\(\int \frac {\sqrt [4]{-b x+a x^4}}{x^2} \, dx\) [1479]

Optimal result

Integrand size = 18, antiderivative size = 104 \[ \int \frac {\sqrt [4]{-b x+a x^4}}{x^2} \, dx=-\frac {4 \sqrt [4]{-b x+a x^4}}{3 x}-\frac {2}{3} \sqrt [4]{a} \arctan \left (\frac {\sqrt [4]{a} \left (-b x+a x^4\right )^{3/4}}{-b+a x^3}\right )+\frac {2}{3} \sqrt [4]{a} \text {arctanh}\left (\frac {\sqrt [4]{a} \left (-b x+a x^4\right )^{3/4}}{-b+a x^3}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.48, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {2045, 2057, 335, 281, 338, 304, 209, 212} \[ \int \frac {\sqrt [4]{-b x+a x^4}}{x^2} \, dx=-\frac {2 \sqrt [4]{a} x^{3/4} \left (a x^3-b\right )^{3/4} \arctan \left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3-b}}\right )}{3 \left (a x^4-b x\right )^{3/4}}+\frac {2 \sqrt [4]{a} x^{3/4} \left (a x^3-b\right )^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x^3-b}}\right )}{3 \left (a x^4-b x\right )^{3/4}}-\frac {4 \sqrt [4]{a x^4-b x}}{3 x} \]

[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 209

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

Rule 212

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

Rule 281

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 304

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 335

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 338

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 2045

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 2057

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[c^IntPart[m]*(c*x)^FracPa
rt[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*} \text {integral}& = -\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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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} \arctan \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} \text {arctanh}\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 [A] (verified)

Time = 4.26 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.18 \[ \int \frac {\sqrt [4]{-b x+a x^4}}{x^2} \, dx=-\frac {2 \sqrt [4]{-b x+a x^4} \left (2 \sqrt [4]{-b+a x^3}+\sqrt [4]{a} x^{3/4} \arctan \left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{-b+a x^3}}\right )-\sqrt [4]{a} x^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{-b+a x^3}}\right )\right )}{3 x \sqrt [4]{-b+a x^3}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.93

method	result	size
pseudoelliptic	\(\frac {\ln \left (\frac {a^{\frac {1}{4}} x +{\left (x \left (a \,x^{3}-b \right )\right )}^{\frac {1}{4}}}{-a^{\frac {1}{4}} x +{\left (x \left (a \,x^{3}-b \right )\right )}^{\frac {1}{4}}}\right ) a^{\frac {1}{4}} x +2 \arctan \left (\frac {{\left (x \left (a \,x^{3}-b \right )\right )}^{\frac {1}{4}}}{x \,a^{\frac {1}{4}}}\right ) a^{\frac {1}{4}} x -4 {\left (x \left (a \,x^{3}-b \right )\right )}^{\frac {1}{4}}}{3 x}\)	\(97\)

[In]

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

[Out]

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

Fricas [F(-1)]

Timed out. \[ \int \frac {\sqrt [4]{-b x+a x^4}}{x^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F]

\[ \int \frac {\sqrt [4]{-b x+a x^4}}{x^2} \, dx=\int \frac {\sqrt [4]{x \left (a x^{3} - b\right )}}{x^{2}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\sqrt [4]{-b x+a x^4}}{x^2} \, dx=\int { \frac {{\left (a x^{4} - b x\right )}^{\frac {1}{4}}}{x^{2}} \,d x } \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 192 vs. \(2 (84) = 168\).

Time = 0.27 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.85 \[ \int \frac {\sqrt [4]{-b x+a x^4}}{x^2} \, dx=\frac {1}{3} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + \frac {1}{3} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + \frac {1}{6} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x^{3}}}\right ) - \frac {1}{6} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x^{3}}}\right ) - \frac {4}{3} \, {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}} \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [4]{-b x+a x^4}}{x^2} \, dx=\int \frac {{\left (a\,x^4-b\,x\right )}^{1/4}}{x^2} \,d x \]

[In]

int((a*x^4 - b*x)^(1/4)/x^2,x)

[Out]

int((a*x^4 - b*x)^(1/4)/x^2, x)