3.29.71 \(\int \frac {a b-2 b x+x^2}{\sqrt [3]{x (-a+x) (-b+x)^2} (b d-(a+d) x+x^2)} \, dx\)

Optimal. Leaf size=306 \[ -\frac {\log \left (\sqrt [3]{x^2 \left (2 a b+b^2\right )-a b^2 x+x^3 (-a-2 b)+x^4} \left (d^{2/3} x-b d^{2/3}\right )+\sqrt [3]{d} \left (x^2 \left (2 a b+b^2\right )-a b^2 x+x^3 (-a-2 b)+x^4\right )^{2/3}+b^2 d-2 b d x+d x^2\right )}{2 \sqrt [3]{d}}+\frac {\log \left (\sqrt [6]{d} \sqrt [3]{x^2 \left (2 a b+b^2\right )-a b^2 x+x^3 (-a-2 b)+x^4}+b \sqrt {d}-\sqrt {d} x\right )}{\sqrt [3]{d}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{x^2 \left (2 a b+b^2\right )-a b^2 x+x^3 (-a-2 b)+x^4}}{\sqrt [3]{x^2 \left (2 a b+b^2\right )-a b^2 x+x^3 (-a-2 b)+x^4}-2 b \sqrt [3]{d}+2 \sqrt [3]{d} x}\right )}{\sqrt [3]{d}} \]

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Rubi [F]  time = 5.66, 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 {a b-2 b x+x^2}{\sqrt [3]{x (-a+x) (-b+x)^2} \left (b d-(a+d) x+x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

(3*x*(1 - x/a)^(1/3)*(1 - x/b)^(2/3)*AppellF1[2/3, 1/3, 2/3, 5/3, x/a, x/b])/(2*(-((a - x)*(b - x)^2*x))^(1/3)
) + ((a - 2*b + d + Sqrt[a^2 + 2*a*d - 4*b*d + d^2])*x^(1/3)*(-a + x)^(1/3)*(-b + x)^(2/3)*Defer[Int][1/(x^(1/
3)*(-a + x)^(1/3)*(-b + x)^(2/3)*(-a - d - Sqrt[a^2 + 2*a*d - 4*b*d + d^2] + 2*x)), x])/(-((a - x)*(b - x)^2*x
))^(1/3) + ((a - 2*b + d - Sqrt[a^2 + 2*a*d - 4*b*d + d^2])*x^(1/3)*(-a + x)^(1/3)*(-b + x)^(2/3)*Defer[Int][1
/(x^(1/3)*(-a + x)^(1/3)*(-b + x)^(2/3)*(-a - d + Sqrt[a^2 + 2*a*d - 4*b*d + d^2] + 2*x)), x])/(-((a - x)*(b -
x)^2*x))^(1/3)

Rubi steps

\begin {align*} \int \frac {a b-2 b x+x^2}{\sqrt [3]{x (-a+x) (-b+x)^2} \left (b d-(a+d) x+x^2\right )} \, dx &=\frac {\left (\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \frac {a b-2 b x+x^2}{\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3} \left (b d-(a+d) x+x^2\right )} \, dx}{\sqrt [3]{x (-a+x) (-b+x)^2}}\\ &=\frac {\left (\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \left (\frac {1}{\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3}}+\frac {b (a-d)+(a-2 b+d) x}{\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3} \left (b d+(-a-d) x+x^2\right )}\right ) \, dx}{\sqrt [3]{x (-a+x) (-b+x)^2}}\\ &=\frac {\left (\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \frac {1}{\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3}} \, dx}{\sqrt [3]{x (-a+x) (-b+x)^2}}+\frac {\left (\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \frac {b (a-d)+(a-2 b+d) x}{\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3} \left (b d+(-a-d) x+x^2\right )} \, dx}{\sqrt [3]{x (-a+x) (-b+x)^2}}\\ &=\frac {\left (\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \left (\frac {a-2 b+d+\sqrt {a^2+2 a d-4 b d+d^2}}{\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3} \left (-a-d-\sqrt {a^2+2 a d-4 b d+d^2}+2 x\right )}+\frac {a-2 b+d-\sqrt {a^2+2 a d-4 b d+d^2}}{\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3} \left (-a-d+\sqrt {a^2+2 a d-4 b d+d^2}+2 x\right )}\right ) \, dx}{\sqrt [3]{x (-a+x) (-b+x)^2}}+\frac {\left (\sqrt [3]{x} (-b+x)^{2/3} \sqrt [3]{1-\frac {x}{a}}\right ) \int \frac {1}{\sqrt [3]{x} (-b+x)^{2/3} \sqrt [3]{1-\frac {x}{a}}} \, dx}{\sqrt [3]{x (-a+x) (-b+x)^2}}\\ &=\frac {\left (\left (a-2 b+d-\sqrt {a^2+2 a d-4 b d+d^2}\right ) \sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \frac {1}{\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3} \left (-a-d+\sqrt {a^2+2 a d-4 b d+d^2}+2 x\right )} \, dx}{\sqrt [3]{x (-a+x) (-b+x)^2}}+\frac {\left (\left (a-2 b+d+\sqrt {a^2+2 a d-4 b d+d^2}\right ) \sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \frac {1}{\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3} \left (-a-d-\sqrt {a^2+2 a d-4 b d+d^2}+2 x\right )} \, dx}{\sqrt [3]{x (-a+x) (-b+x)^2}}+\frac {\left (\sqrt [3]{x} \sqrt [3]{1-\frac {x}{a}} \left (1-\frac {x}{b}\right )^{2/3}\right ) \int \frac {1}{\sqrt [3]{x} \sqrt [3]{1-\frac {x}{a}} \left (1-\frac {x}{b}\right )^{2/3}} \, dx}{\sqrt [3]{x (-a+x) (-b+x)^2}}\\ &=\frac {3 x \sqrt [3]{1-\frac {x}{a}} \left (1-\frac {x}{b}\right )^{2/3} F_1\left (\frac {2}{3};\frac {1}{3},\frac {2}{3};\frac {5}{3};\frac {x}{a},\frac {x}{b}\right )}{2 \sqrt [3]{-\left ((a-x) (b-x)^2 x\right )}}+\frac {\left (\left (a-2 b+d-\sqrt {a^2+2 a d-4 b d+d^2}\right ) \sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \frac {1}{\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3} \left (-a-d+\sqrt {a^2+2 a d-4 b d+d^2}+2 x\right )} \, dx}{\sqrt [3]{x (-a+x) (-b+x)^2}}+\frac {\left (\left (a-2 b+d+\sqrt {a^2+2 a d-4 b d+d^2}\right ) \sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \frac {1}{\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3} \left (-a-d-\sqrt {a^2+2 a d-4 b d+d^2}+2 x\right )} \, dx}{\sqrt [3]{x (-a+x) (-b+x)^2}}\\ \end {align*}

