3.29.68 \(\int \frac {a b-(a+b) x+x^2}{((-a+x) (-b+x)^2)^{2/3} (-b^2+a^2 d+2 (b-a d) x+(-1+d) x^2)} \, dx\)

Optimal. Leaf size=305 \[ -\frac {\log \left (a^2 d^{2/3}+\sqrt [3]{x \left (2 a b+b^2\right )-a b^2+x^2 (-a-2 b)+x^3} \left (\sqrt [3]{d} x-a \sqrt [3]{d}\right )+\left (x \left (2 a b+b^2\right )-a b^2+x^2 (-a-2 b)+x^3\right )^{2/3}-2 a d^{2/3} x+d^{2/3} x^2\right )}{4 d^{2/3} (a-b)}+\frac {\log \left (\sqrt [3]{x \left (2 a b+b^2\right )-a b^2+x^2 (-a-2 b)+x^3}+a \sqrt [3]{d}-\sqrt [3]{d} x\right )}{2 d^{2/3} (a-b)}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} a \sqrt [3]{d}-\sqrt {3} \sqrt [3]{d} x}{-2 \sqrt [3]{x \left (2 a b+b^2\right )-a b^2+x^2 (-a-2 b)+x^3}+a \sqrt [3]{d}-\sqrt [3]{d} x}\right )}{2 d^{2/3} (a-b)} \]

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Rubi [A]  time = 0.86, antiderivative size = 513, normalized size of antiderivative = 1.68, number of steps used = 10, number of rules used = 6, integrand size = 60, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6719, 24, 911, 105, 59, 91} \begin {gather*} -\frac {(x-a)^{2/3} (x-b)^{4/3} \log \left (2 \left (\sqrt {d}+1\right ) \left (b-a \sqrt {d}\right )-2 (1-d) x\right )}{4 d^{2/3} (a-b) \left (-\left ((a-x) (b-x)^2\right )\right )^{2/3}}-\frac {(x-a)^{2/3} (x-b)^{4/3} \log \left (2 (1-d) x-2 \left (1-\sqrt {d}\right ) \left (a \sqrt {d}+b\right )\right )}{4 d^{2/3} (a-b) \left (-\left ((a-x) (b-x)^2\right )\right )^{2/3}}+\frac {3 (x-a)^{2/3} (x-b)^{4/3} \log \left (-\sqrt [3]{x-a}-\frac {\sqrt [3]{x-b}}{\sqrt [6]{d}}\right )}{4 d^{2/3} (a-b) \left (-\left ((a-x) (b-x)^2\right )\right )^{2/3}}+\frac {3 (x-a)^{2/3} (x-b)^{4/3} \log \left (\frac {\sqrt [3]{x-b}}{\sqrt [6]{d}}-\sqrt [3]{x-a}\right )}{4 d^{2/3} (a-b) \left (-\left ((a-x) (b-x)^2\right )\right )^{2/3}}+\frac {\sqrt {3} (x-a)^{2/3} (x-b)^{4/3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{x-b}}{\sqrt {3} \sqrt [6]{d} \sqrt [3]{x-a}}\right )}{2 d^{2/3} (a-b) \left (-\left ((a-x) (b-x)^2\right )\right )^{2/3}}+\frac {\sqrt {3} (x-a)^{2/3} (x-b)^{4/3} \tan ^{-1}\left (\frac {2 \sqrt [3]{x-b}}{\sqrt {3} \sqrt [6]{d} \sqrt [3]{x-a}}+\frac {1}{\sqrt {3}}\right )}{2 d^{2/3} (a-b) \left (-\left ((a-x) (b-x)^2\right )\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 24

Int[(u_.)*((a_) + (b_.)*(v_))^(m_)*((A_.) + (B_.)*(v_) + (C_.)*(v_)^2), x_Symbol] :> Dist[1/b^2, Int[u*(a + b*
v)^(m + 1)*Simp[b*B - a*C + b*C*v, x], x], x] /; FreeQ[{a, b, A, B, C}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0] &&
LeQ[m, -1]

Rule 59

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[d/b, 3]}, -Simp[(Sqrt
[3]*q*ArcTan[(2*q*(a + b*x)^(1/3))/(Sqrt[3]*(c + d*x)^(1/3)) + 1/Sqrt[3]])/d, x] + (-Simp[(3*q*Log[(q*(a + b*x
)^(1/3))/(c + d*x)^(1/3) - 1])/(2*d), x] - Simp[(q*Log[c + d*x])/(2*d), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[
b*c - a*d, 0] && PosQ[d/b]

Rule 91

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)*((e_.) + (f_.)*(x_))), x_Symbol] :> With[{q = Rt[
(d*e - c*f)/(b*e - a*f), 3]}, -Simp[(Sqrt[3]*q*ArcTan[1/Sqrt[3] + (2*q*(a + b*x)^(1/3))/(Sqrt[3]*(c + d*x)^(1/
3))])/(d*e - c*f), x] + (Simp[(q*Log[e + f*x])/(2*(d*e - c*f)), x] - Simp[(3*q*Log[q*(a + b*x)^(1/3) - (c + d*
x)^(1/3)])/(2*(d*e - c*f)), x])] /; FreeQ[{a, b, c, d, e, f}, x]

