3.53.72 \(\int \frac {2 x^2+24 x^5 \log ^4(4)-16 x^9 \log ^8(4)+e^{25 x^2} (2-100 x^2-24 x^3 \log ^4(4)+16 x^7 \log ^8(4))+(8 e^{25 x^2} x^4 \log ^4(4)-8 x^6 \log ^4(4)) \log (\frac {e^{25 x^2}-x^2}{x})+(e^{25 x^2} x-x^3) \log ^2(\frac {e^{25 x^2}-x^2}{x})}{16 e^{25 x^2} x^7 \log ^8(4)-16 x^9 \log ^8(4)+(8 e^{25 x^2} x^4 \log ^4(4)-8 x^6 \log ^4(4)) \log (\frac {e^{25 x^2}-x^2}{x})+(e^{25 x^2} x-x^3) \log ^2(\frac {e^{25 x^2}-x^2}{x})} \, dx\)

Optimal. Leaf size=32 \[ x+\frac {2}{4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )} \]

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Rubi [F]  time = 4.23, 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 {2 x^2+24 x^5 \log ^4(4)-16 x^9 \log ^8(4)+e^{25 x^2} \left (2-100 x^2-24 x^3 \log ^4(4)+16 x^7 \log ^8(4)\right )+\left (8 e^{25 x^2} x^4 \log ^4(4)-8 x^6 \log ^4(4)\right ) \log \left (\frac {e^{25 x^2}-x^2}{x}\right )+\left (e^{25 x^2} x-x^3\right ) \log ^2\left (\frac {e^{25 x^2}-x^2}{x}\right )}{16 e^{25 x^2} x^7 \log ^8(4)-16 x^9 \log ^8(4)+\left (8 e^{25 x^2} x^4 \log ^4(4)-8 x^6 \log ^4(4)\right ) \log \left (\frac {e^{25 x^2}-x^2}{x}\right )+\left (e^{25 x^2} x-x^3\right ) \log ^2\left (\frac {e^{25 x^2}-x^2}{x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2*x^2 + 24*x^5*Log[4]^4 - 16*x^9*Log[4]^8 + E^(25*x^2)*(2 - 100*x^2 - 24*x^3*Log[4]^4 + 16*x^7*Log[4]^8)
+ (8*E^(25*x^2)*x^4*Log[4]^4 - 8*x^6*Log[4]^4)*Log[(E^(25*x^2) - x^2)/x] + (E^(25*x^2)*x - x^3)*Log[(E^(25*x^2
) - x^2)/x]^2)/(16*E^(25*x^2)*x^7*Log[4]^8 - 16*x^9*Log[4]^8 + (8*E^(25*x^2)*x^4*Log[4]^4 - 8*x^6*Log[4]^4)*Lo
g[(E^(25*x^2) - x^2)/x] + (E^(25*x^2)*x - x^3)*Log[(E^(25*x^2) - x^2)/x]^2),x]

[Out]

