3.48.99 \(\int \frac {e^{2 x} (-e^{10}+16 x+15 x^2+e^5 (8+14 x))+e^{2 x} (x^2+2 x^3+e^{10} (1+2 x)+e^5 (2 x+4 x^2)) \log (x)}{64+(16 e^5+16 x) \log (x)+(e^{10}+2 e^5 x+x^2) \log ^2(x)} \, dx\)

Optimal. Leaf size=21 \[ \frac {e^{2 x} x}{\frac {8}{e^5+x}+\log (x)} \]

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Rubi [F]  time = 1.52, 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^{2 x} \left (-e^{10}+16 x+15 x^2+e^5 (8+14 x)\right )+e^{2 x} \left (x^2+2 x^3+e^{10} (1+2 x)+e^5 \left (2 x+4 x^2\right )\right ) \log (x)}{64+\left (16 e^5+16 x\right ) \log (x)+\left (e^{10}+2 e^5 x+x^2\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(2*x)*(-E^10 + 16*x + 15*x^2 + E^5*(8 + 14*x)) + E^(2*x)*(x^2 + 2*x^3 + E^10*(1 + 2*x) + E^5*(2*x + 4*x
^2))*Log[x])/(64 + (16*E^5 + 16*x)*Log[x] + (E^10 + 2*E^5*x + x^2)*Log[x]^2),x]

[Out]

-Defer[Int][E^(10 + 2*x)/(8 + E^5*Log[x] + x*Log[x])^2, x] + 2*(4 - E^5)*Defer[Int][(E^(2*x)*x)/(8 + E^5*Log[x
] + x*Log[x])^2, x] - Defer[Int][(E^(2*x)*x^2)/(8 + E^5*Log[x] + x*Log[x])^2, x] + Defer[Int][E^(5 + 2*x)/(8 +
E^5*Log[x] + x*Log[x]), x] + (1 + 2*E^5)*Defer[Int][(E^(2*x)*x)/(8 + E^5*Log[x] + x*Log[x]), x] + 2*Defer[Int
][(E^(2*x)*x^2)/(8 + E^5*Log[x] + x*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{2 x} \left (-e^{10}+2 e^5 (4+7 x)+x (16+15 x)+\left (e^5+x\right )^2 (1+2 x) \log (x)\right )}{\left (8+\left (e^5+x\right ) \log (x)\right )^2} \, dx\\ &=\int \left (\frac {e^{2 x} \left (-e^{10}+2 \left (4-e^5\right ) x-x^2\right )}{\left (8+e^5 \log (x)+x \log (x)\right )^2}+\frac {e^{2 x} \left (e^5+\left (1+2 e^5\right ) x+2 x^2\right )}{8+e^5 \log (x)+x \log (x)}\right ) \, dx\\ &=\int \frac {e^{2 x} \left (-e^{10}+2 \left (4-e^5\right ) x-x^2\right )}{\left (8+e^5 \log (x)+x \log (x)\right )^2} \, dx+\int \frac {e^{2 x} \left (e^5+\left (1+2 e^5\right ) x+2 x^2\right )}{8+e^5 \log (x)+x \log (x)} \, dx\\ &=\int \left (-\frac {e^{10+2 x}}{\left (8+e^5 \log (x)+x \log (x)\right )^2}-\frac {2 e^{2 x} \left (-4+e^5\right ) x}{\left (8+e^5 \log (x)+x \log (x)\right )^2}-\frac {e^{2 x} x^2}{\left (8+e^5 \log (x)+x \log (x)\right )^2}\right ) \, dx+\int \left (\frac {e^{5+2 x}}{8+e^5 \log (x)+x \log (x)}+\frac {e^{2 x} \left (1+2 e^5\right ) x}{8+e^5 \log (x)+x \log (x)}+\frac {2 e^{2 x} x^2}{8+e^5 \log (x)+x \log (x)}\right ) \, dx\\ &=2 \int \frac {e^{2 x} x^2}{8+e^5 \log (x)+x \log (x)} \, dx+\left (2 \left (4-e^5\right )\right ) \int \frac {e^{2 x} x}{\left (8+e^5 \log (x)+x \log (x)\right )^2} \, dx+\left (1+2 e^5\right ) \int \frac {e^{2 x} x}{8+e^5 \log (x)+x \log (x)} \, dx-\int \frac {e^{10+2 x}}{\left (8+e^5 \log (x)+x \log (x)\right )^2} \, dx-\int \frac {e^{2 x} x^2}{\left (8+e^5 \log (x)+x \log (x)\right )^2} \, dx+\int \frac {e^{5+2 x}}{8+e^5 \log (x)+x \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.45, size = 26, normalized size = 1.24 \begin {gather*} \frac {e^{2 x} x \left (e^5+x\right )}{8+e^5 \log (x)+x \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*x)*(-E^10 + 16*x + 15*x^2 + E^5*(8 + 14*x)) + E^(2*x)*(x^2 + 2*x^3 + E^10*(1 + 2*x) + E^5*(2*x
+ 4*x^2))*Log[x])/(64 + (16*E^5 + 16*x)*Log[x] + (E^10 + 2*E^5*x + x^2)*Log[x]^2),x]

