Integrand size = 17, antiderivative size = 153 \[ \int \frac {1}{x^5 \left (-b+a x^4\right )^{3/4}} \, dx=\frac {\sqrt [4]{-b+a x^4}}{4 b x^4}-\frac {3 a \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^4}}{-\sqrt {b}+\sqrt {-b+a x^4}}\right )}{8 \sqrt {2} b^{7/4}}+\frac {3 a \text {arctanh}\left (\frac {\frac {\sqrt [4]{b}}{\sqrt {2}}+\frac {\sqrt {-b+a x^4}}{\sqrt {2} \sqrt [4]{b}}}{\sqrt [4]{-b+a x^4}}\right )}{8 \sqrt {2} b^{7/4}} \]
1/4*(a*x^4-b)^(1/4)/b/x^4-3/16*a*arctan(2^(1/2)*b^(1/4)*(a*x^4-b)^(1/4)/(- b^(1/2)+(a*x^4-b)^(1/2)))*2^(1/2)/b^(7/4)+3/16*a*arctanh((1/2*b^(1/4)*2^(1 /2)+1/2*(a*x^4-b)^(1/2)*2^(1/2)/b^(1/4))/(a*x^4-b)^(1/4))*2^(1/2)/b^(7/4)
Time = 0.22 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.96 \[ \int \frac {1}{x^5 \left (-b+a x^4\right )^{3/4}} \, dx=\frac {4 b^{3/4} \sqrt [4]{-b+a x^4}+3 \sqrt {2} a x^4 \arctan \left (\frac {-\sqrt {b}+\sqrt {-b+a x^4}}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^4}}\right )+3 \sqrt {2} a x^4 \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^4}}{\sqrt {b}+\sqrt {-b+a x^4}}\right )}{16 b^{7/4} x^4} \]
(4*b^(3/4)*(-b + a*x^4)^(1/4) + 3*Sqrt[2]*a*x^4*ArcTan[(-Sqrt[b] + Sqrt[-b + a*x^4])/(Sqrt[2]*b^(1/4)*(-b + a*x^4)^(1/4))] + 3*Sqrt[2]*a*x^4*ArcTanh [(Sqrt[2]*b^(1/4)*(-b + a*x^4)^(1/4))/(Sqrt[b] + Sqrt[-b + a*x^4])])/(16*b ^(7/4)*x^4)
Time = 0.38 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.49, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.706, Rules used = {798, 52, 73, 755, 27, 1476, 1082, 217, 1479, 25, 27, 1103}
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.
| \(\displaystyle \int \frac {1}{x^5 \left (a x^4-b\right )^{3/4}} \, dx\) |
| \(\Big \downarrow \) 798 |
| \(\displaystyle \frac {1}{4} \int \frac {1}{x^8 \left (a x^4-b\right )^{3/4}}dx^4\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {1}{4} \left (\frac {3 a \int \frac {1}{x^4 \left (a x^4-b\right )^{3/4}}dx^4}{4 b}+\frac {\sqrt [4]{a x^4-b}}{b x^4}\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{4} \left (\frac {3 \int \frac {1}{\frac {x^{16}}{a}+\frac {b}{a}}d\sqrt [4]{a x^4-b}}{b}+\frac {\sqrt [4]{a x^4-b}}{b x^4}\right )\) |
| \(\Big \downarrow \) 755 |
| \(\displaystyle \frac {1}{4} \left (\frac {3 \left (\frac {\int \frac {a \left (\sqrt {b}-x^8\right )}{x^{16}+b}d\sqrt [4]{a x^4-b}}{2 \sqrt {b}}+\frac {\int \frac {a \left (x^8+\sqrt {b}\right )}{x^{16}+b}d\sqrt [4]{a x^4-b}}{2 \sqrt {b}}\right )}{b}+\frac {\sqrt [4]{a x^4-b}}{b x^4}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \left (\frac {3 \left (\frac {a \int \frac {\sqrt {b}-x^8}{x^{16}+b}d\sqrt [4]{a x^4-b}}{2 \sqrt {b}}+\frac {a \int \frac {x^8+\sqrt {b}}{x^{16}+b}d\sqrt [4]{a x^4-b}}{2 \sqrt {b}}\right )}{b}+\frac {\sqrt [4]{a x^4-b}}{b x^4}\right )\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {1}{4} \left (\frac {3 \left (\frac {a \left (\frac {1}{2} \int \frac {1}{x^8+\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^4-b}}d\sqrt [4]{a x^4-b}+\frac {1}{2} \int \frac {1}{x^8+\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^4-b}}d\sqrt [4]{a x^4-b}\right )}{2 \sqrt {b}}+\frac {a \int \frac {\sqrt {b}-x^8}{x^{16}+b}d\sqrt [4]{a x^4-b}}{2 \sqrt {b}}\right )}{b}+\frac {\sqrt [4]{a x^4-b}}{b x^4}\right )\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {1}{4} \left (\frac {3 \left (\frac {a \left (\frac {\int \frac {1}{-x^8-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{a x^4-b}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b}}-\frac {\int \frac {1}{-x^8-1}d\left (\frac {\sqrt {2} \sqrt [4]{a x^4-b}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b}}\right )}{2 \sqrt {b}}+\frac {a \int \frac {\sqrt {b}-x^8}{x^{16}+b}d\sqrt [4]{a x^4-b}}{2 \sqrt {b}}\right )}{b}+\frac {\sqrt [4]{a x^4-b}}{b x^4}\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{4} \left (\frac {3 \left (\frac {a \int \frac {\sqrt {b}-x^8}{x^{16}+b}d\sqrt [4]{a x^4-b}}{2 \sqrt {b}}+\frac {a \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a x^4-b}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x^4-b}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b}}\right )}{2 \sqrt {b}}\right )}{b}+\frac {\sqrt [4]{a x^4-b}}{b x^4}\right )\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {1}{4} \left (\frac {3 \left (\frac {a \left (-\frac {\int -\frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{a x^4-b}}{x^8+\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^4-b}}d\sqrt [4]{a x^4-b}}{2 \sqrt {2} \sqrt [4]{b}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt [4]{b}+\sqrt {2} \sqrt [4]{a x^4-b}\right )}{x^8+\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^4-b}}d\sqrt [4]{a x^4-b}}{2 \sqrt {2} \sqrt [4]{b}}\right )}{2 \sqrt {b}}+\frac {a \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a x^4-b}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x^4-b}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b}}\right )}{2 \sqrt {b}}\right )}{b}+\frac {\sqrt [4]{a x^4-b}}{b x^4}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{4} \left (\frac {3 \left (\frac {a \left (\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{a x^4-b}}{x^8+\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^4-b}}d\sqrt [4]{a x^4-b}}{2 \sqrt {2} \sqrt [4]{b}}+\frac {\int \frac {\sqrt {2} \left (\sqrt [4]{b}+\sqrt {2} \sqrt [4]{a x^4-b}\right )}{x^8+\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^4-b}}d\sqrt [4]{a x^4-b}}{2 \sqrt {2} \sqrt [4]{b}}\right )}{2 \sqrt {b}}+\frac {a \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a x^4-b}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x^4-b}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b}}\right )}{2 \sqrt {b}}\right )}{b}+\frac {\sqrt [4]{a x^4-b}}{b x^4}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} \left (\frac {3 \left (\frac {a \left (\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{a x^4-b}}{x^8+\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^4-b}}d\sqrt [4]{a x^4-b}}{2 \sqrt {2} \sqrt [4]{b}}+\frac {\int \frac {\sqrt [4]{b}+\sqrt {2} \sqrt [4]{a x^4-b}}{x^8+\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^4-b}}d\sqrt [4]{a x^4-b}}{2 \sqrt [4]{b}}\right )}{2 \sqrt {b}}+\frac {a \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a x^4-b}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x^4-b}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b}}\right )}{2 \sqrt {b}}\right )}{b}+\frac {\sqrt [4]{a x^4-b}}{b x^4}\right )\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {1}{4} \left (\frac {3 \left (\frac {a \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a x^4-b}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x^4-b}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b}}\right )}{2 \sqrt {b}}+\frac {a \left (\frac {\log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^4-b}+\sqrt {b}+x^8\right )}{2 \sqrt {2} \sqrt [4]{b}}-\frac {\log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^4-b}+\sqrt {b}+x^8\right )}{2 \sqrt {2} \sqrt [4]{b}}\right )}{2 \sqrt {b}}\right )}{b}+\frac {\sqrt [4]{a x^4-b}}{b x^4}\right )\) |
