3.15.36 \(\int \frac {25 x+5 x^2+e^x (-25 x-5 x^2)+(100+15 x-10 x^2+e^x (25+10 x)+(25+10 x) \log ^2(5)) \log (4+e^x-x+\log ^2(5))+(8-2 x+12 x^2-3 x^3+e^x (2+3 x^2)+(2+3 x^2) \log ^2(5)) \log ^2(4+e^x-x+\log ^2(5))}{(4+e^x-x+\log ^2(5)) \log ^2(4+e^x-x+\log ^2(5))} \, dx\)

Optimal. Leaf size=27 \[ x \left (2+x^2+\frac {5 (5+x)}{\log \left (4+e^x-x+\log ^2(5)\right )}\right ) \]

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Rubi [F]  time = 4.10, 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 {25 x+5 x^2+e^x \left (-25 x-5 x^2\right )+\left (100+15 x-10 x^2+e^x (25+10 x)+(25+10 x) \log ^2(5)\right ) \log \left (4+e^x-x+\log ^2(5)\right )+\left (8-2 x+12 x^2-3 x^3+e^x \left (2+3 x^2\right )+\left (2+3 x^2\right ) \log ^2(5)\right ) \log ^2\left (4+e^x-x+\log ^2(5)\right )}{\left (4+e^x-x+\log ^2(5)\right ) \log ^2\left (4+e^x-x+\log ^2(5)\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(25*x + 5*x^2 + E^x*(-25*x - 5*x^2) + (100 + 15*x - 10*x^2 + E^x*(25 + 10*x) + (25 + 10*x)*Log[5]^2)*Log[4
+ E^x - x + Log[5]^2] + (8 - 2*x + 12*x^2 - 3*x^3 + E^x*(2 + 3*x^2) + (2 + 3*x^2)*Log[5]^2)*Log[4 + E^x - x +
Log[5]^2]^2)/((4 + E^x - x + Log[5]^2)*Log[4 + E^x - x + Log[5]^2]^2),x]

[Out]

