3.23.70 \(\int \frac {(a b-2 b x+x^2) (b^2-2 b x+x^2)}{(x (-a+x) (-b+x)^3)^{3/4} (b d-(a+d) x+x^2)} \, dx\)

Optimal. Leaf size=173 \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \left (a b^3 x+x^3 \left (3 a b+3 b^2\right )+x^2 \left (-3 a b^2-b^3\right )+x^4 (-a-3 b)+x^5\right )^{3/4}}{x (x-a) (b-x)^2}\right )}{d^{3/4}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{d} \left (a b^3 x+x^3 \left (3 a b+3 b^2\right )+x^2 \left (-3 a b^2-b^3\right )+x^4 (-a-3 b)+x^5\right )^{3/4}}{x (x-a) (b-x)^2}\right )}{d^{3/4}} \]

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Rubi [F]  time = 6.83, 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 (a b-2 b x+x^2\right ) \left (b^2-2 b x+x^2\right )}{\left (x (-a+x) (-b+x)^3\right )^{3/4} \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)*(b^2 - 2*b*x + x^2))/((x*(-a + x)*(-b + x)^3)^(3/4)*(b*d - (a + d)*x + x^2)),x]

[Out]

(4*(b - x)^2*x*(1 - x/a)^(3/4)*(1 - x/b)^(1/4)*AppellF1[1/4, 3/4, 1/4, 5/4, x/a, x/b])/((a - x)*(b - x)^3*x)^(
3/4) + ((a - 2*b + d + Sqrt[a^2 + 2*a*d - 4*b*d + d^2])*x^(3/4)*(-a + x)^(3/4)*(-b + x)^(9/4)*Defer[Int][1/(x^
(3/4)*(-a + x)^(3/4)*(-b + x)^(1/4)*(-a - d - Sqrt[a^2 + 2*a*d - 4*b*d + d^2] + 2*x)), x])/((a - x)*(b - x)^3*
x)^(3/4) + ((a - 2*b + d - Sqrt[a^2 + 2*a*d - 4*b*d + d^2])*x^(3/4)*(-a + x)^(3/4)*(-b + x)^(9/4)*Defer[Int][1
/(x^(3/4)*(-a + x)^(3/4)*(-b + x)^(1/4)*(-a - d + Sqrt[a^2 + 2*a*d - 4*b*d + d^2] + 2*x)), x])/((a - x)*(b - x
)^3*x)^(3/4)

Rubi steps

\begin {align*} \int \frac {\left (a b-2 b x+x^2\right ) \left (b^2-2 b x+x^2\right )}{\left (x (-a+x) (-b+x)^3\right )^{3/4} \left (b d-(a+d) x+x^2\right )} \, dx &=\int \frac {(-b+x)^2 \left (a b-2 b x+x^2\right )}{\left (x (-a+x) (-b+x)^3\right )^{3/4} \left (b d-(a+d) x+x^2\right )} \, dx\\ &=\frac {\left (x^{3/4} (-a+x)^{3/4} (-b+x)^{9/4}\right ) \int \frac {a b-2 b x+x^2}{x^{3/4} (-a+x)^{3/4} \sqrt [4]{-b+x} \left (b d-(a+d) x+x^2\right )} \, dx}{\left (x (-a+x) (-b+x)^3\right )^{3/4}}\\ &=\frac {\left (x^{3/4} (-a+x)^{3/4} (-b+x)^{9/4}\right ) \int \left (\frac {1}{x^{3/4} (-a+x)^{3/4} \sqrt [4]{-b+x}}+\frac {b (a-d)+(a-2 b+d) x}{x^{3/4} (-a+x)^{3/4} \sqrt [4]{-b+x} \left (b d+(-a-d) x+x^2\right )}\right ) \, dx}{\left (x (-a+x) (-b+x)^3\right )^{3/4}}\\ &=\frac {\left (x^{3/4} (-a+x)^{3/4} (-b+x)^{9/4}\right ) \int \frac {1}{x^{3/4} (-a+x)^{3/4} \sqrt [4]{-b+x}} \, dx}{\left (x (-a+x) (-b+x)^3\right )^{3/4}}+\frac {\left (x^{3/4} (-a+x)^{3/4} (-b+x)^{9/4}\right ) \int \frac {b (a-d)+(a-2 b+d) x}{x^{3/4} (-a+x)^{3/4} \sqrt [4]{-b+x} \left (b d+(-a-d) x+x^2\right )} \, dx}{\left (x (-a+x) (-b+x)^3\right )^{3/4}}\\ &=\frac {\left (x^{3/4} (-a+x)^{3/4} (-b+x)^{9/4}\right ) \int \left (\frac {a-2 b+d+\sqrt {a^2+2 a d-4 b d+d^2}}{x^{3/4} (-a+x)^{3/4} \sqrt [4]{-b+x} \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}}{x^{3/4} (-a+x)^{3/4} \sqrt [4]{-b+x} \left (-a-d+\sqrt {a^2+2 a d-4 b d+d^2}+2 x\right )}\right ) \, dx}{\left (x (-a+x) (-b+x)^3\right )^{3/4}}+\frac {\left (x^{3/4} (-b+x)^{9/4} \left (1-\frac {x}{a}\right )^{3/4}\right ) \int \frac {1}{x^{3/4} \sqrt [4]{-b+x} \left (1-\frac {x}{a}\right )^{3/4}} \, dx}{\left (x (-a+x) (-b+x)^3\right )^{3/4}}\\ &=\frac {\left (\left (a-2 b+d-\sqrt {a^2+2 a d-4 b d+d^2}\right ) x^{3/4} (-a+x)^{3/4} (-b+x)^{9/4}\right ) \int \frac {1}{x^{3/4} (-a+x)^{3/4} \sqrt [4]{-b+x} \left (-a-d+\sqrt {a^2+2 a d-4 b d+d^2}+2 x\right )} \, dx}{\left (x (-a+x) (-b+x)^3\right )^{3/4}}+\frac {\left (\left (a-2 b+d+\sqrt {a^2+2 a d-4 b d+d^2}\right ) x^{3/4} (-a+x)^{3/4} (-b+x)^{9/4}\right ) \int \frac {1}{x^{3/4} (-a+x)^{3/4} \sqrt [4]{-b+x} \left (-a-d-\sqrt {a^2+2 a d-4 b d+d^2}+2 x\right )} \, dx}{\left (x (-a+x) (-b+x)^3\right )^{3/4}}+\frac {\left (x^{3/4} (-b+x)^2 \left (1-\frac {x}{a}\right )^{3/4} \sqrt [4]{1-\frac {x}{b}}\right ) \int \frac {1}{x^{3/4} \left (1-\frac {x}{a}\right )^{3/4} \sqrt [4]{1-\frac {x}{b}}} \, dx}{\left (x (-a+x) (-b+x)^3\right )^{3/4}}\\ &=\frac {4 (b-x)^2 x \left (1-\frac {x}{a}\right )^{3/4} \sqrt [4]{1-\frac {x}{b}} F_1\left (\frac {1}{4};\frac {3}{4},\frac {1}{4};\frac {5}{4};\frac {x}{a},\frac {x}{b}\right )}{\left ((a-x) (b-x)^3 x\right )^{3/4}}+\frac {\left (\left (a-2 b+d-\sqrt {a^2+2 a d-4 b d+d^2}\right ) x^{3/4} (-a+x)^{3/4} (-b+x)^{9/4}\right ) \int \frac {1}{x^{3/4} (-a+x)^{3/4} \sqrt [4]{-b+x} \left (-a-d+\sqrt {a^2+2 a d-4 b d+d^2}+2 x\right )} \, dx}{\left (x (-a+x) (-b+x)^3\right )^{3/4}}+\frac {\left (\left (a-2 b+d+\sqrt {a^2+2 a d-4 b d+d^2}\right ) x^{3/4} (-a+x)^{3/4} (-b+x)^{9/4}\right ) \int \frac {1}{x^{3/4} (-a+x)^{3/4} \sqrt [4]{-b+x} \left (-a-d-\sqrt {a^2+2 a d-4 b d+d^2}+2 x\right )} \, dx}{\left (x (-a+x) (-b+x)^3\right )^{3/4}}\\ \end {align*}

