3.17.25 \(\int \frac {1}{(1-2 x)^2 (3+5 x)^3} \, dx\) [1625]

Optimal. Leaf size=54 \[ \frac {4}{1331 (1-2 x)}-\frac {5}{242 (3+5 x)^2}-\frac {20}{1331 (3+5 x)}-\frac {60 \log (1-2 x)}{14641}+\frac {60 \log (3+5 x)}{14641} \]

[Out]

4/1331/(1-2*x)-5/242/(3+5*x)^2-20/1331/(3+5*x)-60/14641*ln(1-2*x)+60/14641*ln(3+5*x)

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Rubi [A]
time = 0.02, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {46} \begin {gather*} \frac {4}{1331 (1-2 x)}-\frac {20}{1331 (5 x+3)}-\frac {5}{242 (5 x+3)^2}-\frac {60 \log (1-2 x)}{14641}+\frac {60 \log (5 x+3)}{14641} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((1 - 2*x)^2*(3 + 5*x)^3),x]

[Out]

4/(1331*(1 - 2*x)) - 5/(242*(3 + 5*x)^2) - 20/(1331*(3 + 5*x)) - (60*Log[1 - 2*x])/14641 + (60*Log[3 + 5*x])/1
4641

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rubi steps

\begin {align*} \int \frac {1}{(1-2 x)^2 (3+5 x)^3} \, dx &=\int \left (\frac {8}{1331 (-1+2 x)^2}-\frac {120}{14641 (-1+2 x)}+\frac {25}{121 (3+5 x)^3}+\frac {100}{1331 (3+5 x)^2}+\frac {300}{14641 (3+5 x)}\right ) \, dx\\ &=\frac {4}{1331 (1-2 x)}-\frac {5}{242 (3+5 x)^2}-\frac {20}{1331 (3+5 x)}-\frac {60 \log (1-2 x)}{14641}+\frac {60 \log (3+5 x)}{14641}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 47, normalized size = 0.87 \begin {gather*} \frac {-\frac {11 \left (-103+390 x+600 x^2\right )}{(-1+2 x) (3+5 x)^2}-120 \log (1-2 x)+120 \log (6+10 x)}{29282} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-11*(-103 + 390*x + 600*x^2))/((-1 + 2*x)*(3 + 5*x)^2) - 120*Log[1 - 2*x] + 120*Log[6 + 10*x])/29282

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Maple [A]
time = 0.10, size = 45, normalized size = 0.83

method	result	size
risch	\(\frac {-\frac {300}{1331} x^{2}-\frac {195}{1331} x +\frac {103}{2662}}{\left (-1+2 x \right ) \left (3+5 x \right )^{2}}-\frac {60 \ln \left (-1+2 x \right )}{14641}+\frac {60 \ln \left (3+5 x \right )}{14641}\)	\(44\)
default	\(-\frac {4}{1331 \left (-1+2 x \right )}-\frac {60 \ln \left (-1+2 x \right )}{14641}-\frac {5}{242 \left (3+5 x \right )^{2}}-\frac {20}{1331 \left (3+5 x \right )}+\frac {60 \ln \left (3+5 x \right )}{14641}\)	\(45\)
norman	\(\frac {\frac {2575}{11979} x^{3}-\frac {1795}{23958} x^{2}-\frac {791}{3993} x}{\left (-1+2 x \right ) \left (3+5 x \right )^{2}}-\frac {60 \ln \left (-1+2 x \right )}{14641}+\frac {60 \ln \left (3+5 x \right )}{14641}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/1331/(-1+2*x)-60/14641*ln(-1+2*x)-5/242/(3+5*x)^2-20/1331/(3+5*x)+60/14641*ln(3+5*x)

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Maxima [A]
time = 0.50, size = 46, normalized size = 0.85 \begin {gather*} -\frac {600 \, x^{2} + 390 \, x - 103}{2662 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )}} + \frac {60}{14641} \, \log \left (5 \, x + 3\right ) - \frac {60}{14641} \, \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/(3+5*x)^3,x, algorithm="maxima")

[Out]

-1/2662*(600*x^2 + 390*x - 103)/(50*x^3 + 35*x^2 - 12*x - 9) + 60/14641*log(5*x + 3) - 60/14641*log(2*x - 1)

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Fricas [A]
time = 0.37, size = 75, normalized size = 1.39 \begin {gather*} -\frac {6600 \, x^{2} - 120 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )} \log \left (5 \, x + 3\right ) + 120 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )} \log \left (2 \, x - 1\right ) + 4290 \, x - 1133}{29282 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/29282*(6600*x^2 - 120*(50*x^3 + 35*x^2 - 12*x - 9)*log(5*x + 3) + 120*(50*x^3 + 35*x^2 - 12*x - 9)*log(2*x
- 1) + 4290*x - 1133)/(50*x^3 + 35*x^2 - 12*x - 9)

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Sympy [A]
time = 0.07, size = 44, normalized size = 0.81 \begin {gather*} \frac {- 600 x^{2} - 390 x + 103}{133100 x^{3} + 93170 x^{2} - 31944 x - 23958} - \frac {60 \log {\left (x - \frac {1}{2} \right )}}{14641} + \frac {60 \log {\left (x + \frac {3}{5} \right )}}{14641} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-600*x**2 - 390*x + 103)/(133100*x**3 + 93170*x**2 - 31944*x - 23958) - 60*log(x - 1/2)/14641 + 60*log(x + 3/
5)/14641

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Giac [A]
time = 0.62, size = 51, normalized size = 0.94 \begin {gather*} -\frac {4}{1331 \, {\left (2 \, x - 1\right )}} + \frac {50 \, {\left (\frac {66}{2 \, x - 1} + 25\right )}}{14641 \, {\left (\frac {11}{2 \, x - 1} + 5\right )}^{2}} + \frac {60}{14641} \, \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/(3+5*x)^3,x, algorithm="giac")

[Out]

-4/1331/(2*x - 1) + 50/14641*(66/(2*x - 1) + 25)/(11/(2*x - 1) + 5)^2 + 60/14641*log(abs(-11/(2*x - 1) - 5))

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Mupad [B]
time = 1.06, size = 37, normalized size = 0.69 \begin {gather*} \frac {120\,\mathrm {atanh}\left (\frac {20\,x}{11}+\frac {1}{11}\right )}{14641}+\frac {\frac {6\,x^2}{1331}+\frac {39\,x}{13310}-\frac {103}{133100}}{-x^3-\frac {7\,x^2}{10}+\frac {6\,x}{25}+\frac {9}{50}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((2*x - 1)^2*(5*x + 3)^3),x)

[Out]

(120*atanh((20*x)/11 + 1/11))/14641 + ((39*x)/13310 + (6*x^2)/1331 - 103/133100)/((6*x)/25 - (7*x^2)/10 - x^3
+ 9/50)

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