Integrand size = 17, antiderivative size = 145 \[ \int \frac {\left (-b+a x^4\right )^{3/4}}{x} \, dx=\frac {1}{3} \left (-b+a x^4\right )^{3/4}+\frac {b^{3/4} \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^4}}{-\sqrt {b}+\sqrt {-b+a x^4}}\right )}{2 \sqrt {2}}+\frac {b^{3/4} \text {arctanh}\left (\frac {\frac {\sqrt [4]{b}}{\sqrt {2}}+\frac {\sqrt {-b+a x^4}}{\sqrt {2} \sqrt [4]{b}}}{\sqrt [4]{-b+a x^4}}\right )}{2 \sqrt {2}} \]
1/3*(a*x^4-b)^(3/4)+1/4*b^(3/4)*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)+1/4*b^(3/4)*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)
Time = 0.15 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.94 \[ \int \frac {\left (-b+a x^4\right )^{3/4}}{x} \, dx=\frac {1}{12} \left (4 \left (-b+a x^4\right )^{3/4}-3 \sqrt {2} b^{3/4} \arctan \left (\frac {-\sqrt {b}+\sqrt {-b+a x^4}}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^4}}\right )+3 \sqrt {2} b^{3/4} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^4}}{\sqrt {b}+\sqrt {-b+a x^4}}\right )\right ) \]
(4*(-b + a*x^4)^(3/4) - 3*Sqrt[2]*b^(3/4)*ArcTan[(-Sqrt[b] + Sqrt[-b + a*x ^4])/(Sqrt[2]*b^(1/4)*(-b + a*x^4)^(1/4))] + 3*Sqrt[2]*b^(3/4)*ArcTanh[(Sq rt[2]*b^(1/4)*(-b + a*x^4)^(1/4))/(Sqrt[b] + Sqrt[-b + a*x^4])])/12
Time = 0.37 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.46, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.706, Rules used = {798, 60, 73, 27, 826, 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.
\(\displaystyle \int \frac {\left (a x^4-b\right )^{3/4}}{x} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{4} \int \frac {\left (a x^4-b\right )^{3/4}}{x^4}dx^4\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{4} \left (\frac {4}{3} \left (a x^4-b\right )^{3/4}-b \int \frac {1}{x^4 \sqrt [4]{a x^4-b}}dx^4\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{4} \left (\frac {4}{3} \left (a x^4-b\right )^{3/4}-\frac {4 b \int \frac {a x^8}{x^{16}+b}d\sqrt [4]{a x^4-b}}{a}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \left (\frac {4}{3} \left (a x^4-b\right )^{3/4}-4 b \int \frac {x^8}{x^{16}+b}d\sqrt [4]{a x^4-b}\right )\) |
\(\Big \downarrow \) 826 |
\(\displaystyle \frac {1}{4} \left (\frac {4}{3} \left (a x^4-b\right )^{3/4}-4 b \left (\frac {1}{2} \int \frac {x^8+\sqrt {b}}{x^{16}+b}d\sqrt [4]{a x^4-b}-\frac {1}{2} \int \frac {\sqrt {b}-x^8}{x^{16}+b}d\sqrt [4]{a x^4-b}\right )\right )\) |
\(\Big \downarrow \) 1476 |
\(\displaystyle \frac {1}{4} \left (\frac {4}{3} \left (a x^4-b\right )^{3/4}-4 b \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{x^8+\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^4-b}}d\sqrt [4]{a x^4-b}+\frac {1}{2} \int \frac {1}{x^8+\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^4-b}}d\sqrt [4]{a x^4-b}\right )-\frac {1}{2} \int \frac {\sqrt {b}-x^8}{x^{16}+b}d\sqrt [4]{a x^4-b}\right )\right )\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {1}{4} \left (\frac {4}{3} \left (a x^4-b\right )^{3/4}-4 b \left (\frac {1}{2} \left (\frac {\int \frac {1}{-x^8-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{a x^4-b}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b}}-\frac {\int \frac {1}{-x^8-1}d\left (\frac {\sqrt {2} \sqrt [4]{a x^4-b}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b}}\right )-\frac {1}{2} \int \frac {\sqrt {b}-x^8}{x^{16}+b}d\sqrt [4]{a x^4-b}\right )\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{4} \left (\frac {4}{3} \left (a x^4-b\right )^{3/4}-4 b \left (\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a x^4-b}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x^4-b}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b}}\right )-\frac {1}{2} \int \frac {\sqrt {b}-x^8}{x^{16}+b}d\sqrt [4]{a x^4-b}\right )\right )\) |
\(\Big \downarrow \) 1479 |
