3.96.60 \(\int \frac {64+5 e^{\frac {8+x}{4}}+64 x}{4 e^{\frac {8+x}{4}}+64 x} \, dx\)

Optimal. Leaf size=19 \[ -\frac {1}{2}+x+\log \left (e^{2+\frac {x}{4}}+16 x\right ) \]

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Rubi [F]  time = 0.27, 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 {64+5 e^{\frac {8+x}{4}}+64 x}{4 e^{\frac {8+x}{4}}+64 x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(64 + 5*E^((8 + x)/4) + 64*x)/(4*E^((8 + x)/4) + 64*x),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {64+5 e^{\frac {8+x}{4}}+64 x}{4 \left (e^{2+\frac {x}{4}}+16 x\right )} \, dx\\ &=\frac {1}{4} \int \frac {64+5 e^{\frac {8+x}{4}}+64 x}{e^{2+\frac {x}{4}}+16 x} \, dx\\ &=\frac {1}{4} \int \left (5-\frac {16 (-4+x)}{e^{2+\frac {x}{4}}+16 x}\right ) \, dx\\ &=\frac {5 x}{4}-4 \int \frac {-4+x}{e^{2+\frac {x}{4}}+16 x} \, dx\\ &=\frac {5 x}{4}-4 \int \left (-\frac {4}{e^{2+\frac {x}{4}}+16 x}+\frac {x}{e^{2+\frac {x}{4}}+16 x}\right ) \, dx\\ &=\frac {5 x}{4}-4 \int \frac {x}{e^{2+\frac {x}{4}}+16 x} \, dx+16 \int \frac {1}{e^{2+\frac {x}{4}}+16 x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.10, size = 16, normalized size = 0.84 \begin {gather*} x+\log \left (e^{2+\frac {x}{4}}+16 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(64 + 5*E^((8 + x)/4) + 64*x)/(4*E^((8 + x)/4) + 64*x),x]

[Out]

x + Log[E^(2 + x/4) + 16*x]

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fricas [A]  time = 0.53, size = 13, normalized size = 0.68 \begin {gather*} x + \log \left (16 \, x + e^{\left (\frac {1}{4} \, x + 2\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*exp(2+1/4*x)+64*x+64)/(4*exp(2+1/4*x)+64*x),x, algorithm="fricas")

[Out]

x + log(16*x + e^(1/4*x + 2))

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giac [A]  time = 0.16, size = 13, normalized size = 0.68 \begin {gather*} x + \log \left (16 \, x + e^{\left (\frac {1}{4} \, x + 2\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*exp(2+1/4*x)+64*x+64)/(4*exp(2+1/4*x)+64*x),x, algorithm="giac")

[Out]

x + log(16*x + e^(1/4*x + 2))

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maple [A]  time = 0.10, size = 15, normalized size = 0.79

method	result	size
risch	\(x -2+\ln \left ({\mathrm e}^{2+\frac {x}{4}}+16 x \right )\)	\(15\)
norman	\(x +\ln \left (4 \,{\mathrm e}^{2+\frac {x}{4}}+64 x \right )\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*exp(2+1/4*x)+64*x+64)/(4*exp(2+1/4*x)+64*x),x,method=_RETURNVERBOSE)

[Out]

x-2+ln(exp(2+1/4*x)+16*x)

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maxima [A]  time = 0.40, size = 16, normalized size = 0.84 \begin {gather*} x + \log \left ({\left (16 \, x + e^{\left (\frac {1}{4} \, x + 2\right )}\right )} e^{\left (-2\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*exp(2+1/4*x)+64*x+64)/(4*exp(2+1/4*x)+64*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.12, size = 13, normalized size = 0.68 \begin {gather*} x+\ln \left (16\,x+{\mathrm {e}}^{\frac {x}{4}+2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((64*x + 5*exp(x/4 + 2) + 64)/(64*x + 4*exp(x/4 + 2)),x)

[Out]

x + log(16*x + exp(x/4 + 2))

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sympy [A]  time = 0.11, size = 17, normalized size = 0.89 \begin {gather*} \frac {19 x}{16} + \frac {\log {\left (16 x + e^{\frac {x}{4} + 2} \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*exp(2+1/4*x)+64*x+64)/(4*exp(2+1/4*x)+64*x),x)

[Out]

19*x/16 + log(16*x + exp(x/4 + 2))/4

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