Integrand size = 17, antiderivative size = 145 \[ \int \frac {\sqrt [4]{-b+a x^3}}{x} \, dx=\frac {4}{3} \sqrt [4]{-b+a x^3}+\frac {1}{3} \sqrt {2} \sqrt [4]{b} \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}}{-\sqrt {b}+\sqrt {-b+a x^3}}\right )-\frac {1}{3} \sqrt {2} \sqrt [4]{b} \text {arctanh}\left (\frac {\frac {\sqrt [4]{b}}{\sqrt {2}}+\frac {\sqrt {-b+a x^3}}{\sqrt {2} \sqrt [4]{b}}}{\sqrt [4]{-b+a x^3}}\right ) \] Output:
4/3*(a*x^3-b)^(1/4)+1/3*2^(1/2)*b^(1/4)*arctan(2^(1/2)*b^(1/4)*(a*x^3-b)^( 1/4)/(-b^(1/2)+(a*x^3-b)^(1/2)))-1/3*2^(1/2)*b^(1/4)*arctanh((1/2*b^(1/4)* 2^(1/2)+1/2*(a*x^3-b)^(1/2)*2^(1/2)/b^(1/4))/(a*x^3-b)^(1/4))
Time = 0.20 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt [4]{-b+a x^3}}{x} \, dx=\frac {1}{3} \left (4 \sqrt [4]{-b+a x^3}-\sqrt {2} \sqrt [4]{b} \arctan \left (\frac {-\sqrt {b}+\sqrt {-b+a x^3}}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}}\right )-\sqrt {2} \sqrt [4]{b} \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^3}}{\sqrt {b}+\sqrt {-b+a x^3}}\right )\right ) \] Input:
Integrate[(-b + a*x^3)^(1/4)/x,x]
Output:
(4*(-b + a*x^3)^(1/4) - Sqrt[2]*b^(1/4)*ArcTan[(-Sqrt[b] + Sqrt[-b + a*x^3 ])/(Sqrt[2]*b^(1/4)*(-b + a*x^3)^(1/4))] - Sqrt[2]*b^(1/4)*ArcTanh[(Sqrt[2 ]*b^(1/4)*(-b + a*x^3)^(1/4))/(Sqrt[b] + Sqrt[-b + a*x^3])])/3
Time = 0.39 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.54, 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, 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.
\(\displaystyle \int \frac {\sqrt [4]{a x^3-b}}{x} \, dx\) |
\(\Big \downarrow \) 798 |
\(\displaystyle \frac {1}{3} \int \frac {\sqrt [4]{a x^3-b}}{x^3}dx^3\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {1}{3} \left (4 \sqrt [4]{a x^3-b}-b \int \frac {1}{x^3 \left (a x^3-b\right )^{3/4}}dx^3\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {1}{3} \left (4 \sqrt [4]{a x^3-b}-\frac {4 b \int \frac {1}{\frac {x^{12}}{a}+\frac {b}{a}}d\sqrt [4]{a x^3-b}}{a}\right )\) |
\(\Big \downarrow \) 755 |
\(\displaystyle \frac {1}{3} \left (4 \sqrt [4]{a x^3-b}-\frac {4 b \left (\frac {\int \frac {a \left (\sqrt {b}-x^6\right )}{x^{12}+b}d\sqrt [4]{a x^3-b}}{2 \sqrt {b}}+\frac {\int \frac {a \left (x^6+\sqrt {b}\right )}{x^{12}+b}d\sqrt [4]{a x^3-b}}{2 \sqrt {b}}\right )}{a}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \left (4 \sqrt [4]{a x^3-b}-\frac {4 b \left (\frac {a \int \frac {\sqrt {b}-x^6}{x^{12}+b}d\sqrt [4]{a x^3-b}}{2 \sqrt {b}}+\frac {a \int \frac {x^6+\sqrt {b}}{x^{12}+b}d\sqrt [4]{a x^3-b}}{2 \sqrt {b}}\right )}{a}\right )\) |
\(\Big \downarrow \) 1476 |
