3.16.97 \(\int \frac {1}{(1-2 x)^2 (2+3 x) (3+5 x)} \, dx\) [1597]

Optimal. Leaf size=42 \[ \frac {2}{77 (1-2 x)}-\frac {136 \log (1-2 x)}{5929}-\frac {9}{49} \log (2+3 x)+\frac {25}{121} \log (3+5 x) \]

[Out]

2/77/(1-2*x)-136/5929*ln(1-2*x)-9/49*ln(2+3*x)+25/121*ln(3+5*x)

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Rubi [A]
time = 0.01, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {84} \begin {gather*} \frac {2}{77 (1-2 x)}-\frac {136 \log (1-2 x)}{5929}-\frac {9}{49} \log (3 x+2)+\frac {25}{121} \log (5 x+3) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

2/(77*(1 - 2*x)) - (136*Log[1 - 2*x])/5929 - (9*Log[2 + 3*x])/49 + (25*Log[3 + 5*x])/121

Rule 84

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {1}{(1-2 x)^2 (2+3 x) (3+5 x)} \, dx &=\int \left (\frac {4}{77 (-1+2 x)^2}-\frac {272}{5929 (-1+2 x)}-\frac {27}{49 (2+3 x)}+\frac {125}{121 (3+5 x)}\right ) \, dx\\ &=\frac {2}{77 (1-2 x)}-\frac {136 \log (1-2 x)}{5929}-\frac {9}{49} \log (2+3 x)+\frac {25}{121} \log (3+5 x)\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 40, normalized size = 0.95 \begin {gather*} \frac {\frac {154}{1-2 x}-136 \log (3-6 x)-1089 \log (2+3 x)+1225 \log (-3 (3+5 x))}{5929} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(154/(1 - 2*x) - 136*Log[3 - 6*x] - 1089*Log[2 + 3*x] + 1225*Log[-3*(3 + 5*x)])/5929

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Maple [A]
time = 0.12, size = 35, normalized size = 0.83

method	result	size
risch	\(-\frac {1}{77 \left (-\frac {1}{2}+x \right )}-\frac {136 \ln \left (-1+2 x \right )}{5929}-\frac {9 \ln \left (2+3 x \right )}{49}+\frac {25 \ln \left (3+5 x \right )}{121}\)	\(33\)
default	\(-\frac {2}{77 \left (-1+2 x \right )}-\frac {136 \ln \left (-1+2 x \right )}{5929}-\frac {9 \ln \left (2+3 x \right )}{49}+\frac {25 \ln \left (3+5 x \right )}{121}\)	\(35\)
norman	\(-\frac {4 x}{77 \left (-1+2 x \right )}-\frac {136 \ln \left (-1+2 x \right )}{5929}-\frac {9 \ln \left (2+3 x \right )}{49}+\frac {25 \ln \left (3+5 x \right )}{121}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/77/(-1+2*x)-136/5929*ln(-1+2*x)-9/49*ln(2+3*x)+25/121*ln(3+5*x)

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Maxima [A]
time = 0.31, size = 34, normalized size = 0.81 \begin {gather*} -\frac {2}{77 \, {\left (2 \, x - 1\right )}} + \frac {25}{121} \, \log \left (5 \, x + 3\right ) - \frac {9}{49} \, \log \left (3 \, x + 2\right ) - \frac {136}{5929} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/77/(2*x - 1) + 25/121*log(5*x + 3) - 9/49*log(3*x + 2) - 136/5929*log(2*x - 1)

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Fricas [A]
time = 0.39, size = 50, normalized size = 1.19 \begin {gather*} \frac {1225 \, {\left (2 \, x - 1\right )} \log \left (5 \, x + 3\right ) - 1089 \, {\left (2 \, x - 1\right )} \log \left (3 \, x + 2\right ) - 136 \, {\left (2 \, x - 1\right )} \log \left (2 \, x - 1\right ) - 154}{5929 \, {\left (2 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5929*(1225*(2*x - 1)*log(5*x + 3) - 1089*(2*x - 1)*log(3*x + 2) - 136*(2*x - 1)*log(2*x - 1) - 154)/(2*x - 1
)

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Sympy [A]
time = 0.07, size = 36, normalized size = 0.86 \begin {gather*} - \frac {136 \log {\left (x - \frac {1}{2} \right )}}{5929} + \frac {25 \log {\left (x + \frac {3}{5} \right )}}{121} - \frac {9 \log {\left (x + \frac {2}{3} \right )}}{49} - \frac {2}{154 x - 77} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-136*log(x - 1/2)/5929 + 25*log(x + 3/5)/121 - 9*log(x + 2/3)/49 - 2/(154*x - 77)

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Giac [A]
time = 0.57, size = 40, normalized size = 0.95 \begin {gather*} -\frac {2}{77 \, {\left (2 \, x - 1\right )}} - \frac {9}{49} \, \log \left ({\left | -\frac {7}{2 \, x - 1} - 3 \right |}\right ) + \frac {25}{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(1/(1-2*x)^2/(2+3*x)/(3+5*x),x, algorithm="giac")

[Out]

-2/77/(2*x - 1) - 9/49*log(abs(-7/(2*x - 1) - 3)) + 25/121*log(abs(-11/(2*x - 1) - 5))

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Mupad [B]
time = 1.08, size = 28, normalized size = 0.67 \begin {gather*} \frac {25\,\ln \left (x+\frac {3}{5}\right )}{121}-\frac {9\,\ln \left (x+\frac {2}{3}\right )}{49}-\frac {136\,\ln \left (x-\frac {1}{2}\right )}{5929}-\frac {1}{77\,\left (x-\frac {1}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(25*log(x + 3/5))/121 - (9*log(x + 2/3))/49 - (136*log(x - 1/2))/5929 - 1/(77*(x - 1/2))

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