3.90 \(\int \frac{x^3}{1+2 x} \, dx\)

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

[Out]

x/8 - x^2/8 + x^3/6 - Log[1 + 2*x]/16

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Rubi [A]  time = 0.0113922, antiderivative size = 30, normalized size of antiderivative = 1., 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 = {43} \[ \frac{x^3}{6}-\frac{x^2}{8}+\frac{x}{8}-\frac{1}{16} \log (2 x+1) \]

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 43

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*}

Mathematica [A]  time = 0.0063221, size = 27, normalized size = 0.9 \[ \frac{1}{96} \left (16 x^3-12 x^2+12 x-6 \log (2 x+1)+11\right ) \]

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 = 0.003, size = 23, normalized size = 0.8 \begin{align*}{\frac{x}{8}}-{\frac{{x}^{2}}{8}}+{\frac{{x}^{3}}{6}}-{\frac{\ln \left ( 1+2\,x \right ) }{16}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.929777, size = 30, normalized size = 1. \begin{align*} \frac{1}{6} \, x^{3} - \frac{1}{8} \, x^{2} + \frac{1}{8} \, x - \frac{1}{16} \, \log \left (2 \, x + 1\right ) \end{align*}

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 = 1.7782, size = 63, normalized size = 2.1 \begin{align*} \frac{1}{6} \, x^{3} - \frac{1}{8} \, x^{2} + \frac{1}{8} \, x - \frac{1}{16} \, \log \left (2 \, x + 1\right ) \end{align*}

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 = 0.069682, size = 20, normalized size = 0.67 \begin{align*} \frac{x^{3}}{6} - \frac{x^{2}}{8} + \frac{x}{8} - \frac{\log{\left (2 x + 1 \right )}}{16} \end{align*}

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 = 1.05833, size = 31, normalized size = 1.03 \begin{align*} \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{align*}

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))