3.29.42 \(\int \frac {a b-2 b x+x^2}{\sqrt [3]{x (-a+x) (-b+x)^2} (b-(1+a d) x+d x^2)} \, dx\)

Optimal. Leaf size=291 \[ \frac {\log \left (\sqrt [3]{d} \sqrt [3]{x^2 \left (2 a b+b^2\right )-a b^2 x+x^3 (-a-2 b)+x^4}+b-x\right )}{d^{2/3}}-\frac {\log \left (d^{2/3} \left (x^2 \left (2 a b+b^2\right )-a b^2 x+x^3 (-a-2 b)+x^4\right )^{2/3}+\sqrt [3]{x^2 \left (2 a b+b^2\right )-a b^2 x+x^3 (-a-2 b)+x^4} \left (\sqrt [3]{d} x-b \sqrt [3]{d}\right )+b^2-2 b x+x^2\right )}{2 d^{2/3}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{d} \sqrt [3]{x^2 \left (2 a b+b^2\right )-a b^2 x+x^3 (-a-2 b)+x^4}}{\sqrt [3]{d} \sqrt [3]{x^2 \left (2 a b+b^2\right )-a b^2 x+x^3 (-a-2 b)+x^4}-2 b+2 x}\right )}{d^{2/3}} \]

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Rubi [F]  time = 7.25, 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-(1+a d) x+d 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 - (1 + a*d)*x + d*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*d*(-((a - x)*(b - x)^2*x))^(1/
3)) + ((1 + a*d - 2*b*d + Sqrt[1 + 2*a*d - 4*b*d + a^2*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)*(-1 - a*d - Sqrt[1 + 2*a*d - 4*b*d + a^2*d^2] + 2*d*x)), x])/(d*(-((a
- x)*(b - x)^2*x))^(1/3)) + ((1 + a*d - 2*b*d - Sqrt[1 + 2*a*d - 4*b*d + a^2*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)*(-1 - a*d + Sqrt[1 + 2*a*d - 4*b*d + a^2*d^2]
+ 2*d*x)), x])/(d*(-((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-(1+a d) x+d 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-(1+a d) x+d 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}{d \sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3}}-\frac {b-a b d-(1+a d-2 b d) x}{d \sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3} \left (b+(-1-a d) x+d 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}{d \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 b d-(1+a d-2 b d) x}{\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3} \left (b+(-1-a d) x+d x^2\right )} \, dx}{d \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-a d+2 b d-\sqrt {1+2 a d-4 b d+a^2 d^2}}{\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3} \left (-1-a d-\sqrt {1+2 a d-4 b d+a^2 d^2}+2 d x\right )}+\frac {-1-a d+2 b d+\sqrt {1+2 a d-4 b d+a^2 d^2}}{\sqrt [3]{x} \sqrt [3]{-a+x} (-b+x)^{2/3} \left (-1-a d+\sqrt {1+2 a d-4 b d+a^2 d^2}+2 d x\right )}\right ) \, dx}{d \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}{d \sqrt [3]{x (-a+x) (-b+x)^2}}\\ &=-\frac {\left (\left (-1-a d+2 b d-\sqrt {1+2 a d-4 b d+a^2 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 (-1-a d-\sqrt {1+2 a d-4 b d+a^2 d^2}+2 d x\right )} \, dx}{d \sqrt [3]{x (-a+x) (-b+x)^2}}-\frac {\left (\left (-1-a d+2 b d+\sqrt {1+2 a d-4 b d+a^2 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 (-1-a d+\sqrt {1+2 a d-4 b d+a^2 d^2}+2 d x\right )} \, dx}{d \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}{d \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 d \sqrt [3]{-\left ((a-x) (b-x)^2 x\right )}}-\frac {\left (\left (-1-a d+2 b d-\sqrt {1+2 a d-4 b d+a^2 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 (-1-a d-\sqrt {1+2 a d-4 b d+a^2 d^2}+2 d x\right )} \, dx}{d \sqrt [3]{x (-a+x) (-b+x)^2}}-\frac {\left (\left (-1-a d+2 b d+\sqrt {1+2 a d-4 b d+a^2 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 (-1-a d+\sqrt {1+2 a d-4 b d+a^2 d^2}+2 d x\right )} \, dx}{d \sqrt [3]{x (-a+x) (-b+x)^2}}\\ \end {align*}

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Mathematica [F]  time = 5.77, 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-(1+a d) x+d 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 - (1 + a*d)*x + d*x^2)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.56, size = 291, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{d} \sqrt [3]{-a b^2 x+\left (2 a b+b^2\right ) x^2+(-a-2 b) x^3+x^4}}{-2 b+2 x+\sqrt [3]{d} \sqrt [3]{-a b^2 x+\left (2 a b+b^2\right ) x^2+(-a-2 b) x^3+x^4}}\right )}{d^{2/3}}+\frac {\log \left (b-x+\sqrt [3]{d} \sqrt [3]{-a b^2 x+\left (2 a b+b^2\right ) x^2+(-a-2 b) x^3+x^4}\right )}{d^{2/3}}-\frac {\log \left (b^2-2 b x+x^2+\left (-b \sqrt [3]{d}+\sqrt [3]{d} x\right ) \sqrt [3]{-a b^2 x+\left (2 a b+b^2\right ) x^2+(-a-2 b) x^3+x^4}+d^{2/3} \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 d^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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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-(a*d+1)*x+d*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 (d x^{2} - {\left (a d + 1\right )} x + b\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-(a*d+1)*x+d*x^2),x, algorithm="giac")

[Out]

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

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

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-(a*d+1)*x+d*x^2),x)

[Out]

int((a*b-2*b*x+x^2)/(x*(-a+x)*(-b+x)^2)^(1/3)/(b-(a*d+1)*x+d*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 (d x^{2} - {\left (a d + 1\right )} x + b\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-(a*d+1)*x+d*x^2),x, algorithm="maxima")

[Out]

integrate((a*b - 2*b*x + x^2)/((-(a - x)*(b - x)^2*x)^(1/3)*(d*x^2 - (a*d + 1)*x + b)), 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\,\left (a-x\right )\,{\left (b-x\right )}^2\right )}^{1/3}\,\left (d\,x^2+\left (-a\,d-1\right )\,x+b\right )} \,d x \end {gather*}

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 - x*(a*d + 1) + d*x^2)),x)

[Out]

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

[Out]

Timed out

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