3.33.75 \(\int \frac {e^4 (x^3+x^4)+e^{4+x^2} (2 x^2+2 x^3-2 x^4-2 x^5)+(e^{4+x^2} (1+x)+e^4 (x+x^2)) \log (1+x)+\log (3 x) (e^{4+x^2} x+e^4 x^2+(e^4 (-x-x^2)+e^{4+x^2} (-2 x^2-2 x^3)) \log (1+x))}{x^3+x^4+e^{2 x^2} (x+x^2)+e^{x^2} (2 x^2+2 x^3)} \, dx\)

Optimal. Leaf size=26 \[ \frac {e^4 \left (x^2+\log (3 x) \log (1+x)\right )}{e^{x^2}+x} \]

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Rubi [F]  time = 16.86, 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^4 \left (x^3+x^4\right )+e^{4+x^2} \left (2 x^2+2 x^3-2 x^4-2 x^5\right )+\left (e^{4+x^2} (1+x)+e^4 \left (x+x^2\right )\right ) \log (1+x)+\log (3 x) \left (e^{4+x^2} x+e^4 x^2+\left (e^4 \left (-x-x^2\right )+e^{4+x^2} \left (-2 x^2-2 x^3\right )\right ) \log (1+x)\right )}{x^3+x^4+e^{2 x^2} \left (x+x^2\right )+e^{x^2} \left (2 x^2+2 x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

-(E^4*Log[3*x]*Log[1 + x]*Defer[Int][(E^x^2 + x)^(-2), x]) - E^4*Defer[Int][x^2/(E^x^2 + x)^2, x] + 2*E^4*Log[
3*x]*Log[1 + x]*Defer[Int][x^2/(E^x^2 + x)^2, x] + 2*E^4*Defer[Int][x^4/(E^x^2 + x)^2, x] + E^4*Log[1 + x]*Def
er[Int][1/(x*(E^x^2 + x)), x] + 2*E^4*Defer[Int][x/(E^x^2 + x), x] - 2*E^4*Log[3*x]*Log[1 + x]*Defer[Int][x/(E
^x^2 + x), x] - 2*E^4*Defer[Int][x^3/(E^x^2 + x), x] + E^4*Log[3*x]*Defer[Int][1/((1 + x)*(E^x^2 + x)), x] + E
^4*Log[1 + x]*Defer[Int][Defer[Int][(E^x^2 + x)^(-2), x]/x, x] + E^4*Log[3*x]*Defer[Int][Defer[Int][(E^x^2 + x
)^(-2), x]/(1 + x), x] - 2*E^4*Log[1 + x]*Defer[Int][Defer[Int][x^2/(E^x^2 + x)^2, x]/x, x] - 2*E^4*Log[3*x]*D
efer[Int][Defer[Int][x^2/(E^x^2 + x)^2, x]/(1 + x), x] + 2*E^4*Log[1 + x]*Defer[Int][Defer[Int][x/(E^x^2 + x),
x]/x, x] + 2*E^4*Log[3*x]*Defer[Int][Defer[Int][x/(E^x^2 + x), x]/(1 + x), x] - E^4*Defer[Int][Defer[Int][1/(
(1 + x)*(E^x^2 + x)), x]/x, x] - E^4*Defer[Int][Defer[Int][(E^x^2*x + x^2)^(-1), x]/(1 + x), x] - E^4*Defer[In
t][Defer[Int][Defer[Int][(E^x^2 + x)^(-2), x]/x, x]/(1 + x), x] - E^4*Defer[Int][Defer[Int][Defer[Int][(E^x^2
+ x)^(-2), x]/(1 + x), x]/x, x] + 2*E^4*Defer[Int][Defer[Int][Defer[Int][x^2/(E^x^2 + x)^2, x]/x, x]/(1 + x),
x] + 2*E^4*Defer[Int][Defer[Int][Defer[Int][x^2/(E^x^2 + x)^2, x]/(1 + x), x]/x, x] - 2*E^4*Defer[Int][Defer[I
nt][Defer[Int][x/(E^x^2 + x), x]/x, x]/(1 + x), x] - 2*E^4*Defer[Int][Defer[Int][Defer[Int][x/(E^x^2 + x), x]/
(1 + x), x]/x, x]

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 26, normalized size = 1.00 \begin {gather*} \frac {e^4 \left (x^2+\log (3 x) \log (1+x)\right )}{e^{x^2}+x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.90, size = 32, normalized size = 1.23 \begin {gather*} \frac {x^{2} e^{8} + e^{8} \log \left (3 \, x\right ) \log \left (x + 1\right )}{x e^{4} + e^{\left (x^{2} + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-2*x^3-2*x^2)*exp(4)*exp(x^2)+(-x^2-x)*exp(4))*log(x+1)+x*exp(4)*exp(x^2)+x^2*exp(4))*log(3*x)+(
(x+1)*exp(4)*exp(x^2)+(x^2+x)*exp(4))*log(x+1)+(-2*x^5-2*x^4+2*x^3+2*x^2)*exp(4)*exp(x^2)+(x^4+x^3)*exp(4))/((
x^2+x)*exp(x^2)^2+(2*x^3+2*x^2)*exp(x^2)+x^4+x^3),x, algorithm="fricas")

