\(\int \frac {\sqrt {-x+x^4}}{-b+a x^3} \, dx\) [1162]

Optimal result

Integrand size = 23, antiderivative size = 86 \[ \int \frac {\sqrt {-x+x^4}}{-b+a x^3} \, dx=\frac {2 \sqrt {a-b} \arctan \left (\frac {\sqrt {a-b} x \sqrt {-x+x^4}}{\sqrt {b} (-1+x) \left (1+x+x^2\right )}\right )}{3 a \sqrt {b}}+\frac {2 \text {arctanh}\left (\frac {x^2}{\sqrt {-x+x^4}}\right )}{3 a} \] Output:

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

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.29 \[ \int \frac {\sqrt {-x+x^4}}{-b+a x^3} \, dx=\frac {2 \sqrt {x} \sqrt {-1+x^3} \left (\sqrt {a-b} \arctan \left (\frac {b-a x^{3/2} \left (x^{3/2}+\sqrt {-1+x^3}\right )}{\sqrt {a-b} \sqrt {b}}\right )+\sqrt {b} \log \left (x^{3/2}+\sqrt {-1+x^3}\right )\right )}{3 a \sqrt {b} \sqrt {x \left (-1+x^3\right )}} \] Input:

Integrate[Sqrt[-x + x^4]/(-b + a*x^3),x]
 

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.32 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {2467, 25, 966, 965, 301, 224, 219, 291, 218}

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.

\(\displaystyle \int \frac {\sqrt {x^4-x}}{a x^3-b} \, dx\)

\(\Big \downarrow \) 2467

\(\displaystyle \frac {\sqrt {x^4-x} \int -\frac {\sqrt {x} \sqrt {x^3-1}}{b-a x^3}dx}{\sqrt {x} \sqrt {x^3-1}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\sqrt {x^4-x} \int \frac {\sqrt {x} \sqrt {x^3-1}}{b-a x^3}dx}{\sqrt {x} \sqrt {x^3-1}}\)

\(\Big \downarrow \) 966

\(\displaystyle -\frac {2 \sqrt {x^4-x} \int \frac {x \sqrt {x^3-1}}{b-a x^3}d\sqrt {x}}{\sqrt {x} \sqrt {x^3-1}}\)

\(\Big \downarrow \) 965

\(\displaystyle -\frac {2 \sqrt {x^4-x} \int \frac {\sqrt {x-1}}{b-a x}dx^{3/2}}{3 \sqrt {x} \sqrt {x^3-1}}\)

\(\Big \downarrow \) 301

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

\(\displaystyle -\frac {2 \sqrt {x^4-x} \left (-\frac {(a-b) \int \frac {1}{\sqrt {x-1} (b-a x)}dx^{3/2}}{a}-\frac {\text {arctanh}\left (\frac {x^{3/2}}{\sqrt {x-1}}\right )}{a}\right )}{3 \sqrt {x} \sqrt {x^3-1}}\)

\(\Big \downarrow \) 291

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

\(\Big \downarrow \) 218

\(\displaystyle -\frac {2 \sqrt {x^4-x} \left (-\frac {\sqrt {a-b} \arctan \left (\frac {x^{3/2} \sqrt {a-b}}{\sqrt {b} \sqrt {x-1}}\right )}{a \sqrt {b}}-\frac {\text {arctanh}\left (\frac {x^{3/2}}{\sqrt {x-1}}\right )}{a}\right )}{3 \sqrt {x} \sqrt {x^3-1}}\)

Input:

Int[Sqrt[-x + x^4]/(-b + a*x^3),x]
 

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 218

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

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 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 291

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst 
[Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, 
d}, x] && NeQ[b*c - a*d, 0]
 

rule 301

Int[((a_) + (b_.)*(x_)^2)^(p_.)/((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[b/ 
d   Int[(a + b*x^2)^(p - 1), x], x] - Simp[(b*c - a*d)/d   Int[(a + b*x^2)^ 
(p - 1)/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] 
&& GtQ[p, 0] && (EqQ[p, 1/2] || EqQ[Denominator[p], 4] || (EqQ[p, 2/3] && E 
qQ[b*c + 3*a*d, 0]))
 

rule 965

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), 
 x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 
 1)*(a + b*x^(n/k))^p*(c + d*x^(n/k))^q, x], x, x^k], x] /; k != 1] /; Free 
Q[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IntegerQ[m]
 

rule 966

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_) 
)^(q_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k/e   Subst[Int[x^(k*( 
m + 1) - 1)*(a + b*(x^(k*n)/e^n))^p*(c + d*(x^(k*n)/e^n))^q, x], x, (e*x)^( 
1/k)], x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[ 
n, 0] && FractionQ[m] && IntegerQ[p]
 

rule 2467

Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F 
racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p])   Int[x^( 
p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P 
olyQ[Px, x] &&  !IntegerQ[p] &&  !MonomialQ[Px, x] &&  !PolyQ[Fx, x]
 

