Integrand size = 13, antiderivative size = 52 \[ \int \frac {1}{x^7 \sqrt [4]{1+x^3}} \, dx=\frac {\left (1+x^3\right )^{3/4} \left (-4+5 x^3\right )}{24 x^6}+\frac {5}{48} \arctan \left (\sqrt [4]{1+x^3}\right )-\frac {5}{48} \text {arctanh}\left (\sqrt [4]{1+x^3}\right ) \]
Time = 0.07 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^7 \sqrt [4]{1+x^3}} \, dx=\frac {\left (1+x^3\right )^{3/4} \left (-4+5 x^3\right )}{24 x^6}+\frac {5}{48} \arctan \left (\sqrt [4]{1+x^3}\right )-\frac {5}{48} \text {arctanh}\left (\sqrt [4]{1+x^3}\right ) \]
((1 + x^3)^(3/4)*(-4 + 5*x^3))/(24*x^6) + (5*ArcTan[(1 + x^3)^(1/4)])/48 - (5*ArcTanh[(1 + x^3)^(1/4)])/48
Time = 0.19 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.31, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {798, 52, 52, 73, 25, 827, 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.
\(\displaystyle \int \frac {1}{x^7 \sqrt [4]{x^3+1}} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{3} \int \frac {1}{x^9 \sqrt [4]{x^3+1}}dx^3\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{3} \left (-\frac {5}{8} \int \frac {1}{x^6 \sqrt [4]{x^3+1}}dx^3-\frac {\left (x^3+1\right )^{3/4}}{2 x^6}\right )\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {1}{3} \left (-\frac {5}{8} \left (-\frac {1}{4} \int \frac {1}{x^3 \sqrt [4]{x^3+1}}dx^3-\frac {\left (x^3+1\right )^{3/4}}{x^3}\right )-\frac {\left (x^3+1\right )^{3/4}}{2 x^6}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} \left (-\frac {5}{8} \left (-\int -\frac {x^6}{1-x^{12}}d\sqrt [4]{x^3+1}-\frac {\left (x^3+1\right )^{3/4}}{x^3}\right )-\frac {\left (x^3+1\right )^{3/4}}{2 x^6}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{3} \left (-\frac {5}{8} \left (\int \frac {x^6}{1-x^{12}}d\sqrt [4]{x^3+1}-\frac {\left (x^3+1\right )^{3/4}}{x^3}\right )-\frac {\left (x^3+1\right )^{3/4}}{2 x^6}\right )\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {1}{3} \left (-\frac {5}{8} \left (\frac {1}{2} \int \frac {1}{1-x^6}d\sqrt [4]{x^3+1}-\frac {1}{2} \int \frac {1}{x^6+1}d\sqrt [4]{x^3+1}-\frac {\left (x^3+1\right )^{3/4}}{x^3}\right )-\frac {\left (x^3+1\right )^{3/4}}{2 x^6}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{3} \left (-\frac {5}{8} \left (\frac {1}{2} \int \frac {1}{1-x^6}d\sqrt [4]{x^3+1}-\frac {1}{2} \arctan \left (\sqrt [4]{x^3+1}\right )-\frac {\left (x^3+1\right )^{3/4}}{x^3}\right )-\frac {\left (x^3+1\right )^{3/4}}{2 x^6}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{3} \left (-\frac {5}{8} \left (-\frac {1}{2} \arctan \left (\sqrt [4]{x^3+1}\right )+\frac {1}{2} \text {arctanh}\left (\sqrt [4]{x^3+1}\right )-\frac {\left (x^3+1\right )^{3/4}}{x^3}\right )-\frac {\left (x^3+1\right )^{3/4}}{2 x^6}\right )\) |
(-1/2*(1 + x^3)^(3/4)/x^6 - (5*(-((1 + x^3)^(3/4)/x^3) - ArcTan[(1 + x^3)^ (1/4)]/2 + ArcTanh[(1 + x^3)^(1/4)]/2))/8)/3
3.7.54.3.1 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 + 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[(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]]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Result contains higher order function than in optimal. Order 5 vs. order 3.
