\(\int \frac {1}{9} (9+9 e^{\frac {1}{9} (36 x^2+16 x^5)}+(9+e^{\frac {1}{9} (36 x^2+16 x^5)} (9+72 x^2+80 x^5)) \log (x)) \, dx\) [9491]

Optimal result

Integrand size = 60, antiderivative size = 22 \[ \int \frac {1}{9} \left (9+9 e^{\frac {1}{9} \left (36 x^2+16 x^5\right )}+\left (9+e^{\frac {1}{9} \left (36 x^2+16 x^5\right )} \left (9+72 x^2+80 x^5\right )\right ) \log (x)\right ) \, dx=1+\left (1+e^{4 x \left (x+\frac {4 x^4}{9}\right )}\right ) x \log (x) \]

[Out]

1+ln(x)*(exp(4*(4/9*x^4+x)*x)+1)*x

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(47\) vs. \(2(22)=44\).

Time = 0.08 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.14, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {12, 6873, 2326, 2634} \[ \int \frac {1}{9} \left (9+9 e^{\frac {1}{9} \left (36 x^2+16 x^5\right )}+\left (9+e^{\frac {1}{9} \left (36 x^2+16 x^5\right )} \left (9+72 x^2+80 x^5\right )\right ) \log (x)\right ) \, dx=\frac {e^{\frac {4}{9} \left (4 x^5+9 x^2\right )} \left (10 x^5+9 x^2\right ) \log (x)}{10 x^4+9 x}+x \log (x) \]

[In]

Int[(9 + 9*E^((36*x^2 + 16*x^5)/9) + (9 + E^((36*x^2 + 16*x^5)/9)*(9 + 72*x^2 + 80*x^5))*Log[x])/9,x]

[Out]

x*Log[x] + (E^((4*(9*x^2 + 4*x^5))/9)*(9*x^2 + 10*x^5)*Log[x])/(9*x + 10*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 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 2634

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 6873

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{9} \int \left (9+9 e^{\frac {1}{9} \left (36 x^2+16 x^5\right )}+\left (9+e^{\frac {1}{9} \left (36 x^2+16 x^5\right )} \left (9+72 x^2+80 x^5\right )\right ) \log (x)\right ) \, dx \\ & = x+\frac {1}{9} \int \left (9+e^{\frac {1}{9} \left (36 x^2+16 x^5\right )} \left (9+72 x^2+80 x^5\right )\right ) \log (x) \, dx+\int e^{\frac {1}{9} \left (36 x^2+16 x^5\right )} \, dx \\ & = x+x \log (x)+\frac {e^{\frac {4}{9} \left (9 x^2+4 x^5\right )} \left (9 x^2+10 x^5\right ) \log (x)}{9 x+10 x^4}-\frac {1}{9} \int 9 \left (1+e^{4 x^2+\frac {16 x^5}{9}}\right ) \, dx+\int e^{\frac {4}{9} x^2 \left (9+4 x^3\right )} \, dx \\ & = x+x \log (x)+\frac {e^{\frac {4}{9} \left (9 x^2+4 x^5\right )} \left (9 x^2+10 x^5\right ) \log (x)}{9 x+10 x^4}+\int e^{\frac {4}{9} x^2 \left (9+4 x^3\right )} \, dx-\int \left (1+e^{4 x^2+\frac {16 x^5}{9}}\right ) \, dx \\ & = x \log (x)+\frac {e^{\frac {4}{9} \left (9 x^2+4 x^5\right )} \left (9 x^2+10 x^5\right ) \log (x)}{9 x+10 x^4}+\int e^{\frac {4}{9} x^2 \left (9+4 x^3\right )} \, dx-\int e^{4 x^2+\frac {16 x^5}{9}} \, dx \\ & = x \log (x)+\frac {e^{\frac {4}{9} \left (9 x^2+4 x^5\right )} \left (9 x^2+10 x^5\right ) \log (x)}{9 x+10 x^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {1}{9} \left (9+9 e^{\frac {1}{9} \left (36 x^2+16 x^5\right )}+\left (9+e^{\frac {1}{9} \left (36 x^2+16 x^5\right )} \left (9+72 x^2+80 x^5\right )\right ) \log (x)\right ) \, dx=\left (1+e^{4 x^2+\frac {16 x^5}{9}}\right ) x \log (x) \]

[In]

Integrate[(9 + 9*E^((36*x^2 + 16*x^5)/9) + (9 + E^((36*x^2 + 16*x^5)/9)*(9 + 72*x^2 + 80*x^5))*Log[x])/9,x]

[Out]

(1 + E^(4*x^2 + (16*x^5)/9))*x*Log[x]

