3.31.67 \(\int \frac {-a+x}{\sqrt [3]{(-a+x) (-b+x)^2} (-b^2+a^2 d+2 (b-a d) x+(-1+d) x^2)} \, dx\) [3067]
Optimal. Leaf size=481 \[ \frac {\left (1+i \sqrt {3}\right ) (-b+x)^{2/3} \left (\sqrt [3]{d} (a-x)^{2/3}-\sqrt [6]{d} \sqrt [3]{a-x} \sqrt [3]{-b+x}+(-b+x)^{2/3}\right ) \left (\sqrt [3]{d} (a-x)^{2/3}+\sqrt [6]{d} \sqrt [3]{a-x} \sqrt [3]{-b+x}+(-b+x)^{2/3}\right ) \left (-a \sqrt [3]{d}+\sqrt [3]{d} x+\sqrt [3]{a-x} (-b+x)^{2/3}\right ) \left (\frac {\left (-3 i+\sqrt {3}\right ) \text {ArcTan}\left (\frac {\sqrt {3} \sqrt [3]{-b+x}}{-2 \sqrt [6]{d} \sqrt [3]{a-x}+\sqrt [3]{-b+x}}\right )}{4 (a-b) d^{5/6}}+\frac {\left (3 i-\sqrt {3}\right ) \text {ArcTan}\left (\frac {\sqrt {3} \sqrt [3]{-b+x}}{2 \sqrt [6]{d} \sqrt [3]{a-x}+\sqrt [3]{-b+x}}\right )}{4 (a-b) d^{5/6}}+\frac {\left (1-i \sqrt {3}\right ) \tanh ^{-1}\left (\frac {(a-x)^{2/3} \sqrt [3]{-b+x}}{\sqrt [6]{d} (-a+x)}\right )}{2 (a-b) d^{5/6}}+\frac {i \left (i+\sqrt {3}\right ) \tanh ^{-1}\left (\frac {\sqrt [3]{d} (a-x)^{2/3}+(-b+x)^{2/3}}{\sqrt [6]{d} \sqrt [3]{a-x} \sqrt [3]{-b+x}}\right )}{4 (a-b) d^{5/6}}\right )}{2 \sqrt [3]{(b-x)^2 (-a+x)} \left (b^2-a^2 d-2 b x+2 a d x-(-1+d) x^2\right )} \]
[Out]
1/2*(1+I*3^(1/2))*(-b+x)^(2/3)*(d^(1/3)*(a-x)^(2/3)-d^(1/6)*(a-x)^(1/3)*(-b+x)^(1/3)+(-b+x)^(2/3))*(d^(1/3)*(a
-x)^(2/3)+d^(1/6)*(a-x)^(1/3)*(-b+x)^(1/3)+(-b+x)^(2/3))*(-a*d^(1/3)+d^(1/3)*x+(a-x)^(1/3)*(-b+x)^(2/3))*(1/4*
(-3*I+3^(1/2))*arctan(3^(1/2)*(-b+x)^(1/3)/(-2*d^(1/6)*(a-x)^(1/3)+(-b+x)^(1/3)))/(a-b)/d^(5/6)+1/4*(3*I-3^(1/
2))*arctan(3^(1/2)*(-b+x)^(1/3)/(2*d^(1/6)*(a-x)^(1/3)+(-b+x)^(1/3)))/(a-b)/d^(5/6)+1/2*(1-I*3^(1/2))*arctanh(
(a-x)^(2/3)*(-b+x)^(1/3)/d^(1/6)/(-a+x))/(a-b)/d^(5/6)+1/4*I*(3^(1/2)+I)*arctanh((d^(1/3)*(a-x)^(2/3)+(-b+x)^(
2/3))/d^(1/6)/(a-x)^(1/3)/(-b+x)^(1/3))/(a-b)/d^(5/6))/((b-x)^2*(-a+x))^(1/3)/(b^2-a^2*d-2*b*x+2*a*d*x-(-1+d)*
x^2)
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Rubi [A]
time = 0.92, antiderivative size = 513, normalized size of antiderivative = 1.07, number of steps
used = 11, number of rules used = 6, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {6851, 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 [6]{d} \sqrt [3]{x-a}}{\sqrt {3} \sqrt [3]{x-b}}\right )}{2 d^{5/6} (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 [6]{d} \sqrt [3]{x-a}}{\sqrt {3} \sqrt [3]{x-b}}+\frac {1}{\sqrt {3}}\right )}{2 d^{5/6} (a-b) \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}-\frac {\sqrt [3]{x-a} (x-b)^{2/3} \log \left (2 \left (\sqrt {d}+1\right ) \left (b-a \sqrt {d}\right )-2 (1-d) x\right )}{4 d^{5/6} (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 (1-\sqrt {d}\right ) \left (a \sqrt {d}+b\right )\right )}{4 d^{5/6} (a-b) \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}-\frac {3 \sqrt [3]{x-a} (x-b)^{2/3} \log \left (-\sqrt [6]{d} \sqrt [3]{x-a}-\sqrt [3]{x-b}\right )}{4 d^{5/6} (a-b) \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}+\frac {3 \sqrt [3]{x-a} (x-b)^{2/3} \log \left (\sqrt [6]{d} \sqrt [3]{x-a}-\sqrt [3]{x-b}\right )}{4 d^{5/6} (a-b) \sqrt [3]{-\left ((a-x) (b-x)^2\right )}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(-a + x)/(((-a + x)*(-b + x)^2)^(1/3)*(-b^2 + a^2*d + 2*(b - a*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*d^(1/6)*(-a + x)^(1/3))/(Sqrt[3]*(-b + x)^(1
/3))])/((a - b)*d^(5/6)*(-((a - x)*(b - x)^2))^(1/3)) + (Sqrt[3]*(-a + x)^(1/3)*(-b + x)^(2/3)*ArcTan[1/Sqrt[3
] + (2*d^(1/6)*(-a + x)^(1/3))/(Sqrt[3]*(-b + x)^(1/3))])/(2*(a - b)*d^(5/6)*(-((a - x)*(b - x)^2))^(1/3)) - (
(-a + x)^(1/3)*(-b + x)^(2/3)*Log[2*(1 + Sqrt[d])*(b - a*Sqrt[d]) - 2*(1 - d)*x])/(4*(a - b)*d^(5/6)*(-((a - x
)*(b - x)^2))^(1/3)) + ((-a + x)^(1/3)*(-b + x)^(2/3)*Log[-2*(1 - Sqrt[d])*(b + a*Sqrt[d]) + 2*(1 - d)*x])/(4*
(a - b)*d^(5/6)*(-((a - x)*(b - x)^2))^(1/3)) - (3*(-a + x)^(1/3)*(-b + x)^(2/3)*Log[-(d^(1/6)*(-a + x)^(1/3))
- (-b + x)^(1/3)])/(4*(a - b)*d^(5/6)*(-((a - x)*(b - x)^2))^(1/3)) + (3*(-a + x)^(1/3)*(-b + x)^(2/3)*Log[d^
(1/6)*(-a + x)^(1/3) - (-b + x)^(1/3)])/(4*(a - b)*d^(5/6)*(-((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 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 {-a+x}{\sqrt [3]{(-a+x) (-b+x)^2} \left (-b^2+a^2 d+2 (b-a d) x+(-1+d) x^2\right )} \, dx &=\frac {\left (\sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \frac {(-a+x)^{2/3}}{(-b+x)^{2/3} \left (-b^2+a^2 d+2 (b-a 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) (-a+x)^{2/3}}{(a-b) \sqrt {d} (-b+x)^{2/3} \left (2 b-2 (a-b) \sqrt {d}-2 a d-2 (1-d) x\right )}+\frac {(-1+d) (-a+x)^{2/3}}{(a-b) \sqrt {d} (-b+x)^{2/3} \left (-2 b-2 (a-b) \sqrt {d}+2 a 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 {(-a+x)^{2/3}}{(-b+x)^{2/3} \left (2 b-2 (a-b) \sqrt {d}-2 a 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 {(-a+x)^{2/3}}{(-b+x)^{2/3} \left (-2 b-2 (a-b) \sqrt {d}+2 a 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 b-2 (a-b) \sqrt {d}-2 a 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 b-2 (a-b) \sqrt {d}+2 a 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 [6]{d} \sqrt [3]{-a+x}}{\sqrt {3} \sqrt [3]{-b+x}}\right )}{2 (a-b) d^{5/6} \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 [6]{d} \sqrt [3]{-a+x}}{\sqrt {3} \sqrt [3]{-b+x}}\right )}{2 (a-b) d^{5/6} \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 (b-a \sqrt {d}\right )-2 (1-d) x\right )}{4 (a-b) d^{5/6} \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 (b+a \sqrt {d}\right )+2 (1-d) x\right )}{4 (a-b) d^{5/6} \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}-\frac {3 \sqrt [3]{-a+x} (-b+x)^{2/3} \log \left (-\sqrt [6]{d} \sqrt [3]{-a+x}-\sqrt [3]{-b+x}\right )}{4 (a-b) d^{5/6} \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}+\frac {3 \sqrt [3]{-a+x} (-b+x)^{2/3} \log \left (\sqrt [6]{d} \sqrt [3]{-a+x}-\sqrt [3]{-b+x}\right )}{4 (a-b) d^{5/6} \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}\\ \end {align*}
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Mathematica [A]
time = 1.07, size = 238, normalized size = 0.49 \begin {gather*} \frac {(b-x)^{2/3} \sqrt [3]{-a+x} \left (\sqrt {3} \left (\text {ArcTan}\left (\frac {\sqrt {3} \sqrt [6]{d} \sqrt [3]{-a+x}}{-2 \sqrt [3]{b-x}+\sqrt [6]{d} \sqrt [3]{-a+x}}\right )-\text {ArcTan}\left (\frac {\sqrt {3} \sqrt [6]{d} \sqrt [3]{-a+x}}{2 \sqrt [3]{b-x}+\sqrt [6]{d} \sqrt [3]{-a+x}}\right )\right )+2 \tanh ^{-1}\left (\frac {\sqrt [6]{d} \sqrt [3]{-a+x}}{\sqrt [3]{b-x}}\right )+\tanh ^{-1}\left (\frac {\sqrt [3]{b-x}}{\sqrt [6]{d} \sqrt [3]{-a+x}}+\frac {\sqrt [6]{d} \sqrt [3]{-a+x}}{\sqrt [3]{b-x}}\right )\right )}{2 (a-b) d^{5/6} \sqrt [3]{(b-x)^2 (-a+x)}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(-a + x)/(((-a + x)*(-b + x)^2)^(1/3)*(-b^2 + a^2*d + 2*(b - a*d)*x + (-1 + d)*x^2)),x]
[Out]
((b - x)^(2/3)*(-a + x)^(1/3)*(Sqrt[3]*(ArcTan[(Sqrt[3]*d^(1/6)*(-a + x)^(1/3))/(-2*(b - x)^(1/3) + d^(1/6)*(-
a + x)^(1/3))] - ArcTan[(Sqrt[3]*d^(1/6)*(-a + x)^(1/3))/(2*(b - x)^(1/3) + d^(1/6)*(-a + x)^(1/3))]) + 2*ArcT
anh[(d^(1/6)*(-a + x)^(1/3))/(b - x)^(1/3)] + ArcTanh[(b - x)^(1/3)/(d^(1/6)*(-a + x)^(1/3)) + (d^(1/6)*(-a +
x)^(1/3))/(b - x)^(1/3)]))/(2*(a - b)*d^(5/6)*((b - x)^2*(-a + x))^(1/3))
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Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {-a +x}{\left (\left (-a +x \right ) \left (-b +x \right )^{2}\right )^{\frac {1}{3}} \left (-b^{2}+a^{2} d +2 \left (-a d +b \right ) x +\left (-1+d \right ) x^{2}\right )}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((-a+x)/((-a+x)*(-b+x)^2)^(1/3)/(-b^2+a^2*d+2*(-a*d+b)*x+(-1+d)*x^2),x)
[Out]
int((-a+x)/((-a+x)*(-b+x)^2)^(1/3)/(-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*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((-a+x)/((-a+x)*(-b+x)^2)^(1/3)/(-b^2+a^2*d+2*(-a*d+b)*x+(-1+d)*x^2),x, algorithm="maxima")
[Out]
-integrate((a - x)/((-(a - x)*(b - x)^2)^(1/3)*(a^2*d + (d - 1)*x^2 - b^2 - 2*(a*d - b)*x)), x)
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 2613 vs.
