Integrand size = 24, antiderivative size = 77 \[ \int \frac {x^2 \left (-b+a x^4\right )}{\left (b+a x^4\right )^{3/4}} \, dx=\frac {1}{4} x^3 \sqrt [4]{b+a x^4}+\frac {7 b \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{3/4}}-\frac {7 b \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{3/4}} \]
1/4*x^3*(a*x^4+b)^(1/4)+7/8*b*arctan(a^(1/4)*x/(a*x^4+b)^(1/4))/a^(3/4)-7/ 8*b*arctanh(a^(1/4)*x/(a*x^4+b)^(1/4))/a^(3/4)
Time = 0.37 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00 \[ \int \frac {x^2 \left (-b+a x^4\right )}{\left (b+a x^4\right )^{3/4}} \, dx=\frac {1}{4} x^3 \sqrt [4]{b+a x^4}+\frac {7 b \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{3/4}}-\frac {7 b \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{3/4}} \]
(x^3*(b + a*x^4)^(1/4))/4 + (7*b*ArcTan[(a^(1/4)*x)/(b + a*x^4)^(1/4)])/(8 *a^(3/4)) - (7*b*ArcTanh[(a^(1/4)*x)/(b + a*x^4)^(1/4)])/(8*a^(3/4))
Time = 0.22 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {959, 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.
\(\displaystyle \int \frac {x^2 \left (a x^4-b\right )}{\left (a x^4+b\right )^{3/4}} \, dx\) |
\(\Big \downarrow \) 959 |
\(\displaystyle \frac {1}{4} x^3 \sqrt [4]{a x^4+b}-\frac {7}{4} b \int \frac {x^2}{\left (a x^4+b\right )^{3/4}}dx\) |
\(\Big \downarrow \) 854 |
\(\displaystyle \frac {1}{4} x^3 \sqrt [4]{a x^4+b}-\frac {7}{4} b \int \frac {x^2}{\sqrt {a x^4+b} \left (1-\frac {a x^4}{a x^4+b}\right )}d\frac {x}{\sqrt [4]{a x^4+b}}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {1}{4} x^3 \sqrt [4]{a x^4+b}-\frac {7}{4} b \left (\frac {\int \frac {1}{1-\frac {\sqrt {a} x^2}{\sqrt {a x^4+b}}}d\frac {x}{\sqrt [4]{a x^4+b}}}{2 \sqrt {a}}-\frac {\int \frac {1}{\frac {\sqrt {a} x^2}{\sqrt {a x^4+b}}+1}d\frac {x}{\sqrt [4]{a x^4+b}}}{2 \sqrt {a}}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{4} x^3 \sqrt [4]{a x^4+b}-\frac {7}{4} b \left (\frac {\int \frac {1}{1-\frac {\sqrt {a} x^2}{\sqrt {a x^4+b}}}d\frac {x}{\sqrt [4]{a x^4+b}}}{2 \sqrt {a}}-\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{2 a^{3/4}}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} x^3 \sqrt [4]{a x^4+b}-\frac {7}{4} b \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{2 a^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{2 a^{3/4}}\right )\) |
(x^3*(b + a*x^4)^(1/4))/4 - (7*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
3.11.16.3.1 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[d*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m + n*(p + 1) + 1)) Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[m + n*(p + 1) + 1, 0]
Time = 1.32 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.10
method | result | size |
pseudoelliptic | \(-\frac {7 \left (-\frac {4 \left (a \,x^{4}+b \right )^{\frac {1}{4}} x^{3} a^{\frac {3}{4}}}{7}+\ln \left (\frac {-a^{\frac {1}{4}} x -\left (a \,x^{4}+b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x -\left (a \,x^{4}+b \right )^{\frac {1}{4}}}\right ) b +2 \arctan \left (\frac {\left (a \,x^{4}+b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right ) b \right )}{16 a^{\frac {3}{4}}}\) | \(85\) |
-7/16/a^(3/4)*(-4/7*(a*x^4+b)^(1/4)*x^3*a^(3/4)+ln((-a^(1/4)*x-(a*x^4+b)^( 1/4))/(a^(1/4)*x-(a*x^4+b)^(1/4)))*b+2*arctan(1/a^(1/4)/x*(a*x^4+b)^(1/4)) *b)
Result contains complex when optimal does not.
