3.16.46 \(\int \frac {3+5 x}{(1-2 x)^2} \, dx\) [1546]

Optimal. Leaf size=22 \[ \frac {11}{4 (1-2 x)}+\frac {5}{4} \log (1-2 x) \]

[Out]

11/4/(1-2*x)+5/4*ln(1-2*x)

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Rubi [A]
time = 0.01, antiderivative size = 22, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {45} \begin {gather*} \frac {11}{4 (1-2 x)}+\frac {5}{4} \log (1-2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 + 5*x)/(1 - 2*x)^2,x]

[Out]

11/(4*(1 - 2*x)) + (5*Log[1 - 2*x])/4

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}{(1-2 x)^2} \, dx &=\int \left (\frac {11}{2 (-1+2 x)^2}+\frac {5}{2 (-1+2 x)}\right ) \, dx\\ &=\frac {11}{4 (1-2 x)}+\frac {5}{4} \log (1-2 x)\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 22, normalized size = 1.00 \begin {gather*} \frac {11}{4 (1-2 x)}+\frac {5}{4} \log (1-2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 + 5*x)/(1 - 2*x)^2,x]

[Out]

11/(4*(1 - 2*x)) + (5*Log[1 - 2*x])/4

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Maple [A]
time = 0.09, size = 19, normalized size = 0.86

method	result	size
risch	\(-\frac {11}{8 \left (-\frac {1}{2}+x \right )}+\frac {5 \ln \left (-1+2 x \right )}{4}\)	\(17\)
default	\(-\frac {11}{4 \left (-1+2 x \right )}+\frac {5 \ln \left (-1+2 x \right )}{4}\)	\(19\)
norman	\(-\frac {11 x}{2 \left (-1+2 x \right )}+\frac {5 \ln \left (-1+2 x \right )}{4}\)	\(20\)
meijerg	\(\frac {11 x}{2 \left (1-2 x \right )}+\frac {5 \ln \left (1-2 x \right )}{4}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-11/4/(-1+2*x)+5/4*ln(-1+2*x)

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Maxima [A]
time = 0.30, size = 18, normalized size = 0.82 \begin {gather*} -\frac {11}{4 \, {\left (2 \, x - 1\right )}} + \frac {5}{4} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-11/4/(2*x - 1) + 5/4*log(2*x - 1)

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Fricas [A]
time = 0.89, size = 24, normalized size = 1.09 \begin {gather*} \frac {5 \, {\left (2 \, x - 1\right )} \log \left (2 \, x - 1\right ) - 11}{4 \, {\left (2 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(5*(2*x - 1)*log(2*x - 1) - 11)/(2*x - 1)

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Sympy [A]
time = 0.03, size = 15, normalized size = 0.68 \begin {gather*} \frac {5 \log {\left (2 x - 1 \right )}}{4} - \frac {11}{8 x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*log(2*x - 1)/4 - 11/(8*x - 4)

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Giac [A]
time = 1.27, size = 28, normalized size = 1.27 \begin {gather*} -\frac {11}{4 \, {\left (2 \, x - 1\right )}} - \frac {5}{4} \, \log \left (\frac {{\left | 2 \, x - 1 \right |}}{2 \, {\left (2 \, x - 1\right )}^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-11/4/(2*x - 1) - 5/4*log(1/2*abs(2*x - 1)/(2*x - 1)^2)

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Mupad [B]
time = 1.13, size = 16, normalized size = 0.73 \begin {gather*} \frac {5\,\ln \left (x-\frac {1}{2}\right )}{4}-\frac {11}{8\,\left (x-\frac {1}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x + 3)/(2*x - 1)^2,x)

[Out]

(5*log(x - 1/2))/4 - 11/(8*(x - 1/2))

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