\(\int \frac {1}{x^4 \sqrt [4]{1+x^3}} \, dx\) [578]

Optimal result

Integrand size = 13, antiderivative size = 45 \[ \int \frac {1}{x^4 \sqrt [4]{1+x^3}} \, dx=-\frac {\left (1+x^3\right )^{3/4}}{3 x^3}-\frac {1}{6} \arctan \left (\sqrt [4]{1+x^3}\right )+\frac {1}{6} \text {arctanh}\left (\sqrt [4]{1+x^3}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {272, 44, 65, 304, 209, 212} \[ \int \frac {1}{x^4 \sqrt [4]{1+x^3}} \, dx=-\frac {1}{6} \arctan \left (\sqrt [4]{x^3+1}\right )+\frac {1}{6} \text {arctanh}\left (\sqrt [4]{x^3+1}\right )-\frac {\left (x^3+1\right )^{3/4}}{3 x^3} \]

[In]

Int[1/(x^4*(1 + x^3)^(1/4)),x]

[Out]

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

Rule 44

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

Rule 65

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 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 212

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

Rule 272

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]]

Rule 304

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \frac {1}{x^2 \sqrt [4]{1+x}} \, dx,x,x^3\right ) \\ & = -\frac {\left (1+x^3\right )^{3/4}}{3 x^3}-\frac {1}{12} \text {Subst}\left (\int \frac {1}{x \sqrt [4]{1+x}} \, dx,x,x^3\right ) \\ & = -\frac {\left (1+x^3\right )^{3/4}}{3 x^3}-\frac {1}{3} \text {Subst}\left (\int \frac {x^2}{-1+x^4} \, dx,x,\sqrt [4]{1+x^3}\right ) \\ & = -\frac {\left (1+x^3\right )^{3/4}}{3 x^3}+\frac {1}{6} \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sqrt [4]{1+x^3}\right )-\frac {1}{6} \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt [4]{1+x^3}\right ) \\ & = -\frac {\left (1+x^3\right )^{3/4}}{3 x^3}-\frac {1}{6} \arctan \left (\sqrt [4]{1+x^3}\right )+\frac {1}{6} \text {arctanh}\left (\sqrt [4]{1+x^3}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.91 \[ \int \frac {1}{x^4 \sqrt [4]{1+x^3}} \, dx=\frac {1}{6} \left (-\frac {2 \left (1+x^3\right )^{3/4}}{x^3}-\arctan \left (\sqrt [4]{1+x^3}\right )+\text {arctanh}\left (\sqrt [4]{1+x^3}\right )\right ) \]

[In]

Integrate[1/(x^4*(1 + x^3)^(1/4)),x]

[Out]

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

Maple [A] (verified)

Time = 1.84 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.29

method	result	size
pseudoelliptic	\(\frac {-\ln \left (\left (x^{3}+1\right )^{\frac {1}{4}}-1\right ) x^{3}-2 \arctan \left (\left (x^{3}+1\right )^{\frac {1}{4}}\right ) x^{3}+\ln \left (\left (x^{3}+1\right )^{\frac {1}{4}}+1\right ) x^{3}-4 \left (x^{3}+1\right )^{\frac {3}{4}}}{12 x^{3}}\)	\(58\)
risch	\(-\frac {\left (x^{3}+1\right )^{\frac {3}{4}}}{3 x^{3}}-\frac {\sqrt {2}\, \Gamma \left (\frac {3}{4}\right ) \left (-\frac {\pi \sqrt {2}\, x^{3} \operatorname {hypergeom}\left (\left [1, 1, \frac {5}{4}\right ], \left [2, 2\right ], -x^{3}\right )}{4 \Gamma \left (\frac {3}{4}\right )}+\frac {\left (-3 \ln \left (2\right )-\frac {\pi }{2}+3 \ln \left (x \right )\right ) \pi \sqrt {2}}{\Gamma \left (\frac {3}{4}\right )}\right )}{24 \pi }\)	\(72\)
meijerg	\(\frac {\sqrt {2}\, \Gamma \left (\frac {3}{4}\right ) \left (\frac {5 \pi \sqrt {2}\, x^{3} \operatorname {hypergeom}\left (\left [1, 1, \frac {9}{4}\right ], \left [2, 3\right ], -x^{3}\right )}{32 \Gamma \left (\frac {3}{4}\right )}-\frac {\left (3-3 \ln \left (2\right )-\frac {\pi }{2}+3 \ln \left (x \right )\right ) \pi \sqrt {2}}{4 \Gamma \left (\frac {3}{4}\right )}-\frac {\pi \sqrt {2}}{\Gamma \left (\frac {3}{4}\right ) x^{3}}\right )}{6 \pi }\)	\(74\)
trager	\(-\frac {\left (x^{3}+1\right )^{\frac {3}{4}}}{3 x^{3}}+\frac {\ln \left (\frac {2 \left (x^{3}+1\right )^{\frac {3}{4}}+x^{3}+2 \sqrt {x^{3}+1}+2 \left (x^{3}+1\right )^{\frac {1}{4}}+2}{x^{3}}\right )}{12}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{3}+1}+2 \left (x^{3}+1\right )^{\frac {3}{4}}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )-2 \left (x^{3}+1\right )^{\frac {1}{4}}}{x^{3}}\right )}{12}\)	\(119\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.27 \[ \int \frac {1}{x^4 \sqrt [4]{1+x^3}} \, dx=-\frac {2 \, x^{3} \arctan \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}}\right ) - x^{3} \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} + 1\right ) + x^{3} \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} - 1\right ) + 4 \, {\left (x^{3} + 1\right )}^{\frac {3}{4}}}{12 \, x^{3}} \]

[In]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.62 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.71 \[ \int \frac {1}{x^4 \sqrt [4]{1+x^3}} \, dx=- \frac {\Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{3}}} \right )}}{3 x^{\frac {15}{4}} \Gamma \left (\frac {9}{4}\right )} \]

[In]

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

[Out]

-gamma(5/4)*hyper((1/4, 5/4), (9/4,), exp_polar(I*pi)/x**3)/(3*x**(15/4)*gamma(9/4))

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.04 \[ \int \frac {1}{x^4 \sqrt [4]{1+x^3}} \, dx=-\frac {{\left (x^{3} + 1\right )}^{\frac {3}{4}}}{3 \, x^{3}} - \frac {1}{6} \, \arctan \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{12} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} + 1\right ) - \frac {1}{12} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} - 1\right ) \]

[In]

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

[Out]

-1/3*(x^3 + 1)^(3/4)/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)

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.07 \[ \int \frac {1}{x^4 \sqrt [4]{1+x^3}} \, dx=-\frac {{\left (x^{3} + 1\right )}^{\frac {3}{4}}}{3 \, x^{3}} - \frac {1}{6} \, \arctan \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{12} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} + 1\right ) - \frac {1}{12} \, \log \left ({\left | {\left (x^{3} + 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \]

[In]

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

[Out]

-1/3*(x^3 + 1)^(3/4)/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))

Mupad [B] (verification not implemented)

Time = 5.63 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.73 \[ \int \frac {1}{x^4 \sqrt [4]{1+x^3}} \, dx=\frac {\mathrm {atanh}\left ({\left (x^3+1\right )}^{1/4}\right )}{6}-\frac {\mathrm {atan}\left ({\left (x^3+1\right )}^{1/4}\right )}{6}-\frac {{\left (x^3+1\right )}^{3/4}}{3\,x^3} \]

[In]

int(1/(x^4*(x^3 + 1)^(1/4)),x)

[Out]

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