3.92.82 \(\int \frac {e^x (-9 x^3-3 x^4)+(-3 x^4+e^x (36 x^3+21 x^4+3 x^5)) \log (x)+(18 x^2+6 x^3) \log ^2(x)+(9 x^3+3 x^4+(-36 x^3-12 x^4) \log (x)) \log (3+x)}{(3+x) \log ^2(x)} \, dx\)

Optimal. Leaf size=23 \[ x^3 \left (2+\frac {3 x \left (e^x-\log (3+x)\right )}{\log (x)}\right ) \]

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Rubi [F]  time = 1.25, 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^x \left (-9 x^3-3 x^4\right )+\left (-3 x^4+e^x \left (36 x^3+21 x^4+3 x^5\right )\right ) \log (x)+\left (18 x^2+6 x^3\right ) \log ^2(x)+\left (9 x^3+3 x^4+\left (-36 x^3-12 x^4\right ) \log (x)\right ) \log (3+x)}{(3+x) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(-9*x^3 - 3*x^4) + (-3*x^4 + E^x*(36*x^3 + 21*x^4 + 3*x^5))*Log[x] + (18*x^2 + 6*x^3)*Log[x]^2 + (9*x
^3 + 3*x^4 + (-36*x^3 - 12*x^4)*Log[x])*Log[3 + x])/((3 + x)*Log[x]^2),x]

[Out]

2*x^3 + (3*E^x*x^4)/Log[x] - 3*Defer[Int][x^4/((3 + x)*Log[x]), x] + 3*Defer[Int][(x^3*Log[3 + x])/Log[x]^2, x
] - 12*Defer[Int][(x^3*Log[3 + x])/Log[x], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int 3 x^2 \left (2+\frac {x \left (-e^x+\log (3+x)\right )}{\log ^2(x)}+\frac {x \left (-x+e^x \left (12+7 x+x^2\right )-4 (3+x) \log (3+x)\right )}{(3+x) \log (x)}\right ) \, dx\\ &=3 \int x^2 \left (2+\frac {x \left (-e^x+\log (3+x)\right )}{\log ^2(x)}+\frac {x \left (-x+e^x \left (12+7 x+x^2\right )-4 (3+x) \log (3+x)\right )}{(3+x) \log (x)}\right ) \, dx\\ &=3 \int \left (\frac {e^x x^3 (-1+4 \log (x)+x \log (x))}{\log ^2(x)}-\frac {x^2 \left (x^2 \log (x)-6 \log ^2(x)-2 x \log ^2(x)-3 x \log (3+x)-x^2 \log (3+x)+12 x \log (x) \log (3+x)+4 x^2 \log (x) \log (3+x)\right )}{(3+x) \log ^2(x)}\right ) \, dx\\ &=3 \int \frac {e^x x^3 (-1+4 \log (x)+x \log (x))}{\log ^2(x)} \, dx-3 \int \frac {x^2 \left (x^2 \log (x)-6 \log ^2(x)-2 x \log ^2(x)-3 x \log (3+x)-x^2 \log (3+x)+12 x \log (x) \log (3+x)+4 x^2 \log (x) \log (3+x)\right )}{(3+x) \log ^2(x)} \, dx\\ &=\frac {3 e^x x^4}{\log (x)}-3 \int x^2 \left (-2-\frac {x \log (3+x)}{\log ^2(x)}+\frac {x (x+4 (3+x) \log (3+x))}{(3+x) \log (x)}\right ) \, dx\\ &=\frac {3 e^x x^4}{\log (x)}-3 \int \left (\frac {x^4-6 x^2 \log (x)-2 x^3 \log (x)}{(3+x) \log (x)}+\frac {x^3 (-1+4 \log (x)) \log (3+x)}{\log ^2(x)}\right ) \, dx\\ &=\frac {3 e^x x^4}{\log (x)}-3 \int \frac {x^4-6 x^2 \log (x)-2 x^3 \log (x)}{(3+x) \log (x)} \, dx-3 \int \frac {x^3 (-1+4 \log (x)) \log (3+x)}{\log ^2(x)} \, dx\\ &=\frac {3 e^x x^4}{\log (x)}-3 \int \left (-2 x^2+\frac {x^4}{(3+x) \log (x)}\right ) \, dx-3 \int \left (-\frac {x^3 \log (3+x)}{\log ^2(x)}+\frac {4 x^3 \log (3+x)}{\log (x)}\right ) \, dx\\ &=2 x^3+\frac {3 e^x x^4}{\log (x)}-3 \int \frac {x^4}{(3+x) \log (x)} \, dx+3 \int \frac {x^3 \log (3+x)}{\log ^2(x)} \, dx-12 \int \frac {x^3 \log (3+x)}{\log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.27, size = 28, normalized size = 1.22 \begin {gather*} 3 \left (\frac {2 x^3}{3}+\frac {x^4 \left (e^x-\log (3+x)\right )}{\log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-9*x^3 - 3*x^4) + (-3*x^4 + E^x*(36*x^3 + 21*x^4 + 3*x^5))*Log[x] + (18*x^2 + 6*x^3)*Log[x]^2
+ (9*x^3 + 3*x^4 + (-36*x^3 - 12*x^4)*Log[x])*Log[3 + x])/((3 + x)*Log[x]^2),x]

