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

Optimal. Leaf size=74 \[ \frac {6 x^{5/6}}{5}+2 \sqrt {x}-3 \sqrt [3]{x}-4 \log \left (\sqrt [6]{x}+1\right )-\log \left (\sqrt [3]{x}-\sqrt [6]{x}+1\right )-2 \sqrt {3} \tan ^{-1}\left (\frac {1-2 \sqrt [6]{x}}{\sqrt {3}}\right ) \]

[Out]

-3*x^(1/3)+6/5*x^(5/6)-4*ln(1+x^(1/6))-ln(1-x^(1/6)+x^(1/3))-2*arctan(1/3*(1-2*x^(1/6))*3^(1/2))*3^(1/2)+2*x^(
1/2)

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Rubi [A]  time = 0.11, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {1593, 1887, 1874, 31, 634, 618, 204, 628} \[ \frac {6 x^{5/6}}{5}+2 \sqrt {x}-3 \sqrt [3]{x}-4 \log \left (\sqrt [6]{x}+1\right )-\log \left (\sqrt [3]{x}-\sqrt [6]{x}+1\right )-2 \sqrt {3} \tan ^{-1}\left (\frac {1-2 \sqrt [6]{x}}{\sqrt {3}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(1 + x^(1/3))/(1 + Sqrt[x]),x]

[Out]

-3*x^(1/3) + 2*Sqrt[x] + (6*x^(5/6))/5 - 2*Sqrt[3]*ArcTan[(1 - 2*x^(1/6))/Sqrt[3]] - 4*Log[1 + x^(1/6)] - Log[
1 - x^(1/6) + x^(1/3)]

Rule 31

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

Rule 204

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

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 1593

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 1874

Int[(P2_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{A = Coeff[P2, x, 0], B = Coeff[P2, x, 1], C = Coeff[P2, x,
2], q = (a/b)^(1/3)}, Dist[(q*(A - B*q + C*q^2))/(3*a), Int[1/(q + x), x], x] + Dist[q/(3*a), Int[(q*(2*A + B
*q - C*q^2) - (A - B*q - 2*C*q^2)*x)/(q^2 - q*x + x^2), x], x] /; NeQ[a*B^3 - b*A^3, 0] && NeQ[A - B*q + C*q^2
, 0]] /; FreeQ[{a, b}, x] && PolyQ[P2, x, 2] && GtQ[a/b, 0]

Rule 1887

Int[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[Pq/(a + b*x^n), x], x] /; FreeQ[{a, b}, x
] && PolyQ[Pq, x] && IntegerQ[n]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 74, normalized size = 1.00 \[ \frac {6 x^{5/6}}{5}+2 \sqrt {x}-3 \sqrt [3]{x}-4 \log \left (\sqrt [6]{x}+1\right )-\log \left (\sqrt [3]{x}-\sqrt [6]{x}+1\right )-2 \sqrt {3} \tan ^{-1}\left (\frac {1-2 \sqrt [6]{x}}{\sqrt {3}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-3*x^(1/3) + 2*Sqrt[x] + (6*x^(5/6))/5 - 2*Sqrt[3]*ArcTan[(1 - 2*x^(1/6))/Sqrt[3]] - 4*Log[1 + x^(1/6)] - Log[
1 - x^(1/6) + x^(1/3)]

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fricas [A]  time = 0.94, size = 57, normalized size = 0.77 \[ 2 \, \sqrt {3} \arctan \left (\frac {2}{3} \, \sqrt {3} x^{\frac {1}{6}} - \frac {1}{3} \, \sqrt {3}\right ) + \frac {6}{5} \, x^{\frac {5}{6}} + 2 \, \sqrt {x} - 3 \, x^{\frac {1}{3}} - \log \left (x^{\frac {1}{3}} - x^{\frac {1}{6}} + 1\right ) - 4 \, \log \left (x^{\frac {1}{6}} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(3)*arctan(2/3*sqrt(3)*x^(1/6) - 1/3*sqrt(3)) + 6/5*x^(5/6) + 2*sqrt(x) - 3*x^(1/3) - log(x^(1/3) - x^(1
/6) + 1) - 4*log(x^(1/6) + 1)

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giac [A]  time = 0.46, size = 55, normalized size = 0.74 \[ 2 \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{\frac {1}{6}} - 1\right )}\right ) + \frac {6}{5} \, x^{\frac {5}{6}} + 2 \, \sqrt {x} - 3 \, x^{\frac {1}{3}} - \log \left (x^{\frac {1}{3}} - x^{\frac {1}{6}} + 1\right ) - 4 \, \log \left (x^{\frac {1}{6}} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^(1/6) - 1)) + 6/5*x^(5/6) + 2*sqrt(x) - 3*x^(1/3) - log(x^(1/3) - x^(1/6) +
1) - 4*log(x^(1/6) + 1)

