3.17.38 \(\int \frac {3+5 x}{(1-2 x)^3} \, dx\) [1638]

Optimal. Leaf size=18 \[ \frac {(3+5 x)^2}{22 (1-2 x)^2} \]

[Out]

1/22*(3+5*x)^2/(1-2*x)^2

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Rubi [A]
time = 0.00, antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {37} \begin {gather*} \frac {(5 x+3)^2}{22 (1-2 x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 + 5*x)/(1 - 2*x)^3,x]

[Out]

(3 + 5*x)^2/(22*(1 - 2*x)^2)

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

\begin {align*} \int \frac {3+5 x}{(1-2 x)^3} \, dx &=\frac {(3+5 x)^2}{22 (1-2 x)^2}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 16, normalized size = 0.89 \begin {gather*} \frac {1+20 x}{8 (1-2 x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 + 5*x)/(1 - 2*x)^3,x]

[Out]

(1 + 20*x)/(8*(1 - 2*x)^2)

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Maple [A]
time = 0.09, size = 20, normalized size = 1.11

method result size
gosper \(\frac {1+20 x}{8 \left (-1+2 x \right )^{2}}\) \(15\)
risch \(\frac {\frac {5 x}{2}+\frac {1}{8}}{\left (-1+2 x \right )^{2}}\) \(15\)
norman \(\frac {3 x -\frac {1}{2} x^{2}}{\left (-1+2 x \right )^{2}}\) \(18\)
default \(\frac {5}{4 \left (-1+2 x \right )}+\frac {11}{8 \left (-1+2 x \right )^{2}}\) \(20\)
meijerg \(\frac {3 x \left (2-2 x \right )}{2 \left (1-2 x \right )^{2}}+\frac {5 x^{2}}{2 \left (1-2 x \right )^{2}}\) \(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/4/(-1+2*x)+11/8/(-1+2*x)^2

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Maxima [A]
time = 0.35, size = 19, normalized size = 1.06 \begin {gather*} \frac {20 \, x + 1}{8 \, {\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)/(1-2*x)^3,x, algorithm="maxima")

[Out]

1/8*(20*x + 1)/(4*x^2 - 4*x + 1)

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Fricas [A]
time = 0.40, size = 19, normalized size = 1.06 \begin {gather*} \frac {20 \, x + 1}{8 \, {\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)/(1-2*x)^3,x, algorithm="fricas")

[Out]

1/8*(20*x + 1)/(4*x^2 - 4*x + 1)

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Sympy [A]
time = 0.04, size = 17, normalized size = 0.94 \begin {gather*} - \frac {- 20 x - 1}{32 x^{2} - 32 x + 8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-20*x - 1)/(32*x**2 - 32*x + 8)

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Giac [A]
time = 1.72, size = 14, normalized size = 0.78 \begin {gather*} \frac {20 \, x + 1}{8 \, {\left (2 \, x - 1\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(20*x + 1)/(2*x - 1)^2

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Mupad [B]
time = 0.03, size = 14, normalized size = 0.78 \begin {gather*} \frac {20\,x+1}{8\,{\left (2\,x-1\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*x + 3)/(2*x - 1)^3,x)

[Out]

(20*x + 1)/(8*(2*x - 1)^2)

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