\(\displaystyle \frac {1}{4} \left (\frac {4}{3} \left (a x^4-b\right )^{3/4}-4 b \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{a x^4-b}}{x^8+\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^4-b}}d\sqrt [4]{a x^4-b}}{2 \sqrt {2} \sqrt [4]{b}}+\frac {\int -\frac {\sqrt {2} \left (\sqrt [4]{b}+\sqrt {2} \sqrt [4]{a x^4-b}\right )}{x^8+\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^4-b}}d\sqrt [4]{a x^4-b}}{2 \sqrt {2} \sqrt [4]{b}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a x^4-b}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x^4-b}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b}}\right )\right )\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{4} \left (\frac {4}{3} \left (a x^4-b\right )^{3/4}-4 b \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{a x^4-b}}{x^8+\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^4-b}}d\sqrt [4]{a x^4-b}}{2 \sqrt {2} \sqrt [4]{b}}-\frac {\int \frac {\sqrt {2} \left (\sqrt [4]{b}+\sqrt {2} \sqrt [4]{a x^4-b}\right )}{x^8+\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^4-b}}d\sqrt [4]{a x^4-b}}{2 \sqrt {2} \sqrt [4]{b}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a x^4-b}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x^4-b}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b}}\right )\right )\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{4} \left (\frac {4}{3} \left (a x^4-b\right )^{3/4}-4 b \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{a x^4-b}}{x^8+\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^4-b}}d\sqrt [4]{a x^4-b}}{2 \sqrt {2} \sqrt [4]{b}}-\frac {\int \frac {\sqrt [4]{b}+\sqrt {2} \sqrt [4]{a x^4-b}}{x^8+\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^4-b}}d\sqrt [4]{a x^4-b}}{2 \sqrt [4]{b}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a x^4-b}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x^4-b}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b}}\right )\right )\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{4} \left (\frac {4}{3} \left (a x^4-b\right )^{3/4}-4 b \left (\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a x^4-b}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x^4-b}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b}}\right )+\frac {1}{2} \left (\frac {\log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^4-b}+\sqrt {b}+x^8\right )}{2 \sqrt {2} \sqrt [4]{b}}-\frac {\log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^4-b}+\sqrt {b}+x^8\right )}{2 \sqrt {2} \sqrt [4]{b}}\right )\right )\right )\) |
((4*(-b + a*x^4)^(3/4))/3 - 4*b*((-(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 + (Log[Sqrt[b] + x^8 - Sqrt[2]*b^(1/4)*(-b + a *x^4)^(1/4)]/(2*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))/4
3.21.32.3.1 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/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
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[(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[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s) Int[(r + s*x^2)/(a + b*x^ 4), x], x] - Simp[1/(2*s) 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[((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 = 0.34 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.10
method | result | size |
pseudoelliptic | \(\frac {\left (a \,x^{4}-b \right )^{\frac {3}{4}}}{3}-\frac {\ln \left (\frac {\sqrt {a \,x^{4}-b}-b^{\frac {1}{4}} \left (a \,x^{4}-b \right )^{\frac {1}{4}} \sqrt {2}+\sqrt {b}}{\sqrt {a \,x^{4}-b}+b^{\frac {1}{4}} \left (a \,x^{4}-b \right )^{\frac {1}{4}} \sqrt {2}+\sqrt {b}}\right ) b^{\frac {3}{4}} \sqrt {2}}{8}-\frac {\arctan \left (\frac {\sqrt {2}\, \left (a \,x^{4}-b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right ) b^{\frac {3}{4}} \sqrt {2}}{4}+\frac {\arctan \left (\frac {-\sqrt {2}\, \left (a \,x^{4}-b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right ) b^{\frac {3}{4}} \sqrt {2}}{4}\) | \(159\) |
1/3*(a*x^4-b)^(3/4)-1/8*ln(((a*x^4-b)^(1/2)-b^(1/4)*(a*x^4-b)^(1/4)*2^(1/2 )+b^(1/2))/((a*x^4-b)^(1/2)+b^(1/4)*(a*x^4-b)^(1/4)*2^(1/2)+b^(1/2)))*b^(3 /4)*2^(1/2)-1/4*arctan((2^(1/2)*(a*x^4-b)^(1/4)+b^(1/4))/b^(1/4))*b^(3/4)* 2^(1/2)+1/4*arctan((-2^(1/2)*(a*x^4-b)^(1/4)+b^(1/4))/b^(1/4))*b^(3/4)*2^( 1/2)
Result contains complex when optimal does not.
