Integrand size = 32, antiderivative size = 193 \[ \int \frac {x^4 \left (2+x^5\right )}{\sqrt {1+x^5} \left (-1-x^5+a x^{10}\right )} \, dx=\frac {\sqrt {2} \left (1+4 a+\sqrt {1+4 a}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {a} \sqrt {1+x^5}}{\sqrt {-1-2 a-\sqrt {1+4 a}}}\right )}{5 \sqrt {a} \sqrt {1+4 a} \sqrt {-1-2 a-\sqrt {1+4 a}}}+\frac {\sqrt {2} \left (-1-4 a+\sqrt {1+4 a}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {a} \sqrt {1+x^5}}{\sqrt {-1-2 a+\sqrt {1+4 a}}}\right )}{5 \sqrt {a} \sqrt {1+4 a} \sqrt {-1-2 a+\sqrt {1+4 a}}} \]
1/5*2^(1/2)*(1+4*a+(1+4*a)^(1/2))*arctan(2^(1/2)*a^(1/2)*(x^5+1)^(1/2)/(-1 -2*a-(1+4*a)^(1/2))^(1/2))/a^(1/2)/(1+4*a)^(1/2)/(-1-2*a-(1+4*a)^(1/2))^(1 /2)+1/5*2^(1/2)*(-1-4*a+(1+4*a)^(1/2))*arctan(2^(1/2)*a^(1/2)*(x^5+1)^(1/2 )/(-1-2*a+(1+4*a)^(1/2))^(1/2))/a^(1/2)/(1+4*a)^(1/2)/(-1-2*a+(1+4*a)^(1/2 ))^(1/2)
Time = 0.26 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.89 \[ \int \frac {x^4 \left (2+x^5\right )}{\sqrt {1+x^5} \left (-1-x^5+a x^{10}\right )} \, dx=\frac {\sqrt {2} \left (\frac {\left (1+4 a+\sqrt {1+4 a}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {a} \sqrt {1+x^5}}{\sqrt {-1-2 a-\sqrt {1+4 a}}}\right )}{\sqrt {-1-2 a-\sqrt {1+4 a}}}+\frac {\left (-1-4 a+\sqrt {1+4 a}\right ) \arctan \left (\frac {\sqrt {2} \sqrt {a} \sqrt {1+x^5}}{\sqrt {-1-2 a+\sqrt {1+4 a}}}\right )}{\sqrt {-1-2 a+\sqrt {1+4 a}}}\right )}{5 \sqrt {a} \sqrt {1+4 a}} \]
(Sqrt[2]*(((1 + 4*a + Sqrt[1 + 4*a])*ArcTan[(Sqrt[2]*Sqrt[a]*Sqrt[1 + x^5] )/Sqrt[-1 - 2*a - Sqrt[1 + 4*a]]])/Sqrt[-1 - 2*a - Sqrt[1 + 4*a]] + ((-1 - 4*a + Sqrt[1 + 4*a])*ArcTan[(Sqrt[2]*Sqrt[a]*Sqrt[1 + x^5])/Sqrt[-1 - 2*a + Sqrt[1 + 4*a]]])/Sqrt[-1 - 2*a + Sqrt[1 + 4*a]]))/(5*Sqrt[a]*Sqrt[1 + 4 *a])
Time = 0.59 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.45, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {7266, 25, 1197, 25, 1475, 1083, 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.
\(\displaystyle \int \frac {x^4 \left (x^5+2\right )}{\sqrt {x^5+1} \left (a x^{10}-x^5-1\right )} \, dx\) |
\(\Big \downarrow \) 7266 |
\(\displaystyle \frac {1}{5} \int -\frac {x^5+2}{\sqrt {x^5+1} \left (-a x^{10}+x^5+1\right )}dx^5\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{5} \int \frac {x^5+2}{\sqrt {x^5+1} \left (-a x^{10}+x^5+1\right )}dx^5\) |
\(\Big \downarrow \) 1197 |
\(\displaystyle -\frac {2}{5} \int -\frac {x^{10}+1}{a x^{20}-(2 a+1) x^{10}+a}d\sqrt {x^5+1}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {2}{5} \int \frac {x^{10}+1}{a x^{20}-(2 a+1) x^{10}+a}d\sqrt {x^5+1}\) |
\(\Big \downarrow \) 1475 |
\(\displaystyle -\frac {2}{5} \left (-\frac {\int \frac {1}{x^{10}-\frac {\sqrt {4 a+1} \sqrt {x^5+1}}{\sqrt {a}}+1}d\sqrt {x^5+1}}{2 a}-\frac {\int \frac {1}{x^{10}+\frac {\sqrt {4 a+1} \sqrt {x^5+1}}{\sqrt {a}}+1}d\sqrt {x^5+1}}{2 a}\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle -\frac {2}{5} \left (\frac {\int \frac {1}{\frac {1}{a}-x^{10}}d\left (2 \sqrt {x^5+1}-\frac {\sqrt {4 a+1}}{\sqrt {a}}\right )}{a}+\frac {\int \frac {1}{\frac {1}{a}-x^{10}}d\left (\frac {\sqrt {4 a+1}}{\sqrt {a}}+2 \sqrt {x^5+1}\right )}{a}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {2}{5} \left (\frac {\text {arctanh}\left (\sqrt {a} \left (2 \sqrt {x^5+1}-\frac {\sqrt {4 a+1}}{\sqrt {a}}\right )\right )}{\sqrt {a}}+\frac {\text {arctanh}\left (\sqrt {a} \left (\frac {\sqrt {4 a+1}}{\sqrt {a}}+2 \sqrt {x^5+1}\right )\right )}{\sqrt {a}}\right )\) |
