3.31.93 \(\int \frac {b-x}{\sqrt [3]{(-a+x) (-b+x)^2} (a^2-b^2 d-2 (a-b d) x+(1-d) x^2)} \, dx\) [3093]

Optimal. Leaf size=541 \[ -\frac {\sqrt {3} \text {ArcTan}\left (\frac {\sqrt {3} (a-b)^{2/3} \sqrt [3]{d} \sqrt [3]{-a b^2+\left (2 a b+b^2\right ) x+(-a-2 b) x^2+x^3}}{-2 a (-a+b)^{2/3}+2 (-a+b)^{2/3} x+(a-b)^{2/3} \sqrt [3]{d} \sqrt [3]{-a b^2+\left (2 a b+b^2\right ) x+(-a-2 b) x^2+x^3}}\right )}{2 \sqrt [3]{a-b} (-a+b)^{2/3} d^{2/3}}-\frac {\log \left (a (-a+b)^{2/3}-(-a+b)^{2/3} x+(a-b)^{2/3} \sqrt [3]{d} \sqrt [3]{-a b^2+\left (2 a b+b^2\right ) x+(-a-2 b) x^2+x^3}\right )}{2 \sqrt [3]{a-b} (-a+b)^{2/3} d^{2/3}}+\frac {\log \left (a^3 \sqrt [3]{-a+b}-a^2 b \sqrt [3]{-a+b}-2 a^2 \sqrt [3]{-a+b} x+2 a b \sqrt [3]{-a+b} x+a \sqrt [3]{-a+b} x^2-b \sqrt [3]{-a+b} x^2+\left (a (a-b)^{2/3} (-a+b)^{2/3} \sqrt [3]{d}-(a-b)^{2/3} (-a+b)^{2/3} \sqrt [3]{d} x\right ) \sqrt [3]{-a b^2+\left (2 a b+b^2\right ) x+(-a-2 b) x^2+x^3}+\left (-a \sqrt [3]{a-b} d^{2/3}+\sqrt [3]{a-b} b d^{2/3}\right ) \left (-a b^2+\left (2 a b+b^2\right ) x+(-a-2 b) x^2+x^3\right )^{2/3}\right )}{4 \sqrt [3]{a-b} (-a+b)^{2/3} d^{2/3}} \]

[Out]

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

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Rubi [A]
time = 0.73, antiderivative size = 513, normalized size of antiderivative = 0.95, number of steps used = 12, number of rules used = 7, integrand size = 53, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {6851, 21, 925, 132, 61, 12, 93} \begin {gather*} -\frac {\sqrt {3} \sqrt [3]{x-a} (x-b)^{2/3} \text {ArcTan}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{x-a}}{\sqrt {3} \sqrt [6]{d} \sqrt [3]{x-b}}\right )}{2 d^{2/3} (a-b) \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}-\frac {\sqrt {3} \sqrt [3]{x-a} (x-b)^{2/3} \text {ArcTan}\left (\frac {2 \sqrt [3]{x-a}}{\sqrt {3} \sqrt [6]{d} \sqrt [3]{x-b}}+\frac {1}{\sqrt {3}}\right )}{2 d^{2/3} (a-b) \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}+\frac {\sqrt [3]{x-a} (x-b)^{2/3} \log \left (2 \left (1-\sqrt {d}\right ) \left (a+b \sqrt {d}\right )-2 (1-d) x\right )}{4 d^{2/3} (a-b) \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}+\frac {\sqrt [3]{x-a} (x-b)^{2/3} \log \left (2 (1-d) x-2 \left (\sqrt {d}+1\right ) \left (a-b \sqrt {d}\right )\right )}{4 d^{2/3} (a-b) \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}-\frac {3 \sqrt [3]{x-a} (x-b)^{2/3} \log \left (-\frac {\sqrt [3]{x-a}}{\sqrt [6]{d}}-\sqrt [3]{x-b}\right )}{4 d^{2/3} (a-b) \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}-\frac {3 \sqrt [3]{x-a} (x-b)^{2/3} \log \left (\frac {\sqrt [3]{x-a}}{\sqrt [6]{d}}-\sqrt [3]{x-b}\right )}{4 d^{2/3} (a-b) \sqrt [3]{-\left ((a-x) (b-x)^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 61

