 - x)*x)/(2 + 5*Log[2*x]) + (8000*E^5*x^2
)/(2 + 5*Log[2*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 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 2299

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

Rule 2306

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

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 2320

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

Rule 2330

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = Expand
Integrand[(a + b*Log[c*x^n])^p, (d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n, p, q, r}
, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[r]))

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 {4 e^5 (6-142 x-15 (-1+7 x) \log (2 x)) \left (216-920 \log (2 x)-525 \log ^2(2 x)\right )}{(2+5 \log (2 x))^3} \, dx\\ &=\left (4 e^5\right ) \int \frac {(6-142 x-15 (-1+7 x) \log (2 x)) \left (216-920 \log (2 x)-525 \log ^2(2 x)\right )}{(2+5 \log (2 x))^3} \, dx\\ &=\left (4 e^5\right ) \int \left (63 (-1+7 x)-\frac {50000 x}{(2+5 \log (2 x))^3}-\frac {500 (-3+x)}{(2+5 \log (2 x))^2}+\frac {300 (-1+14 x)}{2+5 \log (2 x)}\right ) \, dx\\ &=18 e^5 (1-7 x)^2+\left (1200 e^5\right ) \int \frac {-1+14 x}{2+5 \log (2 x)} \, dx-\left (2000 e^5\right ) \int \frac {-3+x}{(2+5 \log (2 x))^2} \, dx-\left (200000 e^5\right ) \int \frac {x}{(2+5 \log (2 x))^3} \, dx\\ &=18 e^5 (1-7 x)^2+\frac {20000 e^5 x^2}{(2+5 \log (2 x))^2}-\frac {400 e^5 (3-x) x}{2+5 \log (2 x)}-\left (800 e^5\right ) \int \frac {-3+x}{2+5 \log (2 x)} \, dx-\left (1200 e^5\right ) \int \frac {1}{2+5 \log (2 x)} \, dx+\left (1200 e^5\right ) \int \left (-\frac {1}{2+5 \log (2 x)}+\frac {14 x}{2+5 \log (2 x)}\right ) \, dx-\left (40000 e^5\right ) \int \frac {x}{(2+5 \log (2 x))^2} \, dx\\ &=18 e^5 (1-7 x)^2+\frac {20000 e^5 x^2}{(2+5 \log (2 x))^2}-\frac {400 e^5 (3-x) x}{2+5 \log (2 x)}+\frac {8000 e^5 x^2}{2+5 \log (2 x)}-\left (600 e^5\right ) \operatorname {Subst}\left (\int \frac {e^x}{2+5 x} \, dx,x,\log (2 x)\right )-\left (800 e^5\right ) \int \left (-\frac {3}{2+5 \log (2 x)}+\frac {x}{2+5 \log (2 x)}\right ) \, dx-\left (1200 e^5\right ) \int \frac {1}{2+5 \log (2 x)} \, dx-\left (16000 e^5\right ) \int \frac {x}{2+5 \log (2 x)} \, dx+\left (16800 e^5\right ) \int \frac {x}{2+5 \log (2 x)} \, dx\\ &=18 e^5 (1-7 x)^2-120 e^{23/5} \text {Ei}\left (\frac {1}{5} (2+5 \log (2 x))\right )+\frac {20000 e^5 x^2}{(2+5 \log (2 x))^2}-\frac {400 e^5 (3-x) x}{2+5 \log (2 x)}+\frac {8000 e^5 x^2}{2+5 \log (2 x)}-\left (600 e^5\right ) \operatorname {Subst}\left (\int \frac {e^x}{2+5 x} \, dx,x,\log (2 x)\right )-\left (800 e^5\right ) \int \frac {x}{2+5 \log (2 x)} \, dx+\left (2400 e^5\right ) \int \frac {1}{2+5 \log (2 x)} \, dx-\left (4000 e^5\right ) \operatorname {Subst}\left (\int \frac {e^{2 x}}{2+5 x} \, dx,x,\log (2 x)\right )+\left (4200 e^5\right ) \operatorname {Subst}\left (\int \frac {e^{2 x}}{2+5 x} \, dx,x,\log (2 x)\right )\\ &=18 e^5 (1-7 x)^2-240 e^{23/5} \text {Ei}\left (\frac {1}{5} (2+5 \log (2 x))\right )+40 e^{21/5} \text {Ei}\left (\frac {2}{5} (2+5 \log (2 x))\right )+\frac {20000 e^5 x^2}{(2+5 \log (2 x))^2}-\frac {400 e^5 (3-x) x}{2+5 \log (2 x)}+\frac {8000 e^5 x^2}{2+5 \log (2 x)}-\left (200 e^5\right ) \operatorname {Subst}\left (\int \frac {e^{2 x}}{2+5 x} \, dx,x,\log (2 x)\right )+\left (1200 e^5\right ) \operatorname {Subst}\left (\int \frac {e^x}{2+5 x} \, dx,x,\log (2 x)\right )\\ &=18 e^5 (1-7 x)^2+\frac {20000 e^5 x^2}{(2+5 \log (2 x))^2}-\frac {400 e^5 (3-x) x}{2+5 \log (2 x)}+\frac {8000 e^5 x^2}{2+5 \log (2 x)}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.30, size = 49, normalized size = 1.75 \begin {gather*} 4 e^5 \left (-63 x+\frac {441 x^2}{2}+\frac {5000 x^2}{(2+5 \log (2 x))^2}+\frac {150 x (-2+14 x)}{2+5 \log (2 x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^5*(5184 - 122688*x) + E^5*(-9120 + 431840*x)*Log[2*x] + E^5*(-67800 + 684600*x)*Log[2*x]^2 + E^5*
(-31500 + 220500*x)*Log[2*x]^3)/(8 + 60*Log[2*x] + 150*Log[2*x]^2 + 125*Log[2*x]^3),x]