x + 2*Defer[Int][1/(x*(4*x^3*Log[4]^4 + Log[E^(25*x^2)/x - x])^2), x] - 100*Defer[Int][x/(4*x^3*Log[4]^4 + Log
[E^(25*x^2)/x - x])^2, x] - 24*Log[4]^4*Defer[Int][x^2/(4*x^3*Log[4]^4 + Log[E^(25*x^2)/x - x])^2, x] - 4*Defe
r[Int][x/((-E^(25*x^2) + x^2)*(4*x^3*Log[4]^4 + Log[E^(25*x^2)/x - x])^2), x] + 100*Defer[Int][x^3/((-E^(25*x^
2) + x^2)*(4*x^3*Log[4]^4 + Log[E^(25*x^2)/x - x])^2), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (x^2+12 x^5 \log ^4(4)-8 x^9 \log ^8(4)+e^{25 x^2} \left (1-50 x^2-12 x^3 \log ^4(4)+8 x^7 \log ^8(4)\right )\right )+8 x^4 \left (e^{25 x^2}-x^2\right ) \log ^4(4) \log \left (\frac {e^{25 x^2}}{x}-x\right )-\left (-e^{25 x^2} x+x^3\right ) \log ^2\left (\frac {e^{25 x^2}}{x}-x\right )}{\left (e^{25 x^2} x-x^3\right ) \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2} \, dx\\ &=\int \left (\frac {4 x \left (-1+25 x^2\right )}{\left (-e^{25 x^2}+x^2\right ) \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2}+\frac {2-100 x^2-24 x^3 \log ^4(4)+16 x^7 \log ^8(4)+8 x^4 \log ^4(4) \log \left (\frac {e^{25 x^2}}{x}-x\right )+x \log ^2\left (\frac {e^{25 x^2}}{x}-x\right )}{x \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2}\right ) \, dx\\ &=4 \int \frac {x \left (-1+25 x^2\right )}{\left (-e^{25 x^2}+x^2\right ) \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2} \, dx+\int \frac {2-100 x^2-24 x^3 \log ^4(4)+16 x^7 \log ^8(4)+8 x^4 \log ^4(4) \log \left (\frac {e^{25 x^2}}{x}-x\right )+x \log ^2\left (\frac {e^{25 x^2}}{x}-x\right )}{x \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2} \, dx\\ &=4 \int \left (-\frac {x}{\left (-e^{25 x^2}+x^2\right ) \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2}+\frac {25 x^3}{\left (-e^{25 x^2}+x^2\right ) \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2}\right ) \, dx+\int \left (1-\frac {2 \left (-1+50 x^2+12 x^3 \log ^4(4)\right )}{x \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2}\right ) \, dx\\ &=x-2 \int \frac {-1+50 x^2+12 x^3 \log ^4(4)}{x \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2} \, dx-4 \int \frac {x}{\left (-e^{25 x^2}+x^2\right ) \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2} \, dx+100 \int \frac {x^3}{\left (-e^{25 x^2}+x^2\right ) \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2} \, dx\\ &=x-2 \int \left (-\frac {1}{x \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2}+\frac {50 x}{\left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2}+\frac {12 x^2 \log ^4(4)}{\left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2}\right ) \, dx-4 \int \frac {x}{\left (-e^{25 x^2}+x^2\right ) \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2} \, dx+100 \int \frac {x^3}{\left (-e^{25 x^2}+x^2\right ) \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2} \, dx\\ &=x+2 \int \frac {1}{x \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2} \, dx-4 \int \frac {x}{\left (-e^{25 x^2}+x^2\right ) \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2} \, dx-100 \int \frac {x}{\left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2} \, dx+100 \int \frac {x^3}{\left (-e^{25 x^2}+x^2\right ) \left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2} \, dx-\left (24 \log ^4(4)\right ) \int \frac {x^2}{\left (4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 58, normalized size = 1.81 \begin {gather*} \frac {2+4 x^4 \log ^4(4)+x \log \left (\frac {e^{25 x^2}}{x}-x\right )}{4 x^3 \log ^4(4)+\log \left (\frac {e^{25 x^2}}{x}-x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x^2 + 24*x^5*Log[4]^4 - 16*x^9*Log[4]^8 + E^(25*x^2)*(2 - 100*x^2 - 24*x^3*Log[4]^4 + 16*x^7*Log[
4]^8) + (8*E^(25*x^2)*x^4*Log[4]^4 - 8*x^6*Log[4]^4)*Log[(E^(25*x^2) - x^2)/x] + (E^(25*x^2)*x - x^3)*Log[(E^(
25*x^2) - x^2)/x]^2)/(16*E^(25*x^2)*x^7*Log[4]^8 - 16*x^9*Log[4]^8 + (8*E^(25*x^2)*x^4*Log[4]^4 - 8*x^6*Log[4]
^4)*Log[(E^(25*x^2) - x^2)/x] + (E^(25*x^2)*x - x^3)*Log[(E^(25*x^2) - x^2)/x]^2),x]

[Out]

(2 + 4*x^4*Log[4]^4 + x*Log[E^(25*x^2)/x - x])/(4*x^3*Log[4]^4 + Log[E^(25*x^2)/x - x])

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fricas [A]  time = 0.50, size = 62, normalized size = 1.94 \begin {gather*} \frac {64 \, x^{4} \log \relax (2)^{4} + x \log \left (-\frac {x^{2} - e^{\left (25 \, x^{2}\right )}}{x}\right ) + 2}{64 \, x^{3} \log \relax (2)^{4} + \log \left (-\frac {x^{2} - e^{\left (25 \, x^{2}\right )}}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(25*x^2)-x^3)*log((exp(25*x^2)-x^2)/x)^2+(128*x^4*log(2)^4*exp(25*x^2)-128*x^6*log(2)^4)*log(
(exp(25*x^2)-x^2)/x)+(4096*x^7*log(2)^8-384*x^3*log(2)^4-100*x^2+2)*exp(25*x^2)-4096*x^9*log(2)^8+384*x^5*log(
2)^4+2*x^2)/((x*exp(25*x^2)-x^3)*log((exp(25*x^2)-x^2)/x)^2+(128*x^4*log(2)^4*exp(25*x^2)-128*x^6*log(2)^4)*lo
g((exp(25*x^2)-x^2)/x)+4096*x^7*log(2)^8*exp(25*x^2)-4096*x^9*log(2)^8),x, algorithm="fricas")

