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

Optimal. Leaf size=45 \[ -\frac {\sqrt [4]{x^3+1}}{3 x^3}-\frac {1}{6} \tan ^{-1}\left (\sqrt [4]{x^3+1}\right )-\frac {1}{6} \tanh ^{-1}\left (\sqrt [4]{x^3+1}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 45, 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, 47, 63, 212, 206, 203} \begin {gather*} -\frac {\sqrt [4]{x^3+1}}{3 x^3}-\frac {1}{6} \tan ^{-1}\left (\sqrt [4]{x^3+1}\right )-\frac {1}{6} \tanh ^{-1}\left (\sqrt [4]{x^3+1}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + x^3)^(1/4)/x^4,x]

[Out]

-1/3*(1 + x^3)^(1/4)/x^3 - ArcTan[(1 + x^3)^(1/4)]/6 - ArcTanh[(1 + x^3)^(1/4)]/6

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 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^4} \, dx &=\frac {1}{3} \operatorname {Subst}\left (\int \frac {\sqrt [4]{1+x}}{x^2} \, dx,x,x^3\right )\\ &=-\frac {\sqrt [4]{1+x^3}}{3 x^3}+\frac {1}{12} \operatorname {Subst}\left (\int \frac {1}{x (1+x)^{3/4}} \, dx,x,x^3\right )\\ &=-\frac {\sqrt [4]{1+x^3}}{3 x^3}+\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{-1+x^4} \, dx,x,\sqrt [4]{1+x^3}\right )\\ &=-\frac {\sqrt [4]{1+x^3}}{3 x^3}-\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt [4]{1+x^3}\right )-\frac {1}{6} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [4]{1+x^3}\right )\\ &=-\frac {\sqrt [4]{1+x^3}}{3 x^3}-\frac {1}{6} \tan ^{-1}\left (\sqrt [4]{1+x^3}\right )-\frac {1}{6} \tanh ^{-1}\left (\sqrt [4]{1+x^3}\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.01, size = 26, normalized size = 0.58 \begin {gather*} \frac {4}{15} \left (x^3+1\right )^{5/4} \, _2F_1\left (\frac {5}{4},2;\frac {9}{4};x^3+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + x^3)^(1/4)/x^4,x]

[Out]

(4*(1 + x^3)^(5/4)*Hypergeometric2F1[5/4, 2, 9/4, 1 + x^3])/15

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.03, size = 45, normalized size = 1.00 \begin {gather*} -\frac {\sqrt [4]{1+x^3}}{3 x^3}-\frac {1}{6} \tan ^{-1}\left (\sqrt [4]{1+x^3}\right )-\frac {1}{6} \tanh ^{-1}\left (\sqrt [4]{1+x^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(1 + x^3)^(1/4)/x^3 - ArcTan[(1 + x^3)^(1/4)]/6 - ArcTanh[(1 + x^3)^(1/4)]/6

________________________________________________________________________________________

fricas [A]  time = 0.45, size = 57, normalized size = 1.27 \begin {gather*} -\frac {2 \, x^{3} \arctan \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}}\right ) + x^{3} \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} + 1\right ) - x^{3} \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} - 1\right ) + 4 \, {\left (x^{3} + 1\right )}^{\frac {1}{4}}}{12 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(2*x^3*arctan((x^3 + 1)^(1/4)) + x^3*log((x^3 + 1)^(1/4) + 1) - x^3*log((x^3 + 1)^(1/4) - 1) + 4*(x^3 +
1)^(1/4))/x^3

________________________________________________________________________________________

giac [A]  time = 0.22, size = 48, normalized size = 1.07 \begin {gather*} -\frac {{\left (x^{3} + 1\right )}^{\frac {1}{4}}}{3 \, x^{3}} - \frac {1}{6} \, \arctan \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{12} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{12} \, \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^4,x, algorithm="giac")

[Out]

-1/3*(x^3 + 1)^(1/4)/x^3 - 1/6*arctan((x^3 + 1)^(1/4)) - 1/12*log((x^3 + 1)^(1/4) + 1) + 1/12*log(abs((x^3 + 1
)^(1/4) - 1))

