3.47.5 \(\int \frac {25 e^{4 x}+256 x^2+e^{2 x} (160 x+(8-16 x) \log (2))}{25 e^{4 x}+160 e^{2 x} x+256 x^2} \, dx\)

Optimal. Leaf size=19 \[ 30+x-\frac {\log (2)}{10+32 e^{-2 x} x} \]

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Rubi [F]  time = 0.75, 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 {25 e^{4 x}+256 x^2+e^{2 x} (160 x+(8-16 x) \log (2))}{25 e^{4 x}+160 e^{2 x} x+256 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(25*E^(4*x) + 256*x^2 + E^(2*x)*(160*x + (8 - 16*x)*Log[2]))/(25*E^(4*x) + 160*E^(2*x)*x + 256*x^2),x]

[Out]

x - (128*Log[2]*Defer[Int][x/(5*E^(2*x) + 16*x)^2, x])/5 + (256*Log[2]*Defer[Int][x^2/(5*E^(2*x) + 16*x)^2, x]
)/5 + (4*Log[2]*Defer[Subst][Defer[Int][(5*E^x + 8*x)^(-1), x], x, 2*x])/5 - (4*Log[2]*Defer[Subst][Defer[Int]
[x/(5*E^x + 8*x), x], x, 2*x])/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {25 e^{4 x}+256 x^2+e^{2 x} (160 x+(8-16 x) \log (2))}{\left (5 e^{2 x}+16 x\right )^2} \, dx\\ &=\int \left (1+\frac {128 x (-1+2 x) \log (2)}{5 \left (5 e^{2 x}+16 x\right )^2}-\frac {8 (-1+2 x) \log (2)}{5 \left (5 e^{2 x}+16 x\right )}\right ) \, dx\\ &=x-\frac {1}{5} (8 \log (2)) \int \frac {-1+2 x}{5 e^{2 x}+16 x} \, dx+\frac {1}{5} (128 \log (2)) \int \frac {x (-1+2 x)}{\left (5 e^{2 x}+16 x\right )^2} \, dx\\ &=x-\frac {1}{5} (4 \log (2)) \operatorname {Subst}\left (\int \frac {-1+x}{5 e^x+8 x} \, dx,x,2 x\right )+\frac {1}{5} (128 \log (2)) \int \left (-\frac {x}{\left (5 e^{2 x}+16 x\right )^2}+\frac {2 x^2}{\left (5 e^{2 x}+16 x\right )^2}\right ) \, dx\\ &=x-\frac {1}{5} (4 \log (2)) \operatorname {Subst}\left (\int \left (-\frac {1}{5 e^x+8 x}+\frac {x}{5 e^x+8 x}\right ) \, dx,x,2 x\right )-\frac {1}{5} (128 \log (2)) \int \frac {x}{\left (5 e^{2 x}+16 x\right )^2} \, dx+\frac {1}{5} (256 \log (2)) \int \frac {x^2}{\left (5 e^{2 x}+16 x\right )^2} \, dx\\ &=x+\frac {1}{5} (4 \log (2)) \operatorname {Subst}\left (\int \frac {1}{5 e^x+8 x} \, dx,x,2 x\right )-\frac {1}{5} (4 \log (2)) \operatorname {Subst}\left (\int \frac {x}{5 e^x+8 x} \, dx,x,2 x\right )-\frac {1}{5} (128 \log (2)) \int \frac {x}{\left (5 e^{2 x}+16 x\right )^2} \, dx+\frac {1}{5} (256 \log (2)) \int \frac {x^2}{\left (5 e^{2 x}+16 x\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 22, normalized size = 1.16 \begin {gather*} x+\frac {8 x \log (2)}{5 \left (5 e^{2 x}+16 x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25*E^(4*x) + 256*x^2 + E^(2*x)*(160*x + (8 - 16*x)*Log[2]))/(25*E^(4*x) + 160*E^(2*x)*x + 256*x^2),
x]

