\(\int \frac {1}{25} (45+5 e^{28 x/5}+(-5+e^{28 x/5} (5+28 x)) \log (x)) \, dx\) [3925]

Optimal result

Integrand size = 33, antiderivative size = 22 \[ \int \frac {1}{25} \left (45+5 e^{28 x/5}+\left (-5+e^{28 x/5} (5+28 x)\right ) \log (x)\right ) \, dx=2 x-\frac {1}{5} \left (1-e^{28 x/5}\right ) x \log (x) \]

[Out]

2*x-1/5*x*ln(x)*(1-exp(28/5*x))

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.91, number of steps used = 7, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {12, 2225, 2207, 2634} \[ \int \frac {1}{25} \left (45+5 e^{28 x/5}+\left (-5+e^{28 x/5} (5+28 x)\right ) \log (x)\right ) \, dx=2 x-\frac {1}{5} x \log (x)-\frac {1}{28} e^{28 x/5} \log (x)+\frac {1}{140} e^{28 x/5} (28 x+5) \log (x) \]

[In]

Int[(45 + 5*E^((28*x)/5) + (-5 + E^((28*x)/5)*(5 + 28*x))*Log[x])/25,x]

[Out]

2*x - (E^((28*x)/5)*Log[x])/28 - (x*Log[x])/5 + (E^((28*x)/5)*(5 + 28*x)*Log[x])/140

Rule 12

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

Rule 2207

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(c + d*x)^m*
((b*F^(g*(e + f*x)))^n/(f*g*n*Log[F])), x] - Dist[d*(m/(f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !TrueQ[$UseGamma]

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, 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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{25} \int \left (45+5 e^{28 x/5}+\left (-5+e^{28 x/5} (5+28 x)\right ) \log (x)\right ) \, dx \\ & = \frac {9 x}{5}+\frac {1}{25} \int \left (-5+e^{28 x/5} (5+28 x)\right ) \log (x) \, dx+\frac {1}{5} \int e^{28 x/5} \, dx \\ & = \frac {1}{28} e^{28 x/5}+\frac {9 x}{5}-\frac {1}{28} e^{28 x/5} \log (x)-\frac {1}{5} x \log (x)+\frac {1}{140} e^{28 x/5} (5+28 x) \log (x)-\frac {1}{25} \int 5 \left (-1+e^{28 x/5}\right ) \, dx \\ & = \frac {1}{28} e^{28 x/5}+\frac {9 x}{5}-\frac {1}{28} e^{28 x/5} \log (x)-\frac {1}{5} x \log (x)+\frac {1}{140} e^{28 x/5} (5+28 x) \log (x)-\frac {1}{5} \int \left (-1+e^{28 x/5}\right ) \, dx \\ & = \frac {1}{28} e^{28 x/5}+2 x-\frac {1}{28} e^{28 x/5} \log (x)-\frac {1}{5} x \log (x)+\frac {1}{140} e^{28 x/5} (5+28 x) \log (x)-\frac {1}{5} \int e^{28 x/5} \, dx \\ & = 2 x-\frac {1}{28} e^{28 x/5} \log (x)-\frac {1}{5} x \log (x)+\frac {1}{140} e^{28 x/5} (5+28 x) \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {1}{25} \left (45+5 e^{28 x/5}+\left (-5+e^{28 x/5} (5+28 x)\right ) \log (x)\right ) \, dx=\frac {1}{25} \left (50 x+5 \left (-1+e^{28 x/5}\right ) x \log (x)\right ) \]

[In]

Integrate[(45 + 5*E^((28*x)/5) + (-5 + E^((28*x)/5)*(5 + 28*x))*Log[x])/25,x]

[Out]

