Integrand size = 23, antiderivative size = 48 \[ \int \frac {(-1+x) \sqrt [4]{x^3+x^4}}{x (1+x)} \, dx=\sqrt [4]{x^3+x^4}+\frac {7}{2} \arctan \left (\frac {x}{\sqrt [4]{x^3+x^4}}\right )-\frac {7}{2} \text {arctanh}\left (\frac {x}{\sqrt [4]{x^3+x^4}}\right ) \]
Time = 0.22 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.52 \[ \int \frac {(-1+x) \sqrt [4]{x^3+x^4}}{x (1+x)} \, dx=\frac {x^{9/4} \left (2 x^{3/4} (1+x)+7 (1+x)^{3/4} \arctan \left (\sqrt [4]{\frac {x}{1+x}}\right )-7 (1+x)^{3/4} \text {arctanh}\left (\sqrt [4]{\frac {x}{1+x}}\right )\right )}{2 \left (x^3 (1+x)\right )^{3/4}} \]
(x^(9/4)*(2*x^(3/4)*(1 + x) + 7*(1 + x)^(3/4)*ArcTan[(x/(1 + x))^(1/4)] - 7*(1 + x)^(3/4)*ArcTanh[(x/(1 + x))^(1/4)]))/(2*(x^3*(1 + x))^(3/4))
Time = 0.32 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.65, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {2467, 25, 90, 73, 854, 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.
| \(\displaystyle \int \frac {(x-1) \sqrt [4]{x^4+x^3}}{x (x+1)} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt [4]{x^4+x^3} \int -\frac {1-x}{\sqrt [4]{x} (x+1)^{3/4}}dx}{x^{3/4} \sqrt [4]{x+1}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\sqrt [4]{x^4+x^3} \int \frac {1-x}{\sqrt [4]{x} (x+1)^{3/4}}dx}{x^{3/4} \sqrt [4]{x+1}}\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle -\frac {\sqrt [4]{x^4+x^3} \left (\frac {7}{4} \int \frac {1}{\sqrt [4]{x} (x+1)^{3/4}}dx-x^{3/4} \sqrt [4]{x+1}\right )}{x^{3/4} \sqrt [4]{x+1}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {\sqrt [4]{x^4+x^3} \left (7 \int \frac {\sqrt {x}}{(x+1)^{3/4}}d\sqrt [4]{x}-x^{3/4} \sqrt [4]{x+1}\right )}{x^{3/4} \sqrt [4]{x+1}}\) |
| \(\Big \downarrow \) 854 |
| \(\displaystyle -\frac {\sqrt [4]{x^4+x^3} \left (7 \int \frac {\sqrt {x}}{1-x}d\frac {\sqrt [4]{x}}{\sqrt [4]{x+1}}-x^{3/4} \sqrt [4]{x+1}\right )}{x^{3/4} \sqrt [4]{x+1}}\) |
| \(\Big \downarrow \) 827 |
| \(\displaystyle -\frac {\sqrt [4]{x^4+x^3} \left (7 \left (\frac {1}{2} \int \frac {1}{1-\sqrt {x}}d\frac {\sqrt [4]{x}}{\sqrt [4]{x+1}}-\frac {1}{2} \int \frac {1}{\sqrt {x}+1}d\frac {\sqrt [4]{x}}{\sqrt [4]{x+1}}\right )-x^{3/4} \sqrt [4]{x+1}\right )}{x^{3/4} \sqrt [4]{x+1}}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {\sqrt [4]{x^4+x^3} \left (7 \left (\frac {1}{2} \int \frac {1}{1-\sqrt {x}}d\frac {\sqrt [4]{x}}{\sqrt [4]{x+1}}-\frac {1}{2} \arctan \left (\frac {\sqrt [4]{x}}{\sqrt [4]{x+1}}\right )\right )-x^{3/4} \sqrt [4]{x+1}\right )}{x^{3/4} \sqrt [4]{x+1}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {\sqrt [4]{x^4+x^3} \left (7 \left (\frac {1}{2} \text {arctanh}\left (\frac {\sqrt [4]{x}}{\sqrt [4]{x+1}}\right )-\frac {1}{2} \arctan \left (\frac {\sqrt [4]{x}}{\sqrt [4]{x+1}}\right )\right )-x^{3/4} \sqrt [4]{x+1}\right )}{x^{3/4} \sqrt [4]{x+1}}\) |
-(((x^3 + x^4)^(1/4)*(-(x^(3/4)*(1 + x)^(1/4)) + 7*(-1/2*ArcTan[x^(1/4)/(1 + x)^(1/4)] + ArcTanh[x^(1/4)/(1 + x)^(1/4)]/2)))/(x^(3/4)*(1 + x)^(1/4)) )
3.7.9.3.1 Defintions of rubi rules used
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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
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_)^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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + (m + 1)/n) Subst[Int[x^m/(1 - b*x^n)^(p + (m + 1)/n + 1), x], x, x/(a + b*x^n )^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, - 2^(-1)] && IntegersQ[m, p + (m + 1)/n]
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
Result contains higher order function than in optimal. Order 5 vs. order 3.
