Integrand size = 27, antiderivative size = 169 \[ \int \frac {\left (1+x^2\right ) \sqrt [4]{x^3+x^5}}{x^2 \left (-1+x^2\right )} \, dx=\frac {4 \sqrt [4]{x^3+x^5}}{x}+\sqrt [4]{2} \arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^3+x^5}}\right )-\frac {\arctan \left (\frac {2^{3/4} x \sqrt [4]{x^3+x^5}}{\sqrt {2} x^2-\sqrt {x^3+x^5}}\right )}{\sqrt [4]{2}}-\sqrt [4]{2} \text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^3+x^5}}\right )-\frac {\text {arctanh}\left (\frac {\frac {x^2}{\sqrt [4]{2}}+\frac {\sqrt {x^3+x^5}}{2^{3/4}}}{x \sqrt [4]{x^3+x^5}}\right )}{\sqrt [4]{2}} \]
4*(x^5+x^3)^(1/4)/x+2^(1/4)*arctan(2^(1/4)*x/(x^5+x^3)^(1/4))-1/2*arctan(2 ^(3/4)*x*(x^5+x^3)^(1/4)/(2^(1/2)*x^2-(x^5+x^3)^(1/2)))*2^(3/4)-2^(1/4)*ar ctanh(2^(1/4)*x/(x^5+x^3)^(1/4))-1/2*arctanh((1/2*x^2*2^(3/4)+1/2*(x^5+x^3 )^(1/2)*2^(1/4))/x/(x^5+x^3)^(1/4))*2^(3/4)
Time = 0.64 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.30 \[ \int \frac {\left (1+x^2\right ) \sqrt [4]{x^3+x^5}}{x^2 \left (-1+x^2\right )} \, dx=-\frac {\sqrt [4]{x^3+x^5} \left (-8 \sqrt [4]{1+x^2}-2 \sqrt [4]{2} \sqrt [4]{x} \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{x}}{\sqrt [4]{1+x^2}}\right )+2^{3/4} \sqrt [4]{x} \arctan \left (\frac {2^{3/4} \sqrt [4]{x} \sqrt [4]{1+x^2}}{\sqrt {2} \sqrt {x}-\sqrt {1+x^2}}\right )+2 \sqrt [4]{2} \sqrt [4]{x} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{x}}{\sqrt [4]{1+x^2}}\right )+2^{3/4} \sqrt [4]{x} \text {arctanh}\left (\frac {2 \sqrt [4]{2} \sqrt [4]{x} \sqrt [4]{1+x^2}}{2 \sqrt {x}+\sqrt {2} \sqrt {1+x^2}}\right )\right )}{2 x \sqrt [4]{1+x^2}} \]
-1/2*((x^3 + x^5)^(1/4)*(-8*(1 + x^2)^(1/4) - 2*2^(1/4)*x^(1/4)*ArcTan[(2^ (1/4)*x^(1/4))/(1 + x^2)^(1/4)] + 2^(3/4)*x^(1/4)*ArcTan[(2^(3/4)*x^(1/4)* (1 + x^2)^(1/4))/(Sqrt[2]*Sqrt[x] - Sqrt[1 + x^2])] + 2*2^(1/4)*x^(1/4)*Ar cTanh[(2^(1/4)*x^(1/4))/(1 + x^2)^(1/4)] + 2^(3/4)*x^(1/4)*ArcTanh[(2*2^(1 /4)*x^(1/4)*(1 + x^2)^(1/4))/(2*Sqrt[x] + Sqrt[2]*Sqrt[1 + x^2])]))/(x*(1 + x^2)^(1/4))
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 0.30 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.26, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2467, 25, 368, 1012}
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.
