\(\int \frac {e^{\log ^2(30 x^3+30 e^8 x^3+60 x^4+30 x^5+e^4 (-60 x^3-60 x^4))} (-6+6 e^4-10 x) \log (30 x^3+30 e^8 x^3+60 x^4+30 x^5+e^4 (-60 x^3-60 x^4))}{-x+e^4 x-x^2} \, dx\) [5404]

Optimal result

Integrand size = 111, antiderivative size = 20 \[ \int \frac {e^{\log ^2\left (30 x^3+30 e^8 x^3+60 x^4+30 x^5+e^4 \left (-60 x^3-60 x^4\right )\right )} \left (-6+6 e^4-10 x\right ) \log \left (30 x^3+30 e^8 x^3+60 x^4+30 x^5+e^4 \left (-60 x^3-60 x^4\right )\right )}{-x+e^4 x-x^2} \, dx=e^{\log ^2\left (30 \left (-1+e^4-x\right )^2 x^3\right )} \]

[Out]

exp(ln(30*x^3*(exp(4)-x-1)^2)^2)

Rubi [A] (verified)

Time = 1.97 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {6, 1607, 6820, 12, 6874, 6838} \[ \int \frac {e^{\log ^2\left (30 x^3+30 e^8 x^3+60 x^4+30 x^5+e^4 \left (-60 x^3-60 x^4\right )\right )} \left (-6+6 e^4-10 x\right ) \log \left (30 x^3+30 e^8 x^3+60 x^4+30 x^5+e^4 \left (-60 x^3-60 x^4\right )\right )}{-x+e^4 x-x^2} \, dx=e^{\log ^2\left (30 x^3 \left (x-e^4+1\right )^2\right )} \]

[In]

Int[(E^Log[30*x^3 + 30*E^8*x^3 + 60*x^4 + 30*x^5 + E^4*(-60*x^3 - 60*x^4)]^2*(-6 + 6*E^4 - 10*x)*Log[30*x^3 +
30*E^8*x^3 + 60*x^4 + 30*x^5 + E^4*(-60*x^3 - 60*x^4)])/(-x + E^4*x - x^2),x]

[Out]

E^Log[30*x^3*(1 - E^4 + x)^2]^2

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 1607

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 6820

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

Rule 6838

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[q*(F^v/Log[F]), x] /;  !FalseQ[q]
] /; FreeQ[F, 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 {\exp \left (\log ^2\left (30 x^3+30 e^8 x^3+60 x^4+30 x^5+e^4 \left (-60 x^3-60 x^4\right )\right )\right ) \left (-6+6 e^4-10 x\right ) \log \left (30 x^3+30 e^8 x^3+60 x^4+30 x^5+e^4 \left (-60 x^3-60 x^4\right )\right )}{\left (-1+e^4\right ) x-x^2} \, dx \\ & = \int \frac {\exp \left (\log ^2\left (30 x^3+30 e^8 x^3+60 x^4+30 x^5+e^4 \left (-60 x^3-60 x^4\right )\right )\right ) \left (-6+6 e^4-10 x\right ) \log \left (30 x^3+30 e^8 x^3+60 x^4+30 x^5+e^4 \left (-60 x^3-60 x^4\right )\right )}{\left (-1+e^4-x\right ) x} \, dx \\ & = \int \frac {2 e^{\log ^2\left (30 x^3 \left (1-e^4+x\right )^2\right )} \left (3-3 e^4+5 x\right ) \log \left (30 x^3 \left (1-e^4+x\right )^2\right )}{x \left (1-e^4+x\right )} \, dx \\ & = 2 \int \frac {e^{\log ^2\left (30 x^3 \left (1-e^4+x\right )^2\right )} \left (3-3 e^4+5 x\right ) \log \left (30 x^3 \left (1-e^4+x\right )^2\right )}{x \left (1-e^4+x\right )} \, dx \\ & = e^{\log ^2\left (30 x^3 \left (1-e^4+x\right )^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\log ^2\left (30 x^3+30 e^8 x^3+60 x^4+30 x^5+e^4 \left (-60 x^3-60 x^4\right )\right )} \left (-6+6 e^4-10 x\right ) \log \left (30 x^3+30 e^8 x^3+60 x^4+30 x^5+e^4 \left (-60 x^3-60 x^4\right )\right )}{-x+e^4 x-x^2} \, dx=e^{\log ^2\left (30 x^3 \left (1-e^4+x\right )^2\right )} \]

[In]

Integrate[(E^Log[30*x^3 + 30*E^8*x^3 + 60*x^4 + 30*x^5 + E^4*(-60*x^3 - 60*x^4)]^2*(-6 + 6*E^4 - 10*x)*Log[30*
x^3 + 30*E^8*x^3 + 60*x^4 + 30*x^5 + E^4*(-60*x^3 - 60*x^4)])/(-x + E^4*x - x^2),x]

