3.38.18 \(\int \frac {2 x-3 x^2-5 x^3-2 x^4+(-3-5 x-2 x^2) \log (3+2 x)}{6 x^2+7 x^3+2 x^4+(3 x+2 x^2) \log (3+2 x)} \, dx\)

Optimal. Leaf size=23 \[ -x+\log \left (3+3 \left (x+\frac {x+\log (3+2 x)}{x}\right )\right ) \]

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Rubi [A]  time = 0.56, antiderivative size = 22, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 4, integrand size = 71, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {6741, 6742, 43, 6684} \begin {gather*} \log \left (x^2+2 x+\log (2 x+3)\right )-x-\log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*x - 3*x^2 - 5*x^3 - 2*x^4 + (-3 - 5*x - 2*x^2)*Log[3 + 2*x])/(6*x^2 + 7*x^3 + 2*x^4 + (3*x + 2*x^2)*Log
[3 + 2*x]),x]

[Out]

-x - Log[x] + Log[2*x + x^2 + Log[3 + 2*x]]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]  time = 0.24, size = 22, normalized size = 0.96 \begin {gather*} -x-\log (x)+\log \left (2 x+x^2+\log (3+2 x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x - 3*x^2 - 5*x^3 - 2*x^4 + (-3 - 5*x - 2*x^2)*Log[3 + 2*x])/(6*x^2 + 7*x^3 + 2*x^4 + (3*x + 2*x^
2)*Log[3 + 2*x]),x]

[Out]

-x - Log[x] + Log[2*x + x^2 + Log[3 + 2*x]]

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fricas [A]  time = 0.76, size = 22, normalized size = 0.96 \begin {gather*} -x + \log \left (x^{2} + 2 \, x + \log \left (2 \, x + 3\right )\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2-5*x-3)*log(2*x+3)-2*x^4-5*x^3-3*x^2+2*x)/((2*x^2+3*x)*log(2*x+3)+2*x^4+7*x^3+6*x^2),x, algo
rithm="fricas")

[Out]

-x + log(x^2 + 2*x + log(2*x + 3)) - log(x)

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giac [A]  time = 0.13, size = 26, normalized size = 1.13 \begin {gather*} -x + \log \left (-x^{2} - 2 \, x - \log \left (2 \, x + 3\right )\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2-5*x-3)*log(2*x+3)-2*x^4-5*x^3-3*x^2+2*x)/((2*x^2+3*x)*log(2*x+3)+2*x^4+7*x^3+6*x^2),x, algo
rithm="giac")

[Out]

-x + log(-x^2 - 2*x - log(2*x + 3)) - log(x)

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maple [A]  time = 0.06, size = 22, normalized size = 0.96

method	result	size
risch	\(-x -\ln \relax (x )+\ln \left (\ln \left (2 x +3\right )+x \left (2+x \right )\right )\)	\(22\)
norman	\(-x -\ln \relax (x )+\ln \left (x^{2}+\ln \left (2 x +3\right )+2 x \right )\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x^2-5*x-3)*ln(2*x+3)-2*x^4-5*x^3-3*x^2+2*x)/((2*x^2+3*x)*ln(2*x+3)+2*x^4+7*x^3+6*x^2),x,method=_RETUR
NVERBOSE)

[Out]

-x-ln(x)+ln(ln(2*x+3)+x*(2+x))

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maxima [A]  time = 0.40, size = 22, normalized size = 0.96 \begin {gather*} -x + \log \left (x^{2} + 2 \, x + \log \left (2 \, x + 3\right )\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x^2-5*x-3)*log(2*x+3)-2*x^4-5*x^3-3*x^2+2*x)/((2*x^2+3*x)*log(2*x+3)+2*x^4+7*x^3+6*x^2),x, algo
rithm="maxima")

[Out]

-x + log(x^2 + 2*x + log(2*x + 3)) - log(x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {\ln \left (2\,x+3\right )\,\left (2\,x^2+5\,x+3\right )-2\,x+3\,x^2+5\,x^3+2\,x^4}{\ln \left (2\,x+3\right )\,\left (2\,x^2+3\,x\right )+6\,x^2+7\,x^3+2\,x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(2*x + 3)*(5*x + 2*x^2 + 3) - 2*x + 3*x^2 + 5*x^3 + 2*x^4)/(log(2*x + 3)*(3*x + 2*x^2) + 6*x^2 + 7*x^
3 + 2*x^4),x)

[Out]

int(-(log(2*x + 3)*(5*x + 2*x^2 + 3) - 2*x + 3*x^2 + 5*x^3 + 2*x^4)/(log(2*x + 3)*(3*x + 2*x^2) + 6*x^2 + 7*x^
3 + 2*x^4), x)

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sympy [A]  time = 0.23, size = 19, normalized size = 0.83 \begin {gather*} - x - \log {\relax (x )} + \log {\left (x^{2} + 2 x + \log {\left (2 x + 3 \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x**2-5*x-3)*ln(2*x+3)-2*x**4-5*x**3-3*x**2+2*x)/((2*x**2+3*x)*ln(2*x+3)+2*x**4+7*x**3+6*x**2),x
)

[Out]

-x - log(x) + log(x**2 + 2*x + log(2*x + 3))

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