3.22.20 \(\int \frac {-10+e^x (-2-x^2)+(5+e^x (1+x)) \log (5 x^2)}{100 x^2+40 e^x x^2+4 e^{2 x} x^2} \, dx\)

Optimal. Leaf size=26 \[ 9-\frac {-x+\log \left (5 x^2\right )}{4 \left (5+e^x\right ) x} \]

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Rubi [F]  time = 1.47, 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 {-10+e^x \left (-2-x^2\right )+\left (5+e^x (1+x)\right ) \log \left (5 x^2\right )}{100 x^2+40 e^x x^2+4 e^{2 x} x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-10 + E^x*(-2 - x^2) + (5 + E^x*(1 + x))*Log[5*x^2])/(100*x^2 + 40*E^x*x^2 + 4*E^(2*x)*x^2),x]

[Out]

1/(4*(5 + E^x)) - Defer[Int][1/((5 + E^x)*x^2), x]/2 + (Log[5*x^2]*Defer[Int][1/((5 + E^x)*x^2), x])/4 - (5*Lo
g[5*x^2]*Defer[Int][1/((5 + E^x)^2*x), x])/4 + (Log[5*x^2]*Defer[Int][1/((5 + E^x)*x), x])/4 - Defer[Int][Defe
r[Int][1/((5 + E^x)*x^2), x]/x, x]/2 + (5*Defer[Int][Defer[Int][1/((5 + E^x)^2*x), x]/x, x])/2 - Defer[Int][De
fer[Int][(5*x + E^x*x)^(-1), x]/x, x]/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-10+e^x \left (-2-x^2\right )+\left (5+e^x (1+x)\right ) \log \left (5 x^2\right )}{4 \left (5+e^x\right )^2 x^2} \, dx\\ &=\frac {1}{4} \int \frac {-10+e^x \left (-2-x^2\right )+\left (5+e^x (1+x)\right ) \log \left (5 x^2\right )}{\left (5+e^x\right )^2 x^2} \, dx\\ &=\frac {1}{4} \int \left (\frac {5 \left (x-\log \left (5 x^2\right )\right )}{\left (5+e^x\right )^2 x}-\frac {2+x^2-\log \left (5 x^2\right )-x \log \left (5 x^2\right )}{\left (5+e^x\right ) x^2}\right ) \, dx\\ &=-\left (\frac {1}{4} \int \frac {2+x^2-\log \left (5 x^2\right )-x \log \left (5 x^2\right )}{\left (5+e^x\right ) x^2} \, dx\right )+\frac {5}{4} \int \frac {x-\log \left (5 x^2\right )}{\left (5+e^x\right )^2 x} \, dx\\ &=-\left (\frac {1}{4} \int \frac {2+x^2-(1+x) \log \left (5 x^2\right )}{\left (5+e^x\right ) x^2} \, dx\right )+\frac {5}{4} \int \left (\frac {1}{\left (5+e^x\right )^2}-\frac {\log \left (5 x^2\right )}{\left (5+e^x\right )^2 x}\right ) \, dx\\ &=-\left (\frac {1}{4} \int \left (\frac {1}{5+e^x}+\frac {2}{\left (5+e^x\right ) x^2}-\frac {\log \left (5 x^2\right )}{\left (5+e^x\right ) x^2}-\frac {\log \left (5 x^2\right )}{\left (5+e^x\right ) x}\right ) \, dx\right )+\frac {5}{4} \int \frac {1}{\left (5+e^x\right )^2} \, dx-\frac {5}{4} \int \frac {\log \left (5 x^2\right )}{\left (5+e^x\right )^2 x} \, dx\\ &=-\left (\frac {1}{4} \int \frac {1}{5+e^x} \, dx\right )+\frac {1}{4} \int \frac {\log \left (5 x^2\right )}{\left (5+e^x\right ) x^2} \, dx+\frac {1}{4} \int \frac {\log \left (5 x^2\right )}{\left (5+e^x\right ) x} \, dx-\frac {1}{2} \int \frac {1}{\left (5+e^x\right ) x^2} \, dx+\frac {5}{4} \int \frac {2 \int \frac {1}{\left (5+e^x\right )^2 