\(\int \frac {(-b+a x^2)^{3/4}}{x^3} \, dx\) [2070]

Optimal result

Integrand size = 17, antiderivative size = 150 \[ \int \frac {\left (-b+a x^2\right )^{3/4}}{x^3} \, dx=-\frac {\left (-b+a x^2\right )^{3/4}}{2 x^2}-\frac {3 a \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^2}}{-\sqrt {b}+\sqrt {-b+a x^2}}\right )}{4 \sqrt {2} \sqrt [4]{b}}-\frac {3 a \text {arctanh}\left (\frac {\frac {\sqrt [4]{b}}{\sqrt {2}}+\frac {\sqrt {-b+a x^2}}{\sqrt {2} \sqrt [4]{b}}}{\sqrt [4]{-b+a x^2}}\right )}{4 \sqrt {2} \sqrt [4]{b}} \]

[Out]

-1/2*(a*x^2-b)^(3/4)/x^2-3/8*a*arctan(2^(1/2)*b^(1/4)*(a*x^2-b)^(1/4)/(-b^(1/2)+(a*x^2-b)^(1/2)))*2^(1/2)/b^(1
/4)-3/8*a*arctanh((1/2*b^(1/4)*2^(1/2)+1/2*(a*x^2-b)^(1/2)*2^(1/2)/b^(1/4))/(a*x^2-b)^(1/4))*2^(1/2)/b^(1/4)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.50, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {272, 43, 65, 303, 1176, 631, 210, 1179, 642} \[ \int \frac {\left (-b+a x^2\right )^{3/4}}{x^3} \, dx=-\frac {3 a \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a x^2-b}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} \sqrt [4]{b}}+\frac {3 a \arctan \left (\frac {\sqrt {2} \sqrt [4]{a x^2-b}}{\sqrt [4]{b}}+1\right )}{4 \sqrt {2} \sqrt [4]{b}}-\frac {\left (a x^2-b\right )^{3/4}}{2 x^2}+\frac {3 a \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^2-b}+\sqrt {a x^2-b}+\sqrt {b}\right )}{8 \sqrt {2} \sqrt [4]{b}}-\frac {3 a \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^2-b}+\sqrt {a x^2-b}+\sqrt {b}\right )}{8 \sqrt {2} \sqrt [4]{b}} \]

[In]

Int[(-b + a*x^2)^(3/4)/x^3,x]

[Out]

-1/2*(-b + a*x^2)^(3/4)/x^2 - (3*a*ArcTan[1 - (Sqrt[2]*(-b + a*x^2)^(1/4))/b^(1/4)])/(4*Sqrt[2]*b^(1/4)) + (3*
a*ArcTan[1 + (Sqrt[2]*(-b + a*x^2)^(1/4))/b^(1/4)])/(4*Sqrt[2]*b^(1/4)) + (3*a*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*(
-b + a*x^2)^(1/4) + Sqrt[-b + a*x^2]])/(8*Sqrt[2]*b^(1/4)) - (3*a*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*(-b + a*x^2)^(
1/4) + Sqrt[-b + a*x^2]])/(8*Sqrt[2]*b^(1/4))

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 210

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])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[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]]

Rule 303

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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]

Rule 1176

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[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]