\(\displaystyle \frac {1}{3} \left (4 \sqrt [4]{a x^3-b}-\frac {4 b \left (\frac {a \left (\frac {1}{2} \int \frac {1}{x^6+\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}}d\sqrt [4]{a x^3-b}+\frac {1}{2} \int \frac {1}{x^6+\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}}d\sqrt [4]{a x^3-b}\right )}{2 \sqrt {b}}+\frac {a \int \frac {\sqrt {b}-x^6}{x^{12}+b}d\sqrt [4]{a x^3-b}}{2 \sqrt {b}}\right )}{a}\right )\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle \frac {1}{3} \left (4 \sqrt [4]{a x^3-b}-\frac {4 b \left (\frac {a \left (\frac {\int \frac {1}{-x^6-1}d\left (1-\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b}}-\frac {\int \frac {1}{-x^6-1}d\left (\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b}}\right )}{2 \sqrt {b}}+\frac {a \int \frac {\sqrt {b}-x^6}{x^{12}+b}d\sqrt [4]{a x^3-b}}{2 \sqrt {b}}\right )}{a}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{3} \left (4 \sqrt [4]{a x^3-b}-\frac {4 b \left (\frac {a \int \frac {\sqrt {b}-x^6}{x^{12}+b}d\sqrt [4]{a x^3-b}}{2 \sqrt {b}}+\frac {a \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b}}\right )}{2 \sqrt {b}}\right )}{a}\right )\) |
\(\Big \downarrow \) 1479 |
\(\displaystyle \frac {1}{3} \left (4 \sqrt [4]{a x^3-b}-\frac {4 b \left (\frac {a \left (-\frac {\int -\frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{a x^3-b}}{x^6+\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}}d\sqrt [4]{a x^3-b}}{2 \sqrt {2} \sqrt [4]{b}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt [4]{b}+\sqrt {2} \sqrt [4]{a x^3-b}\right )}{x^6+\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}}d\sqrt [4]{a x^3-b}}{2 \sqrt {2} \sqrt [4]{b}}\right )}{2 \sqrt {b}}+\frac {a \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b}}\right )}{2 \sqrt {b}}\right )}{a}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{3} \left (4 \sqrt [4]{a x^3-b}-\frac {4 b \left (\frac {a \left (\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{a x^3-b}}{x^6+\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}}d\sqrt [4]{a x^3-b}}{2 \sqrt {2} \sqrt [4]{b}}+\frac {\int \frac {\sqrt {2} \left (\sqrt [4]{b}+\sqrt {2} \sqrt [4]{a x^3-b}\right )}{x^6+\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}}d\sqrt [4]{a x^3-b}}{2 \sqrt {2} \sqrt [4]{b}}\right )}{2 \sqrt {b}}+\frac {a \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b}}\right )}{2 \sqrt {b}}\right )}{a}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{3} \left (4 \sqrt [4]{a x^3-b}-\frac {4 b \left (\frac {a \left (\frac {\int \frac {\sqrt {2} \sqrt [4]{b}-2 \sqrt [4]{a x^3-b}}{x^6+\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}}d\sqrt [4]{a x^3-b}}{2 \sqrt {2} \sqrt [4]{b}}+\frac {\int \frac {\sqrt [4]{b}+\sqrt {2} \sqrt [4]{a x^3-b}}{x^6+\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}}d\sqrt [4]{a x^3-b}}{2 \sqrt [4]{b}}\right )}{2 \sqrt {b}}+\frac {a \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b}}\right )}{2 \sqrt {b}}\right )}{a}\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{3} \left (4 \sqrt [4]{a x^3-b}-\frac {4 b \left (\frac {a \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}+1\right )}{\sqrt {2} \sqrt [4]{b}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x^3-b}}{\sqrt [4]{b}}\right )}{\sqrt {2} \sqrt [4]{b}}\right )}{2 \sqrt {b}}+\frac {a \left (\frac {\log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}+\sqrt {b}+x^6\right )}{2 \sqrt {2} \sqrt [4]{b}}-\frac {\log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^3-b}+\sqrt {b}+x^6\right )}{2 \sqrt {2} \sqrt [4]{b}}\right )}{2 \sqrt {b}}\right )}{a}\right )\) |
