3.12.99 \(\int \frac {10 e^x x-10 x^2+(-25 x-10 x^2+e^x (25+10 x)) \log (5+2 x)+e^{-x^3+x^2 \log (-5 e^x+5 x)} (2 e^x-2 x+(-5 x^2+13 x^3+6 x^4+e^x (-10 x^2-4 x^3)) \log (5+2 x)+(-10 x^2-4 x^3+e^x (10 x+4 x^2)) \log (5+2 x) \log (-5 e^x+5 x))}{-5 x-2 x^2+e^x (5+2 x)} \, dx\)

Optimal. Leaf size=31 \[ \left (e^{x^2 \left (-x+\log \left (5 \left (-e^x+x\right )\right )\right )}+5 x\right ) \log (5+2 x) \]

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Rubi [F]  time = 180.00, 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*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

Int[(10*E^x*x - 10*x^2 + (-25*x - 10*x^2 + E^x*(25 + 10*x))*Log[5 + 2*x] + E^(-x^3 + x^2*Log[-5*E^x + 5*x])*(2
*E^x - 2*x + (-5*x^2 + 13*x^3 + 6*x^4 + E^x*(-10*x^2 - 4*x^3))*Log[5 + 2*x] + (-10*x^2 - 4*x^3 + E^x*(10*x + 4
*x^2))*Log[5 + 2*x]*Log[-5*E^x + 5*x]))/(-5*x - 2*x^2 + E^x*(5 + 2*x)),x]

[Out]

$Aborted

Rubi steps

Aborted

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Mathematica [A]  time = 0.27, size = 37, normalized size = 1.19 \begin {gather*} 5 x \log (5+2 x)+e^{-x^3} \left (-5 e^x+5 x\right )^{x^2} \log (5+2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10*E^x*x - 10*x^2 + (-25*x - 10*x^2 + E^x*(25 + 10*x))*Log[5 + 2*x] + E^(-x^3 + x^2*Log[-5*E^x + 5*
x])*(2*E^x - 2*x + (-5*x^2 + 13*x^3 + 6*x^4 + E^x*(-10*x^2 - 4*x^3))*Log[5 + 2*x] + (-10*x^2 - 4*x^3 + E^x*(10
*x + 4*x^2))*Log[5 + 2*x]*Log[-5*E^x + 5*x]))/(-5*x - 2*x^2 + E^x*(5 + 2*x)),x]

[Out]

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

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fricas [A]  time = 0.81, size = 37, normalized size = 1.19 \begin {gather*} 5 \, x \log \left (2 \, x + 5\right ) + e^{\left (-x^{3} + x^{2} \log \left (5 \, x - 5 \, e^{x}\right )\right )} \log \left (2 \, x + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((4*x^2+10*x)*exp(x)-4*x^3-10*x^2)*log(5+2*x)*log(-5*exp(x)+5*x)+((-4*x^3-10*x^2)*exp(x)+6*x^4+13*
x^3-5*x^2)*log(5+2*x)+2*exp(x)-2*x)*exp(x^2*log(-5*exp(x)+5*x)-x^3)+((10*x+25)*exp(x)-10*x^2-25*x)*log(5+2*x)+
10*exp(x)*x-10*x^2)/((5+2*x)*exp(x)-2*x^2-5*x),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {10 \, x^{2} + {\left (2 \, {\left (2 \, x^{3} + 5 \, x^{2} - {\left (2 \, x^{2} + 5 \, x\right )} e^{x}\right )} \log \left (5 \, x - 5 \, e^{x}\right ) \log \left (2 \, x + 5\right ) - {\left (6 \, x^{4} + 13 \, x^{3} - 5 \, x^{2} - 2 \, {\left (2 \, x^{3} + 5 \, x^{2}\right )} e^{x}\right )} \log \left (2 \, x + 5\right ) + 2 \, x - 2 \, e^{x}\right )} e^{\left (-x^{3} + x^{2} \log \left (5 \, x - 5 \, e^{x}\right )\right )} - 10 \, x e^{x} + 5 \, {\left (2 \, x^{2} - {\left (2 \, x + 5\right )} e^{x} + 5 \, x\right )} \log \left (2 \, x + 5\right )}{2 \, x^{2} - {\left (2 \, x + 5\right )} e^{x} + 5 \, x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((4*x^2+10*x)*exp(x)-4*x^3-10*x^2)*log(5+2*x)*log(-5*exp(x)+5*x)+((-4*x^3-10*x^2)*exp(x)+6*x^4+13*
x^3-5*x^2)*log(5+2*x)+2*exp(x)-2*x)*exp(x^2*log(-5*exp(x)+5*x)-x^3)+((10*x+25)*exp(x)-10*x^2-25*x)*log(5+2*x)+
10*exp(x)*x-10*x^2)/((5+2*x)*exp(x)-2*x^2-5*x),x, algorithm="giac")

[Out]

integrate((10*x^2 + (2*(2*x^3 + 5*x^2 - (2*x^2 + 5*x)*e^x)*log(5*x - 5*e^x)*log(2*x + 5) - (6*x^4 + 13*x^3 - 5
*x^2 - 2*(2*x^3 + 5*x^2)*e^x)*log(2*x + 5) + 2*x - 2*e^x)*e^(-x^3 + x^2*log(5*x - 5*e^x)) - 10*x*e^x + 5*(2*x^
2 - (2*x + 5)*e^x + 5*x)*log(2*x + 5))/(2*x^2 - (2*x + 5)*e^x + 5*x), x)

