3.47.24 \(\int \frac {125+e^{1+\frac {1}{25} (-25 x+e x)}+150 x}{-25+25 e^{\frac {1}{25} (-25 x+e x)}+150 x} \, dx\)

Optimal. Leaf size=20 \[ x+\log \left (-1+e^{-x+\frac {e x}{25}}+6 x\right ) \]

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Rubi [F]  time = 0.62, 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 {125+e^{1+\frac {1}{25} (-25 x+e x)}+150 x}{-25+25 e^{\frac {1}{25} (-25 x+e x)}+150 x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(125 + E^(1 + (-25*x + E*x)/25) + 150*x)/(-25 + 25*E^((-25*x + E*x)/25) + 150*x),x]

[Out]

(E*x)/25 + (125 + E)*Defer[Subst][Defer[Int][(-1 + E^((-25 + E)*x) + 150*x)^(-1), x], x, x/25] + 150*(25 - E)*
Defer[Subst][Defer[Int][x/(-1 + E^((-25 + E)*x) + 150*x), x], x, x/25]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-125-e^{1+\frac {1}{25} (-25 x+e x)}-150 x}{25 \left (1-e^{\frac {1}{25} (-25+e) x}-6 x\right )} \, dx\\ &=\frac {1}{25} \int \frac {-125-e^{1+\frac {1}{25} (-25 x+e x)}-150 x}{1-e^{\frac {1}{25} (-25+e) x}-6 x} \, dx\\ &=\operatorname {Subst}\left (\int \frac {-125-e^{1+(-25+e) x}-3750 x}{1-e^{(-25+e) x}-150 x} \, dx,x,\frac {x}{25}\right )\\ &=\operatorname {Subst}\left (\int \left (e+\frac {-125-e-150 (25-e) x}{1-e^{(-25+e) x}-150 x}\right ) \, dx,x,\frac {x}{25}\right )\\ &=\frac {e x}{25}+\operatorname {Subst}\left (\int \frac {-125-e-150 (25-e) x}{1-e^{(-25+e) x}-150 x} \, dx,x,\frac {x}{25}\right )\\ &=\frac {e x}{25}+\operatorname {Subst}\left (\int \left (\frac {125 \left (1+\frac {e}{125}\right )}{-1+e^{(-25+e) x}+150 x}-\frac {150 (-25+e) x}{-1+e^{(-25+e) x}+150 x}\right ) \, dx,x,\frac {x}{25}\right )\\ &=\frac {e x}{25}+(150 (25-e)) \operatorname {Subst}\left (\int \frac {x}{-1+e^{(-25+e) x}+150 x} \, dx,x,\frac {x}{25}\right )+(125+e) \operatorname {Subst}\left (\int \frac {1}{-1+e^{(-25+e) x}+150 x} \, dx,x,\frac {x}{25}\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.30, size = 21, normalized size = 1.05 \begin {gather*} \log \left (-e^x+e^{\frac {e x}{25}}+6 e^x x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(125 + E^(1 + (-25*x + E*x)/25) + 150*x)/(-25 + 25*E^((-25*x + E*x)/25) + 150*x),x]

[Out]

Log[-E^x + E^((E*x)/25) + 6*E^x*x]

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fricas [A]  time = 0.65, size = 23, normalized size = 1.15 \begin {gather*} x + \log \left ({\left (6 \, x - 1\right )} e + e^{\left (\frac {1}{25} \, x e - x + 1\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(1)*exp(1/25*x*exp(1)-x)+150*x+125)/(25*exp(1/25*x*exp(1)-x)+150*x-25),x, algorithm="fricas")

[Out]

x + log((6*x - 1)*e + e^(1/25*x*e - x + 1))

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giac [A]  time = 0.16, size = 18, normalized size = 0.90 \begin {gather*} x + \log \left (6 \, x + e^{\left (\frac {1}{25} \, x e - x\right )} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(1)*exp(1/25*x*exp(1)-x)+150*x+125)/(25*exp(1/25*x*exp(1)-x)+150*x-25),x, algorithm="giac")

[Out]

x + log(6*x + e^(1/25*x*e - x) - 1)

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maple [A]  time = 0.10, size = 17, normalized size = 0.85

method	result	size
risch	\(x +\ln \left (-1+6 x +{\mathrm e}^{\frac {x \left ({\mathrm e}-25\right )}{25}}\right )\)	\(17\)
norman	\(x +\ln \left (25 \,{\mathrm e}^{\frac {x \,{\mathrm e}}{25}-x}+150 x -25\right )\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(1)*exp(1/25*x*exp(1)-x)+150*x+125)/(25*exp(1/25*x*exp(1)-x)+150*x-25),x,method=_RETURNVERBOSE)

[Out]

x+ln(-1+6*x+exp(1/25*x*(exp(1)-25)))

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maxima [A]  time = 0.42, size = 16, normalized size = 0.80 \begin {gather*} \log \left ({\left (6 \, x - 1\right )} e^{x} + e^{\left (\frac {1}{25} \, x e\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(1)*exp(1/25*x*exp(1)-x)+150*x+125)/(25*exp(1/25*x*exp(1)-x)+150*x-25),x, algorithm="maxima")

[Out]

log((6*x - 1)*e^x + e^(1/25*x*e))

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mupad [B]  time = 0.18, size = 16, normalized size = 0.80 \begin {gather*} x+\ln \left (x+\frac {{\mathrm {e}}^{\frac {x\,\left (\mathrm {e}-25\right )}{25}}}{6}-\frac {1}{6}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((150*x + exp(1)*exp((x*exp(1))/25 - x) + 125)/(150*x + 25*exp((x*exp(1))/25 - x) - 25),x)

[Out]

x + log(x + exp((x*(exp(1) - 25))/25)/6 - 1/6)

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sympy [A]  time = 0.15, size = 29, normalized size = 1.45 \begin {gather*} x \left (\frac {1}{25} + \frac {24 e}{625}\right ) + \frac {\log {\left (6 x + e^{- x + \frac {e x}{25}} - 1 \right )}}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(1)*exp(1/25*x*exp(1)-x)+150*x+125)/(25*exp(1/25*x*exp(1)-x)+150*x-25),x)

[Out]

x*(1/25 + 24*E/625) + log(6*x + exp(-x + E*x/25) - 1)/25

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