\(\int \frac {7500 x+9000 x^2+e^2 (-25000-30000 x) \log ^3(x)-7500 e^2 x \log ^4(x)+25000 e^4 \log ^7(x)}{-216 x^4+540 e^2 x^3 \log ^4(x)-450 e^4 x^2 \log ^8(x)+125 e^6 x \log ^{12}(x)} \, dx\) [7541]

Optimal result

Integrand size = 84, antiderivative size = 22 \[ \int \frac {7500 x+9000 x^2+e^2 (-25000-30000 x) \log ^3(x)-7500 e^2 x \log ^4(x)+25000 e^4 \log ^7(x)}{-216 x^4+540 e^2 x^3 \log ^4(x)-450 e^4 x^2 \log ^8(x)+125 e^6 x \log ^{12}(x)} \, dx=\left (-5+\frac {5}{-\frac {6 x}{5}+e^2 \log ^4(x)}\right )^2 \]

[Out]

(5/(exp(1)^2*ln(x)^4-6/5*x)-5)^2

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.59, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {6820, 12, 6874, 2624} \[ \int \frac {7500 x+9000 x^2+e^2 (-25000-30000 x) \log ^3(x)-7500 e^2 x \log ^4(x)+25000 e^4 \log ^7(x)}{-216 x^4+540 e^2 x^3 \log ^4(x)-450 e^4 x^2 \log ^8(x)+125 e^6 x \log ^{12}(x)} \, dx=\frac {250}{6 x-5 e^2 \log ^4(x)}+\frac {625}{\left (6 x-5 e^2 \log ^4(x)\right )^2} \]

[In]

Int[(7500*x + 9000*x^2 + E^2*(-25000 - 30000*x)*Log[x]^3 - 7500*E^2*x*Log[x]^4 + 25000*E^4*Log[x]^7)/(-216*x^4
+ 540*E^2*x^3*Log[x]^4 - 450*E^4*x^2*Log[x]^8 + 125*E^6*x*Log[x]^12),x]

[Out]

625/(6*x - 5*E^2*Log[x]^4)^2 + 250/(6*x - 5*E^2*Log[x]^4)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2624

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

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {500 \left (-3 x (5+6 x)+10 e^2 (5+6 x) \log ^3(x)+15 e^2 x \log ^4(x)-50 e^4 \log ^7(x)\right )}{x \left (6 x-5 e^2 \log ^4(x)\right )^3} \, dx \\ & = 500 \int \frac {-3 x (5+6 x)+10 e^2 (5+6 x) \log ^3(x)+15 e^2 x \log ^4(x)-50 e^4 \log ^7(x)}{x \left (6 x-5 e^2 \log ^4(x)\right )^3} \, dx \\ & = 500 \int \left (-\frac {5 \left (3 x-10 e^2 \log ^3(x)\right )}{x \left (6 x-5 e^2 \log ^4(x)\right )^3}+\frac {-3 x+10 e^2 \log ^3(x)}{x \left (6 x-5 e^2 \log ^4(x)\right )^2}\right ) \, dx \\ & = 500 \int \frac {-3 x+10 e^2 \log ^3(x)}{x \left (6 x-5 e^2 \log ^4(x)\right )^2} \, dx-2500 \int \frac {3 x-10 e^2 \log ^3(x)}{x \left (6 x-5 e^2 \log ^4(x)\right )^3} \, dx \\ & = \frac {625}{\left (6 x-5 e^2 \log ^4(x)\right )^2}+\frac {250}{6 x-5 e^2 \log ^4(x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.41 \[ \int \frac {7500 x+9000 x^2+e^2 (-25000-30000 x) \log ^3(x)-7500 e^2 x \log ^4(x)+25000 e^4 \log ^7(x)}{-216 x^4+540 e^2 x^3 \log ^4(x)-450 e^4 x^2 \log ^8(x)+125 e^6 x \log ^{12}(x)} \, dx=-\frac {125 \left (-5-12 x+10 e^2 \log ^4(x)\right )}{\left (6 x-5 e^2 \log ^4(x)\right )^2} \]

[In]

