\(\int \frac {-1+4 e^6 x+4 e^{6+x} x}{e^5 x} \, dx\) [5578]

Optimal result

Integrand size = 23, antiderivative size = 19 \[ \int \frac {-1+4 e^6 x+4 e^{6+x} x}{e^5 x} \, dx=-5+4 e \left (e^x+x\right )-\frac {\log (4 x)}{e^5} \]

[Out]

4*exp(1)*(exp(x)+x)-5-ln(4*x)/exp(5)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {12, 14, 2225, 45} \[ \int \frac {-1+4 e^6 x+4 e^{6+x} x}{e^5 x} \, dx=4 e x+4 e^{x+1}-\frac {\log (x)}{e^5} \]

[In]

Int[(-1 + 4*E^6*x + 4*E^(6 + x)*x)/(E^5*x),x]

[Out]

4*E^(1 + x) + 4*E*x - Log[x]/E^5

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {-1+4 e^6 x+4 e^{6+x} x}{x} \, dx}{e^5} \\ & = \frac {\int \left (4 e^{6+x}+\frac {-1+4 e^6 x}{x}\right ) \, dx}{e^5} \\ & = \frac {\int \frac {-1+4 e^6 x}{x} \, dx}{e^5}+\frac {4 \int e^{6+x} \, dx}{e^5} \\ & = 4 e^{1+x}+\frac {\int \left (4 e^6-\frac {1}{x}\right ) \, dx}{e^5} \\ & = 4 e^{1+x}+4 e x-\frac {\log (x)}{e^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-1+4 e^6 x+4 e^{6+x} x}{e^5 x} \, dx=4 e^{1+x}+4 e x-\frac {\log (x)}{e^5} \]

[In]

Integrate[(-1 + 4*E^6*x + 4*E^(6 + x)*x)/(E^5*x),x]

[Out]

4*E^(1 + x) + 4*E*x - Log[x]/E^5

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00

method	result	size
risch	\(4 x \,{\mathrm e}-{\mathrm e}^{-5} \ln \left (x \right )+4 \,{\mathrm e}^{1+x}\)	\(19\)
norman	\(4 x \,{\mathrm e}+4 \,{\mathrm e} \,{\mathrm e}^{x}-{\mathrm e}^{-5} \ln \left (x \right )\)	\(21\)
parts	\({\mathrm e}^{-5} \left (4 x \,{\mathrm e}^{6}-\ln \left (x \right )\right )+4 \,{\mathrm e} \,{\mathrm e}^{x}\)	\(23\)
default	\({\mathrm e}^{-5} \left (-\ln \left (x \right )+4 x \,{\mathrm e} \,{\mathrm e}^{5}+4 \,{\mathrm e} \,{\mathrm e}^{5} {\mathrm e}^{x}\right )\)	\(26\)
parallelrisch	\({\mathrm e}^{-5} \left (-\ln \left (x \right )+4 x \,{\mathrm e} \,{\mathrm e}^{5}+4 \,{\mathrm e} \,{\mathrm e}^{5} {\mathrm e}^{x}\right )\)	\(26\)

[In]

int((4*x*exp(1)*exp(5)*exp(x)+4*x*exp(1)*exp(5)-1)/x/exp(5),x,method=_RETURNVERBOSE)

[Out]

4*x*exp(1)-exp(-5)*ln(x)+4*exp(1+x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-1+4 e^6 x+4 e^{6+x} x}{e^5 x} \, dx={\left (4 \, x e^{6} + 4 \, e^{\left (x + 6\right )} - \log \left (x\right )\right )} e^{\left (-5\right )} \]

[In]

integrate((4*x*exp(1)*exp(5)*exp(x)+4*x*exp(1)*exp(5)-1)/x/exp(5),x, algorithm="fricas")

[Out]

(4*x*e^6 + 4*e^(x + 6) - log(x))*e^(-5)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.05 \[ \int \frac {-1+4 e^6 x+4 e^{6+x} x}{e^5 x} \, dx=\frac {4 x e^{6} - \log {\left (x \right )}}{e^{5}} + 4 e e^{x} \]

[In]

integrate((4*x*exp(1)*exp(5)*exp(x)+4*x*exp(1)*exp(5)-1)/x/exp(5),x)

[Out]

(4*x*exp(6) - log(x))*exp(-5) + 4*E*exp(x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-1+4 e^6 x+4 e^{6+x} x}{e^5 x} \, dx={\left (4 \, x e^{6} + 4 \, e^{\left (x + 6\right )} - \log \left (x\right )\right )} e^{\left (-5\right )} \]

[In]

integrate((4*x*exp(1)*exp(5)*exp(x)+4*x*exp(1)*exp(5)-1)/x/exp(5),x, algorithm="maxima")

[Out]

(4*x*e^6 + 4*e^(x + 6) - log(x))*e^(-5)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {-1+4 e^6 x+4 e^{6+x} x}{e^5 x} \, dx={\left (4 \, x e^{6} + 4 \, e^{\left (x + 6\right )} - \log \left (x\right )\right )} e^{\left (-5\right )} \]

[In]

integrate((4*x*exp(1)*exp(5)*exp(x)+4*x*exp(1)*exp(5)-1)/x/exp(5),x, algorithm="giac")

[Out]

(4*x*e^6 + 4*e^(x + 6) - log(x))*e^(-5)

Mupad [B] (verification not implemented)

Time = 10.65 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {-1+4 e^6 x+4 e^{6+x} x}{e^5 x} \, dx=4\,x\,\mathrm {e}+4\,\mathrm {e}\,{\mathrm {e}}^x-{\mathrm {e}}^{-5}\,\ln \left (x\right ) \]

[In]

int((exp(-5)*(4*x*exp(6) + 4*x*exp(6)*exp(x) - 1))/x,x)

[Out]

4*x*exp(1) + 4*exp(1)*exp(x) - exp(-5)*log(x)