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

Optimal. Leaf size=201 \[ -\frac {\log \left (d^{2/3} \left (x (-a-b)+a b+x^2\right )^{4/3}+\left (x (-a-b)+a b+x^2\right )^{2/3} \left (\sqrt [3]{d} x-b \sqrt [3]{d}\right )+b^2-2 b x+x^2\right )}{2 d^{2/3}}+\frac {\log \left (\sqrt [3]{d} \left (x (-a-b)+a b+x^2\right )^{2/3}+b-x\right )}{d^{2/3}}+\frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{d} \left (x (-a-b)+a b+x^2\right )^{2/3}}{\sqrt [3]{d} \left (x (-a-b)+a b+x^2\right )^{2/3}-2 b+2 x}\right )}{d^{2/3}} \]

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Rubi [C]  time = 1.56, antiderivative size = 191, normalized size of antiderivative = 0.95, number of steps used = 8, number of rules used = 5, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6719, 1586, 6728, 137, 136} \begin {gather*} -\frac {3 (a-x)^2 \left (-\frac {b-x}{a-b}\right )^{2/3} F_1\left (\frac {4}{3};\frac {2}{3},1;\frac {7}{3};\frac {a-x}{a-b},-\frac {2 d (a-x)}{1-\sqrt {4 a d-4 b d+1}}\right )}{4 ((a-x) (b-x))^{2/3}}-\frac {3 (a-x)^2 \left (-\frac {b-x}{a-b}\right )^{2/3} F_1\left (\frac {4}{3};\frac {2}{3},1;\frac {7}{3};\frac {a-x}{a-b},-\frac {2 d (a-x)}{\sqrt {4 a d-4 b d+1}+1}\right )}{4 ((a-x) (b-x))^{2/3}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-3*(a - x)^2*(-((b - x)/(a - b)))^(2/3)*AppellF1[4/3, 2/3, 1, 7/3, (a - x)/(a - b), (-2*d*(a - x))/(1 - Sqrt[
1 + 4*a*d - 4*b*d])])/(4*((a - x)*(b - x))^(2/3)) - (3*(a - x)^2*(-((b - x)/(a - b)))^(2/3)*AppellF1[4/3, 2/3,
1, 7/3, (a - x)/(a - b), (-2*d*(a - x))/(1 + Sqrt[1 + 4*a*d - 4*b*d])])/(4*((a - x)*(b - x))^(2/3))

Rule 136

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((b*e - a*
f)^p*(a + b*x)^(m + 1)*AppellF1[m + 1, -n, -p, m + 2, -((d*(a + b*x))/(b*c - a*d)), -((f*(a + b*x))/(b*e - a*f
))])/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] && IntegerQ[p] && GtQ[b/(b*c - a*d), 0] &&  !(GtQ[d/(d*a - c*b), 0] && SimplerQ[c + d*x, a + b*x])

Rule 137

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

Rule 1586

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

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]