[Out]

4*E^5*(-63*x + (441*x^2)/2 + (5000*x^2)/(2 + 5*Log[2*x])^2 + (150*x*(-2 + 14*x))/(2 + 5*Log[2*x]))

________________________________________________________________________________________

fricas [B]  time = 0.56, size = 70, normalized size = 2.50 \begin {gather*} \frac {2 \, {\left (1575 \, {\left (7 \, x^{2} - 2 \, x\right )} e^{5} \log \left (2 \, x\right )^{2} + 60 \, {\left (497 \, x^{2} - 92 \, x\right )} e^{5} \log \left (2 \, x\right ) + 284 \, {\left (71 \, x^{2} - 6 \, x\right )} e^{5}\right )}}{25 \, \log \left (2 \, x\right )^{2} + 20 \, \log \left (2 \, x\right ) + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((220500*x-31500)*exp(5)*log(2*x)^3+(684600*x-67800)*exp(5)*log(2*x)^2+(431840*x-9120)*exp(5)*log(2*
x)+(-122688*x+5184)*exp(5))/(125*log(2*x)^3+150*log(2*x)^2+60*log(2*x)+8),x, algorithm="fricas")

[Out]

2*(1575*(7*x^2 - 2*x)*e^5*log(2*x)^2 + 60*(497*x^2 - 92*x)*e^5*log(2*x) + 284*(71*x^2 - 6*x)*e^5)/(25*log(2*x)
^2 + 20*log(2*x) + 4)

