3.17.52 \(\int \frac {(3+5 x)^2}{(1-2 x)^3} \, dx\) [1652]

Optimal. Leaf size=33 \[ \frac {121}{16 (1-2 x)^2}-\frac {55}{4 (1-2 x)}-\frac {25}{8} \log (1-2 x) \]

[Out]

121/16/(1-2*x)^2-55/4/(1-2*x)-25/8*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 = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {45} \begin {gather*} -\frac {55}{4 (1-2 x)}+\frac {121}{16 (1-2 x)^2}-\frac {25}{8} \log (1-2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 + 5*x)^2/(1 - 2*x)^3,x]

[Out]

121/(16*(1 - 2*x)^2) - 55/(4*(1 - 2*x)) - (25*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)^3} \, dx &=\int \left (-\frac {121}{4 (-1+2 x)^3}-\frac {55}{2 (-1+2 x)^2}-\frac {25}{4 (-1+2 x)}\right ) \, dx\\ &=\frac {121}{16 (1-2 x)^2}-\frac {55}{4 (1-2 x)}-\frac {25}{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 {121}{16 (1-2 x)^2}-\frac {55}{4 (1-2 x)}-\frac {25}{8} \log (1-2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 + 5*x)^2/(1 - 2*x)^3,x]

[Out]

121/(16*(1 - 2*x)^2) - 55/(4*(1 - 2*x)) - (25*Log[1 - 2*x])/8

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Maple [A]
time = 0.10, size = 28, normalized size = 0.85

method	result	size
risch	\(\frac {\frac {55 x}{2}-\frac {99}{16}}{\left (-1+2 x \right )^{2}}-\frac {25 \ln \left (-1+2 x \right )}{8}\)	\(24\)
norman	\(\frac {\frac {11}{4} x +\frac {99}{4} x^{2}}{\left (-1+2 x \right )^{2}}-\frac {25 \ln \left (-1+2 x \right )}{8}\)	\(27\)
default	\(\frac {55}{4 \left (-1+2 x \right )}+\frac {121}{16 \left (-1+2 x \right )^{2}}-\frac {25 \ln \left (-1+2 x \right )}{8}\)	\(28\)
meijerg	\(\frac {9 x \left (2-2 x \right )}{2 \left (1-2 x \right )^{2}}+\frac {15 x^{2}}{\left (1-2 x \right )^{2}}-\frac {25 x \left (-18 x +6\right )}{24 \left (1-2 x \right )^{2}}-\frac {25 \ln \left (1-2 x \right )}{8}\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

55/4/(-1+2*x)+121/16/(-1+2*x)^2-25/8*ln(-1+2*x)

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Maxima [A]
time = 0.28, size = 28, normalized size = 0.85 \begin {gather*} \frac {11 \, {\left (40 \, x - 9\right )}}{16 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} - \frac {25}{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)^3,x, algorithm="maxima")

[Out]

11/16*(40*x - 9)/(4*x^2 - 4*x + 1) - 25/8*log(2*x - 1)

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Fricas [A]
time = 0.39, size = 37, normalized size = 1.12 \begin {gather*} -\frac {50 \, {\left (4 \, x^{2} - 4 \, x + 1\right )} \log \left (2 \, x - 1\right ) - 440 \, x + 99}{16 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(50*(4*x^2 - 4*x + 1)*log(2*x - 1) - 440*x + 99)/(4*x^2 - 4*x + 1)

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Sympy [A]
time = 0.04, size = 26, normalized size = 0.79 \begin {gather*} - \frac {99 - 440 x}{64 x^{2} - 64 x + 16} - \frac {25 \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)**3,x)

[Out]

-(99 - 440*x)/(64*x**2 - 64*x + 16) - 25*log(2*x - 1)/8

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Giac [A]
time = 0.46, size = 24, normalized size = 0.73 \begin {gather*} \frac {11 \, {\left (40 \, x - 9\right )}}{16 \, {\left (2 \, x - 1\right )}^{2}} - \frac {25}{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)^3,x, algorithm="giac")

[Out]

11/16*(40*x - 9)/(2*x - 1)^2 - 25/8*log(abs(2*x - 1))

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Mupad [B]
time = 0.03, size = 23, normalized size = 0.70 \begin {gather*} \frac {\frac {55\,x}{8}-\frac {99}{64}}{x^2-x+\frac {1}{4}}-\frac {25\,\ln \left (x-\frac {1}{2}\right )}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*x + 3)^2/(2*x - 1)^3,x)

[Out]

((55*x)/8 - 99/64)/(x^2 - x + 1/4) - (25*log(x - 1/2))/8

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