2*x + x^3 - 25*Defer[Int][x/Log[E^x - x + 4*(1 + Log[5]^2/4)]^2, x] - 5*Defer[Int][x^2/Log[E^x - x + 4*(1 + Lo
g[5]^2/4)]^2, x] + 5*Defer[Int][x^3/((-E^x + x - 4*(1 + Log[5]^2/4))*Log[E^x - x + 4*(1 + Log[5]^2/4)]^2), x]
+ 25*(5 + Log[5]^2)*Defer[Int][x/((E^x - x + 4*(1 + Log[5]^2/4))*Log[E^x - x + 4*(1 + Log[5]^2/4)]^2), x] + 5*
Log[5]^2*Defer[Int][x^2/((E^x - x + 4*(1 + Log[5]^2/4))*Log[E^x - x + 4*(1 + Log[5]^2/4)]^2), x] + 25*Defer[In
t][Log[E^x - x + 4*(1 + Log[5]^2/4)]^(-1), x] + 10*Defer[Int][x/Log[E^x - x + 4*(1 + Log[5]^2/4)], x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {25 x+5 x^2+e^x \left (-25 x-5 x^2\right )+\left (100+15 x-10 x^2+e^x (25+10 x)+(25+10 x) \log ^2(5)\right ) \log \left (4+e^x-x+\log ^2(5)\right )+\left (8-2 x+12 x^2-3 x^3+e^x \left (2+3 x^2\right )+\left (2+3 x^2\right ) \log ^2(5)\right ) \log ^2\left (4+e^x-x+\log ^2(5)\right )}{\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right ) \log ^2\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )} \, dx\\ &=\int \frac {25 x+5 x^2-5 e^x x (5+x)+5 (5+2 x) \left (4+e^x-x+\log ^2(5)\right ) \log \left (4+e^x-x+\log ^2(5)\right )+\left (2+3 x^2\right ) \left (4+e^x-x+\log ^2(5)\right ) \log ^2\left (4+e^x-x+\log ^2(5)\right )}{\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right ) \log ^2\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )} \, dx\\ &=\int \left (\frac {5 x (5+x) \left (5-x+\log ^2(5)\right )}{\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right ) \log ^2\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )}+\frac {-25 x-5 x^2+25 \log \left (4+e^x-x+\log ^2(5)\right )+10 x \log \left (4+e^x-x+\log ^2(5)\right )+2 \log ^2\left (4+e^x-x+\log ^2(5)\right )+3 x^2 \log ^2\left (4+e^x-x+\log ^2(5)\right )}{\log ^2\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )}\right ) \, dx\\ &=5 \int \frac {x (5+x) \left (5-x+\log ^2(5)\right )}{\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right ) \log ^2\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )} \, dx+\int \frac {-25 x-5 x^2+25 \log \left (4+e^x-x+\log ^2(5)\right )+10 x \log \left (4+e^x-x+\log ^2(5)\right )+2 \log ^2\left (4+e^x-x+\log ^2(5)\right )+3 x^2 \log ^2\left (4+e^x-x+\log ^2(5)\right )}{\log ^2\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )} \, dx\\ &=5 \int \left (\frac {x^3}{\left (-e^x+x-4 \left (1+\frac {\log ^2(5)}{4}\right )\right ) \log ^2\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )}+\frac {x^2 \log ^2(5)}{\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right ) \log ^2\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )}+\frac {5 x \left (5+\log ^2(5)\right )}{\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right ) \log ^2\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )}\right ) \, dx+\int \left (2+3 x^2-\frac {5 x (5+x)}{\log ^2\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )}+\frac {5 (5+2 x)}{\log \left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )}\right ) \, dx\\ &=2 x+x^3-5 \int \frac {x (5+x)}{\log ^2\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )} \, dx+5 \int \frac {x^3}{\left (-e^x+x-4 \left (1+\frac {\log ^2(5)}{4}\right )\right ) \log ^2\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )} \, dx+5 \int \frac {5+2 x}{\log \left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )} \, dx+\left (5 \log ^2(5)\right ) \int \frac {x^2}{\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right ) \log ^2\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )} \, dx+\left (25 \left (5+\log ^2(5)\right )\right ) \int \frac {x}{\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right ) \log ^2\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )} \, dx\\ &=2 x+x^3-5 \int \left (\frac {5 x}{\log ^2\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )}+\frac {x^2}{\log ^2\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )}\right ) \, dx+5 \int \left (\frac {5}{\log \left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )}+\frac {2 x}{\log \left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )}\right ) \, dx+5 \int \frac {x^3}{\left (-e^x+x-4 \left (1+\frac {\log ^2(5)}{4}\right )\right ) \log ^2\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )} \, dx+\left (5 \log ^2(5)\right ) \int \frac {x^2}{\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right ) \log ^2\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )} \, dx+\left (25 \left (5+\log ^2(5)\right )\right ) \int \frac {x}{\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right ) \log ^2\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )} \, dx\\ &=2 x+x^3-5 \int \frac {x^2}{\log ^2\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )} \, dx+5 \int \frac {x^3}{\left (-e^x+x-4 \left (1+\frac {\log ^2(5)}{4}\right )\right ) \log ^2\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )} \, dx+10 \int \frac {x}{\log \left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )} \, dx-25 \int \frac {x}{\log ^2\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )} \, dx+25 \int \frac {1}{\log \left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )} \, dx+\left (5 \log ^2(5)\right ) \int \frac {x^2}{\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right ) \log ^2\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )} \, dx+\left (25 \left (5+\log ^2(5)\right )\right ) \int \frac {x}{\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right ) \log ^2\left (e^x-x+4 \left (1+\frac {\log ^2(5)}{4}\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 28, normalized size = 1.04 \begin {gather*} 2 x+x^3+\frac {5 x (5+x)}{\log \left (4+e^x-x+\log ^2(5)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25*x + 5*x^2 + E^x*(-25*x - 5*x^2) + (100 + 15*x - 10*x^2 + E^x*(25 + 10*x) + (25 + 10*x)*Log[5]^2)
*Log[4 + E^x - x + Log[5]^2] + (8 - 2*x + 12*x^2 - 3*x^3 + E^x*(2 + 3*x^2) + (2 + 3*x^2)*Log[5]^2)*Log[4 + E^x
- x + Log[5]^2]^2)/((4 + E^x - x + Log[5]^2)*Log[4 + E^x - x + Log[5]^2]^2),x]