Integrate[(7500*x + 9000*x^2 + E^2*(-25000 - 30000*x)*Log[x]^3 - 7500*E^2*x*Log[x]^4 + 25000*E^4*Log[x]^7)/(-2
16*x^4 + 540*E^2*x^3*Log[x]^4 - 450*E^4*x^2*Log[x]^8 + 125*E^6*x*Log[x]^12),x]

[Out]

(-125*(-5 - 12*x + 10*E^2*Log[x]^4))/(6*x - 5*E^2*Log[x]^4)^2

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.36

method	result	size
default	\(-\frac {125 \left (10 \,{\mathrm e}^{2} \ln \left (x \right )^{4}-12 x -5\right )}{\left (5 \,{\mathrm e}^{2} \ln \left (x \right )^{4}-6 x \right )^{2}}\)	\(30\)
risch	\(-\frac {125 \left (10 \,{\mathrm e}^{2} \ln \left (x \right )^{4}-12 x -5\right )}{\left (5 \,{\mathrm e}^{2} \ln \left (x \right )^{4}-6 x \right )^{2}}\)	\(30\)
parallelrisch	\(\frac {22500-45000 \,{\mathrm e}^{2} \ln \left (x \right )^{4}+54000 x}{900 \,{\mathrm e}^{4} \ln \left (x \right )^{8}-2160 x \,{\mathrm e}^{2} \ln \left (x \right )^{4}+1296 x^{2}}\)	\(47\)

[In]

int((25000*exp(1)^4*ln(x)^7-7500*x*exp(1)^2*ln(x)^4+(-30000*x-25000)*exp(1)^2*ln(x)^3+9000*x^2+7500*x)/(125*x*
exp(1)^6*ln(x)^12-450*x^2*exp(1)^4*ln(x)^8+540*x^3*exp(1)^2*ln(x)^4-216*x^4),x,method=_RETURNVERBOSE)

[Out]

-125*(10*exp(2)*ln(x)^4-12*x-5)/(5*exp(2)*ln(x)^4-6*x)^2

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.82 \[ \int \frac {7500 x+9000 x^2+e^2 (-25000-30000 x) \log ^3(x)-7500 e^2 x \log ^4(x)+25000 e^4 \log ^7(x)}{-216 x^4+540 e^2 x^3 \log ^4(x)-450 e^4 x^2 \log ^8(x)+125 e^6 x \log ^{12}(x)} \, dx=-\frac {125 \, {\left (10 \, e^{2} \log \left (x\right )^{4} - 12 \, x - 5\right )}}{25 \, e^{4} \log \left (x\right )^{8} - 60 \, x e^{2} \log \left (x\right )^{4} + 36 \, x^{2}} \]

[In]

integrate((25000*exp(1)^4*log(x)^7-7500*x*exp(1)^2*log(x)^4+(-30000*x-25000)*exp(1)^2*log(x)^3+9000*x^2+7500*x
)/(125*x*exp(1)^6*log(x)^12-450*x^2*exp(1)^4*log(x)^8+540*x^3*exp(1)^2*log(x)^4-216*x^4),x, algorithm="fricas"
)

[Out]

-125*(10*e^2*log(x)^4 - 12*x - 5)/(25*e^4*log(x)^8 - 60*x*e^2*log(x)^4 + 36*x^2)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (17) = 34\).

Time = 0.08 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.86 \[ \int \frac {7500 x+9000 x^2+e^2 (-25000-30000 x) \log ^3(x)-7500 e^2 x \log ^4(x)+25000 e^4 \log ^7(x)}{-216 x^4+540 e^2 x^3 \log ^4(x)-450 e^4 x^2 \log ^8(x)+125 e^6 x \log ^{12}(x)} \, dx=\frac {1500 x - 1250 e^{2} \log {\left (x \right )}^{4} + 625}{36 x^{2} - 60 x e^{2} \log {\left (x \right )}^{4} + 25 e^{4} \log {\left (x \right )}^{8}} \]

[In]

integrate((25000*exp(1)**4*ln(x)**7-7500*x*exp(1)**2*ln(x)**4+(-30000*x-25000)*exp(1)**2*ln(x)**3+9000*x**2+75
00*x)/(125*x*exp(1)**6*ln(x)**12-450*x**2*exp(1)**4*ln(x)**8+540*x**3*exp(1)**2*ln(x)**4-216*x**4),x)

