3.17.37 \(\int \frac {(2+3 x) (3+5 x)}{(1-2 x)^3} \, dx\) [1637]

Optimal. Leaf size=33 \[ \frac {77}{16 (1-2 x)^2}-\frac {17}{2 (1-2 x)}-\frac {15}{8} \log (1-2 x) \]

[Out]

77/16/(1-2*x)^2-17/2/(1-2*x)-15/8*ln(1-2*x)

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Rubi [A]
time = 0.01, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {78} \begin {gather*} -\frac {17}{2 (1-2 x)}+\frac {77}{16 (1-2 x)^2}-\frac {15}{8} \log (1-2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

77/(16*(1 - 2*x)^2) - 17/(2*(1 - 2*x)) - (15*Log[1 - 2*x])/8

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) (3+5 x)}{(1-2 x)^3} \, dx &=\int \left (-\frac {77}{4 (-1+2 x)^3}-\frac {17}{(-1+2 x)^2}-\frac {15}{4 (-1+2 x)}\right ) \, dx\\ &=\frac {77}{16 (1-2 x)^2}-\frac {17}{2 (1-2 x)}-\frac {15}{8} \log (1-2 x)\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 33, normalized size = 1.00 \begin {gather*} \frac {77}{16 (1-2 x)^2}-\frac {17}{2 (1-2 x)}-\frac {15}{8} \log (1-2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

77/(16*(1 - 2*x)^2) - 17/(2*(1 - 2*x)) - (15*Log[1 - 2*x])/8

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Maple [A]
time = 0.12, size = 28, normalized size = 0.85

method	result	size
risch	\(\frac {17 x -\frac {59}{16}}{\left (-1+2 x \right )^{2}}-\frac {15 \ln \left (-1+2 x \right )}{8}\)	\(24\)
norman	\(\frac {\frac {9}{4} x +\frac {59}{4} x^{2}}{\left (-1+2 x \right )^{2}}-\frac {15 \ln \left (-1+2 x \right )}{8}\)	\(27\)
default	\(\frac {17}{2 \left (-1+2 x \right )}+\frac {77}{16 \left (-1+2 x \right )^{2}}-\frac {15 \ln \left (-1+2 x \right )}{8}\)	\(28\)
meijerg	\(\frac {3 x \left (2-2 x \right )}{\left (1-2 x \right )^{2}}+\frac {19 x^{2}}{2 \left (1-2 x \right )^{2}}-\frac {5 x \left (-18 x +6\right )}{8 \left (1-2 x \right )^{2}}-\frac {15 \ln \left (1-2 x \right )}{8}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

17/2/(-1+2*x)+77/16/(-1+2*x)^2-15/8*ln(-1+2*x)

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Maxima [A]
time = 0.28, size = 28, normalized size = 0.85 \begin {gather*} \frac {272 \, x - 59}{16 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} - \frac {15}{8} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(272*x - 59)/(4*x^2 - 4*x + 1) - 15/8*log(2*x - 1)

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Fricas [A]
time = 0.43, size = 37, normalized size = 1.12 \begin {gather*} -\frac {30 \, {\left (4 \, x^{2} - 4 \, x + 1\right )} \log \left (2 \, x - 1\right ) - 272 \, x + 59}{16 \, {\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)*(3+5*x)/(1-2*x)^3,x, algorithm="fricas")

[Out]

-1/16*(30*(4*x^2 - 4*x + 1)*log(2*x - 1) - 272*x + 59)/(4*x^2 - 4*x + 1)

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Sympy [A]
time = 0.05, size = 26, normalized size = 0.79 \begin {gather*} - \frac {59 - 272 x}{64 x^{2} - 64 x + 16} - \frac {15 \log {\left (2 x - 1 \right )}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(59 - 272*x)/(64*x**2 - 64*x + 16) - 15*log(2*x - 1)/8

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Giac [A]
time = 0.98, size = 24, normalized size = 0.73 \begin {gather*} \frac {272 \, x - 59}{16 \, {\left (2 \, x - 1\right )}^{2}} - \frac {15}{8} \, \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)*(3+5*x)/(1-2*x)^3,x, algorithm="giac")

[Out]

1/16*(272*x - 59)/(2*x - 1)^2 - 15/8*log(abs(2*x - 1))

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Mupad [B]
time = 1.06, size = 23, normalized size = 0.70 \begin {gather*} \frac {\frac {17\,x}{4}-\frac {59}{64}}{x^2-x+\frac {1}{4}}-\frac {15\,\ln \left (x-\frac {1}{2}\right )}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((17*x)/4 - 59/64)/(x^2 - x + 1/4) - (15*log(x - 1/2))/8

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