3.160 \(\int \frac {-3 x+x^4}{(1+2 x)^5} \, dx\)

Optimal. Leaf size=55 \[ \frac {1}{8 (2 x+1)}-\frac {3}{32 (2 x+1)^2}+\frac {7}{24 (2 x+1)^3}-\frac {25}{128 (2 x+1)^4}+\frac {1}{32} \log (2 x+1) \]

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Rubi [A]  time = 0.04, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {1593, 1620} \[ \frac {1}{8 (2 x+1)}-\frac {3}{32 (2 x+1)^2}+\frac {7}{24 (2 x+1)^3}-\frac {25}{128 (2 x+1)^4}+\frac {1}{32} \log (2 x+1) \]

Antiderivative was successfully verified.

[In]

Int[(-3*x + x^4)/(1 + 2*x)^5,x]

[Out]

-25/(128*(1 + 2*x)^4) + 7/(24*(1 + 2*x)^3) - 3/(32*(1 + 2*x)^2) + 1/(8*(1 + 2*x)) + Log[1 + 2*x]/32

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps

\begin {align*} \int \frac {-3 x+x^4}{(1+2 x)^5} \, dx &=\int \frac {x \left (-3+x^3\right )}{(1+2 x)^5} \, dx\\ &=\int \left (\frac {25}{16 (1+2 x)^5}-\frac {7}{4 (1+2 x)^4}+\frac {3}{8 (1+2 x)^3}-\frac {1}{4 (1+2 x)^2}+\frac {1}{16 (1+2 x)}\right ) \, dx\\ &=-\frac {25}{128 (1+2 x)^4}+\frac {7}{24 (1+2 x)^3}-\frac {3}{32 (1+2 x)^2}+\frac {1}{8 (1+2 x)}+\frac {1}{32} \log (1+2 x)\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 41, normalized size = 0.75 \[ \frac {384 x^3+432 x^2+368 x+12 (2 x+1)^4 \log (2 x+1)+49}{384 (2 x+1)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(-3*x + x^4)/(1 + 2*x)^5,x]

[Out]

(49 + 368*x + 432*x^2 + 384*x^3 + 12*(1 + 2*x)^4*Log[1 + 2*x])/(384*(1 + 2*x)^4)

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IntegrateAlgebraic [A]  time = 0.02, size = 37, normalized size = 0.67 \[ \frac {384 x^3+432 x^2+368 x+49}{384 (2 x+1)^4}+\frac {1}{32} \log (2 x+1) \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-3*x + x^4)/(1 + 2*x)^5,x]

[Out]

(49 + 368*x + 432*x^2 + 384*x^3)/(384*(1 + 2*x)^4) + Log[1 + 2*x]/32

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fricas [A]  time = 1.24, size = 67, normalized size = 1.22 \[ \frac {384 \, x^{3} + 432 \, x^{2} + 12 \, {\left (16 \, x^{4} + 32 \, x^{3} + 24 \, x^{2} + 8 \, x + 1\right )} \log \left (2 \, x + 1\right ) + 368 \, x + 49}{384 \, {\left (16 \, x^{4} + 32 \, x^{3} + 24 \, x^{2} + 8 \, x + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-3*x)/(1+2*x)^5,x, algorithm="fricas")

[Out]

1/384*(384*x^3 + 432*x^2 + 12*(16*x^4 + 32*x^3 + 24*x^2 + 8*x + 1)*log(2*x + 1) + 368*x + 49)/(16*x^4 + 32*x^3
+ 24*x^2 + 8*x + 1)

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giac [A]  time = 0.89, size = 55, normalized size = 1.00 \[ \frac {1}{8 \, {\left (2 \, x + 1\right )}} - \frac {3}{32 \, {\left (2 \, x + 1\right )}^{2}} + \frac {7}{24 \, {\left (2 \, x + 1\right )}^{3}} - \frac {25}{128 \, {\left (2 \, x + 1\right )}^{4}} - \frac {1}{32} \, \log \left (\frac {{\left | 2 \, x + 1 \right |}}{2 \, {\left (2 \, x + 1\right )}^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-3*x)/(1+2*x)^5,x, algorithm="giac")

