Integrand size = 26, antiderivative size = 81 \[ \int \frac {-2 b+a x^4}{x^4 \sqrt [4]{-b+a x^4}} \, dx=-\frac {2 \left (-b+a x^4\right )^{3/4}}{3 x^3}+\frac {1}{2} a^{3/4} \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )+\frac {1}{2} a^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right ) \]
-2/3*(a*x^4-b)^(3/4)/x^3+1/2*a^(3/4)*arctan(a^(1/4)*x/(a*x^4-b)^(1/4))+1/2 *a^(3/4)*arctanh(a^(1/4)*x/(a*x^4-b)^(1/4))
Time = 0.27 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00 \[ \int \frac {-2 b+a x^4}{x^4 \sqrt [4]{-b+a x^4}} \, dx=-\frac {2 \left (-b+a x^4\right )^{3/4}}{3 x^3}+\frac {1}{2} a^{3/4} \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )+\frac {1}{2} a^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right ) \]
(-2*(-b + a*x^4)^(3/4))/(3*x^3) + (a^(3/4)*ArcTan[(a^(1/4)*x)/(-b + a*x^4) ^(1/4)])/2 + (a^(3/4)*ArcTanh[(a^(1/4)*x)/(-b + a*x^4)^(1/4)])/2
Time = 0.20 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {953, 770, 756, 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.
\(\displaystyle \int \frac {a x^4-2 b}{x^4 \sqrt [4]{a x^4-b}} \, dx\) |
\(\Big \downarrow \) 953 |
\(\displaystyle a \int \frac {1}{\sqrt [4]{a x^4-b}}dx-\frac {2 \left (a x^4-b\right )^{3/4}}{3 x^3}\) |
\(\Big \downarrow \) 770 |
\(\displaystyle a \int \frac {1}{1-\frac {a x^4}{a x^4-b}}d\frac {x}{\sqrt [4]{a x^4-b}}-\frac {2 \left (a x^4-b\right )^{3/4}}{3 x^3}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle a \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {a} x^2}{\sqrt {a x^4-b}}}d\frac {x}{\sqrt [4]{a x^4-b}}+\frac {1}{2} \int \frac {1}{\frac {\sqrt {a} x^2}{\sqrt {a x^4-b}}+1}d\frac {x}{\sqrt [4]{a x^4-b}}\right )-\frac {2 \left (a x^4-b\right )^{3/4}}{3 x^3}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle a \left (\frac {1}{2} \int \frac {1}{1-\frac {\sqrt {a} x^2}{\sqrt {a x^4-b}}}d\frac {x}{\sqrt [4]{a x^4-b}}+\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a}}\right )-\frac {2 \left (a x^4-b\right )^{3/4}}{3 x^3}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle a \left (\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a}}\right )-\frac {2 \left (a x^4-b\right )^{3/4}}{3 x^3}\) |
(-2*(-b + a*x^4)^(3/4))/(3*x^3) + a*(ArcTan[(a^(1/4)*x)/(-b + a*x^4)^(1/4) ]/(2*a^(1/4)) + ArcTanh[(a^(1/4)*x)/(-b + a*x^4)^(1/4)]/(2*a^(1/4)))
3.11.84.3.1 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[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n) Subst[In t[1/(1 - b*x^n)^(p + 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)] && IntegerQ[p + 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 = 1.44 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.21
method | result | size |
pseudoelliptic | \(\frac {-6 \arctan \left (\frac {\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right ) a^{\frac {3}{4}} x^{3}+3 \ln \left (\frac {-a^{\frac {1}{4}} x -\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x -\left (a \,x^{4}-b \right )^{\frac {1}{4}}}\right ) a^{\frac {3}{4}} x^{3}-8 \left (a \,x^{4}-b \right )^{\frac {3}{4}}}{12 x^{3}}\) | \(98\) |
1/12*(-6*arctan(1/a^(1/4)/x*(a*x^4-b)^(1/4))*a^(3/4)*x^3+3*ln((-a^(1/4)*x- (a*x^4-b)^(1/4))/(a^(1/4)*x-(a*x^4-b)^(1/4)))*a^(3/4)*x^3-8*(a*x^4-b)^(3/4 ))/x^3
Timed out. \[ \int \frac {-2 b+a x^4}{x^4 \sqrt [4]{-b+a x^4}} \, dx=\text {Timed out} \]
Result contains complex when optimal does not.