Time = 0.26 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.05 \[ \int \frac {\left (-b+a x^4\right )^{3/4}}{x} \, dx=-\frac {1}{4} \, \left (-b^{3}\right )^{\frac {1}{4}} \log \left ({\left (a x^{4} - b\right )}^{\frac {1}{4}} b^{2} + \left (-b^{3}\right )^{\frac {3}{4}}\right ) + \frac {1}{4} i \, \left (-b^{3}\right )^{\frac {1}{4}} \log \left ({\left (a x^{4} - b\right )}^{\frac {1}{4}} b^{2} + i \, \left (-b^{3}\right )^{\frac {3}{4}}\right ) - \frac {1}{4} i \, \left (-b^{3}\right )^{\frac {1}{4}} \log \left ({\left (a x^{4} - b\right )}^{\frac {1}{4}} b^{2} - i \, \left (-b^{3}\right )^{\frac {3}{4}}\right ) + \frac {1}{4} \, \left (-b^{3}\right )^{\frac {1}{4}} \log \left ({\left (a x^{4} - b\right )}^{\frac {1}{4}} b^{2} - \left (-b^{3}\right )^{\frac {3}{4}}\right ) + \frac {1}{3} \, {\left (a x^{4} - b\right )}^{\frac {3}{4}} \]
-1/4*(-b^3)^(1/4)*log((a*x^4 - b)^(1/4)*b^2 + (-b^3)^(3/4)) + 1/4*I*(-b^3) ^(1/4)*log((a*x^4 - b)^(1/4)*b^2 + I*(-b^3)^(3/4)) - 1/4*I*(-b^3)^(1/4)*lo g((a*x^4 - b)^(1/4)*b^2 - I*(-b^3)^(3/4)) + 1/4*(-b^3)^(1/4)*log((a*x^4 - b)^(1/4)*b^2 - (-b^3)^(3/4)) + 1/3*(a*x^4 - b)^(3/4)
Result contains complex when optimal does not.
Time = 0.85 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.32 \[ \int \frac {\left (-b+a x^4\right )^{3/4}}{x} \, dx=- \frac {a^{\frac {3}{4}} x^{3} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {3}{4} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b e^{2 i \pi }}{a x^{4}}} \right )}}{4 \Gamma \left (\frac {1}{4}\right )} \]
-a**(3/4)*x**3*gamma(-3/4)*hyper((-3/4, -3/4), (1/4,), b*exp_polar(2*I*pi) /(a*x**4))/(4*gamma(1/4))
Time = 0.29 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.23 \[ \int \frac {\left (-b+a x^4\right )^{3/4}}{x} \, dx=-\frac {1}{8} \, {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{4} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {1}{4}}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{4} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {1}{4}}} - \frac {\sqrt {2} \log \left (\sqrt {2} {\left (a x^{4} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{4} - b} + \sqrt {b}\right )}{b^{\frac {1}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} {\left (a x^{4} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{4} - b} + \sqrt {b}\right )}{b^{\frac {1}{4}}}\right )} b + \frac {1}{3} \, {\left (a x^{4} - b\right )}^{\frac {3}{4}} \]
-1/8*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4) + 2*(a*x^4 - b)^(1/4)) /b^(1/4))/b^(1/4) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4) - 2*(a* x^4 - b)^(1/4))/b^(1/4))/b^(1/4) - sqrt(2)*log(sqrt(2)*(a*x^4 - b)^(1/4)*b ^(1/4) + sqrt(a*x^4 - b) + sqrt(b))/b^(1/4) + sqrt(2)*log(-sqrt(2)*(a*x^4 - b)^(1/4)*b^(1/4) + sqrt(a*x^4 - b) + sqrt(b))/b^(1/4))*b + 1/3*(a*x^4 - b)^(3/4)
Time = 0.28 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.21 \[ \int \frac {\left (-b+a x^4\right )^{3/4}}{x} \, dx=-\frac {1}{4} \, \sqrt {2} b^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{4} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right ) - \frac {1}{4} \, \sqrt {2} b^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{4} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right ) + \frac {1}{8} \, \sqrt {2} b^{\frac {3}{4}} \log \left (\sqrt {2} {\left (a x^{4} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{4} - b} + \sqrt {b}\right ) - \frac {1}{8} \, \sqrt {2} b^{\frac {3}{4}} \log \left (-\sqrt {2} {\left (a x^{4} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{4} - b} + \sqrt {b}\right ) + \frac {1}{3} \, {\left (a x^{4} - b\right )}^{\frac {3}{4}} \]
-1/4*sqrt(2)*b^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4) + 2*(a*x^4 - b)^( 1/4))/b^(1/4)) - 1/4*sqrt(2)*b^(3/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4) - 2*(a*x^4 - b)^(1/4))/b^(1/4)) + 1/8*sqrt(2)*b^(3/4)*log(sqrt(2)*(a*x^4 - b)^(1/4)*b^(1/4) + sqrt(a*x^4 - b) + sqrt(b)) - 1/8*sqrt(2)*b^(3/4)*log(- sqrt(2)*(a*x^4 - b)^(1/4)*b^(1/4) + sqrt(a*x^4 - b) + sqrt(b)) + 1/3*(a*x^ 4 - b)^(3/4)
Time = 6.08 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.44 \[ \int \frac {\left (-b+a x^4\right )^{3/4}}{x} \, dx=\frac {{\left (a\,x^4-b\right )}^{3/4}}{3}+\frac {{\left (-b\right )}^{3/4}\,\mathrm {atan}\left (\frac {{\left (a\,x^4-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right )}{2}-\frac {{\left (-b\right )}^{3/4}\,\mathrm {atanh}\left (\frac {{\left (a\,x^4-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right )}{2} \]