3.17.39 \(\int \frac {x^3 (3-2 (1+k) x+k x^2)}{(-1+x) \sqrt [4]{(1-x) x (1-k x)} (-1+k x) (-d+d (1+k) x-d k x^2+x^3)} \, dx\) [1639]

3.17.39.1 Optimal result

Integrand size = 68, antiderivative size = 111 \[ \int \frac {x^3 \left (3-2 (1+k) x+k x^2\right )}{(-1+x) \sqrt [4]{(1-x) x (1-k x)} (-1+k x) \left (-d+d (1+k) x-d k x^2+x^3\right )} \, dx=\frac {4 \left (x-x^2-k x^2+k x^3\right )^{3/4}}{(-1+x) (-1+k x)}+2 \sqrt [4]{d} \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{x+(-1-k) x^2+k x^3}}{x}\right )-2 \sqrt [4]{d} \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{x+(-1-k) x^2+k x^3}}{x}\right ) \]

output

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

3.17.39.2 Mathematica [F]

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

input

Integrate[(x^3*(3 - 2*(1 + k)*x + k*x^2))/((-1 + x)*((1 - x)*x*(1 - k*x))^ 
(1/4)*(-1 + k*x)*(-d + d*(1 + k)*x - d*k*x^2 + x^3)),x]
 

output

Integrate[(x^3*(3 - 2*(1 + k)*x + k*x^2))/((-1 + x)*((1 - x)*x*(1 - k*x))^ 
(1/4)*(-1 + k*x)*(-d + d*(1 + k)*x - d*k*x^2 + x^3)), x]
 

3.17.39.3 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.

\(\displaystyle \int \frac {x^3 \left (k x^2-2 (k+1) x+3\right )}{(x-1) \sqrt [4]{(1-x) x (1-k x)} (k x-1) \left (-d k x^2+d (k+1) x-d+x^3\right )} \, dx\)

\(\Big \downarrow \) 2467

\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{k x^2-(k+1) x+1} \int -\frac {x^{11/4} \left (k x^2-2 (k+1) x+3\right )}{(1-x) (1-k x) \sqrt [4]{k x^2-(k+1) x+1} \left (-x^3+d k x^2-d (k+1) x+d\right )}dx}{\sqrt [4]{(1-x) x (1-k x)}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\sqrt [4]{x} \sqrt [4]{k x^2-(k+1) x+1} \int \frac {x^{11/4} \left (k x^2-2 (k+1) x+3\right )}{(1-x) (1-k x) \sqrt [4]{k x^2-(k+1) x+1} \left (-x^3+d k x^2-d (k+1) x+d\right )}dx}{\sqrt [4]{(1-x) x (1-k x)}}\)

\(\Big \downarrow \) 2035

\(\displaystyle -\frac {4 \sqrt [4]{x} \sqrt [4]{k x^2-(k+1) x+1} \int \frac {x^{7/2} \left (k x^2-2 (k+1) x+3\right )}{(1-x) (1-k x) \sqrt [4]{k x^2-(k+1) x+1} \left (-x^3+d k x^2-d (k+1) x+d\right )}d\sqrt [4]{x}}{\sqrt [4]{(1-x) x (1-k x)}}\)

\(\Big \downarrow \) 1395

\(\displaystyle -\frac {4 \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x} \int \frac {x^{7/2} \left (k x^2-2 (k+1) x+3\right )}{(1-x)^{5/4} (1-k x)^{5/4} \left (-x^3+d k x^2-d (k+1) x+d\right )}d\sqrt [4]{x}}{\sqrt [4]{(1-x) x (1-k x)}}\)

