Integrand size = 13, antiderivative size = 42 \[ \int \frac {\sqrt [4]{1+x^3}}{x} \, dx=\frac {4}{3} \sqrt [4]{1+x^3}-\frac {2}{3} \arctan \left (\sqrt [4]{1+x^3}\right )-\frac {2}{3} \text {arctanh}\left (\sqrt [4]{1+x^3}\right ) \] Output:
4/3*(x^3+1)^(1/4)-2/3*arctan((x^3+1)^(1/4))-2/3*arctanh((x^3+1)^(1/4))
Time = 0.02 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt [4]{1+x^3}}{x} \, dx=\frac {4}{3} \sqrt [4]{1+x^3}-\frac {2}{3} \arctan \left (\sqrt [4]{1+x^3}\right )-\frac {2}{3} \text {arctanh}\left (\sqrt [4]{1+x^3}\right ) \] Input:
Integrate[(1 + x^3)^(1/4)/x,x]
Output:
(4*(1 + x^3)^(1/4))/3 - (2*ArcTan[(1 + x^3)^(1/4)])/3 - (2*ArcTanh[(1 + x^ 3)^(1/4)])/3
Time = 0.16 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.12, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {798, 60, 73, 756, 216, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sqrt [4]{x^3+1}}{x} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{3} \int \frac {\sqrt [4]{x^3+1}}{x^3}dx^3\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{3} \left (\int \frac {1}{x^3 \left (x^3+1\right )^{3/4}}dx^3+4 \sqrt [4]{x^3+1}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} \left (4 \int \frac {1}{x^{12}-1}d\sqrt [4]{x^3+1}+4 \sqrt [4]{x^3+1}\right )\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {1}{3} \left (4 \left (-\frac {1}{2} \int \frac {1}{1-x^6}d\sqrt [4]{x^3+1}-\frac {1}{2} \int \frac {1}{x^6+1}d\sqrt [4]{x^3+1}\right )+4 \sqrt [4]{x^3+1}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{3} \left (4 \left (-\frac {1}{2} \int \frac {1}{1-x^6}d\sqrt [4]{x^3+1}-\frac {1}{2} \arctan \left (\sqrt [4]{x^3+1}\right )\right )+4 \sqrt [4]{x^3+1}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{3} \left (4 \left (-\frac {1}{2} \arctan \left (\sqrt [4]{x^3+1}\right )-\frac {1}{2} \text {arctanh}\left (\sqrt [4]{x^3+1}\right )\right )+4 \sqrt [4]{x^3+1}\right )\) |
Input:
Int[(1 + x^3)^(1/4)/x,x]
Output:
(4*(1 + x^3)^(1/4) + 4*(-1/2*ArcTan[(1 + x^3)^(1/4)] - ArcTanh[(1 + x^3)^( 1/4)]/2))/3
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] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[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]]
Result contains higher order function than in optimal. Order 5 vs. order 3.
