3.35.8 \(\int \frac {e^{2+3 x} (1+\log (x^2))^3 (30+15 x+(15+15 x) \log (x^2)+15 \log ^2(x^2))}{(-4-4 x-4 \log (x^2))^3 (1+x+(2+x) \log (x^2)+\log ^2(x^2))} \, dx\)

Optimal. Leaf size=29 \[ 5 e^{2+3 \left (x-\log \left (4 \left (-1+\frac {x}{-1-\log \left (x^2\right )}\right )\right )\right )} \]

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Rubi [F]  time = 2.81, 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 {e^{2+3 x} \left (1+\log \left (x^2\right )\right )^3 \left (30+15 x+(15+15 x) \log \left (x^2\right )+15 \log ^2\left (x^2\right )\right )}{\left (-4-4 x-4 \log \left (x^2\right )\right )^3 \left (1+x+(2+x) \log \left (x^2\right )+\log ^2\left (x^2\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(2 + 3*x)*(1 + Log[x^2])^3*(30 + 15*x + (15 + 15*x)*Log[x^2] + 15*Log[x^2]^2))/((-4 - 4*x - 4*Log[x^2])
^3*(1 + x + (2 + x)*Log[x^2] + Log[x^2]^2)),x]

[Out]

(-5*E^(2 + 3*x))/64 - (15*Defer[Int][(E^(2 + 3*x)*x^2)/(1 + x + Log[x^2])^4, x])/32 - (15*Defer[Int][(E^(2 + 3
*x)*x^3)/(1 + x + Log[x^2])^4, x])/64 + (15*Defer[Int][(E^(2 + 3*x)*x)/(1 + x + Log[x^2])^3, x])/16 + (45*Defe
r[Int][(E^(2 + 3*x)*x^2)/(1 + x + Log[x^2])^3, x])/64 + (15*Defer[Int][(E^(2 + 3*x)*x^3)/(1 + x + Log[x^2])^3,
x])/64 - (15*Defer[Int][E^(2 + 3*x)/(1 + x + Log[x^2])^2, x])/32 - (45*Defer[Int][(E^(2 + 3*x)*x)/(1 + x + Lo
g[x^2])^2, x])/64 - (45*Defer[Int][(E^(2 + 3*x)*x^2)/(1 + x + Log[x^2])^2, x])/64 + (15*Defer[Int][E^(2 + 3*x)
/(1 + x + Log[x^2]), x])/64 + (45*Defer[Int][(E^(2 + 3*x)*x)/(1 + x + Log[x^2]), x])/64