(-2*(ArcTanh[Sqrt[a]*(-(Sqrt[1 + 4*a]/Sqrt[a]) + 2*Sqrt[1 + x^5])]/Sqrt[a] + ArcTanh[Sqrt[a]*(Sqrt[1 + 4*a]/Sqrt[a] + 2*Sqrt[1 + x^5])]/Sqrt[a]))/5
3.25.6.3.1 Defintions of rubi rules used
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_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)), x_Symbol] :> Simp[2 Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; Fr eeQ[{a, b, c, d, e, f, g}, x]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[2*(d/e) - b/c, 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^ 2, x], x], x] + Simp[e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; F reeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - a*e^2, 0] && (GtQ[2*(d/e) - b/c, 0] || ( !LtQ[2*(d/e) - b/c, 0] && EqQ[d - e*Rt[a/c, 2] , 0]))
Int[(u_)*(x_)^(m_.), x_Symbol] :> Simp[1/(m + 1) Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /; FreeQ[m, x] && NeQ[m, -1] && Function OfQ[x^(m + 1), u, x]
Time = 0.52 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.68
method | result | size |
pseudoelliptic | \(\frac {\sqrt {2}\, \left (-\frac {\left (1+4 a +\sqrt {1+4 a}\right ) \operatorname {arctanh}\left (\frac {a \sqrt {x^{5}+1}\, \sqrt {2}}{\sqrt {\left (\sqrt {1+4 a}+2 a +1\right ) a}}\right )}{\sqrt {\left (\sqrt {1+4 a}+2 a +1\right ) a}}+\frac {\left (-1-4 a +\sqrt {1+4 a}\right ) \arctan \left (\frac {a \sqrt {x^{5}+1}\, \sqrt {2}}{\sqrt {\left (-1-2 a +\sqrt {1+4 a}\right ) a}}\right )}{\sqrt {\left (-1-2 a +\sqrt {1+4 a}\right ) a}}\right )}{5 \sqrt {1+4 a}}\) | \(131\) |
1/5*2^(1/2)/(1+4*a)^(1/2)*(-(1+4*a+(1+4*a)^(1/2))/(((1+4*a)^(1/2)+2*a+1)*a )^(1/2)*arctanh(a*(x^5+1)^(1/2)*2^(1/2)/(((1+4*a)^(1/2)+2*a+1)*a)^(1/2))+( -1-4*a+(1+4*a)^(1/2))/((-1-2*a+(1+4*a)^(1/2))*a)^(1/2)*arctan(a*(x^5+1)^(1 /2)*2^(1/2)/((-1-2*a+(1+4*a)^(1/2))*a)^(1/2)))
Time = 0.26 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.38 \[ \int \frac {x^4 \left (2+x^5\right )}{\sqrt {1+x^5} \left (-1-x^5+a x^{10}\right )} \, dx=\left [\frac {\log \left (\frac {a x^{10} - 2 \, \sqrt {x^{5} + 1} \sqrt {a} x^{5} + x^{5} + 1}{a x^{10} - x^{5} - 1}\right )}{5 \, \sqrt {a}}, \frac {2 \, \sqrt {-a} \arctan \left (\frac {\sqrt {-a} x^{5}}{\sqrt {x^{5} + 1}}\right )}{5 \, a}\right ] \]
[1/5*log((a*x^10 - 2*sqrt(x^5 + 1)*sqrt(a)*x^5 + x^5 + 1)/(a*x^10 - x^5 - 1))/sqrt(a), 2/5*sqrt(-a)*arctan(sqrt(-a)*x^5/sqrt(x^5 + 1))/a]
Timed out. \[ \int \frac {x^4 \left (2+x^5\right )}{\sqrt {1+x^5} \left (-1-x^5+a x^{10}\right )} \, dx=\text {Timed out} \]
\[ \int \frac {x^4 \left (2+x^5\right )}{\sqrt {1+x^5} \left (-1-x^5+a x^{10}\right )} \, dx=\int { \frac {{\left (x^{5} + 2\right )} x^{4}}{{\left (a x^{10} - x^{5} - 1\right )} \sqrt {x^{5} + 1}} \,d x } \]
\[ \int \frac {x^4 \left (2+x^5\right )}{\sqrt {1+x^5} \left (-1-x^5+a x^{10}\right )} \, dx=\int { \frac {{\left (x^{5} + 2\right )} x^{4}}{{\left (a x^{10} - x^{5} - 1\right )} \sqrt {x^{5} + 1}} \,d x } \]
Time = 6.77 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.24 \[ \int \frac {x^4 \left (2+x^5\right )}{\sqrt {1+x^5} \left (-1-x^5+a x^{10}\right )} \, dx=\frac {\ln \left (\frac {a\,x^{10}+x^5-2\,\sqrt {a}\,x^5\,\sqrt {x^5+1}+1}{-4\,a\,x^{10}+4\,x^5+4}\right )}{5\,\sqrt {a}} \]