[Out]

x + (8*x*Log[2])/(5*(5*E^(2*x) + 16*x))

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fricas [A]  time = 0.56, size = 32, normalized size = 1.68 \begin {gather*} \frac {80 \, x^{2} + 25 \, x e^{\left (2 \, x\right )} + 8 \, x \log \relax (2)}{5 \, {\left (16 \, x + 5 \, e^{\left (2 \, x\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*exp(2*x)^2+((-16*x+8)*log(2)+160*x)*exp(2*x)+256*x^2)/(25*exp(2*x)^2+160*x*exp(2*x)+256*x^2),x,
algorithm="fricas")

[Out]

1/5*(80*x^2 + 25*x*e^(2*x) + 8*x*log(2))/(16*x + 5*e^(2*x))

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giac [A]  time = 0.16, size = 32, normalized size = 1.68 \begin {gather*} \frac {80 \, x^{2} + 25 \, x e^{\left (2 \, x\right )} + 8 \, x \log \relax (2)}{5 \, {\left (16 \, x + 5 \, e^{\left (2 \, x\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*exp(2*x)^2+((-16*x+8)*log(2)+160*x)*exp(2*x)+256*x^2)/(25*exp(2*x)^2+160*x*exp(2*x)+256*x^2),x,
algorithm="giac")

[Out]

1/5*(80*x^2 + 25*x*e^(2*x) + 8*x*log(2))/(16*x + 5*e^(2*x))

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maple [A]  time = 0.08, size = 20, normalized size = 1.05

method	result	size
risch	\(x +\frac {8 x \ln \relax (2)}{5 \left (5 \,{\mathrm e}^{2 x}+16 x \right )}\)	\(20\)
norman	\(\frac {-\frac {\ln \relax (2) {\mathrm e}^{2 x}}{2}+16 x^{2}+5 x \,{\mathrm e}^{2 x}}{5 \,{\mathrm e}^{2 x}+16 x}\)	\(35\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((25*exp(2*x)^2+((-16*x+8)*ln(2)+160*x)*exp(2*x)+256*x^2)/(25*exp(2*x)^2+160*x*exp(2*x)+256*x^2),x,method=_
RETURNVERBOSE)

[Out]

x+8/5*x*ln(2)/(5*exp(2*x)+16*x)

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maxima [A]  time = 0.48, size = 32, normalized size = 1.68 \begin {gather*} \frac {80 \, x^{2} + 25 \, x e^{\left (2 \, x\right )} + 8 \, x \log \relax (2)}{5 \, {\left (16 \, x + 5 \, e^{\left (2 \, x\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*exp(2*x)^2+((-16*x+8)*log(2)+160*x)*exp(2*x)+256*x^2)/(25*exp(2*x)^2+160*x*exp(2*x)+256*x^2),x,
algorithm="maxima")

[Out]

1/5*(80*x^2 + 25*x*e^(2*x) + 8*x*log(2))/(16*x + 5*e^(2*x))

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mupad [B]  time = 3.24, size = 19, normalized size = 1.00 \begin {gather*} x+\frac {8\,x\,\ln \relax (2)}{5\,\left (16\,x+5\,{\mathrm {e}}^{2\,x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((25*exp(4*x) + exp(2*x)*(160*x - log(2)*(16*x - 8)) + 256*x^2)/(25*exp(4*x) + 160*x*exp(2*x) + 256*x^2),x)

[Out]

x + (8*x*log(2))/(5*(16*x + 5*exp(2*x)))

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sympy [A]  time = 0.14, size = 17, normalized size = 0.89 \begin {gather*} x + \frac {8 x \log {\relax (2 )}}{80 x + 25 e^{2 x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((25*exp(2*x)**2+((-16*x+8)*ln(2)+160*x)*exp(2*x)+256*x**2)/(25*exp(2*x)**2+160*x*exp(2*x)+256*x**2),
x)

[Out]

x + 8*x*log(2)/(80*x + 25*exp(2*x))

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