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

Optimal. Leaf size=42 \[ \frac {4}{3} \sqrt [4]{x^3+1}-\frac {2}{3} \tan ^{-1}\left (\sqrt [4]{x^3+1}\right )-\frac {2}{3} \tanh ^{-1}\left (\sqrt [4]{x^3+1}\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {266, 50, 63, 212, 206, 203} \begin {gather*} \frac {4}{3} \sqrt [4]{x^3+1}-\frac {2}{3} \tan ^{-1}\left (\sqrt [4]{x^3+1}\right )-\frac {2}{3} \tanh ^{-1}\left (\sqrt [4]{x^3+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + x^3)^(1/4)/x,x]

[Out]

(4*(1 + x^3)^(1/4))/3 - (2*ArcTan[(1 + x^3)^(1/4)])/3 - (2*ArcTanh[(1 + x^3)^(1/4)])/3

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 206

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

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 266

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

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 42, normalized size = 1.00 \begin {gather*} \frac {4}{3} \sqrt [4]{x^3+1}-\frac {2}{3} \tan ^{-1}\left (\sqrt [4]{x^3+1}\right )-\frac {2}{3} \tanh ^{-1}\left (\sqrt [4]{x^3+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^3)^(1/4)/x,x]

[Out]

(4*(1 + x^3)^(1/4))/3 - (2*ArcTan[(1 + x^3)^(1/4)])/3 - (2*ArcTanh[(1 + x^3)^(1/4)])/3

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IntegrateAlgebraic [A]  time = 0.03, size = 42, normalized size = 1.00 \begin {gather*} \frac {4}{3} \sqrt [4]{1+x^3}-\frac {2}{3} \tan ^{-1}\left (\sqrt [4]{1+x^3}\right )-\frac {2}{3} \tanh ^{-1}\left (\sqrt [4]{1+x^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(1 + x^3)^(1/4)/x,x]

[Out]

(4*(1 + x^3)^(1/4))/3 - (2*ArcTan[(1 + x^3)^(1/4)])/3 - (2*ArcTanh[(1 + x^3)^(1/4)])/3

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fricas [A]  time = 0.57, size = 44, normalized size = 1.05 \begin {gather*} \frac {4}{3} \, {\left (x^{3} + 1\right )}^{\frac {1}{4}} - \frac {2}{3} \, \arctan \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{3} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{3} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.33, size = 45, normalized size = 1.07 \begin {gather*} \frac {4}{3} \, {\left (x^{3} + 1\right )}^{\frac {1}{4}} - \frac {2}{3} \, \arctan \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{3} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{3} \, \log \left ({\left | {\left (x^{3} + 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/3*(x^3 + 1)^(1/4) - 2/3*arctan((x^3 + 1)^(1/4)) - 1/3*log((x^3 + 1)^(1/4) + 1) + 1/3*log(abs((x^3 + 1)^(1/4)
- 1))

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maple [C]  time = 1.56, size = 45, normalized size = 1.07

method	result	size
meijerg	\(-\frac {-4 \left (4-3 \ln \relax (2)+\frac {\pi }{2}+3 \ln \relax (x )\right ) \Gamma \left (\frac {3}{4}\right )-\hypergeom \left (\left [\frac {3}{4}, 1, 1\right ], \left [2, 2\right ], -x^{3}\right ) \Gamma \left (\frac {3}{4}\right ) x^{3}}{12 \Gamma \left (\frac {3}{4}\right )}\)	\(45\)
trager	\(\frac {4 \left (x^{3}+1\right )^{\frac {1}{4}}}{3}+\frac {\ln \left (\frac {2 \left (x^{3}+1\right )^{\frac {3}{4}}-x^{3}-2 \sqrt {x^{3}+1}+2 \left (x^{3}+1\right )^{\frac {1}{4}}-2}{x^{3}}\right )}{3}-\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {-\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{3}+1}+2 \left (x^{3}+1\right )^{\frac {3}{4}}-2 \RootOf \left (\textit {\_Z}^{2}+1\right )-2 \left (x^{3}+1\right )^{\frac {1}{4}}}{x^{3}}\right )}{3}\)	\(118\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3+1)^(1/4)/x,x,method=_RETURNVERBOSE)

[Out]

-1/12/GAMMA(3/4)*(-4*(4-3*ln(2)+1/2*Pi+3*ln(x))*GAMMA(3/4)-hypergeom([3/4,1,1],[2,2],-x^3)*GAMMA(3/4)*x^3)

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maxima [A]  time = 0.59, size = 44, normalized size = 1.05 \begin {gather*} \frac {4}{3} \, {\left (x^{3} + 1\right )}^{\frac {1}{4}} - \frac {2}{3} \, \arctan \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{3} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{3} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.28, size = 30, normalized size = 0.71 \begin {gather*} \frac {4\,{\left (x^3+1\right )}^{1/4}}{3}-\frac {2\,\mathrm {atanh}\left ({\left (x^3+1\right )}^{1/4}\right )}{3}-\frac {2\,\mathrm {atan}\left ({\left (x^3+1\right )}^{1/4}\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3 + 1)^(1/4)/x,x)

[Out]

(4*(x^3 + 1)^(1/4))/3 - (2*atanh((x^3 + 1)^(1/4)))/3 - (2*atan((x^3 + 1)^(1/4)))/3

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sympy [C]  time = 0.79, size = 37, normalized size = 0.88 \begin {gather*} - \frac {x^{\frac {3}{4}} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{3}}} \right )}}{3 \Gamma \left (\frac {3}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**(3/4)*gamma(-1/4)*hyper((-1/4, -1/4), (3/4,), exp_polar(I*pi)/x**3)/(3*gamma(3/4))

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