x} \, dx}{x} \, dx+\frac {5}{4} \operatorname {Subst}\left (\int \frac {1}{x (5+x)^2} \, dx,x,e^x\right )-\frac {1}{4} \left (5 \log \left (5 x^2\right )\right ) \int \frac {1}{\left (5+e^x\right )^2 x} \, dx\\ &=-\left (\frac {1}{4} \int \frac {2 \int \frac {1}{\left (5+e^x\right ) x^2} \, dx}{x} \, dx\right )-\frac {1}{4} \int \frac {2 \int \frac {1}{5 x+e^x x} \, dx}{x} \, dx-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{x (5+x)} \, dx,x,e^x\right )-\frac {1}{2} \int \frac {1}{\left (5+e^x\right ) x^2} \, dx+\frac {5}{4} \operatorname {Subst}\left (\int \left (\frac {1}{25 x}-\frac {1}{5 (5+x)^2}-\frac {1}{25 (5+x)}\right ) \, dx,x,e^x\right )+\frac {5}{2} \int \frac {\int \frac {1}{\left (5+e^x\right )^2 x} \, dx}{x} \, dx+\frac {1}{4} \log \left (5 x^2\right ) \int \frac {1}{\left (5+e^x\right ) x^2} \, dx+\frac {1}{4} \log \left (5 x^2\right ) \int \frac {1}{\left (5+e^x\right ) x} \, dx-\frac {1}{4} \left (5 \log \left (5 x^2\right )\right ) \int \frac {1}{\left (5+e^x\right )^2 x} \, dx\\ &=\frac {1}{4 \left (5+e^x\right )}+\frac {x}{20}-\frac {1}{20} \log \left (5+e^x\right )-\frac {1}{20} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )+\frac {1}{20} \operatorname {Subst}\left (\int \frac {1}{5+x} \, dx,x,e^x\right )-\frac {1}{2} \int \frac {1}{\left (5+e^x\right ) x^2} \, dx-\frac {1}{2} \int \frac {\int \frac {1}{\left (5+e^x\right ) x^2} \, dx}{x} \, dx-\frac {1}{2} \int \frac {\int \frac {1}{5 x+e^x x} \, dx}{x} \, dx+\frac {5}{2} \int \frac {\int \frac {1}{\left (5+e^x\right )^2 x} \, dx}{x} \, dx+\frac {1}{4} \log \left (5 x^2\right ) \int \frac {1}{\left (5+e^x\right ) x^2} \, dx+\frac {1}{4} \log \left (5 x^2\right ) \int \frac {1}{\left (5+e^x\right ) x} \, dx-\frac {1}{4} \left (5 \log \left (5 x^2\right )\right ) \int \frac {1}{\left (5+e^x\right )^2 x} \, dx\\ &=\frac {1}{4 \left (5+e^x\right )}-\frac {1}{2} \int \frac {1}{\left (5+e^x\right ) x^2} \, dx-\frac {1}{2} \int \frac {\int \frac {1}{\left (5+e^x\right ) x^2} \, dx}{x} \, dx-\frac {1}{2} \int \frac {\int \frac {1}{5 x+e^x x} \, dx}{x} \, dx+\frac {5}{2} \int \frac {\int \frac {1}{\left (5+e^x\right )^2 x} \, dx}{x} \, dx+\frac {1}{4} \log \left (5 x^2\right ) \int \frac {1}{\left (5+e^x\right ) x^2} \, dx+\frac {1}{4} \log \left (5 x^2\right ) \int \frac {1}{\left (5+e^x\right ) x} \, dx-\frac {1}{4} \left (5 \log \left (5 x^2\right )\right ) \int \frac {1}{\left (5+e^x\right )^2 x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.27, size = 25, normalized size = 0.96 \begin {gather*} \frac {x-\log \left (5 x^2\right )}{4 \left (5 x+e^x x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10 + E^x*(-2 - x^2) + (5 + E^x*(1 + x))*Log[5*x^2])/(100*x^2 + 40*E^x*x^2 + 4*E^(2*x)*x^2),x]

