3.2.71 \(\int \frac {-30+384 e^4-165 x-80 x^2}{-25+384 e^4-5 x-80 x^2} \, dx\)

Optimal. Leaf size=25 \[ x+\log \left (1+\frac {1+x}{4}+4 \left (-\frac {24 e^4}{5}+x^2\right )\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 18, normalized size of antiderivative = 0.72, number of steps used = 3, number of rules used = 2, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {1657, 628} \begin {gather*} \log \left (80 x^2+5 x-384 e^4+25\right )+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-30 + 384*E^4 - 165*x - 80*x^2)/(-25 + 384*E^4 - 5*x - 80*x^2),x]

[Out]

x + Log[25 - 384*E^4 + 5*x + 80*x^2]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1657

Int[(Pq_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x + c*x^2)^p, x
], x] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (1-\frac {5 (1+32 x)}{-25+384 e^4-5 x-80 x^2}\right ) \, dx\\ &=x-5 \int \frac {1+32 x}{-25+384 e^4-5 x-80 x^2} \, dx\\ &=x+\log \left (25-384 e^4+5 x+80 x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 0.72 \begin {gather*} x+\log \left (25-384 e^4+5 x+80 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-30 + 384*E^4 - 165*x - 80*x^2)/(-25 + 384*E^4 - 5*x - 80*x^2),x]

[Out]

x + Log[25 - 384*E^4 + 5*x + 80*x^2]

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fricas [A]  time = 0.90, size = 17, normalized size = 0.68 \begin {gather*} x + \log \left (80 \, x^{2} + 5 \, x - 384 \, e^{4} + 25\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((384*exp(4)-80*x^2-165*x-30)/(384*exp(4)-80*x^2-5*x-25),x, algorithm="fricas")

[Out]

x + log(80*x^2 + 5*x - 384*e^4 + 25)

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giac [A]  time = 0.27, size = 18, normalized size = 0.72 \begin {gather*} x + \log \left ({\left | 80 \, x^{2} + 5 \, x - 384 \, e^{4} + 25 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((384*exp(4)-80*x^2-165*x-30)/(384*exp(4)-80*x^2-5*x-25),x, algorithm="giac")

[Out]

x + log(abs(80*x^2 + 5*x - 384*e^4 + 25))

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maple [A]  time = 0.15, size = 18, normalized size = 0.72

method	result	size
default	\(x +\ln \left (-384 \,{\mathrm e}^{4}+80 x^{2}+5 x +25\right )\)	\(18\)
norman	\(x +\ln \left (384 \,{\mathrm e}^{4}-80 x^{2}-5 x -25\right )\)	\(18\)
risch	\(x +\ln \left (-384 \,{\mathrm e}^{4}+80 x^{2}+5 x +25\right )\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((384*exp(4)-80*x^2-165*x-30)/(384*exp(4)-80*x^2-5*x-25),x,method=_RETURNVERBOSE)

[Out]

x+ln(-384*exp(4)+80*x^2+5*x+25)

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maxima [A]  time = 0.43, size = 17, normalized size = 0.68 \begin {gather*} x + \log \left (80 \, x^{2} + 5 \, x - 384 \, e^{4} + 25\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((384*exp(4)-80*x^2-165*x-30)/(384*exp(4)-80*x^2-5*x-25),x, algorithm="maxima")

[Out]

x + log(80*x^2 + 5*x - 384*e^4 + 25)

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mupad [B]  time = 0.08, size = 17, normalized size = 0.68 \begin {gather*} x+\ln \left (80\,x^2+5\,x-384\,{\mathrm {e}}^4+25\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((165*x - 384*exp(4) + 80*x^2 + 30)/(5*x - 384*exp(4) + 80*x^2 + 25),x)

[Out]

x + log(5*x - 384*exp(4) + 80*x^2 + 25)

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sympy [A]  time = 0.17, size = 17, normalized size = 0.68 \begin {gather*} x + \log {\left (80 x^{2} + 5 x - 384 e^{4} + 25 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((384*exp(4)-80*x**2-165*x-30)/(384*exp(4)-80*x**2-5*x-25),x)

[Out]

x + log(80*x**2 + 5*x - 384*exp(4) + 25)

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