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

Optimal. Leaf size=19 \[ \frac {16 e^{2+4 x+4 x^2} x}{\log ^8(x)} \]

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Rubi [F]  time = 0.60, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

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

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Mathematica [A]  time = 0.13, size = 19, normalized size = 1.00 \begin {gather*} \frac {16 e^{2+4 x+4 x^2} x}{\log ^8(x)} \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.76, size = 18, normalized size = 0.95 \begin {gather*} \frac {16 \, x e^{\left (4 \, x^{2} + 4 \, x + 2\right )}}{\log \relax (x)^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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giac [A]  time = 0.19, size = 18, normalized size = 0.95 \begin {gather*} \frac {16 \, x e^{\left (4 \, x^{2} + 4 \, x + 2\right )}}{\log \relax (x)^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [F]  time = 0.10, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (8 x^{2}+4 x +1\right ) \ln \relax (x )-8\right ) {\mathrm e}^{-\ln \left (\frac {\ln \relax (x )^{8} {\mathrm e}^{-4 x} {\mathrm e}^{-4 x^{2}}}{16}\right )+2}}{\ln \relax (x )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[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)

[Out]

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)

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maxima [A]  time = 0.49, size = 18, normalized size = 0.95 \begin {gather*} \frac {16 \, x e^{\left (4 \, x^{2} + 4 \, x + 2\right )}}{\log \relax (x)^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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mupad [B]  time = 1.52, size = 19, normalized size = 1.00 \begin {gather*} \frac {16\,x\,{\mathrm {e}}^{4\,x}\,{\mathrm {e}}^2\,{\mathrm {e}}^{4\,x^2}}{{\ln \relax (x)}^8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.34, size = 22, normalized size = 1.16 \begin {gather*} \frac {16 x e^{2} e^{4 x} e^{4 x^{2}}}{\log {\relax (x )}^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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