\(2 (364) = 728\).
time = 0.55, size = 2613, normalized size = 5.43 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((-a+x)/((-a+x)*(-b+x)^2)^(1/3)/(-b^2+a^2*d+2*(-a*d+b)*x+(-1+d)*x^2),x, algorithm="fricas")
[Out]
sqrt(3)*(1/((a^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3 + 15*a^2*b^4 - 6*a*b^5 + b^6)*d^5))^(1/6)*arctan(-1/3*(2*
sqrt(3)*(-a*b^2 - (a + 2*b)*x^2 + x^3 + (2*a*b + b^2)*x)^(1/3)*(a - b)*d*(1/((a^6 - 6*a^5*b + 15*a^4*b^2 - 20*
a^3*b^3 + 15*a^2*b^4 - 6*a*b^5 + b^6)*d^5))^(1/6) - 2*sqrt(3)*((a - b)*d*x - (a*b - b^2)*d)*sqrt((((a^5 - 5*a^
4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d^4*x - (a^5*b - 5*a^4*b^2 + 10*a^3*b^3 - 10*a^2*b^4 + 5*a*b^5
- b^6)*d^4)*(-a*b^2 - (a + 2*b)*x^2 + x^3 + (2*a*b + b^2)*x)^(1/3)*(1/((a^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^
3 + 15*a^2*b^4 - 6*a*b^5 + b^6)*d^5))^(5/6) + ((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*d^3*x^2 - 2*(a^4*b
- 4*a^3*b^2 + 6*a^2*b^3 - 4*a*b^4 + b^5)*d^3*x + (a^4*b^2 - 4*a^3*b^3 + 6*a^2*b^4 - 4*a*b^5 + b^6)*d^3)*(1/((a
^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3 + 15*a^2*b^4 - 6*a*b^5 + b^6)*d^5))^(2/3) + (-a*b^2 - (a + 2*b)*x^2 + x
^3 + (2*a*b + b^2)*x)^(2/3))/(b^2 - 2*b*x + x^2))*(1/((a^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3 + 15*a^2*b^4 -
6*a*b^5 + b^6)*d^5))^(1/6) - sqrt(3)*(b - x))/(b - x)) + sqrt(3)*(1/((a^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3
+ 15*a^2*b^4 - 6*a*b^5 + b^6)*d^5))^(1/6)*arctan(-1/3*(2*sqrt(3)*(-a*b^2 - (a + 2*b)*x^2 + x^3 + (2*a*b + b^2)
*x)^(1/3)*(a - b)*d*(1/((a^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3 + 15*a^2*b^4 - 6*a*b^5 + b^6)*d^5))^(1/6) - 2
*sqrt(3)*((a - b)*d*x - (a*b - b^2)*d)*sqrt(-(((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d^4*x
- (a^5*b - 5*a^4*b^2 + 10*a^3*b^3 - 10*a^2*b^4 + 5*a*b^5 - b^6)*d^4)*(-a*b^2 - (a + 2*b)*x^2 + x^3 + (2*a*b +
b^2)*x)^(1/3)*(1/((a^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3 + 15*a^2*b^4 - 6*a*b^5 + b^6)*d^5))^(5/6) - ((a^4
- 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*d^3*x^2 - 2*(a^4*b - 4*a^3*b^2 + 6*a^2*b^3 - 4*a*b^4 + b^5)*d^3*x + (a^
4*b^2 - 4*a^3*b^3 + 6*a^2*b^4 - 4*a*b^5 + b^6)*d^3)*(1/((a^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3 + 15*a^2*b^4
- 6*a*b^5 + b^6)*d^5))^(2/3) - (-a*b^2 - (a + 2*b)*x^2 + x^3 + (2*a*b + b^2)*x)^(2/3))/(b^2 - 2*b*x + x^2))*(1
/((a^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3 + 15*a^2*b^4 - 6*a*b^5 + b^6)*d^5))^(1/6) + sqrt(3)*(b - x))/(b - x
)) - 1/4*(1/((a^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3 + 15*a^2*b^4 - 6*a*b^5 + b^6)*d^5))^(1/6)*log((((a^5 - 5
*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d^4*x - (a^5*b - 5*a^4*b^2 + 10*a^3*b^3 - 10*a^2*b^4 + 5*a*b
^5 - b^6)*d^4)*(-a*b^2 - (a + 2*b)*x^2 + x^3 + (2*a*b + b^2)*x)^(1/3)*(1/((a^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3
