3.16.47 \(\int \frac {3+5 x}{(1-2 x)^2 (2+3 x)} \, dx\) [1547]

Optimal. Leaf size=32 \[ \frac {11}{14 (1-2 x)}+\frac {1}{49} \log (1-2 x)-\frac {1}{49} \log (2+3 x) \]

[Out]

11/14/(1-2*x)+1/49*ln(1-2*x)-1/49*ln(2+3*x)

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Rubi [A]
time = 0.01, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {78} \begin {gather*} \frac {11}{14 (1-2 x)}+\frac {1}{49} \log (1-2 x)-\frac {1}{49} \log (3 x+2) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 + 5*x)/((1 - 2*x)^2*(2 + 3*x)),x]

[Out]

11/(14*(1 - 2*x)) + Log[1 - 2*x]/49 - Log[2 + 3*x]/49

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin {align*} \int \frac {3+5 x}{(1-2 x)^2 (2+3 x)} \, dx &=\int \left (\frac {11}{7 (-1+2 x)^2}+\frac {2}{49 (-1+2 x)}-\frac {3}{49 (2+3 x)}\right ) \, dx\\ &=\frac {11}{14 (1-2 x)}+\frac {1}{49} \log (1-2 x)-\frac {1}{49} \log (2+3 x)\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 37, normalized size = 1.16 \begin {gather*} \frac {-77+(-2+4 x) \log (1-2 x)+(2-4 x) \log (4+6 x)}{98 (-1+2 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 + 5*x)/((1 - 2*x)^2*(2 + 3*x)),x]

[Out]

(-77 + (-2 + 4*x)*Log[1 - 2*x] + (2 - 4*x)*Log[4 + 6*x])/(98*(-1 + 2*x))

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Maple [A]
time = 0.11, size = 27, normalized size = 0.84

method	result	size
risch	\(-\frac {11}{28 \left (-\frac {1}{2}+x \right )}+\frac {\ln \left (-1+2 x \right )}{49}-\frac {\ln \left (2+3 x \right )}{49}\)	\(25\)
default	\(-\frac {11}{14 \left (-1+2 x \right )}+\frac {\ln \left (-1+2 x \right )}{49}-\frac {\ln \left (2+3 x \right )}{49}\)	\(27\)
norman	\(-\frac {11 x}{7 \left (-1+2 x \right )}+\frac {\ln \left (-1+2 x \right )}{49}-\frac {\ln \left (2+3 x \right )}{49}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3+5*x)/(1-2*x)^2/(2+3*x),x,method=_RETURNVERBOSE)

[Out]

-11/14/(-1+2*x)+1/49*ln(-1+2*x)-1/49*ln(2+3*x)

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Maxima [A]
time = 0.30, size = 26, normalized size = 0.81 \begin {gather*} -\frac {11}{14 \, {\left (2 \, x - 1\right )}} - \frac {1}{49} \, \log \left (3 \, x + 2\right ) + \frac {1}{49} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3+5*x)/(1-2*x)^2/(2+3*x),x, algorithm="maxima")

[Out]

-11/14/(2*x - 1) - 1/49*log(3*x + 2) + 1/49*log(2*x - 1)

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Fricas [A]
time = 0.78, size = 37, normalized size = 1.16 \begin {gather*} -\frac {2 \, {\left (2 \, x - 1\right )} \log \left (3 \, x + 2\right ) - 2 \, {\left (2 \, x - 1\right )} \log \left (2 \, x - 1\right ) + 77}{98 \, {\left (2 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3+5*x)/(1-2*x)^2/(2+3*x),x, algorithm="fricas")

[Out]

-1/98*(2*(2*x - 1)*log(3*x + 2) - 2*(2*x - 1)*log(2*x - 1) + 77)/(2*x - 1)

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Sympy [A]
time = 0.04, size = 22, normalized size = 0.69 \begin {gather*} \frac {\log {\left (x - \frac {1}{2} \right )}}{49} - \frac {\log {\left (x + \frac {2}{3} \right )}}{49} - \frac {11}{28 x - 14} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3+5*x)/(1-2*x)**2/(2+3*x),x)

[Out]

log(x - 1/2)/49 - log(x + 2/3)/49 - 11/(28*x - 14)

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Giac [A]
time = 1.76, size = 25, normalized size = 0.78 \begin {gather*} -\frac {11}{14 \, {\left (2 \, x - 1\right )}} - \frac {1}{49} \, \log \left ({\left | -\frac {7}{2 \, x - 1} - 3 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3+5*x)/(1-2*x)^2/(2+3*x),x, algorithm="giac")

[Out]

-11/14/(2*x - 1) - 1/49*log(abs(-7/(2*x - 1) - 3))

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Mupad [B]
time = 0.04, size = 18, normalized size = 0.56 \begin {gather*} -\frac {2\,\mathrm {atanh}\left (\frac {12\,x}{7}+\frac {1}{7}\right )}{49}-\frac {11}{28\,\left (x-\frac {1}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x + 3)/((2*x - 1)^2*(3*x + 2)),x)

[Out]

- (2*atanh((12*x)/7 + 1/7))/49 - 11/(28*(x - 1/2))

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