3.16.95 \(\int \frac {2+3 x}{(1-2 x)^2 (3+5 x)} \, dx\) [1595]

Optimal. Leaf size=32 \[ \frac {7}{22 (1-2 x)}-\frac {1}{121} \log (1-2 x)+\frac {1}{121} \log (3+5 x) \]

[Out]

7/22/(1-2*x)-1/121*ln(1-2*x)+1/121*ln(3+5*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 {7}{22 (1-2 x)}-\frac {1}{121} \log (1-2 x)+\frac {1}{121} \log (5 x+3) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

7/(22*(1 - 2*x)) - Log[1 - 2*x]/121 + Log[3 + 5*x]/121

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)^2 (3+5 x)} \, dx &=\int \left (\frac {7}{11 (-1+2 x)^2}-\frac {2}{121 (-1+2 x)}+\frac {5}{121 (3+5 x)}\right ) \, dx\\ &=\frac {7}{22 (1-2 x)}-\frac {1}{121} \log (1-2 x)+\frac {1}{121} \log (3+5 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 (6+10 x)}{242 (-1+2 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-77 + (2 - 4*x)*Log[1 - 2*x] + (-2 + 4*x)*Log[6 + 10*x])/(242*(-1 + 2*x))

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Maple [A]
time = 0.10, size = 27, normalized size = 0.84

method	result	size
risch	\(-\frac {7}{44 \left (-\frac {1}{2}+x \right )}-\frac {\ln \left (-1+2 x \right )}{121}+\frac {\ln \left (3+5 x \right )}{121}\)	\(25\)
default	\(-\frac {7}{22 \left (-1+2 x \right )}-\frac {\ln \left (-1+2 x \right )}{121}+\frac {\ln \left (3+5 x \right )}{121}\)	\(27\)
norman	\(-\frac {7 x}{11 \left (-1+2 x \right )}-\frac {\ln \left (-1+2 x \right )}{121}+\frac {\ln \left (3+5 x \right )}{121}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/22/(-1+2*x)-1/121*ln(-1+2*x)+1/121*ln(3+5*x)

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Maxima [A]
time = 0.33, size = 26, normalized size = 0.81 \begin {gather*} -\frac {7}{22 \, {\left (2 \, x - 1\right )}} + \frac {1}{121} \, \log \left (5 \, x + 3\right ) - \frac {1}{121} \, \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)^2/(3+5*x),x, algorithm="maxima")

[Out]

-7/22/(2*x - 1) + 1/121*log(5*x + 3) - 1/121*log(2*x - 1)

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Fricas [A]
time = 0.39, size = 37, normalized size = 1.16 \begin {gather*} \frac {2 \, {\left (2 \, x - 1\right )} \log \left (5 \, x + 3\right ) - 2 \, {\left (2 \, x - 1\right )} \log \left (2 \, x - 1\right ) - 77}{242 \, {\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)^2/(3+5*x),x, algorithm="fricas")

[Out]

1/242*(2*(2*x - 1)*log(5*x + 3) - 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 )}}{121} + \frac {\log {\left (x + \frac {3}{5} \right )}}{121} - \frac {7}{44 x - 22} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(x - 1/2)/121 + log(x + 3/5)/121 - 7/(44*x - 22)

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Giac [A]
time = 0.62, size = 25, normalized size = 0.78 \begin {gather*} -\frac {7}{22 \, {\left (2 \, x - 1\right )}} + \frac {1}{121} \, \log \left ({\left | -\frac {11}{2 \, x - 1} - 5 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/22/(2*x - 1) + 1/121*log(abs(-11/(2*x - 1) - 5))

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Mupad [B]
time = 1.07, size = 18, normalized size = 0.56 \begin {gather*} \frac {2\,\mathrm {atanh}\left (\frac {20\,x}{11}+\frac {1}{11}\right )}{121}-\frac {7}{44\,\left (x-\frac {1}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*atanh((20*x)/11 + 1/11))/121 - 7/(44*(x - 1/2))

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