| \(\displaystyle \int \frac {\left (x^2+1\right ) \sqrt [4]{x^5+x^3}}{x^2 \left (x^2-1\right )} \, dx\) |
| \(\Big \downarrow \) 2467 |
| \(\displaystyle \frac {\sqrt [4]{x^5+x^3} \int -\frac {\left (x^2+1\right )^{5/4}}{x^{5/4} \left (1-x^2\right )}dx}{x^{3/4} \sqrt [4]{x^2+1}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\sqrt [4]{x^5+x^3} \int \frac {\left (x^2+1\right )^{5/4}}{x^{5/4} \left (1-x^2\right )}dx}{x^{3/4} \sqrt [4]{x^2+1}}\) |
| \(\Big \downarrow \) 368 |
| \(\displaystyle -\frac {4 \sqrt [4]{x^5+x^3} \int \frac {\left (x^2+1\right )^{5/4}}{\sqrt {x} \left (1-x^2\right )}d\sqrt [4]{x}}{x^{3/4} \sqrt [4]{x^2+1}}\) |
| \(\Big \downarrow \) 1012 |
| \(\displaystyle \frac {4 \sqrt [4]{x^5+x^3} \operatorname {AppellF1}\left (-\frac {1}{8},1,-\frac {5}{4},\frac {7}{8},x^2,-x^2\right )}{x \sqrt [4]{x^2+1}}\) |
3.23.53.3.1 Defintions of rubi rules used
Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_) , x_Symbol] :> With[{k = Denominator[m]}, Simp[k/e Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*2)/e^2))^p*(c + d*(x^(k*2)/e^2))^q, x], x, (e*x)^(1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && FractionQ[m ] && IntegerQ[p]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
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]
Time = 26.58 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.38
| method | result | size |
| pseudoelliptic | \(\frac {-\ln \left (\frac {2^{\frac {3}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}} x +\sqrt {2}\, x^{2}+\sqrt {x^{3} \left (x^{2}+1\right )}}{-2^{\frac {3}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}} x +\sqrt {2}\, x^{2}+\sqrt {x^{3} \left (x^{2}+1\right )}}\right ) 2^{\frac {3}{4}} x -2 \arctan \left (\frac {2^{\frac {1}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}+x}{x}\right ) 2^{\frac {3}{4}} x -2 \arctan \left (\frac {2^{\frac {1}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}-x}{x}\right ) 2^{\frac {3}{4}} x -2 \ln \left (\frac {-2^{\frac {1}{4}} x -\left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{2^{\frac {1}{4}} x -\left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}\right ) 2^{\frac {1}{4}} x -4 \arctan \left (\frac {2^{\frac {3}{4}} \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{2 x}\right ) 2^{\frac {1}{4}} x +16 \left (x^{3} \left (x^{2}+1\right )\right )^{\frac {1}{4}}}{4 x}\) | \(234\) |
| trager | \(\text {Expression too large to display}\) | \(731\) |
| risch | \(\text {Expression too large to display}\) | \(1754\) |
1/4*(-ln((2^(3/4)*(x^3*(x^2+1))^(1/4)*x+2^(1/2)*x^2+(x^3*(x^2+1))^(1/2))/( -2^(3/4)*(x^3*(x^2+1))^(1/4)*x+2^(1/2)*x^2+(x^3*(x^2+1))^(1/2)))*2^(3/4)*x -2*arctan((2^(1/4)*(x^3*(x^2+1))^(1/4)+x)/x)*2^(3/4)*x-2*arctan((2^(1/4)*( x^3*(x^2+1))^(1/4)-x)/x)*2^(3/4)*x-2*ln((-2^(1/4)*x-(x^3*(x^2+1))^(1/4))/( 2^(1/4)*x-(x^3*(x^2+1))^(1/4)))*2^(1/4)*x-4*arctan(1/2*2^(3/4)/x*(x^3*(x^2 +1))^(1/4))*2^(1/4)*x+16*(x^3*(x^2+1))^(1/4))/x
Result contains complex when optimal does not.