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {15 e^{2+3 x} \left (1+\log \left (x^2\right )\right )^2 \left (-2-x-(1+x) \log \left (x^2\right )-\log ^2\left (x^2\right )\right )}{64 \left (1+x+\log \left (x^2\right )\right )^4} \, dx\\ &=\frac {15}{64} \int \frac {e^{2+3 x} \left (1+\log \left (x^2\right )\right )^2 \left (-2-x-(1+x) \log \left (x^2\right )-\log ^2\left (x^2\right )\right )}{\left (1+x+\log \left (x^2\right )\right )^4} \, dx\\ &=\frac {15}{64} \int \left (-e^{2+3 x}-\frac {e^{2+3 x} x^2 (2+x)}{\left (1+x+\log \left (x^2\right )\right )^4}+\frac {e^{2+3 x} x \left (4+3 x+x^2\right )}{\left (1+x+\log \left (x^2\right )\right )^3}+\frac {e^{2+3 x} \left (-2-3 x-3 x^2\right )}{\left (1+x+\log \left (x^2\right )\right )^2}+\frac {e^{2+3 x} (1+3 x)}{1+x+\log \left (x^2\right )}\right ) \, dx\\ &=-\left (\frac {15}{64} \int e^{2+3 x} \, dx\right )-\frac {15}{64} \int \frac {e^{2+3 x} x^2 (2+x)}{\left (1+x+\log \left (x^2\right )\right )^4} \, dx+\frac {15}{64} \int \frac {e^{2+3 x} x \left (4+3 x+x^2\right )}{\left (1+x+\log \left (x^2\right )\right )^3} \, dx+\frac {15}{64} \int \frac {e^{2+3 x} \left (-2-3 x-3 x^2\right )}{\left (1+x+\log \left (x^2\right )\right )^2} \, dx+\frac {15}{64} \int \frac {e^{2+3 x} (1+3 x)}{1+x+\log \left (x^2\right )} \, dx\\ &=-\frac {5}{64} e^{2+3 x}-\frac {15}{64} \int \left (\frac {2 e^{2+3 x} x^2}{\left (1+x+\log \left (x^2\right )\right )^4}+\frac {e^{2+3 x} x^3}{\left (1+x+\log \left (x^2\right )\right )^4}\right ) \, dx+\frac {15}{64} \int \left (\frac {4 e^{2+3 x} x}{\left (1+x+\log \left (x^2\right )\right )^3}+\frac {3 e^{2+3 x} x^2}{\left (1+x+\log \left (x^2\right )\right )^3}+\frac {e^{2+3 x} x^3}{\left (1+x+\log \left (x^2\right )\right )^3}\right ) \, dx+\frac {15}{64} \int \left (-\frac {2 e^{2+3 x}}{\left (1+x+\log \left (x^2\right )\right )^2}-\frac {3 e^{2+3 x} x}{\left (1+x+\log \left (x^2\right )\right )^2}-\frac {3 e^{2+3 x} x^2}{\left (1+x+\log \left (x^2\right )\right )^2}\right ) \, dx+\frac {15}{64} \int \left (\frac {e^{2+3 x}}{1+x+\log \left (x^2\right )}+\frac {3 e^{2+3 x} x}{1+x+\log \left (x^2\right )}\right ) \, dx\\ &=-\frac {5}{64} e^{2+3 x}-\frac {15}{64} \int \frac {e^{2+3 x} x^3}{\left (1+x+\log \left (x^2\right )\right )^4} \, dx+\frac {15}{64} \int \frac {e^{2+3 x} x^3}{\left (1+x+\log \left (x^2\right )\right )^3} \, dx+\frac {15}{64} \int \frac {e^{2+3 x}}{1+x+\log \left (x^2\right )} \, dx-\frac {15}{32} \int \frac {e^{2+3 x} x^2}{\left (1+x+\log \left (x^2\right )\right )^4} \, dx-\frac {15}{32} \int \frac {e^{2+3 x}}{\left (1+x+\log \left (x^2\right )\right )^2} \, dx+\frac {45}{64} \int \frac {e^{2+3 x} x^2}{\left (1+x+\log \left (x^2\right )\right )^3} \, dx-\frac {45}{64} \int \frac {e^{2+3 x} x}{\left (1+x+\log \left (x^2\right )\right )^2} \, dx-\frac {45}{64} \int \frac {e^{2+3 x} x^2}{\left (1+x+\log \left (x^2\right )\right )^2} \, dx+\frac {45}{64} \int \frac {e^{2+3 x} x}{1+x+\log \left (x^2\right )} \, dx+\frac {15}{16} \int \frac {e^{2+3 x} x}{\left (1+x+\log \left (x^2\right )\right )^3} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.40, size = 28, normalized size = 0.97 \begin {gather*} -\frac {5 e^{2+3 x} \left (1+\log \left (x^2\right )\right )^3}{64 \left (1+x+\log \left (x^2\right )\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2 + 3*x)*(1 + Log[x^2])^3*(30 + 15*x + (15 + 15*x)*Log[x^2] + 15*Log[x^2]^2))/((-4 - 4*x - 4*Log
[x^2])^3*(1 + x + (2 + x)*Log[x^2] + Log[x^2]^2)),x]

[Out]

(-5*E^(2 + 3*x)*(1 + Log[x^2])^3)/(64*(1 + x + Log[x^2])^3)

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fricas [A]  time = 0.51, size = 28, normalized size = 0.97 \begin {gather*} 5 \, e^{\left (3 \, x - 3 \, \log \left (-\frac {4 \, {\left (x + \log \left (x^{2}\right ) + 1\right )}}{\log \left (x^{2}\right ) + 1}\right ) + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((15*log(x^2)^2+(15*x+15)*log(x^2)+15*x+30)*exp(-3*log((-4*log(x^2)-4*x-4)/(log(x^2)+1))+3*x+2)/(log(
x^2)^2+(2+x)*log(x^2)+x+1),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.83, size = 47, normalized size = 1.62 \begin {gather*} 5 \, e^{\left (3 \, x - 3 \, \log \left (-\frac {4 \, x}{\log \left (x^{2}\right ) + 1} - \frac {4 \, \log \left (x^{2}\right )}{\log \left (x^{2}\right ) + 1} - \frac {4}{\log \left (x^{2}\right ) + 1}\right ) + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((15*log(x^2)^2+(15*x+15)*log(x^2)+15*x+30)*exp(-3*log((-4*log(x^2)-4*x-4)/(log(x^2)+1))+3*x+2)/(log(
x^2)^2+(2+x)*log(x^2)+x+1),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {\left (15 \ln \left (x^{2}\right )^{2}+\left (15 x +15\right ) \ln \left (x^{2}\right )+15 x +30\right ) {\mathrm e}^{-3 \ln \left (\frac {-4 \ln \left (x^{2}\right )-4 x -4}{\ln \left (x^{2}\right )+1}\right )+3 x +2}}{\ln \left (x^{2}\right )^{2}+\left (2+x \right ) \ln \left (x^{2}\right )+x +1}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((15*ln(x^2)^2+(15*x+15)*ln(x^2)+15*x+30)*exp(-3*ln((-4*ln(x^2)-4*x-4)/(ln(x^2)+1))+3*x+2)/(ln(x^2)^2+(2+x)
*ln(x^2)+x+1),x)

