3.17.77 \(\int \frac {(2+3 x)^2}{(1-2 x)^3 (3+5 x)} \, dx\) [1677]

Optimal. Leaf size=43 \[ \frac {49}{88 (1-2 x)^2}-\frac {217}{484 (1-2 x)}-\frac {\log (1-2 x)}{1331}+\frac {\log (3+5 x)}{1331} \]

[Out]

49/88/(1-2*x)^2-217/484/(1-2*x)-1/1331*ln(1-2*x)+1/1331*ln(3+5*x)

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Rubi [A]
time = 0.01, antiderivative size = 43, 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 = {90} \begin {gather*} -\frac {217}{484 (1-2 x)}+\frac {49}{88 (1-2 x)^2}-\frac {\log (1-2 x)}{1331}+\frac {\log (5 x+3)}{1331} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

49/(88*(1 - 2*x)^2) - 217/(484*(1 - 2*x)) - Log[1 - 2*x]/1331 + Log[3 + 5*x]/1331

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin {align*} \int \frac {(2+3 x)^2}{(1-2 x)^3 (3+5 x)} \, dx &=\int \left (-\frac {49}{22 (-1+2 x)^3}-\frac {217}{242 (-1+2 x)^2}-\frac {2}{1331 (-1+2 x)}+\frac {5}{1331 (3+5 x)}\right ) \, dx\\ &=\frac {49}{88 (1-2 x)^2}-\frac {217}{484 (1-2 x)}-\frac {\log (1-2 x)}{1331}+\frac {\log (3+5 x)}{1331}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 35, normalized size = 0.81 \begin {gather*} \frac {\frac {77 (15+124 x)}{(1-2 x)^2}-8 \log (5-10 x)+8 \log (3+5 x)}{10648} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((77*(15 + 124*x))/(1 - 2*x)^2 - 8*Log[5 - 10*x] + 8*Log[3 + 5*x])/10648

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Maple [A]
time = 0.10, size = 36, normalized size = 0.84

method	result	size
risch	\(\frac {\frac {217 x}{242}+\frac {105}{968}}{\left (-1+2 x \right )^{2}}-\frac {\ln \left (-1+2 x \right )}{1331}+\frac {\ln \left (3+5 x \right )}{1331}\)	\(32\)
norman	\(\frac {\frac {161}{121} x -\frac {105}{242} x^{2}}{\left (-1+2 x \right )^{2}}-\frac {\ln \left (-1+2 x \right )}{1331}+\frac {\ln \left (3+5 x \right )}{1331}\)	\(35\)
default	\(\frac {49}{88 \left (-1+2 x \right )^{2}}+\frac {217}{484 \left (-1+2 x \right )}-\frac {\ln \left (-1+2 x \right )}{1331}+\frac {\ln \left (3+5 x \right )}{1331}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

49/88/(-1+2*x)^2+217/484/(-1+2*x)-1/1331*ln(-1+2*x)+1/1331*ln(3+5*x)

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Maxima [A]
time = 0.32, size = 36, normalized size = 0.84 \begin {gather*} \frac {7 \, {\left (124 \, x + 15\right )}}{968 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} + \frac {1}{1331} \, \log \left (5 \, x + 3\right ) - \frac {1}{1331} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7/968*(124*x + 15)/(4*x^2 - 4*x + 1) + 1/1331*log(5*x + 3) - 1/1331*log(2*x - 1)

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Fricas [A]
time = 0.46, size = 55, normalized size = 1.28 \begin {gather*} \frac {8 \, {\left (4 \, x^{2} - 4 \, x + 1\right )} \log \left (5 \, x + 3\right ) - 8 \, {\left (4 \, x^{2} - 4 \, x + 1\right )} \log \left (2 \, x - 1\right ) + 9548 \, x + 1155}{10648 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10648*(8*(4*x^2 - 4*x + 1)*log(5*x + 3) - 8*(4*x^2 - 4*x + 1)*log(2*x - 1) + 9548*x + 1155)/(4*x^2 - 4*x + 1
)

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Sympy [A]
time = 0.07, size = 32, normalized size = 0.74 \begin {gather*} - \frac {- 868 x - 105}{3872 x^{2} - 3872 x + 968} - \frac {\log {\left (x - \frac {1}{2} \right )}}{1331} + \frac {\log {\left (x + \frac {3}{5} \right )}}{1331} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-868*x - 105)/(3872*x**2 - 3872*x + 968) - log(x - 1/2)/1331 + log(x + 3/5)/1331

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Giac [A]
time = 2.71, size = 33, normalized size = 0.77 \begin {gather*} \frac {7 \, {\left (124 \, x + 15\right )}}{968 \, {\left (2 \, x - 1\right )}^{2}} + \frac {1}{1331} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - \frac {1}{1331} \, \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)^2/(1-2*x)^3/(3+5*x),x, algorithm="giac")

[Out]

7/968*(124*x + 15)/(2*x - 1)^2 + 1/1331*log(abs(5*x + 3)) - 1/1331*log(abs(2*x - 1))

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Mupad [B]
time = 1.09, size = 25, normalized size = 0.58 \begin {gather*} \frac {2\,\mathrm {atanh}\left (\frac {20\,x}{11}+\frac {1}{11}\right )}{1331}+\frac {\frac {217\,x}{968}+\frac {105}{3872}}{x^2-x+\frac {1}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*atanh((20*x)/11 + 1/11))/1331 + ((217*x)/968 + 105/3872)/(x^2 - x + 1/4)

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