3.92.36 \(\int \frac {15 e^{32}+40 e^{16} x+15 x^2+e^x (12 e^{32}+12 x^2+e^{16} (8+24 x))}{15 e^{32} x+30 e^{16} x^2+15 x^3+e^x (12 e^{32}+24 e^{16} x+12 x^2)} \, dx\)

Optimal. Leaf size=23 \[ \frac {2 x}{3 \left (e^{16}+x\right )}+\log \left (\frac {4 e^x}{5}+x\right ) \]

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Rubi [F]  time = 0.54, 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 {15 e^{32}+40 e^{16} x+15 x^2+e^x \left (12 e^{32}+12 x^2+e^{16} (8+24 x)\right )}{15 e^{32} x+30 e^{16} x^2+15 x^3+e^x \left (12 e^{32}+24 e^{16} x+12 x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(15*E^32 + 40*E^16*x + 15*x^2 + E^x*(12*E^32 + 12*x^2 + E^16*(8 + 24*x)))/(15*E^32*x + 30*E^16*x^2 + 15*x^
3 + E^x*(12*E^32 + 24*E^16*x + 12*x^2)),x]

[Out]

x - (2*E^16)/(3*(E^16 + x)) + 5*Defer[Int][(4*E^x + 5*x)^(-1), x] - 5*Defer[Int][x/(4*E^x + 5*x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {15 e^{32}+40 e^{16} x+15 x^2+e^x \left (12 e^{32}+12 x^2+e^{16} (8+24 x)\right )}{3 \left (e^{16}+x\right )^2 \left (4 e^x+5 x\right )} \, dx\\ &=\frac {1}{3} \int \frac {15 e^{32}+40 e^{16} x+15 x^2+e^x \left (12 e^{32}+12 x^2+e^{16} (8+24 x)\right )}{\left (e^{16}+x\right )^2 \left (4 e^x+5 x\right )} \, dx\\ &=\frac {1}{3} \int \left (-\frac {15 (-1+x)}{4 e^x+5 x}+\frac {e^{16} \left (2+3 e^{16}\right )+6 e^{16} x+3 x^2}{\left (e^{16}+x\right )^2}\right ) \, dx\\ &=\frac {1}{3} \int \frac {e^{16} \left (2+3 e^{16}\right )+6 e^{16} x+3 x^2}{\left (e^{16}+x\right )^2} \, dx-5 \int \frac {-1+x}{4 e^x+5 x} \, dx\\ &=\frac {1}{3} \int \left (3+\frac {2 e^{16}}{\left (e^{16}+x\right )^2}\right ) \, dx-5 \int \left (-\frac {1}{4 e^x+5 x}+\frac {x}{4 e^x+5 x}\right ) \, dx\\ &=x-\frac {2 e^{16}}{3 \left (e^{16}+x\right )}+5 \int \frac {1}{4 e^x+5 x} \, dx-5 \int \frac {x}{4 e^x+5 x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 29, normalized size = 1.26 \begin {gather*} \frac {1}{3} \left (-\frac {2 e^{16}}{e^{16}+x}+3 \log \left (4 e^x+5 x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(15*E^32 + 40*E^16*x + 15*x^2 + E^x*(12*E^32 + 12*x^2 + E^16*(8 + 24*x)))/(15*E^32*x + 30*E^16*x^2 +
15*x^3 + E^x*(12*E^32 + 24*E^16*x + 12*x^2)),x]

[Out]

((-2*E^16)/(E^16 + x) + 3*Log[4*E^x + 5*x])/3

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fricas [A]  time = 0.52, size = 28, normalized size = 1.22 \begin {gather*} \frac {3 \, {\left (x + e^{16}\right )} \log \left (5 \, x + 4 \, e^{x}\right ) - 2 \, e^{16}}{3 \, {\left (x + e^{16}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*exp(16)^2+(24*x+8)*exp(16)+12*x^2)*exp(x)+15*exp(16)^2+40*x*exp(16)+15*x^2)/((12*exp(16)^2+24*x
*exp(16)+12*x^2)*exp(x)+15*x*exp(16)^2+30*x^2*exp(16)+15*x^3),x, algorithm="fricas")

