3.15.6 \(\int \frac {(2+3 x) (3+5 x)}{(1-2 x)^2} \, dx\)

Optimal. Leaf size=27 \[ \frac {15 x}{4}+\frac {77}{8 (1-2 x)}+\frac {17}{2} \log (1-2 x) \]

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Rubi [A]  time = 0.01, antiderivative size = 27, 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 = {77} \begin {gather*} \frac {15 x}{4}+\frac {77}{8 (1-2 x)}+\frac {17}{2} \log (1-2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

77/(8*(1 - 2*x)) + (15*x)/4 + (17*Log[1 - 2*x])/2

Rule 77

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

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Mathematica [A]  time = 0.02, size = 26, normalized size = 0.96 \begin {gather*} \frac {1}{8} \left (30 x+\frac {77}{1-2 x}+68 \log (1-2 x)-15\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-15 + 77/(1 - 2*x) + 30*x + 68*Log[1 - 2*x])/8

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(2+3 x) (3+5 x)}{(1-2 x)^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.41, size = 32, normalized size = 1.19 \begin {gather*} \frac {60 \, x^{2} + 68 \, {\left (2 \, x - 1\right )} \log \left (2 \, x - 1\right ) - 30 \, x - 77}{8 \, {\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)^2,x, algorithm="fricas")

[Out]

1/8*(60*x^2 + 68*(2*x - 1)*log(2*x - 1) - 30*x - 77)/(2*x - 1)

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giac [A]  time = 0.98, size = 32, normalized size = 1.19 \begin {gather*} \frac {15}{4} \, x - \frac {77}{8 \, {\left (2 \, x - 1\right )}} - \frac {17}{2} \, \log \left (\frac {{\left | 2 \, x - 1 \right |}}{2 \, {\left (2 \, x - 1\right )}^{2}}\right ) - \frac {15}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15/4*x - 77/8/(2*x - 1) - 17/2*log(1/2*abs(2*x - 1)/(2*x - 1)^2) - 15/8

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maple [A]  time = 0.01, size = 22, normalized size = 0.81 \begin {gather*} \frac {15 x}{4}+\frac {17 \ln \left (2 x -1\right )}{2}-\frac {77}{8 \left (2 x -1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15/4*x-77/8/(2*x-1)+17/2*ln(2*x-1)

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maxima [A]  time = 0.65, size = 21, normalized size = 0.78 \begin {gather*} \frac {15}{4} \, x - \frac {77}{8 \, {\left (2 \, x - 1\right )}} + \frac {17}{2} \, \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)^2,x, algorithm="maxima")

[Out]

15/4*x - 77/8/(2*x - 1) + 17/2*log(2*x - 1)

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mupad [B]  time = 0.03, size = 19, normalized size = 0.70 \begin {gather*} \frac {15\,x}{4}+\frac {17\,\ln \left (x-\frac {1}{2}\right )}{2}-\frac {77}{16\,\left (x-\frac {1}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(15*x)/4 + (17*log(x - 1/2))/2 - 77/(16*(x - 1/2))

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sympy [A]  time = 0.10, size = 20, normalized size = 0.74 \begin {gather*} \frac {15 x}{4} + \frac {17 \log {\left (2 x - 1 \right )}}{2} - \frac {77}{16 x - 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15*x/4 + 17*log(2*x - 1)/2 - 77/(16*x - 8)

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