3.96.47 \(\int \frac {x^2+4 x^3+e^3 (-21 x-8 x^2) (x+4 x^2)+(5+20 x) \log (5)}{x^2+4 x^3} \, dx\)

Optimal. Leaf size=27 \[ \frac {(5+x) \left (x-4 e^3 \left (\frac {x}{4}+x^2\right )-\log (5)\right )}{x} \]

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Rubi [A]  time = 0.07, antiderivative size = 25, normalized size of antiderivative = 0.93, number of steps used = 4, number of rules used = 3, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {1593, 1586, 14} \begin {gather*} -4 e^3 x^2+\left (1-21 e^3\right ) x-\frac {5 \log (5)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^2 + 4*x^3 + E^3*(-21*x - 8*x^2)*(x + 4*x^2) + (5 + 20*x)*Log[5])/(x^2 + 4*x^3),x]

[Out]

(1 - 21*E^3)*x - 4*E^3*x^2 - (5*Log[5])/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 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x^2+4 x^3+e^3 \left (-21 x-8 x^2\right ) \left (x+4 x^2\right )+(5+20 x) \log (5)}{x^2 (1+4 x)} \, dx\\ &=\int \frac {\left (1-21 e^3\right ) x^2-8 e^3 x^3+5 \log (5)}{x^2} \, dx\\ &=\int \left (1-21 e^3-8 e^3 x+\frac {5 \log (5)}{x^2}\right ) \, dx\\ &=\left (1-21 e^3\right ) x-4 e^3 x^2-\frac {5 \log (5)}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 23, normalized size = 0.85 \begin {gather*} x-21 e^3 x-4 e^3 x^2-\frac {5 \log (5)}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^2 + 4*x^3 + E^3*(-21*x - 8*x^2)*(x + 4*x^2) + (5 + 20*x)*Log[5])/(x^2 + 4*x^3),x]

[Out]

x - 21*E^3*x - 4*E^3*x^2 - (5*Log[5])/x

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fricas [A]  time = 0.51, size = 27, normalized size = 1.00 \begin {gather*} \frac {x^{2} - {\left (4 \, x^{3} + 21 \, x^{2}\right )} e^{3} - 5 \, \log \relax (5)}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^2-21*x)*exp(log(4*x^2+x)+3)+(20*x+5)*log(5)+4*x^3+x^2)/(4*x^3+x^2),x, algorithm="fricas")

[Out]

(x^2 - (4*x^3 + 21*x^2)*e^3 - 5*log(5))/x

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giac [A]  time = 0.12, size = 21, normalized size = 0.78 \begin {gather*} -4 \, x^{2} e^{3} - 21 \, x e^{3} + x - \frac {5 \, \log \relax (5)}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^2-21*x)*exp(log(4*x^2+x)+3)+(20*x+5)*log(5)+4*x^3+x^2)/(4*x^3+x^2),x, algorithm="giac")

[Out]

-4*x^2*e^3 - 21*x*e^3 + x - 5*log(5)/x

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maple [A]  time = 0.06, size = 22, normalized size = 0.81

method	result	size
risch	\(-4 x^{2} {\mathrm e}^{3}-21 x \,{\mathrm e}^{3}+x -\frac {5 \ln \relax (5)}{x}\)	\(22\)
default	\(x -\frac {5 \ln \relax (5)}{x}-{\mathrm e}^{3} \left (4 x^{2}+21 x \right )\)	\(23\)
norman	\(\frac {\left (-21 \,{\mathrm e}^{3}+1\right ) x^{2}-4 x^{3} {\mathrm e}^{3}-5 \ln \relax (5)}{x}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-8*x^2-21*x)*exp(ln(4*x^2+x)+3)+(20*x+5)*ln(5)+4*x^3+x^2)/(4*x^3+x^2),x,method=_RETURNVERBOSE)

[Out]

-4*x^2*exp(3)-21*x*exp(3)+x-5*ln(5)/x

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maxima [A]  time = 0.34, size = 24, normalized size = 0.89 \begin {gather*} -4 \, x^{2} e^{3} - x {\left (21 \, e^{3} - 1\right )} - \frac {5 \, \log \relax (5)}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^2-21*x)*exp(log(4*x^2+x)+3)+(20*x+5)*log(5)+4*x^3+x^2)/(4*x^3+x^2),x, algorithm="maxima")

[Out]

-4*x^2*e^3 - x*(21*e^3 - 1) - 5*log(5)/x

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mupad [B]  time = 0.15, size = 24, normalized size = 0.89 \begin {gather*} -4\,x^2\,{\mathrm {e}}^3-\frac {5\,\ln \relax (5)}{x}-x\,\left (21\,{\mathrm {e}}^3-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(5)*(20*x + 5) - exp(log(x + 4*x^2) + 3)*(21*x + 8*x^2) + x^2 + 4*x^3)/(x^2 + 4*x^3),x)

[Out]

- 4*x^2*exp(3) - (5*log(5))/x - x*(21*exp(3) - 1)

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sympy [A]  time = 0.10, size = 24, normalized size = 0.89 \begin {gather*} - 4 x^{2} e^{3} - x \left (-1 + 21 e^{3}\right ) - \frac {5 \log {\relax (5 )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x**2-21*x)*exp(ln(4*x**2+x)+3)+(20*x+5)*ln(5)+4*x**3+x**2)/(4*x**3+x**2),x)

[Out]

-4*x**2*exp(3) - x*(-1 + 21*exp(3)) - 5*log(5)/x

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