[Out]

1/8/(2*x + 1) - 3/32/(2*x + 1)^2 + 7/24/(2*x + 1)^3 - 25/128/(2*x + 1)^4 - 1/32*log(1/2*abs(2*x + 1)/(2*x + 1)
^2)

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maple [A]  time = 0.26, size = 34, normalized size = 0.62

method	result	size
risch	\(\frac {x^{3}+\frac {9}{8} x^{2}+\frac {23}{24} x +\frac {49}{384}}{\left (1+2 x \right )^{4}}+\frac {\ln \left (1+2 x \right )}{32}\)	\(34\)
norman	\(\frac {-\frac {37}{12} x^{3}-\frac {31}{16} x^{2}-\frac {1}{16} x -\frac {49}{24} x^{4}}{\left (1+2 x \right )^{4}}+\frac {\ln \left (1+2 x \right )}{32}\)	\(37\)
default	\(-\frac {25}{128 \left (1+2 x \right )^{4}}+\frac {7}{24 \left (1+2 x \right )^{3}}-\frac {3}{32 \left (1+2 x \right )^{2}}+\frac {1}{8+16 x}+\frac {\ln \left (1+2 x \right )}{32}\)	\(46\)
meijerg	\(-\frac {x \left (1000 x^{3}+1040 x^{2}+420 x +60\right )}{960 \left (1+2 x \right )^{4}}+\frac {\ln \left (1+2 x \right )}{32}-\frac {x^{2} \left (4 x^{2}+8 x +6\right )}{4 \left (1+2 x \right )^{4}}\)	\(57\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4-3*x)/(1+2*x)^5,x,method=_RETURNVERBOSE)

[Out]

16*(1/16*x^3+9/128*x^2+23/384*x+49/6144)/(1+2*x)^4+1/32*ln(1+2*x)

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maxima [A]  time = 0.46, size = 48, normalized size = 0.87 \[ \frac {384 \, x^{3} + 432 \, x^{2} + 368 \, x + 49}{384 \, {\left (16 \, x^{4} + 32 \, x^{3} + 24 \, x^{2} + 8 \, x + 1\right )}} + \frac {1}{32} \, \log \left (2 \, x + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^4-3*x)/(1+2*x)^5,x, algorithm="maxima")

[Out]

1/384*(384*x^3 + 432*x^2 + 368*x + 49)/(16*x^4 + 32*x^3 + 24*x^2 + 8*x + 1) + 1/32*log(2*x + 1)

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mupad [B]  time = 0.05, size = 43, normalized size = 0.78 \[ \frac {\ln \left (x+\frac {1}{2}\right )}{32}+\frac {\frac {x^3}{16}+\frac {9\,x^2}{128}+\frac {23\,x}{384}+\frac {49}{6144}}{x^4+2\,x^3+\frac {3\,x^2}{2}+\frac {x}{2}+\frac {1}{16}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x - x^4)/(2*x + 1)^5,x)

[Out]

log(x + 1/2)/32 + ((23*x)/384 + (9*x^2)/128 + x^3/16 + 49/6144)/(x/2 + (3*x^2)/2 + 2*x^3 + x^4 + 1/16)

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sympy [A]  time = 0.13, size = 42, normalized size = 0.76 \[ \frac {384 x^{3} + 432 x^{2} + 368 x + 49}{6144 x^{4} + 12288 x^{3} + 9216 x^{2} + 3072 x + 384} + \frac {\log {\left (2 x + 1 \right )}}{32} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**4-3*x)/(1+2*x)**5,x)

[Out]

(384*x**3 + 432*x**2 + 368*x + 49)/(6144*x**4 + 12288*x**3 + 9216*x**2 + 3072*x + 384) + log(2*x + 1)/32

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