3.607 \(\int \frac{2+\sqrt [3]{1-5 x}}{3+\sqrt [3]{1-5 x}} \, dx\)

Optimal. Leaf size=44 \[ x+\frac{3}{10} (1-5 x)^{2/3}-\frac{9}{5} \sqrt [3]{1-5 x}+\frac{27}{5} \log \left (\sqrt [3]{1-5 x}+3\right ) \]

[Out]

(-9*(1 - 5*x)^(1/3))/5 + (3*(1 - 5*x)^(2/3))/10 + x + (27*Log[3 + (1 - 5*x)^(1/3)])/5

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Rubi [A]  time = 0.0243852, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {431, 376, 77} \[ x+\frac{3}{10} (1-5 x)^{2/3}-\frac{9}{5} \sqrt [3]{1-5 x}+\frac{27}{5} \log \left (\sqrt [3]{1-5 x}+3\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*(1 - 5*x)^(1/3))/5 + (3*(1 - 5*x)^(2/3))/10 + x + (27*Log[3 + (1 - 5*x)^(1/3)])/5

Rule 431

Int[((a_.) + (b_.)*(u_)^(n_))^(p_.)*((c_.) + (d_.)*(u_)^(n_))^(q_.), x_Symbol] :> Dist[1/Coefficient[u, x, 1],
Subst[Int[(a + b*x^n)^p*(c + d*x^n)^q, x], x, u], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && LinearQ[u, x] && N
eQ[u, x]

Rule 376

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{g = Denominator[n]}, Dis
t[g, Subst[Int[x^(g - 1)*(a + b*x^(g*n))^p*(c + d*x^(g*n))^q, x], x, x^(1/g)], x]] /; FreeQ[{a, b, c, d, p, q}
, x] && NeQ[b*c - a*d, 0] && FractionQ[n]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int \frac{2+\sqrt [3]{1-5 x}}{3+\sqrt [3]{1-5 x}} \, dx &=-\left (\frac{1}{5} \operatorname{Subst}\left (\int \frac{2+\sqrt [3]{x}}{3+\sqrt [3]{x}} \, dx,x,1-5 x\right )\right )\\ &=-\left (\frac{3}{5} \operatorname{Subst}\left (\int \frac{x^2 (2+x)}{3+x} \, dx,x,\sqrt [3]{1-5 x}\right )\right )\\ &=-\left (\frac{3}{5} \operatorname{Subst}\left (\int \left (3-x+x^2-\frac{9}{3+x}\right ) \, dx,x,\sqrt [3]{1-5 x}\right )\right )\\ &=-\frac{9}{5} \sqrt [3]{1-5 x}+\frac{3}{10} (1-5 x)^{2/3}+x+\frac{27}{5} \log \left (3+\sqrt [3]{1-5 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0207475, size = 44, normalized size = 1. \[ x+\frac{3}{10} (1-5 x)^{2/3}-\frac{9}{5} \sqrt [3]{1-5 x}+\frac{27}{5} \log \left (\sqrt [3]{1-5 x}+3\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*(1 - 5*x)^(1/3))/5 + (3*(1 - 5*x)^(2/3))/10 + x + (27*Log[3 + (1 - 5*x)^(1/3)])/5

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Maple [A]  time = 0.003, size = 34, normalized size = 0.8 \begin{align*} -{\frac{1}{5}}+x+{\frac{3}{10} \left ( 1-5\,x \right ) ^{{\frac{2}{3}}}}-{\frac{9}{5}\sqrt [3]{1-5\,x}}+{\frac{27}{5}\ln \left ( 3+\sqrt [3]{1-5\,x} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5+x+3/10*(1-5*x)^(2/3)-9/5*(1-5*x)^(1/3)+27/5*ln(3+(1-5*x)^(1/3))

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Maxima [A]  time = 1.11109, size = 45, normalized size = 1.02 \begin{align*} x + \frac{3}{10} \,{\left (-5 \, x + 1\right )}^{\frac{2}{3}} - \frac{9}{5} \,{\left (-5 \, x + 1\right )}^{\frac{1}{3}} + \frac{27}{5} \, \log \left ({\left (-5 \, x + 1\right )}^{\frac{1}{3}} + 3\right ) - \frac{1}{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 3/10*(-5*x + 1)^(2/3) - 9/5*(-5*x + 1)^(1/3) + 27/5*log((-5*x + 1)^(1/3) + 3) - 1/5

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Fricas [A]  time = 1.59752, size = 112, normalized size = 2.55 \begin{align*} x + \frac{3}{10} \,{\left (-5 \, x + 1\right )}^{\frac{2}{3}} - \frac{9}{5} \,{\left (-5 \, x + 1\right )}^{\frac{1}{3}} + \frac{27}{5} \, \log \left ({\left (-5 \, x + 1\right )}^{\frac{1}{3}} + 3\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 3/10*(-5*x + 1)^(2/3) - 9/5*(-5*x + 1)^(1/3) + 27/5*log((-5*x + 1)^(1/3) + 3)

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Sympy [A]  time = 0.175741, size = 39, normalized size = 0.89 \begin{align*} x + \frac{3 \left (1 - 5 x\right )^{\frac{2}{3}}}{10} - \frac{9 \sqrt [3]{1 - 5 x}}{5} + \frac{27 \log{\left (\sqrt [3]{1 - 5 x} + 3 \right )}}{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 3*(1 - 5*x)**(2/3)/10 - 9*(1 - 5*x)**(1/3)/5 + 27*log((1 - 5*x)**(1/3) + 3)/5

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Giac [A]  time = 1.12737, size = 45, normalized size = 1.02 \begin{align*} x + \frac{3}{10} \,{\left (-5 \, x + 1\right )}^{\frac{2}{3}} - \frac{9}{5} \,{\left (-5 \, x + 1\right )}^{\frac{1}{3}} + \frac{27}{5} \, \log \left ({\left (-5 \, x + 1\right )}^{\frac{1}{3}} + 3\right ) - \frac{1}{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + 3/10*(-5*x + 1)^(2/3) - 9/5*(-5*x + 1)^(1/3) + 27/5*log((-5*x + 1)^(1/3) + 3) - 1/5