Time = 3.17 (sec) , antiderivative size = 689, normalized size of antiderivative = 4.08 \[ \int \frac {\left (1+x^2\right ) \sqrt [4]{x^3+x^5}}{x^2 \left (-1+x^2\right )} \, dx=-\frac {2^{\frac {1}{4}} x \log \left (\frac {4 \, \sqrt {2} {\left (x^{5} + x^{3}\right )}^{\frac {1}{4}} x^{2} + 2^{\frac {3}{4}} {\left (x^{4} + 2 \, x^{3} + x^{2}\right )} + 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{5} + x^{3}} x + 4 \, {\left (x^{5} + x^{3}\right )}^{\frac {3}{4}}}{x^{4} - 2 \, x^{3} + x^{2}}\right ) - 2^{\frac {1}{4}} x \log \left (\frac {4 \, \sqrt {2} {\left (x^{5} + x^{3}\right )}^{\frac {1}{4}} x^{2} - 2^{\frac {3}{4}} {\left (x^{4} + 2 \, x^{3} + x^{2}\right )} - 4 \cdot 2^{\frac {1}{4}} \sqrt {x^{5} + x^{3}} x + 4 \, {\left (x^{5} + x^{3}\right )}^{\frac {3}{4}}}{x^{4} - 2 \, x^{3} + x^{2}}\right ) - i \cdot 2^{\frac {1}{4}} x \log \left (-\frac {4 \, \sqrt {2} {\left (x^{5} + x^{3}\right )}^{\frac {1}{4}} x^{2} - 2^{\frac {3}{4}} {\left (i \, x^{4} + 2 i \, x^{3} + i \, x^{2}\right )} + 4 i \cdot 2^{\frac {1}{4}} \sqrt {x^{5} + x^{3}} x - 4 \, {\left (x^{5} + x^{3}\right )}^{\frac {3}{4}}}{x^{4} - 2 \, x^{3} + x^{2}}\right ) + i \cdot 2^{\frac {1}{4}} x \log \left (-\frac {4 \, \sqrt {2} {\left (x^{5} + x^{3}\right )}^{\frac {1}{4}} x^{2} - 2^{\frac {3}{4}} {\left (-i \, x^{4} - 2 i \, x^{3} - i \, x^{2}\right )} - 4 i \cdot 2^{\frac {1}{4}} \sqrt {x^{5} + x^{3}} x - 4 \, {\left (x^{5} + x^{3}\right )}^{\frac {3}{4}}}{x^{4} - 2 \, x^{3} + x^{2}}\right ) - \left (-2\right )^{\frac {1}{4}} x \log \left (-\frac {4 \, \sqrt {-2} {\left (x^{5} + x^{3}\right )}^{\frac {1}{4}} x^{2} + \left (-2\right )^{\frac {3}{4}} {\left (x^{4} - 2 \, x^{3} + x^{2}\right )} - 4 \, \left (-2\right )^{\frac {1}{4}} \sqrt {x^{5} + x^{3}} x + 4 \, {\left (x^{5} + x^{3}\right )}^{\frac {3}{4}}}{x^{4} + 2 \, x^{3} + x^{2}}\right ) - i \, \left (-2\right )^{\frac {1}{4}} x \log \left (\frac {4 \, \sqrt {-2} {\left (x^{5} + x^{3}\right )}^{\frac {1}{4}} x^{2} + i \, \left (-2\right )^{\frac {3}{4}} {\left (x^{4} - 2 \, x^{3} + x^{2}\right )} + 4 i \, \left (-2\right )^{\frac {1}{4}} \sqrt {x^{5} + x^{3}} x - 4 \, {\left (x^{5} + x^{3}\right )}^{\frac {3}{4}}}{x^{4} + 2 \, x^{3} + x^{2}}\right ) + i \, \left (-2\right )^{\frac {1}{4}} x \log \left (\frac {4 \, \sqrt {-2} {\left (x^{5} + x^{3}\right )}^{\frac {1}{4}} x^{2} - i \, \left (-2\right )^{\frac {3}{4}} {\left (x^{4} - 2 \, x^{3} + x^{2}\right )} - 4 i \, \left (-2\right )^{\frac {1}{4}} \sqrt {x^{5} + x^{3}} x - 4 \, {\left (x^{5} + x^{3}\right )}^{\frac {3}{4}}}{x^{4} + 2 \, x^{3} + x^{2}}\right ) + \left (-2\right )^{\frac {1}{4}} x \log \left (-\frac {4 \, \sqrt {-2} {\left (x^{5} + x^{3}\right )}^{\frac {1}{4}} x^{2} - \left (-2\right )^{\frac {3}{4}} {\left (x^{4} - 2 \, x^{3} + x^{2}\right )} + 4 \, \left (-2\right )^{\frac {1}{4}} \sqrt {x^{5} + x^{3}} x + 4 \, {\left (x^{5} + x^{3}\right )}^{\frac {3}{4}}}{x^{4} + 2 \, x^{3} + x^{2}}\right ) - 16 \, {\left (x^{5} + x^{3}\right )}^{\frac {1}{4}}}{4 \, x} \]
-1/4*(2^(1/4)*x*log((4*sqrt(2)*(x^5 + x^3)^(1/4)*x^2 + 2^(3/4)*(x^4 + 2*x^ 3 + x^2) + 4*2^(1/4)*sqrt(x^5 + x^3)*x + 4*(x^5 + x^3)^(3/4))/(x^4 - 2*x^3 + x^2)) - 2^(1/4)*x*log((4*sqrt(2)*(x^5 + x^3)^(1/4)*x^2 - 2^(3/4)*(x^4 + 2*x^3 + x^2) - 4*2^(1/4)*sqrt(x^5 + x^3)*x + 4*(x^5 + x^3)^(3/4))/(x^4 - 2*x^3 + x^2)) - I*2^(1/4)*x*log(-(4*sqrt(2)*(x^5 + x^3)^(1/4)*x^2 - 2^(3/4 )*(I*x^4 + 2*I*x^3 + I*x^2) + 4*I*2^(1/4)*sqrt(x^5 + x^3)*x - 4*(x^5 + x^3 )^(3/4))/(x^4 - 2*x^3 + x^2)) + I*2^(1/4)*x*log(-(4*sqrt(2)*(x^5 + x^3)^(1 /4)*x^2 - 2^(3/4)*(-I*x^4 - 2*I*x^3 - I*x^2) - 4*I*2^(1/4)*sqrt(x^5 + x^3) *x - 4*(x^5 + x^3)^(3/4))/(x^4 - 2*x^3 + x^2)) - (-2)^(1/4)*x*log(-(4*sqrt (-2)*(x^5 + x^3)^(1/4)*x^2 + (-2)^(3/4)*(x^4 - 2*x^3 + x^2) - 4*(-2)^(1/4) *sqrt(x^5 + x^3)*x + 4*(x^5 + x^3)^(3/4))/(x^4 + 2*x^3 + x^2)) - I*(-2)^(1 /4)*x*log((4*sqrt(-2)*(x^5 + x^3)^(1/4)*x^2 + I*(-2)^(3/4)*(x^4 - 2*x^3 + x^2) + 4*I*(-2)^(1/4)*sqrt(x^5 + x^3)*x - 4*(x^5 + x^3)^(3/4))/(x^4 + 2*x^ 3 + x^2)) + I*(-2)^(1/4)*x*log((4*sqrt(-2)*(x^5 + x^3)^(1/4)*x^2 - I*(-2)^ (3/4)*(x^4 - 2*x^3 + x^2) - 4*I*(-2)^(1/4)*sqrt(x^5 + x^3)*x - 4*(x^5 + x^ 3)^(3/4))/(x^4 + 2*x^3 + x^2)) + (-2)^(1/4)*x*log(-(4*sqrt(-2)*(x^5 + x^3) ^(1/4)*x^2 - (-2)^(3/4)*(x^4 - 2*x^3 + x^2) + 4*(-2)^(1/4)*sqrt(x^5 + x^3) *x + 4*(x^5 + x^3)^(3/4))/(x^4 + 2*x^3 + x^2)) - 16*(x^5 + x^3)^(1/4))/x
\[ \int \frac {\left (1+x^2\right ) \sqrt [4]{x^3+x^5}}{x^2 \left (-1+x^2\right )} \, dx=\int \frac {\sqrt [4]{x^{3} \left (x^{2} + 1\right )} \left (x^{2} + 1\right )}{x^{2} \left (x - 1\right ) \left (x + 1\right )}\, dx \]
\[ \int \frac {\left (1+x^2\right ) \sqrt [4]{x^3+x^5}}{x^2 \left (-1+x^2\right )} \, dx=\int { \frac {{\left (x^{5} + x^{3}\right )}^{\frac {1}{4}} {\left (x^{2} + 1\right )}}{{\left (x^{2} - 1\right )} x^{2}} \,d x } \]
\[ \int \frac {\left (1+x^2\right ) \sqrt [4]{x^3+x^5}}{x^2 \left (-1+x^2\right )} \, dx=\int { \frac {{\left (x^{5} + x^{3}\right )}^{\frac {1}{4}} {\left (x^{2} + 1\right )}}{{\left (x^{2} - 1\right )} x^{2}} \,d x } \]
Timed out. \[ \int \frac {\left (1+x^2\right ) \sqrt [4]{x^3+x^5}}{x^2 \left (-1+x^2\right )} \, dx=\int \frac {{\left (x^5+x^3\right )}^{1/4}\,\left (x^2+1\right )}{x^2\,\left (x^2-1\right )} \,d x \]