Time = 1.16 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.07
method | result | size |
meijerg | \(-\frac {-4 \left (4-3 \ln \left (2\right )+\frac {\pi }{2}+3 \ln \left (x \right )\right ) \Gamma \left (\frac {3}{4}\right )-\Gamma \left (\frac {3}{4}\right ) x^{3} \operatorname {hypergeom}\left (\left [\frac {3}{4}, 1, 1\right ], \left [2, 2\right ], -x^{3}\right )}{12 \Gamma \left (\frac {3}{4}\right )}\) | \(45\) |
pseudoelliptic | \(\frac {4 \left (x^{3}+1\right )^{\frac {1}{4}}}{3}+\frac {\ln \left (\left (x^{3}+1\right )^{\frac {1}{4}}-1\right )}{3}-\frac {\ln \left (1+\left (x^{3}+1\right )^{\frac {1}{4}}\right )}{3}-\frac {2 \arctan \left (\left (x^{3}+1\right )^{\frac {1}{4}}\right )}{3}\) | \(45\) |
trager | \(\frac {4 \left (x^{3}+1\right )^{\frac {1}{4}}}{3}-\frac {\ln \left (\frac {2 \left (x^{3}+1\right )^{\frac {3}{4}}+x^{3}+2 \sqrt {x^{3}+1}+2 \left (x^{3}+1\right )^{\frac {1}{4}}+2}{x^{3}}\right )}{3}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \left (x^{3}+1\right )^{\frac {3}{4}}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{3}+1}-2 \left (x^{3}+1\right )^{\frac {1}{4}}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )}{x^{3}}\right )}{3}\) | \(115\) |
Input:
int((x^3+1)^(1/4)/x,x,method=_RETURNVERBOSE)
Output:
-1/12/GAMMA(3/4)*(-4*(4-3*ln(2)+1/2*Pi+3*ln(x))*GAMMA(3/4)-GAMMA(3/4)*x^3* hypergeom([3/4,1,1],[2,2],-x^3))
Time = 0.10 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.05 \[ \int \frac {\sqrt [4]{1+x^3}}{x} \, dx=\frac {4}{3} \, {\left (x^{3} + 1\right )}^{\frac {1}{4}} - \frac {2}{3} \, \arctan \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{3} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{3} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} - 1\right ) \] Input:
integrate((x^3+1)^(1/4)/x,x, algorithm="fricas")
Output:
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)
Result contains complex when optimal does not.
Time = 0.62 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.88 \[ \int \frac {\sqrt [4]{1+x^3}}{x} \, dx=- \frac {x^{\frac {3}{4}} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{3}}} \right )}}{3 \Gamma \left (\frac {3}{4}\right )} \] Input:
integrate((x**3+1)**(1/4)/x,x)
Output:
-x**(3/4)*gamma(-1/4)*hyper((-1/4, -1/4), (3/4,), exp_polar(I*pi)/x**3)/(3 *gamma(3/4))
Time = 0.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.05 \[ \int \frac {\sqrt [4]{1+x^3}}{x} \, dx=\frac {4}{3} \, {\left (x^{3} + 1\right )}^{\frac {1}{4}} - \frac {2}{3} \, \arctan \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{3} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{3} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} - 1\right ) \] Input:
integrate((x^3+1)^(1/4)/x,x, algorithm="maxima")
Output:
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)
Time = 0.42 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.07 \[ \int \frac {\sqrt [4]{1+x^3}}{x} \, dx=\frac {4}{3} \, {\left (x^{3} + 1\right )}^{\frac {1}{4}} - \frac {2}{3} \, \arctan \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}}\right ) - \frac {1}{3} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {1}{3} \, \log \left ({\left | {\left (x^{3} + 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \] Input:
integrate((x^3+1)^(1/4)/x,x, algorithm="giac")
Output:
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))
Time = 7.77 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.71 \[ \int \frac {\sqrt [4]{1+x^3}}{x} \, dx=\frac {4\,{\left (x^3+1\right )}^{1/4}}{3}-\frac {2\,\mathrm {atanh}\left ({\left (x^3+1\right )}^{1/4}\right )}{3}-\frac {2\,\mathrm {atan}\left ({\left (x^3+1\right )}^{1/4}\right )}{3} \] Input:
int((x^3 + 1)^(1/4)/x,x)
Output:
(4*(x^3 + 1)^(1/4))/3 - (2*atanh((x^3 + 1)^(1/4)))/3 - (2*atan((x^3 + 1)^( 1/4)))/3
\[ \int \frac {\sqrt [4]{1+x^3}}{x} \, dx=\frac {4 \left (x^{3}+1\right )^{\frac {1}{4}}}{3}+\int \frac {\left (x^{3}+1\right )^{\frac {1}{4}}}{x^{4}+x}d x \] Input:
int((x^3+1)^(1/4)/x,x)
Output:
(4*(x**3 + 1)**(1/4) + 3*int((x**3 + 1)**(1/4)/(x**4 + x),x))/3