3.15.92 \(\int \frac {2+3 x}{(1-2 x) (3+5 x)} \, dx\) [1492]

Optimal. Leaf size=21 \[ -\frac {7}{22} \log (1-2 x)+\frac {1}{55} \log (3+5 x) \]

[Out]

-7/22*ln(1-2*x)+1/55*ln(3+5*x)

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Rubi [A]
time = 0.01, antiderivative size = 21, 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 {1}{55} \log (5 x+3)-\frac {7}{22} \log (1-2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*Log[1 - 2*x])/22 + Log[3 + 5*x]/55

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 {2+3 x}{(1-2 x) (3+5 x)} \, dx &=\int \left (-\frac {7}{11 (-1+2 x)}+\frac {1}{11 (3+5 x)}\right ) \, dx\\ &=-\frac {7}{22} \log (1-2 x)+\frac {1}{55} \log (3+5 x)\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 21, normalized size = 1.00 \begin {gather*} -\frac {7}{22} \log (1-2 x)+\frac {1}{55} \log (3+5 x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-7*Log[1 - 2*x])/22 + Log[3 + 5*x]/55

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Maple [A]
time = 0.11, size = 18, normalized size = 0.86

method	result	size
default	\(-\frac {7 \ln \left (-1+2 x \right )}{22}+\frac {\ln \left (3+5 x \right )}{55}\)	\(18\)
norman	\(-\frac {7 \ln \left (-1+2 x \right )}{22}+\frac {\ln \left (3+5 x \right )}{55}\)	\(18\)
risch	\(-\frac {7 \ln \left (-1+2 x \right )}{22}+\frac {\ln \left (3+5 x \right )}{55}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/22*ln(-1+2*x)+1/55*ln(3+5*x)

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Maxima [A]
time = 0.31, size = 17, normalized size = 0.81 \begin {gather*} \frac {1}{55} \, \log \left (5 \, x + 3\right ) - \frac {7}{22} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/55*log(5*x + 3) - 7/22*log(2*x - 1)

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Fricas [A]
time = 0.90, size = 17, normalized size = 0.81 \begin {gather*} \frac {1}{55} \, \log \left (5 \, x + 3\right ) - \frac {7}{22} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/55*log(5*x + 3) - 7/22*log(2*x - 1)

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Sympy [A]
time = 0.04, size = 17, normalized size = 0.81 \begin {gather*} - \frac {7 \log {\left (x - \frac {1}{2} \right )}}{22} + \frac {\log {\left (x + \frac {3}{5} \right )}}{55} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7*log(x - 1/2)/22 + log(x + 3/5)/55

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Giac [A]
time = 1.31, size = 19, normalized size = 0.90 \begin {gather*} \frac {1}{55} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - \frac {7}{22} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/55*log(abs(5*x + 3)) - 7/22*log(abs(2*x - 1))

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Mupad [B]
time = 1.15, size = 13, normalized size = 0.62 \begin {gather*} \frac {\ln \left (x+\frac {3}{5}\right )}{55}-\frac {7\,\ln \left (x-\frac {1}{2}\right )}{22} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x + 3/5)/55 - (7*log(x - 1/2))/22

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