________________________________________________________________________________________

giac [B]  time = 0.37, size = 165, normalized size = 5.89 \begin {gather*} \frac {22050 \, x^{2} e^{5} \log \left (2 \, x\right )^{2}}{25 \, \log \left (2 \, x\right )^{2} + 20 \, \log \left (2 \, x\right ) + 4} + \frac {59640 \, x^{2} e^{5} \log \left (2 \, x\right )}{25 \, \log \left (2 \, x\right )^{2} + 20 \, \log \left (2 \, x\right ) + 4} - \frac {6300 \, x e^{5} \log \left (2 \, x\right )^{2}}{25 \, \log \left (2 \, x\right )^{2} + 20 \, \log \left (2 \, x\right ) + 4} + \frac {40328 \, x^{2} e^{5}}{25 \, \log \left (2 \, x\right )^{2} + 20 \, \log \left (2 \, x\right ) + 4} - \frac {11040 \, x e^{5} \log \left (2 \, x\right )}{25 \, \log \left (2 \, x\right )^{2} + 20 \, \log \left (2 \, x\right ) + 4} - \frac {3408 \, x e^{5}}{25 \, \log \left (2 \, x\right )^{2} + 20 \, \log \left (2 \, x\right ) + 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((220500*x-31500)*exp(5)*log(2*x)^3+(684600*x-67800)*exp(5)*log(2*x)^2+(431840*x-9120)*exp(5)*log(2*
x)+(-122688*x+5184)*exp(5))/(125*log(2*x)^3+150*log(2*x)^2+60*log(2*x)+8),x, algorithm="giac")

[Out]

22050*x^2*e^5*log(2*x)^2/(25*log(2*x)^2 + 20*log(2*x) + 4) + 59640*x^2*e^5*log(2*x)/(25*log(2*x)^2 + 20*log(2*
x) + 4) - 6300*x*e^5*log(2*x)^2/(25*log(2*x)^2 + 20*log(2*x) + 4) + 40328*x^2*e^5/(25*log(2*x)^2 + 20*log(2*x)
+ 4) - 11040*x*e^5*log(2*x)/(25*log(2*x)^2 + 20*log(2*x) + 4) - 3408*x*e^5/(25*log(2*x)^2 + 20*log(2*x) + 4)

________________________________________________________________________________________

maple [A]  time = 0.07, size = 45, normalized size = 1.61

method	result	size
risch	\(126 \,{\mathrm e}^{5} x \left (7 x -2\right )+\frac {400 x \,{\mathrm e}^{5} \left (105 x \ln \left (2 x \right )+92 x -15 \ln \left (2 x \right )-6\right )}{\left (5 \ln \left (2 x \right )+2\right )^{2}}\)	\(45\)
norman	\(\frac {-3408 x \,{\mathrm e}^{5}+40328 x^{2} {\mathrm e}^{5}-11040 x \,{\mathrm e}^{5} \ln \left (2 x \right )-6300 x \,{\mathrm e}^{5} \ln \left (2 x \right )^{2}+59640 x^{2} {\mathrm e}^{5} \ln \left (2 x \right )+22050 x^{2} {\mathrm e}^{5} \ln \left (2 x \right )^{2}}{\left (5 \ln \left (2 x \right )+2\right )^{2}}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((220500*x-31500)*exp(5)*ln(2*x)^3+(684600*x-67800)*exp(5)*ln(2*x)^2+(431840*x-9120)*exp(5)*ln(2*x)+(-1226
88*x+5184)*exp(5))/(125*ln(2*x)^3+150*ln(2*x)^2+60*ln(2*x)+8),x,method=_RETURNVERBOSE)

[Out]

126*exp(5)*x*(7*x-2)+400*x*exp(5)*(105*x*ln(2*x)+92*x-15*ln(2*x)-6)/(5*ln(2*x)+2)^2

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {2 \, {\left ({\left (275625 \, \log \relax (2)^{2} + 438780 \, \log \relax (2) + 228052\right )} x^{2} e^{5} - 6 \, {\left (13125 \, \log \relax (2)^{2} + 21920 \, \log \relax (2) + 5588\right )} x e^{5} + 39375 \, {\left (7 \, x^{2} e^{5} - 2 \, x e^{5}\right )} \log \relax (x)^{2} + 30 \, {\left (x^{2} {\left (18375 \, \log \relax (2) + 14626\right )} e^{5} - 2 \, x {\left (2625 \, \log \relax (2) + 2192\right )} e^{5}\right )} \log \relax (x)\right )}}{25 \, {\left (25 \, \log \relax (2)^{2} + 10 \, {\left (5 \, \log \relax (2) + 2\right )} \log \relax (x) + 25 \, \log \relax (x)^{2} + 20 \, \log \relax (2) + 4\right )}} - \frac {2592 \, e^{\frac {23}{5}} E_{3}\left (-\log \left (2 \, x\right ) - \frac {2}{5}\right )}{5 \, {\left (5 \, \log \left (2 \, x\right ) + 2\right )}^{2}} + \frac {30672 \, e^{\frac {21}{5}} E_{3}\left (-2 \, \log \left (2 \, x\right ) - \frac {4}{5}\right )}{5 \, {\left (5 \, \log \left (2 \, x\right ) + 2\right )}^{2}} + 4 \, \int \frac {216 \, {\left (284 \, x e^{5} - 3 \, e^{5}\right )}}{25 \, {\left (5 \, \log \relax (2) + 5 \, \log \relax (x) + 2\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((220500*x-31500)*exp(5)*log(2*x)^3+(684600*x-67800)*exp(5)*log(2*x)^2+(431840*x-9120)*exp(5)*log(2*
x)+(-122688*x+5184)*exp(5))/(125*log(2*x)^3+150*log(2*x)^2+60*log(2*x)+8),x, algorithm="maxima")