(50*x + 5*(-1 + E^((28*x)/5))*x*Log[x])/25

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86

method	result	size
default	\(2 x +\frac {{\mathrm e}^{\frac {28 x}{5}} \ln \left (x \right ) x}{5}-\frac {x \ln \left (x \right )}{5}\)	\(19\)
norman	\(2 x +\frac {{\mathrm e}^{\frac {28 x}{5}} \ln \left (x \right ) x}{5}-\frac {x \ln \left (x \right )}{5}\)	\(19\)
parallelrisch	\(2 x +\frac {{\mathrm e}^{\frac {28 x}{5}} \ln \left (x \right ) x}{5}-\frac {x \ln \left (x \right )}{5}\)	\(19\)
parts	\(2 x +\frac {{\mathrm e}^{\frac {28 x}{5}} \ln \left (x \right ) x}{5}-\frac {x \ln \left (x \right )}{5}\)	\(19\)
risch	\(\frac {\left (5 \,{\mathrm e}^{\frac {28 x}{5}} x -5 x \right ) \ln \left (x \right )}{25}+2 x\)	\(20\)

[In]

int(1/25*((28*x+5)*exp(28/5*x)-5)*ln(x)+1/5*exp(28/5*x)+9/5,x,method=_RETURNVERBOSE)

[Out]

2*x+1/5*exp(28/5*x)*ln(x)*x-1/5*x*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {1}{25} \left (45+5 e^{28 x/5}+\left (-5+e^{28 x/5} (5+28 x)\right ) \log (x)\right ) \, dx=\frac {1}{5} \, {\left (x e^{\left (\frac {28}{5} \, x\right )} - x\right )} \log \left (x\right ) + 2 \, x \]

[In]

integrate(1/25*((28*x+5)*exp(28/5*x)-5)*log(x)+1/5*exp(28/5*x)+9/5,x, algorithm="fricas")

[Out]

1/5*(x*e^(28/5*x) - x)*log(x) + 2*x

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {1}{25} \left (45+5 e^{28 x/5}+\left (-5+e^{28 x/5} (5+28 x)\right ) \log (x)\right ) \, dx=\frac {x e^{\frac {28 x}{5}} \log {\left (x \right )}}{5} - \frac {x \log {\left (x \right )}}{5} + 2 x \]

[In]

integrate(1/25*((28*x+5)*exp(28/5*x)-5)*ln(x)+1/5*exp(28/5*x)+9/5,x)

[Out]

x*exp(28*x/5)*log(x)/5 - x*log(x)/5 + 2*x

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {1}{25} \left (45+5 e^{28 x/5}+\left (-5+e^{28 x/5} (5+28 x)\right ) \log (x)\right ) \, dx=\frac {1}{5} \, {\left (x e^{\left (\frac {28}{5} \, x\right )} - x\right )} \log \left (x\right ) + 2 \, x \]

[In]

integrate(1/25*((28*x+5)*exp(28/5*x)-5)*log(x)+1/5*exp(28/5*x)+9/5,x, algorithm="maxima")

[Out]

1/5*(x*e^(28/5*x) - x)*log(x) + 2*x

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {1}{25} \left (45+5 e^{28 x/5}+\left (-5+e^{28 x/5} (5+28 x)\right ) \log (x)\right ) \, dx=\frac {1}{5} \, {\left (x e^{\left (\frac {28}{5} \, x\right )} - x\right )} \log \left (x\right ) + 2 \, x \]

[In]

integrate(1/25*((28*x+5)*exp(28/5*x)-5)*log(x)+1/5*exp(28/5*x)+9/5,x, algorithm="giac")

[Out]

1/5*(x*e^(28/5*x) - x)*log(x) + 2*x

Mupad [B] (verification not implemented)

Time = 9.54 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73 \[ \int \frac {1}{25} \left (45+5 e^{28 x/5}+\left (-5+e^{28 x/5} (5+28 x)\right ) \log (x)\right ) \, dx=\frac {x\,\left ({\mathrm {e}}^{\frac {28\,x}{5}}\,\ln \left (x\right )-\ln \left (x\right )+10\right )}{5} \]

[In]

int(exp((28*x)/5)/5 + (log(x)*(exp((28*x)/5)*(28*x + 5) - 5))/25 + 9/5,x)

[Out]

(x*(exp((28*x)/5)*log(x) - log(x) + 10))/5