[Out]

(64*x^4*log(2)^4 + x*log(-(x^2 - e^(25*x^2))/x) + 2)/(64*x^3*log(2)^4 + log(-(x^2 - e^(25*x^2))/x))

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giac [A]  time = 0.95, size = 61, normalized size = 1.91 \begin {gather*} \frac {64 \, x^{4} \log \relax (2)^{4} + x \log \left (-x^{2} + e^{\left (25 \, x^{2}\right )}\right ) - x \log \relax (x) + 2}{64 \, x^{3} \log \relax (2)^{4} + \log \left (-x^{2} + e^{\left (25 \, x^{2}\right )}\right ) - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(25*x^2)-x^3)*log((exp(25*x^2)-x^2)/x)^2+(128*x^4*log(2)^4*exp(25*x^2)-128*x^6*log(2)^4)*log(
(exp(25*x^2)-x^2)/x)+(4096*x^7*log(2)^8-384*x^3*log(2)^4-100*x^2+2)*exp(25*x^2)-4096*x^9*log(2)^8+384*x^5*log(
2)^4+2*x^2)/((x*exp(25*x^2)-x^3)*log((exp(25*x^2)-x^2)/x)^2+(128*x^4*log(2)^4*exp(25*x^2)-128*x^6*log(2)^4)*lo
g((exp(25*x^2)-x^2)/x)+4096*x^7*log(2)^8*exp(25*x^2)-4096*x^9*log(2)^8),x, algorithm="giac")

[Out]

(64*x^4*log(2)^4 + x*log(-x^2 + e^(25*x^2)) - x*log(x) + 2)/(64*x^3*log(2)^4 + log(-x^2 + e^(25*x^2)) - log(x)
)

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maple [C]  time = 0.18, size = 206, normalized size = 6.44

method	result	size
risch	\(x +\frac {4 i}{128 i x^{3} \ln \relax (2)^{4}+\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (-{\mathrm e}^{25 x^{2}}+x^{2}\right )\right ) \mathrm {csgn}\left (\frac {i \left (-{\mathrm e}^{25 x^{2}}+x^{2}\right )}{x}\right )-\pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (-{\mathrm e}^{25 x^{2}}+x^{2}\right )}{x}\right )^{2}+2 \pi \mathrm {csgn}\left (\frac {i \left (-{\mathrm e}^{25 x^{2}}+x^{2}\right )}{x}\right )^{2}-\pi \,\mathrm {csgn}\left (i \left (-{\mathrm e}^{25 x^{2}}+x^{2}\right )\right ) \mathrm {csgn}\left (\frac {i \left (-{\mathrm e}^{25 x^{2}}+x^{2}\right )}{x}\right )^{2}-\pi \mathrm {csgn}\left (\frac {i \left (-{\mathrm e}^{25 x^{2}}+x^{2}\right )}{x}\right )^{3}-2 \pi -2 i \ln \relax (x )+2 i \ln \left (-{\mathrm e}^{25 x^{2}}+x^{2}\right )}\)	\(206\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x*exp(25*x^2)-x^3)*ln((exp(25*x^2)-x^2)/x)^2+(128*x^4*ln(2)^4*exp(25*x^2)-128*x^6*ln(2)^4)*ln((exp(25*x^
2)-x^2)/x)+(4096*x^7*ln(2)^8-384*x^3*ln(2)^4-100*x^2+2)*exp(25*x^2)-4096*x^9*ln(2)^8+384*x^5*ln(2)^4+2*x^2)/((
x*exp(25*x^2)-x^3)*ln((exp(25*x^2)-x^2)/x)^2+(128*x^4*ln(2)^4*exp(25*x^2)-128*x^6*ln(2)^4)*ln((exp(25*x^2)-x^2
)/x)+4096*x^7*ln(2)^8*exp(25*x^2)-4096*x^9*ln(2)^8),x,method=_RETURNVERBOSE)

