3.20.54 \(\int \frac {\sqrt [4]{-b+a x^2}}{x} \, dx\)

Optimal. Leaf size=138 \[ 2 \sqrt [4]{a x^2-b}+\frac {\sqrt [4]{b} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^2-b}}{\sqrt {a x^2-b}-\sqrt {b}}\right )}{\sqrt {2}}-\frac {\sqrt [4]{b} \tanh ^{-1}\left (\frac {\frac {\sqrt {a x^2-b}}{\sqrt {2} \sqrt [4]{b}}+\frac {\sqrt [4]{b}}{\sqrt {2}}}{\sqrt [4]{a x^2-b}}\right )}{\sqrt {2}} \]

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Rubi [A]  time = 0.21, antiderivative size = 211, normalized size of antiderivative = 1.53, number of steps used = 12, number of rules used = 9, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {266, 50, 63, 211, 1165, 628, 1162, 617, 204} \begin {gather*} 2 \sqrt [4]{a x^2-b}+\frac {\sqrt [4]{b} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^2-b}+\sqrt {a x^2-b}+\sqrt {b}\right )}{2 \sqrt {2}}-\frac {\sqrt [4]{b} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^2-b}+\sqrt {a x^2-b}+\sqrt {b}\right )}{2 \sqrt {2}}+\frac {\sqrt [4]{b} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{a x^2-b}}{\sqrt [4]{b}}\right )}{\sqrt {2}}-\frac {\sqrt [4]{b} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a x^2-b}}{\sqrt [4]{b}}+1\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-b + a*x^2)^(1/4)/x,x]

[Out]

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

Rule 50

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] + Dist[(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] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[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
]]))

Rule 266

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 617

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 628

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 1162

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 1165

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*} \int \frac {\sqrt [4]{-b+a x^2}}{x} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {\sqrt [4]{-b+a x}}{x} \, dx,x,x^2\right )\\ &=2 \sqrt [4]{-b+a x^2}-\frac {1}{2} b \operatorname {Subst}\left (\int \frac {1}{x (-b+a x)^{3/4}} \, dx,x,x^2\right )\\ &=2 \sqrt [4]{-b+a x^2}-\frac {(2 b) \operatorname {Subst}\left (\int \frac {1}{\frac {b}{a}+\frac {x^4}{a}} \, dx,x,\sqrt [4]{-b+a x^2}\right )}{a}\\ &=2 \sqrt [4]{-b+a x^2}-\frac {\sqrt {b} \operatorname {Subst}\left (\int \frac {\sqrt {b}-x^2}{\frac {b}{a}+\frac {x^4}{a}} \, dx,x,\sqrt [4]{-b+a x^2}\right )}{a}-\frac {\sqrt {b} \operatorname {Subst}\left (\int \frac {\sqrt {b}+x^2}{\frac {b}{a}+\frac {x^4}{a}} \, dx,x,\sqrt [4]{-b+a x^2}\right )}{a}\\ &=2 \sqrt [4]{-b+a x^2}+\frac {\sqrt [4]{b} \operatorname {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 )}{2 \sqrt {2}}+\frac {\sqrt [4]{b} \operatorname {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 )}{2 \sqrt {2}}-\frac {1}{2} \sqrt {b} \operatorname {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}{2} \sqrt {b} \operatorname {Subst}\left (\int \frac {1}{\sqrt {b}+\sqrt {2} \sqrt [4]{b} x+x^2} \, dx,x,\sqrt [4]{-b+a x^2}\right )\\ &=2 \sqrt [4]{-b+a x^2}+\frac {\sqrt [4]{b} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^2}+\sqrt {-b+a x^2}\right )}{2 \sqrt {2}}-\frac {\sqrt [4]{b} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^2}+\sqrt {-b+a x^2}\right )}{2 \sqrt {2}}-\frac {\sqrt [4]{b} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{-b+a x^2}}{\sqrt [4]{b}}\right )}{\sqrt {2}}+\frac {\sqrt [4]{b} \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{-b+a x^2}}{\sqrt [4]{b}}\right )}{\sqrt {2}}\\ &=2 \sqrt [4]{-b+a x^2}+\frac {\sqrt [4]{b} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{-b+a x^2}}{\sqrt [4]{b}}\right )}{\sqrt {2}}-\frac {\sqrt [4]{b} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{-b+a x^2}}{\sqrt [4]{b}}\right )}{\sqrt {2}}+\frac {\sqrt [4]{b} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^2}+\sqrt {-b+a x^2}\right )}{2 \sqrt {2}}-\frac {\sqrt [4]{b} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^2}+\sqrt {-b+a x^2}\right )}{2 \sqrt {2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 211, normalized size = 1.53 \begin {gather*} 2 \sqrt [4]{a x^2-b}+\frac {\sqrt [4]{b} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^2-b}+\sqrt {a x^2-b}+\sqrt {b}\right )}{2 \sqrt {2}}-\frac {\sqrt [4]{b} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{a x^2-b}+\sqrt {a x^2-b}+\sqrt {b}\right )}{2 \sqrt {2}}+\frac {\sqrt [4]{b} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{a x^2-b}}{\sqrt [4]{b}}\right )}{\sqrt {2}}-\frac {\sqrt [4]{b} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a x^2-b}}{\sqrt [4]{b}}+1\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-b + a*x^2)^(1/4)/x,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.19, size = 138, normalized size = 1.00 \begin {gather*} 2 \sqrt [4]{-b+a x^2}-\frac {\sqrt [4]{b} \tan ^{-1}\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 )}{\sqrt {2}}-\frac {\sqrt [4]{b} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{-b+a x^2}}{\sqrt {b}+\sqrt {-b+a x^2}}\right )}{\sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(-b + a*x^2)^(1/4)/x,x]

