Integrand size = 13, antiderivative size = 52 \[ \int \frac {\sqrt [4]{1+x^3}}{x^7} \, dx=\frac {\left (-4-x^3\right ) \sqrt [4]{1+x^3}}{24 x^6}+\frac {1}{16} \arctan \left (\sqrt [4]{1+x^3}\right )+\frac {1}{16} \text {arctanh}\left (\sqrt [4]{1+x^3}\right ) \]
Time = 0.06 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.92 \[ \int \frac {\sqrt [4]{1+x^3}}{x^7} \, dx=\frac {1}{48} \left (-\frac {2 \sqrt [4]{1+x^3} \left (4+x^3\right )}{x^6}+3 \arctan \left (\sqrt [4]{1+x^3}\right )+3 \text {arctanh}\left (\sqrt [4]{1+x^3}\right )\right ) \]
((-2*(1 + x^3)^(1/4)*(4 + x^3))/x^6 + 3*ArcTan[(1 + x^3)^(1/4)] + 3*ArcTan h[(1 + x^3)^(1/4)])/48
Time = 0.18 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.37, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {798, 51, 52, 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.
\(\displaystyle \int \frac {\sqrt [4]{x^3+1}}{x^7} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{3} \int \frac {\sqrt [4]{x^3+1}}{x^9}dx^3\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{8} \int \frac {1}{x^6 \left (x^3+1\right )^{3/4}}dx^3-\frac {\sqrt [4]{x^3+1}}{2 x^6}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{8} \left (-\frac {3}{4} \int \frac {1}{x^3 \left (x^3+1\right )^{3/4}}dx^3-\frac {\sqrt [4]{x^3+1}}{x^3}\right )-\frac {\sqrt [4]{x^3+1}}{2 x^6}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{8} \left (-3 \int \frac {1}{x^{12}-1}d\sqrt [4]{x^3+1}-\frac {\sqrt [4]{x^3+1}}{x^3}\right )-\frac {\sqrt [4]{x^3+1}}{2 x^6}\right )\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{8} \left (-3 \left (-\frac {1}{2} \int \frac {1}{1-x^6}d\sqrt [4]{x^3+1}-\frac {1}{2} \int \frac {1}{x^6+1}d\sqrt [4]{x^3+1}\right )-\frac {\sqrt [4]{x^3+1}}{x^3}\right )-\frac {\sqrt [4]{x^3+1}}{2 x^6}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{8} \left (-3 \left (-\frac {1}{2} \int \frac {1}{1-x^6}d\sqrt [4]{x^3+1}-\frac {1}{2} \arctan \left (\sqrt [4]{x^3+1}\right )\right )-\frac {\sqrt [4]{x^3+1}}{x^3}\right )-\frac {\sqrt [4]{x^3+1}}{2 x^6}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{3} \left (\frac {1}{8} \left (-3 \left (-\frac {1}{2} \arctan \left (\sqrt [4]{x^3+1}\right )-\frac {1}{2} \text {arctanh}\left (\sqrt [4]{x^3+1}\right )\right )-\frac {\sqrt [4]{x^3+1}}{x^3}\right )-\frac {\sqrt [4]{x^3+1}}{2 x^6}\right )\) |
(-1/2*(1 + x^3)^(1/4)/x^6 + (-((1 + x^3)^(1/4)/x^3) - 3*(-1/2*ArcTan[(1 + x^3)^(1/4)] - ArcTanh[(1 + x^3)^(1/4)]/2))/8)/3
3.7.55.3.1 Defintions of rubi rules used
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] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
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] - Simp[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] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
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 = 1.74 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.12
method | result | size |
meijerg | \(-\frac {-\frac {7 \Gamma \left (\frac {3}{4}\right ) x^{3} \operatorname {hypergeom}\left (\left [1, 1, \frac {11}{4}\right ], \left [2, 4\right ], -x^{3}\right )}{32}+\frac {3 \left (-\frac {1}{6}-3 \ln \left (2\right )+\frac {\pi }{2}+3 \ln \left (x \right )\right ) \Gamma \left (\frac {3}{4}\right )}{8}+\frac {2 \Gamma \left (\frac {3}{4}\right )}{x^{6}}+\frac {\Gamma \left (\frac {3}{4}\right )}{x^{3}}}{12 \Gamma \left (\frac {3}{4}\right )}\) | \(58\) |
risch | \(-\frac {x^{6}+5 x^{3}+4}{24 x^{6} \left (x^{3}+1\right )^{\frac {3}{4}}}-\frac {-\frac {3 \Gamma \left (\frac {3}{4}\right ) x^{3} \operatorname {hypergeom}\left (\left [1, 1, \frac {7}{4}\right ], \left [2, 2\right ], -x^{3}\right )}{4}+\left (-3 \ln \left (2\right )+\frac {\pi }{2}+3 \ln \left (x \right )\right ) \Gamma \left (\frac {3}{4}\right )}{32 \Gamma \left (\frac {3}{4}\right )}\) | \(66\) |
