3.47.41 \(\int \frac {e^3 (31+9 x)+(12+28 x+8 x^2) \log (5)}{e^3 (12+4 x)} \, dx\)

Optimal. Leaf size=22 \[ \frac {9 x}{4}+\frac {\left (3+x+x^2\right ) \log (5)}{e^3}+\log (3+x) \]

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Rubi [A]  time = 0.03, antiderivative size = 33, normalized size of antiderivative = 1.50, number of steps used = 3, number of rules used = 2, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {12, 1850} \begin {gather*} \frac {x^2 \log (25)}{2 e^3}+\frac {x \left (9 e^3+\log (625)\right )}{4 e^3}+\log (x+3) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^3*(31 + 9*x) + (12 + 28*x + 8*x^2)*Log[5])/(E^3*(12 + 4*x)),x]

[Out]

(x^2*Log[25])/(2*E^3) + (x*(9*E^3 + Log[625]))/(4*E^3) + Log[3 + 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 1850

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n)^p, x], x] /; FreeQ[
{a, b, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {e^3 (31+9 x)+\left (12+28 x+8 x^2\right ) \log (5)}{12+4 x} \, dx}{e^3}\\ &=\frac {\int \left (\frac {e^3}{3+x}+x \log (25)+\frac {1}{4} \left (9 e^3+\log (625)\right )\right ) \, dx}{e^3}\\ &=\frac {x^2 \log (25)}{2 e^3}+\frac {x \left (9 e^3+\log (625)\right )}{4 e^3}+\log (3+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 34, normalized size = 1.55 \begin {gather*} \frac {(3+x) \left (9 e^3+4 (-2+x) \log (5)\right )+4 e^3 \log (3+x)}{4 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^3*(31 + 9*x) + (12 + 28*x + 8*x^2)*Log[5])/(E^3*(12 + 4*x)),x]

[Out]

((3 + x)*(9*E^3 + 4*(-2 + x)*Log[5]) + 4*E^3*Log[3 + x])/(4*E^3)

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fricas [A]  time = 0.51, size = 27, normalized size = 1.23 \begin {gather*} \frac {1}{4} \, {\left (9 \, x e^{3} + 4 \, {\left (x^{2} + x\right )} \log \relax (5) + 4 \, e^{3} \log \left (x + 3\right )\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2+28*x+12)*log(5)+(9*x+31)*exp(3))/(4*x+12)/exp(3),x, algorithm="fricas")

[Out]

1/4*(9*x*e^3 + 4*(x^2 + x)*log(5) + 4*e^3*log(x + 3))*e^(-3)

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giac [A]  time = 0.14, size = 31, normalized size = 1.41 \begin {gather*} \frac {1}{4} \, {\left (4 \, x^{2} \log \relax (5) + 9 \, x e^{3} + 4 \, x \log \relax (5) + 4 \, e^{3} \log \left ({\left | x + 3 \right |}\right )\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2+28*x+12)*log(5)+(9*x+31)*exp(3))/(4*x+12)/exp(3),x, algorithm="giac")

[Out]

1/4*(4*x^2*log(5) + 9*x*e^3 + 4*x*log(5) + 4*e^3*log(abs(x + 3)))*e^(-3)

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maple [A]  time = 0.08, size = 23, normalized size = 1.05

method	result	size
risch	\({\mathrm e}^{-3} \ln \relax (5) x^{2}+\frac {9 x}{4}+{\mathrm e}^{-3} x \ln \relax (5)+\ln \left (3+x \right )\)	\(23\)
norman	\({\mathrm e}^{-3} \ln \relax (5) x^{2}+\frac {\left (9 \,{\mathrm e}^{3}+4 \ln \relax (5)\right ) {\mathrm e}^{-3} x}{4}+\ln \left (3+x \right )\)	\(32\)
default	\(\frac {{\mathrm e}^{-3} \left (4 x^{2} \ln \relax (5)+9 x \,{\mathrm e}^{3}+4 x \ln \relax (5)+4 \,{\mathrm e}^{3} \ln \left (3+x \right )\right )}{4}\)	\(33\)
meijerg	\(18 \,{\mathrm e}^{-3} \ln \relax (5) \left (-\frac {x \left (-x +6\right )}{18}+\ln \left (1+\frac {x}{3}\right )\right )+9 \left (\frac {3 \,{\mathrm e}^{3}}{4}+\frac {7 \ln \relax (5)}{3}\right ) {\mathrm e}^{-3} \left (\frac {x}{3}-\ln \left (1+\frac {x}{3}\right )\right )+\frac {31 \ln \left (1+\frac {x}{3}\right )}{4}+3 \,{\mathrm e}^{-3} \ln \relax (5) \ln \left (1+\frac {x}{3}\right )\)	\(68\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((8*x^2+28*x+12)*ln(5)+(9*x+31)*exp(3))/(4*x+12)/exp(3),x,method=_RETURNVERBOSE)

[Out]

exp(-3)*ln(5)*x^2+9/4*x+exp(-3)*x*ln(5)+ln(3+x)

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maxima [A]  time = 0.38, size = 31, normalized size = 1.41 \begin {gather*} \frac {1}{4} \, {\left (4 \, x^{2} \log \relax (5) + x {\left (9 \, e^{3} + 4 \, \log \relax (5)\right )} + 4 \, e^{3} \log \left (x + 3\right )\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x^2+28*x+12)*log(5)+(9*x+31)*exp(3))/(4*x+12)/exp(3),x, algorithm="maxima")

[Out]

1/4*(4*x^2*log(5) + x*(9*e^3 + 4*log(5)) + 4*e^3*log(x + 3))*e^(-3)

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mupad [B]  time = 0.08, size = 22, normalized size = 1.00 \begin {gather*} \frac {9\,x}{4}+\ln \left (x+3\right )+x^2\,{\mathrm {e}}^{-3}\,\ln \relax (5)+x\,{\mathrm {e}}^{-3}\,\ln \relax (5) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-3)*(log(5)*(28*x + 8*x^2 + 12) + exp(3)*(9*x + 31)))/(4*x + 12),x)

[Out]

(9*x)/4 + log(x + 3) + x^2*exp(-3)*log(5) + x*exp(-3)*log(5)

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sympy [A]  time = 0.15, size = 26, normalized size = 1.18 \begin {gather*} \frac {x^{2} \log {\relax (5 )}}{e^{3}} + x \left (\frac {\log {\relax (5 )}}{e^{3}} + \frac {9}{4}\right ) + \log {\left (x + 3 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((8*x**2+28*x+12)*ln(5)+(9*x+31)*exp(3))/(4*x+12)/exp(3),x)

[Out]

x**2*exp(-3)*log(5) + x*(exp(-3)*log(5) + 9/4) + log(x + 3)

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