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Mathematica [F]  time = 8.17, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a b-2 b x+x^2}{\sqrt [3]{x (-a+x) (-b+x)^2} \left (b d-(a+d) x+x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.58, size = 306, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{-a b^2 x+\left (2 a b+b^2\right ) x^2+(-a-2 b) x^3+x^4}}{-2 b \sqrt [3]{d}+2 \sqrt [3]{d} x+\sqrt [3]{-a b^2 x+\left (2 a b+b^2\right ) x^2+(-a-2 b) x^3+x^4}}\right )}{\sqrt [3]{d}}+\frac {\log \left (b \sqrt {d}-\sqrt {d} x+\sqrt [6]{d} \sqrt [3]{-a b^2 x+\left (2 a b+b^2\right ) x^2+(-a-2 b) x^3+x^4}\right )}{\sqrt [3]{d}}-\frac {\log \left (b^2 d-2 b d x+d x^2+\left (-b d^{2/3}+d^{2/3} x\right ) \sqrt [3]{-a b^2 x+\left (2 a b+b^2\right ) x^2+(-a-2 b) x^3+x^4}+\sqrt [3]{d} \left (-a b^2 x+\left (2 a b+b^2\right ) x^2+(-a-2 b) x^3+x^4\right )^{2/3}\right )}{2 \sqrt [3]{d}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a*b - 2*b*x + x^2)/((x*(-a + x)*(-b + x)^2)^(1/3)*(b*d - (a + d)*x + x^2)),x]

[Out]

(Sqrt[3]*ArcTan[(Sqrt[3]*(-(a*b^2*x) + (2*a*b + b^2)*x^2 + (-a - 2*b)*x^3 + x^4)^(1/3))/(-2*b*d^(1/3) + 2*d^(1
/3)*x + (-(a*b^2*x) + (2*a*b + b^2)*x^2 + (-a - 2*b)*x^3 + x^4)^(1/3))])/d^(1/3) + Log[b*Sqrt[d] - Sqrt[d]*x +
d^(1/6)*(-(a*b^2*x) + (2*a*b + b^2)*x^2 + (-a - 2*b)*x^3 + x^4)^(1/3)]/d^(1/3) - Log[b^2*d - 2*b*d*x + d*x^2
+ (-(b*d^(2/3)) + d^(2/3)*x)*(-(a*b^2*x) + (2*a*b + b^2)*x^2 + (-a - 2*b)*x^3 + x^4)^(1/3) + d^(1/3)*(-(a*b^2*
x) + (2*a*b + b^2)*x^2 + (-a - 2*b)*x^3 + x^4)^(2/3)]/(2*d^(1/3))

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.17, size = 0, normalized size = 0.00 \[\int \frac {a b -2 b x +x^{2}}{\left (x \left (-a +x \right ) \left (-b +x \right )^{2}\right )^{\frac {1}{3}} \left (b d -\left (a +d \right ) x +x^{2}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [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((a*b-2*b*x+x**2)/(x*(-a+x)*(-b+x)**2)**(1/3)/(b*d-(a+d)*x+x**2),x)

[Out]

Timed out

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