Maple [A] (verified)

Time = 0.83 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.15

method	result	size
default	\(-\frac {\left (2 a -2 b \right ) \arctan \left (\frac {\sqrt {x^{4}-x}\, b}{x^{2} \sqrt {\left (a -b \right ) b}}\right )+\left (\ln \left (\frac {-x^{2}+\sqrt {x^{4}-x}}{x^{2}}\right )-\ln \left (\frac {x^{2}+\sqrt {x^{4}-x}}{x^{2}}\right )\right ) \sqrt {\left (a -b \right ) b}}{3 \sqrt {\left (a -b \right ) b}\, a}\)	\(99\)
pseudoelliptic	\(-\frac {\left (2 a -2 b \right ) \arctan \left (\frac {\sqrt {x^{4}-x}\, b}{x^{2} \sqrt {\left (a -b \right ) b}}\right )+\left (\ln \left (\frac {-x^{2}+\sqrt {x^{4}-x}}{x^{2}}\right )-\ln \left (\frac {x^{2}+\sqrt {x^{4}-x}}{x^{2}}\right )\right ) \sqrt {\left (a -b \right ) b}}{3 \sqrt {\left (a -b \right ) b}\, a}\)	\(99\)
elliptic	\(\text {Expression too large to display}\)	\(636\)

Input:

int((x^4-x)^(1/2)/(a*x^3-b),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 219, normalized size of antiderivative = 2.55 \[ \int \frac {\sqrt {-x+x^4}}{-b+a x^3} \, dx=\left [\frac {\sqrt {-\frac {a - b}{b}} \log \left (-\frac {{\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} x^{6} + 2 \, {\left (3 \, a b - 4 \, b^{2}\right )} x^{3} + b^{2} + 4 \, {\left ({\left (a b - 2 \, b^{2}\right )} x^{4} + b^{2} x\right )} \sqrt {x^{4} - x} \sqrt {-\frac {a - b}{b}}}{a^{2} x^{6} - 2 \, a b x^{3} + b^{2}}\right ) + 2 \, \log \left (-2 \, x^{3} - 2 \, \sqrt {x^{4} - x} x + 1\right )}{6 \, a}, \frac {\sqrt {\frac {a - b}{b}} \arctan \left (-\frac {2 \, \sqrt {x^{4} - x} b x \sqrt {\frac {a - b}{b}}}{{\left (a - 2 \, b\right )} x^{3} + b}\right ) + \log \left (-2 \, x^{3} - 2 \, \sqrt {x^{4} - x} x + 1\right )}{3 \, a}\right ] \] Input:

integrate((x^4-x)^(1/2)/(a*x^3-b),x, algorithm="fricas")
 

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

\[ \int \frac {\sqrt {-x+x^4}}{-b+a x^3} \, dx=\int { \frac {\sqrt {x^{4} - x}}{a x^{3} - b} \,d x } \] Input:

integrate((x^4-x)^(1/2)/(a*x^3-b),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.93 \[ \int \frac {\sqrt {-x+x^4}}{-b+a x^3} \, dx=-\frac {2 \, {\left (a - b\right )} \arctan \left (\frac {b \sqrt {-\frac {1}{x^{3}} + 1}}{\sqrt {a b - b^{2}}}\right )}{3 \, \sqrt {a b - b^{2}} a} + \frac {\log \left (\sqrt {-\frac {1}{x^{3}} + 1} + 1\right )}{3 \, a} - \frac {\log \left ({\left | \sqrt {-\frac {1}{x^{3}} + 1} - 1 \right |}\right )}{3 \, a} \] Input:

integrate((x^4-x)^(1/2)/(a*x^3-b),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {-x+x^4}}{-b+a x^3} \, dx=-\int \frac {\sqrt {x^4-x}}{b-a\,x^3} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt {-x+x^4}}{-b+a x^3} \, dx=\int \frac {\sqrt {x}\, \sqrt {x^{3}-1}}{a \,x^{3}-b}d x \] Input:

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

Output:

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