3.67.32 \(\int \frac {5 e^{16 x^2-4 x^3}+e^{16 x^2-4 x^3} (-5+160 x^2-60 x^3) \log (\frac {1}{x})+20 \log ^2(\frac {1}{x})}{e^{32 x^2-8 x^3}+e^{16 x^2-4 x^3} (-8-2 x) \log (\frac {1}{x})+(16+8 x+x^2) \log ^2(\frac {1}{x})} \, dx\)

Optimal. Leaf size=30 \[ -3+\frac {5 x}{4+x-\frac {e^{4 (4-x) x^2}}{\log \left (\frac {1}{x}\right )}} \]

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Rubi [F]  time = 93.66, 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 {5 e^{16 x^2-4 x^3}+e^{16 x^2-4 x^3} \left (-5+160 x^2-60 x^3\right ) \log \left (\frac {1}{x}\right )+20 \log ^2\left (\frac {1}{x}\right )}{e^{32 x^2-8 x^3}+e^{16 x^2-4 x^3} (-8-2 x) \log \left (\frac {1}{x}\right )+\left (16+8 x+x^2\right ) \log ^2\left (\frac {1}{x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(5*E^(16*x^2 - 4*x^3) + E^(16*x^2 - 4*x^3)*(-5 + 160*x^2 - 60*x^3)*Log[x^(-1)] + 20*Log[x^(-1)]^2)/(E^(32*
x^2 - 8*x^3) + E^(16*x^2 - 4*x^3)*(-8 - 2*x)*Log[x^(-1)] + (16 + 8*x + x^2)*Log[x^(-1)]^2),x]

[Out]