Input:
Int[(-b + a*x^3)^(1/4)/x,x]
Output:
(4*(-b + a*x^3)^(1/4) - (4*b*((a*(-(ArcTan[1 - (Sqrt[2]*(-b + a*x^3)^(1/4) )/b^(1/4)]/(Sqrt[2]*b^(1/4))) + ArcTan[1 + (Sqrt[2]*(-b + a*x^3)^(1/4))/b^ (1/4)]/(Sqrt[2]*b^(1/4))))/(2*Sqrt[b]) + (a*(-1/2*Log[Sqrt[b] + x^6 - Sqrt [2]*b^(1/4)*(-b + a*x^3)^(1/4)]/(Sqrt[2]*b^(1/4)) + Log[Sqrt[b] + x^6 + Sq rt[2]*b^(1/4)*(-b + a*x^3)^(1/4)]/(2*Sqrt[2]*b^(1/4))))/(2*Sqrt[b])))/a)/3
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[((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 = 0.38 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.11
method | result | size |
pseudoelliptic | \(\frac {4 \left (a \,x^{3}-b \right )^{\frac {1}{4}}}{3}-\frac {\ln \left (\frac {-b^{\frac {1}{4}} \left (a \,x^{3}-b \right )^{\frac {1}{4}} \sqrt {2}-\sqrt {b}-\sqrt {a \,x^{3}-b}}{b^{\frac {1}{4}} \left (a \,x^{3}-b \right )^{\frac {1}{4}} \sqrt {2}-\sqrt {b}-\sqrt {a \,x^{3}-b}}\right ) b^{\frac {1}{4}} \sqrt {2}}{6}-\frac {\arctan \left (\frac {\sqrt {2}\, \left (a \,x^{3}-b \right )^{\frac {1}{4}}}{b^{\frac {1}{4}}}+1\right ) b^{\frac {1}{4}} \sqrt {2}}{3}+\frac {\arctan \left (-\frac {\sqrt {2}\, \left (a \,x^{3}-b \right )^{\frac {1}{4}}}{b^{\frac {1}{4}}}+1\right ) b^{\frac {1}{4}} \sqrt {2}}{3}\) | \(161\) |
Input:
int((a*x^3-b)^(1/4)/x,x,method=_RETURNVERBOSE)
Output:
4/3*(a*x^3-b)^(1/4)-1/6*ln((-b^(1/4)*(a*x^3-b)^(1/4)*2^(1/2)-b^(1/2)-(a*x^ 3-b)^(1/2))/(b^(1/4)*(a*x^3-b)^(1/4)*2^(1/2)-b^(1/2)-(a*x^3-b)^(1/2)))*b^( 1/4)*2^(1/2)-1/3*arctan(2^(1/2)/b^(1/4)*(a*x^3-b)^(1/4)+1)*b^(1/4)*2^(1/2) +1/3*arctan(-2^(1/2)/b^(1/4)*(a*x^3-b)^(1/4)+1)*b^(1/4)*2^(1/2)
Result contains complex when optimal does not.
Time = 0.08 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.83 \[ \int \frac {\sqrt [4]{-b+a x^3}}{x} \, dx=-\frac {1}{3} \, \left (-b\right )^{\frac {1}{4}} \log \left ({\left (a x^{3} - b\right )}^{\frac {1}{4}} + \left (-b\right )^{\frac {1}{4}}\right ) - \frac {1}{3} i \, \left (-b\right )^{\frac {1}{4}} \log \left ({\left (a x^{3} - b\right )}^{\frac {1}{4}} + i \, \left (-b\right )^{\frac {1}{4}}\right ) + \frac {1}{3} i \, \left (-b\right )^{\frac {1}{4}} \log \left ({\left (a x^{3} - b\right )}^{\frac {1}{4}} - i \, \left (-b\right )^{\frac {1}{4}}\right ) + \frac {1}{3} \, \left (-b\right )^{\frac {1}{4}} \log \left ({\left (a x^{3} - b\right )}^{\frac {1}{4}} - \left (-b\right )^{\frac {1}{4}}\right ) + \frac {4}{3} \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} \] Input:
integrate((a*x^3-b)^(1/4)/x,x, algorithm="fricas")
Output:
-1/3*(-b)^(1/4)*log((a*x^3 - b)^(1/4) + (-b)^(1/4)) - 1/3*I*(-b)^(1/4)*log ((a*x^3 - b)^(1/4) + I*(-b)^(1/4)) + 1/3*I*(-b)^(1/4)*log((a*x^3 - b)^(1/4 ) - I*(-b)^(1/4)) + 1/3*(-b)^(1/4)*log((a*x^3 - b)^(1/4) - (-b)^(1/4)) + 4 /3*(a*x^3 - b)^(1/4)
Result contains complex when optimal does not.