Rule 105

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Dist[b/f, Int[(a
+ b*x)^(m - 1)*(c + d*x)^n, x], x] - Dist[(b*e - a*f)/f, Int[((a + b*x)^(m - 1)*(c + d*x)^n)/(e + f*x), x], x]
/; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[Simplify[m + n + 1], 0] && (GtQ[m, 0] || ( !RationalQ[m] && (Su
mSimplerQ[m, -1] ||  !SumSimplerQ[n, -1])))

Rule 911

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> In
t[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n, 1/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x
] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[m] &&  !IntegerQ[n]

Rule 6719

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m*w^n)^FracPart[p])/(v^(m*F
racPart[p])*w^(n*FracPart[p])), Int[u*v^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] &&  !IntegerQ[p] &&  !
FreeQ[v, x] &&  !FreeQ[w, x]

Rubi steps

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

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Mathematica [C]  time = 0.27, size = 91, normalized size = 0.30 \begin {gather*} -\frac {3 \sqrt [3]{(x-a) (b-x)^2} \left (\, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {b-x}{\sqrt {d} (x-a)}\right )+\, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {x-b}{\sqrt {d} (x-a)}\right )\right )}{4 d (a-b) (x-a)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*((b - x)^2*(-a + x))^(1/3)*(Hypergeometric2F1[2/3, 1, 5/3, (b - x)/(Sqrt[d]*(-a + x))] + Hypergeometric2F1
[2/3, 1, 5/3, (-b + x)/(Sqrt[d]*(-a + x))]))/(4*(a - b)*d*(-a + x))

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IntegrateAlgebraic [A]  time = 6.39, size = 305, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} a \sqrt [3]{d}-\sqrt {3} \sqrt [3]{d} x}{a \sqrt [3]{d}-\sqrt [3]{d} x-2 \sqrt [3]{-a b^2+\left (2 a b+b^2\right ) x+(-a-2 b) x^2+x^3}}\right )}{2 (a-b) d^{2/3}}+\frac {\log \left (a \sqrt [3]{d}-\sqrt [3]{d} x+\sqrt [3]{-a b^2+\left (2 a b+b^2\right ) x+(-a-2 b) x^2+x^3}\right )}{2 (a-b) d^{2/3}}-\frac {\log \left (a^2 d^{2/3}-2 a d^{2/3} x+d^{2/3} x^2+\left (-a \sqrt [3]{d}+\sqrt [3]{d} x\right ) \sqrt [3]{-a b^2+\left (2 a b+b^2\right ) x+(-a-2 b) x^2+x^3}+\left (-a b^2+\left (2 a b+b^2\right ) x+(-a-2 b) x^2+x^3\right )^{2/3}\right )}{4 (a-b) d^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.12, size = 331, normalized size = 1.09 \begin {gather*} \frac {2 \, \sqrt {3} d \sqrt {-\left (-d^{2}\right )^{\frac {1}{3}}} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (-a b^{2} - {\left (a + 2 \, b\right )} x^{2} + x^{3} + {\left (2 \, a b + b^{2}\right )} x\right )}^{\frac {2}{3}} d - {\left (b^{2} - 2 \, b x + x^{2}\right )} \left (-d^{2}\right )^{\frac {1}{3}}\right )} \sqrt {-\left (-d^{2}\right )^{\frac {1}{3}}}}{3 \, {\left (b^{2} d - 2 \, b d x + d x^{2}\right )}}\right ) - \left (-d^{2}\right )^{\frac {2}{3}} \log \left (-\frac {{\left (-a b^{2} - {\left (a + 2 \, b\right )} x^{2} + x^{3} + {\left (2 \, a b + b^{2}\right )} x\right )}^{\frac {2}{3}} \left (-d^{2}\right )^{\frac {1}{3}} d - {\left (b^{2} - 2 \, b x + x^{2}\right )} \left (-d^{2}\right )^{\frac {2}{3}} + {\left (-a b^{2} - {\left (a + 2 \, b\right )} x^{2} + x^{3} + {\left (2 \, a b + b^{2}\right )} x\right )}^{\frac {1}{3}} {\left (a d^{2} - d^{2} x\right )}}{b^{2} - 2 \, b x + x^{2}}\right ) + 2 \, \left (-d^{2}\right )^{\frac {2}{3}} \log \left (\frac {{\left (-a b^{2} - {\left (a + 2 \, b\right )} x^{2} + x^{3} + {\left (2 \, a b + b^{2}\right )} x\right )}^{\frac {2}{3}} d + {\left (b^{2} - 2 \, b x + x^{2}\right )} \left (-d^{2}\right )^{\frac {1}{3}}}{b^{2} - 2 \, b x + x^{2}}\right )}{4 \, {\left (a - b\right )} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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