Integrand size = 27, antiderivative size = 74 \[ \int \frac {-b+2 a x^4}{x^2 \left (-b+a x^4\right )^{3/4}} \, dx=-\frac {\sqrt [4]{-b+a x^4}}{x}-\sqrt [4]{a} \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )+\sqrt [4]{a} \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right ) \]
-(a*x^4-b)^(1/4)/x-a^(1/4)*arctan(a^(1/4)*x/(a*x^4-b)^(1/4))+a^(1/4)*arcta nh(a^(1/4)*x/(a*x^4-b)^(1/4))
Time = 0.34 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00 \[ \int \frac {-b+2 a x^4}{x^2 \left (-b+a x^4\right )^{3/4}} \, dx=-\frac {\sqrt [4]{-b+a x^4}}{x}-\sqrt [4]{a} \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )+\sqrt [4]{a} \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right ) \]
-((-b + a*x^4)^(1/4)/x) - a^(1/4)*ArcTan[(a^(1/4)*x)/(-b + a*x^4)^(1/4)] + a^(1/4)*ArcTanh[(a^(1/4)*x)/(-b + a*x^4)^(1/4)]
Time = 0.22 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {953, 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.
\(\displaystyle \int \frac {2 a x^4-b}{x^2 \left (a x^4-b\right )^{3/4}} \, dx\) |
\(\Big \downarrow \) 953 |
\(\displaystyle 2 a \int \frac {x^2}{\left (a x^4-b\right )^{3/4}}dx-\frac {\sqrt [4]{a x^4-b}}{x}\) |
\(\Big \downarrow \) 854 |
\(\displaystyle 2 a \int \frac {x^2}{\sqrt {a x^4-b} \left (1-\frac {a x^4}{a x^4-b}\right )}d\frac {x}{\sqrt [4]{a x^4-b}}-\frac {\sqrt [4]{a x^4-b}}{x}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle 2 a \left (\frac {\int \frac {1}{1-\frac {\sqrt {a} x^2}{\sqrt {a x^4-b}}}d\frac {x}{\sqrt [4]{a x^4-b}}}{2 \sqrt {a}}-\frac {\int \frac {1}{\frac {\sqrt {a} x^2}{\sqrt {a x^4-b}}+1}d\frac {x}{\sqrt [4]{a x^4-b}}}{2 \sqrt {a}}\right )-\frac {\sqrt [4]{a x^4-b}}{x}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle 2 a \left (\frac {\int \frac {1}{1-\frac {\sqrt {a} x^2}{\sqrt {a x^4-b}}}d\frac {x}{\sqrt [4]{a x^4-b}}}{2 \sqrt {a}}-\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 a^{3/4}}\right )-\frac {\sqrt [4]{a x^4-b}}{x}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle 2 a \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 a^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 a^{3/4}}\right )-\frac {\sqrt [4]{a x^4-b}}{x}\) |
-((-b + a*x^4)^(1/4)/x) + 2*a*(-1/2*ArcTan[(a^(1/4)*x)/(-b + a*x^4)^(1/4)] /a^(3/4) + ArcTanh[(a^(1/4)*x)/(-b + a*x^4)^(1/4)]/(2*a^(3/4)))
3.10.79.3.1 Defintions of rubi rules used
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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Simp[d/e^n Int[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && G tQ[m + n, -1]))
Time = 1.26 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.26
method | result | size |
pseudoelliptic | \(\frac {\ln \left (\frac {-a^{\frac {1}{4}} x -\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x -\left (a \,x^{4}-b \right )^{\frac {1}{4}}}\right ) a^{\frac {1}{4}} x +2 \arctan \left (\frac {\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right ) a^{\frac {1}{4}} x -2 \left (a \,x^{4}-b \right )^{\frac {1}{4}}}{2 x}\) | \(93\) |
1/2*(ln((-a^(1/4)*x-(a*x^4-b)^(1/4))/(a^(1/4)*x-(a*x^4-b)^(1/4)))*a^(1/4)* x+2*arctan(1/a^(1/4)/x*(a*x^4-b)^(1/4))*a^(1/4)*x-2*(a*x^4-b)^(1/4))/x
Timed out. \[ \int \frac {-b+2 a x^4}{x^2 \left (-b+a x^4\right )^{3/4}} \, dx=\text {Timed out} \]
Result contains complex when optimal does not.
Time = 1.27 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.70 \[ \int \frac {-b+2 a x^4}{x^2 \left (-b+a x^4\right )^{3/4}} \, dx=\frac {a x^{3} e^{- \frac {3 i \pi }{4}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {a x^{4}}{b}} \right )}}{2 b^{\frac {3}{4}} \Gamma \left (\frac {7}{4}\right )} - b \left (\begin {cases} - \frac {\sqrt [4]{a} \sqrt [4]{-1 + \frac {b}{a x^{4}}} e^{\frac {i \pi }{4}} \Gamma \left (- \frac {1}{4}\right )}{4 b \Gamma \left (\frac {3}{4}\right )} & \text {for}\: \left |{\frac {b}{a x^{4}}}\right | > 1 \\- \frac {\sqrt [4]{a} \sqrt [4]{1 - \frac {b}{a x^{4}}} \Gamma \left (- \frac {1}{4}\right )}{4 b \Gamma \left (\frac {3}{4}\right )} & \text {otherwise} \end {cases}\right ) \]
a*x**3*exp(-3*I*pi/4)*gamma(3/4)*hyper((3/4, 3/4), (7/4,), a*x**4/b)/(2*b* *(3/4)*gamma(7/4)) - b*Piecewise((-a**(1/4)*(-1 + b/(a*x**4))**(1/4)*exp(I *pi/4)*gamma(-1/4)/(4*b*gamma(3/4)), Abs(b/(a*x**4)) > 1), (-a**(1/4)*(1 - b/(a*x**4))**(1/4)*gamma(-1/4)/(4*b*gamma(3/4)), True))
Time = 0.28 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.27 \[ \int \frac {-b+2 a x^4}{x^2 \left (-b+a x^4\right )^{3/4}} \, dx=\frac {1}{2} \, a {\left (\frac {2 \, \arctan \left (\frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )}{a^{\frac {3}{4}}} - \frac {\log \left (-\frac {a^{\frac {1}{4}} - \frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x}}{a^{\frac {1}{4}} + \frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x}}\right )}{a^{\frac {3}{4}}}\right )} - \frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x} \]
1/2*a*(2*arctan((a*x^4 - b)^(1/4)/(a^(1/4)*x))/a^(3/4) - log(-(a^(1/4) - ( a*x^4 - b)^(1/4)/x)/(a^(1/4) + (a*x^4 - b)^(1/4)/x))/a^(3/4)) - (a*x^4 - b )^(1/4)/x
\[ \int \frac {-b+2 a x^4}{x^2 \left (-b+a x^4\right )^{3/4}} \, dx=\int { \frac {2 \, a x^{4} - b}{{\left (a x^{4} - b\right )}^{\frac {3}{4}} x^{2}} \,d x } \]
Timed out. \[ \int \frac {-b+2 a x^4}{x^2 \left (-b+a x^4\right )^{3/4}} \, dx=-\int \frac {b-2\,a\,x^4}{x^2\,{\left (a\,x^4-b\right )}^{3/4}} \,d x \]