Rule 1179

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {(-b+a x)^{3/4}}{x^2} \, dx,x,x^2\right ) \\ & = -\frac {\left (-b+a x^2\right )^{3/4}}{2 x^2}+\frac {1}{8} (3 a) \text {Subst}\left (\int \frac {1}{x \sqrt [4]{-b+a x}} \, dx,x,x^2\right ) \\ & = -\frac {\left (-b+a x^2\right )^{3/4}}{2 x^2}+\frac {3}{2} \text {Subst}\left (\int \frac {x^2}{\frac {b}{a}+\frac {x^4}{a}} \, dx,x,\sqrt [4]{-b+a x^2}\right ) \\ & = -\frac {\left (-b+a x^2\right )^{3/4}}{2 x^2}-\frac {3}{4} \text {Subst}\left (\int \frac {\sqrt {b}-x^2}{\frac {b}{a}+\frac {x^4}{a}} \, dx,x,\sqrt [4]{-b+a x^2}\right )+\frac {3}{4} \text {Subst}\left (\int \frac {\sqrt {b}+x^2}{\frac {b}{a}+\frac {x^4}{a}} \, dx,x,\sqrt [4]{-b+a x^2}\right ) \\ & = -\frac {\left (-b+a x^2\right )^{3/4}}{2 x^2}+\frac {1}{8} (3 a) \text {Subst}\left (\int \frac {1}{\sqrt {b}-\sqrt {2} \sqrt [4]{b} x+x^2} \, dx,x,\sqrt [4]{-b+a x^2}\right )+\frac {1}{8} (3 a) \text {Subst}\left (\int \frac {1}{\sqrt {b}+\sqrt {2} \sqrt [4]{b} x+x^2} \, dx,x,\sqrt [4]{-b+a x^2}\right )+\frac {(3 a) \text {Subst}\left (\int \frac {\sqrt {2} \sqrt [4]{b}+2 x}{-\sqrt {b}-\sqrt {2} \sqrt [4]{b} x-x^2} \, dx,x,\sqrt [4]{-b+a x^2}\right )}{8 \sqrt {2} \sqrt [4]{b}}+\frac {(3 a) \text {Subst}\left (\int \frac {\sqrt {2} \sqrt [4]{b}-2 x}{-\sqrt {b}+\sqrt {2} \sqrt [4]{b} x-x^2} \, dx,x,\sqrt [4]{-b+a x^2}\right )}{8 \sqrt {2} \sqrt [4]{b}} \\ & = -\frac {\left (-b+a x^2\right )^{3/4}}{2 x^2}+\frac {3 a \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^2}+\sqrt {-b+a x^2}\right )}{8 \sqrt {2} \sqrt [4]{b}}-\frac {3 a \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^2}+\sqrt {-b+a x^2}\right )}{8 \sqrt {2} \sqrt [4]{b}}+\frac {(3 a) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{-b+a x^2}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} \sqrt [4]{b}}-\frac {(3 a) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{-b+a x^2}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} \sqrt [4]{b}} \\ & = -\frac {\left (-b+a x^2\right )^{3/4}}{2 x^2}-\frac {3 a \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{-b+a x^2}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} \sqrt [4]{b}}+\frac {3 a \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{-b+a x^2}}{\sqrt [4]{b}}\right )}{4 \sqrt {2} \sqrt [4]{b}}+\frac {3 a \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^2}+\sqrt {-b+a x^2}\right )}{8 \sqrt {2} \sqrt [4]{b}}-\frac {3 a \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^2}+\sqrt {-b+a x^2}\right )}{8 \sqrt {2} \sqrt [4]{b}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.94 \[ \int \frac {\left (-b+a x^2\right )^{3/4}}{x^3} \, dx=\frac {1}{8} \left (-\frac {4 \left (-b+a x^2\right )^{3/4}}{x^2}+\frac {3 \sqrt {2} a \arctan \left (\frac {-\sqrt {b}+\sqrt {-b+a x^2}}{\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^2}}\right )}{\sqrt [4]{b}}-\frac {3 \sqrt {2} a \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^2}}{\sqrt {b}+\sqrt {-b+a x^2}}\right )}{\sqrt [4]{b}}\right ) \]

[In]

Integrate[(-b + a*x^2)^(3/4)/x^3,x]

[Out]

((-4*(-b + a*x^2)^(3/4))/x^2 + (3*Sqrt[2]*a*ArcTan[(-Sqrt[b] + Sqrt[-b + a*x^2])/(Sqrt[2]*b^(1/4)*(-b + a*x^2)
^(1/4))])/b^(1/4) - (3*Sqrt[2]*a*ArcTanh[(Sqrt[2]*b^(1/4)*(-b + a*x^2)^(1/4))/(Sqrt[b] + Sqrt[-b + a*x^2])])/b
^(1/4))/8

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.15

method	result	size
pseudoelliptic	\(-\frac {3 \left (\arctan \left (\frac {-\sqrt {2}\, \left (a \,x^{2}-b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right ) \sqrt {2}\, a \,x^{2}-\arctan \left (\frac {\sqrt {2}\, \left (a \,x^{2}-b \right )^{\frac {1}{4}}+b^{\frac {1}{4}}}{b^{\frac {1}{4}}}\right ) \sqrt {2}\, a \,x^{2}-\frac {\ln \left (\frac {\sqrt {a \,x^{2}-b}-b^{\frac {1}{4}} \left (a \,x^{2}-b \right )^{\frac {1}{4}} \sqrt {2}+\sqrt {b}}{\sqrt {a \,x^{2}-b}+b^{\frac {1}{4}} \left (a \,x^{2}-b \right )^{\frac {1}{4}} \sqrt {2}+\sqrt {b}}\right ) \sqrt {2}\, a \,x^{2}}{2}+\frac {4 \left (a \,x^{2}-b \right )^{\frac {3}{4}} b^{\frac {1}{4}}}{3}\right )}{8 b^{\frac {1}{4}} x^{2}}\)	\(172\)

