3.27.9 \(\int \frac {(-a+x) (-b+x) (-2 a b+(a+b) x)}{\sqrt [3]{x (-a+x) (-b+x)} (a^2 b^2 d-2 a b (a+b) d x+(a^2+4 a b+b^2) d x^2-2 (a+b) d x^3+(-1+d) x^4)} \, dx\)

Optimal. Leaf size=227 \[ \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{x-2 \sqrt [6]{d} \sqrt [3]{x^2 (-a-b)+a b x+x^3}}\right )}{2 d^{5/6}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{2 \sqrt [6]{d} \sqrt [3]{x^2 (-a-b)+a b x+x^3}+x}\right )}{2 d^{5/6}}-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt [6]{d} \sqrt [3]{x^2 (-a-b)+a b x+x^3}}\right )}{d^{5/6}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [6]{d} \left (x^2 (-a-b)+a b x+x^3\right )^{2/3}+\frac {x^2}{\sqrt [6]{d}}}{x \sqrt [3]{x^2 (-a-b)+a b x+x^3}}\right )}{2 d^{5/6}} \]

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Rubi [F]  time = 41.62, 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+x) (-b+x) (-2 a b+(a+b) x)}{\sqrt [3]{x (-a+x) (-b+x)} \left (a^2 b^2 d-2 a b (a+b) d x+\left (a^2+4 a b+b^2\right ) d x^2-2 (a+b) d x^3+(-1+d) x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

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

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Mathematica [F]  time = 3.71, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(-a+x) (-b+x) (-2 a b+(a+b) x)}{\sqrt [3]{x (-a+x) (-b+x)} \left (a^2 b^2 d-2 a b (a+b) d x+\left (a^2+4 a b+b^2\right ) d x^2-2 (a+b) d x^3+(-1+d) x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 1.21, size = 227, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{x-2 \sqrt [6]{d} \sqrt [3]{a b x+(-a-b) x^2+x^3}}\right )}{2 d^{5/6}}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} x}{x+2 \sqrt [6]{d} \sqrt [3]{a b x+(-a-b) x^2+x^3}}\right )}{2 d^{5/6}}-\frac {\tanh ^{-1}\left (\frac {x}{\sqrt [6]{d} \sqrt [3]{a b x+(-a-b) x^2+x^3}}\right )}{d^{5/6}}-\frac {\tanh ^{-1}\left (\frac {\frac {x^2}{\sqrt [6]{d}}+\sqrt [6]{d} \left (a b x+(-a-b) x^2+x^3\right )^{2/3}}{x \sqrt [3]{a b x+(-a-b) x^2+x^3}}\right )}{2 d^{5/6}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[3]*ArcTan[(Sqrt[3]*x)/(x - 2*d^(1/6)*(a*b*x + (-a - b)*x^2 + x^3)^(1/3))])/(2*d^(5/6)) - (Sqrt[3]*ArcTan
[(Sqrt[3]*x)/(x + 2*d^(1/6)*(a*b*x + (-a - b)*x^2 + x^3)^(1/3))])/(2*d^(5/6)) - ArcTanh[x/(d^(1/6)*(a*b*x + (-
a - b)*x^2 + x^3)^(1/3))]/d^(5/6) - ArcTanh[(x^2/d^(1/6) + d^(1/6)*(a*b*x + (-a - b)*x^2 + x^3)^(2/3))/(x*(a*b
*x + (-a - b)*x^2 + x^3)^(1/3))]/(2*d^(5/6))

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

[Out]