(5*E^(8*x^3 - 4*x^2*(4 + x))*(8*x^2*Log[x^(-1)] - 3*x^3*Log[x^(-1)]))/(5*x^2 - 2*x*(4 + x)) + 20*Defer[Int][(E
^(8*x^3)*Log[x^(-1)])/(E^(16*x^2) - E^(4*x^3)*(4 + x)*Log[x^(-1)])^2, x] + 5*Defer[Int][(E^(8*x^3)*x*Log[x^(-1
)])/(E^(16*x^2) - E^(4*x^3)*(4 + x)*Log[x^(-1)])^2, x] - 5*Defer[Int][(E^(8*x^3)*x*Log[x^(-1)]^2)/(E^(16*x^2)
- E^(4*x^3)*(4 + x)*Log[x^(-1)])^2, x] + 640*Defer[Int][(E^(8*x^3)*x^2*Log[x^(-1)]^2)/(E^(16*x^2) - E^(4*x^3)*
(4 + x)*Log[x^(-1)])^2, x] - 80*Defer[Int][(E^(8*x^3)*x^3*Log[x^(-1)]^2)/(E^(16*x^2) - E^(4*x^3)*(4 + x)*Log[x
^(-1)])^2, x] - 60*Defer[Int][(E^(8*x^3)*x^4*Log[x^(-1)]^2)/(E^(16*x^2) - E^(4*x^3)*(4 + x)*Log[x^(-1)])^2, x]
- 20*Defer[Int][(E^(8*x^3)*Log[x^(-1)])/(-E^(32*x^2) + E^(4*x^2*(4 + x))*(4 + x)*Log[x^(-1)]), x] - 5*Defer[I
nt][(E^(8*x^3)*x*Log[x^(-1)])/(-E^(32*x^2) + E^(4*x^2*(4 + x))*(4 + x)*Log[x^(-1)]), x] + 20*Defer[Int][(E^(8*
x^3)*Log[x^(-1)]^2)/(-E^(32*x^2) + E^(4*x^2*(4 + x))*(4 + x)*Log[x^(-1)]), x] + 5*Defer[Int][(E^(8*x^3)*x*Log[
x^(-1)]^2)/(-E^(32*x^2) + E^(4*x^2*(4 + x))*(4 + x)*Log[x^(-1)]), x] - 640*Defer[Int][(E^(8*x^3)*x^2*Log[x^(-1
)]^2)/(-E^(32*x^2) + E^(4*x^2*(4 + x))*(4 + x)*Log[x^(-1)]), x] + 80*Defer[Int][(E^(8*x^3)*x^3*Log[x^(-1)]^2)/
(-E^(32*x^2) + E^(4*x^2*(4 + x))*(4 + x)*Log[x^(-1)]), x] + 60*Defer[Int][(E^(8*x^3)*x^4*Log[x^(-1)]^2)/(-E^(3
2*x^2) + E^(4*x^2*(4 + x))*(4 + x)*Log[x^(-1)]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{8 x^3} \left (5 e^{16 x^2-4 x^3}+e^{16 x^2-4 x^3} \left (-5+160 x^2-60 x^3\right ) \log \left (\frac {1}{x}\right )+20 \log ^2\left (\frac {1}{x}\right )\right )}{\left (e^{16 x^2}-4 e^{4 x^3} \log \left (\frac {1}{x}\right )-e^{4 x^3} x \log \left (\frac {1}{x}\right )\right )^2} \, dx\\ &=\int \left (-5 e^{8 x^3-4 x^2 (4+x)} \left (-1+\log \left (\frac {1}{x}\right )-32 x^2 \log \left (\frac {1}{x}\right )+12 x^3 \log \left (\frac {1}{x}\right )\right )+\frac {5 e^{8 x^3} (4+x) \log \left (\frac {1}{x}\right ) \left (-1+\log \left (\frac {1}{x}\right )-32 x^2 \log \left (\frac {1}{x}\right )+12 x^3 \log \left (\frac {1}{x}\right )\right )}{-e^{32 x^2}+4 e^{4 x^2 (4+x)} \log \left (\frac {1}{x}\right )+e^{4 x^2 (4+x)} x \log \left (\frac {1}{x}\right )}-\frac {5 e^{8 x^3} \log \left (\frac {1}{x}\right ) \left (-4-x+x \log \left (\frac {1}{x}\right )-128 x^2 \log \left (\frac {1}{x}\right )+16 x^3 \log \left (\frac {1}{x}\right )+12 x^4 \log \left (\frac {1}{x}\right )\right )}{\left (-e^{16 x^2}+4 e^{4 x^3} \log \left (\frac {1}{x}\right )+e^{4 x^3} x \log \left (\frac {1}{x}\right )\right )^2}\right ) \, dx\\ &=-\left (5 \int e^{8 x^3-4 x^2 (4+x)} \left (-1+\log \left (\frac {1}{x}\right )-32 x^2 \log \left (\frac {1}{x}\right )+12 x^3 \log \left (\frac {1}{x}\right )\right ) \, dx\right )+5 \int \frac {e^{8 x^3} (4+x) \log \left (\frac {1}{x}\right ) \left (-1+\log \left (\frac {1}{x}\right )-32 x^2 \log \left (\frac {1}{x}\right )+12 x^3 \log \left (\frac {1}{x}\right )\right )}{-e^{32 x^2}+4 e^{4 x^2 (4+x)} \log \left (\frac {1}{x}\right )+e^{4 x^2 (4+x)} x \log \left (\frac {1}{x}\right )} \, dx-5 \int \frac {e^{8 x^3} \log \left (\frac {1}{x}\right ) \left (-4-x+x \log \left (\frac {1}{x}\right )-128 x^2 \log \left (\frac {1}{x}\right )+16 x^3 \log \left (\frac {1}{x}\right )+12 x^4 \log \left (\frac {1}{x}\right )\right )}{\left (-e^{16 x^2}+4 e^{4 x^3} \log \left (\frac {1}{x}\right )+e^{4 x^3} x \log \left (\frac {1}{x}\right )\right )^2} \, dx\\ &=\frac {5 e^{8 x^3-4 x^2 (4+x)} \left (8 x^2 \log \left (\frac {1}{x}\right )-3 x^3 \log \left (\frac {1}{x}\right )\right )}{5 x^2-2 x (4+x)}+5 \int \frac {e^{8 x^3} (4+x) \log \left (\frac {1}{x}\right ) \left (1-\left (1-32 x^2+12 x^3\right ) \log \left (\frac {1}{x}\right )\right )}{e^{32 x^2}-e^{4 x^2 (4+x)} (4+x) \log \left (\frac {1}{x}\right )} \, dx-5 \int \frac {e^{8 x^3} \log \left (\frac {1}{x}\right ) \left (-4-x+x \left (1-128 x+16 x^2+12 x^3\right ) \log \left (\frac {1}{x}\right )\right )}{\left (e^{16 x^2}-e^{4 x^3} (4+x) \log \left (\frac {1}{x}\right )\right )^2} \, dx\\ &=\frac {5 e^{8 x^3-4 x^2 (4+x)} \left (8 x^2 \log \left (\frac {1}{x}\right )-3 x^3 \log \left (\frac {1}{x}\right )\right )}{5 x^2-2 x (4+x)}-5 \int \left (-\frac {4 e^{8 x^3} \log \left (\frac {1}{x}\right )}{\left (-e^{16 x^2}+4 e^{4 x^3} \log \left (\frac {1}{x}\right )+e^{4 x^3} x \log \left (\frac {1}{x}\right )\right )^2}-\frac {e^{8 x^3} x \log \left (\frac {1}{x}\right )}{\left (-e^{16 x^2}+4 e^{4 x^3} \log \left (\frac {1}{x}\right )+e^{4 x^3} x \log \left (\frac {1}{x}\right )\right )^2}+\frac {e^{8 x^3} x \log ^2\left (\frac {1}{x}\right )}{\left (-e^{16 x^2}+4 e^{4 x^3} \log \left (\frac {1}{x}\right )+e^{4 x^3} x \log \left (\frac {1}{x}\right )\right )^2}-\frac {128 e^{8 x^3} x^2 \log ^2\left (\frac {1}{x}\right )}{\left (-e^{16 x^2}+4 e^{4 x^3} \log \left (\frac {1}{x}\right )+e^{4 x^3} x \log \left (\frac {1}{x}\right )\right )^2}+\frac {16 e^{8 x^3} x^3 \log ^2\left (\frac {1}{x}\right )}{\left (-e^{16 x^2}+4 e^{4 x^3} \log \left (\frac {1}{x}\right )+e^{4 x^3} x \log \left (\frac {1}{x}\right )\right )^2}+\frac {12 e^{8 x^3} x^4 \log ^2\left (\frac {1}{x}\right )}{\left (-e^{16 x^2}+4 e^{4 x^3} \log \left (\frac {1}{x}\right )+e^{4 x^3} x \log \left (\frac {1}{x}\right )\right )^2}\right ) \, dx+5 \int \left (\frac {4 e^{8 x^3} \log \left (\frac {1}{x}\right ) \left (-1+\log \left (\frac {1}{x}\right )-32 x^2 \log \left (\frac {1}{x}\right )+12 x^3 \log \left (\frac {1}{x}\right )\right )}{-e^{32 x^2}+4 e^{4 x^2 (4+x)} \log \left (\frac {1}{x}\right )+e^{4 x^2 (4+x)} x \log \left (\frac {1}{x}\right )}+\frac {e^{8 x^3} x \log \left (\frac {1}{x}\right ) \left (-1+\log \left (\frac {1}{x}\right )-32 x^2 \log \left (\frac {1}{x}\right )+12 x^3 \log \left (\frac {1}{x}\right )\right )}{-e^{32 x^2}+4 e^{4 x^2 (4+x)} \log \left (\frac {1}{x}\right )+e^{4 x^2 (4+x)} x \log \left (\frac {1}{x}\right )}\right ) \, dx\\ &=\frac {5 e^{8 x^3-4 x^2 (4+x)} \left (8 x^2 \log \left (\frac {1}{x}\right )-3 x^3 \log \left (\frac {1}{x}\right )\right )}{5 x^2-2 x (4+x)}+5 \int \frac {e^{8 x^3} x \log \left (\frac {1}{x}\right )}{\left (-e^{16 x^2}+4 e^{4 x^3} \log \left (\frac {1}{x}\right )+e^{4 x^3} x \log \left (\frac {1}{x}\right )\right )^2} \, dx-5 \int \frac {e^{8 x^3} x \log ^2\left (\frac {1}{x}\right )}{\left (-e^{16 x^2}+4 e^{4 x^3} \log \left (\frac {1}{x}\right )+e^{4 x^3} x \log \left (\frac {1}{x}\right )\right )^2} \, dx+5 \int \frac {e^{8 x^3} x \log \left (\frac {1}{x}\right ) \left (-1+\log \left (\frac {1}{x}\right )-32 x^2 \log \left (\frac {1}{x}\right )+12 x^3 \log \left (\frac {1}{x}\right )\right )}{-e^{32 x^2}+4 e^{4 x^2 (4+x)} \log \left (\frac {1}{x}\right )+e^{4 x^2 (4+x)} x \log \left (\frac {1}{x}\right )} \, dx+20 \int \frac {e^{8 x^3} \log \left (\frac {1}{x}\right )}{\left (-e^{16 x^2}+4 e^{4 x^3} \log \left (\frac {1}{x}\right )+e^{4 x^3} x \log \left (\frac {1}{x}\right )\right )^2} \, dx+20 \int \frac {e^{8 x^3} \log \left (\frac {1}{x}\right ) \left (-1+\log \left (\frac {1}{x}\right )-32 x^2 \log \left (\frac {1}{x}\right )+12 x^3 \log \left (\frac {1}{x}\right )\right )}{-e^{32 x^2}+4 e^{4 x^2 (4+x)} \log \left (\frac {1}{x}\right )+e^{4 x^2 (4+x)} x \log \left (\frac {1}{x}\right )} \, dx-60 \int \frac {e^{8 x^3} x^4 \log ^2\left (\frac {1}{x}\right )}{\left (-e^{16 x^2}+4 e^{4 x^3} \log \left (\frac {1}{x}\right )+e^{4 x^3} x \log \left (\frac {1}{x}\right )\right )^2} \, dx-80 \int \frac {e^{8 x^3} x^3 \log ^2\left (\frac {1}{x}\right )}{\left (-e^{16 x^2}+4 e^{4 x^3} \log \left (\frac {1}{x}\right )+e^{4 x^3} x \log \left (\frac {1}{x}\right )\right )^2} \, dx+640 \int \frac {e^{8 x^3} x^2 \log ^2\left (\frac {1}{x}\right )}{\left (-e^{16 x^2}+4 e^{4 x^3} \log \left (\frac {1}{x}\right )+e^{4 x^3} x \log \left (\frac {1}{x}\right )\right )^2} \, dx\\ &=\frac {5 e^{8 x^3-4 x^2 (4+x)} \left (8 x^2 \log \left (\frac {1}{x}\right )-3 x^3 \log \left (\frac {1}{x}\right )\right )}{5 x^2-2 x (4+x)}+5 \int \frac {e^{8 x^3} x \log \left (\frac {1}{x}\right )}{\left (e^{16 x^2}-e^{4 x^3} (4+x) \log \left (\frac {1}{x}\right )\right )^2} \, dx-5 \int \frac {e^{8 x^3} x \log ^2\left (\frac {1}{x}\right )}{\left (e^{16 x^2}-e^{4 x^3} (4+x) \log \left (\frac {1}{x}\right )\right )^2} \, dx+5 \int \frac {e^{8 x^3} x \log \left (\frac {1}{x}\right ) \left (1-\left (1-32 x^2+12 x^3\right ) \log \left (\frac {1}{x}\right )\right )}{e^{32 x^2}-e^{4 x^2 (4+x)} (4+x) \log \left (\frac {1}{x}\right )} \, dx+20 \int \frac {e^{8 x^3} \log \left (\frac {1}{x}\right )}{\left (e^{16 x^2}-e^{4 x^3} (4+x) \log \left (\frac {1}{x}\right )\right )^2} \, dx+20 \int \frac {e^{8 x^3} \log \left (\frac {1}{x}\right ) \left (1-\left (1-32 x^2+12 x^3\right ) \log \left (\frac {1}{x}\right )\right )}{e^{32 x^2}-e^{4 x^2 (4+x)} (4+x) \log \left (\frac {1}{x}\right )} \, dx-60 \int \frac {e^{8 x^3} x^4 \log ^2\left (\frac {1}{x}\right )}{\left (e^{16 x^2}-e^{4 x^3} (4+x) \log \left (\frac {1}{x}\right )\right )^2} \, dx-80 \int \frac {e^{8 x^3} x^3 \log ^2\left (\frac {1}{x}\right )}{\left (e^{16 x^2}-e^{4 x^3} (4+x) \log \left (\frac {1}{x}\right )\right )^2} \, dx+640 \int \frac {e^{8 x^3} x^2 \log ^2\left (\frac {1}{x}\right )}{\left (e^{16 x^2}-e^{4 x^3} (4+x) \log \left (\frac {1}{x}\right )\right )^2} \, dx\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 51, normalized size = 1.70 \begin {gather*} \frac {5 \left (-e^{16 x^2}+4 e^{4 x^3} \log \left (\frac {1}{x}\right )\right )}{e^{16 x^2}-e^{4 x^3} (4+x) \log \left (\frac {1}{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(5*E^(16*x^2 - 4*x^3) + E^(16*x^2 - 4*x^3)*(-5 + 160*x^2 - 60*x^3)*Log[x^(-1)] + 20*Log[x^(-1)]^2)/(
E^(32*x^2 - 8*x^3) + E^(16*x^2 - 4*x^3)*(-8 - 2*x)*Log[x^(-1)] + (16 + 8*x + x^2)*Log[x^(-1)]^2),x]

