3.15.59 \(\int \frac {(3+5 x)^2}{1-2 x} \, dx\) [1459]

Optimal. Leaf size=23 \[ -\frac {85 x}{4}-\frac {25 x^2}{4}-\frac {121}{8} \log (1-2 x) \]

[Out]

-85/4*x-25/4*x^2-121/8*ln(1-2*x)

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Rubi [A]
time = 0.01, antiderivative size = 23, 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 = {45} \begin {gather*} -\frac {25 x^2}{4}-\frac {85 x}{4}-\frac {121}{8} \log (1-2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 + 5*x)^2/(1 - 2*x),x]

[Out]

(-85*x)/4 - (25*x^2)/4 - (121*Log[1 - 2*x])/8

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin {align*} \int \frac {(3+5 x)^2}{1-2 x} \, dx &=\int \left (-\frac {85}{4}-\frac {25 x}{2}-\frac {121}{4 (-1+2 x)}\right ) \, dx\\ &=-\frac {85 x}{4}-\frac {25 x^2}{4}-\frac {121}{8} \log (1-2 x)\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 25, normalized size = 1.09 \begin {gather*} \frac {1}{16} \left (-5 \left (-39+68 x+20 x^2\right )-242 \log (1-2 x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 + 5*x)^2/(1 - 2*x),x]

[Out]

(-5*(-39 + 68*x + 20*x^2) - 242*Log[1 - 2*x])/16

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Maple [A]
time = 0.10, size = 18, normalized size = 0.78

method	result	size
default	\(-\frac {25 x^{2}}{4}-\frac {85 x}{4}-\frac {121 \ln \left (-1+2 x \right )}{8}\)	\(18\)
norman	\(-\frac {25 x^{2}}{4}-\frac {85 x}{4}-\frac {121 \ln \left (-1+2 x \right )}{8}\)	\(18\)
risch	\(-\frac {25 x^{2}}{4}-\frac {85 x}{4}-\frac {121 \ln \left (-1+2 x \right )}{8}\)	\(18\)
meijerg	\(-\frac {121 \ln \left (1-2 x \right )}{8}-15 x -\frac {25 x \left (6 x +6\right )}{24}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/4*x^2-85/4*x-121/8*ln(-1+2*x)

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Maxima [A]
time = 0.53, size = 17, normalized size = 0.74 \begin {gather*} -\frac {25}{4} \, x^{2} - \frac {85}{4} \, x - \frac {121}{8} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/4*x^2 - 85/4*x - 121/8*log(2*x - 1)

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Fricas [A]
time = 0.42, size = 17, normalized size = 0.74 \begin {gather*} -\frac {25}{4} \, x^{2} - \frac {85}{4} \, x - \frac {121}{8} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/4*x^2 - 85/4*x - 121/8*log(2*x - 1)

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Sympy [A]
time = 0.02, size = 22, normalized size = 0.96 \begin {gather*} - \frac {25 x^{2}}{4} - \frac {85 x}{4} - \frac {121 \log {\left (2 x - 1 \right )}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25*x**2/4 - 85*x/4 - 121*log(2*x - 1)/8

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Giac [A]
time = 0.69, size = 18, normalized size = 0.78 \begin {gather*} -\frac {25}{4} \, x^{2} - \frac {85}{4} \, x - \frac {121}{8} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/4*x^2 - 85/4*x - 121/8*log(abs(2*x - 1))

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Mupad [B]
time = 0.02, size = 15, normalized size = 0.65 \begin {gather*} -\frac {85\,x}{4}-\frac {121\,\ln \left (x-\frac {1}{2}\right )}{8}-\frac {25\,x^2}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*x + 3)^2/(2*x - 1),x)

[Out]

- (85*x)/4 - (121*log(x - 1/2))/8 - (25*x^2)/4

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