3.28.51 \(\int \frac {1+(3-2 k) x-(4+k) x^2+3 k x^3}{\sqrt [3]{(1-x) x (1-k x)} (-b+(1+5 b) x-(10 b+k) x^2+10 b x^3-5 b x^4+b x^5)} \, dx\) [2751]

3.28.51.1 Optimal result

Integrand size = 81, antiderivative size = 257 \[ \int \frac {1+(3-2 k) x-(4+k) x^2+3 k x^3}{\sqrt [3]{(1-x) x (1-k x)} \left (-b+(1+5 b) x-(10 b+k) x^2+10 b x^3-5 b x^4+b x^5\right )} \, dx=\frac {\sqrt {3} \arctan \left (\frac {\sqrt {3} \sqrt [3]{x+(-1-k) x^2+k x^3}}{2 \sqrt [3]{b}-4 \sqrt [3]{b} x+2 \sqrt [3]{b} x^2+\sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{\sqrt [3]{b}}+\frac {\log \left (-\sqrt [3]{b}+2 \sqrt [3]{b} x-\sqrt [3]{b} x^2+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{\sqrt [3]{b}}-\frac {\log \left (b^{2/3}-4 b^{2/3} x+6 b^{2/3} x^2-4 b^{2/3} x^3+b^{2/3} x^4+\left (\sqrt [3]{b}-2 \sqrt [3]{b} x+\sqrt [3]{b} x^2\right ) \sqrt [3]{x+(-1-k) x^2+k x^3}+\left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{2 \sqrt [3]{b}} \]

output

3^(1/2)*arctan(3^(1/2)*(x+(-1-k)*x^2+k*x^3)^(1/3)/(2*b^(1/3)-4*b^(1/3)*x+2 
*b^(1/3)*x^2+(x+(-1-k)*x^2+k*x^3)^(1/3)))/b^(1/3)+ln(-b^(1/3)+2*b^(1/3)*x- 
b^(1/3)*x^2+(x+(-1-k)*x^2+k*x^3)^(1/3))/b^(1/3)-1/2*ln(b^(2/3)-4*b^(2/3)*x 
+6*b^(2/3)*x^2-4*b^(2/3)*x^3+b^(2/3)*x^4+(b^(1/3)-2*b^(1/3)*x+b^(1/3)*x^2) 
*(x+(-1-k)*x^2+k*x^3)^(1/3)+(x+(-1-k)*x^2+k*x^3)^(2/3))/b^(1/3)
 

3.28.51.2 Mathematica [F]

\[ \int \frac {1+(3-2 k) x-(4+k) x^2+3 k x^3}{\sqrt [3]{(1-x) x (1-k x)} \left (-b+(1+5 b) x-(10 b+k) x^2+10 b x^3-5 b x^4+b x^5\right )} \, dx=\int \frac {1+(3-2 k) x-(4+k) x^2+3 k x^3}{\sqrt [3]{(1-x) x (1-k x)} \left (-b+(1+5 b) x-(10 b+k) x^2+10 b x^3-5 b x^4+b x^5\right )} \, dx \]

input

Integrate[(1 + (3 - 2*k)*x - (4 + k)*x^2 + 3*k*x^3)/(((1 - x)*x*(1 - k*x)) 
^(1/3)*(-b + (1 + 5*b)*x - (10*b + k)*x^2 + 10*b*x^3 - 5*b*x^4 + b*x^5)),x 
]
 

output

Integrate[(1 + (3 - 2*k)*x - (4 + k)*x^2 + 3*k*x^3)/(((1 - x)*x*(1 - k*x)) 
^(1/3)*(-b + (1 + 5*b)*x - (10*b + k)*x^2 + 10*b*x^3 - 5*b*x^4 + b*x^5)), 
x]
 

3.28.51.3 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.

\(\displaystyle \int \frac {3 k x^3-(k+4) x^2+(3-2 k) x+1}{\sqrt [3]{(1-x) x (1-k x)} \left (-x^2 (10 b+k)+b x^5-5 b x^4+10 b x^3+(5 b+1) x-b\right )} \, dx\)

\(\Big \downarrow \) 2467

\(\displaystyle \frac {\sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int -\frac {3 k x^3-(k+4) x^2+(3-2 k) x+1}{\sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \left (-b x^5+5 b x^4-10 b x^3+(10 b+k) x^2-(5 b+1) x+b\right )}dx}{\sqrt [3]{(1-x) x (1-k x)}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \frac {3 k x^3-(k+4) x^2+(3-2 k) x+1}{\sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \left (-b x^5+5 b x^4-10 b x^3+(10 b+k) x^2-(5 b+1) x+b\right )}dx}{\sqrt [3]{(1-x) x (1-k x)}}\)