Time = 1.05 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.56 \[ \int \frac {-2 b+a x^4}{x^4 \sqrt [4]{-b+a x^4}} \, dx=\frac {a x e^{- \frac {i \pi }{4}} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {a x^{4}}{b}} \right )}}{4 \sqrt [4]{b} \Gamma \left (\frac {5}{4}\right )} - 2 b \left (\begin {cases} - \frac {a^{\frac {3}{4}} \left (-1 + \frac {b}{a x^{4}}\right )^{\frac {3}{4}} e^{\frac {3 i \pi }{4}} \Gamma \left (- \frac {3}{4}\right )}{4 b \Gamma \left (\frac {1}{4}\right )} & \text {for}\: \left |{\frac {b}{a x^{4}}}\right | > 1 \\- \frac {a^{\frac {3}{4}} \left (1 - \frac {b}{a x^{4}}\right )^{\frac {3}{4}} \Gamma \left (- \frac {3}{4}\right )}{4 b \Gamma \left (\frac {1}{4}\right )} & \text {otherwise} \end {cases}\right ) \]
a*x*exp(-I*pi/4)*gamma(1/4)*hyper((1/4, 1/4), (5/4,), a*x**4/b)/(4*b**(1/4 )*gamma(5/4)) - 2*b*Piecewise((-a**(3/4)*(-1 + b/(a*x**4))**(3/4)*exp(3*I* pi/4)*gamma(-3/4)/(4*b*gamma(1/4)), Abs(b/(a*x**4)) > 1), (-a**(3/4)*(1 - b/(a*x**4))**(3/4)*gamma(-3/4)/(4*b*gamma(1/4)), True))
Time = 0.27 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.15 \[ \int \frac {-2 b+a x^4}{x^4 \sqrt [4]{-b+a x^4}} \, dx=-\frac {1}{4} \, a {\left (\frac {2 \, \arctan \left (\frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )}{a^{\frac {1}{4}}} + \frac {\log \left (-\frac {a^{\frac {1}{4}} - \frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x}}{a^{\frac {1}{4}} + \frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}}}{x}}\right )}{a^{\frac {1}{4}}}\right )} - \frac {2 \, {\left (a x^{4} - b\right )}^{\frac {3}{4}}}{3 \, x^{3}} \]
-1/4*a*(2*arctan((a*x^4 - b)^(1/4)/(a^(1/4)*x))/a^(1/4) + log(-(a^(1/4) - (a*x^4 - b)^(1/4)/x)/(a^(1/4) + (a*x^4 - b)^(1/4)/x))/a^(1/4)) - 2/3*(a*x^ 4 - b)^(3/4)/x^3
\[ \int \frac {-2 b+a x^4}{x^4 \sqrt [4]{-b+a x^4}} \, dx=\int { \frac {a x^{4} - 2 \, b}{{\left (a x^{4} - b\right )}^{\frac {1}{4}} x^{4}} \,d x } \]
Time = 6.64 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.70 \[ \int \frac {-2 b+a x^4}{x^4 \sqrt [4]{-b+a x^4}} \, dx=\frac {a\,x\,{\left (1-\frac {a\,x^4}{b}\right )}^{1/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {1}{4};\ \frac {5}{4};\ \frac {a\,x^4}{b}\right )}{{\left (a\,x^4-b\right )}^{1/4}}-\frac {2\,{\left (a\,x^4-b\right )}^{3/4}}{3\,x^3} \]