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

Optimal. Leaf size=423 \[ \frac {\log \left (\sqrt [3]{d} (a-b)^{2/3} \sqrt [3]{x \left (2 a b+b^2\right )-a b^2+x^2 (-a-2 b)+x^3}-x (a-b)^{2/3}+b (a-b)^{2/3}\right )}{d^{2/3} (a-b)}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{d} \sqrt [3]{x \left (2 a b+b^2\right )-a b^2+x^2 (-a-2 b)+x^3}}{\sqrt [3]{d} \sqrt [3]{x \left (2 a b+b^2\right )-a b^2+x^2 (-a-2 b)+x^3}-2 b+2 x}\right )}{d^{2/3} (a-b)}-\frac {\log \left (b^3 \left (-\sqrt [3]{a-b}\right )+d^{2/3} (a-b)^{4/3} \left (x \left (2 a b+b^2\right )-a b^2+x^2 (-a-2 b)+x^3\right )^{2/3}+\sqrt [3]{d} \sqrt [3]{a-b} \left (-a b+a x+b^2-b x\right ) \sqrt [3]{x \left (2 a b+b^2\right )-a b^2+x^2 (-a-2 b)+x^3}+2 b^2 x \sqrt [3]{a-b}+a b^2 \sqrt [3]{a-b}-b x^2 \sqrt [3]{a-b}+a x^2 \sqrt [3]{a-b}-2 a b x \sqrt [3]{a-b}\right )}{2 d^{2/3} (a-b)} \]

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Rubi [A]  time = 0.68, antiderivative size = 376, normalized size of antiderivative = 0.89, number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2081, 2077, 91} \begin {gather*} -\frac {\left ((a-b)^2 (b-x)\right )^{2/3} \sqrt [3]{(a-b)^2 (x-a)} \log (-a d+b+(d-1) x)}{2 d^{2/3} (a-b)^3 \sqrt [3]{-a b^2+x^2 (-a-2 b)+b x (2 a+b)+x^3}}+\frac {3 \left ((a-b)^2 (b-x)\right )^{2/3} \sqrt [3]{(a-b)^2 (x-a)} \log \left (-\sqrt [3]{\frac {2}{3}} \sqrt [3]{d} \sqrt [3]{(a-b)^2 (x-a)}-\sqrt [3]{\frac {2}{3}} \sqrt [3]{(a-b)^2 (b-x)}\right )}{2 d^{2/3} (a-b)^3 \sqrt [3]{-a b^2+x^2 (-a-2 b)+b x (2 a+b)+x^3}}+\frac {\sqrt {3} \left ((a-b)^2 (b-x)\right )^{2/3} \sqrt [3]{(a-b)^2 (x-a)} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{d} \sqrt [3]{(a-b)^2 (x-a)}}{\sqrt {3} \sqrt [3]{(a-b)^2 (b-x)}}\right )}{d^{2/3} (a-b)^3 \sqrt [3]{-a b^2+x^2 (-a-2 b)+b x (2 a+b)+x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2077

Int[((e_.) + (f_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (d_.)*(x_)^3)^(p_), x_Symbol] :> Dist[(a + b*x + d*x^3)^p/
((3*a - b*x)^p*(3*a + 2*b*x)^(2*p)), Int[(e + f*x)^m*(3*a - b*x)^p*(3*a + 2*b*x)^(2*p), x], x] /; FreeQ[{a, b,
d, e, f, m, p}, x] && EqQ[4*b^3 + 27*a^2*d, 0] &&  !IntegerQ[p]