[Out]

2/25*((275625*log(2)^2 + 438780*log(2) + 228052)*x^2*e^5 - 6*(13125*log(2)^2 + 21920*log(2) + 5588)*x*e^5 + 39
375*(7*x^2*e^5 - 2*x*e^5)*log(x)^2 + 30*(x^2*(18375*log(2) + 14626)*e^5 - 2*x*(2625*log(2) + 2192)*e^5)*log(x)
)/(25*log(2)^2 + 10*(5*log(2) + 2)*log(x) + 25*log(x)^2 + 20*log(2) + 4) - 2592/5*e^(23/5)*exp_integral_e(3, -
log(2*x) - 2/5)/(5*log(2*x) + 2)^2 + 30672/5*e^(21/5)*exp_integral_e(3, -2*log(2*x) - 4/5)/(5*log(2*x) + 2)^2
+ 4*integrate(216/25*(284*x*e^5 - 3*e^5)/(5*log(2) + 5*log(x) + 2), x)

________________________________________________________________________________________

mupad [B]  time = 0.64, size = 41, normalized size = 1.46 \begin {gather*} \frac {2\,x\,{\mathrm {e}}^5\,\left (105\,\ln \left (2\,x\right )+142\right )\,\left (142\,x-30\,\ln \left (2\,x\right )+105\,x\,\ln \left (2\,x\right )-12\right )}{{\left (5\,\ln \left (2\,x\right )+2\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(2*x)*exp(5)*(431840*x - 9120) - exp(5)*(122688*x - 5184) + log(2*x)^3*exp(5)*(220500*x - 31500) + log
(2*x)^2*exp(5)*(684600*x - 67800))/(60*log(2*x) + 150*log(2*x)^2 + 125*log(2*x)^3 + 8),x)

[Out]

(2*x*exp(5)*(105*log(2*x) + 142)*(142*x - 30*log(2*x) + 105*x*log(2*x) - 12))/(5*log(2*x) + 2)^2

________________________________________________________________________________________

sympy [B]  time = 0.20, size = 68, normalized size = 2.43 \begin {gather*} 882 x^{2} e^{5} - 252 x e^{5} + \frac {1472 x^{2} e^{5} - 96 x e^{5} + \left (1680 x^{2} e^{5} - 240 x e^{5}\right ) \log {\left (2 x \right )}}{\log {\left (2 x \right )}^{2} + \frac {4 \log {\left (2 x \right )}}{5} + \frac {4}{25}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((220500*x-31500)*exp(5)*ln(2*x)**3+(684600*x-67800)*exp(5)*ln(2*x)**2+(431840*x-9120)*exp(5)*ln(2*x
)+(-122688*x+5184)*exp(5))/(125*ln(2*x)**3+150*ln(2*x)**2+60*ln(2*x)+8),x)

[Out]

882*x**2*exp(5) - 252*x*exp(5) + (1472*x**2*exp(5) - 96*x*exp(5) + (1680*x**2*exp(5) - 240*x*exp(5))*log(2*x))
/(log(2*x)**2 + 4*log(2*x)/5 + 4/25)

________________________________________________________________________________________