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

Optimal. Leaf size=301 \[ -\frac {\log \left (\sqrt [3]{x \left (2 a b+b^2\right )-a b^2+x^2 (-a-2 b)+x^3}+b \sqrt [3]{d}-\sqrt [3]{d} x\right )}{d^{2/3} (a-b)}+\frac {\log \left (\sqrt [3]{x \left (2 a b+b^2\right )-a b^2+x^2 (-a-2 b)+x^3} \left (\sqrt [3]{d} x-b \sqrt [3]{d}\right )+\left (x \left (2 a b+b^2\right )-a b^2+x^2 (-a-2 b)+x^3\right )^{2/3}+b^2 d^{2/3}-2 b d^{2/3} x+d^{2/3} x^2\right )}{2 d^{2/3} (a-b)}-\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} b \sqrt [3]{d}-\sqrt {3} \sqrt [3]{d} x}{-2 \sqrt [3]{x \left (2 a b+b^2\right )-a b^2+x^2 (-a-2 b)+x^3}+b \sqrt [3]{d}-\sqrt [3]{d} x}\right )}{d^{2/3} (a-b)} \]

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Rubi [A]  time = 0.60, antiderivative size = 240, normalized size of antiderivative = 0.80, number of steps used = 2, number of rules used = 2, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {6719, 91} \begin {gather*} \frac {(x-a)^{2/3} (x-b)^{4/3} \log (a-b d-(1-d) x)}{2 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]{d} \sqrt [3]{x-b}-\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]{d} \sqrt [3]{x-b}}{\sqrt {3} \sqrt [3]{x-a}}+\frac {1}{\sqrt {3}}\right )}{d^{2/3} (a-b) \left (-\left ((a-x) (b-x)^2\right )\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 {-b+x}{\left ((-a+x) (-b+x)^2\right )^{2/3} (a-b d+(-1+d) x)} \, dx &=\frac {\left ((-a+x)^{2/3} (-b+x)^{4/3}\right ) \int \frac {1}{(-a+x)^{2/3} \sqrt [3]{-b+x} (a-b d+(-1+d) x)} \, dx}{\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]{d} \sqrt [3]{-b+x}}{\sqrt {3} \sqrt [3]{-a+x}}\right )}{(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 (a-b d-(1-d) x)}{2 (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}+\sqrt [3]{d} \sqrt [3]{-b+x}\right )}{2 (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 = 0.06, size = 57, normalized size = 0.19 \begin {gather*} \frac {3 (b-x)^2 \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {d (b-x)}{a-x}\right )}{2 (a-b) \left ((x-a) (b-x)^2\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 5.84, size = 301, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} b \sqrt [3]{d}-\sqrt {3} \sqrt [3]{d} x}{b \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 )}{(a-b) d^{2/3}}-\frac {\log \left (b \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 )}{(a-b) d^{2/3}}+\frac {\log \left (b^2 d^{2/3}-2 b d^{2/3} x+d^{2/3} x^2+\left (-b \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 )}{2 (a-b) d^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.49, size = 290, normalized size = 0.96 \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 d - d x\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 {1}{3}} {\left (d^{2}\right )}^{\frac {2}{3}}\right )}}{3 \, {\left (b d^{2} - d^{2} 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 {1}{3}} {\left (d^{2}\right )}^{\frac {2}{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 {2}{3}} d - {\left (b^{2} d - 2 \, b d x + d x^{2}\right )} {\left (d^{2}\right )}^{\frac {1}{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 {2}{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((-b+x)/((-a+x)*(-b+x)^2)^(2/3)/(a-b*d+(-1+d)*x),x, algorithm="fricas")

[Out]

1/2*(2*sqrt(3)*(d^2)^(1/6)*d*arctan(1/3*sqrt(3)*(d^2)^(1/6)*((b*d - d*x)*(d^2)^(1/3) - 2*(-a*b^2 - (a + 2*b)*x
^2 + x^3 + (2*a*b + b^2)*x)^(1/3)*(d^2)^(2/3))/(b*d^2 - d^2*x)) + (d^2)^(2/3)*log(-((-a*b^2 - (a + 2*b)*x^2 +
x^3 + (2*a*b + b^2)*x)^(1/3)*(d^2)^(2/3)*(b - x) - (-a*b^2 - (a + 2*b)*x^2 + x^3 + (2*a*b + b^2)*x)^(2/3)*d -
(b^2*d - 2*b*d*x + d*x^2)*(d^2)^(1/3))/(b^2 - 2*b*x + x^2)) - 2*(d^2)^(2/3)*log(-((d^2)^(2/3)*(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 {b - x}{\left (-{\left (a - x\right )} {\left (b - x\right )}^{2}\right )^{\frac {2}{3}} {\left (b d - {\left (d - 1\right )} x - a\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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