3.47.31 \(\int \frac {289 x-289 x^2+e^{2 x} (-1156+2312 x)}{4 x^2} \, dx\)

Optimal. Leaf size=24 \[ 3-\frac {289}{4} \left (-\frac {4 e^{2 x}}{x}+x-\log (3 x)\right ) \]

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Rubi [A]  time = 0.05, antiderivative size = 22, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {12, 14, 43, 2197} \begin {gather*} -\frac {289 x}{4}+\frac {289 e^{2 x}}{x}+\frac {289 \log (x)}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(289*x - 289*x^2 + E^(2*x)*(-1156 + 2312*x))/(4*x^2),x]

[Out]

(289*E^(2*x))/x - (289*x)/4 + (289*Log[x])/4

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 2197

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> With[{b = Coefficient[v, x, 1], d = Coefficient[u, x, 0],
e = Coefficient[u, x, 1], f = Coefficient[w, x, 0], g = Coefficient[w, x, 1]}, Simp[(g*u^(m + 1)*F^(c*v))/(b*c
*e*Log[F]), x] /; EqQ[e*g*(m + 1) - b*c*(e*f - d*g)*Log[F], 0]] /; FreeQ[{F, c, m}, x] && LinearQ[{u, v, w}, x
]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \frac {289 x-289 x^2+e^{2 x} (-1156+2312 x)}{x^2} \, dx\\ &=\frac {1}{4} \int \left (-\frac {289 (-1+x)}{x}+\frac {1156 e^{2 x} (-1+2 x)}{x^2}\right ) \, dx\\ &=-\left (\frac {289}{4} \int \frac {-1+x}{x} \, dx\right )+289 \int \frac {e^{2 x} (-1+2 x)}{x^2} \, dx\\ &=\frac {289 e^{2 x}}{x}-\frac {289}{4} \int \left (1-\frac {1}{x}\right ) \, dx\\ &=\frac {289 e^{2 x}}{x}-\frac {289 x}{4}+\frac {289 \log (x)}{4}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 20, normalized size = 0.83 \begin {gather*} \frac {289}{4} \left (\frac {4 e^{2 x}}{x}-x+\log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(289*x - 289*x^2 + E^(2*x)*(-1156 + 2312*x))/(4*x^2),x]

[Out]

(289*((4*E^(2*x))/x - x + Log[x]))/4

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fricas [A]  time = 1.01, size = 20, normalized size = 0.83 \begin {gather*} -\frac {289 \, {\left (x^{2} - x \log \relax (x) - 4 \, e^{\left (2 \, x\right )}\right )}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((2312*x-1156)*exp(x)^2-289*x^2+289*x)/x^2,x, algorithm="fricas")

[Out]

-289/4*(x^2 - x*log(x) - 4*e^(2*x))/x

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giac [A]  time = 0.15, size = 20, normalized size = 0.83 \begin {gather*} -\frac {289 \, {\left (x^{2} - x \log \relax (x) - 4 \, e^{\left (2 \, x\right )}\right )}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((2312*x-1156)*exp(x)^2-289*x^2+289*x)/x^2,x, algorithm="giac")

[Out]

-289/4*(x^2 - x*log(x) - 4*e^(2*x))/x

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maple [A]  time = 0.06, size = 18, normalized size = 0.75

method	result	size
default	\(-\frac {289 x}{4}+\frac {289 \ln \relax (x )}{4}+\frac {289 \,{\mathrm e}^{2 x}}{x}\)	\(18\)
risch	\(-\frac {289 x}{4}+\frac {289 \ln \relax (x )}{4}+\frac {289 \,{\mathrm e}^{2 x}}{x}\)	\(18\)
norman	\(\frac {-\frac {289 x^{2}}{4}+289 \,{\mathrm e}^{2 x}}{x}+\frac {289 \ln \relax (x )}{4}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((2312*x-1156)*exp(x)^2-289*x^2+289*x)/x^2,x,method=_RETURNVERBOSE)

[Out]

-289/4*x+289/4*ln(x)+289*exp(x)^2/x

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maxima [C]  time = 0.39, size = 21, normalized size = 0.88 \begin {gather*} -\frac {289}{4} \, x + 578 \, {\rm Ei}\left (2 \, x\right ) - 578 \, \Gamma \left (-1, -2 \, x\right ) + \frac {289}{4} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((2312*x-1156)*exp(x)^2-289*x^2+289*x)/x^2,x, algorithm="maxima")

[Out]

-289/4*x + 578*Ei(2*x) - 578*gamma(-1, -2*x) + 289/4*log(x)

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mupad [B]  time = 3.14, size = 22, normalized size = 0.92 \begin {gather*} \frac {289\,\ln \relax (x)}{4}+\frac {1156\,{\mathrm {e}}^{2\,x}-289\,x^2}{4\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((289*x)/4 + (exp(2*x)*(2312*x - 1156))/4 - (289*x^2)/4)/x^2,x)

[Out]

(289*log(x))/4 + (1156*exp(2*x) - 289*x^2)/(4*x)

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sympy [A]  time = 0.15, size = 19, normalized size = 0.79 \begin {gather*} - \frac {289 x}{4} + \frac {289 \log {\relax (x )}}{4} + \frac {289 e^{2 x}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((2312*x-1156)*exp(x)**2-289*x**2+289*x)/x**2,x)

[Out]

-289*x/4 + 289*log(x)/4 + 289*exp(2*x)/x

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