\(\int \frac {e^{e^{3+x}} (8 \log ^7(4+x)+e^{3+x} (4+x) \log ^8(4+x))}{4+x} \, dx\) [6630]

Optimal result

Integrand size = 37, antiderivative size = 16 \[ \int \frac {e^{e^{3+x}} \left (8 \log ^7(4+x)+e^{3+x} (4+x) \log ^8(4+x)\right )}{4+x} \, dx=6+e^{e^{3+x}} \log ^8(4+x) \]

[Out]

6+ln(4+x)^8*exp(exp(3+x))

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {2326} \[ \int \frac {e^{e^{3+x}} \left (8 \log ^7(4+x)+e^{3+x} (4+x) \log ^8(4+x)\right )}{4+x} \, dx=e^{e^{x+3}} \log ^8(x+4) \]

[In]

Int[(E^E^(3 + x)*(8*Log[4 + x]^7 + E^(3 + x)*(4 + x)*Log[4 + x]^8))/(4 + x),x]

[Out]

E^E^(3 + x)*Log[4 + x]^8

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]

Rubi steps \begin{align*} \text {integral}& = e^{e^{3+x}} \log ^8(4+x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {e^{e^{3+x}} \left (8 \log ^7(4+x)+e^{3+x} (4+x) \log ^8(4+x)\right )}{4+x} \, dx=e^{e^{3+x}} \log ^8(4+x) \]

[In]

Integrate[(E^E^(3 + x)*(8*Log[4 + x]^7 + E^(3 + x)*(4 + x)*Log[4 + x]^8))/(4 + x),x]

[Out]

E^E^(3 + x)*Log[4 + x]^8

Maple [A] (verified)

Time = 1.92 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.81

method	result	size
risch	\(\ln \left (4+x \right )^{8} {\mathrm e}^{{\mathrm e}^{3+x}}\)	\(13\)
parallelrisch	\(\ln \left (4+x \right )^{8} {\mathrm e}^{{\mathrm e}^{3+x}}\)	\(13\)

[In]

int(((4+x)*exp(3+x)*ln(4+x)^8+8*ln(4+x)^7)*exp(exp(3+x))/(4+x),x,method=_RETURNVERBOSE)

[Out]

ln(4+x)^8*exp(exp(3+x))

Fricas [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {e^{e^{3+x}} \left (8 \log ^7(4+x)+e^{3+x} (4+x) \log ^8(4+x)\right )}{4+x} \, dx=e^{\left (e^{\left (x + 3\right )}\right )} \log \left (x + 4\right )^{8} \]

[In]

integrate(((4+x)*exp(3+x)*log(4+x)^8+8*log(4+x)^7)*exp(exp(3+x))/(4+x),x, algorithm="fricas")

[Out]

e^(e^(x + 3))*log(x + 4)^8

Sympy [F(-1)]

Timed out. \[ \int \frac {e^{e^{3+x}} \left (8 \log ^7(4+x)+e^{3+x} (4+x) \log ^8(4+x)\right )}{4+x} \, dx=\text {Timed out} \]

[In]

integrate(((4+x)*exp(3+x)*ln(4+x)**8+8*ln(4+x)**7)*exp(exp(3+x))/(4+x),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {e^{e^{3+x}} \left (8 \log ^7(4+x)+e^{3+x} (4+x) \log ^8(4+x)\right )}{4+x} \, dx=e^{\left (e^{\left (x + 3\right )}\right )} \log \left (x + 4\right )^{8} \]

[In]

integrate(((4+x)*exp(3+x)*log(4+x)^8+8*log(4+x)^7)*exp(exp(3+x))/(4+x),x, algorithm="maxima")

[Out]

e^(e^(x + 3))*log(x + 4)^8

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {e^{e^{3+x}} \left (8 \log ^7(4+x)+e^{3+x} (4+x) \log ^8(4+x)\right )}{4+x} \, dx=e^{\left (e^{\left (x + 3\right )}\right )} \log \left (x + 4\right )^{8} \]

[In]

integrate(((4+x)*exp(3+x)*log(4+x)^8+8*log(4+x)^7)*exp(exp(3+x))/(4+x),x, algorithm="giac")

[Out]

e^(e^(x + 3))*log(x + 4)^8

Mupad [F(-1)]

Timed out. \[ \int \frac {e^{e^{3+x}} \left (8 \log ^7(4+x)+e^{3+x} (4+x) \log ^8(4+x)\right )}{4+x} \, dx=\int \frac {{\mathrm {e}}^{{\mathrm {e}}^{x+3}}\,\left ({\mathrm {e}}^{x+3}\,\left (x+4\right )\,{\ln \left (x+4\right )}^8+8\,{\ln \left (x+4\right )}^7\right )}{x+4} \,d x \]

[In]

int((exp(exp(x + 3))*(8*log(x + 4)^7 + log(x + 4)^8*exp(x + 3)*(x + 4)))/(x + 4),x)

[Out]

int((exp(exp(x + 3))*(8*log(x + 4)^7 + log(x + 4)^8*exp(x + 3)*(x + 4)))/(x + 4), x)