[Out]

E^Log[30*x^3*(1 - E^4 + x)^2]^2

Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.55

method	result	size
parallelrisch	\({\mathrm e}^{\ln \left (30 x^{3} \left (-2 x \,{\mathrm e}^{4}+x^{2}+{\mathrm e}^{8}-2 \,{\mathrm e}^{4}+2 x +1\right )\right )^{2}}\)	\(31\)
risch	\({\mathrm e}^{\ln \left (30 x^{3} {\mathrm e}^{8}+\left (-60 x^{4}-60 x^{3}\right ) {\mathrm e}^{4}+30 x^{5}+60 x^{4}+30 x^{3}\right )^{2}}\)	\(42\)
norman	\({\mathrm e}^{\ln \left (30 x^{3} {\mathrm e}^{8}+\left (-60 x^{4}-60 x^{3}\right ) {\mathrm e}^{4}+30 x^{5}+60 x^{4}+30 x^{3}\right )^{2}}\)	\(44\)

[In]

int((6*exp(4)-10*x-6)*ln(30*x^3*exp(4)^2+(-60*x^4-60*x^3)*exp(4)+30*x^5+60*x^4+30*x^3)*exp(ln(30*x^3*exp(4)^2+
(-60*x^4-60*x^3)*exp(4)+30*x^5+60*x^4+30*x^3)^2)/(x*exp(4)-x^2-x),x,method=_RETURNVERBOSE)

[Out]

exp(ln(30*x^3*(exp(4)^2-2*x*exp(4)+x^2-2*exp(4)+2*x+1))^2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 38 vs. \(2 (18) = 36\).

Time = 0.25 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.90 \[ \int \frac {e^{\log ^2\left (30 x^3+30 e^8 x^3+60 x^4+30 x^5+e^4 \left (-60 x^3-60 x^4\right )\right )} \left (-6+6 e^4-10 x\right ) \log \left (30 x^3+30 e^8 x^3+60 x^4+30 x^5+e^4 \left (-60 x^3-60 x^4\right )\right )}{-x+e^4 x-x^2} \, dx=e^{\left (\log \left (30 \, x^{5} + 60 \, x^{4} + 30 \, x^{3} e^{8} + 30 \, x^{3} - 60 \, {\left (x^{4} + x^{3}\right )} e^{4}\right )^{2}\right )} \]

[In]

integrate((6*exp(4)-10*x-6)*log(30*x^3*exp(4)^2+(-60*x^4-60*x^3)*exp(4)+30*x^5+60*x^4+30*x^3)*exp(log(30*x^3*e
xp(4)^2+(-60*x^4-60*x^3)*exp(4)+30*x^5+60*x^4+30*x^3)^2)/(x*exp(4)-x^2-x),x, algorithm="fricas")

[Out]

e^(log(30*x^5 + 60*x^4 + 30*x^3*e^8 + 30*x^3 - 60*(x^4 + x^3)*e^4)^2)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (17) = 34\).

Time = 0.16 (sec) , antiderivative size = 42, normalized size of antiderivative = 2.10 \[ \int \frac {e^{\log ^2\left (30 x^3+30 e^8 x^3+60 x^4+30 x^5+e^4 \left (-60 x^3-60 x^4\right )\right )} \left (-6+6 e^4-10 x\right ) \log \left (30 x^3+30 e^8 x^3+60 x^4+30 x^5+e^4 \left (-60 x^3-60 x^4\right )\right )}{-x+e^4 x-x^2} \, dx=e^{\log {\left (30 x^{5} + 60 x^{4} + 30 x^{3} + 30 x^{3} e^{8} + \left (- 60 x^{4} - 60 x^{3}\right ) e^{4} \right )}^{2}} \]

[In]

integrate((6*exp(4)-10*x-6)*ln(30*x**3*exp(4)**2+(-60*x**4-60*x**3)*exp(4)+30*x**5+60*x**4+30*x**3)*exp(ln(30*
x**3*exp(4)**2+(-60*x**4-60*x**3)*exp(4)+30*x**5+60*x**4+30*x**3)**2)/(x*exp(4)-x**2-x),x)

[Out]

exp(log(30*x**5 + 60*x**4 + 30*x**3 + 30*x**3*exp(8) + (-60*x**4 - 60*x**3)*exp(4))**2)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 116 vs. \(2 (18) = 36\).

