\(\int \frac {16 e^{2+4 x+4 x^2} (-8+(1+4 x+8 x^2) \log (x))}{\log ^9(x)} \, dx\) [2324]

Optimal result

Integrand size = 33, antiderivative size = 19 \[ \int \frac {16 e^{2+4 x+4 x^2} \left (-8+\left (1+4 x+8 x^2\right ) \log (x)\right )}{\log ^9(x)} \, dx=\frac {16 e^{2+4 x+4 x^2} x}{\log ^8(x)} \]

[Out]

x*exp(-ln(1/16*ln(x)^8/exp(x)^4/exp(x^2)^4)+2)

Rubi [F]

\[ \int \frac {16 e^{2+4 x+4 x^2} \left (-8+\left (1+4 x+8 x^2\right ) \log (x)\right )}{\log ^9(x)} \, dx=\int \frac {16 e^{2+4 x+4 x^2} \left (-8+\left (1+4 x+8 x^2\right ) \log (x)\right )}{\log ^9(x)} \, dx \]

[In]

Int[(16*E^(2 + 4*x + 4*x^2)*(-8 + (1 + 4*x + 8*x^2)*Log[x]))/Log[x]^9,x]

[Out]

-128*Defer[Int][E^(2 + 4*x + 4*x^2)/Log[x]^9, x] + 16*Defer[Int][E^(2 + 4*x + 4*x^2)/Log[x]^8, x] + 64*Defer[I
nt][(E^(2 + 4*x + 4*x^2)*x)/Log[x]^8, x] + 128*Defer[Int][(E^(2 + 4*x + 4*x^2)*x^2)/Log[x]^8, x]

Rubi steps \begin{align*} \text {integral}& = 16 \int \frac {e^{2+4 x+4 x^2} \left (-8+\left (1+4 x+8 x^2\right ) \log (x)\right )}{\log ^9(x)} \, dx \\ & = 16 \int \left (-\frac {8 e^{2+4 x+4 x^2}}{\log ^9(x)}+\frac {e^{2+4 x+4 x^2} \left (1+4 x+8 x^2\right )}{\log ^8(x)}\right ) \, dx \\ & = 16 \int \frac {e^{2+4 x+4 x^2} \left (1+4 x+8 x^2\right )}{\log ^8(x)} \, dx-128 \int \frac {e^{2+4 x+4 x^2}}{\log ^9(x)} \, dx \\ & = 16 \int \left (\frac {e^{2+4 x+4 x^2}}{\log ^8(x)}+\frac {4 e^{2+4 x+4 x^2} x}{\log ^8(x)}+\frac {8 e^{2+4 x+4 x^2} x^2}{\log ^8(x)}\right ) \, dx-128 \int \frac {e^{2+4 x+4 x^2}}{\log ^9(x)} \, dx \\ & = 16 \int \frac {e^{2+4 x+4 x^2}}{\log ^8(x)} \, dx+64 \int \frac {e^{2+4 x+4 x^2} x}{\log ^8(x)} \, dx-128 \int \frac {e^{2+4 x+4 x^2}}{\log ^9(x)} \, dx+128 \int \frac {e^{2+4 x+4 x^2} x^2}{\log ^8(x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {16 e^{2+4 x+4 x^2} \left (-8+\left (1+4 x+8 x^2\right ) \log (x)\right )}{\log ^9(x)} \, dx=\frac {16 e^{2+4 x+4 x^2} x}{\log ^8(x)} \]

[In]

Integrate[(16*E^(2 + 4*x + 4*x^2)*(-8 + (1 + 4*x + 8*x^2)*Log[x]))/Log[x]^9,x]

[Out]

(16*E^(2 + 4*x + 4*x^2)*x)/Log[x]^8

Maple [A] (verified)

Time = 1.02 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.32

method	result	size
parallelrisch	\(x \,{\mathrm e}^{-\ln \left (\frac {\ln \left (x \right )^{8} {\mathrm e}^{-4 x} {\mathrm e}^{-4 x^{2}}}{16}\right )+2}\)	\(25\)

[In]

int(((8*x^2+4*x+1)*ln(x)-8)*exp(-ln(1/16*ln(x)^8/exp(x)^4/exp(x^2)^4)+2)/ln(x),x,method=_RETURNVERBOSE)

[Out]

x*exp(-ln(1/16*ln(x)^8/exp(x)^4/exp(x^2)^4)+2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {16 e^{2+4 x+4 x^2} \left (-8+\left (1+4 x+8 x^2\right ) \log (x)\right )}{\log ^9(x)} \, dx=\frac {16 \, x e^{\left (4 \, x^{2} + 4 \, x + 2\right )}}{\log \left (x\right )^{8}} \]

[In]

integrate(((8*x^2+4*x+1)*log(x)-8)*exp(-log(1/16*log(x)^8/exp(x)^4/exp(x^2)^4)+2)/log(x),x, algorithm="fricas"
)

[Out]

16*x*e^(4*x^2 + 4*x + 2)/log(x)^8

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.16 \[ \int \frac {16 e^{2+4 x+4 x^2} \left (-8+\left (1+4 x+8 x^2\right ) \log (x)\right )}{\log ^9(x)} \, dx=\frac {16 x e^{2} e^{4 x} e^{4 x^{2}}}{\log {\left (x \right )}^{8}} \]

[In]

integrate(((8*x**2+4*x+1)*ln(x)-8)*exp(-ln(1/16*ln(x)**8/exp(x)**4/exp(x**2)**4)+2)/ln(x),x)

[Out]

16*x*exp(2)*exp(4*x)*exp(4*x**2)/log(x)**8

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {16 e^{2+4 x+4 x^2} \left (-8+\left (1+4 x+8 x^2\right ) \log (x)\right )}{\log ^9(x)} \, dx=\frac {16 \, x e^{\left (4 \, x^{2} + 4 \, x + 2\right )}}{\log \left (x\right )^{8}} \]

[In]

integrate(((8*x^2+4*x+1)*log(x)-8)*exp(-log(1/16*log(x)^8/exp(x)^4/exp(x^2)^4)+2)/log(x),x, algorithm="maxima"
)

[Out]

16*x*e^(4*x^2 + 4*x + 2)/log(x)^8

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.95 \[ \int \frac {16 e^{2+4 x+4 x^2} \left (-8+\left (1+4 x+8 x^2\right ) \log (x)\right )}{\log ^9(x)} \, dx=\frac {16 \, x e^{\left (4 \, x^{2} + 4 \, x + 2\right )}}{\log \left (x\right )^{8}} \]

[In]

integrate(((8*x^2+4*x+1)*log(x)-8)*exp(-log(1/16*log(x)^8/exp(x)^4/exp(x^2)^4)+2)/log(x),x, algorithm="giac")

[Out]

16*x*e^(4*x^2 + 4*x + 2)/log(x)^8

Mupad [B] (verification not implemented)

Time = 9.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {16 e^{2+4 x+4 x^2} \left (-8+\left (1+4 x+8 x^2\right ) \log (x)\right )}{\log ^9(x)} \, dx=\frac {16\,x\,{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^2\,{\mathrm {e}}^{4\,x^2}}{{\ln \left (x\right )}^8} \]

[In]

int((exp(2 - log((exp(-4*x)*exp(-4*x^2)*log(x)^8)/16))*(log(x)*(4*x + 8*x^2 + 1) - 8))/log(x),x)

[Out]

(16*x*exp(4*x)*exp(2)*exp(4*x^2))/log(x)^8