3.1.90 \(\int \frac {x^3}{1+2 x} \, dx\) [90]

Optimal. Leaf size=30 \[ \frac {x}{8}-\frac {x^2}{8}+\frac {x^3}{6}-\frac {1}{16} \log (1+2 x) \]

[Out]

1/8*x-1/8*x^2+1/6*x^3-1/16*ln(1+2*x)

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Rubi [A]
time = 0.01, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {45} \begin {gather*} \frac {x^3}{6}-\frac {x^2}{8}+\frac {x}{8}-\frac {1}{16} \log (2 x+1) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/(1 + 2*x),x]

[Out]

x/8 - x^2/8 + x^3/6 - Log[1 + 2*x]/16

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 {x^3}{1+2 x} \, dx &=\int \left (\frac {1}{8}-\frac {x}{4}+\frac {x^2}{2}-\frac {1}{8 (1+2 x)}\right ) \, dx\\ &=\frac {x}{8}-\frac {x^2}{8}+\frac {x^3}{6}-\frac {1}{16} \log (1+2 x)\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 27, normalized size = 0.90 \begin {gather*} \frac {1}{96} \left (11+12 x-12 x^2+16 x^3-6 \log (1+2 x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/(1 + 2*x),x]

[Out]

(11 + 12*x - 12*x^2 + 16*x^3 - 6*Log[1 + 2*x])/96

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Maple [A]
time = 0.05, size = 23, normalized size = 0.77

method	result	size
default	\(\frac {x}{8}-\frac {x^{2}}{8}+\frac {x^{3}}{6}-\frac {\ln \left (1+2 x \right )}{16}\)	\(23\)
norman	\(\frac {x}{8}-\frac {x^{2}}{8}+\frac {x^{3}}{6}-\frac {\ln \left (1+2 x \right )}{16}\)	\(23\)
meijerg	\(\frac {x \left (16 x^{2}-12 x +12\right )}{96}-\frac {\ln \left (1+2 x \right )}{16}\)	\(23\)
risch	\(\frac {x}{8}-\frac {x^{2}}{8}+\frac {x^{3}}{6}-\frac {\ln \left (1+2 x \right )}{16}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*x-1/8*x^2+1/6*x^3-1/16*ln(1+2*x)

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Maxima [A]
time = 0.90, size = 22, normalized size = 0.73 \begin {gather*} \frac {1}{6} \, x^{3} - \frac {1}{8} \, x^{2} + \frac {1}{8} \, x - \frac {1}{16} \, \log \left (2 \, x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3 - 1/8*x^2 + 1/8*x - 1/16*log(2*x + 1)

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Fricas [A]
time = 0.39, size = 22, normalized size = 0.73 \begin {gather*} \frac {1}{6} \, x^{3} - \frac {1}{8} \, x^{2} + \frac {1}{8} \, x - \frac {1}{16} \, \log \left (2 \, x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3 - 1/8*x^2 + 1/8*x - 1/16*log(2*x + 1)

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Sympy [A]
time = 0.02, size = 20, normalized size = 0.67 \begin {gather*} \frac {x^{3}}{6} - \frac {x^{2}}{8} + \frac {x}{8} - \frac {\log {\left (2 x + 1 \right )}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**3/6 - x**2/8 + x/8 - log(2*x + 1)/16

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Giac [A]
time = 0.48, size = 23, normalized size = 0.77 \begin {gather*} \frac {1}{6} \, x^{3} - \frac {1}{8} \, x^{2} + \frac {1}{8} \, x - \frac {1}{16} \, \log \left ({\left | 2 \, x + 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3 - 1/8*x^2 + 1/8*x - 1/16*log(abs(2*x + 1))

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Mupad [B]
time = 0.03, size = 20, normalized size = 0.67 \begin {gather*} \frac {x}{8}-\frac {\ln \left (x+\frac {1}{2}\right )}{16}-\frac {x^2}{8}+\frac {x^3}{6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(2*x + 1),x)

[Out]

x/8 - log(x + 1/2)/16 - x^2/8 + x^3/6

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