[Out]

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

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fricas [A]  time = 0.54, size = 29, normalized size = 1.26 \begin {gather*} \frac {3 \, x^{4} e^{x} - 3 \, x^{4} \log \left (x + 3\right ) + 2 \, x^{3} \log \relax (x)}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-12*x^4-36*x^3)*log(x)+3*x^4+9*x^3)*log(3+x)+(6*x^3+18*x^2)*log(x)^2+((3*x^5+21*x^4+36*x^3)*exp(x
)-3*x^4)*log(x)+(-3*x^4-9*x^3)*exp(x))/(3+x)/log(x)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.14, size = 29, normalized size = 1.26 \begin {gather*} \frac {3 \, x^{4} e^{x} - 3 \, x^{4} \log \left (x + 3\right ) + 2 \, x^{3} \log \relax (x)}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-12*x^4-36*x^3)*log(x)+3*x^4+9*x^3)*log(3+x)+(6*x^3+18*x^2)*log(x)^2+((3*x^5+21*x^4+36*x^3)*exp(x
)-3*x^4)*log(x)+(-3*x^4-9*x^3)*exp(x))/(3+x)/log(x)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.50, size = 33, normalized size = 1.43

method	result	size
risch	\(-\frac {3 x^{4} \ln \left (3+x \right )}{\ln \relax (x )}+\frac {x^{3} \left (3 \,{\mathrm e}^{x} x +2 \ln \relax (x )\right )}{\ln \relax (x )}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-12*x^4-36*x^3)*ln(x)+3*x^4+9*x^3)*ln(3+x)+(6*x^3+18*x^2)*ln(x)^2+((3*x^5+21*x^4+36*x^3)*exp(x)-3*x^4)*
ln(x)+(-3*x^4-9*x^3)*exp(x))/(3+x)/ln(x)^2,x,method=_RETURNVERBOSE)

[Out]

-3*x^4/ln(x)*ln(3+x)+x^3*(3*exp(x)*x+2*ln(x))/ln(x)

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maxima [A]  time = 0.40, size = 29, normalized size = 1.26 \begin {gather*} \frac {3 \, x^{4} e^{x} - 3 \, x^{4} \log \left (x + 3\right ) + 2 \, x^{3} \log \relax (x)}{\log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-12*x^4-36*x^3)*log(x)+3*x^4+9*x^3)*log(3+x)+(6*x^3+18*x^2)*log(x)^2+((3*x^5+21*x^4+36*x^3)*exp(x
)-3*x^4)*log(x)+(-3*x^4-9*x^3)*exp(x))/(3+x)/log(x)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 8.14, size = 90, normalized size = 3.91 \begin {gather*} {\mathrm {e}}^x\,\left (3\,x^5+12\,x^4\right )+2\,x^3+\frac {3\,x^4\,{\mathrm {e}}^x-3\,x^4\,{\mathrm {e}}^x\,\ln \relax (x)\,\left (x+4\right )}{\ln \relax (x)}+\frac {\ln \left (x+3\right )\,\left (\ln \relax (x)\,\left (12\,x^4-\frac {12\,x^6+36\,x^5}{x\,\left (x+3\right )}\right )-3\,x^4\right )}{\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)*(exp(x)*(36*x^3 + 21*x^4 + 3*x^5) - 3*x^4) - exp(x)*(9*x^3 + 3*x^4) + log(x)^2*(18*x^2 + 6*x^3) +
log(x + 3)*(9*x^3 - log(x)*(36*x^3 + 12*x^4) + 3*x^4))/(log(x)^2*(x + 3)),x)

[Out]

exp(x)*(12*x^4 + 3*x^5) + 2*x^3 + (3*x^4*exp(x) - 3*x^4*exp(x)*log(x)*(x + 4))/log(x) + (log(x + 3)*(log(x)*(1
2*x^4 - (36*x^5 + 12*x^6)/(x*(x + 3))) - 3*x^4))/log(x)

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sympy [A]  time = 0.52, size = 29, normalized size = 1.26 \begin {gather*} \frac {3 x^{4} e^{x}}{\log {\relax (x )}} - \frac {3 x^{4} \log {\left (x + 3 \right )}}{\log {\relax (x )}} + 2 x^{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-12*x**4-36*x**3)*ln(x)+3*x**4+9*x**3)*ln(3+x)+(6*x**3+18*x**2)*ln(x)**2+((3*x**5+21*x**4+36*x**3
)*exp(x)-3*x**4)*ln(x)+(-3*x**4-9*x**3)*exp(x))/(3+x)/ln(x)**2,x)

[Out]

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

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