3.13.16 \(\int \frac {(1-2 x+x^2) (-1+2 (-1+k) x+k x^2)}{((1-x) x (1-k x))^{3/4} (-d+(1+3 d) x-(3 d+k) x^2+d x^3)} \, dx\) [1216]

3.13.16.1 Optimal result

Integrand size = 67, antiderivative size = 89 \[ \int \frac {\left (1-2 x+x^2\right ) \left (-1+2 (-1+k) x+k x^2\right )}{((1-x) x (1-k x))^{3/4} \left (-d+(1+3 d) x-(3 d+k) x^2+d x^3\right )} \, dx=\frac {2 \arctan \left (\frac {-\sqrt [4]{d}+\sqrt [4]{d} x}{\sqrt [4]{x+(-1-k) x^2+k x^3}}\right )}{d^{3/4}}-\frac {2 \text {arctanh}\left (\frac {-\sqrt [4]{d}+\sqrt [4]{d} x}{\sqrt [4]{x+(-1-k) x^2+k x^3}}\right )}{d^{3/4}} \]

output

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

3.13.16.2 Mathematica [F]

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

input

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

output

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

3.13.16.3 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.

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

\(\Big \downarrow \) 2467

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

\(\Big \downarrow \) 2035

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

\(\Big \downarrow \) 1380

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

\(\Big \downarrow \) 1395

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

input

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

output

$Aborted
 

3.13.16.3.1 Defintions of rubi rules used

rule 1380

Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S 
imp[1/c^p   Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] 
&& EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
 

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 7293

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

3.13.16.4 Maple [F]

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

input

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

output

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

3.13.16.5 Fricas [F(-1)]

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

input

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

output

Timed out
 

3.13.16.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.13.16.7 Maxima [F]

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

input

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

output

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

3.13.16.8 Giac [F]

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

input

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

output

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

3.13.16.9 Mupad [F(-1)]

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

input

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

output

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