3.19.7 \(\int \frac {-144-289 x+e (1+2 x)}{-144 x+e x} \, dx\)

Optimal. Leaf size=21 \[ 2 x+\frac {1+x}{144-e}+\log \left (\frac {x}{4}\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 17, normalized size of antiderivative = 0.81, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {6, 12, 186, 43} \begin {gather*} \frac {(289-2 e) x}{144-e}+\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-144 - 289*x + E*(1 + 2*x))/(-144*x + E*x),x]

[Out]

((289 - 2*E)*x)/(144 - E) + Log[x]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 186

Int[(u_)^(m_.)*(v_)^(n_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*ExpandToSum[v, x]^n, x] /; FreeQ[{m, n}, x] &&
LinearQ[{u, v}, x] &&  !LinearMatchQ[{u, v}, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-144-289 x+e (1+2 x)}{(-144+e) x} \, dx\\ &=\frac {\int \frac {-144-289 x+e (1+2 x)}{x} \, dx}{-144+e}\\ &=\frac {\int \frac {-144+e-(289-2 e) x}{x} \, dx}{-144+e}\\ &=\frac {\int \left (-289+2 e+\frac {-144+e}{x}\right ) \, dx}{-144+e}\\ &=\frac {(289-2 e) x}{144-e}+\log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 20, normalized size = 0.95 \begin {gather*} \frac {(-289+2 e) x+(-144+e) \log (x)}{-144+e} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-144 - 289*x + E*(1 + 2*x))/(-144*x + E*x),x]

[Out]

((-289 + 2*E)*x + (-144 + E)*Log[x])/(-144 + E)

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fricas [A]  time = 0.78, size = 23, normalized size = 1.10 \begin {gather*} \frac {2 \, x e + {\left (e - 144\right )} \log \relax (x) - 289 \, x}{e - 144} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+1)*exp(1)-289*x-144)/(x*exp(1)-144*x),x, algorithm="fricas")

[Out]

(2*x*e + (e - 144)*log(x) - 289*x)/(e - 144)

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giac [A]  time = 0.29, size = 20, normalized size = 0.95 \begin {gather*} \frac {2 \, x e - 289 \, x}{e - 144} + \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+1)*exp(1)-289*x-144)/(x*exp(1)-144*x),x, algorithm="giac")

[Out]

(2*x*e - 289*x)/(e - 144) + log(abs(x))

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maple [A]  time = 0.04, size = 18, normalized size = 0.86

method	result	size
norman	\(\frac {\left (2 \,{\mathrm e}-289\right ) x}{{\mathrm e}-144}+\ln \relax (x )\)	\(18\)
default	\(\frac {2 x \,{\mathrm e}-289 x +\left ({\mathrm e}-144\right ) \ln \relax (x )}{{\mathrm e}-144}\)	\(24\)
risch	\(\frac {2 x \,{\mathrm e}}{{\mathrm e}-144}-\frac {289 x}{{\mathrm e}-144}+\ln \relax (x )\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x+1)*exp(1)-289*x-144)/(x*exp(1)-144*x),x,method=_RETURNVERBOSE)

[Out]

(2*exp(1)-289)/(exp(1)-144)*x+ln(x)

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maxima [A]  time = 0.37, size = 17, normalized size = 0.81 \begin {gather*} \frac {x {\left (2 \, e - 289\right )}}{e - 144} + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+1)*exp(1)-289*x-144)/(x*exp(1)-144*x),x, algorithm="maxima")

[Out]

x*(2*e - 289)/(e - 144) + log(x)

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mupad [B]  time = 0.06, size = 17, normalized size = 0.81 \begin {gather*} \ln \relax (x)+\frac {x\,\left (2\,\mathrm {e}-289\right )}{\mathrm {e}-144} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((289*x - exp(1)*(2*x + 1) + 144)/(144*x - x*exp(1)),x)

[Out]

log(x) + (x*(2*exp(1) - 289))/(exp(1) - 144)

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sympy [A]  time = 0.11, size = 22, normalized size = 1.05 \begin {gather*} \frac {- x \left (289 - 2 e\right ) - \left (144 - e\right ) \log {\relax (x )}}{-144 + e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x+1)*exp(1)-289*x-144)/(x*exp(1)-144*x),x)

[Out]

(-x*(289 - 2*E) - (144 - E)*log(x))/(-144 + E)

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