pseudoelliptic | \(\frac {-3 \ln \left (\left (x^{3}+1\right )^{\frac {1}{4}}-1\right ) x^{6}+6 \arctan \left (\left (x^{3}+1\right )^{\frac {1}{4}}\right ) x^{6}+3 \ln \left (\left (x^{3}+1\right )^{\frac {1}{4}}+1\right ) x^{6}-4 \left (x^{3}+1\right )^{\frac {1}{4}} x^{3}-16 \left (x^{3}+1\right )^{\frac {1}{4}}}{96 {\left (\left (x^{3}+1\right )^{\frac {1}{4}}-1\right )}^{2} \left (\sqrt {x^{3}+1}+1\right )^{2} {\left (\left (x^{3}+1\right )^{\frac {1}{4}}+1\right )}^{2}}\) | \(101\) |
trager | \(-\frac {\left (x^{3}+4\right ) \left (x^{3}+1\right )^{\frac {1}{4}}}{24 x^{6}}-\frac {\ln \left (\frac {2 \left (x^{3}+1\right )^{\frac {3}{4}}-x^{3}-2 \sqrt {x^{3}+1}+2 \left (x^{3}+1\right )^{\frac {1}{4}}-2}{x^{3}}\right )}{32}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{3}+1}+2 \left (x^{3}+1\right )^{\frac {3}{4}}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )-2 \left (x^{3}+1\right )^{\frac {1}{4}}}{x^{3}}\right )}{32}\) | \(126\) |
-1/12/GAMMA(3/4)*(-7/32*GAMMA(3/4)*x^3*hypergeom([1,1,11/4],[2,4],-x^3)+3/ 8*(-1/6-3*ln(2)+1/2*Pi+3*ln(x))*GAMMA(3/4)+2*GAMMA(3/4)/x^6+GAMMA(3/4)/x^3 )
Time = 0.26 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.21 \[ \int \frac {\sqrt [4]{1+x^3}}{x^7} \, dx=\frac {6 \, x^{6} \arctan \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}}\right ) + 3 \, x^{6} \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} + 1\right ) - 3 \, x^{6} \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} - 1\right ) - 4 \, {\left (x^{3} + 4\right )} {\left (x^{3} + 1\right )}^{\frac {1}{4}}}{96 \, x^{6}} \]
1/96*(6*x^6*arctan((x^3 + 1)^(1/4)) + 3*x^6*log((x^3 + 1)^(1/4) + 1) - 3*x ^6*log((x^3 + 1)^(1/4) - 1) - 4*(x^3 + 4)*(x^3 + 1)^(1/4))/x^6
Result contains complex when optimal does not.
Time = 1.08 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.65 \[ \int \frac {\sqrt [4]{1+x^3}}{x^7} \, dx=- \frac {\Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{3}}} \right )}}{3 x^{\frac {21}{4}} \Gamma \left (\frac {11}{4}\right )} \]
Time = 0.27 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.38 \[ \int \frac {\sqrt [4]{1+x^3}}{x^7} \, dx=\frac {{\left (x^{3} + 1\right )}^{\frac {5}{4}} + 3 \, {\left (x^{3} + 1\right )}^{\frac {1}{4}}}{24 \, {\left (2 \, x^{3} - {\left (x^{3} + 1\right )}^{2} + 1\right )}} + \frac {1}{16} \, \arctan \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{32} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} + 1\right ) - \frac {1}{32} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} - 1\right ) \]
1/24*((x^3 + 1)^(5/4) + 3*(x^3 + 1)^(1/4))/(2*x^3 - (x^3 + 1)^2 + 1) + 1/1 6*arctan((x^3 + 1)^(1/4)) + 1/32*log((x^3 + 1)^(1/4) + 1) - 1/32*log((x^3 + 1)^(1/4) - 1)
Time = 0.26 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.12 \[ \int \frac {\sqrt [4]{1+x^3}}{x^7} \, dx=-\frac {{\left (x^{3} + 1\right )}^{\frac {5}{4}} + 3 \, {\left (x^{3} + 1\right )}^{\frac {1}{4}}}{24 \, x^{6}} + \frac {1}{16} \, \arctan \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}}\right ) + \frac {1}{32} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} + 1\right ) - \frac {1}{32} \, \log \left ({\left | {\left (x^{3} + 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \]
-1/24*((x^3 + 1)^(5/4) + 3*(x^3 + 1)^(1/4))/x^6 + 1/16*arctan((x^3 + 1)^(1 /4)) + 1/32*log((x^3 + 1)^(1/4) + 1) - 1/32*log(abs((x^3 + 1)^(1/4) - 1))
Time = 6.12 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.87 \[ \int \frac {\sqrt [4]{1+x^3}}{x^7} \, dx=\frac {\mathrm {atan}\left ({\left (x^3+1\right )}^{1/4}\right )}{16}+\frac {\mathrm {atanh}\left ({\left (x^3+1\right )}^{1/4}\right )}{16}-\frac {{\left (x^3+1\right )}^{1/4}}{8\,x^6}-\frac {{\left (x^3+1\right )}^{5/4}}{24\,x^6} \]