((-b + a*x^4)^(1/4)/(b*x^4) + (3*((a*(-(ArcTan[1 - (Sqrt[2]*(-b + a*x^4)^( 1/4))/b^(1/4)]/(Sqrt[2]*b^(1/4))) + ArcTan[1 + (Sqrt[2]*(-b + a*x^4)^(1/4) )/b^(1/4)]/(Sqrt[2]*b^(1/4))))/(2*Sqrt[b]) + (a*(-1/2*Log[Sqrt[b] + x^8 - Sqrt[2]*b^(1/4)*(-b + a*x^4)^(1/4)]/(Sqrt[2]*b^(1/4)) + Log[Sqrt[b] + x^8 + Sqrt[2]*b^(1/4)*(-b + a*x^4)^(1/4)]/(2*Sqrt[2]*b^(1/4))))/(2*Sqrt[b])))/ b)/4
3.22.9.3.1 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_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[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[1/(2*r) Int[(r - s*x^2)/(a + b*x^4) , x], x] + Simp[1/(2*r) Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] & & AtomQ[SplitProduct[SumBaseQ, b]]))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n Subst [Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Time = 0.72 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.18
| method | result | size |
| pseudoelliptic | \(-\frac {3 \left (\arctan \left (\frac {-\sqrt {2}\, \left (a \,x^{4}-b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right ) \sqrt {2}\, a \,x^{4}-\arctan \left (\frac {\sqrt {2}\, \left (a \,x^{4}-b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right ) \sqrt {2}\, a \,x^{4}-\frac {\ln \left (\frac {-b^{\frac {1}{4}} \left (a \,x^{4}-b \right )^{\frac {1}{4}} \sqrt {2}-\sqrt {a \,x^{4}-b}-\sqrt {b}}{b^{\frac {1}{4}} \left (a \,x^{4}-b \right )^{\frac {1}{4}} \sqrt {2}-\sqrt {a \,x^{4}-b}-\sqrt {b}}\right ) \sqrt {2}\, a \,x^{4}}{2}-\frac {4 \left (a \,x^{4}-b \right )^{\frac {1}{4}} b^{\frac {3}{4}}}{3}\right )}{16 b^{\frac {7}{4}} x^{4}}\) | \(180\) |
-3/16/b^(7/4)*(arctan((-2^(1/2)*(a*x^4-b)^(1/4)+b^(1/4))/b^(1/4))*2^(1/2)* a*x^4-arctan((2^(1/2)*(a*x^4-b)^(1/4)+b^(1/4))/b^(1/4))*2^(1/2)*a*x^4-1/2* ln((-b^(1/4)*(a*x^4-b)^(1/4)*2^(1/2)-(a*x^4-b)^(1/2)-b^(1/2))/(b^(1/4)*(a* x^4-b)^(1/4)*2^(1/2)-(a*x^4-b)^(1/2)-b^(1/2)))*2^(1/2)*a*x^4-4/3*(a*x^4-b) ^(1/4)*b^(3/4))/x^4
Result contains complex when optimal does not.
Time = 0.26 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.37 \[ \int \frac {1}{x^5 \left (-b+a x^4\right )^{3/4}} \, dx=\frac {3 \, b x^{4} \left (-\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} \log \left (3 \, b^{2} \left (-\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} + 3 \, {\left (a x^{4} - b\right )}^{\frac {1}{4}} a\right ) + 3 i \, b x^{4} \left (-\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} \log \left (3 i \, b^{2} \left (-\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} + 3 \, {\left (a x^{4} - b\right )}^{\frac {1}{4}} a\right ) - 3 i \, b x^{4} \left (-\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} \log \left (-3 i \, b^{2} \left (-\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} + 3 \, {\left (a x^{4} - b\right )}^{\frac {1}{4}} a\right ) - 3 \, b x^{4} \left (-\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} \log \left (-3 \, b^{2} \left (-\frac {a^{4}}{b^{7}}\right )^{\frac {1}{4}} + 3 \, {\left (a x^{4} - b\right )}^{\frac {1}{4}} a\right ) + 4 \, {\left (a x^{4} - b\right )}^{\frac {1}{4}}}{16 \, b x^{4}} \]
1/16*(3*b*x^4*(-a^4/b^7)^(1/4)*log(3*b^2*(-a^4/b^7)^(1/4) + 3*(a*x^4 - b)^ (1/4)*a) + 3*I*b*x^4*(-a^4/b^7)^(1/4)*log(3*I*b^2*(-a^4/b^7)^(1/4) + 3*(a* x^4 - b)^(1/4)*a) - 3*I*b*x^4*(-a^4/b^7)^(1/4)*log(-3*I*b^2*(-a^4/b^7)^(1/ 4) + 3*(a*x^4 - b)^(1/4)*a) - 3*b*x^4*(-a^4/b^7)^(1/4)*log(-3*b^2*(-a^4/b^ 7)^(1/4) + 3*(a*x^4 - b)^(1/4)*a) + 4*(a*x^4 - b)^(1/4))/(b*x^4)
Result contains complex when optimal does not.
