3.64.32 \(\int \frac {e^{\frac {1}{25} (125+x+\log (-14 x+2 x^2))} (-7-5 x+x^2)}{-175 x+25 x^2} \, dx\)

Optimal. Leaf size=26 \[ e^{5+\frac {1}{25} (x+\log (2 (-2 x+(-5+x) x)))}+\log (\log (5)) \]

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Rubi [F]  time = 2.79, 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^{\frac {1}{25} \left (125+x+\log \left (-14 x+2 x^2\right )\right )} \left (-7-5 x+x^2\right )}{-175 x+25 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((125 + x + Log[-14*x + 2*x^2])/25)*(-7 - 5*x + x^2))/(-175*x + 25*x^2),x]

[Out]

(-14*(-7*x + x^2)^(1/25)*Defer[Subst][Defer[Int][E^(5 + x^25/25)/(-14 + 2*x^25)^(24/25), x], x, x^(1/25)])/((-
7 + x)^(1/25)*x^(1/25)) - (10*(-7*x + x^2)^(1/25)*Defer[Subst][Defer[Int][(E^(5 + x^25/25)*x^25)/(-14 + 2*x^25
)^(24/25), x], x, x^(1/25)])/((-7 + x)^(1/25)*x^(1/25)) + (2*(-7*x + x^2)^(1/25)*Defer[Subst][Defer[Int][(E^(5
+ x^25/25)*x^50)/(-14 + 2*x^25)^(24/25), x], x, x^(1/25)])/((-7 + x)^(1/25)*x^(1/25))

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {1}{25} \left (125+x+\log \left (-14 x+2 x^2\right )\right )} \left (-7-5 x+x^2\right )}{x (-175+25 x)} \, dx\\ &=\int \frac {e^{\frac {125+x}{25}} \left (-7-5 x+x^2\right ) \sqrt [25]{-14 x+2 x^2}}{x (-175+25 x)} \, dx\\ &=\frac {\sqrt [25]{-14 x+2 x^2} \int \frac {e^{\frac {125+x}{25}} \sqrt [25]{-14+2 x} \left (-7-5 x+x^2\right )}{x^{24/25} (-175+25 x)} \, dx}{\sqrt [25]{x} \sqrt [25]{-14+2 x}}\\ &=\frac {\left (2 \sqrt [25]{-14 x+2 x^2}\right ) \int \frac {e^{\frac {125+x}{25}} \left (-7-5 x+x^2\right )}{x^{24/25} (-14+2 x)^{24/25}} \, dx}{25 \sqrt [25]{x} \sqrt [25]{-14+2 x}}\\ &=\frac {\left (2 \sqrt [25]{-14 x+2 x^2}\right ) \operatorname {Subst}\left (\int \frac {e^{\frac {1}{25} \left (125+x^{25}\right )} \left (-7-5 x^{25}+x^{50}\right )}{\left (-14+2 x^{25}\right )^{24/25}} \, dx,x,\sqrt [25]{x}\right )}{\sqrt [25]{x} \sqrt [25]{-14+2 x}}\\ &=\frac {\left (2 \sqrt [25]{-14 x+2 x^2}\right ) \operatorname {Subst}\left (\int \frac {e^{5+\frac {x^{25}}{25}} \left (-7-5 x^{25}+x^{50}\right )}{\left (-14+2 x^{25}\right )^{24/25}} \, dx,x,\sqrt [25]{x}\right )}{\sqrt [25]{x} \sqrt [25]{-14+2 x}}\\ &=\frac {\left (2 \sqrt [25]{-14 x+2 x^2}\right ) \operatorname {Subst}\left (\int \left (-\frac {7 e^{5+\frac {x^{25}}{25}}}{\left (-14+2 x^{25}\right )^{24/25}}-\frac {5 e^{5+\frac {x^{25}}{25}} x^{25}}{\left (-14+2 x^{25}\right )^{24/25}}+\frac {e^{5+\frac {x^{25}}{25}} x^{50}}{\left (-14+2 x^{25}\right )^{24/25}}\right ) \, dx,x,\sqrt [25]{x}\right )}{\sqrt [25]{x} \sqrt [25]{-14+2 x}}\\ &=\frac {\left (2 \sqrt [25]{-14 x+2 x^2}\right ) \operatorname {Subst}\left (\int \frac {e^{5+\frac {x^{25}}{25}} x^{50}}{\left (-14+2 x^{25}\right )^{24/25}} \, dx,x,\sqrt [25]{x}\right )}{\sqrt [25]{x} \sqrt [25]{-14+2 x}}-\frac {\left (10 \sqrt [25]{-14 x+2 x^2}\right ) \operatorname {Subst}\left (\int \frac {e^{5+\frac {x^{25}}{25}} x^{25}}{\left (-14+2 x^{25}\right )^{24/25}} \, dx,x,\sqrt [25]{x}\right )}{\sqrt [25]{x} \sqrt [25]{-14+2 x}}-\frac {\left (14 \sqrt [25]{-14 x+2 x^2}\right ) \operatorname {Subst}\left (\int \frac {e^{5+\frac {x^{25}}{25}}}{\left (-14+2 x^{25}\right )^{24/25}} \, dx,x,\sqrt [25]{x}\right )}{\sqrt [25]{x} \sqrt [25]{-14+2 x}}\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 0.27, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {e^{\frac {1}{25} \left (125+x+\log \left (-14 x+2 x^2\right )\right )} \left (-7-5 x+x^2\right )}{-175 x+25 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(E^((125 + x + Log[-14*x + 2*x^2])/25)*(-7 - 5*x + x^2))/(-175*x + 25*x^2),x]