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

Rule 93

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 132

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

Rule 925

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 6851

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a*v^m*w^n)^FracPart[p]/(v^(m*Fr
acPart[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 {b-x}{\sqrt [3]{(-a+x) (-b+x)^2} \left (a^2-b^2 d-2 (a-b d) x+(1-d) x^2\right )} \, dx &=\frac {\left (\sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \frac {b-x}{\sqrt [3]{-a+x} (-b+x)^{2/3} \left (a^2-b^2 d-2 (a-b d) x+(1-d) x^2\right )} \, dx}{\sqrt [3]{(-a+x) (-b+x)^2}}\\ &=-\frac {\left (\sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \frac {\sqrt [3]{-b+x}}{\sqrt [3]{-a+x} \left (a^2-b^2 d-2 (a-b d) x+(1-d) x^2\right )} \, dx}{\sqrt [3]{(-a+x) (-b+x)^2}}\\ &=-\frac {\left (\sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \left (\frac {(1-d) \sqrt [3]{-b+x}}{(a-b) \sqrt {d} \sqrt [3]{-a+x} \left (2 a-2 (a-b) \sqrt {d}-2 b d-2 (1-d) x\right )}+\frac {(1-d) \sqrt [3]{-b+x}}{(a-b) \sqrt {d} \sqrt [3]{-a+x} \left (-2 a-2 (a-b) \sqrt {d}+2 b d+2 (1-d) x\right )}\right ) \, dx}{\sqrt [3]{(-a+x) (-b+x)^2}}\\ &=-\frac {\left ((1-d) \sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \frac {\sqrt [3]{-b+x}}{\sqrt [3]{-a+x} \left (2 a-2 (a-b) \sqrt {d}-2 b d-2 (1-d) x\right )} \, dx}{(a-b) \sqrt {d} \sqrt [3]{(-a+x) (-b+x)^2}}-\frac {\left ((1-d) \sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \frac {\sqrt [3]{-b+x}}{\sqrt [3]{-a+x} \left (-2 a-2 (a-b) \sqrt {d}+2 b d+2 (1-d) x\right )} \, dx}{(a-b) \sqrt {d} \sqrt [3]{(-a+x) (-b+x)^2}}\\ &=-\frac {\left ((1-d) \sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \frac {1}{\sqrt [3]{-a+x} (-b+x)^{2/3} \left (-2 a-2 (a-b) \sqrt {d}+2 b d+2 (1-d) x\right )} \, dx}{\left (1-\sqrt {d}\right ) \sqrt {d} \sqrt [3]{(-a+x) (-b+x)^2}}-\frac {\left ((1-d) \sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \frac {1}{\sqrt [3]{-a+x} (-b+x)^{2/3} \left (2 a-2 (a-b) \sqrt {d}-2 b d-2 (1-d) x\right )} \, dx}{\left (1+\sqrt {d}\right ) \sqrt {d} \sqrt [3]{(-a+x) (-b+x)^2}}\\ &=-\frac {\sqrt {3} \sqrt [3]{-a+x} (-b+x)^{2/3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{-a+x}}{\sqrt {3} \sqrt [6]{d} \sqrt [3]{-b+x}}\right )}{2 (a-b) d^{2/3} \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}-\frac {\sqrt {3} \sqrt [3]{-a+x} (-b+x)^{2/3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{-a+x}}{\sqrt {3} \sqrt [6]{d} \sqrt [3]{-b+x}}\right )}{2 (a-b) d^{2/3} \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}+\frac {\sqrt [3]{-a+x} (-b+x)^{2/3} \log \left (2 \left (1-\sqrt {d}\right ) \left (a+b \sqrt {d}\right )-2 (1-d) x\right )}{4 (a-b) d^{2/3} \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}+\frac {\sqrt [3]{-a+x} (-b+x)^{2/3} \log \left (-2 \left (1+\sqrt {d}\right ) \left (a-b \sqrt {d}\right )+2 (1-d) x\right )}{4 (a-b) d^{2/3} \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}-\frac {3 \sqrt [3]{-a+x} (-b+x)^{2/3} \log \left (-\frac {\sqrt [3]{-a+x}}{\sqrt [6]{d}}-\sqrt [3]{-b+x}\right )}{4 (a-b) d^{2/3} \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}-\frac {3 \sqrt [3]{-a+x} (-b+x)^{2/3} \log \left (\frac {\sqrt [3]{-a+x}}{\sqrt [6]{d}}-\sqrt [3]{-b+x}\right )}{4 (a-b) d^{2/3} \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}\\ \end {align*}

