3.42.56 \(\int \frac {-26-10 x+13 x^2+5 x^3+e^4 (13 x+10 x^2)}{e^4 (13 x^2+5 x^3)} \, dx\)

Optimal. Leaf size=22 \[ 3+\frac {4+\frac {2}{x}+x}{e^4}+\log (x (13+5 x)) \]

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Rubi [A]  time = 0.07, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {12, 1593, 1620} \begin {gather*} \frac {x}{e^4}+\frac {2}{e^4 x}+\log (x)+\log (5 x+13) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-26 - 10*x + 13*x^2 + 5*x^3 + E^4*(13*x + 10*x^2))/(E^4*(13*x^2 + 5*x^3)),x]

[Out]

2/(E^4*x) + x/E^4 + Log[x] + Log[13 + 5*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 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - 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} &=\frac {\int \frac {-26-10 x+13 x^2+5 x^3+e^4 \left (13 x+10 x^2\right )}{13 x^2+5 x^3} \, dx}{e^4}\\ &=\frac {\int \frac {-26-10 x+13 x^2+5 x^3+e^4 \left (13 x+10 x^2\right )}{x^2 (13+5 x)} \, dx}{e^4}\\ &=\frac {\int \left (1-\frac {2}{x^2}+\frac {e^4}{x}+\frac {5 e^4}{13+5 x}\right ) \, dx}{e^4}\\ &=\frac {2}{e^4 x}+\frac {x}{e^4}+\log (x)+\log (13+5 x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 27, normalized size = 1.23 \begin {gather*} \frac {\frac {2}{x}+x+e^4 \log (x)+e^4 \log (13+5 x)}{e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-26 - 10*x + 13*x^2 + 5*x^3 + E^4*(13*x + 10*x^2))/(E^4*(13*x^2 + 5*x^3)),x]

[Out]

(2/x + x + E^4*Log[x] + E^4*Log[13 + 5*x])/E^4

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fricas [A]  time = 0.68, size = 25, normalized size = 1.14 \begin {gather*} \frac {{\left (x e^{4} \log \left (5 \, x^{2} + 13 \, x\right ) + x^{2} + 2\right )} e^{\left (-4\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^2+13*x)*exp(4)+5*x^3+13*x^2-10*x-26)/(5*x^3+13*x^2)/exp(4),x, algorithm="fricas")

[Out]

(x*e^4*log(5*x^2 + 13*x) + x^2 + 2)*e^(-4)/x

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giac [A]  time = 0.19, size = 26, normalized size = 1.18 \begin {gather*} {\left (e^{4} \log \left ({\left | 5 \, x + 13 \right |}\right ) + e^{4} \log \left ({\left | x \right |}\right ) + x + \frac {2}{x}\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^2+13*x)*exp(4)+5*x^3+13*x^2-10*x-26)/(5*x^3+13*x^2)/exp(4),x, algorithm="giac")

[Out]

(e^4*log(abs(5*x + 13)) + e^4*log(abs(x)) + x + 2/x)*e^(-4)

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maple [A]  time = 0.09, size = 23, normalized size = 1.05

method	result	size
risch	\(x \,{\mathrm e}^{-4}+\frac {2 \,{\mathrm e}^{-4}}{x}+\ln \left (5 x^{2}+13 x \right )\)	\(23\)
default	\({\mathrm e}^{-4} \left (x +{\mathrm e}^{4} \ln \relax (x )+\frac {2}{x}+{\mathrm e}^{4} \ln \left (13+5 x \right )\right )\)	\(27\)
norman	\(\frac {x^{2} {\mathrm e}^{-4}+2 \,{\mathrm e}^{-4}}{x}+\ln \relax (x )+\ln \left (13+5 x \right )\)	\(29\)
meijerg	\(-\frac {10 \,{\mathrm e}^{-4} \left (-\frac {13}{5 x}-\ln \relax (x )-\ln \relax (5)+\ln \left (13\right )+\ln \left (1+\frac {5 x}{13}\right )\right )}{13}+\frac {\left (10 \,{\mathrm e}^{4}+13\right ) {\mathrm e}^{-4} \ln \left (1+\frac {5 x}{13}\right )}{5}+\frac {\left (13 \,{\mathrm e}^{4}-10\right ) {\mathrm e}^{-4} \left (\ln \relax (x )+\ln \relax (5)-\ln \left (13\right )-\ln \left (1+\frac {5 x}{13}\right )\right )}{13}+\frac {13 \,{\mathrm e}^{-4} \left (\frac {5 x}{13}-\ln \left (1+\frac {5 x}{13}\right )\right )}{5}\)	\(87\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((10*x^2+13*x)*exp(4)+5*x^3+13*x^2-10*x-26)/(5*x^3+13*x^2)/exp(4),x,method=_RETURNVERBOSE)

[Out]

x*exp(-4)+2*exp(-4)/x+ln(5*x^2+13*x)

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maxima [A]  time = 0.36, size = 24, normalized size = 1.09 \begin {gather*} {\left (e^{4} \log \left (5 \, x + 13\right ) + e^{4} \log \relax (x) + x + \frac {2}{x}\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x^2+13*x)*exp(4)+5*x^3+13*x^2-10*x-26)/(5*x^3+13*x^2)/exp(4),x, algorithm="maxima")

[Out]

(e^4*log(5*x + 13) + e^4*log(x) + x + 2/x)*e^(-4)

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mupad [B]  time = 0.10, size = 20, normalized size = 0.91 \begin {gather*} \ln \left (x\,\left (5\,x+13\right )\right )+x\,{\mathrm {e}}^{-4}+\frac {2\,{\mathrm {e}}^{-4}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-4)*(exp(4)*(13*x + 10*x^2) - 10*x + 13*x^2 + 5*x^3 - 26))/(13*x^2 + 5*x^3),x)

[Out]

log(x*(5*x + 13)) + x*exp(-4) + (2*exp(-4))/x

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sympy [A]  time = 0.27, size = 20, normalized size = 0.91 \begin {gather*} \frac {x}{e^{4}} + \log {\left (5 x^{2} + 13 x \right )} + \frac {2}{x e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((10*x**2+13*x)*exp(4)+5*x**3+13*x**2-10*x-26)/(5*x**3+13*x**2)/exp(4),x)

[Out]

x*exp(-4) + log(5*x**2 + 13*x) + 2*exp(-4)/x

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