Integrand size = 15, antiderivative size = 94 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^2} \, dx=\frac {5}{4} b x^3 \sqrt [4]{a+b x^4}-\frac {\left (a+b x^4\right )^{5/4}}{x}-\frac {5}{8} a \sqrt [4]{b} \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+\frac {5}{8} a \sqrt [4]{b} \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right ) \]
5/4*b*x^3*(b*x^4+a)^(1/4)-(b*x^4+a)^(5/4)/x-5/8*a*b^(1/4)*arctan(b^(1/4)*x /(b*x^4+a)^(1/4))+5/8*a*b^(1/4)*arctanh(b^(1/4)*x/(b*x^4+a)^(1/4))
Time = 0.45 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.91 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^2} \, dx=\frac {\left (-4 a+b x^4\right ) \sqrt [4]{a+b x^4}}{4 x}-\frac {5}{8} a \sqrt [4]{b} \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+\frac {5}{8} a \sqrt [4]{b} \text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right ) \]
((-4*a + b*x^4)*(a + b*x^4)^(1/4))/(4*x) - (5*a*b^(1/4)*ArcTan[(b^(1/4)*x) /(a + b*x^4)^(1/4)])/8 + (5*a*b^(1/4)*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4 )])/8
Time = 0.23 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {809, 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.
\(\displaystyle \int \frac {\left (a+b x^4\right )^{5/4}}{x^2} \, dx\) |
\(\Big \downarrow \) 809 |
\(\displaystyle 5 b \int x^2 \sqrt [4]{b x^4+a}dx-\frac {\left (a+b x^4\right )^{5/4}}{x}\) |
\(\Big \downarrow \) 811 |
\(\displaystyle 5 b \left (\frac {1}{4} a \int \frac {x^2}{\left (b x^4+a\right )^{3/4}}dx+\frac {1}{4} x^3 \sqrt [4]{a+b x^4}\right )-\frac {\left (a+b x^4\right )^{5/4}}{x}\) |
\(\Big \downarrow \) 854 |
\(\displaystyle 5 b \left (\frac {1}{4} a \int \frac {x^2}{\sqrt {b x^4+a} \left (1-\frac {b x^4}{b x^4+a}\right )}d\frac {x}{\sqrt [4]{b x^4+a}}+\frac {1}{4} x^3 \sqrt [4]{a+b x^4}\right )-\frac {\left (a+b x^4\right )^{5/4}}{x}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle 5 b \left (\frac {1}{4} a \left (\frac {\int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {b}}-\frac {\int \frac {1}{\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}+1}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {b}}\right )+\frac {1}{4} x^3 \sqrt [4]{a+b x^4}\right )-\frac {\left (a+b x^4\right )^{5/4}}{x}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle 5 b \left (\frac {1}{4} a \left (\frac {\int \frac {1}{1-\frac {\sqrt {b} x^2}{\sqrt {b x^4+a}}}d\frac {x}{\sqrt [4]{b x^4+a}}}{2 \sqrt {b}}-\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}\right )+\frac {1}{4} x^3 \sqrt [4]{a+b x^4}\right )-\frac {\left (a+b x^4\right )^{5/4}}{x}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle 5 b \left (\frac {1}{4} a \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{2 b^{3/4}}\right )+\frac {1}{4} x^3 \sqrt [4]{a+b x^4}\right )-\frac {\left (a+b x^4\right )^{5/4}}{x}\) |
-((a + b*x^4)^(5/4)/x) + 5*b*((x^3*(a + b*x^4)^(1/4))/4 + (a*(-1/2*ArcTan[ (b^(1/4)*x)/(a + b*x^4)^(1/4)]/b^(3/4) + ArcTanh[(b^(1/4)*x)/(a + b*x^4)^( 1/4)]/(2*b^(3/4))))/4)
3.11.65.3.1 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 + 1))), x] - Simp[b*n*(p/(c^n*(m + 1))) I nt[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && IGtQ [n, 0] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + n*p + n + 1)/n, 0] && IntB inomialQ[a, b, c, n, m, p, x]
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]
Time = 4.42 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00
method | result | size |
pseudoelliptic | \(\frac {5 a x \left (\ln \left (\frac {-b^{\frac {1}{4}} x -\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x -\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )\right ) b^{\frac {1}{4}}-4 \left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (-b \,x^{4}+4 a \right )}{16 x}\) | \(94\) |
1/16*(5*a*x*(ln((-b^(1/4)*x-(b*x^4+a)^(1/4))/(b^(1/4)*x-(b*x^4+a)^(1/4)))+ 2*arctan(1/b^(1/4)/x*(b*x^4+a)^(1/4)))*b^(1/4)-4*(b*x^4+a)^(1/4)*(-b*x^4+4 *a))/x
Timed out. \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^2} \, dx=\text {Timed out} \]
Result contains complex when optimal does not.
Time = 1.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.44 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^2} \, dx=\frac {a^{\frac {5}{4}} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} \]
Time = 0.28 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.27 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^2} \, dx=\frac {5}{16} \, {\left (2 \, b^{\frac {1}{4}} \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right ) - b^{\frac {1}{4}} \log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )\right )} a - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} a}{x} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} a b}{4 \, {\left (b - \frac {b x^{4} + a}{x^{4}}\right )} x} \]
5/16*(2*b^(1/4)*arctan((b*x^4 + a)^(1/4)/(b^(1/4)*x)) - b^(1/4)*log(-(b^(1 /4) - (b*x^4 + a)^(1/4)/x)/(b^(1/4) + (b*x^4 + a)^(1/4)/x)))*a - (b*x^4 + a)^(1/4)*a/x - 1/4*(b*x^4 + a)^(1/4)*a*b/((b - (b*x^4 + a)/x^4)*x)
\[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^2} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {5}{4}}}{x^{2}} \,d x } \]
Time = 6.46 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.43 \[ \int \frac {\left (a+b x^4\right )^{5/4}}{x^2} \, dx=-\frac {{\left (b\,x^4+a\right )}^{5/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},-\frac {1}{4};\ \frac {3}{4};\ -\frac {b\,x^4}{a}\right )}{x\,{\left (\frac {b\,x^4}{a}+1\right )}^{5/4}} \]