\(\Big \downarrow \) 2035

\(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \frac {\sqrt [3]{x} \left (3 k x^3-(k+4) x^2+(3-2 k) x+1\right )}{\sqrt [3]{k x^2-(k+1) x+1} \left (-b x^5+5 b x^4-10 b x^3+(10 b+k) x^2-(5 b+1) x+b\right )}d\sqrt [3]{x}}{\sqrt [3]{(1-x) x (1-k x)}}\)

\(\Big \downarrow \) 7292

\(\displaystyle -\frac {3 \sqrt [3]{x} \sqrt [3]{k x^2-(k+1) x+1} \int \frac {(1-x) \sqrt [3]{x} \left (-3 k x^2+2 (2-k) x+1\right )}{\sqrt [3]{k x^2-(k+1) x+1} \left (-b x^5+5 b x^4-10 b x^3+(10 b+k) x^2-(5 b+1) x+b\right )}d\sqrt [3]{x}}{\sqrt [3]{(1-x) x (1-k x)}}\)

\(\Big \downarrow \) 1395

\(\displaystyle -\frac {3 \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x} \int \frac {(1-x)^{2/3} \sqrt [3]{x} \left (-3 k x^2+2 (2-k) x+1\right )}{\sqrt [3]{1-k x} \left (-b x^5+5 b x^4-10 b x^3+(10 b+k) x^2-(5 b+1) x+b\right )}d\sqrt [3]{x}}{\sqrt [3]{(1-x) x (1-k x)}}\)

\(\Big \downarrow \) 7293

\(\displaystyle -\frac {3 \sqrt [3]{1-x} \sqrt [3]{x} \sqrt [3]{1-k x} \int \left (\frac {3 k (1-x)^{2/3} x^{7/3}}{\sqrt [3]{1-k x} \left (b x^5-5 b x^4+10 b x^3-10 b \left (\frac {k}{10 b}+1\right ) x^2+(5 b+1) x-b\right )}+\frac {2 (2-k) (1-x)^{2/3} x^{4/3}}{\sqrt [3]{1-k x} \left (-b x^5+5 b x^4-10 b x^3+10 b \left (\frac {k}{10 b}+1\right ) x^2-(5 b+1) x+b\right )}+\frac {(1-x)^{2/3} \sqrt [3]{x}}{\sqrt [3]{1-k x} \left (-b x^5+5 b x^4-10 b x^3+10 b \left (\frac {k}{10 b}+1\right ) x^2-(5 b+1) x+b\right )}\right )d\sqrt [3]{x}}{\sqrt [3]{(1-x) x (1-k x)}}\)

\(\Big \downarrow \) 2009

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

input

Int[(1 + (3 - 2*k)*x - (4 + k)*x^2 + 3*k*x^3)/(((1 - x)*x*(1 - k*x))^(1/3) 
*(-b + (1 + 5*b)*x - (10*b + k)*x^2 + 10*b*x^3 - 5*b*x^4 + b*x^5)),x]
 

output

$Aborted
 

3.28.51.3.1 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 7292

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

rule 7293

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

3.28.51.4 Maple [F]

\[\int \frac {1+\left (3-2 k \right ) x -\left (4+k \right ) x^{2}+3 k \,x^{3}}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{3}} \left (-b +\left (1+5 b \right ) x -\left (10 b +k \right ) x^{2}+10 b \,x^{3}-5 b \,x^{4}+b \,x^{5}\right )}d x\]

input

int((1+(3-2*k)*x-(4+k)*x^2+3*k*x^3)/((1-x)*x*(-k*x+1))^(1/3)/(-b+(1+5*b)*x 
-(10*b+k)*x^2+10*b*x^3-5*b*x^4+b*x^5),x)
 

output

int((1+(3-2*k)*x-(4+k)*x^2+3*k*x^3)/((1-x)*x*(-k*x+1))^(1/3)/(-b+(1+5*b)*x 
-(10*b+k)*x^2+10*b*x^3-5*b*x^4+b*x^5),x)
 

3.28.51.5 Fricas [F(-1)]

Timed out. \[ \int \frac {1+(3-2 k) x-(4+k) x^2+3 k x^3}{\sqrt [3]{(1-x) x (1-k x)} \left (-b+(1+5 b) x-(10 b+k) x^2+10 b x^3-5 b x^4+b x^5\right )} \, dx=\text {Timed out} \]

input

integrate((1+(3-2*k)*x-(4+k)*x^2+3*k*x^3)/((1-x)*x*(-k*x+1))^(1/3)/(-b+(1+ 
5*b)*x-(10*b+k)*x^2+10*b*x^3-5*b*x^4+b*x^5),x, algorithm="fricas")
 

output

Timed out
 

3.28.51.6 Sympy [F]

\[ \int \frac {1+(3-2 k) x-(4+k) x^2+3 k x^3}{\sqrt [3]{(1-x) x (1-k x)} \left (-b+(1+5 b) x-(10 b+k) x^2+10 b x^3-5 b x^4+b x^5\right )} \, dx=\int \frac {\left (x - 1\right ) \left (3 k x^{2} + 2 k x - 4 x - 1\right )}{\sqrt [3]{x \left (x - 1\right ) \left (k x - 1\right )} \left (b x^{5} - 5 b x^{4} + 10 b x^{3} - 10 b x^{2} + 5 b x - b - k x^{2} + x\right )}\, dx \]

