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

Optimal. Leaf size=104 \[ -\frac {4 \sqrt [4]{a x^4-b x}}{3 x}-\frac {2}{3} \sqrt [4]{a} \tan ^{-1}\left (\frac {\sqrt [4]{a} \left (a x^4-b x\right )^{3/4}}{a x^3-b}\right )+\frac {2}{3} \sqrt [4]{a} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \left (a x^4-b x\right )^{3/4}}{a x^3-b}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.17, antiderivative size = 154, normalized size of antiderivative = 1.48, number of steps used = 8, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {2020, 2032, 329, 275, 331, 298, 203, 206} \begin {gather*} -\frac {4 \sqrt [4]{a x^4-b x}}{3 x}-\frac {2 \sqrt [4]{a} x^{3/4} \left (a x^3-b\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3-b}}\right )}{3 \left (a x^4-b x\right )^{3/4}}+\frac {2 \sqrt [4]{a} x^{3/4} \left (a x^3-b\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3-b}}\right )}{3 \left (a x^4-b x\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-(b*x) + a*x^4)^(1/4)/x^2,x]

[Out]

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

Rule 203

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

Rule 206

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

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 298

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

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

Rule 2020

Int[((c_.)*(x_))^(m_)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a*x^j + b*
x^n)^p)/(c*(m + j*p + 1)), x] - Dist[(b*p*(n - j))/(c^n*(m + j*p + 1)), Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p -
1), x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[p
, 0] && LtQ[m + j*p + 1, 0]

Rule 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rubi steps

\begin {align*} \int \frac {\sqrt [4]{-b x+a x^4}}{x^2} \, dx &=-\frac {4 \sqrt [4]{-b x+a x^4}}{3 x}+a \int \frac {x^2}{\left (-b x+a x^4\right )^{3/4}} \, dx\\ &=-\frac {4 \sqrt [4]{-b x+a x^4}}{3 x}+\frac {\left (a x^{3/4} \left (-b+a x^3\right )^{3/4}\right ) \int \frac {x^{5/4}}{\left (-b+a x^3\right )^{3/4}} \, dx}{\left (-b x+a x^4\right )^{3/4}}\\ &=-\frac {4 \sqrt [4]{-b x+a x^4}}{3 x}+\frac {\left (4 a x^{3/4} \left (-b+a x^3\right )^{3/4}\right ) \operatorname {Subst}\left (\int \frac {x^8}{\left (-b+a x^{12}\right )^{3/4}} \, dx,x,\sqrt [4]{x}\right )}{\left (-b x+a x^4\right )^{3/4}}\\ &=-\frac {4 \sqrt [4]{-b x+a x^4}}{3 x}+\frac {\left (4 a x^{3/4} \left (-b+a x^3\right )^{3/4}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\left (-b+a x^4\right )^{3/4}} \, dx,x,x^{3/4}\right )}{3 \left (-b x+a x^4\right )^{3/4}}\\ &=-\frac {4 \sqrt [4]{-b x+a x^4}}{3 x}+\frac {\left (4 a x^{3/4} \left (-b+a x^3\right )^{3/4}\right ) \operatorname {Subst}\left (\int \frac {x^2}{1-a x^4} \, dx,x,\frac {x^{3/4}}{\sqrt [4]{-b+a x^3}}\right )}{3 \left (-b x+a x^4\right )^{3/4}}\\ &=-\frac {4 \sqrt [4]{-b x+a x^4}}{3 x}+\frac {\left (2 \sqrt {a} x^{3/4} \left (-b+a x^3\right )^{3/4}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {x^{3/4}}{\sqrt [4]{-b+a x^3}}\right )}{3 \left (-b x+a x^4\right )^{3/4}}-\frac {\left (2 \sqrt {a} x^{3/4} \left (-b+a x^3\right )^{3/4}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {x^{3/4}}{\sqrt [4]{-b+a x^3}}\right )}{3 \left (-b x+a x^4\right )^{3/4}}\\ &=-\frac {4 \sqrt [4]{-b x+a x^4}}{3 x}-\frac {2 \sqrt [4]{a} x^{3/4} \left (-b+a x^3\right )^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{-b+a x^3}}\right )}{3 \left (-b x+a x^4\right )^{3/4}}+\frac {2 \sqrt [4]{a} x^{3/4} \left (-b+a x^3\right )^{3/4} \tanh ^{-1}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{-b+a x^3}}\right )}{3 \left (-b x+a x^4\right )^{3/4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.02, size = 54, normalized size = 0.52 \begin {gather*} -\frac {4 \sqrt [4]{a x^4-b x} \, _2F_1\left (-\frac {1}{4},-\frac {1}{4};\frac {3}{4};\frac {a x^3}{b}\right )}{3 x \sqrt [4]{1-\frac {a x^3}{b}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(-(b*x) + a*x^4)^(1/4)*Hypergeometric2F1[-1/4, -1/4, 3/4, (a*x^3)/b])/(3*x*(1 - (a*x^3)/b)^(1/4))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.31, size = 104, normalized size = 1.00 \begin {gather*} -\frac {4 \sqrt [4]{-b x+a x^4}}{3 x}-\frac {2}{3} \sqrt [4]{a} \tan ^{-1}\left (\frac {\sqrt [4]{a} \left (-b x+a x^4\right )^{3/4}}{-b+a x^3}\right )+\frac {2}{3} \sqrt [4]{a} \tanh ^{-1}\left (\frac {\sqrt [4]{a} \left (-b x+a x^4\right )^{3/4}}{-b+a x^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

giac [B]  time = 0.32, size = 192, normalized size = 1.85 \begin {gather*} \frac {1}{3} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + \frac {1}{3} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + \frac {1}{6} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x^{3}}}\right ) - \frac {1}{6} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x^{3}}}\right ) - \frac {4}{3} \, {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (a \,x^{4}-b x \right )^{\frac {1}{4}}}{x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (a x^{4} - b x\right )}^{\frac {1}{4}}}{x^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*x^4 - b*x)^(1/4)/x^2, x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a\,x^4-b\,x\right )}^{1/4}}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x^4 - b*x)^(1/4)/x^2,x)

[Out]

int((a*x^4 - b*x)^(1/4)/x^2, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt [4]{x \left (a x^{3} - b\right )}}{x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x*(a*x**3 - b))**(1/4)/x**2, x)

________________________________________________________________________________________