3.30.78 \(\int \frac {(-2 x+(1+k) x^2) (1-(1+k) x+(a+k) x^2)}{((1-x) x (1-k x))^{2/3} (1-2 (1+k) x+(1+4 k+k^2) x^2-2 (k+k^2) x^3+(-b+k^2) x^4)} \, dx\) [2978]

Optimal. Leaf size=383 \[ -\frac {\sqrt {3} \left (a-\sqrt {b}\right ) \text {ArcTan}\left (\frac {\sqrt {3} \sqrt [6]{b} x}{\sqrt [6]{b} x-2 \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{2 b^{5/6}}+\frac {\sqrt {3} \left (a+\sqrt {b}\right ) \text {ArcTan}\left (\frac {\sqrt {3} \sqrt [6]{b} x}{\sqrt [6]{b} x+2 \sqrt [3]{x+(-1-k) x^2+k x^3}}\right )}{2 b^{5/6}}+\frac {\left (a+\sqrt {b}\right ) \log \left (-\sqrt [6]{b} x+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{2 b^{5/6}}+\frac {\left (-a+\sqrt {b}\right ) \log \left (\sqrt [6]{b} x+\sqrt [3]{x+(-1-k) x^2+k x^3}\right )}{2 b^{5/6}}+\frac {\left (a-\sqrt {b}\right ) \log \left (\sqrt [3]{b} x^2-\sqrt [6]{b} x \sqrt [3]{x+(-1-k) x^2+k x^3}+\left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{4 b^{5/6}}+\frac {\left (-a-\sqrt {b}\right ) \log \left (\sqrt [3]{b} x^2+\sqrt [6]{b} x \sqrt [3]{x+(-1-k) x^2+k x^3}+\left (x+(-1-k) x^2+k x^3\right )^{2/3}\right )}{4 b^{5/6}} \]

[Out]

-1/2*3^(1/2)*(a-b^(1/2))*arctan(3^(1/2)*b^(1/6)*x/(b^(1/6)*x-2*(x+(-1-k)*x^2+k*x^3)^(1/3)))/b^(5/6)+1/2*3^(1/2
)*(a+b^(1/2))*arctan(3^(1/2)*b^(1/6)*x/(b^(1/6)*x+2*(x+(-1-k)*x^2+k*x^3)^(1/3)))/b^(5/6)+1/2*(a+b^(1/2))*ln(-b
^(1/6)*x+(x+(-1-k)*x^2+k*x^3)^(1/3))/b^(5/6)+1/2*(-a+b^(1/2))*ln(b^(1/6)*x+(x+(-1-k)*x^2+k*x^3)^(1/3))/b^(5/6)
+1/4*(a-b^(1/2))*ln(b^(1/3)*x^2-b^(1/6)*x*(x+(-1-k)*x^2+k*x^3)^(1/3)+(x+(-1-k)*x^2+k*x^3)^(2/3))/b^(5/6)+1/4*(
-a-b^(1/2))*ln(b^(1/3)*x^2+b^(1/6)*x*(x+(-1-k)*x^2+k*x^3)^(1/3)+(x+(-1-k)*x^2+k*x^3)^(2/3))/b^(5/6)

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Rubi [F]
time = 25.95, 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 {\left (-2 x+(1+k) x^2\right ) \left (1-(1+k) x+(a+k) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (1-2 (1+k) x+\left (1+4 k+k^2\right ) x^2-2 \left (k+k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

(-3*(1 + k)*(a + k)*x*((1 - x)/(1 - k*x))^(2/3)*(1 - k*x)*Hypergeometric2F1[1/3, 2/3, 4/3, ((1 - k)*x)/(1 - k*
x)])/((b - k^2)*((1 - x)*x*(1 - k*x))^(2/3)) + (3*(1 + k)*(a + k)*(1 - x)^(2/3)*x^(2/3)*(1 - k*x)^(2/3)*Defer[
Subst][Defer[Int][1/((1 - x^3)^(2/3)*(1 - k*x^3)^(2/3)*(1 - 2*(1 + k)*x^3 + (1 + k*(4 + k))*x^6 - 2*k*(1 + k)*
x^9 - b*(1 - k^2/b)*x^12)), x], x, x^(1/3)])/((b - k^2)*((1 - x)*x*(1 - k*x))^(2/3)) - (6*(b + k + k^2 + k^3 +
a*(1 + k)^2)*(1 - x)^(2/3)*x^(2/3)*(1 - k*x)^(2/3)*Defer[Subst][Defer[Int][x^3/((1 - x^3)^(2/3)*(1 - k*x^3)^(
2/3)*(1 - 2*(1 + k)*x^3 + (1 + k*(4 + k))*x^6 - 2*k*(1 + k)*x^9 - b*(1 - k^2/b)*x^12)), x], x, x^(1/3)])/((b -
k^2)*((1 - x)*x*(1 - k*x))^(2/3)) + (3*(1 + k)*(3*b + k + k^2 + k^3 + a*(1 + 4*k + k^2))*(1 - x)^(2/3)*x^(2/3
)*(1 - k*x)^(2/3)*Defer[Subst][Defer[Int][x^6/((1 - x^3)^(2/3)*(1 - k*x^3)^(2/3)*(1 - 2*(1 + k)*x^3 + (1 + k*(
4 + k))*x^6 - 2*k*(1 + k)*x^9 - b*(1 - k^2/b)*x^12)), x], x, x^(1/3)])/((b - k^2)*((1 - x)*x*(1 - k*x))^(2/3))
- (3*(b*(1 + 2*a + 4*k + k^2) + k*(k + k^3 + 2*a*(1 + k + k^2)))*(1 - x)^(2/3)*x^(2/3)*(1 - k*x)^(2/3)*Defer[
Subst][Defer[Int][x^9/((1 - x^3)^(2/3)*(1 - k*x^3)^(2/3)*(1 - 2*(1 + k)*x^3 + (1 + k*(4 + k))*x^6 - 2*k*(1 + k
)*x^9 - b*(1 - k^2/b)*x^12)), x], x, x^(1/3)])/((b - k^2)*((1 - x)*x*(1 - k*x))^(2/3))