[Out]

x+4*I/(128*I*x^3*ln(2)^4+Pi*csgn(I/x)*csgn(I*(-exp(25*x^2)+x^2))*csgn(I/x*(-exp(25*x^2)+x^2))-Pi*csgn(I/x)*csg
n(I/x*(-exp(25*x^2)+x^2))^2+2*Pi*csgn(I/x*(-exp(25*x^2)+x^2))^2-Pi*csgn(I*(-exp(25*x^2)+x^2))*csgn(I/x*(-exp(2
5*x^2)+x^2))^2-Pi*csgn(I/x*(-exp(25*x^2)+x^2))^3-2*Pi-2*I*ln(x)+2*I*ln(-exp(25*x^2)+x^2))

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maxima [A]  time = 0.51, size = 61, normalized size = 1.91 \begin {gather*} \frac {64 \, x^{4} \log \relax (2)^{4} + x \log \left (-x^{2} + e^{\left (25 \, x^{2}\right )}\right ) - x \log \relax (x) + 2}{64 \, x^{3} \log \relax (2)^{4} + \log \left (-x^{2} + e^{\left (25 \, x^{2}\right )}\right ) - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(25*x^2)-x^3)*log((exp(25*x^2)-x^2)/x)^2+(128*x^4*log(2)^4*exp(25*x^2)-128*x^6*log(2)^4)*log(
(exp(25*x^2)-x^2)/x)+(4096*x^7*log(2)^8-384*x^3*log(2)^4-100*x^2+2)*exp(25*x^2)-4096*x^9*log(2)^8+384*x^5*log(
2)^4+2*x^2)/((x*exp(25*x^2)-x^3)*log((exp(25*x^2)-x^2)/x)^2+(128*x^4*log(2)^4*exp(25*x^2)-128*x^6*log(2)^4)*lo
g((exp(25*x^2)-x^2)/x)+4096*x^7*log(2)^8*exp(25*x^2)-4096*x^9*log(2)^8),x, algorithm="maxima")

[Out]

(64*x^4*log(2)^4 + x*log(-x^2 + e^(25*x^2)) - x*log(x) + 2)/(64*x^3*log(2)^4 + log(-x^2 + e^(25*x^2)) - log(x)
)

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mupad [B]  time = 3.76, size = 33, normalized size = 1.03 \begin {gather*} x+\frac {2}{\ln \left (\frac {{\mathrm {e}}^{25\,x^2}-x^2}{x}\right )+64\,x^3\,{\ln \relax (2)}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(384*x^5*log(2)^4 - 4096*x^9*log(2)^8 - log((exp(25*x^2) - x^2)/x)*(128*x^6*log(2)^4 - 128*x^4*exp(25*x^2
)*log(2)^4) - exp(25*x^2)*(384*x^3*log(2)^4 - 4096*x^7*log(2)^8 + 100*x^2 - 2) + log((exp(25*x^2) - x^2)/x)^2*
(x*exp(25*x^2) - x^3) + 2*x^2)/(4096*x^9*log(2)^8 + log((exp(25*x^2) - x^2)/x)*(128*x^6*log(2)^4 - 128*x^4*exp
(25*x^2)*log(2)^4) - log((exp(25*x^2) - x^2)/x)^2*(x*exp(25*x^2) - x^3) - 4096*x^7*exp(25*x^2)*log(2)^8),x)

[Out]

x + 2/(log((exp(25*x^2) - x^2)/x) + 64*x^3*log(2)^4)

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sympy [A]  time = 0.42, size = 26, normalized size = 0.81 \begin {gather*} x + \frac {2}{64 x^{3} \log {\relax (2 )}^{4} + \log {\left (\frac {- x^{2} + e^{25 x^{2}}}{x} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x*exp(25*x**2)-x**3)*ln((exp(25*x**2)-x**2)/x)**2+(128*x**4*ln(2)**4*exp(25*x**2)-128*x**6*ln(2)**
4)*ln((exp(25*x**2)-x**2)/x)+(4096*x**7*ln(2)**8-384*x**3*ln(2)**4-100*x**2+2)*exp(25*x**2)-4096*x**9*ln(2)**8
+384*x**5*ln(2)**4+2*x**2)/((x*exp(25*x**2)-x**3)*ln((exp(25*x**2)-x**2)/x)**2+(128*x**4*ln(2)**4*exp(25*x**2)
-128*x**6*ln(2)**4)*ln((exp(25*x**2)-x**2)/x)+4096*x**7*ln(2)**8*exp(25*x**2)-4096*x**9*ln(2)**8),x)

[Out]

x + 2/(64*x**3*log(2)**4 + log((-x**2 + exp(25*x**2))/x))

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