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91

method	result	size
risch	\(x \left ({\mathrm e}^{\frac {4 x^{2} \left (4 x^{3}+9\right )}{9}}+1\right ) \ln \left (x \right )\)	\(20\)
parallelrisch	\(x \ln \left (x \right ) {\mathrm e}^{\frac {16}{9} x^{5}+4 x^{2}}+x \ln \left (x \right )\)	\(22\)

[In]

int(1/9*((80*x^5+72*x^2+9)*exp(16/9*x^5+4*x^2)+9)*ln(x)+exp(16/9*x^5+4*x^2)+1,x,method=_RETURNVERBOSE)

[Out]

x*(exp(4/9*x^2*(4*x^3+9))+1)*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {1}{9} \left (9+9 e^{\frac {1}{9} \left (36 x^2+16 x^5\right )}+\left (9+e^{\frac {1}{9} \left (36 x^2+16 x^5\right )} \left (9+72 x^2+80 x^5\right )\right ) \log (x)\right ) \, dx={\left (x e^{\left (\frac {16}{9} \, x^{5} + 4 \, x^{2}\right )} + x\right )} \log \left (x\right ) \]

[In]

integrate(1/9*((80*x^5+72*x^2+9)*exp(16/9*x^5+4*x^2)+9)*log(x)+exp(16/9*x^5+4*x^2)+1,x, algorithm="fricas")

[Out]

(x*e^(16/9*x^5 + 4*x^2) + x)*log(x)

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {1}{9} \left (9+9 e^{\frac {1}{9} \left (36 x^2+16 x^5\right )}+\left (9+e^{\frac {1}{9} \left (36 x^2+16 x^5\right )} \left (9+72 x^2+80 x^5\right )\right ) \log (x)\right ) \, dx=x e^{\frac {16 x^{5}}{9} + 4 x^{2}} \log {\left (x \right )} + x \log {\left (x \right )} \]

[In]

integrate(1/9*((80*x**5+72*x**2+9)*exp(16/9*x**5+4*x**2)+9)*ln(x)+exp(16/9*x**5+4*x**2)+1,x)

[Out]

x*exp(16*x**5/9 + 4*x**2)*log(x) + x*log(x)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {1}{9} \left (9+9 e^{\frac {1}{9} \left (36 x^2+16 x^5\right )}+\left (9+e^{\frac {1}{9} \left (36 x^2+16 x^5\right )} \left (9+72 x^2+80 x^5\right )\right ) \log (x)\right ) \, dx=x e^{\left (\frac {16}{9} \, x^{5} + 4 \, x^{2}\right )} \log \left (x\right ) + x \log \left (x\right ) \]

[In]

integrate(1/9*((80*x^5+72*x^2+9)*exp(16/9*x^5+4*x^2)+9)*log(x)+exp(16/9*x^5+4*x^2)+1,x, algorithm="maxima")

[Out]

x*e^(16/9*x^5 + 4*x^2)*log(x) + x*log(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86 \[ \int \frac {1}{9} \left (9+9 e^{\frac {1}{9} \left (36 x^2+16 x^5\right )}+\left (9+e^{\frac {1}{9} \left (36 x^2+16 x^5\right )} \left (9+72 x^2+80 x^5\right )\right ) \log (x)\right ) \, dx={\left (x e^{\left (\frac {16}{9} \, x^{5} + 4 \, x^{2}\right )} + x\right )} \log \left (x\right ) \]

[In]

integrate(1/9*((80*x^5+72*x^2+9)*exp(16/9*x^5+4*x^2)+9)*log(x)+exp(16/9*x^5+4*x^2)+1,x, algorithm="giac")

[Out]

(x*e^(16/9*x^5 + 4*x^2) + x)*log(x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{9} \left (9+9 e^{\frac {1}{9} \left (36 x^2+16 x^5\right )}+\left (9+e^{\frac {1}{9} \left (36 x^2+16 x^5\right )} \left (9+72 x^2+80 x^5\right )\right ) \log (x)\right ) \, dx=\int {\mathrm {e}}^{\frac {16\,x^5}{9}+4\,x^2}+\frac {\ln \left (x\right )\,\left ({\mathrm {e}}^{\frac {16\,x^5}{9}+4\,x^2}\,\left (80\,x^5+72\,x^2+9\right )+9\right )}{9}+1 \,d x \]

[In]

int(exp(4*x^2 + (16*x^5)/9) + (log(x)*(exp(4*x^2 + (16*x^5)/9)*(72*x^2 + 80*x^5 + 9) + 9))/9 + 1,x)

[Out]

int(exp(4*x^2 + (16*x^5)/9) + (log(x)*(exp(4*x^2 + (16*x^5)/9)*(72*x^2 + 80*x^5 + 9) + 9))/9 + 1, x)