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]

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

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Rubi [A]  time = 0.02, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, 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 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]))))

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 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]

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*}

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Mathematica [A]  time = 0.02, size = 44, normalized size = 1.00 \[ x+\frac {3}{10} (1-5 x)^{2/3}-\frac {9}{5} \sqrt [3]{1-5 x}+\frac {27}{5} \log \left (\sqrt [3]{1-5 x}+3\right ) \]

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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fricas [A]  time = 0.41, size = 32, normalized size = 0.73 \[ x + \frac {3}{10} \, {\left (-5 \, x + 1\right )}^{\frac {2}{3}} - \frac {9}{5} \, {\left (-5 \, x + 1\right )}^{\frac {1}{3}} + \frac {27}{5} \, \log \left ({\left (-5 \, x + 1\right )}^{\frac {1}{3}} + 3\right ) \]

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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giac [A]  time = 0.48, size = 33, normalized size = 0.75 \[ x + \frac {3}{10} \, {\left (-5 \, x + 1\right )}^{\frac {2}{3}} - \frac {9}{5} \, {\left (-5 \, x + 1\right )}^{\frac {1}{3}} + \frac {27}{5} \, \log \left ({\left (-5 \, x + 1\right )}^{\frac {1}{3}} + 3\right ) - \frac {1}{5} \]

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

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maple [A]  time = 0.00, size = 34, normalized size = 0.77 \[ x +\frac {27 \ln \left (3+\left (-5 x +1\right )^{\frac {1}{3}}\right )}{5}-\frac {1}{5}+\frac {3 \left (-5 x +1\right )^{\frac {2}{3}}}{10}-\frac {9 \left (-5 x +1\right )^{\frac {1}{3}}}{5} \]

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 = 0.84, size = 33, normalized size = 0.75 \[ x + \frac {3}{10} \, {\left (-5 \, x + 1\right )}^{\frac {2}{3}} - \frac {9}{5} \, {\left (-5 \, x + 1\right )}^{\frac {1}{3}} + \frac {27}{5} \, \log \left ({\left (-5 \, x + 1\right )}^{\frac {1}{3}} + 3\right ) - \frac {1}{5} \]

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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mupad [B]  time = 0.11, size = 32, normalized size = 0.73 \[ x+\frac {27\,\ln \left ({\left (1-5\,x\right )}^{1/3}+3\right )}{5}-\frac {9\,{\left (1-5\,x\right )}^{1/3}}{5}+\frac {3\,{\left (1-5\,x\right )}^{2/3}}{10} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.20, size = 39, normalized size = 0.89 \[ x + \frac {3 \left (1 - 5 x\right )^{\frac {2}{3}}}{10} - \frac {9 \sqrt [3]{1 - 5 x}}{5} + \frac {27 \log {\left (\sqrt [3]{1 - 5 x} + 3 \right )}}{5} \]

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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