[Out]

(1500*x - 1250*exp(2)*log(x)**4 + 625)/(36*x**2 - 60*x*exp(2)*log(x)**4 + 25*exp(4)*log(x)**8)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.82 \[ \int \frac {7500 x+9000 x^2+e^2 (-25000-30000 x) \log ^3(x)-7500 e^2 x \log ^4(x)+25000 e^4 \log ^7(x)}{-216 x^4+540 e^2 x^3 \log ^4(x)-450 e^4 x^2 \log ^8(x)+125 e^6 x \log ^{12}(x)} \, dx=-\frac {125 \, {\left (10 \, e^{2} \log \left (x\right )^{4} - 12 \, x - 5\right )}}{25 \, e^{4} \log \left (x\right )^{8} - 60 \, x e^{2} \log \left (x\right )^{4} + 36 \, x^{2}} \]

[In]

integrate((25000*exp(1)^4*log(x)^7-7500*x*exp(1)^2*log(x)^4+(-30000*x-25000)*exp(1)^2*log(x)^3+9000*x^2+7500*x
)/(125*x*exp(1)^6*log(x)^12-450*x^2*exp(1)^4*log(x)^8+540*x^3*exp(1)^2*log(x)^4-216*x^4),x, algorithm="maxima"
)

[Out]

-125*(10*e^2*log(x)^4 - 12*x - 5)/(25*e^4*log(x)^8 - 60*x*e^2*log(x)^4 + 36*x^2)

Giac [A] (verification not implemented)

none

Time = 3.83 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.82 \[ \int \frac {7500 x+9000 x^2+e^2 (-25000-30000 x) \log ^3(x)-7500 e^2 x \log ^4(x)+25000 e^4 \log ^7(x)}{-216 x^4+540 e^2 x^3 \log ^4(x)-450 e^4 x^2 \log ^8(x)+125 e^6 x \log ^{12}(x)} \, dx=-\frac {250 \, {\left (10 \, e^{2} \log \left (x\right )^{4} - 12 \, x - 5\right )}}{25 \, e^{4} \log \left (x\right )^{8} - 60 \, x e^{2} \log \left (x\right )^{4} + 36 \, x^{2}} \]

[In]

integrate((25000*exp(1)^4*log(x)^7-7500*x*exp(1)^2*log(x)^4+(-30000*x-25000)*exp(1)^2*log(x)^3+9000*x^2+7500*x
)/(125*x*exp(1)^6*log(x)^12-450*x^2*exp(1)^4*log(x)^8+540*x^3*exp(1)^2*log(x)^4-216*x^4),x, algorithm="giac")

[Out]

-250*(10*e^2*log(x)^4 - 12*x - 5)/(25*e^4*log(x)^8 - 60*x*e^2*log(x)^4 + 36*x^2)

Mupad [B] (verification not implemented)

Time = 13.21 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.32 \[ \int \frac {7500 x+9000 x^2+e^2 (-25000-30000 x) \log ^3(x)-7500 e^2 x \log ^4(x)+25000 e^4 \log ^7(x)}{-216 x^4+540 e^2 x^3 \log ^4(x)-450 e^4 x^2 \log ^8(x)+125 e^6 x \log ^{12}(x)} \, dx=\frac {125\,\left (-10\,{\mathrm {e}}^2\,{\ln \left (x\right )}^4+12\,x+5\right )}{{\left (6\,x-5\,{\mathrm {e}}^2\,{\ln \left (x\right )}^4\right )}^2} \]

[In]

int(-(7500*x + 25000*exp(4)*log(x)^7 + 9000*x^2 - exp(2)*log(x)^3*(30000*x + 25000) - 7500*x*exp(2)*log(x)^4)/
(216*x^4 - 125*x*exp(6)*log(x)^12 - 540*x^3*exp(2)*log(x)^4 + 450*x^2*exp(4)*log(x)^8),x)

[Out]

(125*(12*x - 10*exp(2)*log(x)^4 + 5))/(6*x - 5*exp(2)*log(x)^4)^2