\(\int \frac {1+\sqrt {x}}{x^{5/6}+x^{7/6}} \, dx\) [916]

Optimal result

Integrand size = 21, antiderivative size = 26 \[ \int \frac {1+\sqrt {x}}{x^{5/6}+x^{7/6}} \, dx=3 \sqrt [3]{x}+6 \arctan \left (\sqrt [6]{x}\right )-3 \log \left (1+\sqrt [3]{x}\right ) \]

[Out]

3*x^(1/3)+6*arctan(x^(1/6))-3*ln(1+x^(1/3))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1607, 1833, 1824, 649, 209, 266} \[ \int \frac {1+\sqrt {x}}{x^{5/6}+x^{7/6}} \, dx=6 \arctan \left (\sqrt [6]{x}\right )+3 \sqrt [3]{x}-3 \log \left (\sqrt [3]{x}+1\right ) \]

[In]

Int[(1 + Sqrt[x])/(x^(5/6) + x^(7/6)),x]

[Out]

3*x^(1/3) + 6*ArcTan[x^(1/6)] - 3*Log[1 + x^(1/3)]

Rule 209

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

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 649

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[(-a)*c]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 1824

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 1833

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1)
, Pq, x]*(a + b*x^Simplify[n/(m + 1)])^p, x], x, x^(m + 1)], x] /; FreeQ[{a, b, m, n, p}, x] && NeQ[m, -1] &&
IGtQ[Simplify[n/(m + 1)], 0] && PolyQ[Pq, x^(m + 1)]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1+\sqrt {x}}{\left (1+\sqrt [3]{x}\right ) x^{5/6}} \, dx \\ & = 6 \text {Subst}\left (\int \frac {1+x^3}{1+x^2} \, dx,x,\sqrt [6]{x}\right ) \\ & = 6 \text {Subst}\left (\int \left (x+\frac {1-x}{1+x^2}\right ) \, dx,x,\sqrt [6]{x}\right ) \\ & = 3 \sqrt [3]{x}+6 \text {Subst}\left (\int \frac {1-x}{1+x^2} \, dx,x,\sqrt [6]{x}\right ) \\ & = 3 \sqrt [3]{x}+6 \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [6]{x}\right )-6 \text {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,\sqrt [6]{x}\right ) \\ & = 3 \sqrt [3]{x}+6 \tan ^{-1}\left (\sqrt [6]{x}\right )-3 \log \left (1+\sqrt [3]{x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {1+\sqrt {x}}{x^{5/6}+x^{7/6}} \, dx=3 \sqrt [3]{x}+6 \arctan \left (\sqrt [6]{x}\right )-3 \log \left (1+\sqrt [3]{x}\right ) \]

[In]

Integrate[(1 + Sqrt[x])/(x^(5/6) + x^(7/6)),x]

[Out]

3*x^(1/3) + 6*ArcTan[x^(1/6)] - 3*Log[1 + x^(1/3)]

Maple [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81

\[3 x^{\frac {1}{3}}+6 \arctan \left (x^{\frac {1}{6}}\right )-3 \ln \left (1+x^{\frac {1}{3}}\right )\]

[In]

int((1+x^(1/2))/(x^(5/6)+x^(7/6)),x)

[Out]

3*x^(1/3)+6*arctan(x^(1/6))-3*ln(1+x^(1/3))

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {1+\sqrt {x}}{x^{5/6}+x^{7/6}} \, dx=3 \, x^{\frac {1}{3}} + 6 \, \arctan \left (x^{\frac {1}{6}}\right ) - 3 \, \log \left (x^{\frac {1}{3}} + 1\right ) \]

[In]

integrate((1+x^(1/2))/(x^(5/6)+x^(7/6)),x, algorithm="fricas")

[Out]

3*x^(1/3) + 6*arctan(x^(1/6)) - 3*log(x^(1/3) + 1)

Sympy [A] (verification not implemented)

Time = 0.95 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {1+\sqrt {x}}{x^{5/6}+x^{7/6}} \, dx=3 \sqrt [3]{x} - 3 \log {\left (\sqrt [3]{x} + 1 \right )} + 6 \operatorname {atan}{\left (\sqrt [6]{x} \right )} \]

[In]

integrate((1+x**(1/2))/(x**(5/6)+x**(7/6)),x)

[Out]

3*x**(1/3) - 3*log(x**(1/3) + 1) + 6*atan(x**(1/6))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {1+\sqrt {x}}{x^{5/6}+x^{7/6}} \, dx=3 \, x^{\frac {1}{3}} + 6 \, \arctan \left (x^{\frac {1}{6}}\right ) - 3 \, \log \left (x^{\frac {1}{3}} + 1\right ) \]

[In]

integrate((1+x^(1/2))/(x^(5/6)+x^(7/6)),x, algorithm="maxima")

[Out]

3*x^(1/3) + 6*arctan(x^(1/6)) - 3*log(x^(1/3) + 1)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {1+\sqrt {x}}{x^{5/6}+x^{7/6}} \, dx=3 \, x^{\frac {1}{3}} + 6 \, \arctan \left (x^{\frac {1}{6}}\right ) - 3 \, \log \left (x^{\frac {1}{3}} + 1\right ) \]

[In]

integrate((1+x^(1/2))/(x^(5/6)+x^(7/6)),x, algorithm="giac")

[Out]

3*x^(1/3) + 6*arctan(x^(1/6)) - 3*log(x^(1/3) + 1)

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {1+\sqrt {x}}{x^{5/6}+x^{7/6}} \, dx=6\,\mathrm {atan}\left (x^{1/6}\right )-3\,\ln \left (36\,x^{1/3}+36\right )+3\,x^{1/3} \]

[In]

int((x^(1/2) + 1)/(x^(5/6) + x^(7/6)),x)

[Out]

6*atan(x^(1/6)) - 3*log(36*x^(1/3) + 36) + 3*x^(1/3)