[Out]

Integrate[(E^((125 + x + Log[-14*x + 2*x^2])/25)*(-7 - 5*x + x^2))/(-175*x + 25*x^2), x]

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fricas [A]  time = 0.73, size = 18, normalized size = 0.69 \begin {gather*} e^{\left (\frac {1}{25} \, x + \frac {1}{25} \, \log \left (2 \, x^{2} - 14 \, x\right ) + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-5*x-7)*exp(1/25*log(2*x^2-14*x)+1/25*x+5)/(25*x^2-175*x),x, algorithm="fricas")

[Out]

e^(1/25*x + 1/25*log(2*x^2 - 14*x) + 5)

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giac [A]  time = 0.12, size = 18, normalized size = 0.69 \begin {gather*} e^{\left (\frac {1}{25} \, x + \frac {1}{25} \, \log \left (2 \, x^{2} - 14 \, x\right ) + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-5*x-7)*exp(1/25*log(2*x^2-14*x)+1/25*x+5)/(25*x^2-175*x),x, algorithm="giac")

[Out]

e^(1/25*x + 1/25*log(2*x^2 - 14*x) + 5)

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maple [A]  time = 0.37, size = 19, normalized size = 0.73

method	result	size
gosper	\({\mathrm e}^{\frac {\ln \left (2 x^{2}-14 x \right )}{25}+\frac {x}{25}+5}\)	\(19\)
norman	\({\mathrm e}^{\frac {\ln \left (2 x^{2}-14 x \right )}{25}+\frac {x}{25}+5}\)	\(19\)
risch	\(\left (2 x^{2}-14 x \right )^{\frac {1}{25}} {\mathrm e}^{5+\frac {x}{25}}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2-5*x-7)*exp(1/25*ln(2*x^2-14*x)+1/25*x+5)/(25*x^2-175*x),x,method=_RETURNVERBOSE)

[Out]

exp(1/25*ln(2*x^2-14*x)+1/25*x+5)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {1}{25} \, \int \frac {{\left (x^{2} - 5 \, x - 7\right )} e^{\left (\frac {1}{25} \, x + \frac {1}{25} \, \log \left (2 \, x^{2} - 14 \, x\right ) + 5\right )}}{x^{2} - 7 \, x}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-5*x-7)*exp(1/25*log(2*x^2-14*x)+1/25*x+5)/(25*x^2-175*x),x, algorithm="maxima")

[Out]

1/25*integrate((x^2 - 5*x - 7)*e^(1/25*x + 1/25*log(2*x^2 - 14*x) + 5)/(x^2 - 7*x), x)

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mupad [B]  time = 4.22, size = 19, normalized size = 0.73 \begin {gather*} 2^{1/25}\,{\mathrm {e}}^{x/25}\,{\mathrm {e}}^5\,{\left (x^2-7\,x\right )}^{1/25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x/25 + log(2*x^2 - 14*x)/25 + 5)*(5*x - x^2 + 7))/(175*x - 25*x^2),x)

[Out]

2^(1/25)*exp(x/25)*exp(5)*(x^2 - 7*x)^(1/25)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2-5*x-7)*exp(1/25*ln(2*x**2-14*x)+1/25*x+5)/(25*x**2-175*x),x)

[Out]

Timed out

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