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

Optimal result

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

-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))
 

Mathematica [A] (verified)

Time = 4.25 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.18 \[ \int \frac {\sqrt [4]{-b x+a x^4}}{x^2} \, dx=-\frac {2 \sqrt [4]{-b x+a x^4} \left (2 \sqrt [4]{-b+a x^3}+\sqrt [4]{a} x^{3/4} \arctan \left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{-b+a x^3}}\right )-\sqrt [4]{a} x^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{-b+a x^3}}\right )\right )}{3 x \sqrt [4]{-b+a x^3}} \] Input:

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

Output:

(-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))
 

Rubi [A] (warning: unable to verify)

Time = 0.30 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.19, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {1926, 1938, 851, 807, 854, 827, 216, 219}

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.

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

\(\Big \downarrow \) 1926

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

\(\Big \downarrow \) 1938

\(\displaystyle \frac {a x^{3/4} \left (a x^3-b\right )^{3/4} \int \frac {x^{5/4}}{\left (a x^3-b\right )^{3/4}}dx}{\left (a x^4-b x\right )^{3/4}}-\frac {4 \sqrt [4]{a x^4-b x}}{3 x}\)

\(\Big \downarrow \) 851

\(\displaystyle \frac {4 a x^{3/4} \left (a x^3-b\right )^{3/4} \int \frac {x^2}{\left (a x^3-b\right )^{3/4}}d\sqrt [4]{x}}{\left (a x^4-b x\right )^{3/4}}-\frac {4 \sqrt [4]{a x^4-b x}}{3 x}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {4 a x^{3/4} \left (a x^3-b\right )^{3/4} \int \frac {\sqrt {x}}{(a x-b)^{3/4}}dx^{3/4}}{3 \left (a x^4-b x\right )^{3/4}}-\frac {4 \sqrt [4]{a x^4-b x}}{3 x}\)

\(\Big \downarrow \) 854

\(\displaystyle \frac {4 a x^{3/4} \left (a x^3-b\right )^{3/4} \int \frac {\sqrt {x}}{1-a x}d\frac {x^{3/4}}{\sqrt [4]{a x-b}}}{3 \left (a x^4-b x\right )^{3/4}}-\frac {4 \sqrt [4]{a x^4-b x}}{3 x}\)

\(\Big \downarrow \) 827

\(\displaystyle \frac {4 a x^{3/4} \left (a x^3-b\right )^{3/4} \left (\frac {\int \frac {1}{1-\sqrt {a} \sqrt {x}}d\frac {x^{3/4}}{\sqrt [4]{a x-b}}}{2 \sqrt {a}}-\frac {\int \frac {1}{\sqrt {a} \sqrt {x}+1}d\frac {x^{3/4}}{\sqrt [4]{a x-b}}}{2 \sqrt {a}}\right )}{3 \left (a x^4-b x\right )^{3/4}}-\frac {4 \sqrt [4]{a x^4-b x}}{3 x}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {4 a x^{3/4} \left (a x^3-b\right )^{3/4} \left (\frac {\int \frac {1}{1-\sqrt {a} \sqrt {x}}d\frac {x^{3/4}}{\sqrt [4]{a x-b}}}{2 \sqrt {a}}-\frac {\arctan \left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x-b}}\right )}{2 a^{3/4}}\right )}{3 \left (a x^4-b x\right )^{3/4}}-\frac {4 \sqrt [4]{a x^4-b x}}{3 x}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {4 a x^{3/4} \left (a x^3-b\right )^{3/4} \left (\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x-b}}\right )}{2 a^{3/4}}-\frac {\arctan \left (\frac {\sqrt [4]{a} x^{3/4}}{\sqrt [4]{a x-b}}\right )}{2 a^{3/4}}\right )}{3 \left (a x^4-b x\right )^{3/4}}-\frac {4 \sqrt [4]{a x^4-b x}}{3 x}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 216

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

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[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 827

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

rule 851

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = 
 Denominator[m]}, Simp[k/c   Subst[Int[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 854

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[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 1926

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] - Simp[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] && (Integer 
sQ[j, n] || GtQ[c, 0]) && GtQ[p, 0] && LtQ[m + j*p + 1, 0]
 

rule 1938

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol 
] :> Simp[c^IntPart[m]*(c*x)^FracPart[m]*((a*x^j + b*x^n)^FracPart[p]/(x^(F 
racPart[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] &&  !Inte 
gerQ[p] && NeQ[n, j] && PosQ[n - j]
 

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.97

method	result	size
pseudoelliptic	\(\frac {\ln \left (\frac {-a^{\frac {1}{4}} x -{\left (x \left (a \,x^{3}-b \right )\right )}^{\frac {1}{4}}}{a^{\frac {1}{4}} x -{\left (x \left (a \,x^{3}-b \right )\right )}^{\frac {1}{4}}}\right ) a^{\frac {1}{4}} x +2 \arctan \left (\frac {{\left (x \left (a \,x^{3}-b \right )\right )}^{\frac {1}{4}}}{x \,a^{\frac {1}{4}}}\right ) a^{\frac {1}{4}} x -4 {\left (x \left (a \,x^{3}-b \right )\right )}^{\frac {1}{4}}}{3 x}\)	\(101\)

Input:

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

Output:

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

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

Output:

Timed out
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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 = 0.20 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.85 \[ \int \frac {\sqrt [4]{-b x+a x^4}}{x^2} \, dx=\frac {1}{3} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} + 2 \, {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + \frac {1}{3} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} - 2 \, {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-a\right )^{\frac {1}{4}}}\right ) + \frac {1}{6} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \log \left (\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x^{3}}}\right ) - \frac {1}{6} \, \sqrt {2} \left (-a\right )^{\frac {1}{4}} \log \left (-\sqrt {2} \left (-a\right )^{\frac {1}{4}} {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}} + \sqrt {-a} + \sqrt {a - \frac {b}{x^{3}}}\right ) - \frac {4}{3} \, {\left (a - \frac {b}{x^{3}}\right )}^{\frac {1}{4}} \] Input:

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

Output:

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt [4]{-b x+a x^4}}{x^2} \, dx=\int \frac {\left (a \,x^{3}-b \right )^{\frac {1}{4}}}{x^{\frac {7}{4}}}d x \] Input:

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

Output:

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