Rule 2081

Int[(P3_)^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> With[{a = Coeff[P3, x, 0], b = Coeff[P3, x, 1], c = C
oeff[P3, x, 2], d = Coeff[P3, x, 3]}, Subst[Int[((3*d*e - c*f)/(3*d) + f*x)^m*Simp[(2*c^3 - 9*b*c*d + 27*a*d^2
)/(27*d^2) - ((c^2 - 3*b*d)*x)/(3*d) + d*x^3, x]^p, x], x, x + c/(3*d)] /; NeQ[c, 0]] /; FreeQ[{e, f, m, p}, x
] && PolyQ[P3, x, 3]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt [3]{(-a+x) (-b+x)^2} (b-a d+(-1+d) x)} \, dx &=\operatorname {Subst}\left (\int \frac {1}{\left (\frac {1}{3} (-((-a-2 b) (-1+d))+3 (b-a d))+(-1+d) x\right ) \sqrt [3]{-\frac {2}{27} (a-b)^3-\frac {1}{3} (a-b)^2 x+x^3}} \, dx,x,\frac {1}{3} (-a-2 b)+x\right )\\ &=\frac {\left (2^{2/3} \sqrt [3]{-(a-b)^2 (a-x)} \left ((a-b)^2 (b-x)\right )^{2/3}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (-\frac {2}{9} (a-b)^3-\frac {2}{3} (a-b)^2 x\right )^{2/3} \sqrt [3]{-\frac {2}{9} (a-b)^3+\frac {1}{3} (a-b)^2 x} \left (\frac {1}{3} (-((-a-2 b) (-1+d))+3 (b-a d))+(-1+d) x\right )} \, dx,x,\frac {1}{3} (-a-2 b)+x\right )}{3 \sqrt [3]{-a b^2+b (2 a+b) x-(a+2 b) x^2+x^3}}\\ &=\frac {\sqrt {3} \sqrt [3]{-(a-b)^2 (a-x)} \left ((a-b)^2 (b-x)\right )^{2/3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{d} \sqrt [3]{-(a-b)^2 (a-x)}}{\sqrt {3} \sqrt [3]{(a-b)^2 (b-x)}}\right )}{(a-b)^3 d^{2/3} \sqrt [3]{-a b^2+b (2 a+b) x-(a+2 b) x^2+x^3}}+\frac {3 \sqrt [3]{-(a-b)^2 (a-x)} \left ((a-b)^2 (b-x)\right )^{2/3} \log \left (\sqrt [3]{d} \sqrt [3]{-(a-b)^2 (a-x)}+\sqrt [3]{(a-b)^2 (b-x)}\right )}{2 (a-b)^3 d^{2/3} \sqrt [3]{-a b^2+b (2 a+b) x-(a+2 b) x^2+x^3}}-\frac {\sqrt [3]{-(a-b)^2 (a-x)} \left ((a-b)^2 (b-x)\right )^{2/3} \log (b-a d-(1-d) x)}{2 (a-b)^3 d^{2/3} \sqrt [3]{-a b^2+b (2 a+b) x-(a+2 b) x^2+x^3}}\\ \end {align*}

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Mathematica [C]  time = 0.06, size = 58, normalized size = 0.14 \begin {gather*} -\frac {3 (x-b) \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};\frac {b-x}{a d-d x}\right )}{d (a-b) \sqrt [3]{(x-a) (b-x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 3.21, size = 423, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {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 b+2 x+\sqrt [3]{d} \sqrt [3]{-a b^2+\left (2 a b+b^2\right ) x+(-a-2 b) x^2+x^3}}\right )}{(a-b) d^{2/3}}+\frac {\log \left ((a-b)^{2/3} b-(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 )}{(a-b) d^{2/3}}-\frac {\log \left (a \sqrt [3]{a-b} b^2-\sqrt [3]{a-b} b^3-2 a \sqrt [3]{a-b} b x+2 \sqrt [3]{a-b} b^2 x+a \sqrt [3]{a-b} x^2-\sqrt [3]{a-b} b x^2+\sqrt [3]{a-b} \sqrt [3]{d} \left (-a b+b^2+a x-b x\right ) \sqrt [3]{-a b^2+\left (2 a b+b^2\right ) x+(-a-2 b) x^2+x^3}+(a-b)^{4/3} d^{2/3} \left (-a b^2+\left (2 a b+b^2\right ) x+(-a-2 b) x^2+x^3\right )^{2/3}\right )}{2 (a-b) d^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.61, size = 301, normalized size = 0.71 \begin {gather*} \frac {2 \, \sqrt {3} d \sqrt {-\left (-d^{2}\right )^{\frac {1}{3}}} \arctan \left (-\frac {\sqrt {3} {\left (\left (-d^{2}\right )^{\frac {1}{3}} {\left (b - x\right )} + 2 \, {\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}} d\right )} \sqrt {-\left (-d^{2}\right )^{\frac {1}{3}}}}{3 \, {\left (b d - d x\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}} d^{2} + {\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 (b d - d x\right )} \left (-d^{2}\right )^{\frac {1}{3}} + {\left (b^{2} - 2 \, b x + x^{2}\right )} \left (-d^{2}\right )^{\frac {2}{3}}}{b^{2} - 2 \, b x + x^{2}}\right ) + 2 \, \left (-d^{2}\right )^{\frac {2}{3}} \log \left (\frac {\left (-d^{2}\right )^{\frac {1}{3}} {\left (b - x\right )} - {\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}} d}{b - x}\right )}{2 \, {\left (a - b\right )} d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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