3.915 \(\int \frac {-4+x}{(1+\sqrt [3]{x}) \sqrt {x}} \, dx\)

Optimal. Leaf size=41 \[ \frac {6 x^{7/6}}{7}-\frac {6 x^{5/6}}{5}+2 \sqrt {x}-30 \sqrt [6]{x}+30 \tan ^{-1}\left (\sqrt [6]{x}\right ) \]

[Out]

-30*x^(1/6)-6/5*x^(5/6)+6/7*x^(7/6)+30*arctan(x^(1/6))+2*x^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {1840, 1620, 50, 63, 203} \[ \frac {6 x^{7/6}}{7}-\frac {6 x^{5/6}}{5}+2 \sqrt {x}-30 \sqrt [6]{x}+30 \tan ^{-1}\left (\sqrt [6]{x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(-4 + x)/((1 + x^(1/3))*Sqrt[x]),x]

[Out]

-30*x^(1/6) + 2*Sqrt[x] - (6*x^(5/6))/5 + (6*x^(7/6))/7 + 30*ArcTan[x^(1/6)]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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]

Rule 1840

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{g = Denominator[n]}, Dist[g, Subst[Int[
x^(g*(m + 1) - 1)*(Pq /. x -> x^g)*(a + b*x^(g*n))^p, x], x, x^(1/g)], x]] /; FreeQ[{a, b, m, p}, x] && PolyQ[
Pq, x] && FractionQ[n]

Rubi steps

\begin {align*} \int \frac {-4+x}{\left (1+\sqrt [3]{x}\right ) \sqrt {x}} \, dx &=3 \operatorname {Subst}\left (\int \frac {\sqrt {x} \left (-4+x^3\right )}{1+x} \, dx,x,\sqrt [3]{x}\right )\\ &=3 \operatorname {Subst}\left (\int \left (\sqrt {x}-x^{3/2}+x^{5/2}-\frac {5 \sqrt {x}}{1+x}\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=2 \sqrt {x}-\frac {6 x^{5/6}}{5}+\frac {6 x^{7/6}}{7}-15 \operatorname {Subst}\left (\int \frac {\sqrt {x}}{1+x} \, dx,x,\sqrt [3]{x}\right )\\ &=-30 \sqrt [6]{x}+2 \sqrt {x}-\frac {6 x^{5/6}}{5}+\frac {6 x^{7/6}}{7}+15 \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} (1+x)} \, dx,x,\sqrt [3]{x}\right )\\ &=-30 \sqrt [6]{x}+2 \sqrt {x}-\frac {6 x^{5/6}}{5}+\frac {6 x^{7/6}}{7}+30 \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [6]{x}\right )\\ &=-30 \sqrt [6]{x}+2 \sqrt {x}-\frac {6 x^{5/6}}{5}+\frac {6 x^{7/6}}{7}+30 \tan ^{-1}\left (\sqrt [6]{x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 41, normalized size = 1.00 \[ \frac {6 x^{7/6}}{7}-\frac {6 x^{5/6}}{5}+2 \sqrt {x}-30 \sqrt [6]{x}+30 \tan ^{-1}\left (\sqrt [6]{x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(-4 + x)/((1 + x^(1/3))*Sqrt[x]),x]

[Out]

-30*x^(1/6) + 2*Sqrt[x] - (6*x^(5/6))/5 + (6*x^(7/6))/7 + 30*ArcTan[x^(1/6)]

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fricas [A]  time = 0.95, size = 25, normalized size = 0.61 \[ \frac {6}{7} \, {\left (x - 35\right )} x^{\frac {1}{6}} - \frac {6}{5} \, x^{\frac {5}{6}} + 2 \, \sqrt {x} + 30 \, \arctan \left (x^{\frac {1}{6}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6/7*(x - 35)*x^(1/6) - 6/5*x^(5/6) + 2*sqrt(x) + 30*arctan(x^(1/6))

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giac [A]  time = 0.40, size = 27, normalized size = 0.66 \[ \frac {6}{7} \, x^{\frac {7}{6}} - \frac {6}{5} \, x^{\frac {5}{6}} + 2 \, \sqrt {x} - 30 \, x^{\frac {1}{6}} + 30 \, \arctan \left (x^{\frac {1}{6}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6/7*x^(7/6) - 6/5*x^(5/6) + 2*sqrt(x) - 30*x^(1/6) + 30*arctan(x^(1/6))

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maple [A]  time = 0.00, size = 28, normalized size = 0.68 \[ \frac {6 x^{\frac {7}{6}}}{7}+30 \arctan \left (x^{\frac {1}{6}}\right )-\frac {6 x^{\frac {5}{6}}}{5}+2 \sqrt {x}-30 x^{\frac {1}{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x-4)/(1+x^(1/3))/x^(1/2),x)

[Out]

-30*x^(1/6)-6/5*x^(5/6)+6/7*x^(7/6)+30*arctan(x^(1/6))+2*x^(1/2)

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maxima [A]  time = 2.34, size = 27, normalized size = 0.66 \[ \frac {6}{7} \, x^{\frac {7}{6}} - \frac {6}{5} \, x^{\frac {5}{6}} + 2 \, \sqrt {x} - 30 \, x^{\frac {1}{6}} + 30 \, \arctan \left (x^{\frac {1}{6}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6/7*x^(7/6) - 6/5*x^(5/6) + 2*sqrt(x) - 30*x^(1/6) + 30*arctan(x^(1/6))

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mupad [B]  time = 3.37, size = 27, normalized size = 0.66 \[ 30\,\mathrm {atan}\left (x^{1/6}\right )+2\,\sqrt {x}-30\,x^{1/6}-\frac {6\,x^{5/6}}{5}+\frac {6\,x^{7/6}}{7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x - 4)/(x^(1/2)*(x^(1/3) + 1)),x)

[Out]

30*atan(x^(1/6)) + 2*x^(1/2) - 30*x^(1/6) - (6*x^(5/6))/5 + (6*x^(7/6))/7

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sympy [A]  time = 11.87, size = 37, normalized size = 0.90 \[ \frac {6 x^{\frac {7}{6}}}{7} - \frac {6 x^{\frac {5}{6}}}{5} - 30 \sqrt [6]{x} + 2 \sqrt {x} + 30 \operatorname {atan}{\left (\sqrt [6]{x} \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6*x**(7/6)/7 - 6*x**(5/6)/5 - 30*x**(1/6) + 2*sqrt(x) + 30*atan(x**(1/6))

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