Timed out

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giac [A]  time = 0.59, size = 318, normalized size = 1.40 \begin {gather*} \frac {\sqrt {3} \left (-d^{5}\right )^{\frac {5}{6}} \log \left (\sqrt {3} {\left (\frac {a b}{x^{2}} - \frac {a}{x} - \frac {b}{x} + 1\right )}^{\frac {1}{3}} \left (-\frac {1}{d}\right )^{\frac {1}{6}} + {\left (\frac {a b}{x^{2}} - \frac {a}{x} - \frac {b}{x} + 1\right )}^{\frac {2}{3}} + \left (-\frac {1}{d}\right )^{\frac {1}{3}}\right )}{4 \, d^{5}} - \frac {\sqrt {3} \left (-d^{5}\right )^{\frac {5}{6}} \log \left (-\sqrt {3} {\left (\frac {a b}{x^{2}} - \frac {a}{x} - \frac {b}{x} + 1\right )}^{\frac {1}{3}} \left (-\frac {1}{d}\right )^{\frac {1}{6}} + {\left (\frac {a b}{x^{2}} - \frac {a}{x} - \frac {b}{x} + 1\right )}^{\frac {2}{3}} + \left (-\frac {1}{d}\right )^{\frac {1}{3}}\right )}{4 \, d^{5}} - \frac {\left (-d^{5}\right )^{\frac {5}{6}} \arctan \left (\frac {\sqrt {3} \left (-\frac {1}{d}\right )^{\frac {1}{6}} + 2 \, {\left (\frac {a b}{x^{2}} - \frac {a}{x} - \frac {b}{x} + 1\right )}^{\frac {1}{3}}}{\left (-\frac {1}{d}\right )^{\frac {1}{6}}}\right )}{2 \, d^{5}} - \frac {\left (-d^{5}\right )^{\frac {5}{6}} \arctan \left (-\frac {\sqrt {3} \left (-\frac {1}{d}\right )^{\frac {1}{6}} - 2 \, {\left (\frac {a b}{x^{2}} - \frac {a}{x} - \frac {b}{x} + 1\right )}^{\frac {1}{3}}}{\left (-\frac {1}{d}\right )^{\frac {1}{6}}}\right )}{2 \, d^{5}} - \frac {\left (-d^{5}\right )^{\frac {5}{6}} \arctan \left (\frac {{\left (\frac {a b}{x^{2}} - \frac {a}{x} - \frac {b}{x} + 1\right )}^{\frac {1}{3}}}{\left (-\frac {1}{d}\right )^{\frac {1}{6}}}\right )}{d^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(3)*(-d^5)^(5/6)*log(sqrt(3)*(a*b/x^2 - a/x - b/x + 1)^(1/3)*(-1/d)^(1/6) + (a*b/x^2 - a/x - b/x + 1)^
(2/3) + (-1/d)^(1/3))/d^5 - 1/4*sqrt(3)*(-d^5)^(5/6)*log(-sqrt(3)*(a*b/x^2 - a/x - b/x + 1)^(1/3)*(-1/d)^(1/6)
+ (a*b/x^2 - a/x - b/x + 1)^(2/3) + (-1/d)^(1/3))/d^5 - 1/2*(-d^5)^(5/6)*arctan((sqrt(3)*(-1/d)^(1/6) + 2*(a*
b/x^2 - a/x - b/x + 1)^(1/3))/(-1/d)^(1/6))/d^5 - 1/2*(-d^5)^(5/6)*arctan(-(sqrt(3)*(-1/d)^(1/6) - 2*(a*b/x^2
- a/x - b/x + 1)^(1/3))/(-1/d)^(1/6))/d^5 - (-d^5)^(5/6)*arctan((a*b/x^2 - a/x - b/x + 1)^(1/3)/(-1/d)^(1/6))/
d^5

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {\left (-a +x \right ) \left (-b +x \right ) \left (-2 a b +\left (a +b \right ) x \right )}{\left (x \left (-a +x \right ) \left (-b +x \right )\right )^{\frac {1}{3}} \left (a^{2} b^{2} d -2 a b \left (a +b \right ) d x +\left (a^{2}+4 a b +b^{2}\right ) d \,x^{2}-2 \left (a +b \right ) d \,x^{3}+\left (-1+d \right ) x^{4}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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