[In]

int((a*x^2-b)^(3/4)/x^3,x,method=_RETURNVERBOSE)

[Out]

-3/8/b^(1/4)*(arctan((-2^(1/2)*(a*x^2-b)^(1/4)+b^(1/4))/b^(1/4))*2^(1/2)*a*x^2-arctan((2^(1/2)*(a*x^2-b)^(1/4)
+b^(1/4))/b^(1/4))*2^(1/2)*a*x^2-1/2*ln(((a*x^2-b)^(1/2)-b^(1/4)*(a*x^2-b)^(1/4)*2^(1/2)+b^(1/2))/((a*x^2-b)^(
1/2)+b^(1/4)*(a*x^2-b)^(1/4)*2^(1/2)+b^(1/2)))*2^(1/2)*a*x^2+4/3*(a*x^2-b)^(3/4)*b^(1/4))/x^2

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.35 \[ \int \frac {\left (-b+a x^2\right )^{3/4}}{x^3} \, dx=\frac {3 \, \left (-\frac {a^{4}}{b}\right )^{\frac {1}{4}} x^{2} \log \left (27 \, {\left (a x^{2} - b\right )}^{\frac {1}{4}} a^{3} + 27 \, \left (-\frac {a^{4}}{b}\right )^{\frac {3}{4}} b\right ) - 3 i \, \left (-\frac {a^{4}}{b}\right )^{\frac {1}{4}} x^{2} \log \left (27 \, {\left (a x^{2} - b\right )}^{\frac {1}{4}} a^{3} + 27 i \, \left (-\frac {a^{4}}{b}\right )^{\frac {3}{4}} b\right ) + 3 i \, \left (-\frac {a^{4}}{b}\right )^{\frac {1}{4}} x^{2} \log \left (27 \, {\left (a x^{2} - b\right )}^{\frac {1}{4}} a^{3} - 27 i \, \left (-\frac {a^{4}}{b}\right )^{\frac {3}{4}} b\right ) - 3 \, \left (-\frac {a^{4}}{b}\right )^{\frac {1}{4}} x^{2} \log \left (27 \, {\left (a x^{2} - b\right )}^{\frac {1}{4}} a^{3} - 27 \, \left (-\frac {a^{4}}{b}\right )^{\frac {3}{4}} b\right ) - 4 \, {\left (a x^{2} - b\right )}^{\frac {3}{4}}}{8 \, x^{2}} \]

[In]

integrate((a*x^2-b)^(3/4)/x^3,x, algorithm="fricas")

[Out]

1/8*(3*(-a^4/b)^(1/4)*x^2*log(27*(a*x^2 - b)^(1/4)*a^3 + 27*(-a^4/b)^(3/4)*b) - 3*I*(-a^4/b)^(1/4)*x^2*log(27*
(a*x^2 - b)^(1/4)*a^3 + 27*I*(-a^4/b)^(3/4)*b) + 3*I*(-a^4/b)^(1/4)*x^2*log(27*(a*x^2 - b)^(1/4)*a^3 - 27*I*(-
a^4/b)^(3/4)*b) - 3*(-a^4/b)^(1/4)*x^2*log(27*(a*x^2 - b)^(1/4)*a^3 - 27*(-a^4/b)^(3/4)*b) - 4*(a*x^2 - b)^(3/
4))/x^2

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.84 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.29 \[ \int \frac {\left (-b+a x^2\right )^{3/4}}{x^3} \, dx=- \frac {a^{\frac {3}{4}} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b e^{2 i \pi }}{a x^{2}}} \right )}}{2 \sqrt {x} \Gamma \left (\frac {5}{4}\right )} \]

[In]

integrate((a*x**2-b)**(3/4)/x**3,x)

[Out]

