3.68.82 \(\int \frac {1+2 x+x^2+e^3 (-3-7 x+4 x^2+20 x^3+15 x^4+3 x^5)+e^3 (-60-50 x-25 x^2-5 x^3) \log (3)}{3+7 x+5 x^2+x^3} \, dx\)

Optimal. Leaf size=32 \[ -e^3 \left (x+x \left (-x^2+\frac {5 (4+x) \log (3)}{1+x}\right )\right )+\log (3+x) \]

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Rubi [A]  time = 0.07, antiderivative size = 34, normalized size of antiderivative = 1.06, number of steps used = 2, number of rules used = 1, integrand size = 74, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {2074} \begin {gather*} e^3 x^3-e^3 x (1+\log (243))+\log (x+3)+\frac {15 e^3 \log (3)}{x+1} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + 2*x + x^2 + E^3*(-3 - 7*x + 4*x^2 + 20*x^3 + 15*x^4 + 3*x^5) + E^3*(-60 - 50*x - 25*x^2 - 5*x^3)*Log[
3])/(3 + 7*x + 5*x^2 + x^3),x]

[Out]

E^3*x^3 + (15*E^3*Log[3])/(1 + x) - E^3*x*(1 + Log[243]) + Log[3 + x]

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (3 e^3 x^2+\frac {1}{3+x}-\frac {15 e^3 \log (3)}{(1+x)^2}-e^3 (1+\log (243))\right ) \, dx\\ &=e^3 x^3+\frac {15 e^3 \log (3)}{1+x}-e^3 x (1+\log (243))+\log (3+x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 55, normalized size = 1.72 \begin {gather*} -3 e^3 (1+x)^2+e^3 (1+x)^3+e^3 (1+x) (2-5 \log (3))+\frac {e^3 (25 \log (3)-2 \log (243))}{1+x}+\log (3+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + 2*x + x^2 + E^3*(-3 - 7*x + 4*x^2 + 20*x^3 + 15*x^4 + 3*x^5) + E^3*(-60 - 50*x - 25*x^2 - 5*x^3
)*Log[3])/(3 + 7*x + 5*x^2 + x^3),x]

[Out]

-3*E^3*(1 + x)^2 + E^3*(1 + x)^3 + E^3*(1 + x)*(2 - 5*Log[3]) + (E^3*(25*Log[3] - 2*Log[243]))/(1 + x) + Log[3
+ x]

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fricas [A]  time = 0.50, size = 48, normalized size = 1.50 \begin {gather*} -\frac {5 \, {\left (x^{2} + x - 3\right )} e^{3} \log \relax (3) - {\left (x^{4} + x^{3} - x^{2} - x\right )} e^{3} - {\left (x + 1\right )} \log \left (x + 3\right )}{x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^3-25*x^2-50*x-60)*exp(3)*log(3)+(3*x^5+15*x^4+20*x^3+4*x^2-7*x-3)*exp(3)+x^2+2*x+1)/(x^3+5*x^
2+7*x+3),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.20, size = 35, normalized size = 1.09 \begin {gather*} x^{3} e^{3} - 5 \, x e^{3} \log \relax (3) - x e^{3} + \frac {15 \, e^{3} \log \relax (3)}{x + 1} + \log \left ({\left | x + 3 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^3-25*x^2-50*x-60)*exp(3)*log(3)+(3*x^5+15*x^4+20*x^3+4*x^2-7*x-3)*exp(3)+x^2+2*x+1)/(x^3+5*x^
2+7*x+3),x, algorithm="giac")

[Out]

x^3*e^3 - 5*x*e^3*log(3) - x*e^3 + 15*e^3*log(3)/(x + 1) + log(abs(x + 3))

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maple [A]  time = 0.07, size = 35, normalized size = 1.09

method	result	size
default	\(x^{3} {\mathrm e}^{3}-5 x \,{\mathrm e}^{3} \ln \relax (3)-x \,{\mathrm e}^{3}+\frac {15 \,{\mathrm e}^{3} \ln \relax (3)}{x +1}+\ln \left (3+x \right )\)	\(35\)
risch	\(x^{3} {\mathrm e}^{3}-5 x \,{\mathrm e}^{3} \ln \relax (3)-x \,{\mathrm e}^{3}+\frac {15 \,{\mathrm e}^{3} \ln \relax (3)}{x +1}+\ln \left (3+x \right )\)	\(35\)
norman	\(\frac {x^{3} {\mathrm e}^{3}+x^{4} {\mathrm e}^{3}+\left (-5 \,{\mathrm e}^{3} \ln \relax (3)-{\mathrm e}^{3}\right ) x^{2}+20 \,{\mathrm e}^{3} \ln \relax (3)+{\mathrm e}^{3}}{x +1}+\ln \left (3+x \right )\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-5*x^3-25*x^2-50*x-60)*exp(3)*ln(3)+(3*x^5+15*x^4+20*x^3+4*x^2-7*x-3)*exp(3)+x^2+2*x+1)/(x^3+5*x^2+7*x+3
),x,method=_RETURNVERBOSE)

[Out]

x^3*exp(3)-5*x*exp(3)*ln(3)-x*exp(3)+15*exp(3)*ln(3)/(x+1)+ln(3+x)

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maxima [A]  time = 0.46, size = 34, normalized size = 1.06 \begin {gather*} x^{3} e^{3} - {\left (5 \, e^{3} \log \relax (3) + e^{3}\right )} x + \frac {15 \, e^{3} \log \relax (3)}{x + 1} + \log \left (x + 3\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x^3-25*x^2-50*x-60)*exp(3)*log(3)+(3*x^5+15*x^4+20*x^3+4*x^2-7*x-3)*exp(3)+x^2+2*x+1)/(x^3+5*x^
2+7*x+3),x, algorithm="maxima")

[Out]

x^3*e^3 - (5*e^3*log(3) + e^3)*x + 15*e^3*log(3)/(x + 1) + log(x + 3)

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mupad [B]  time = 0.15, size = 34, normalized size = 1.06 \begin {gather*} \ln \left (x+3\right )+x^3\,{\mathrm {e}}^3-x\,\left ({\mathrm {e}}^3+5\,{\mathrm {e}}^3\,\ln \relax (3)\right )+\frac {15\,{\mathrm {e}}^3\,\ln \relax (3)}{x+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x + exp(3)*(4*x^2 - 7*x + 20*x^3 + 15*x^4 + 3*x^5 - 3) + x^2 - exp(3)*log(3)*(50*x + 25*x^2 + 5*x^3 + 6
0) + 1)/(7*x + 5*x^2 + x^3 + 3),x)

[Out]

log(x + 3) + x^3*exp(3) - x*(exp(3) + 5*exp(3)*log(3)) + (15*exp(3)*log(3))/(x + 1)

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sympy [A]  time = 0.31, size = 37, normalized size = 1.16 \begin {gather*} x^{3} e^{3} + x \left (- 5 e^{3} \log {\relax (3 )} - e^{3}\right ) + \log {\left (x + 3 \right )} + \frac {15 e^{3} \log {\relax (3 )}}{x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x**3-25*x**2-50*x-60)*exp(3)*ln(3)+(3*x**5+15*x**4+20*x**3+4*x**2-7*x-3)*exp(3)+x**2+2*x+1)/(x*
*3+5*x**2+7*x+3),x)

[Out]

x**3*exp(3) + x*(-5*exp(3)*log(3) - exp(3)) + log(x + 3) + 15*exp(3)*log(3)/(x + 1)

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