input

integrate((1+(3-2*k)*x-(4+k)*x**2+3*k*x**3)/((1-x)*x*(-k*x+1))**(1/3)/(-b+ 
(1+5*b)*x-(10*b+k)*x**2+10*b*x**3-5*b*x**4+b*x**5),x)
 

output

Integral((x - 1)*(3*k*x**2 + 2*k*x - 4*x - 1)/((x*(x - 1)*(k*x - 1))**(1/3 
)*(b*x**5 - 5*b*x**4 + 10*b*x**3 - 10*b*x**2 + 5*b*x - b - k*x**2 + x)), x 
)
 

3.28.51.7 Maxima [F]

\[ \int \frac {1+(3-2 k) x-(4+k) x^2+3 k x^3}{\sqrt [3]{(1-x) x (1-k x)} \left (-b+(1+5 b) x-(10 b+k) x^2+10 b x^3-5 b x^4+b x^5\right )} \, dx=\int { \frac {3 \, k x^{3} - {\left (k + 4\right )} x^{2} - {\left (2 \, k - 3\right )} x + 1}{{\left (b x^{5} - 5 \, b x^{4} + 10 \, b x^{3} - {\left (10 \, b + k\right )} x^{2} + {\left (5 \, b + 1\right )} x - b\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}}} \,d x } \]

input

integrate((1+(3-2*k)*x-(4+k)*x^2+3*k*x^3)/((1-x)*x*(-k*x+1))^(1/3)/(-b+(1+ 
5*b)*x-(10*b+k)*x^2+10*b*x^3-5*b*x^4+b*x^5),x, algorithm="maxima")
 

output

integrate((3*k*x^3 - (k + 4)*x^2 - (2*k - 3)*x + 1)/((b*x^5 - 5*b*x^4 + 10 
*b*x^3 - (10*b + k)*x^2 + (5*b + 1)*x - b)*((k*x - 1)*(x - 1)*x)^(1/3)), x 
)
 

3.28.51.8 Giac [F]

\[ \int \frac {1+(3-2 k) x-(4+k) x^2+3 k x^3}{\sqrt [3]{(1-x) x (1-k x)} \left (-b+(1+5 b) x-(10 b+k) x^2+10 b x^3-5 b x^4+b x^5\right )} \, dx=\int { \frac {3 \, k x^{3} - {\left (k + 4\right )} x^{2} - {\left (2 \, k - 3\right )} x + 1}{{\left (b x^{5} - 5 \, b x^{4} + 10 \, b x^{3} - {\left (10 \, b + k\right )} x^{2} + {\left (5 \, b + 1\right )} x - b\right )} \left ({\left (k x - 1\right )} {\left (x - 1\right )} x\right )^{\frac {1}{3}}} \,d x } \]

input

integrate((1+(3-2*k)*x-(4+k)*x^2+3*k*x^3)/((1-x)*x*(-k*x+1))^(1/3)/(-b+(1+ 
5*b)*x-(10*b+k)*x^2+10*b*x^3-5*b*x^4+b*x^5),x, algorithm="giac")
 

output

integrate((3*k*x^3 - (k + 4)*x^2 - (2*k - 3)*x + 1)/((b*x^5 - 5*b*x^4 + 10 
*b*x^3 - (10*b + k)*x^2 + (5*b + 1)*x - b)*((k*x - 1)*(x - 1)*x)^(1/3)), x 
)
 

3.28.51.9 Mupad [F(-1)]

Timed out. \[ \int \frac {1+(3-2 k) x-(4+k) x^2+3 k x^3}{\sqrt [3]{(1-x) x (1-k x)} \left (-b+(1+5 b) x-(10 b+k) x^2+10 b x^3-5 b x^4+b x^5\right )} \, dx=-\int -\frac {-3\,k\,x^3+\left (k+4\right )\,x^2+\left (2\,k-3\right )\,x-1}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{1/3}\,\left (-b\,x^5+5\,b\,x^4-10\,b\,x^3+\left (10\,b+k\right )\,x^2+\left (-5\,b-1\right )\,x+b\right )} \,d x \]

input

int((x*(2*k - 3) + x^2*(k + 4) - 3*k*x^3 - 1)/((x*(k*x - 1)*(x - 1))^(1/3) 
*(b - 10*b*x^3 + 5*b*x^4 - b*x^5 + x^2*(10*b + k) - x*(5*b + 1))),x)
 

output

-int(-(x*(2*k - 3) + x^2*(k + 4) - 3*k*x^3 - 1)/((x*(k*x - 1)*(x - 1))^(1/ 
3)*(b - 10*b*x^3 + 5*b*x^4 - b*x^5 + x^2*(10*b + k) - x*(5*b + 1))), x)