3.50.17 \(\int \frac {25 x^2+10 x^3+x^4+(-80+e^2 (-5-2 x)-32 x) \log (3)}{25 x^2+10 x^3+x^4} \, dx\)

Optimal. Leaf size=22 \[ 1+e^4+x+\frac {\left (16+e^2\right ) \log (3)}{x (5+x)} \]

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Rubi [A]  time = 0.08, antiderivative size = 32, normalized size of antiderivative = 1.45, number of steps used = 4, number of rules used = 3, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {1594, 27, 1620} \begin {gather*} x-\frac {\left (16+e^2\right ) \log (3)}{5 (x+5)}+\frac {\left (16+e^2\right ) \log (3)}{5 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(25*x^2 + 10*x^3 + x^4 + (-80 + E^2*(-5 - 2*x) - 32*x)*Log[3])/(25*x^2 + 10*x^3 + x^4),x]

[Out]

x + ((16 + E^2)*Log[3])/(5*x) - ((16 + E^2)*Log[3])/(5*(5 + 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 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {25 x^2+10 x^3+x^4+\left (-80+e^2 (-5-2 x)-32 x\right ) \log (3)}{x^2 \left (25+10 x+x^2\right )} \, dx\\ &=\int \frac {25 x^2+10 x^3+x^4+\left (-80+e^2 (-5-2 x)-32 x\right ) \log (3)}{x^2 (5+x)^2} \, dx\\ &=\int \left (1-\frac {\left (16+e^2\right ) \log (3)}{5 x^2}+\frac {\left (16+e^2\right ) \log (3)}{5 (5+x)^2}\right ) \, dx\\ &=x+\frac {\left (16+e^2\right ) \log (3)}{5 x}-\frac {\left (16+e^2\right ) \log (3)}{5 (5+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 26, normalized size = 1.18 \begin {gather*} \frac {5 x^2+x^3+\left (16+e^2\right ) \log (3)}{x (5+x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(25*x^2 + 10*x^3 + x^4 + (-80 + E^2*(-5 - 2*x) - 32*x)*Log[3])/(25*x^2 + 10*x^3 + x^4),x]

[Out]

(5*x^2 + x^3 + (16 + E^2)*Log[3])/(x*(5 + x))

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fricas [A]  time = 1.15, size = 26, normalized size = 1.18 \begin {gather*} \frac {x^{3} + 5 \, x^{2} + {\left (e^{2} + 16\right )} \log \relax (3)}{x^{2} + 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x-5)*exp(2)-32*x-80)*log(3)+x^4+10*x^3+25*x^2)/(x^4+10*x^3+25*x^2),x, algorithm="fricas")

[Out]

(x^3 + 5*x^2 + (e^2 + 16)*log(3))/(x^2 + 5*x)

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giac [A]  time = 0.12, size = 22, normalized size = 1.00 \begin {gather*} x + \frac {e^{2} \log \relax (3) + 16 \, \log \relax (3)}{x^{2} + 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x-5)*exp(2)-32*x-80)*log(3)+x^4+10*x^3+25*x^2)/(x^4+10*x^3+25*x^2),x, algorithm="giac")

[Out]

x + (e^2*log(3) + 16*log(3))/(x^2 + 5*x)

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maple [A]  time = 0.07, size = 22, normalized size = 1.00

method	result	size
risch	\(x +\frac {{\mathrm e}^{2} \ln \relax (3)+16 \ln \relax (3)}{\left (5+x \right ) x}\)	\(22\)
gosper	\(\frac {x^{3}+{\mathrm e}^{2} \ln \relax (3)+16 \ln \relax (3)-25 x}{x \left (5+x \right )}\)	\(26\)
norman	\(\frac {x^{3}+{\mathrm e}^{2} \ln \relax (3)+16 \ln \relax (3)-25 x}{x \left (5+x \right )}\)	\(26\)
default	\(x -\frac {\ln \relax (3) \left (16+{\mathrm e}^{2}\right )}{5 \left (5+x \right )}+\frac {\ln \relax (3) \left (16+{\mathrm e}^{2}\right )}{5 x}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-2*x-5)*exp(2)-32*x-80)*ln(3)+x^4+10*x^3+25*x^2)/(x^4+10*x^3+25*x^2),x,method=_RETURNVERBOSE)

[Out]

x+(exp(2)*ln(3)+16*ln(3))/(5+x)/x

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maxima [A]  time = 0.35, size = 18, normalized size = 0.82 \begin {gather*} x + \frac {{\left (e^{2} + 16\right )} \log \relax (3)}{x^{2} + 5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x-5)*exp(2)-32*x-80)*log(3)+x^4+10*x^3+25*x^2)/(x^4+10*x^3+25*x^2),x, algorithm="maxima")

[Out]

x + (e^2 + 16)*log(3)/(x^2 + 5*x)

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mupad [B]  time = 3.76, size = 21, normalized size = 0.95 \begin {gather*} x+\frac {16\,\ln \relax (3)+{\mathrm {e}}^2\,\ln \relax (3)}{x\,\left (x+5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((25*x^2 + 10*x^3 + x^4 - log(3)*(32*x + exp(2)*(2*x + 5) + 80))/(25*x^2 + 10*x^3 + x^4),x)

[Out]

x + (16*log(3) + exp(2)*log(3))/(x*(x + 5))

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sympy [A]  time = 0.31, size = 19, normalized size = 0.86 \begin {gather*} x + \frac {e^{2} \log {\relax (3 )} + 16 \log {\relax (3 )}}{x^{2} + 5 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-2*x-5)*exp(2)-32*x-80)*ln(3)+x**4+10*x**3+25*x**2)/(x**4+10*x**3+25*x**2),x)

[Out]

x + (exp(2)*log(3) + 16*log(3))/(x**2 + 5*x)

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