-a**(3/4)*gamma(1/4)*hyper((-3/4, 1/4), (5/4,), b*exp_polar(2*I*pi)/(a*x**2))/(2*sqrt(x)*gamma(5/4))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.21 \[ \int \frac {\left (-b+a x^2\right )^{3/4}}{x^3} \, dx=\frac {3}{16} \, {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{2} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {1}{4}}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{2} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {1}{4}}} - \frac {\sqrt {2} \log \left (\sqrt {2} {\left (a x^{2} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{2} - b} + \sqrt {b}\right )}{b^{\frac {1}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} {\left (a x^{2} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{2} - b} + \sqrt {b}\right )}{b^{\frac {1}{4}}}\right )} a - \frac {{\left (a x^{2} - b\right )}^{\frac {3}{4}}}{2 \, x^{2}} \]

[In]

integrate((a*x^2-b)^(3/4)/x^3,x, algorithm="maxima")

[Out]

3/16*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4) + 2*(a*x^2 - 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^2 - b)^(1/4))/b^(1/4))/b^(1/4) - sqrt(2)*log(sqrt(2)*(a*x^2 - b)^(1/4)
*b^(1/4) + sqrt(a*x^2 - b) + sqrt(b))/b^(1/4) + sqrt(2)*log(-sqrt(2)*(a*x^2 - b)^(1/4)*b^(1/4) + sqrt(a*x^2 -
b) + sqrt(b))/b^(1/4))*a - 1/2*(a*x^2 - b)^(3/4)/x^2

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.31 \[ \int \frac {\left (-b+a x^2\right )^{3/4}}{x^3} \, dx=\frac {\frac {6 \, \sqrt {2} a^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} + 2 \, {\left (a x^{2} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {1}{4}}} + \frac {6 \, \sqrt {2} a^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} - 2 \, {\left (a x^{2} - b\right )}^{\frac {1}{4}}\right )}}{2 \, b^{\frac {1}{4}}}\right )}{b^{\frac {1}{4}}} - \frac {3 \, \sqrt {2} a^{2} \log \left (\sqrt {2} {\left (a x^{2} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{2} - b} + \sqrt {b}\right )}{b^{\frac {1}{4}}} + \frac {3 \, \sqrt {2} a^{2} \log \left (-\sqrt {2} {\left (a x^{2} - b\right )}^{\frac {1}{4}} b^{\frac {1}{4}} + \sqrt {a x^{2} - b} + \sqrt {b}\right )}{b^{\frac {1}{4}}} - \frac {8 \, {\left (a x^{2} - b\right )}^{\frac {3}{4}} a}{x^{2}}}{16 \, a} \]

[In]

integrate((a*x^2-b)^(3/4)/x^3,x, algorithm="giac")

[Out]

1/16*(6*sqrt(2)*a^2*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4) + 2*(a*x^2 - b)^(1/4))/b^(1/4))/b^(1/4) + 6*sqrt(2)*a^
2*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4) - 2*(a*x^2 - b)^(1/4))/b^(1/4))/b^(1/4) - 3*sqrt(2)*a^2*log(sqrt(2)*(a*
x^2 - b)^(1/4)*b^(1/4) + sqrt(a*x^2 - b) + sqrt(b))/b^(1/4) + 3*sqrt(2)*a^2*log(-sqrt(2)*(a*x^2 - b)^(1/4)*b^(
1/4) + sqrt(a*x^2 - b) + sqrt(b))/b^(1/4) - 8*(a*x^2 - b)^(3/4)*a/x^2)/a

Mupad [B] (verification not implemented)

Time = 6.27 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.46 \[ \int \frac {\left (-b+a x^2\right )^{3/4}}{x^3} \, dx=\frac {3\,a\,\mathrm {atan}\left (\frac {{\left (a\,x^2-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right )}{4\,{\left (-b\right )}^{1/4}}-\frac {{\left (a\,x^2-b\right )}^{3/4}}{2\,x^2}-\frac {3\,a\,\mathrm {atanh}\left (\frac {{\left (a\,x^2-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right )}{4\,{\left (-b\right )}^{1/4}} \]

[In]

int((a*x^2 - b)^(3/4)/x^3,x)

[Out]

(3*a*atan((a*x^2 - b)^(1/4)/(-b)^(1/4)))/(4*(-b)^(1/4)) - (a*x^2 - b)^(3/4)/(2*x^2) - (3*a*atanh((a*x^2 - b)^(
1/4)/(-b)^(1/4)))/(4*(-b)^(1/4))