Integrand size = 24, antiderivative size = 82 \[ \int \frac {-b+a x^8}{x^2 \left (b+a x^4\right )^{3/4}} \, dx=\frac {\left (4+x^4\right ) \sqrt [4]{b+a x^4}}{4 x}+\frac {3 b \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{3/4}}-\frac {3 b \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{3/4}} \]
1/4*(x^4+4)*(a*x^4+b)^(1/4)/x+3/8*b*arctan(a^(1/4)*x/(a*x^4+b)^(1/4))/a^(3 /4)-3/8*b*arctanh(a^(1/4)*x/(a*x^4+b)^(1/4))/a^(3/4)
Time = 0.47 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00 \[ \int \frac {-b+a x^8}{x^2 \left (b+a x^4\right )^{3/4}} \, dx=\frac {\left (4+x^4\right ) \sqrt [4]{b+a x^4}}{4 x}+\frac {3 b \arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{3/4}}-\frac {3 b \text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{8 a^{3/4}} \]
((4 + x^4)*(b + a*x^4)^(1/4))/(4*x) + (3*b*ArcTan[(a^(1/4)*x)/(b + a*x^4)^ (1/4)])/(8*a^(3/4)) - (3*b*ArcTanh[(a^(1/4)*x)/(b + a*x^4)^(1/4)])/(8*a^(3 /4))
Time = 0.25 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.26, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {1811, 25, 27, 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.
\(\displaystyle \int \frac {a x^8-b}{x^2 \left (a x^4+b\right )^{3/4}} \, dx\) |
\(\Big \downarrow \) 1811 |
\(\displaystyle \frac {\int -\frac {a b \left (3 x^4+4\right )}{x^2 \left (a x^4+b\right )^{3/4}}dx}{4 a}+\frac {1}{4} x^3 \sqrt [4]{a x^4+b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{4} x^3 \sqrt [4]{a x^4+b}-\frac {\int \frac {a b \left (3 x^4+4\right )}{x^2 \left (a x^4+b\right )^{3/4}}dx}{4 a}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} x^3 \sqrt [4]{a x^4+b}-\frac {1}{4} b \int \frac {3 x^4+4}{x^2 \left (a x^4+b\right )^{3/4}}dx\) |
\(\Big \downarrow \) 953 |
\(\displaystyle \frac {1}{4} x^3 \sqrt [4]{a x^4+b}-\frac {1}{4} b \left (3 \int \frac {x^2}{\left (a x^4+b\right )^{3/4}}dx-\frac {4 \sqrt [4]{a x^4+b}}{b x}\right )\) |
\(\Big \downarrow \) 854 |
\(\displaystyle \frac {1}{4} x^3 \sqrt [4]{a x^4+b}-\frac {1}{4} b \left (3 \int \frac {x^2}{\sqrt {a x^4+b} \left (1-\frac {a x^4}{a x^4+b}\right )}d\frac {x}{\sqrt [4]{a x^4+b}}-\frac {4 \sqrt [4]{a x^4+b}}{b x}\right )\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {1}{4} x^3 \sqrt [4]{a x^4+b}-\frac {1}{4} b \left (3 \left (\frac {\int \frac {1}{1-\frac {\sqrt {a} x^2}{\sqrt {a x^4+b}}}d\frac {x}{\sqrt [4]{a x^4+b}}}{2 \sqrt {a}}-\frac {\int \frac {1}{\frac {\sqrt {a} x^2}{\sqrt {a x^4+b}}+1}d\frac {x}{\sqrt [4]{a x^4+b}}}{2 \sqrt {a}}\right )-\frac {4 \sqrt [4]{a x^4+b}}{b x}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{4} x^3 \sqrt [4]{a x^4+b}-\frac {1}{4} b \left (3 \left (\frac {\int \frac {1}{1-\frac {\sqrt {a} x^2}{\sqrt {a x^4+b}}}d\frac {x}{\sqrt [4]{a x^4+b}}}{2 \sqrt {a}}-\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{2 a^{3/4}}\right )-\frac {4 \sqrt [4]{a x^4+b}}{b x}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} x^3 \sqrt [4]{a x^4+b}-\frac {1}{4} b \left (3 \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{2 a^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{2 a^{3/4}}\right )-\frac {4 \sqrt [4]{a x^4+b}}{b x}\right )\) |
(x^3*(b + a*x^4)^(1/4))/4 - (b*((-4*(b + a*x^4)^(1/4))/(b*x) + 3*(-1/2*Arc Tan[(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
