Integrand size = 24, antiderivative size = 86 \[ \int \frac {\left (-b+a x^4\right ) \sqrt [4]{b+a x^4}}{x^2} \, dx=\frac {\sqrt [4]{b+a x^4} \left (4 b+a x^4\right )}{4 x}+\frac {3}{8} \sqrt [4]{a} b \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )-\frac {3}{8} \sqrt [4]{a} b \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right ) \]
1/4*(a*x^4+b)^(1/4)*(a*x^4+4*b)/x+3/8*a^(1/4)*b*arctan(a^(1/4)*x/(a*x^4+b) ^(1/4))-3/8*a^(1/4)*b*arctanh(a^(1/4)*x/(a*x^4+b)^(1/4))
Time = 0.30 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-b+a x^4\right ) \sqrt [4]{b+a x^4}}{x^2} \, dx=\frac {\sqrt [4]{b+a x^4} \left (4 b+a x^4\right )}{4 x}+\frac {3}{8} \sqrt [4]{a} b \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )-\frac {3}{8} \sqrt [4]{a} b \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right ) \]
((b + a*x^4)^(1/4)*(4*b + a*x^4))/(4*x) + (3*a^(1/4)*b*ArcTan[(a^(1/4)*x)/ (b + a*x^4)^(1/4)])/8 - (3*a^(1/4)*b*ArcTanh[(a^(1/4)*x)/(b + a*x^4)^(1/4) ])/8
Time = 0.23 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.16, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {955, 811, 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.
\(\displaystyle \int \frac {\left (a x^4-b\right ) \sqrt [4]{a x^4+b}}{x^2} \, dx\) |
\(\Big \downarrow \) 955 |
\(\displaystyle \frac {\left (a x^4+b\right )^{5/4}}{x}-3 a \int x^2 \sqrt [4]{a x^4+b}dx\) |
\(\Big \downarrow \) 811 |
\(\displaystyle \frac {\left (a x^4+b\right )^{5/4}}{x}-3 a \left (\frac {1}{4} b \int \frac {x^2}{\left (a x^4+b\right )^{3/4}}dx+\frac {1}{4} x^3 \sqrt [4]{a x^4+b}\right )\) |
\(\Big \downarrow \) 854 |
\(\displaystyle \frac {\left (a x^4+b\right )^{5/4}}{x}-3 a \left (\frac {1}{4} b \int \frac {x^2}{\sqrt {a x^4+b} \left (1-\frac {a x^4}{a x^4+b}\right )}d\frac {x}{\sqrt [4]{a x^4+b}}+\frac {1}{4} x^3 \sqrt [4]{a x^4+b}\right )\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {\left (a x^4+b\right )^{5/4}}{x}-3 a \left (\frac {1}{4} b \left (\frac {\int \frac {1}{1-\frac {\sqrt {a} x^2}{\sqrt {a x^4+b}}}d\frac {x}{\sqrt [4]{a x^4+b}}}{2 \sqrt {a}}-\frac {\int \frac {1}{\frac {\sqrt {a} x^2}{\sqrt {a x^4+b}}+1}d\frac {x}{\sqrt [4]{a x^4+b}}}{2 \sqrt {a}}\right )+\frac {1}{4} x^3 \sqrt [4]{a x^4+b}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\left (a x^4+b\right )^{5/4}}{x}-3 a \left (\frac {1}{4} b \left (\frac {\int \frac {1}{1-\frac {\sqrt {a} x^2}{\sqrt {a x^4+b}}}d\frac {x}{\sqrt [4]{a x^4+b}}}{2 \sqrt {a}}-\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{2 a^{3/4}}\right )+\frac {1}{4} x^3 \sqrt [4]{a x^4+b}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\left (a x^4+b\right )^{5/4}}{x}-3 a \left (\frac {1}{4} b \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{2 a^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{2 a^{3/4}}\right )+\frac {1}{4} x^3 \sqrt [4]{a x^4+b}\right )\) |
(b + a*x^4)^(5/4)/x - 3*a*((x^3*(b + a*x^4)^(1/4))/4 + (b*(-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))))/4)
3.12.67.3.1 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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* x)^(m + 1)*((a + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a*n*(p/(m + n*p + 1 )) Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && I GtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m , p, x]