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Mathematica [F]  time = 5.82, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a b-2 b x+x^2\right ) \left (b^2-2 b x+x^2\right )}{\left (x (-a+x) (-b+x)^3\right )^{3/4} \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)*(b^2 - 2*b*x + x^2))/((x*(-a + x)*(-b + x)^3)^(3/4)*(b*d - (a + d)*x + x^2)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 3.87, size = 173, normalized size = 1.00 \begin {gather*} \frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \left (a b^3 x+\left (-3 a b^2-b^3\right ) x^2+\left (3 a b+3 b^2\right ) x^3+(-a-3 b) x^4+x^5\right )^{3/4}}{(b-x)^2 x (-a+x)}\right )}{d^{3/4}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{d} \left (a b^3 x+\left (-3 a b^2-b^3\right ) x^2+\left (3 a b+3 b^2\right ) x^3+(-a-3 b) x^4+x^5\right )^{3/4}}{(b-x)^2 x (-a+x)}\right )}{d^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcTan[(d^(1/4)*(a*b^3*x + (-3*a*b^2 - b^3)*x^2 + (3*a*b + 3*b^2)*x^3 + (-a - 3*b)*x^4 + x^5)^(3/4))/((b -
x)^2*x*(-a + x))])/d^(3/4) - (2*ArcTanh[(d^(1/4)*(a*b^3*x + (-3*a*b^2 - b^3)*x^2 + (3*a*b + 3*b^2)*x^3 + (-a -
3*b)*x^4 + x^5)^(3/4))/((b - x)^2*x*(-a + x))])/d^(3/4)

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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)*(b^2-2*b*x+x^2)/(x*(-a+x)*(-b+x)^3)^(3/4)/(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 {{\left (a b - 2 \, b x + x^{2}\right )} {\left (b^{2} - 2 \, b x + x^{2}\right )}}{\left ({\left (a - x\right )} {\left (b - x\right )}^{3} x\right )^{\frac {3}{4}} {\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)*(b^2-2*b*x+x^2)/(x*(-a+x)*(-b+x)^3)^(3/4)/(b*d-(a+d)*x+x^2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[\int \frac {\left (a b -2 b x +x^{2}\right ) \left (b^{2}-2 b x +x^{2}\right )}{\left (x \left (-a +x \right ) \left (-b +x \right )^{3}\right )^{\frac {3}{4}} \left (b d -\left (a +d \right ) x +x^{2}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a*b-2*b*x+x^2)*(b^2-2*b*x+x^2)/(x*(-a+x)*(-b+x)^3)^(3/4)/(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 {{\left (a b - 2 \, b x + x^{2}\right )} {\left (b^{2} - 2 \, b x + x^{2}\right )}}{\left ({\left (a - x\right )} {\left (b - x\right )}^{3} x\right )^{\frac {3}{4}} {\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)*(b^2-2*b*x+x^2)/(x*(-a+x)*(-b+x)^3)^(3/4)/(b*d-(a+d)*x+x^2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (b^2-2\,b\,x+x^2\right )\,\left (x^2-2\,b\,x+a\,b\right )}{\left (x^2+\left (-a-d\right )\,x+b\,d\right )\,{\left (x\,\left (a-x\right )\,{\left (b-x\right )}^3\right )}^{3/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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