[Out]

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

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fricas [A]  time = 1.13, size = 44, normalized size = 1.63 \begin {gather*} \frac {5 \, x^{2} + {\left (x^{3} + 2 \, x\right )} \log \left (\log \relax (5)^{2} - x + e^{x} + 4\right ) + 25 \, x}{\log \left (\log \relax (5)^{2} - x + e^{x} + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^2+2)*exp(x)+(3*x^2+2)*log(5)^2-3*x^3+12*x^2-2*x+8)*log(exp(x)+log(5)^2-x+4)^2+((10*x+25)*exp(
x)+(10*x+25)*log(5)^2-10*x^2+15*x+100)*log(exp(x)+log(5)^2-x+4)+(-5*x^2-25*x)*exp(x)+5*x^2+25*x)/(exp(x)+log(5
)^2-x+4)/log(exp(x)+log(5)^2-x+4)^2,x, algorithm="fricas")

[Out]

(5*x^2 + (x^3 + 2*x)*log(log(5)^2 - x + e^x + 4) + 25*x)/log(log(5)^2 - x + e^x + 4)

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giac [B]  time = 1.02, size = 55, normalized size = 2.04 \begin {gather*} \frac {x^{3} \log \left (\log \relax (5)^{2} - x + e^{x} + 4\right ) + 5 \, x^{2} + 2 \, x \log \left (\log \relax (5)^{2} - x + e^{x} + 4\right ) + 25 \, x}{\log \left (\log \relax (5)^{2} - x + e^{x} + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^2+2)*exp(x)+(3*x^2+2)*log(5)^2-3*x^3+12*x^2-2*x+8)*log(exp(x)+log(5)^2-x+4)^2+((10*x+25)*exp(
x)+(10*x+25)*log(5)^2-10*x^2+15*x+100)*log(exp(x)+log(5)^2-x+4)+(-5*x^2-25*x)*exp(x)+5*x^2+25*x)/(exp(x)+log(5
)^2-x+4)/log(exp(x)+log(5)^2-x+4)^2,x, algorithm="giac")

[Out]

(x^3*log(log(5)^2 - x + e^x + 4) + 5*x^2 + 2*x*log(log(5)^2 - x + e^x + 4) + 25*x)/log(log(5)^2 - x + e^x + 4)

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maple [A]  time = 0.04, size = 28, normalized size = 1.04

method	result	size
risch	\(x^{3}+2 x +\frac {5 \left (5+x \right ) x}{\ln \left ({\mathrm e}^{x}+\ln \relax (5)^{2}-x +4\right )}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((3*x^2+2)*exp(x)+(3*x^2+2)*ln(5)^2-3*x^3+12*x^2-2*x+8)*ln(exp(x)+ln(5)^2-x+4)^2+((10*x+25)*exp(x)+(10*x+
25)*ln(5)^2-10*x^2+15*x+100)*ln(exp(x)+ln(5)^2-x+4)+(-5*x^2-25*x)*exp(x)+5*x^2+25*x)/(exp(x)+ln(5)^2-x+4)/ln(e
xp(x)+ln(5)^2-x+4)^2,x,method=_RETURNVERBOSE)