Rubi steps

\begin {align*} \int \frac {\left (-2 x+(1+k) x^2\right ) \left (1-(1+k) x+(a+k) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (1-2 (1+k) x+\left (1+4 k+k^2\right ) x^2-2 \left (k+k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx &=\int \frac {x (-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (1-2 (1+k) x+\left (1+4 k+k^2\right ) x^2-2 \left (k+k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx\\ &=\frac {\left ((1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \int \frac {\sqrt [3]{x} (-2+(1+k) x) \left (1-(1+k) x+(a+k) x^2\right )}{(1-x)^{2/3} (1-k x)^{2/3} \left (1-2 (1+k) x+\left (1+4 k+k^2\right ) x^2-2 \left (k+k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx}{((1-x) x (1-k x))^{2/3}}\\ &=\frac {\left (3 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \text {Subst}\left (\int \frac {x^3 \left (-2+(1+k) x^3\right ) \left (1-(1+k) x^3+(a+k) x^6\right )}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (1-2 (1+k) x^3+\left (1+4 k+k^2\right ) x^6-2 \left (k+k^2\right ) x^9+\left (-b+k^2\right ) x^{12}\right )} \, dx,x,\sqrt [3]{x}\right )}{((1-x) x (1-k x))^{2/3}}\\ &=\frac {\left (3 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \text {Subst}\left (\int \left (-\frac {(1+k) (a+k)}{\left (b-k^2\right ) \left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3}}+\frac {(1+k) (a+k)-2 \left (b+k+k^2+k^3+a (1+k)^2\right ) x^3+(1+k) \left (3 b+k+k^2+k^3+a \left (1+4 k+k^2\right )\right ) x^6-\left (b \left (1+2 a+4 k+k^2\right )+k \left (k+k^3+2 a \left (1+k+k^2\right )\right )\right ) x^9}{\left (b-k^2\right ) \left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (1-2 (1+k) x^3+\left (1+4 k+k^2\right ) x^6-2 \left (k+k^2\right ) x^9+\left (-b+k^2\right ) x^{12}\right )}\right ) \, dx,x,\sqrt [3]{x}\right )}{((1-x) x (1-k x))^{2/3}}\\ &=\frac {\left (3 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \text {Subst}\left (\int \frac {(1+k) (a+k)-2 \left (b+k+k^2+k^3+a (1+k)^2\right ) x^3+(1+k) \left (3 b+k+k^2+k^3+a \left (1+4 k+k^2\right )\right ) x^6-\left (b \left (1+2 a+4 k+k^2\right )+k \left (k+k^3+2 a \left (1+k+k^2\right )\right )\right ) x^9}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (1-2 (1+k) x^3+\left (1+4 k+k^2\right ) x^6-2 \left (k+k^2\right ) x^9+\left (-b+k^2\right ) x^{12}\right )} \, dx,x,\sqrt [3]{x}\right )}{\left (b-k^2\right ) ((1-x) x (1-k x))^{2/3}}-\frac {\left (3 (1+k) (a+k) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \text {Subst}\left (\int \frac {1}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3}} \, dx,x,\sqrt [3]{x}\right )}{\left (b-k^2\right ) ((1-x) x (1-k x))^{2/3}}\\ &=-\frac {3 (1+k) (a+k) x \left (\frac {1-x}{1-k x}\right )^{2/3} (1-k x) \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};\frac {(1-k) x}{1-k x}\right )}{\left (b-k^2\right ) ((1-x) x (1-k x))^{2/3}}+\frac {\left (3 (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \text {Subst}\left (\int \left (\frac {(1+k) (a+k)}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (1-2 (1+k) x^3+(1+k (4+k)) x^6-2 k (1+k) x^9-b \left (1-\frac {k^2}{b}\right ) x^{12}\right )}+\frac {2 \left (-b-k-k^2-k^3-a (1+k)^2\right ) x^3}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (1-2 (1+k) x^3+(1+k (4+k)) x^6-2 k (1+k) x^9-b \left (1-\frac {k^2}{b}\right ) x^{12}\right )}+\frac {(1+k) \left (3 b+k+k^2+k^3+a \left (1+4 k+k^2\right )\right ) x^6}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (1-2 (1+k) x^3+(1+k (4+k)) x^6-2 k (1+k) x^9-b \left (1-\frac {k^2}{b}\right ) x^{12}\right )}+\frac {\left (-b \left (1+2 a+4 k+k^2\right )-k \left (k+k^3+2 a \left (1+k+k^2\right )\right )\right ) x^9}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (1-2 (1+k) x^3+(1+k (4+k)) x^6-2 k (1+k) x^9-b \left (1-\frac {k^2}{b}\right ) x^{12}\right )}\right ) \, dx,x,\sqrt [3]{x}\right )}{\left (b-k^2\right ) ((1-x) x (1-k x))^{2/3}}\\ &=-\frac {3 (1+k) (a+k) x \left (\frac {1-x}{1-k x}\right )^{2/3} (1-k x) \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {4}{3};\frac {(1-k) x}{1-k x}\right )}{\left (b-k^2\right ) ((1-x) x (1-k x))^{2/3}}+\frac {\left (3 (1+k) (a+k) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \text {Subst}\left (\int \frac {1}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (1-2 (1+k) x^3+(1+k (4+k)) x^6-2 k (1+k) x^9-b \left (1-\frac {k^2}{b}\right ) x^{12}\right )} \, dx,x,\sqrt [3]{x}\right )}{\left (b-k^2\right ) ((1-x) x (1-k x))^{2/3}}-\frac {\left (6 \left (b+k+k^2+k^3+a (1+k)^2\right ) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \text {Subst}\left (\int \frac {x^3}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (1-2 (1+k) x^3+(1+k (4+k)) x^6-2 k (1+k) x^9-b \left (1-\frac {k^2}{b}\right ) x^{12}\right )} \, dx,x,\sqrt [3]{x}\right )}{\left (b-k^2\right ) ((1-x) x (1-k x))^{2/3}}+\frac {\left (3 (1+k) \left (3 b+k+k^2+k^3+a \left (1+4 k+k^2\right )\right ) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \text {Subst}\left (\int \frac {x^6}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (1-2 (1+k) x^3+(1+k (4+k)) x^6-2 k (1+k) x^9-b \left (1-\frac {k^2}{b}\right ) x^{12}\right )} \, dx,x,\sqrt [3]{x}\right )}{\left (b-k^2\right ) ((1-x) x (1-k x))^{2/3}}+\frac {\left (3 \left (-b \left (1+2 a+4 k+k^2\right )-k \left (k+k^3+2 a \left (1+k+k^2\right )\right )\right ) (1-x)^{2/3} x^{2/3} (1-k x)^{2/3}\right ) \text {Subst}\left (\int \frac {x^9}{\left (1-x^3\right )^{2/3} \left (1-k x^3\right )^{2/3} \left (1-2 (1+k) x^3+(1+k (4+k)) x^6-2 k (1+k) x^9-b \left (1-\frac {k^2}{b}\right ) x^{12}\right )} \, dx,x,\sqrt [3]{x}\right )}{\left (b-k^2\right ) ((1-x) x (1-k x))^{2/3}}\\ \end {align*}