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maple [A]  time = 0.01, size = 56, normalized size = 0.76 \[ 2 \sqrt {3}\, \arctan \left (\frac {\left (2 x^{\frac {1}{6}}-1\right ) \sqrt {3}}{3}\right )-4 \ln \left (x^{\frac {1}{6}}+1\right )-\ln \left (x^{\frac {1}{3}}-x^{\frac {1}{6}}+1\right )+\frac {6 x^{\frac {5}{6}}}{5}+2 \sqrt {x}-3 x^{\frac {1}{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6/5*x^(5/6)+2*x^(1/2)-3*x^(1/3)-ln(x^(1/3)-x^(1/6)+1)+2*3^(1/2)*arctan(1/3*(2*x^(1/6)-1)*3^(1/2))-4*ln(x^(1/6)
+1)

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maxima [A]  time = 0.97, size = 55, normalized size = 0.74 \[ 2 \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left (2 \, x^{\frac {1}{6}} - 1\right )}\right ) + \frac {6}{5} \, x^{\frac {5}{6}} + 2 \, \sqrt {x} - 3 \, x^{\frac {1}{3}} - \log \left (x^{\frac {1}{3}} - x^{\frac {1}{6}} + 1\right ) - 4 \, \log \left (x^{\frac {1}{6}} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(3)*arctan(1/3*sqrt(3)*(2*x^(1/6) - 1)) + 6/5*x^(5/6) + 2*sqrt(x) - 3*x^(1/3) - log(x^(1/3) - x^(1/6) +
1) - 4*log(x^(1/6) + 1)

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mupad [B]  time = 3.41, size = 95, normalized size = 1.28 \[ 2\,\sqrt {x}+\ln \left (\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (27+\sqrt {3}\,9{}\mathrm {i}\right )+36\,x^{1/6}+36\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )-\ln \left (\left (1+\sqrt {3}\,1{}\mathrm {i}\right )\,\left (-27+\sqrt {3}\,9{}\mathrm {i}\right )+36\,x^{1/6}+36\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )-4\,\ln \left (36\,x^{1/6}+36\right )-3\,x^{1/3}+\frac {6\,x^{5/6}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(1/3) + 1)/(x^(1/2) + 1),x)

[Out]

log((3^(1/2)*1i - 1)*(3^(1/2)*9i + 27) + 36*x^(1/6) + 36)*(3^(1/2)*1i - 1) - 4*log(36*x^(1/6) + 36) - log((3^(
1/2)*1i + 1)*(3^(1/2)*9i - 27) + 36*x^(1/6) + 36)*(3^(1/2)*1i + 1) + 2*x^(1/2) - 3*x^(1/3) + (6*x^(5/6))/5

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sympy [C]  time = 3.78, size = 155, normalized size = 2.09 \[ \frac {16 x^{\frac {5}{6}} \Gamma \left (\frac {8}{3}\right )}{5 \Gamma \left (\frac {11}{3}\right )} - \frac {8 \sqrt [3]{x} \Gamma \left (\frac {8}{3}\right )}{\Gamma \left (\frac {11}{3}\right )} + 2 \sqrt {x} - 2 \log {\left (\sqrt {x} + 1 \right )} - \frac {16 e^{- \frac {2 i \pi }{3}} \log {\left (- \sqrt [6]{x} e^{\frac {i \pi }{3}} + 1 \right )} \Gamma \left (\frac {8}{3}\right )}{3 \Gamma \left (\frac {11}{3}\right )} - \frac {16 \log {\left (- \sqrt [6]{x} e^{i \pi } + 1 \right )} \Gamma \left (\frac {8}{3}\right )}{3 \Gamma \left (\frac {11}{3}\right )} - \frac {16 e^{\frac {2 i \pi }{3}} \log {\left (- \sqrt [6]{x} e^{\frac {5 i \pi }{3}} + 1 \right )} \Gamma \left (\frac {8}{3}\right )}{3 \Gamma \left (\frac {11}{3}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16*x**(5/6)*gamma(8/3)/(5*gamma(11/3)) - 8*x**(1/3)*gamma(8/3)/gamma(11/3) + 2*sqrt(x) - 2*log(sqrt(x) + 1) -
16*exp(-2*I*pi/3)*log(-x**(1/6)*exp_polar(I*pi/3) + 1)*gamma(8/3)/(3*gamma(11/3)) - 16*log(-x**(1/6)*exp_polar
(I*pi) + 1)*gamma(8/3)/(3*gamma(11/3)) - 16*exp(2*I*pi/3)*log(-x**(1/6)*exp_polar(5*I*pi/3) + 1)*gamma(8/3)/(3
*gamma(11/3))

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