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {-a (a-2 b)-2 b x+x^2}{((-a+x) (-b+x))^{2/3} \left (b+a^2 d-(1+2 a d) x+d x^2\right )} \, dx &=\frac {\left ((-a+x)^{2/3} (-b+x)^{2/3}\right ) \int \frac {-a (a-2 b)-2 b x+x^2}{(-a+x)^{2/3} (-b+x)^{2/3} \left (b+a^2 d-(1+2 a d) x+d x^2\right )} \, dx}{((-a+x) (-b+x))^{2/3}}\\ &=\frac {\left ((-a+x)^{2/3} (-b+x)^{2/3}\right ) \int \frac {\sqrt [3]{-a+x} (a-2 b+x)}{(-b+x)^{2/3} \left (b+a^2 d-(1+2 a d) x+d x^2\right )} \, dx}{((-a+x) (-b+x))^{2/3}}\\ &=\frac {\left ((-a+x)^{2/3} (-b+x)^{2/3}\right ) \int \left (\frac {\left (1+\sqrt {1+4 a d-4 b d}\right ) \sqrt [3]{-a+x}}{(-b+x)^{2/3} \left (-1-2 a d-\sqrt {1+4 a d-4 b d}+2 d x\right )}+\frac {\left (1-\sqrt {1+4 a d-4 b d}\right ) \sqrt [3]{-a+x}}{(-b+x)^{2/3} \left (-1-2 a d+\sqrt {1+4 a d-4 b d}+2 d x\right )}\right ) \, dx}{((-a+x) (-b+x))^{2/3}}\\ &=\frac {\left (\left (1-\sqrt {1+4 a d-4 b d}\right ) (-a+x)^{2/3} (-b+x)^{2/3}\right ) \int \frac {\sqrt [3]{-a+x}}{(-b+x)^{2/3} \left (-1-2 a d+\sqrt {1+4 a d-4 b d}+2 d x\right )} \, dx}{((-a+x) (-b+x))^{2/3}}+\frac {\left (\left (1+\sqrt {1+4 a d-4 b d}\right ) (-a+x)^{2/3} (-b+x)^{2/3}\right ) \int \frac {\sqrt [3]{-a+x}}{(-b+x)^{2/3} \left (-1-2 a d-\sqrt {1+4 a d-4 b d}+2 d x\right )} \, dx}{((-a+x) (-b+x))^{2/3}}\\ &=\frac {\left (\left (1-\sqrt {1+4 a d-4 b d}\right ) (-a+x)^{2/3} \left (\frac {-b+x}{a-b}\right )^{2/3}\right ) \int \frac {\sqrt [3]{-a+x}}{\left (-\frac {b}{a-b}+\frac {x}{a-b}\right )^{2/3} \left (-1-2 a d+\sqrt {1+4 a d-4 b d}+2 d x\right )} \, dx}{((-a+x) (-b+x))^{2/3}}+\frac {\left (\left (1+\sqrt {1+4 a d-4 b d}\right ) (-a+x)^{2/3} \left (\frac {-b+x}{a-b}\right )^{2/3}\right ) \int \frac {\sqrt [3]{-a+x}}{\left (-\frac {b}{a-b}+\frac {x}{a-b}\right )^{2/3} \left (-1-2 a d-\sqrt {1+4 a d-4 b d}+2 d x\right )} \, dx}{((-a+x) (-b+x))^{2/3}}\\ &=-\frac {3 (a-x)^2 \left (-\frac {b-x}{a-b}\right )^{2/3} F_1\left (\frac {4}{3};\frac {2}{3},1;\frac {7}{3};\frac {a-x}{a-b},-\frac {2 d (a-x)}{1-\sqrt {1+4 a d-4 b d}}\right )}{4 ((a-x) (b-x))^{2/3}}-\frac {3 (a-x)^2 \left (-\frac {b-x}{a-b}\right )^{2/3} F_1\left (\frac {4}{3};\frac {2}{3},1;\frac {7}{3};\frac {a-x}{a-b},-\frac {2 d (a-x)}{1+\sqrt {1+4 a d-4 b d}}\right )}{4 ((a-x) (b-x))^{2/3}}\\ \end {align*}

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Mathematica [F]  time = 1.07, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-a (a-2 b)-2 b x+x^2}{((-a+x) (-b+x))^{2/3} \left (b+a^2 d-(1+2 a d) x+d x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.70, size = 201, normalized size = 1.00 \begin {gather*} \frac {\sqrt {3} \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{d} \left (a b+(-a-b) x+x^2\right )^{2/3}}{-2 b+2 x+\sqrt [3]{d} \left (a b+(-a-b) x+x^2\right )^{2/3}}\right )}{d^{2/3}}+\frac {\log \left (b-x+\sqrt [3]{d} \left (a b+(-a-b) x+x^2\right )^{2/3}\right )}{d^{2/3}}-\frac {\log \left (b^2-2 b x+x^2+\left (-b \sqrt [3]{d}+\sqrt [3]{d} x\right ) \left (a b+(-a-b) x+x^2\right )^{2/3}+d^{2/3} \left (a b+(-a-b) x+x^2\right )^{4/3}\right )}{2 d^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.25, size = 0, normalized size = 0.00 \[\int \frac {-a \left (a -2 b \right )-2 b x +x^{2}}{\left (\left (-a +x \right ) \left (-b +x \right )\right )^{\frac {2}{3}} \left (b +a^{2} d -\left (2 a d +1\right ) x +d \,x^{2}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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