[Out]

x^3+2*x+5*(5+x)*x/ln(exp(x)+ln(5)^2-x+4)

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maxima [A]  time = 0.61, size = 44, normalized size = 1.63 \begin {gather*} \frac {5 \, x^{2} + {\left (x^{3} + 2 \, x\right )} \log \left (\log \relax (5)^{2} - x + e^{x} + 4\right ) + 25 \, x}{\log \left (\log \relax (5)^{2} - x + e^{x} + 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x^2+2)*exp(x)+(3*x^2+2)*log(5)^2-3*x^3+12*x^2-2*x+8)*log(exp(x)+log(5)^2-x+4)^2+((10*x+25)*exp(
x)+(10*x+25)*log(5)^2-10*x^2+15*x+100)*log(exp(x)+log(5)^2-x+4)+(-5*x^2-25*x)*exp(x)+5*x^2+25*x)/(exp(x)+log(5
)^2-x+4)/log(exp(x)+log(5)^2-x+4)^2,x, algorithm="maxima")

[Out]

(5*x^2 + (x^3 + 2*x)*log(log(5)^2 - x + e^x + 4) + 25*x)/log(log(5)^2 - x + e^x + 4)

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mupad [B]  time = 1.17, size = 95, normalized size = 3.52 \begin {gather*} 12\,x+\frac {25\,x+10\,x\,{\ln \relax (5)}^2+25\,{\ln \relax (5)}^2-10\,x^2+125}{{\mathrm {e}}^x-1}+\frac {5\,x\,\left (x+5\right )-\frac {5\,\ln \left ({\mathrm {e}}^x-x+{\ln \relax (5)}^2+4\right )\,\left (2\,x+5\right )\,\left ({\mathrm {e}}^x-x+{\ln \relax (5)}^2+4\right )}{{\mathrm {e}}^x-1}}{\ln \left ({\mathrm {e}}^x-x+{\ln \relax (5)}^2+4\right )}+x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((25*x + log(exp(x) - x + log(5)^2 + 4)^2*(log(5)^2*(3*x^2 + 2) - 2*x + exp(x)*(3*x^2 + 2) + 12*x^2 - 3*x^3
+ 8) - exp(x)*(25*x + 5*x^2) + 5*x^2 + log(exp(x) - x + log(5)^2 + 4)*(15*x + log(5)^2*(10*x + 25) + exp(x)*(
10*x + 25) - 10*x^2 + 100))/(log(exp(x) - x + log(5)^2 + 4)^2*(exp(x) - x + log(5)^2 + 4)),x)

[Out]

12*x + (25*x + 10*x*log(5)^2 + 25*log(5)^2 - 10*x^2 + 125)/(exp(x) - 1) + (5*x*(x + 5) - (5*log(exp(x) - x + l
og(5)^2 + 4)*(2*x + 5)*(exp(x) - x + log(5)^2 + 4))/(exp(x) - 1))/log(exp(x) - x + log(5)^2 + 4) + x^3

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sympy [A]  time = 0.23, size = 27, normalized size = 1.00 \begin {gather*} x^{3} + 2 x + \frac {5 x^{2} + 25 x}{\log {\left (- x + e^{x} + \log {\relax (5 )}^{2} + 4 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((3*x**2+2)*exp(x)+(3*x**2+2)*ln(5)**2-3*x**3+12*x**2-2*x+8)*ln(exp(x)+ln(5)**2-x+4)**2+((10*x+25)*
exp(x)+(10*x+25)*ln(5)**2-10*x**2+15*x+100)*ln(exp(x)+ln(5)**2-x+4)+(-5*x**2-25*x)*exp(x)+5*x**2+25*x)/(exp(x)
+ln(5)**2-x+4)/ln(exp(x)+ln(5)**2-x+4)**2,x)

[Out]

x**3 + 2*x + (5*x**2 + 25*x)/log(-x + exp(x) + log(5)**2 + 4)

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