[Out]

1/3*(3*(x + e^16)*log(5*x + 4*e^x) - 2*e^16)/(x + e^16)

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giac [B]  time = 0.20, size = 38, normalized size = 1.65 \begin {gather*} \frac {3 \, x \log \left (5 \, x + 4 \, e^{x}\right ) + 3 \, e^{16} \log \left (5 \, x + 4 \, e^{x}\right ) - 2 \, e^{16}}{3 \, {\left (x + e^{16}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*exp(16)^2+(24*x+8)*exp(16)+12*x^2)*exp(x)+15*exp(16)^2+40*x*exp(16)+15*x^2)/((12*exp(16)^2+24*x
*exp(16)+12*x^2)*exp(x)+15*x*exp(16)^2+30*x^2*exp(16)+15*x^3),x, algorithm="giac")

[Out]

1/3*(3*x*log(5*x + 4*e^x) + 3*e^16*log(5*x + 4*e^x) - 2*e^16)/(x + e^16)

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maple [A]  time = 0.13, size = 19, normalized size = 0.83

method	result	size
risch	\(-\frac {2 \,{\mathrm e}^{16}}{3 \left (x +{\mathrm e}^{16}\right )}+\ln \left (\frac {5 x}{4}+{\mathrm e}^{x}\right )\)	\(19\)
norman	\(-\frac {2 \,{\mathrm e}^{16}}{3 \left (x +{\mathrm e}^{16}\right )}+\ln \left (5 x +4 \,{\mathrm e}^{x}\right )\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((12*exp(16)^2+(24*x+8)*exp(16)+12*x^2)*exp(x)+15*exp(16)^2+40*x*exp(16)+15*x^2)/((12*exp(16)^2+24*x*exp(1
6)+12*x^2)*exp(x)+15*x*exp(16)^2+30*x^2*exp(16)+15*x^3),x,method=_RETURNVERBOSE)

[Out]

-2/3*exp(16)/(x+exp(16))+ln(5/4*x+exp(x))

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maxima [A]  time = 0.42, size = 18, normalized size = 0.78 \begin {gather*} -\frac {2 \, e^{16}}{3 \, {\left (x + e^{16}\right )}} + \log \left (\frac {5}{4} \, x + e^{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*exp(16)^2+(24*x+8)*exp(16)+12*x^2)*exp(x)+15*exp(16)^2+40*x*exp(16)+15*x^2)/((12*exp(16)^2+24*x
*exp(16)+12*x^2)*exp(x)+15*x*exp(16)^2+30*x^2*exp(16)+15*x^3),x, algorithm="maxima")

[Out]

-2/3*e^16/(x + e^16) + log(5/4*x + e^x)

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mupad [B]  time = 0.39, size = 21, normalized size = 0.91 \begin {gather*} \ln \left (x+\frac {4\,{\mathrm {e}}^x}{5}\right )+\frac {2\,x}{3\,x+3\,{\mathrm {e}}^{16}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((15*exp(32) + 40*x*exp(16) + exp(x)*(12*exp(32) + 12*x^2 + exp(16)*(24*x + 8)) + 15*x^2)/(15*x*exp(32) + 3
0*x^2*exp(16) + exp(x)*(12*exp(32) + 24*x*exp(16) + 12*x^2) + 15*x^3),x)

[Out]

log(x + (4*exp(x))/5) + (2*x)/(3*x + 3*exp(16))

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sympy [A]  time = 0.22, size = 22, normalized size = 0.96 \begin {gather*} \log {\left (\frac {5 x}{4} + e^{x} \right )} - \frac {2 e^{16}}{3 x + 3 e^{16}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((12*exp(16)**2+(24*x+8)*exp(16)+12*x**2)*exp(x)+15*exp(16)**2+40*x*exp(16)+15*x**2)/((12*exp(16)**2
+24*x*exp(16)+12*x**2)*exp(x)+15*x*exp(16)**2+30*x**2*exp(16)+15*x**3),x)

[Out]

log(5*x/4 + exp(x)) - 2*exp(16)/(3*x + 3*exp(16))

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