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

Optimal. Leaf size=257 \[ -\frac {\log \left (b^{2/3} x^4-4 b^{2/3} x^3+6 b^{2/3} x^2-4 b^{2/3} x+b^{2/3}+\left (\sqrt [3]{b} x^2-2 \sqrt [3]{b} x+\sqrt [3]{b}\right ) \sqrt [3]{k x^3+(-k-1) x^2+x}+\left (k x^3+(-k-1) x^2+x\right )^{2/3}\right )}{2 \sqrt [3]{b}}+\frac {\log \left (-\sqrt [3]{b} x^2+2 \sqrt [3]{b} x-\sqrt [3]{b}+\sqrt [3]{k x^3+(-k-1) x^2+x}\right )}{\sqrt [3]{b}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{k x^3+(-k-1) x^2+x}}{2 \sqrt [3]{b} x^2-4 \sqrt [3]{b} x+2 \sqrt [3]{b}+\sqrt [3]{k x^3+(-k-1) x^2+x}}\right )}{\sqrt [3]{b}} \]

________________________________________________________________________________________

Rubi [F]  time = 11.28, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[In]

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]

[Out]

(6*(2 - k)*(-1 + x)^(1/3)*x^(1/3)*(-1 + k*x)^(1/3)*Defer[Subst][Defer[Int][(x^4*(-1 + x^3)^(2/3))/((-1 + k*x^3
)^(1/3)*(-x^3 + k*x^6 - b*(-1 + x^3)^5)), x], x, x^(1/3)])/((1 - x)*x*(1 - k*x))^(1/3) - (9*k*(-1 + x)^(1/3)*x
^(1/3)*(-1 + k*x)^(1/3)*Defer[Subst][Defer[Int][(x^7*(-1 + x^3)^(2/3))/((-1 + k*x^3)^(1/3)*(-x^3 + k*x^6 - b*(
-1 + x^3)^5)), x], x, x^(1/3)])/((1 - x)*x*(1 - k*x))^(1/3) - (3*(-1 + x)^(1/3)*x^(1/3)*(-1 + k*x)^(1/3)*Defer
[Subst][Defer[Int][(x*(-1 + x^3)^(2/3))/((-1 + k*x^3)^(1/3)*(x^3 - k*x^6 + b*(-1 + x^3)^5)), x], x, x^(1/3)])/
((1 - x)*x*(1 - k*x))^(1/3)