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[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)) Int[(e *x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b* c - a*d, 0] && (IntegerQ[n] || GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) && !ILtQ[p, -1]
Time = 1.22 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.08
method | result | size |
pseudoelliptic | \(\frac {-3 b x \left (2 \arctan \left (\frac {\left (a \,x^{4}+b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )+\ln \left (\frac {-a^{\frac {1}{4}} x -\left (a \,x^{4}+b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x -\left (a \,x^{4}+b \right )^{\frac {1}{4}}}\right )\right ) a^{\frac {1}{4}}+4 \left (a \,x^{4}+b \right )^{\frac {1}{4}} \left (a \,x^{4}+4 b \right )}{16 x}\) | \(93\) |
1/16*(-3*b*x*(2*arctan(1/a^(1/4)/x*(a*x^4+b)^(1/4))+ln((-a^(1/4)*x-(a*x^4+ b)^(1/4))/(a^(1/4)*x-(a*x^4+b)^(1/4))))*a^(1/4)+4*(a*x^4+b)^(1/4)*(a*x^4+4 *b))/x
Timed out. \[ \int \frac {\left (-b+a x^4\right ) \sqrt [4]{b+a x^4}}{x^2} \, dx=\text {Timed out} \]
Result contains complex when optimal does not.
Time = 1.54 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.97 \[ \int \frac {\left (-b+a x^4\right ) \sqrt [4]{b+a x^4}}{x^2} \, dx=\frac {a \sqrt [4]{b} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {a x^{4} e^{i \pi }}{b}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} - \frac {b^{\frac {5}{4}} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {a x^{4} e^{i \pi }}{b}} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} \]
a*b**(1/4)*x**3*gamma(3/4)*hyper((-1/4, 3/4), (7/4,), a*x**4*exp_polar(I*p i)/b)/(4*gamma(7/4)) - b**(5/4)*gamma(-1/4)*hyper((-1/4, -1/4), (3/4,), a* x**4*exp_polar(I*pi)/b)/(4*x*gamma(3/4))
Leaf count of result is larger than twice the leaf count of optimal. 190 vs. \(2 (66) = 132\).
Time = 0.29 (sec) , antiderivative size = 190, normalized size of antiderivative = 2.21 \[ \int \frac {\left (-b+a x^4\right ) \sqrt [4]{b+a x^4}}{x^2} \, dx=\frac {1}{16} \, {\left (\frac {2 \, b \arctan \left (\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )}{a^{\frac {3}{4}}} - \frac {b \log \left (-\frac {a^{\frac {1}{4}} - \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}{a^{\frac {1}{4}} + \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}\right )}{a^{\frac {3}{4}}} - \frac {4 \, {\left (a x^{4} + b\right )}^{\frac {1}{4}} b}{{\left (a - \frac {a x^{4} + b}{x^{4}}\right )} x}\right )} a - \frac {1}{4} \, {\left (2 \, a^{\frac {1}{4}} \arctan \left (\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right ) - a^{\frac {1}{4}} \log \left (-\frac {a^{\frac {1}{4}} - \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}{a^{\frac {1}{4}} + \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}\right ) - \frac {4 \, {\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}\right )} b \]
1/16*(2*b*arctan((a*x^4 + b)^(1/4)/(a^(1/4)*x))/a^(3/4) - b*log(-(a^(1/4) - (a*x^4 + b)^(1/4)/x)/(a^(1/4) + (a*x^4 + b)^(1/4)/x))/a^(3/4) - 4*(a*x^4 + b)^(1/4)*b/((a - (a*x^4 + b)/x^4)*x))*a - 1/4*(2*a^(1/4)*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)) - 4*(a*x^4 + b)^(1/4)/x)*b
\[ \int \frac {\left (-b+a x^4\right ) \sqrt [4]{b+a x^4}}{x^2} \, dx=\int { \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}} {\left (a x^{4} - b\right )}}{x^{2}} \,d x } \]
Timed out. \[ \int \frac {\left (-b+a x^4\right ) \sqrt [4]{b+a x^4}}{x^2} \, dx=-\int \frac {{\left (a\,x^4+b\right )}^{1/4}\,\left (b-a\,x^4\right )}{x^2} \,d x \]