\(\Big \downarrow \) 7293

\(\displaystyle -\frac {4 \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x} \int \left (-\frac {k x^{5/2}}{(1-x)^{5/4} (1-k x)^{5/4}}+\frac {\left (-d k^2+2 k+2\right ) x^{3/2}}{(1-x)^{5/4} (1-k x)^{5/4}}+\frac {\left (d \left (d^2 k^4-4 d k^3+(2-4 d) k^2+8 k+2\right ) x^2-d \left (d^2 k^4-(3-d) d k^3-7 d k^2+(5-3 d) k+5\right ) x+d \left (d^2 k^3-3 d (k+1) k+3\right )\right ) \sqrt {x}}{(1-x)^{5/4} (1-k x)^{5/4} \left (-x^3+d k x^2-d (k+1) x+d\right )}-\frac {\left (d^2 k^3-3 d (k+1) k+3\right ) \sqrt {x}}{(1-x)^{5/4} (1-k x)^{5/4}}\right )d\sqrt [4]{x}}{\sqrt [4]{(1-x) x (1-k x)}}\)

\(\Big \downarrow \) 7239

\(\displaystyle -\frac {4 \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x} \int \frac {x^{7/2} \left (-k x^2+2 (k+1) x-3\right )}{(1-x)^{5/4} (1-k x)^{5/4} \left (x^3-d (x-1) (k x-1)\right )}d\sqrt [4]{x}}{\sqrt [4]{(1-x) x (1-k x)}}\)

\(\Big \downarrow \) 7293

\(\displaystyle -\frac {4 \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x} \int \left (-\frac {k x^{5/2}}{(1-x)^{5/4} (1-k x)^{5/4}}+\frac {\left (-d k^2+2 k+2\right ) x^{3/2}}{(1-x)^{5/4} (1-k x)^{5/4}}+\frac {\left (-d \left (d^2 k^4-4 d k^3+(2-4 d) k^2+8 k+2\right ) x^2+d \left (d^2 k^4-(3-d) d k^3-7 d k^2+(5-3 d) k+5\right ) x-d \left (d^2 k^3-3 d (k+1) k+3\right )\right ) \sqrt {x}}{(1-x)^{5/4} (1-k x)^{5/4} \left (x^3-d (x-1) (k x-1)\right )}-\frac {\left (d^2 k^3-3 d (k+1) k+3\right ) \sqrt {x}}{(1-x)^{5/4} (1-k x)^{5/4}}\right )d\sqrt [4]{x}}{\sqrt [4]{(1-x) x (1-k x)}}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {4 \sqrt [4]{1-x} \sqrt [4]{x} \sqrt [4]{1-k x} \left (d \left (d^2 k^3-3 d (k+1) k+3\right ) \int \frac {\sqrt {x}}{(1-x)^{5/4} (1-k x)^{5/4} \left (-x^3+d k x^2-d (k+1) x+d\right )}d\sqrt [4]{x}-d \left (d^2 k^4-(3-d) d k^3-7 d k^2+(5-3 d) k+5\right ) \int \frac {x^{3/2}}{(1-x)^{5/4} (1-k x)^{5/4} \left (-x^3+d k x^2-d (k+1) x+d\right )}d\sqrt [4]{x}+d \left (d^2 k^4-4 d k^3+(2-4 d) k^2+8 k+2\right ) \int \frac {x^{5/2}}{(1-x)^{5/4} (1-k x)^{5/4} \left (-x^3+d k x^2-d (k+1) x+d\right )}d\sqrt [4]{x}-\frac {1}{3} x^{3/4} \left (d^2 k^3-3 d (k+1) k+3\right ) \operatorname {AppellF1}\left (\frac {3}{4},\frac {5}{4},\frac {5}{4},\frac {7}{4},x,k x\right )+\frac {1}{7} x^{7/4} \left (-d k^2+2 k+2\right ) \operatorname {AppellF1}\left (\frac {7}{4},\frac {5}{4},\frac {5}{4},\frac {11}{4},x,k x\right )-\frac {1}{11} k x^{11/4} \operatorname {AppellF1}\left (\frac {11}{4},\frac {5}{4},\frac {5}{4},\frac {15}{4},x,k x\right )\right )}{\sqrt [4]{(1-x) x (1-k x)}}\)

input

Int[(x^3*(3 - 2*(1 + k)*x + k*x^2))/((-1 + x)*((1 - x)*x*(1 - k*x))^(1/4)* 
(-1 + k*x)*(-d + d*(1 + k)*x - d*k*x^2 + x^3)),x]
 

output

$Aborted
 

3.17.39.3.1 Defintions of rubi rules used

rule 25

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

rule 1395

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_)*((d_) + (e_.)*( 
x_)^(n_))^(q_.), x_Symbol] :> Simp[(a + b*x^n + c*x^(2*n))^FracPart[p]/((d 
+ e*x^n)^FracPart[p]*(a/d + c*(x^n/e))^FracPart[p])   Int[u*(d + e*x^n)^(p 
+ q)*(a/d + (c/e)*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && E 
qQ[n2, 2*n] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] &&  !(EqQ[q, 
1] && EqQ[n, 2])
 