*b^3 + 15*a^2*b^4 - 6*a*b^5 + b^6)*d^5))^(5/6) + ((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*d^3*x^2 - 2*(a^4
*b - 4*a^3*b^2 + 6*a^2*b^3 - 4*a*b^4 + b^5)*d^3*x + (a^4*b^2 - 4*a^3*b^3 + 6*a^2*b^4 - 4*a*b^5 + b^6)*d^3)*(1/
((a^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3 + 15*a^2*b^4 - 6*a*b^5 + b^6)*d^5))^(2/3) + (-a*b^2 - (a + 2*b)*x^2
+ x^3 + (2*a*b + b^2)*x)^(2/3))/(b^2 - 2*b*x + x^2)) + 1/4*(1/((a^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3 + 15*a
^2*b^4 - 6*a*b^5 + b^6)*d^5))^(1/6)*log(-(((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d^4*x - (
a^5*b - 5*a^4*b^2 + 10*a^3*b^3 - 10*a^2*b^4 + 5*a*b^5 - b^6)*d^4)*(-a*b^2 - (a + 2*b)*x^2 + x^3 + (2*a*b + b^2
)*x)^(1/3)*(1/((a^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3 + 15*a^2*b^4 - 6*a*b^5 + b^6)*d^5))^(5/6) - ((a^4 - 4*
a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*d^3*x^2 - 2*(a^4*b - 4*a^3*b^2 + 6*a^2*b^3 - 4*a*b^4 + b^5)*d^3*x + (a^4*b^
2 - 4*a^3*b^3 + 6*a^2*b^4 - 4*a*b^5 + b^6)*d^3)*(1/((a^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3 + 15*a^2*b^4 - 6*
a*b^5 + b^6)*d^5))^(2/3) - (-a*b^2 - (a + 2*b)*x^2 + x^3 + (2*a*b + b^2)*x)^(2/3))/(b^2 - 2*b*x + x^2)) - 1/2*
(1/((a^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3 + 15*a^2*b^4 - 6*a*b^5 + b^6)*d^5))^(1/6)*log(-(((a^5 - 5*a^4*b +
10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d^4*x - (a^5*b - 5*a^4*b^2 + 10*a^3*b^3 - 10*a^2*b^4 + 5*a*b^5 - b^6
)*d^4)*(1/((a^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3 + 15*a^2*b^4 - 6*a*b^5 + b^6)*d^5))^(5/6) + (-a*b^2 - (a +
2*b)*x^2 + x^3 + (2*a*b + b^2)*x)^(1/3))/(b - x)) + 1/2*(1/((a^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3 + 15*a^2
*b^4 - 6*a*b^5 + b^6)*d^5))^(1/6)*log((((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d^4*x - (a^5
*b - 5*a^4*b^2 + 10*a^3*b^3 - 10*a^2*b^4 + 5*a*b^5 - b^6)*d^4)*(1/((a^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3 +
15*a^2*b^4 - 6*a*b^5 + b^6)*d^5))^(5/6) - (-a*b^2 - (a + 2*b)*x^2 + x^3 + (2*a*b + b^2)*x)^(1/3))/(b - x))
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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((-a+x)/((-a+x)*(-b+x)**2)**(1/3)/(-b**2+a**2*d+2*(-a*d+b)*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((-a+x)/((-a+x)*(-b+x)^2)^(1/3)/(-b^2+a^2*d+2*(-a*d+b)*x+(-1+d)*x^2),x, algorithm="giac")
[Out]
integrate(-(a - x)/((-(a - x)*(b - x)^2)^(1/3)*(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 {a-x}{{\left (-\left (a-x\right )\,{\left (b-x\right )}^2\right )}^{1/3}\,\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 - x)/((-(a - x)*(b - x)^2)^(1/3)*(a^2*d + 2*x*(b - a*d) - b^2 + x^2*(d - 1))),x)
[Out]
int(-(a - x)/((-(a - x)*(b - x)^2)^(1/3)*(a^2*d + 2*x*(b - a*d) - b^2 + x^2*(d - 1))), x)
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