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maple [A]  time = 0.05, size = 36, normalized size = 1.16

method	result	size
risch	\(5 x \ln \left (5+2 x \right )+\ln \left (5+2 x \right ) \left (-5 \,{\mathrm e}^{x}+5 x \right )^{x^{2}} {\mathrm e}^{-x^{3}}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((4*x^2+10*x)*exp(x)-4*x^3-10*x^2)*ln(5+2*x)*ln(-5*exp(x)+5*x)+((-4*x^3-10*x^2)*exp(x)+6*x^4+13*x^3-5*x^
2)*ln(5+2*x)+2*exp(x)-2*x)*exp(x^2*ln(-5*exp(x)+5*x)-x^3)+((10*x+25)*exp(x)-10*x^2-25*x)*ln(5+2*x)+10*exp(x)*x
-10*x^2)/((5+2*x)*exp(x)-2*x^2-5*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 5 \, x \log \left (2 \, x + 5\right ) + \int \frac {{\left (2 \, {\left (2 \, x^{3} + 5 \, x^{2} - {\left (2 \, x^{2} + 5 \, x\right )} e^{x}\right )} 5^{\left (x^{2}\right )} \log \left (2 \, x + 5\right ) \log \left (-x + e^{x}\right ) - {\left ({\left ({\left (-4 i \, \pi - 4 \, \log \relax (5) + 13\right )} x^{3} + 6 \, x^{4} - 5 \, {\left (2 i \, \pi + 2 \, \log \relax (5) + 1\right )} x^{2} - 2 \, {\left ({\left (-2 i \, \pi - 2 \, \log \relax (5) + 5\right )} x^{2} + 2 \, x^{3} + 5 \, {\left (-i \, \pi - \log \relax (5)\right )} x\right )} e^{x}\right )} \log \left (2 \, x + 5\right ) - 2 \, x + 2 \, e^{x}\right )} 5^{\left (x^{2}\right )}\right )} e^{\left (-x^{3} + x^{2} \log \left (x - e^{x}\right )\right )}}{2 \, x^{2} - {\left (2 \, x + 5\right )} e^{x} + 5 \, x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((4*x^2+10*x)*exp(x)-4*x^3-10*x^2)*log(5+2*x)*log(-5*exp(x)+5*x)+((-4*x^3-10*x^2)*exp(x)+6*x^4+13*
x^3-5*x^2)*log(5+2*x)+2*exp(x)-2*x)*exp(x^2*log(-5*exp(x)+5*x)-x^3)+((10*x+25)*exp(x)-10*x^2-25*x)*log(5+2*x)+
10*exp(x)*x-10*x^2)/((5+2*x)*exp(x)-2*x^2-5*x),x, algorithm="maxima")

[Out]

5*x*log(2*x + 5) + integrate((2*(2*x^3 + 5*x^2 - (2*x^2 + 5*x)*e^x)*5^(x^2)*log(2*x + 5)*log(-x + e^x) - (((-4
*I*pi - 4*log(5) + 13)*x^3 + 6*x^4 - 5*(2*I*pi + 2*log(5) + 1)*x^2 - 2*((-2*I*pi - 2*log(5) + 5)*x^2 + 2*x^3 +
5*(-I*pi - log(5))*x)*e^x)*log(2*x + 5) - 2*x + 2*e^x)*5^(x^2))*e^(-x^3 + x^2*log(x - e^x))/(2*x^2 - (2*x + 5
)*e^x + 5*x), x)

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mupad [B]  time = 1.17, size = 30, normalized size = 0.97 \begin {gather*} \ln \left (2\,x+5\right )\,\left (5\,x+{\mathrm {e}}^{-x^3}\,{\left (5\,x-5\,{\mathrm {e}}^x\right )}^{x^2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x^2*log(5*x - 5*exp(x)) - x^3)*(2*x - 2*exp(x) + log(2*x + 5)*(exp(x)*(10*x^2 + 4*x^3) + 5*x^2 - 13*x
^3 - 6*x^4) + log(2*x + 5)*log(5*x - 5*exp(x))*(10*x^2 - exp(x)*(10*x + 4*x^2) + 4*x^3)) - 10*x*exp(x) + 10*x^
2 + log(2*x + 5)*(25*x - exp(x)*(10*x + 25) + 10*x^2))/(5*x - exp(x)*(2*x + 5) + 2*x^2),x)

[Out]

log(2*x + 5)*(5*x + exp(-x^3)*(5*x - 5*exp(x))^(x^2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((4*x**2+10*x)*exp(x)-4*x**3-10*x**2)*ln(5+2*x)*ln(-5*exp(x)+5*x)+((-4*x**3-10*x**2)*exp(x)+6*x**4
+13*x**3-5*x**2)*ln(5+2*x)+2*exp(x)-2*x)*exp(x**2*ln(-5*exp(x)+5*x)-x**3)+((10*x+25)*exp(x)-10*x**2-25*x)*ln(5
+2*x)+10*exp(x)*x-10*x**2)/((5+2*x)*exp(x)-2*x**2-5*x),x)

[Out]

Timed out

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