3.11.71 \(\int \frac {-2 a b+(a+b) x}{\sqrt [4]{x^2 (-a+x) (-b+x)} (a b d-(a+b) d x+(-1+d) x^2)} \, dx\) [1071]

3.11.71.1 Optimal result

Integrand size = 50, antiderivative size = 81 \[ \int \frac {-2 a b+(a+b) x}{\sqrt [4]{x^2 (-a+x) (-b+x)} \left (a b d-(a+b) d x+(-1+d) x^2\right )} \, dx=-\frac {2 \arctan \left (\frac {x}{\sqrt [4]{d} \sqrt [4]{a b x^2+(-a-b) x^3+x^4}}\right )}{d^{3/4}}-\frac {2 \text {arctanh}\left (\frac {x}{\sqrt [4]{d} \sqrt [4]{a b x^2+(-a-b) x^3+x^4}}\right )}{d^{3/4}} \]

output

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

3.11.71.2 Mathematica [A] (verified)

Time = 15.28 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.74 \[ \int \frac {-2 a b+(a+b) x}{\sqrt [4]{x^2 (-a+x) (-b+x)} \left (a b d-(a+b) d x+(-1+d) x^2\right )} \, dx=-\frac {2 \left (\arctan \left (\frac {x}{\sqrt [4]{d} \sqrt [4]{x^2 (-a+x) (-b+x)}}\right )+\text {arctanh}\left (\frac {x}{\sqrt [4]{d} \sqrt [4]{x^2 (-a+x) (-b+x)}}\right )\right )}{d^{3/4}} \]

input

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

output

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

3.11.71.3 Rubi [F]

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.

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

\(\Big \downarrow \) 2467

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2035

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

\(\Big \downarrow \) 7279

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

\(\Big \downarrow \) 2009

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

input

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

output

$Aborted
 

3.11.71.3.1 Defintions of rubi rules used

rule 25

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

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2035

Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k   Subst 
[Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti 
onQ[m] && AlgebraicFunctionQ[Fx, x]
 

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]
 

rule 7279

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ 
{v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su 
mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
 

3.11.71.4 Maple [A] (verified)

Time = 3.63 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.21

method	result	size
pseudoelliptic	\(\frac {2 \arctan \left (\frac {\left (x^{2} \left (a -x \right ) \left (b -x \right )\right )^{\frac {1}{4}}}{x \left (\frac {1}{d}\right )^{\frac {1}{4}}}\right )-\ln \left (\frac {-\left (\frac {1}{d}\right )^{\frac {1}{4}} x -\left (x^{2} \left (a -x \right ) \left (b -x \right )\right )^{\frac {1}{4}}}{\left (\frac {1}{d}\right )^{\frac {1}{4}} x -\left (x^{2} \left (a -x \right ) \left (b -x \right )\right )^{\frac {1}{4}}}\right )}{\left (\frac {1}{d}\right )^{\frac {1}{4}} d}\)	\(98\)

input

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

output

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

3.11.71.5 Fricas [F(-1)]

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

input

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

output

Timed out
 

3.11.71.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.11.71.7 Maxima [F]

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

input

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

output

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

3.11.71.8 Giac [F]

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

input

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

output

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

3.11.71.9 Mupad [F(-1)]

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

input

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

output

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