Time = 0.66 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.33 \[ \int \frac {\sqrt [4]{-b+a x^3}}{x} \, dx=- \frac {\sqrt [4]{a} x^{\frac {3}{4}} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b e^{2 i \pi }}{a x^{3}}} \right )}}{3 \Gamma \left (\frac {3}{4}\right )} \] Input:
integrate((a*x**3-b)**(1/4)/x,x)
Output:
-a**(1/4)*x**(3/4)*gamma(-1/4)*hyper((-1/4, -1/4), (3/4,), b*exp_polar(2*I *pi)/(a*x**3))/(3*gamma(3/4))
Time = 0.14 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.21 \[ \int \frac {\sqrt [4]{-b+a x^3}}{x} \, dx=-\frac {1}{3} \, \sqrt {2} b^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right ) - \frac {1}{3} \, \sqrt {2} b^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right ) - \frac {1}{6} \, \sqrt {2} b^{\frac {1}{4}} \log \left (\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right ) + \frac {1}{6} \, \sqrt {2} b^{\frac {1}{4}} \log \left (-\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right ) + \frac {4}{3} \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} \] Input:
integrate((a*x^3-b)^(1/4)/x,x, algorithm="maxima")
Output:
-1/3*sqrt(2)*b^(1/4)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4) + 2*(a*x^3 - b)^( 1/4))/b^(1/4)) - 1/3*sqrt(2)*b^(1/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4) - 2*(a*x^3 - b)^(1/4))/b^(1/4)) - 1/6*sqrt(2)*b^(1/4)*log(sqrt(2)*(a*x^3 - b)^(1/4)*b^(1/4) + sqrt(a*x^3 - b) + sqrt(b)) + 1/6*sqrt(2)*b^(1/4)*log(- sqrt(2)*(a*x^3 - b)^(1/4)*b^(1/4) + sqrt(a*x^3 - b) + sqrt(b)) + 4/3*(a*x^ 3 - b)^(1/4)
Time = 0.13 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.21 \[ \int \frac {\sqrt [4]{-b+a x^3}}{x} \, dx=-\frac {1}{3} \, \sqrt {2} b^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right ) - \frac {1}{3} \, \sqrt {2} b^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{3} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right ) - \frac {1}{6} \, \sqrt {2} b^{\frac {1}{4}} \log \left (\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right ) + \frac {1}{6} \, \sqrt {2} b^{\frac {1}{4}} \log \left (-\sqrt {2} {\left (a x^{3} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{3} - b} + \sqrt {b}\right ) + \frac {4}{3} \, {\left (a x^{3} - b\right )}^{\frac {1}{4}} \] Input:
integrate((a*x^3-b)^(1/4)/x,x, algorithm="giac")
Output:
-1/3*sqrt(2)*b^(1/4)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4) + 2*(a*x^3 - b)^( 1/4))/b^(1/4)) - 1/3*sqrt(2)*b^(1/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4) - 2*(a*x^3 - b)^(1/4))/b^(1/4)) - 1/6*sqrt(2)*b^(1/4)*log(sqrt(2)*(a*x^3 - b)^(1/4)*b^(1/4) + sqrt(a*x^3 - b) + sqrt(b)) + 1/6*sqrt(2)*b^(1/4)*log(- sqrt(2)*(a*x^3 - b)^(1/4)*b^(1/4) + sqrt(a*x^3 - b) + sqrt(b)) + 4/3*(a*x^ 3 - b)^(1/4)
Time = 8.32 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.44 \[ \int \frac {\sqrt [4]{-b+a x^3}}{x} \, dx=\frac {4\,{\left (a\,x^3-b\right )}^{1/4}}{3}-\frac {2\,{\left (-b\right )}^{1/4}\,\mathrm {atan}\left (\frac {{\left (a\,x^3-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right )}{3}-\frac {2\,{\left (-b\right )}^{1/4}\,\mathrm {atanh}\left (\frac {{\left (a\,x^3-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right )}{3} \] Input:
int((a*x^3 - b)^(1/4)/x,x)
Output:
(4*(a*x^3 - b)^(1/4))/3 - (2*(-b)^(1/4)*atan((a*x^3 - b)^(1/4)/(-b)^(1/4)) )/3 - (2*(-b)^(1/4)*atanh((a*x^3 - b)^(1/4)/(-b)^(1/4)))/3
\[ \int \frac {\sqrt [4]{-b+a x^3}}{x} \, dx=\frac {4 \left (a \,x^{3}-b \right )^{\frac {1}{4}}}{3}-\left (\int \frac {\left (a \,x^{3}-b \right )^{\frac {1}{4}}}{a \,x^{4}-b x}d x \right ) b \] Input:
int((a*x^3-b)^(1/4)/x,x)
Output:
(4*(a*x**3 - b)**(1/4) - 3*int((a*x**3 - b)**(1/4)/(a*x**4 - b*x),x)*b)/3