[Out]

(5*(-E^(16*x^2) + 4*E^(4*x^3)*Log[x^(-1)]))/(E^(16*x^2) - E^(4*x^3)*(4 + x)*Log[x^(-1)])

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fricas [A]  time = 0.51, size = 46, normalized size = 1.53 \begin {gather*} \frac {5 \, {\left (e^{\left (-4 \, x^{3} + 16 \, x^{2}\right )} - 4 \, \log \left (\frac {1}{x}\right )\right )}}{{\left (x + 4\right )} \log \left (\frac {1}{x}\right ) - e^{\left (-4 \, x^{3} + 16 \, x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*log(1/x)^2+(-60*x^3+160*x^2-5)*exp(-4*x^3+16*x^2)*log(1/x)+5*exp(-4*x^3+16*x^2))/((x^2+8*x+16)*l
og(1/x)^2+(-2*x-8)*exp(-4*x^3+16*x^2)*log(1/x)+exp(-4*x^3+16*x^2)^2),x, algorithm="fricas")

[Out]

5*(e^(-4*x^3 + 16*x^2) - 4*log(1/x))/((x + 4)*log(1/x) - e^(-4*x^3 + 16*x^2))

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giac [A]  time = 0.23, size = 28, normalized size = 0.93 \begin {gather*} \frac {5 \, x \log \relax (x)}{x \log \relax (x) + e^{\left (-4 \, x^{3} + 16 \, x^{2}\right )} + 4 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*log(1/x)^2+(-60*x^3+160*x^2-5)*exp(-4*x^3+16*x^2)*log(1/x)+5*exp(-4*x^3+16*x^2))/((x^2+8*x+16)*l
og(1/x)^2+(-2*x-8)*exp(-4*x^3+16*x^2)*log(1/x)+exp(-4*x^3+16*x^2)^2),x, algorithm="giac")

[Out]

5*x*log(x)/(x*log(x) + e^(-4*x^3 + 16*x^2) + 4*log(x))