Time = 1.86 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.58
method | result | size |
risch | \(\frac {5 x^{6}+x^{3}-4}{24 x^{6} \left (x^{3}+1\right )^{\frac {1}{4}}}+\frac {5 \sqrt {2}\, \Gamma \left (\frac {3}{4}\right ) \left (-\frac {\pi \sqrt {2}\, x^{3} \operatorname {hypergeom}\left (\left [1, 1, \frac {5}{4}\right ], \left [2, 2\right ], -x^{3}\right )}{4 \Gamma \left (\frac {3}{4}\right )}+\frac {\left (-3 \ln \left (2\right )-\frac {\pi }{2}+3 \ln \left (x \right )\right ) \pi \sqrt {2}}{\Gamma \left (\frac {3}{4}\right )}\right )}{192 \pi }\) | \(82\) |
meijerg | \(\frac {\sqrt {2}\, \Gamma \left (\frac {3}{4}\right ) \left (-\frac {15 \pi \sqrt {2}\, x^{3} \operatorname {hypergeom}\left (\left [1, 1, \frac {13}{4}\right ], \left [2, 4\right ], -x^{3}\right )}{128 \Gamma \left (\frac {3}{4}\right )}+\frac {5 \left (\frac {33}{10}-3 \ln \left (2\right )-\frac {\pi }{2}+3 \ln \left (x \right )\right ) \pi \sqrt {2}}{32 \Gamma \left (\frac {3}{4}\right )}-\frac {\pi \sqrt {2}}{2 \Gamma \left (\frac {3}{4}\right ) x^{6}}+\frac {\pi \sqrt {2}}{4 \Gamma \left (\frac {3}{4}\right ) x^{3}}\right )}{6 \pi }\) | \(87\) |
pseudoelliptic | \(\frac {5 \ln \left (\left (x^{3}+1\right )^{\frac {1}{4}}-1\right ) x^{6}+10 \arctan \left (\left (x^{3}+1\right )^{\frac {1}{4}}\right ) x^{6}-5 \ln \left (\left (x^{3}+1\right )^{\frac {1}{4}}+1\right ) x^{6}+20 x^{3} \left (x^{3}+1\right )^{\frac {3}{4}}-16 \left (x^{3}+1\right )^{\frac {3}{4}}}{96 {\left (\left (x^{3}+1\right )^{\frac {1}{4}}-1\right )}^{2} \left (\sqrt {x^{3}+1}+1\right )^{2} {\left (\left (x^{3}+1\right )^{\frac {1}{4}}+1\right )}^{2}}\) | \(101\) |
trager | \(\frac {\left (x^{3}+1\right )^{\frac {3}{4}} \left (5 x^{3}-4\right )}{24 x^{6}}+\frac {5 \ln \left (\frac {2 \left (x^{3}+1\right )^{\frac {3}{4}}-x^{3}-2 \sqrt {x^{3}+1}+2 \left (x^{3}+1\right )^{\frac {1}{4}}-2}{x^{3}}\right )}{96}+\frac {5 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \sqrt {x^{3}+1}+2 \left (x^{3}+1\right )^{\frac {3}{4}}-2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )-2 \left (x^{3}+1\right )^{\frac {1}{4}}}{x^{3}}\right )}{96}\) | \(128\) |
1/24*(5*x^6+x^3-4)/x^6/(x^3+1)^(1/4)+5/192/Pi*2^(1/2)*GAMMA(3/4)*(-1/4*Pi* 2^(1/2)/GAMMA(3/4)*x^3*hypergeom([1,1,5/4],[2,2],-x^3)+(-3*ln(2)-1/2*Pi+3* ln(x))*Pi*2^(1/2)/GAMMA(3/4))
Time = 0.26 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.25 \[ \int \frac {1}{x^7 \sqrt [4]{1+x^3}} \, dx=\frac {10 \, x^{6} \arctan \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}}\right ) - 5 \, x^{6} \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} + 1\right ) + 5 \, x^{6} \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} - 1\right ) + 4 \, {\left (5 \, x^{3} - 4\right )} {\left (x^{3} + 1\right )}^{\frac {3}{4}}}{96 \, x^{6}} \]
1/96*(10*x^6*arctan((x^3 + 1)^(1/4)) - 5*x^6*log((x^3 + 1)^(1/4) + 1) + 5* x^6*log((x^3 + 1)^(1/4) - 1) + 4*(5*x^3 - 4)*(x^3 + 1)^(3/4))/x^6
Result contains complex when optimal does not.
Time = 1.42 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.62 \[ \int \frac {1}{x^7 \sqrt [4]{1+x^3}} \, dx=- \frac {\Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {9}{4} \\ \frac {13}{4} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{3}}} \right )}}{3 x^{\frac {27}{4}} \Gamma \left (\frac {13}{4}\right )} \]
Time = 0.28 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.42 \[ \int \frac {1}{x^7 \sqrt [4]{1+x^3}} \, dx=-\frac {5 \, {\left (x^{3} + 1\right )}^{\frac {7}{4}} - 9 \, {\left (x^{3} + 1\right )}^{\frac {3}{4}}}{24 \, {\left (2 \, x^{3} - {\left (x^{3} + 1\right )}^{2} + 1\right )}} + \frac {5}{48} \, \arctan \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}}\right ) - \frac {5}{96} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {5}{96} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} - 1\right ) \]
-1/24*(5*(x^3 + 1)^(7/4) - 9*(x^3 + 1)^(3/4))/(2*x^3 - (x^3 + 1)^2 + 1) + 5/48*arctan((x^3 + 1)^(1/4)) - 5/96*log((x^3 + 1)^(1/4) + 1) + 5/96*log((x ^3 + 1)^(1/4) - 1)
Time = 0.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.15 \[ \int \frac {1}{x^7 \sqrt [4]{1+x^3}} \, dx=\frac {5 \, {\left (x^{3} + 1\right )}^{\frac {7}{4}} - 9 \, {\left (x^{3} + 1\right )}^{\frac {3}{4}}}{24 \, x^{6}} + \frac {5}{48} \, \arctan \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}}\right ) - \frac {5}{96} \, \log \left ({\left (x^{3} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {5}{96} \, \log \left ({\left | {\left (x^{3} + 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \]
1/24*(5*(x^3 + 1)^(7/4) - 9*(x^3 + 1)^(3/4))/x^6 + 5/48*arctan((x^3 + 1)^( 1/4)) - 5/96*log((x^3 + 1)^(1/4) + 1) + 5/96*log(abs((x^3 + 1)^(1/4) - 1))
Time = 5.96 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.87 \[ \int \frac {1}{x^7 \sqrt [4]{1+x^3}} \, dx=\frac {5\,\mathrm {atan}\left ({\left (x^3+1\right )}^{1/4}\right )}{48}-\frac {5\,\mathrm {atanh}\left ({\left (x^3+1\right )}^{1/4}\right )}{48}-\frac {3\,{\left (x^3+1\right )}^{3/4}}{8\,x^6}+\frac {5\,{\left (x^3+1\right )}^{7/4}}{24\,x^6} \]