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Mathematica [F]
time = 20.87, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (-2 x+(1+k) x^2\right ) \left (1-(1+k) x+(a+k) x^2\right )}{((1-x) x (1-k x))^{2/3} \left (1-2 (1+k) x+\left (1+4 k+k^2\right ) x^2-2 \left (k+k^2\right ) x^3+\left (-b+k^2\right ) x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (-2 x +\left (1+k \right ) x^{2}\right ) \left (1-\left (1+k \right ) x +\left (a +k \right ) x^{2}\right )}{\left (\left (1-x \right ) x \left (-k x +1\right )\right )^{\frac {2}{3}} \left (1-2 \left (1+k \right ) x +\left (k^{2}+4 k +1\right ) x^{2}-2 \left (k^{2}+k \right ) x^{3}+\left (k^{2}-b \right ) x^{4}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (tr
ace 0)

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Sympy [F(-1)] Timed out
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((-2*x+(1+k)*x**2)*(1-(1+k)*x+(a+k)*x**2)/((1-x)*x*(-k*x+1))**(2/3)/(1-2*(1+k)*x+(k**2+4*k+1)*x**2-2*
(k**2+k)*x**3+(k**2-b)*x**4),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (2\,x-x^2\,\left (k+1\right )\right )\,\left (\left (a+k\right )\,x^2+\left (-k-1\right )\,x+1\right )}{{\left (x\,\left (k\,x-1\right )\,\left (x-1\right )\right )}^{2/3}\,\left (x^4\,\left (b-k^2\right )-x^2\,\left (k^2+4\,k+1\right )+2\,x\,\left (k+1\right )+2\,x^3\,\left (k^2+k\right )-1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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