Time = 0.88 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.27 \[ \int \frac {1}{x^5 \left (-b+a x^4\right )^{3/4}} \, dx=- \frac {\Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b e^{2 i \pi }}{a x^{4}}} \right )}}{4 a^{\frac {3}{4}} x^{7} \Gamma \left (\frac {11}{4}\right )} \]
-gamma(7/4)*hyper((3/4, 7/4), (11/4,), b*exp_polar(2*I*pi)/(a*x**4))/(4*a* *(3/4)*x**7*gamma(11/4))
Time = 0.30 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.32 \[ \int \frac {1}{x^5 \left (-b+a x^4\right )^{3/4}} \, dx=\frac {{\left (a x^{4} - b\right )}^{\frac {1}{4}} a}{4 \, {\left ({\left (a x^{4} - b\right )} b + b^{2}\right )}} + \frac {3 \, {\left (\frac {2 \, \sqrt {2} a \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{4} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {3}{4}}} + \frac {2 \, \sqrt {2} a \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{4} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {3}{4}}} + \frac {\sqrt {2} a \log \left (\sqrt {2} {\left (a x^{4} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{4} - b} + \sqrt {b}\right )}{b^{\frac {3}{4}}} - \frac {\sqrt {2} a \log \left (-\sqrt {2} {\left (a x^{4} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{4} - b} + \sqrt {b}\right )}{b^{\frac {3}{4}}}\right )}}{32 \, b} \]
1/4*(a*x^4 - b)^(1/4)*a/((a*x^4 - b)*b + b^2) + 3/32*(2*sqrt(2)*a*arctan(1 /2*sqrt(2)*(sqrt(2)*b^(1/4) + 2*(a*x^4 - b)^(1/4))/b^(1/4))/b^(3/4) + 2*sq rt(2)*a*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4) - 2*(a*x^4 - b)^(1/4))/b^(1/4 ))/b^(3/4) + sqrt(2)*a*log(sqrt(2)*(a*x^4 - b)^(1/4)*b^(1/4) + sqrt(a*x^4 - b) + sqrt(b))/b^(3/4) - sqrt(2)*a*log(-sqrt(2)*(a*x^4 - b)^(1/4)*b^(1/4) + sqrt(a*x^4 - b) + sqrt(b))/b^(3/4))/b
Time = 0.28 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.30 \[ \int \frac {1}{x^5 \left (-b+a x^4\right )^{3/4}} \, dx=\frac {\frac {6 \, \sqrt {2} a^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{4} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {7}{4}}} + \frac {6 \, \sqrt {2} a^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{4} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {7}{4}}} + \frac {3 \, \sqrt {2} a^{2} \log \left (\sqrt {2} {\left (a x^{4} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{4} - b} + \sqrt {b}\right )}{b^{\frac {7}{4}}} - \frac {3 \, \sqrt {2} a^{2} \log \left (-\sqrt {2} {\left (a x^{4} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{4} - b} + \sqrt {b}\right )}{b^{\frac {7}{4}}} + \frac {8 \, {\left (a x^{4} - b\right )}^{\frac {1}{4}} a}{b x^{4}}}{32 \, a} \]
1/32*(6*sqrt(2)*a^2*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4) + 2*(a*x^4 - b)^(1 /4))/b^(1/4))/b^(7/4) + 6*sqrt(2)*a^2*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4) - 2*(a*x^4 - b)^(1/4))/b^(1/4))/b^(7/4) + 3*sqrt(2)*a^2*log(sqrt(2)*(a*x^ 4 - b)^(1/4)*b^(1/4) + sqrt(a*x^4 - b) + sqrt(b))/b^(7/4) - 3*sqrt(2)*a^2* log(-sqrt(2)*(a*x^4 - b)^(1/4)*b^(1/4) + sqrt(a*x^4 - b) + sqrt(b))/b^(7/4 ) + 8*(a*x^4 - b)^(1/4)*a/(b*x^4))/a
Time = 6.24 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.47 \[ \int \frac {1}{x^5 \left (-b+a x^4\right )^{3/4}} \, dx=\frac {{\left (a\,x^4-b\right )}^{1/4}}{4\,b\,x^4}+\frac {3\,a\,\mathrm {atan}\left (\frac {{\left (a\,x^4-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right )}{8\,{\left (-b\right )}^{7/4}}+\frac {3\,a\,\mathrm {atanh}\left (\frac {{\left (a\,x^4-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right )}{8\,{\left (-b\right )}^{7/4}} \]