Time = 0.27 (sec) , antiderivative size = 184, normalized size of antiderivative = 2.39 \[ \int \frac {x^2 \left (-b+a x^4\right )}{\left (b+a x^4\right )^{3/4}} \, dx=\frac {1}{4} \, {\left (a x^{4} + b\right )}^{\frac {1}{4}} x^{3} - \frac {7}{16} \, \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} \log \left (\frac {7 \, {\left (a \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} x + {\left (a x^{4} + b\right )}^{\frac {1}{4}} b\right )}}{x}\right ) + \frac {7}{16} \, \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} \log \left (-\frac {7 \, {\left (a \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} x - {\left (a x^{4} + b\right )}^{\frac {1}{4}} b\right )}}{x}\right ) + \frac {7}{16} i \, \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} \log \left (-\frac {7 \, {\left (i \, a \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} x - {\left (a x^{4} + b\right )}^{\frac {1}{4}} b\right )}}{x}\right ) - \frac {7}{16} i \, \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} \log \left (-\frac {7 \, {\left (-i \, a \left (\frac {b^{4}}{a^{3}}\right )^{\frac {1}{4}} x - {\left (a x^{4} + b\right )}^{\frac {1}{4}} b\right )}}{x}\right ) \]
1/4*(a*x^4 + b)^(1/4)*x^3 - 7/16*(b^4/a^3)^(1/4)*log(7*(a*(b^4/a^3)^(1/4)* x + (a*x^4 + b)^(1/4)*b)/x) + 7/16*(b^4/a^3)^(1/4)*log(-7*(a*(b^4/a^3)^(1/ 4)*x - (a*x^4 + b)^(1/4)*b)/x) + 7/16*I*(b^4/a^3)^(1/4)*log(-7*(I*a*(b^4/a ^3)^(1/4)*x - (a*x^4 + b)^(1/4)*b)/x) - 7/16*I*(b^4/a^3)^(1/4)*log(-7*(-I* a*(b^4/a^3)^(1/4)*x - (a*x^4 + b)^(1/4)*b)/x)
Result contains complex when optimal does not.
Time = 1.66 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.01 \[ \int \frac {x^2 \left (-b+a x^4\right )}{\left (b+a x^4\right )^{3/4}} \, dx=\frac {a x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {a x^{4} e^{i \pi }}{b}} \right )}}{4 b^{\frac {3}{4}} \Gamma \left (\frac {11}{4}\right )} - \frac {\sqrt [4]{b} x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {a x^{4} e^{i \pi }}{b}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} \]
a*x**7*gamma(7/4)*hyper((3/4, 7/4), (11/4,), a*x**4*exp_polar(I*pi)/b)/(4* b**(3/4)*gamma(11/4)) - b**(1/4)*x**3*gamma(3/4)*hyper((3/4, 3/4), (7/4,), a*x**4*exp_polar(I*pi)/b)/(4*gamma(7/4))
Leaf count of result is larger than twice the leaf count of optimal. 185 vs. \(2 (57) = 114\).
Time = 0.27 (sec) , antiderivative size = 185, normalized size of antiderivative = 2.40 \[ \int \frac {x^2 \left (-b+a x^4\right )}{\left (b+a x^4\right )^{3/4}} \, dx=-\frac {1}{4} \, b {\left (\frac {2 \, \arctan \left (\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )}{a^{\frac {3}{4}}} - \frac {\log \left (-\frac {a^{\frac {1}{4}} - \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}{a^{\frac {1}{4}} + \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}\right )}{a^{\frac {3}{4}}}\right )} - \frac {1}{16} \, a {\left (\frac {3 \, {\left (\frac {2 \, b \arctan \left (\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )}{a^{\frac {3}{4}}} - \frac {b \log \left (-\frac {a^{\frac {1}{4}} - \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}{a^{\frac {1}{4}} + \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}\right )}{a^{\frac {3}{4}}}\right )}}{a} + \frac {4 \, {\left (a x^{4} + b\right )}^{\frac {1}{4}} b}{{\left (a^{2} - \frac {{\left (a x^{4} + b\right )} a}{x^{4}}\right )} x}\right )} \]
-1/4*b*(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)) - 1/16*a*(3 *(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))/a + 4*(a*x^4 + b)^(1/4)*b/((a^2 - (a*x^4 + b)*a/x^4)*x))
\[ \int \frac {x^2 \left (-b+a x^4\right )}{\left (b+a x^4\right )^{3/4}} \, dx=\int { \frac {{\left (a x^{4} - b\right )} x^{2}}{{\left (a x^{4} + b\right )}^{\frac {3}{4}}} \,d x } \]
Timed out. \[ \int \frac {x^2 \left (-b+a x^4\right )}{\left (b+a x^4\right )^{3/4}} \, dx=-\int \frac {x^2\,\left (b-a\,x^4\right )}{{\left (a\,x^4+b\right )}^{3/4}} \,d x \]