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

Optimal. Leaf size=22 \[ 1+\left (1+e^{4 x \left (x+\frac {4 x^4}{9}\right )}\right ) x \log (x) \]

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Rubi [B]  time = 0.11, antiderivative size = 47, normalized size of antiderivative = 2.14, number of steps used = 5, number of rules used = 4, integrand size = 60, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {12, 6741, 2288, 2554} \begin {gather*} \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) \end {gather*}

Antiderivative was successfully verified.

[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 2288

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 2554

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 6741

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

Rubi steps

\begin {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 21, normalized size = 0.95 \begin {gather*} \left (1+e^{4 x^2+\frac {16 x^5}{9}}\right ) x \log (x) \end {gather*}

Antiderivative was successfully verified.

[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]

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fricas [A]  time = 0.53, size = 19, normalized size = 0.86 \begin {gather*} {\left (x e^{\left (\frac {16}{9} \, x^{5} + 4 \, x^{2}\right )} + x\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.29, size = 19, normalized size = 0.86 \begin {gather*} {\left (x e^{\left (\frac {16}{9} \, x^{5} + 4 \, x^{2}\right )} + x\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.10, size = 20, normalized size = 0.91

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maxima [A]  time = 0.40, size = 21, normalized size = 0.95 \begin {gather*} x e^{\left (\frac {16}{9} \, x^{5} + 4 \, x^{2}\right )} \log \relax (x) + x \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int {\mathrm {e}}^{\frac {16\,x^5}{9}+4\,x^2}+\frac {\ln \relax (x)\,\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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [A]  time = 0.33, size = 22, normalized size = 1.00 \begin {gather*} x e^{\frac {16 x^{5}}{9} + 4 x^{2}} \log {\relax (x )} + x \log {\relax (x )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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