Time = 0.53 (sec) , antiderivative size = 116, normalized size of antiderivative = 5.80 \[ \int \frac {e^{\log ^2\left (30 x^3+30 e^8 x^3+60 x^4+30 x^5+e^4 \left (-60 x^3-60 x^4\right )\right )} \left (-6+6 e^4-10 x\right ) \log \left (30 x^3+30 e^8 x^3+60 x^4+30 x^5+e^4 \left (-60 x^3-60 x^4\right )\right )}{-x+e^4 x-x^2} \, dx=3^{2 \, \log \left (5\right )} 2^{2 \, \log \left (5\right ) + 2 \, \log \left (3\right )} e^{\left (\log \left (5\right )^{2} + \log \left (3\right )^{2} + \log \left (2\right )^{2} + 4 \, \log \left (5\right ) \log \left (x - e^{4} + 1\right ) + 4 \, \log \left (3\right ) \log \left (x - e^{4} + 1\right ) + 4 \, \log \left (2\right ) \log \left (x - e^{4} + 1\right ) + 4 \, \log \left (x - e^{4} + 1\right )^{2} + 6 \, \log \left (5\right ) \log \left (x\right ) + 6 \, \log \left (3\right ) \log \left (x\right ) + 6 \, \log \left (2\right ) \log \left (x\right ) + 12 \, \log \left (x - e^{4} + 1\right ) \log \left (x\right ) + 9 \, \log \left (x\right )^{2}\right )} \]

[In]

integrate((6*exp(4)-10*x-6)*log(30*x^3*exp(4)^2+(-60*x^4-60*x^3)*exp(4)+30*x^5+60*x^4+30*x^3)*exp(log(30*x^3*e
xp(4)^2+(-60*x^4-60*x^3)*exp(4)+30*x^5+60*x^4+30*x^3)^2)/(x*exp(4)-x^2-x),x, algorithm="maxima")

[Out]

3^(2*log(5))*2^(2*log(5) + 2*log(3))*e^(log(5)^2 + log(3)^2 + log(2)^2 + 4*log(5)*log(x - e^4 + 1) + 4*log(3)*
log(x - e^4 + 1) + 4*log(2)*log(x - e^4 + 1) + 4*log(x - e^4 + 1)^2 + 6*log(5)*log(x) + 6*log(3)*log(x) + 6*lo
g(2)*log(x) + 12*log(x - e^4 + 1)*log(x) + 9*log(x)^2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 41 vs. \(2 (18) = 36\).

Time = 4.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.05 \[ \int \frac {e^{\log ^2\left (30 x^3+30 e^8 x^3+60 x^4+30 x^5+e^4 \left (-60 x^3-60 x^4\right )\right )} \left (-6+6 e^4-10 x\right ) \log \left (30 x^3+30 e^8 x^3+60 x^4+30 x^5+e^4 \left (-60 x^3-60 x^4\right )\right )}{-x+e^4 x-x^2} \, dx=e^{\left (\log \left (30 \, x^{5} - 60 \, x^{4} e^{4} + 60 \, x^{4} + 30 \, x^{3} e^{8} - 60 \, x^{3} e^{4} + 30 \, x^{3}\right )^{2}\right )} \]

[In]

integrate((6*exp(4)-10*x-6)*log(30*x^3*exp(4)^2+(-60*x^4-60*x^3)*exp(4)+30*x^5+60*x^4+30*x^3)*exp(log(30*x^3*e
xp(4)^2+(-60*x^4-60*x^3)*exp(4)+30*x^5+60*x^4+30*x^3)^2)/(x*exp(4)-x^2-x),x, algorithm="giac")

[Out]

e^(log(30*x^5 - 60*x^4*e^4 + 60*x^4 + 30*x^3*e^8 - 60*x^3*e^4 + 30*x^3)^2)

Mupad [B] (verification not implemented)

Time = 13.98 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.05 \[ \int \frac {e^{\log ^2\left (30 x^3+30 e^8 x^3+60 x^4+30 x^5+e^4 \left (-60 x^3-60 x^4\right )\right )} \left (-6+6 e^4-10 x\right ) \log \left (30 x^3+30 e^8 x^3+60 x^4+30 x^5+e^4 \left (-60 x^3-60 x^4\right )\right )}{-x+e^4 x-x^2} \, dx={\mathrm {e}}^{{\ln \left (30\,x^3\,{\mathrm {e}}^8-60\,x^4\,{\mathrm {e}}^4-60\,x^3\,{\mathrm {e}}^4+30\,x^3+60\,x^4+30\,x^5\right )}^2} \]

[In]

int((exp(log(30*x^3*exp(8) - exp(4)*(60*x^3 + 60*x^4) + 30*x^3 + 60*x^4 + 30*x^5)^2)*log(30*x^3*exp(8) - exp(4
)*(60*x^3 + 60*x^4) + 30*x^3 + 60*x^4 + 30*x^5)*(10*x - 6*exp(4) + 6))/(x - x*exp(4) + x^2),x)

[Out]

exp(log(30*x^3*exp(8) - 60*x^4*exp(4) - 60*x^3*exp(4) + 30*x^3 + 60*x^4 + 30*x^5)^2)