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 7239

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl 
erIntegrandQ[v, u, x]]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

3.17.39.4 Maple [F]

\[\int \frac {x^{3} \left (3-2 \left (1+k \right ) x +k \,x^{2}\right )}{\left (-1+x \right ) \left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{4}} \left (k x -1\right ) \left (-d +d \left (1+k \right ) x -d k \,x^{2}+x^{3}\right )}d x\]

input

int(x^3*(3-2*(1+k)*x+k*x^2)/(-1+x)/((1-x)*x*(-k*x+1))^(1/4)/(k*x-1)/(-d+d* 
(1+k)*x-d*k*x^2+x^3),x)
 

output

int(x^3*(3-2*(1+k)*x+k*x^2)/(-1+x)/((1-x)*x*(-k*x+1))^(1/4)/(k*x-1)/(-d+d* 
(1+k)*x-d*k*x^2+x^3),x)
 

3.17.39.5 Fricas [F(-1)]

Timed out. \[ \int \frac {x^3 \left (3-2 (1+k) x+k x^2\right )}{(-1+x) \sqrt [4]{(1-x) x (1-k x)} (-1+k x) \left (-d+d (1+k) x-d k x^2+x^3\right )} \, dx=\text {Timed out} \]

input

integrate(x^3*(3-2*(1+k)*x+k*x^2)/(-1+x)/((1-x)*x*(-k*x+1))^(1/4)/(k*x-1)/ 
(-d+d*(1+k)*x-d*k*x^2+x^3),x, algorithm="fricas")
 

output

Timed out
 

3.17.39.6 Sympy [F(-1)]

Timed out. \[ \int \frac {x^3 \left (3-2 (1+k) x+k x^2\right )}{(-1+x) \sqrt [4]{(1-x) x (1-k x)} (-1+k x) \left (-d+d (1+k) x-d k x^2+x^3\right )} \, dx=\text {Timed out} \]

input

integrate(x**3*(3-2*(1+k)*x+k*x**2)/(-1+x)/((1-x)*x*(-k*x+1))**(1/4)/(k*x- 
1)/(-d+d*(1+k)*x-d*k*x**2+x**3),x)
 

output

Timed out
 

3.17.39.7 Maxima [F]

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

input

integrate(x^3*(3-2*(1+k)*x+k*x^2)/(-1+x)/((1-x)*x*(-k*x+1))^(1/4)/(k*x-1)/ 
(-d+d*(1+k)*x-d*k*x^2+x^3),x, algorithm="maxima")
 

output

-integrate((k*x^2 - 2*(k + 1)*x + 3)*x^3/((d*k*x^2 - d*(k + 1)*x - x^3 + d 
)*((k*x - 1)*(x - 1)*x)^(1/4)*(k*x - 1)*(x - 1)), x)
 

3.17.39.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 312 vs. \(2 (95) = 190\).