[Out]

(x - Log[5*x^2])/(4*(5*x + E^x*x))

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fricas [A]  time = 0.64, size = 22, normalized size = 0.85 \begin {gather*} \frac {x - \log \left (5 \, x^{2}\right )}{4 \, {\left (x e^{x} + 5 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(x)+5)*log(5*x^2)+(-x^2-2)*exp(x)-10)/(4*exp(x)^2*x^2+40*exp(x)*x^2+100*x^2),x, algorithm
="fricas")

[Out]

1/4*(x - log(5*x^2))/(x*e^x + 5*x)

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giac [A]  time = 0.23, size = 22, normalized size = 0.85 \begin {gather*} \frac {x - \log \left (5 \, x^{2}\right )}{4 \, {\left (x e^{x} + 5 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(x)+5)*log(5*x^2)+(-x^2-2)*exp(x)-10)/(4*exp(x)^2*x^2+40*exp(x)*x^2+100*x^2),x, algorithm
="giac")

[Out]

1/4*(x - log(5*x^2))/(x*e^x + 5*x)

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maple [A]  time = 0.14, size = 22, normalized size = 0.85

method	result	size
default	\(\frac {x -\ln \left (5 x^{2}\right )}{4 x \left ({\mathrm e}^{x}+5\right )}\)	\(22\)
norman	\(\frac {\frac {x}{4}-\frac {\ln \left (5 x^{2}\right )}{4}}{\left ({\mathrm e}^{x}+5\right ) x}\)	\(23\)
risch	\(-\frac {\ln \relax (x )}{2 x \left ({\mathrm e}^{x}+5\right )}-\frac {-i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}+2 i \pi \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )-i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 \ln \relax (5)-2 x}{8 x \left ({\mathrm e}^{x}+5\right )}\)	\(83\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((x+1)*exp(x)+5)*ln(5*x^2)+(-x^2-2)*exp(x)-10)/(4*exp(x)^2*x^2+40*exp(x)*x^2+100*x^2),x,method=_RETURNVER
BOSE)

[Out]

1/4*(x-ln(5*x^2))/x/(exp(x)+5)

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maxima [A]  time = 0.58, size = 22, normalized size = 0.85 \begin {gather*} \frac {x - \log \relax (5) - 2 \, \log \relax (x)}{4 \, {\left (x e^{x} + 5 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(x)+5)*log(5*x^2)+(-x^2-2)*exp(x)-10)/(4*exp(x)^2*x^2+40*exp(x)*x^2+100*x^2),x, algorithm
="maxima")

[Out]

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

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mupad [B]  time = 1.29, size = 21, normalized size = 0.81 \begin {gather*} \frac {x-\ln \left (5\,x^2\right )}{4\,x\,\left ({\mathrm {e}}^x+5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(x^2 + 2) - log(5*x^2)*(exp(x)*(x + 1) + 5) + 10)/(40*x^2*exp(x) + 4*x^2*exp(2*x) + 100*x^2),x)

[Out]

(x - log(5*x^2))/(4*x*(exp(x) + 5))

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sympy [A]  time = 0.26, size = 17, normalized size = 0.65 \begin {gather*} \frac {x - \log {\left (5 x^{2} \right )}}{4 x e^{x} + 20 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((x+1)*exp(x)+5)*ln(5*x**2)+(-x**2-2)*exp(x)-10)/(4*exp(x)**2*x**2+40*exp(x)*x**2+100*x**2),x)

[Out]

(x - log(5*x**2))/(4*x*exp(x) + 20*x)

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