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maple [A]  time = 0.11, size = 46, normalized size = 1.53

method	result	size
risch	\(-\frac {20}{4+x}-\frac {5 \,{\mathrm e}^{-4 \left (x -4\right ) x^{2}} x}{\left (4+x \right ) \left (x \ln \relax (x )+{\mathrm e}^{-4 \left (x -4\right ) x^{2}}+4 \ln \relax (x )\right )}\)	\(46\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((20*ln(1/x)^2+(-60*x^3+160*x^2-5)*exp(-4*x^3+16*x^2)*ln(1/x)+5*exp(-4*x^3+16*x^2))/((x^2+8*x+16)*ln(1/x)^2
+(-2*x-8)*exp(-4*x^3+16*x^2)*ln(1/x)+exp(-4*x^3+16*x^2)^2),x,method=_RETURNVERBOSE)

[Out]

-20/(4+x)-5*exp(-4*(x-4)*x^2)*x/(4+x)/(x*ln(x)+exp(-4*(x-4)*x^2)+4*ln(x))

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maxima [A]  time = 0.49, size = 40, normalized size = 1.33 \begin {gather*} -\frac {5 \, {\left (4 \, e^{\left (4 \, x^{3}\right )} \log \relax (x) + e^{\left (16 \, x^{2}\right )}\right )}}{{\left (x + 4\right )} e^{\left (4 \, x^{3}\right )} \log \relax (x) + e^{\left (16 \, x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*log(1/x)^2+(-60*x^3+160*x^2-5)*exp(-4*x^3+16*x^2)*log(1/x)+5*exp(-4*x^3+16*x^2))/((x^2+8*x+16)*l
og(1/x)^2+(-2*x-8)*exp(-4*x^3+16*x^2)*log(1/x)+exp(-4*x^3+16*x^2)^2),x, algorithm="maxima")

[Out]

-5*(4*e^(4*x^3)*log(x) + e^(16*x^2))/((x + 4)*e^(4*x^3)*log(x) + e^(16*x^2))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {20\,{\ln \left (\frac {1}{x}\right )}^2-{\mathrm {e}}^{16\,x^2-4\,x^3}\,\left (60\,x^3-160\,x^2+5\right )\,\ln \left (\frac {1}{x}\right )+5\,{\mathrm {e}}^{16\,x^2-4\,x^3}}{\left (x^2+8\,x+16\right )\,{\ln \left (\frac {1}{x}\right )}^2-{\mathrm {e}}^{16\,x^2-4\,x^3}\,\left (2\,x+8\right )\,\ln \left (\frac {1}{x}\right )+{\mathrm {e}}^{32\,x^2-8\,x^3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*exp(16*x^2 - 4*x^3) + 20*log(1/x)^2 - log(1/x)*exp(16*x^2 - 4*x^3)*(60*x^3 - 160*x^2 + 5))/(exp(32*x^2
- 8*x^3) + log(1/x)^2*(8*x + x^2 + 16) - log(1/x)*exp(16*x^2 - 4*x^3)*(2*x + 8)),x)

[Out]

int((5*exp(16*x^2 - 4*x^3) + 20*log(1/x)^2 - log(1/x)*exp(16*x^2 - 4*x^3)*(60*x^3 - 160*x^2 + 5))/(exp(32*x^2
- 8*x^3) + log(1/x)^2*(8*x + x^2 + 16) - log(1/x)*exp(16*x^2 - 4*x^3)*(2*x + 8)), x)

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sympy [A]  time = 0.32, size = 34, normalized size = 1.13 \begin {gather*} - \frac {5 x \log {\left (\frac {1}{x} \right )}}{- x \log {\left (\frac {1}{x} \right )} + e^{- 4 x^{3} + 16 x^{2}} - 4 \log {\left (\frac {1}{x} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((20*ln(1/x)**2+(-60*x**3+160*x**2-5)*exp(-4*x**3+16*x**2)*ln(1/x)+5*exp(-4*x**3+16*x**2))/((x**2+8*x
+16)*ln(1/x)**2+(-2*x-8)*exp(-4*x**3+16*x**2)*ln(1/x)+exp(-4*x**3+16*x**2)**2),x)

[Out]

-5*x*log(1/x)/(-x*log(1/x) + exp(-4*x**3 + 16*x**2) - 4*log(1/x))

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