Time = 0.40 (sec) , antiderivative size = 312, normalized size of antiderivative = 2.81 \[ \int \frac {x^3 \left (3-2 (1+k) x+k x^2\right )}{(-1+x) \sqrt [4]{(1-x) x (1-k x)} (-1+k x) \left (-d+d (1+k) x-d k x^2+x^3\right )} \, dx=-\frac {\sqrt {2} \left (-d^{3}\right )^{\frac {3}{4}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (-\frac {1}{d}\right )^{\frac {1}{4}} + 2 \, {\left (\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-\frac {1}{d}\right )^{\frac {1}{4}}}\right )}{d^{2}} - \frac {\sqrt {2} \left (-d^{3}\right )^{\frac {3}{4}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (-\frac {1}{d}\right )^{\frac {1}{4}} - 2 \, {\left (\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}\right )}^{\frac {1}{4}}\right )}}{2 \, \left (-\frac {1}{d}\right )^{\frac {1}{4}}}\right )}{d^{2}} + \frac {\sqrt {2} \left (-d^{3}\right )^{\frac {3}{4}} \log \left (\sqrt {2} {\left (\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}\right )}^{\frac {1}{4}} \left (-\frac {1}{d}\right )^{\frac {1}{4}} + \sqrt {\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}} + \sqrt {-\frac {1}{d}}\right )}{2 \, d^{2}} - \frac {\sqrt {2} \left (-d^{3}\right )^{\frac {3}{4}} \log \left (-\sqrt {2} {\left (\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}\right )}^{\frac {1}{4}} \left (-\frac {1}{d}\right )^{\frac {1}{4}} + \sqrt {\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}} + \sqrt {-\frac {1}{d}}\right )}{2 \, d^{2}} + \frac {4}{{\left (\frac {k}{x} - \frac {k}{x^{2}} - \frac {1}{x^{2}} + \frac {1}{x^{3}}\right )}^{\frac {1}{4}}} \]

input

integrate(x^3*(3-2*(1+k)*x+k*x^2)/(-1+x)/((1-x)*x*(-k*x+1))^(1/4)/(k*x-1)/ 
(-d+d*(1+k)*x-d*k*x^2+x^3),x, algorithm="giac")
 

output

-sqrt(2)*(-d^3)^(3/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-1/d)^(1/4) + 2*(k/x - 
k/x^2 - 1/x^2 + 1/x^3)^(1/4))/(-1/d)^(1/4))/d^2 - sqrt(2)*(-d^3)^(3/4)*arc 
tan(-1/2*sqrt(2)*(sqrt(2)*(-1/d)^(1/4) - 2*(k/x - k/x^2 - 1/x^2 + 1/x^3)^( 
1/4))/(-1/d)^(1/4))/d^2 + 1/2*sqrt(2)*(-d^3)^(3/4)*log(sqrt(2)*(k/x - k/x^ 
2 - 1/x^2 + 1/x^3)^(1/4)*(-1/d)^(1/4) + sqrt(k/x - k/x^2 - 1/x^2 + 1/x^3) 
+ sqrt(-1/d))/d^2 - 1/2*sqrt(2)*(-d^3)^(3/4)*log(-sqrt(2)*(k/x - k/x^2 - 1 
/x^2 + 1/x^3)^(1/4)*(-1/d)^(1/4) + sqrt(k/x - k/x^2 - 1/x^2 + 1/x^3) + sqr 
t(-1/d))/d^2 + 4/(k/x - k/x^2 - 1/x^2 + 1/x^3)^(1/4)
 

3.17.39.9 Mupad [F(-1)]

Timed out. \[ \int \frac {x^3 \left (3-2 (1+k) x+k x^2\right )}{(-1+x) \sqrt [4]{(1-x) x (1-k x)} (-1+k x) \left (-d+d (1+k) x-d k x^2+x^3\right )} \, dx=\int -\frac {x^3\,\left (k\,x^2-2\,x\,\left (k+1\right )+3\right )}{\left (k\,x-1\right )\,\left (x-1\right )\,{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/4}\,\left (-x^3+d\,k\,x^2-d\,\left (k+1\right )\,x+d\right )} \,d x \]

input

int(-(x^3*(k*x^2 - 2*x*(k + 1) + 3))/((k*x - 1)*(x - 1)*(x*(k*x - 1)*(x - 
1))^(1/4)*(d - x^3 - d*x*(k + 1) + d*k*x^2)),x)
 

output

int(-(x^3*(k*x^2 - 2*x*(k + 1) + 3))/((k*x - 1)*(x - 1)*(x*(k*x - 1)*(x - 
1))^(1/4)*(d - x^3 - d*x*(k + 1) + d*k*x^2)), x)