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Mathematica [A]
time = 0.26, size = 254, normalized size = 0.47 \begin {gather*} \frac {(b-x)^{2/3} \sqrt [3]{-a+x} \left (-2 \sqrt {3} \text {ArcTan}\left (\frac {1+\frac {2 \sqrt [3]{d} (b-x)^{2/3}}{(-a+x)^{2/3}}}{\sqrt {3}}\right )+\log \left (1+\frac {\sqrt [3]{d} (b-x)^{2/3}}{(-a+x)^{2/3}}-\frac {\sqrt [6]{d} \sqrt [3]{b-x}}{\sqrt [3]{-a+x}}\right )-2 \log \left (-1+\frac {\sqrt [6]{d} \sqrt [3]{b-x}}{\sqrt [3]{-a+x}}\right )-2 \log \left (1+\frac {\sqrt [6]{d} \sqrt [3]{b-x}}{\sqrt [3]{-a+x}}\right )+\log \left (1+\frac {\sqrt [3]{d} (b-x)^{2/3}}{(-a+x)^{2/3}}+\frac {\sqrt [6]{d} \sqrt [3]{b-x}}{\sqrt [3]{-a+x}}\right )\right )}{4 (a-b) d^{2/3} \sqrt [3]{(b-x)^2 (-a+x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {b -x}{\left (\left (-a +x \right ) \left (-b +x \right )^{2}\right )^{\frac {1}{3}} \left (a^{2}-b^{2} d -2 \left (-b d +a \right ) x +\left (1-d \right ) x^{2}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.43, size = 323, normalized size = 0.60 \begin {gather*} \frac {2 \, \sqrt {3} {\left (d^{2}\right )}^{\frac {1}{6}} d \arctan \left (\frac {\sqrt {3} {\left (d^{2}\right )}^{\frac {1}{6}} {\left ({\left (b^{2} d - 2 \, b d x + d x^{2}\right )} {\left (d^{2}\right )}^{\frac {1}{3}} + 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}} {\left (d^{2}\right )}^{\frac {2}{3}}\right )}}{3 \, {\left (b^{2} d^{2} - 2 \, b d^{2} x + d^{2} x^{2}\right )}}\right ) - 2 \, {\left (d^{2}\right )}^{\frac {2}{3}} \log \left (-\frac {{\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 {2}{3}} d}{b^{2} - 2 \, b x + x^{2}}\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 {1}{3}} {\left (a d - d x\right )} - {\left (b^{2} d - 2 \, b d x + d x^{2}\right )} {\left (d^{2}\right )}^{\frac {1}{3}} - {\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 {2}{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((b-x)/((-a+x)*(-b+x)^2)^(1/3)/(a^2-b^2*d-2*(-b*d+a)*x+(1-d)*x^2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)] Timed out
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((b-x)/((-a+x)*(-b+x)**2)**(1/3)/(a**2-b**2*d-2*(-b*d+a)*x+(1-d)*x**2),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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