Rubi steps

\begin {align*} \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-x) \left (1+2 (2-k) x-3 k x^2\right )}{\sqrt [3]{(-1+x) x (-1+k x)} \left (b (-1+x)^5+x-k x^2\right )} \, dx\\ &=\frac {\left (\sqrt [3]{-1+x} \sqrt [3]{x} \sqrt [3]{-1+k x}\right ) \int \frac {(1-x) \left (1+2 (2-k) x-3 k x^2\right )}{\sqrt [3]{-1+x} \sqrt [3]{x} \sqrt [3]{-1+k x} \left (b (-1+x)^5+x-k x^2\right )} \, dx}{\sqrt [3]{(-1+x) x (-1+k x)}}\\ &=-\frac {\left (\sqrt [3]{-1+x} \sqrt [3]{x} \sqrt [3]{-1+k x}\right ) \int \frac {(-1+x)^{2/3} \left (1+2 (2-k) x-3 k x^2\right )}{\sqrt [3]{x} \sqrt [3]{-1+k x} \left (b (-1+x)^5+x-k x^2\right )} \, dx}{\sqrt [3]{(-1+x) x (-1+k x)}}\\ &=-\frac {\left (3 \sqrt [3]{-1+x} \sqrt [3]{x} \sqrt [3]{-1+k x}\right ) \operatorname {Subst}\left (\int \frac {x \left (-1+x^3\right )^{2/3} \left (1+2 (2-k) x^3-3 k x^6\right )}{\sqrt [3]{-1+k x^3} \left (x^3-k x^6+b \left (-1+x^3\right )^5\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(-1+x) x (-1+k x)}}\\ &=-\frac {\left (3 \sqrt [3]{-1+x} \sqrt [3]{x} \sqrt [3]{-1+k x}\right ) \operatorname {Subst}\left (\int \left (\frac {2 (-2+k) x^4 \left (-1+x^3\right )^{2/3}}{\sqrt [3]{-1+k x^3} \left (b-(1+5 b) x^3+10 b \left (1+\frac {k}{10 b}\right ) x^6-10 b x^9+5 b x^{12}-b x^{15}\right )}+\frac {3 k x^7 \left (-1+x^3\right )^{2/3}}{\sqrt [3]{-1+k x^3} \left (b-(1+5 b) x^3+10 b \left (1+\frac {k}{10 b}\right ) x^6-10 b x^9+5 b x^{12}-b x^{15}\right )}+\frac {x \left (-1+x^3\right )^{2/3}}{\sqrt [3]{-1+k x^3} \left (-b+(1+5 b) x^3-10 b \left (1+\frac {k}{10 b}\right ) x^6+10 b x^9-5 b x^{12}+b x^{15}\right )}\right ) \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(-1+x) x (-1+k x)}}\\ &=-\frac {\left (3 \sqrt [3]{-1+x} \sqrt [3]{x} \sqrt [3]{-1+k x}\right ) \operatorname {Subst}\left (\int \frac {x \left (-1+x^3\right )^{2/3}}{\sqrt [3]{-1+k x^3} \left (-b+(1+5 b) x^3-10 b \left (1+\frac {k}{10 b}\right ) x^6+10 b x^9-5 b x^{12}+b x^{15}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(-1+x) x (-1+k x)}}+\frac {\left (6 (2-k) \sqrt [3]{-1+x} \sqrt [3]{x} \sqrt [3]{-1+k x}\right ) \operatorname {Subst}\left (\int \frac {x^4 \left (-1+x^3\right )^{2/3}}{\sqrt [3]{-1+k x^3} \left (b-(1+5 b) x^3+10 b \left (1+\frac {k}{10 b}\right ) x^6-10 b x^9+5 b x^{12}-b x^{15}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(-1+x) x (-1+k x)}}-\frac {\left (9 k \sqrt [3]{-1+x} \sqrt [3]{x} \sqrt [3]{-1+k x}\right ) \operatorname {Subst}\left (\int \frac {x^7 \left (-1+x^3\right )^{2/3}}{\sqrt [3]{-1+k x^3} \left (b-(1+5 b) x^3+10 b \left (1+\frac {k}{10 b}\right ) x^6-10 b x^9+5 b x^{12}-b x^{15}\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(-1+x) x (-1+k x)}}\\ &=-\frac {\left (3 \sqrt [3]{-1+x} \sqrt [3]{x} \sqrt [3]{-1+k x}\right ) \operatorname {Subst}\left (\int \frac {x \left (-1+x^3\right )^{2/3}}{\sqrt [3]{-1+k x^3} \left (x^3-k x^6+b \left (-1+x^3\right )^5\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(-1+x) x (-1+k x)}}+\frac {\left (6 (2-k) \sqrt [3]{-1+x} \sqrt [3]{x} \sqrt [3]{-1+k x}\right ) \operatorname {Subst}\left (\int \frac {x^4 \left (-1+x^3\right )^{2/3}}{\sqrt [3]{-1+k x^3} \left (-x^3+k x^6-b \left (-1+x^3\right )^5\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(-1+x) x (-1+k x)}}-\frac {\left (9 k \sqrt [3]{-1+x} \sqrt [3]{x} \sqrt [3]{-1+k x}\right ) \operatorname {Subst}\left (\int \frac {x^7 \left (-1+x^3\right )^{2/3}}{\sqrt [3]{-1+k x^3} \left (-x^3+k x^6-b \left (-1+x^3\right )^5\right )} \, dx,x,\sqrt [3]{x}\right )}{\sqrt [3]{(-1+x) x (-1+k x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]  time = 8.80, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

Verification is not applicable to the result.

[In]

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]

[Out]

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]

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 3.01, size = 257, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(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]

[Out]

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

________________________________________________________________________________________

fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

Timed out

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

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)

________________________________________________________________________________________

maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {1+\left (3-2 k \right ) x -\left (4+k \right ) x^{2}+3 k \,x^{3}}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {1}{3}} \left (-b +\left (1+5 b \right ) x -\left (10 b +k \right ) x^{2}+10 b \,x^{3}-5 b \,x^{4}+b \,x^{5}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

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)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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")

[Out]

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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} -\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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

-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)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

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 - 1
0*b*x**2 + 5*b*x - b - k*x**2 + x)), x)

________________________________________________________________________________________