________________________________________________________________________________________

maple [C]  time = 1.59, size = 52, normalized size = 1.16

method	result	size
meijerg	\(-\frac {\frac {4 \Gamma \left (\frac {3}{4}\right )}{x^{3}}-\left (-3 \ln \relax (2)+\frac {\pi }{2}-1+3 \ln \relax (x )\right ) \Gamma \left (\frac {3}{4}\right )+\frac {3 \hypergeom \left (\left [1, 1, \frac {7}{4}\right ], \left [2, 3\right ], -x^{3}\right ) \Gamma \left (\frac {3}{4}\right ) x^{3}}{8}}{12 \Gamma \left (\frac {3}{4}\right )}\)	\(52\)
risch	\(-\frac {\left (x^{3}+1\right )^{\frac {1}{4}}}{3 x^{3}}+\frac {\left (-3 \ln \relax (2)+\frac {\pi }{2}+3 \ln \relax (x )\right ) \Gamma \left (\frac {3}{4}\right )-\frac {3 \hypergeom \left (\left [1, 1, \frac {7}{4}\right ], \left [2, 2\right ], -x^{3}\right ) \Gamma \left (\frac {3}{4}\right ) x^{3}}{4}}{12 \Gamma \left (\frac {3}{4}\right )}\)	\(56\)
trager	\(-\frac {\left (x^{3}+1\right )^{\frac {1}{4}}}{3 x^{3}}-\frac {\ln \left (\frac {2 \left (x^{3}+1\right )^{\frac {3}{4}}+x^{3}+2 \sqrt {x^{3}+1}+2 \left (x^{3}+1\right )^{\frac {1}{4}}+2}{x^{3}}\right )}{12}-\frac {\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {-\RootOf \left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \RootOf \left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{3}+1}+2 \left (x^{3}+1\right )^{\frac {3}{4}}-2 \RootOf \left (\textit {\_Z}^{2}+1\right )-2 \left (x^{3}+1\right )^{\frac {1}{4}}}{x^{3}}\right )}{12}\)	\(119\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12/GAMMA(3/4)*(4*GAMMA(3/4)/x^3-(-3*ln(2)+1/2*Pi-1+3*ln(x))*GAMMA(3/4)+3/8*hypergeom([1,1,7/4],[2,3],-x^3)*
GAMMA(3/4)*x^3)

________________________________________________________________________________________

maxima [A]  time = 0.52, size = 47, normalized size = 1.04 \begin {gather*} -\frac {{\left (x^{3} + 1\right )}^{\frac {1}{4}}}{3 \, x^{3}} - \frac {1}{6} \, \arctan \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{12} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{12} \, \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^4,x, algorithm="maxima")

[Out]

-1/3*(x^3 + 1)^(1/4)/x^3 - 1/6*arctan((x^3 + 1)^(1/4)) - 1/12*log((x^3 + 1)^(1/4) + 1) + 1/12*log((x^3 + 1)^(1
/4) - 1)

________________________________________________________________________________________

mupad [B]  time = 0.54, size = 33, normalized size = 0.73 \begin {gather*} -\frac {\mathrm {atan}\left ({\left (x^3+1\right )}^{1/4}\right )}{6}-\frac {\mathrm {atanh}\left ({\left (x^3+1\right )}^{1/4}\right )}{6}-\frac {{\left (x^3+1\right )}^{1/4}}{3\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3 + 1)^(1/4)/x^4,x)

[Out]

- atan((x^3 + 1)^(1/4))/6 - atanh((x^3 + 1)^(1/4))/6 - (x^3 + 1)^(1/4)/(3*x^3)

________________________________________________________________________________________

sympy [C]  time = 0.90, size = 34, normalized size = 0.76 \begin {gather*} - \frac {\Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{3}}} \right )}}{3 x^{\frac {9}{4}} \Gamma \left (\frac {7}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-gamma(3/4)*hyper((-1/4, 3/4), (7/4,), exp_polar(I*pi)/x**3)/(3*x**(9/4)*gamma(7/4))

________________________________________________________________________________________