[Out]

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

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fricas [A]  time = 0.49, size = 122, normalized size = 0.88 \begin {gather*} -2 \, \left (-b\right )^{\frac {1}{4}} \arctan \left (\frac {\left (-b\right )^{\frac {3}{4}} \sqrt {\sqrt {a x^{2} - b} + \sqrt {-b}} - {\left (a x^{2} - b\right )}^{\frac {1}{4}} \left (-b\right )^{\frac {3}{4}}}{b}\right ) - \frac {1}{2} \, \left (-b\right )^{\frac {1}{4}} \log \left ({\left (a x^{2} - b\right )}^{\frac {1}{4}} + \left (-b\right )^{\frac {1}{4}}\right ) + \frac {1}{2} \, \left (-b\right )^{\frac {1}{4}} \log \left ({\left (a x^{2} - b\right )}^{\frac {1}{4}} - \left (-b\right )^{\frac {1}{4}}\right ) + 2 \, {\left (a x^{2} - b\right )}^{\frac {1}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^2-b)^(1/4)/x,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.27, size = 175, normalized size = 1.27 \begin {gather*} -\frac {1}{2} \, \sqrt {2} b^{\frac {1}{4}} \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 ) - \frac {1}{2} \, \sqrt {2} b^{\frac {1}{4}} \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 ) - \frac {1}{4} \, \sqrt {2} b^{\frac {1}{4}} \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 ) + \frac {1}{4} \, \sqrt {2} b^{\frac {1}{4}} \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 ) + 2 \, {\left (a x^{2} - b\right )}^{\frac {1}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^2-b)^(1/4)/x,x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (a \,x^{2}-b \right )^{\frac {1}{4}}}{x}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^2-b)^(1/4)/x,x)

[Out]

int((a*x^2-b)^(1/4)/x,x)

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maxima [A]  time = 0.42, size = 175, normalized size = 1.27 \begin {gather*} -\frac {1}{2} \, \sqrt {2} b^{\frac {1}{4}} \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 ) - \frac {1}{2} \, \sqrt {2} b^{\frac {1}{4}} \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 ) - \frac {1}{4} \, \sqrt {2} b^{\frac {1}{4}} \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 ) + \frac {1}{4} \, \sqrt {2} b^{\frac {1}{4}} \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 ) + 2 \, {\left (a x^{2} - b\right )}^{\frac {1}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x^2-b)^(1/4)/x,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.03, size = 64, normalized size = 0.46 \begin {gather*} 2\,{\left (a\,x^2-b\right )}^{1/4}-{\left (-b\right )}^{1/4}\,\mathrm {atan}\left (\frac {{\left (a\,x^2-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right )-{\left (-b\right )}^{1/4}\,\mathrm {atanh}\left (\frac {{\left (a\,x^2-b\right )}^{1/4}}{{\left (-b\right )}^{1/4}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^2 - b)^(1/4)/x,x)

[Out]

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

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sympy [C]  time = 0.96, size = 48, normalized size = 0.35 \begin {gather*} - \frac {\sqrt [4]{a} \sqrt {x} \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^{2}}} \right )}}{2 \Gamma \left (\frac {3}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*x**2-b)**(1/4)/x,x)

[Out]

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

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