[Out]

int((15*ln(x^2)^2+(15*x+15)*ln(x^2)+15*x+30)*exp(-3*ln((-4*ln(x^2)-4*x-4)/(ln(x^2)+1))+3*x+2)/(ln(x^2)^2+(2+x)
*ln(x^2)+x+1),x)

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maxima [B]  time = 0.87, size = 73, normalized size = 2.52 \begin {gather*} -\frac {5 \, {\left (8 \, e^{2} \log \relax (x)^{3} + 12 \, e^{2} \log \relax (x)^{2} + 6 \, e^{2} \log \relax (x) + e^{2}\right )} e^{\left (3 \, x\right )}}{64 \, {\left (x^{3} + 12 \, {\left (x + 1\right )} \log \relax (x)^{2} + 8 \, \log \relax (x)^{3} + 3 \, x^{2} + 6 \, {\left (x^{2} + 2 \, x + 1\right )} \log \relax (x) + 3 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((15*log(x^2)^2+(15*x+15)*log(x^2)+15*x+30)*exp(-3*log((-4*log(x^2)-4*x-4)/(log(x^2)+1))+3*x+2)/(log(
x^2)^2+(2+x)*log(x^2)+x+1),x, algorithm="maxima")

[Out]

-5/64*(8*e^2*log(x)^3 + 12*e^2*log(x)^2 + 6*e^2*log(x) + e^2)*e^(3*x)/(x^3 + 12*(x + 1)*log(x)^2 + 8*log(x)^3
+ 3*x^2 + 6*(x^2 + 2*x + 1)*log(x) + 3*x + 1)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^{3\,x-3\,\ln \left (-\frac {4\,x+4\,\ln \left (x^2\right )+4}{\ln \left (x^2\right )+1}\right )+2}\,\left (15\,{\ln \left (x^2\right )}^2+\left (15\,x+15\right )\,\ln \left (x^2\right )+15\,x+30\right )}{{\ln \left (x^2\right )}^2+\left (x+2\right )\,\ln \left (x^2\right )+x+1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(3*x - 3*log(-(4*x + 4*log(x^2) + 4)/(log(x^2) + 1)) + 2)*(15*x + 15*log(x^2)^2 + log(x^2)*(15*x + 15)
+ 30))/(x + log(x^2)*(x + 2) + log(x^2)^2 + 1),x)

[Out]

int((exp(3*x - 3*log(-(4*x + 4*log(x^2) + 4)/(log(x^2) + 1)) + 2)*(15*x + 15*log(x^2)^2 + log(x^2)*(15*x + 15)
+ 30))/(x + log(x^2)*(x + 2) + log(x^2)^2 + 1), x)

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sympy [B]  time = 0.53, size = 100, normalized size = 3.45 \begin {gather*} \frac {\left (- 5 \log {\left (x^{2} \right )}^{3} - 15 \log {\left (x^{2} \right )}^{2} - 15 \log {\left (x^{2} \right )} - 5\right ) e^{3 x + 2}}{64 x^{3} + 192 x^{2} \log {\left (x^{2} \right )} + 192 x^{2} + 192 x \log {\left (x^{2} \right )}^{2} + 384 x \log {\left (x^{2} \right )} + 192 x + 64 \log {\left (x^{2} \right )}^{3} + 192 \log {\left (x^{2} \right )}^{2} + 192 \log {\left (x^{2} \right )} + 64} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((15*ln(x**2)**2+(15*x+15)*ln(x**2)+15*x+30)*exp(-3*ln((-4*ln(x**2)-4*x-4)/(ln(x**2)+1))+3*x+2)/(ln(x
**2)**2+(2+x)*ln(x**2)+x+1),x)

[Out]

(-5*log(x**2)**3 - 15*log(x**2)**2 - 15*log(x**2) - 5)*exp(3*x + 2)/(64*x**3 + 192*x**2*log(x**2) + 192*x**2 +
192*x*log(x**2)**2 + 384*x*log(x**2) + 192*x + 64*log(x**2)**3 + 192*log(x**2)**2 + 192*log(x**2) + 64)

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