3.12.7.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]))
Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^ (n_))^(q_.), x_Symbol] :> Simp[c^p*(f*x)^(m + 2*n*p - n + 1)*((d + e*x^n)^( q + 1)/(e*f^(2*n*p - n + 1)*(m + 2*n*p + n*q + 1))), x] + Simp[1/(e*(m + 2* n*p + n*q + 1)) Int[(f*x)^m*(d + e*x^n)^q*ExpandToSum[e*(m + 2*n*p + n*q + 1)*((a + c*x^(2*n))^p - c^p*x^(2*n*p)) - d*c^p*(m + 2*n*p - n + 1)*x^(2*n *p - n), x], x], x] /; FreeQ[{a, c, d, e, f, m, q}, x] && EqQ[n2, 2*n] && I GtQ[n, 0] && IGtQ[p, 0] && GtQ[2*n*p, n - 1] && !IntegerQ[q] && NeQ[m + 2* n*p + n*q + 1, 0]
Time = 1.21 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.11
method | result | size |
pseudoelliptic | \(\frac {\left (a \,x^{4}+b \right )^{\frac {1}{4}} \left (x^{4}+4\right ) a^{\frac {3}{4}}-\frac {3 b x \left (2 \arctan \left (\frac {\left (a \,x^{4}+b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )+\ln \left (\frac {-a^{\frac {1}{4}} x -\left (a \,x^{4}+b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x -\left (a \,x^{4}+b \right )^{\frac {1}{4}}}\right )\right )}{4}}{4 a^{\frac {3}{4}} x}\) | \(91\) |
1/4/a^(3/4)*((a*x^4+b)^(1/4)*(x^4+4)*a^(3/4)-3/4*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)))))/x
Timed out. \[ \int \frac {-b+a x^8}{x^2 \left (b+a x^4\right )^{3/4}} \, dx=\text {Timed out} \]
Result contains complex when optimal does not.
Time = 1.38 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.85 \[ \int \frac {-b+a x^8}{x^2 \left (b+a x^4\right )^{3/4}} \, dx=- \frac {\sqrt [4]{a} \sqrt [4]{1 + \frac {b}{a x^{4}}} \Gamma \left (- \frac {1}{4}\right )}{4 \Gamma \left (\frac {3}{4}\right )} + \frac {a x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {a x^{4} e^{i \pi }}{b}} \right )}}{4 b^{\frac {3}{4}} \Gamma \left (\frac {11}{4}\right )} \]
-a**(1/4)*(1 + b/(a*x**4))**(1/4)*gamma(-1/4)/(4*gamma(3/4)) + a*x**7*gamm a(7/4)*hyper((3/4, 7/4), (11/4,), a*x**4*exp_polar(I*pi)/b)/(4*b**(3/4)*ga mma(11/4))
Leaf count of result is larger than twice the leaf count of optimal. 127 vs. \(2 (62) = 124\).
Time = 0.27 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.55 \[ \int \frac {-b+a x^8}{x^2 \left (b+a x^4\right )^{3/4}} \, dx=-\frac {1}{16} \, a {\left (\frac {3 \, {\left (\frac {2 \, b \arctan \left (\frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )}{a^{\frac {3}{4}}} - \frac {b \log \left (-\frac {a^{\frac {1}{4}} - \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}{a^{\frac {1}{4}} + \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x}}\right )}{a^{\frac {3}{4}}}\right )}}{a} + \frac {4 \, {\left (a x^{4} + b\right )}^{\frac {1}{4}} b}{{\left (a^{2} - \frac {{\left (a x^{4} + b\right )} a}{x^{4}}\right )} x}\right )} + \frac {{\left (a x^{4} + b\right )}^{\frac {1}{4}}}{x} \]
-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)) + (a*x^4 + b)^(1/4)/ x
\[ \int \frac {-b+a x^8}{x^2 \left (b+a x^4\right )^{3/4}} \, dx=\int { \frac {a x^{8} - b}{{\left (a x^{4} + b\right )}^{\frac {3}{4}} x^{2}} \,d x } \]
Timed out. \[ \int \frac {-b+a x^8}{x^2 \left (b+a x^4\right )^{3/4}} \, dx=-\int \frac {b-a\,x^8}{x^2\,{\left (a\,x^4+b\right )}^{3/4}} \,d x \]