[Out]

(E^(2*x)*x*(E^5 + x))/(8 + E^5*Log[x] + x*Log[x])

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fricas [A]  time = 1.44, size = 24, normalized size = 1.14 \begin {gather*} \frac {{\left (x^{2} + x e^{5}\right )} e^{\left (2 \, x\right )}}{{\left (x + e^{5}\right )} \log \relax (x) + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x+1)*exp(5)^2+(4*x^2+2*x)*exp(5)+2*x^3+x^2)*exp(x)^2*log(x)+(-exp(5)^2+(14*x+8)*exp(5)+15*x^2+1
6*x)*exp(x)^2)/((exp(5)^2+2*x*exp(5)+x^2)*log(x)^2+(16*exp(5)+16*x)*log(x)+64),x, algorithm="fricas")

[Out]

(x^2 + x*e^5)*e^(2*x)/((x + e^5)*log(x) + 8)

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giac [A]  time = 0.30, size = 31, normalized size = 1.48 \begin {gather*} \frac {x^{2} e^{\left (2 \, x\right )} + x e^{\left (2 \, x + 5\right )}}{x \log \relax (x) + e^{5} \log \relax (x) + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x+1)*exp(5)^2+(4*x^2+2*x)*exp(5)+2*x^3+x^2)*exp(x)^2*log(x)+(-exp(5)^2+(14*x+8)*exp(5)+15*x^2+1
6*x)*exp(x)^2)/((exp(5)^2+2*x*exp(5)+x^2)*log(x)^2+(16*exp(5)+16*x)*log(x)+64),x, algorithm="giac")

[Out]

(x^2*e^(2*x) + x*e^(2*x + 5))/(x*log(x) + e^5*log(x) + 8)

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maple [A]  time = 0.05, size = 24, normalized size = 1.14

method	result	size
risch	\(\frac {x \,{\mathrm e}^{2 x} \left ({\mathrm e}^{5}+x \right )}{{\mathrm e}^{5} \ln \relax (x )+x \ln \relax (x )+8}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x+1)*exp(5)^2+(4*x^2+2*x)*exp(5)+2*x^3+x^2)*exp(x)^2*ln(x)+(-exp(5)^2+(14*x+8)*exp(5)+15*x^2+16*x)*ex
p(x)^2)/((exp(5)^2+2*x*exp(5)+x^2)*ln(x)^2+(16*exp(5)+16*x)*ln(x)+64),x,method=_RETURNVERBOSE)

[Out]

x*exp(2*x)*(exp(5)+x)/(exp(5)*ln(x)+x*ln(x)+8)

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maxima [A]  time = 0.41, size = 24, normalized size = 1.14 \begin {gather*} \frac {{\left (x^{2} + x e^{5}\right )} e^{\left (2 \, x\right )}}{{\left (x + e^{5}\right )} \log \relax (x) + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x+1)*exp(5)^2+(4*x^2+2*x)*exp(5)+2*x^3+x^2)*exp(x)^2*log(x)+(-exp(5)^2+(14*x+8)*exp(5)+15*x^2+1
6*x)*exp(x)^2)/((exp(5)^2+2*x*exp(5)+x^2)*log(x)^2+(16*exp(5)+16*x)*log(x)+64),x, algorithm="maxima")

[Out]

(x^2 + x*e^5)*e^(2*x)/((x + e^5)*log(x) + 8)

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mupad [B]  time = 4.35, size = 23, normalized size = 1.10 \begin {gather*} \frac {x\,{\mathrm {e}}^{2\,x}\,\left (x+{\mathrm {e}}^5\right )}{{\mathrm {e}}^5\,\ln \relax (x)+x\,\ln \relax (x)+8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x)*(16*x - exp(10) + 15*x^2 + exp(5)*(14*x + 8)) + exp(2*x)*log(x)*(exp(5)*(2*x + 4*x^2) + x^2 + 2*
x^3 + exp(10)*(2*x + 1)))/(log(x)*(16*x + 16*exp(5)) + log(x)^2*(exp(10) + 2*x*exp(5) + x^2) + 64),x)

[Out]

(x*exp(2*x)*(x + exp(5)))/(exp(5)*log(x) + x*log(x) + 8)

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sympy [A]  time = 0.37, size = 26, normalized size = 1.24 \begin {gather*} \frac {\left (x^{2} + x e^{5}\right ) e^{2 x}}{x \log {\relax (x )} + e^{5} \log {\relax (x )} + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x+1)*exp(5)**2+(4*x**2+2*x)*exp(5)+2*x**3+x**2)*exp(x)**2*ln(x)+(-exp(5)**2+(14*x+8)*exp(5)+15*
x**2+16*x)*exp(x)**2)/((exp(5)**2+2*x*exp(5)+x**2)*ln(x)**2+(16*exp(5)+16*x)*ln(x)+64),x)

[Out]

(x**2 + x*exp(5))*exp(2*x)/(x*log(x) + exp(5)*log(x) + 8)

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