3.1.88 \(\int \frac {12+71 e^{25/4}+\log (2)}{e^{25/2}-2 e^{25/4} x+x^2} \, dx\)

Optimal. Leaf size=25 \[ \frac {-x+3 (4+24 x)+\log (2)}{e^{25/4}-x} \]

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Rubi [A]  time = 0.01, antiderivative size = 23, normalized size of antiderivative = 0.92, number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {12, 27, 32} \begin {gather*} \frac {12+71 e^{25/4}+\log (2)}{e^{25/4}-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(12 + 71*E^(25/4) + Log[2])/(E^(25/2) - 2*E^(25/4)*x + x^2),x]

[Out]

(12 + 71*E^(25/4) + Log[2])/(E^(25/4) - x)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\left (12+71 e^{25/4}+\log (2)\right ) \int \frac {1}{e^{25/2}-2 e^{25/4} x+x^2} \, dx\\ &=\left (12+71 e^{25/4}+\log (2)\right ) \int \frac {1}{\left (-e^{25/4}+x\right )^2} \, dx\\ &=\frac {12+71 e^{25/4}+\log (2)}{e^{25/4}-x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 23, normalized size = 0.92 \begin {gather*} \frac {12+71 e^{25/4}+\log (2)}{e^{25/4}-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(12 + 71*E^(25/4) + Log[2])/(E^(25/2) - 2*E^(25/4)*x + x^2),x]

[Out]

(12 + 71*E^(25/4) + Log[2])/(E^(25/4) - x)

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fricas [A]  time = 0.54, size = 18, normalized size = 0.72 \begin {gather*} -\frac {71 \, e^{\frac {25}{4}} + \log \relax (2) + 12}{x - e^{\frac {25}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(2)+71*exp(25/4)+12)/(exp(25/4)^2-2*x*exp(25/4)+x^2),x, algorithm="fricas")

[Out]

-(71*e^(25/4) + log(2) + 12)/(x - e^(25/4))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(2)+71*exp(25/4)+12)/(exp(25/4)^2-2*x*exp(25/4)+x^2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: (71*exp(25/4)+ln(2)+12)*1/2/sqrt(-exp(25
/2)+exp(25/4)^2)*ln(sqrt((2*sageVARx-2*exp(25/4))^2+(-2*sqrt(exp(25/2)-exp(25/4)^2))^2)/sqrt((2*sageVARx-2*exp
(25/4))^2+(2*sqrt(exp

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maple [A]  time = 0.16, size = 18, normalized size = 0.72

method	result	size
gosper	\(\frac {\ln \relax (2)+71 \,{\mathrm e}^{\frac {25}{4}}+12}{{\mathrm e}^{\frac {25}{4}}-x}\)	\(18\)
norman	\(\frac {\ln \relax (2)+71 \,{\mathrm e}^{\frac {25}{4}}+12}{{\mathrm e}^{\frac {25}{4}}-x}\)	\(18\)
risch	\(\frac {\ln \relax (2)}{{\mathrm e}^{\frac {25}{4}}-x}+\frac {71 \,{\mathrm e}^{\frac {25}{4}}}{{\mathrm e}^{\frac {25}{4}}-x}+\frac {12}{{\mathrm e}^{\frac {25}{4}}-x}\)	\(35\)
meijerg	\(\frac {\ln \relax (2) {\mathrm e}^{-\frac {25}{2}} x}{1-x \,{\mathrm e}^{-\frac {25}{4}}}+\frac {71 \,{\mathrm e}^{-\frac {25}{4}} x}{1-x \,{\mathrm e}^{-\frac {25}{4}}}+\frac {12 \,{\mathrm e}^{-\frac {25}{2}} x}{1-x \,{\mathrm e}^{-\frac {25}{4}}}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln(2)+71*exp(25/4)+12)/(exp(25/4)^2-2*x*exp(25/4)+x^2),x,method=_RETURNVERBOSE)

[Out]

(ln(2)+71*exp(25/4)+12)/(exp(25/4)-x)

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maxima [A]  time = 0.56, size = 18, normalized size = 0.72 \begin {gather*} -\frac {71 \, e^{\frac {25}{4}} + \log \relax (2) + 12}{x - e^{\frac {25}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(2)+71*exp(25/4)+12)/(exp(25/4)^2-2*x*exp(25/4)+x^2),x, algorithm="maxima")

[Out]

-(71*e^(25/4) + log(2) + 12)/(x - e^(25/4))

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mupad [B]  time = 0.11, size = 18, normalized size = 0.72 \begin {gather*} -\frac {71\,{\mathrm {e}}^{25/4}+\ln \relax (2)+12}{x-{\mathrm {e}}^{25/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((71*exp(25/4) + log(2) + 12)/(exp(25/2) - 2*x*exp(25/4) + x^2),x)

[Out]

-(71*exp(25/4) + log(2) + 12)/(x - exp(25/4))

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sympy [A]  time = 0.10, size = 19, normalized size = 0.76 \begin {gather*} - \frac {\log {\relax (2 )} + 12 + 71 e^{\frac {25}{4}}}{x - e^{\frac {25}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((ln(2)+71*exp(25/4)+12)/(exp(25/4)**2-2*x*exp(25/4)+x**2),x)

[Out]

-(log(2) + 12 + 71*exp(25/4))/(x - exp(25/4))

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