[Out]

(x^2*e^8 + e^8*log(3*x)*log(x + 1))/(x*e^4 + e^(x^2 + 4))

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giac [A]  time = 0.25, size = 34, normalized size = 1.31 \begin {gather*} \frac {x^{2} e^{4} + e^{4} \log \relax (3) \log \left (x + 1\right ) + e^{4} \log \left (x + 1\right ) \log \relax (x)}{x + e^{\left (x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-2*x^3-2*x^2)*exp(4)*exp(x^2)+(-x^2-x)*exp(4))*log(x+1)+x*exp(4)*exp(x^2)+x^2*exp(4))*log(3*x)+(
(x+1)*exp(4)*exp(x^2)+(x^2+x)*exp(4))*log(x+1)+(-2*x^5-2*x^4+2*x^3+2*x^2)*exp(4)*exp(x^2)+(x^4+x^3)*exp(4))/((
x^2+x)*exp(x^2)^2+(2*x^3+2*x^2)*exp(x^2)+x^4+x^3),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 35, normalized size = 1.35

method	result	size
risch	\(\frac {{\mathrm e}^{4} \ln \left (3 x \right ) \ln \left (x +1\right )}{{\mathrm e}^{x^{2}}+x}+\frac {x^{2} {\mathrm e}^{4}}{{\mathrm e}^{x^{2}}+x}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((-2*x^3-2*x^2)*exp(4)*exp(x^2)+(-x^2-x)*exp(4))*ln(x+1)+x*exp(4)*exp(x^2)+x^2*exp(4))*ln(3*x)+((x+1)*ex
p(4)*exp(x^2)+(x^2+x)*exp(4))*ln(x+1)+(-2*x^5-2*x^4+2*x^3+2*x^2)*exp(4)*exp(x^2)+(x^4+x^3)*exp(4))/((x^2+x)*ex
p(x^2)^2+(2*x^3+2*x^2)*exp(x^2)+x^4+x^3),x,method=_RETURNVERBOSE)

[Out]

exp(4)/(exp(x^2)+x)*ln(3*x)*ln(x+1)+x^2*exp(4)/(exp(x^2)+x)

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maxima [A]  time = 0.85, size = 32, normalized size = 1.23 \begin {gather*} \frac {x^{2} e^{4} + {\left (e^{4} \log \relax (3) + e^{4} \log \relax (x)\right )} \log \left (x + 1\right )}{x + e^{\left (x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-2*x^3-2*x^2)*exp(4)*exp(x^2)+(-x^2-x)*exp(4))*log(x+1)+x*exp(4)*exp(x^2)+x^2*exp(4))*log(3*x)+(
(x+1)*exp(4)*exp(x^2)+(x^2+x)*exp(4))*log(x+1)+(-2*x^5-2*x^4+2*x^3+2*x^2)*exp(4)*exp(x^2)+(x^4+x^3)*exp(4))/((
x^2+x)*exp(x^2)^2+(2*x^3+2*x^2)*exp(x^2)+x^4+x^3),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.33, size = 24, normalized size = 0.92 \begin {gather*} \frac {{\mathrm {e}}^4\,\left (x^2+\ln \left (3\,x\right )\,\ln \left (x+1\right )\right )}{x+{\mathrm {e}}^{x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x + 1)*(exp(4)*(x + x^2) + exp(x^2)*exp(4)*(x + 1)) + log(3*x)*(x^2*exp(4) - log(x + 1)*(exp(4)*(x +
x^2) + exp(x^2)*exp(4)*(2*x^2 + 2*x^3)) + x*exp(x^2)*exp(4)) + exp(4)*(x^3 + x^4) + exp(x^2)*exp(4)*(2*x^2 + 2
*x^3 - 2*x^4 - 2*x^5))/(exp(x^2)*(2*x^2 + 2*x^3) + exp(2*x^2)*(x + x^2) + x^3 + x^4),x)

[Out]

(exp(4)*(x^2 + log(3*x)*log(x + 1)))/(x + exp(x^2))

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sympy [A]  time = 0.76, size = 26, normalized size = 1.00 \begin {gather*} \frac {x^{2} e^{4} + e^{4} \log {\left (3 x \right )} \log {\left (x + 1 \right )}}{x + e^{x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((-2*x**3-2*x**2)*exp(4)*exp(x**2)+(-x**2-x)*exp(4))*ln(x+1)+x*exp(4)*exp(x**2)+x**2*exp(4))*ln(3*
x)+((x+1)*exp(4)*exp(x**2)+(x**2+x)*exp(4))*ln(x+1)+(-2*x**5-2*x**4+2*x**3+2*x**2)*exp(4)*exp(x**2)+(x**4+x**3
)*exp(4))/((x**2+x)*exp(x**2)**2+(2*x**3+2*x**2)*exp(x**2)+x**4+x**3),x)

[Out]

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

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