Time = 2.22 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.62
| method | result | size |
| meijerg | \(-\frac {4 x^{\frac {3}{4}} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {3}{4}\right ], \left [\frac {7}{4}\right ], -x \right )}{3}+\frac {4 x^{\frac {7}{4}} \operatorname {hypergeom}\left (\left [\frac {3}{4}, \frac {7}{4}\right ], \left [\frac {11}{4}\right ], -x \right )}{7}\) | \(30\) |
| pseudoelliptic | \(\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}+\frac {7 \ln \left (\frac {\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}-x}{x}\right )}{4}-\frac {7 \arctan \left (\frac {\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}}{x}\right )}{2}-\frac {7 \ln \left (\frac {\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}+x}{x}\right )}{4}\) | \(65\) |
| trager | \(\left (x^{4}+x^{3}\right )^{\frac {1}{4}}-\frac {7 \ln \left (\frac {2 \left (x^{4}+x^{3}\right )^{\frac {3}{4}}+2 \sqrt {x^{4}+x^{3}}\, x +2 \left (x^{4}+x^{3}\right )^{\frac {1}{4}} x^{2}+2 x^{3}+x^{2}}{x^{2}}\right )}{4}-\frac {7 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (\frac {2 \sqrt {x^{4}+x^{3}}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x -2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \left (x^{4}+x^{3}\right )^{\frac {3}{4}}-2 \left (x^{4}+x^{3}\right )^{\frac {1}{4}} x^{2}-\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}}{x^{2}}\right )}{4}\) | \(143\) |
| risch | \(\left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}}+\frac {\left (-\frac {7 \ln \left (\frac {2 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {3}{4}}+2 \sqrt {x^{4}+3 x^{3}+3 x^{2}+x}\, x +2 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {1}{4}} x^{2}+2 x^{3}+2 \sqrt {x^{4}+3 x^{3}+3 x^{2}+x}+4 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {1}{4}} x +5 x^{2}+2 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {1}{4}}+4 x +1}{\left (1+x \right )^{2}}\right )}{4}-\frac {7 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) \ln \left (-\frac {2 \sqrt {x^{4}+3 x^{3}+3 x^{2}+x}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x -2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{3}+2 \sqrt {x^{4}+3 x^{3}+3 x^{2}+x}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x^{2}-2 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {3}{4}}+2 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {1}{4}} x^{2}-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right ) x +4 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {1}{4}} x -\operatorname {RootOf}\left (\textit {\_Z}^{2}+1\right )+2 \left (x^{4}+3 x^{3}+3 x^{2}+x \right )^{\frac {1}{4}}}{\left (1+x \right )^{2}}\right )}{4}\right ) \left (x^{3} \left (1+x \right )\right )^{\frac {1}{4}} \left (\left (1+x \right )^{3} x \right )^{\frac {1}{4}}}{x \left (1+x \right )}\) | \(369\) |
Time = 0.26 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.35 \[ \int \frac {(-1+x) \sqrt [4]{x^3+x^4}}{x (1+x)} \, dx={\left (x^{4} + x^{3}\right )}^{\frac {1}{4}} - \frac {7}{2} \, \arctan \left (\frac {{\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) - \frac {7}{4} \, \log \left (\frac {x + {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) + \frac {7}{4} \, \log \left (-\frac {x - {\left (x^{4} + x^{3}\right )}^{\frac {1}{4}}}{x}\right ) \]
(x^4 + x^3)^(1/4) - 7/2*arctan((x^4 + x^3)^(1/4)/x) - 7/4*log((x + (x^4 + x^3)^(1/4))/x) + 7/4*log(-(x - (x^4 + x^3)^(1/4))/x)
\[ \int \frac {(-1+x) \sqrt [4]{x^3+x^4}}{x (1+x)} \, dx=\int \frac {\sqrt [4]{x^{3} \left (x + 1\right )} \left (x - 1\right )}{x \left (x + 1\right )}\, dx \]
\[ \int \frac {(-1+x) \sqrt [4]{x^3+x^4}}{x (1+x)} \, dx=\int { \frac {{\left (x^{4} + x^{3}\right )}^{\frac {1}{4}} {\left (x - 1\right )}}{{\left (x + 1\right )} x} \,d x } \]
Time = 0.28 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.94 \[ \int \frac {(-1+x) \sqrt [4]{x^3+x^4}}{x (1+x)} \, dx=x {\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}} - \frac {7}{2} \, \arctan \left ({\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}}\right ) - \frac {7}{4} \, \log \left ({\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}} + 1\right ) + \frac {7}{4} \, \log \left ({\left | {\left (\frac {1}{x} + 1\right )}^{\frac {1}{4}} - 1 \right |}\right ) \]
x*(1/x + 1)^(1/4) - 7/2*arctan((1/x + 1)^(1/4)) - 7/4*log((1/x + 1)^(1/4) + 1) + 7/4*log(abs((1/x + 1)^(1/4) - 1))
Timed out. \[ \int \frac {(-1+x) \sqrt [4]{x^3+x^4}}{x (1+x)} \, dx=\int \frac {{\left (x^4+x^3\right )}^{1/4}\,\left (x-1\right )}{x\,\left (x+1\right )} \,d x \]