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

Optimal. Leaf size=849 \[ \frac {(b-x)^{4/3} (x-a)^{2/3} \left (\frac {\sqrt {3} (d-1) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{x-a}}{\sqrt [3]{x-a}-2 \sqrt [3]{d} \sqrt [3]{b-x}}\right ) a}{(a-b)^2 d^{2/3}}-\frac {(d-1) \log \left (\sqrt [3]{d} \sqrt [3]{b-x}+\sqrt [3]{x-a}\right ) a}{(a-b)^2 d^{2/3}}+\frac {(d-1) \log \left (d^{2/3} (b-x)^{2/3}-\sqrt [3]{d} \sqrt [3]{x-a} \sqrt [3]{b-x}+(x-a)^{2/3}\right ) a}{2 (a-b)^2 d^{2/3}}-\frac {3 (b-x)^{2/3} \sqrt [3]{x-a} a}{(a-b)^2 (x-b)}+\frac {\sqrt {3} c (a-b d) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{x-a}}{\sqrt [3]{x-a}-2 \sqrt [3]{d} \sqrt [3]{b-x}}\right )}{(a-b)^2 d^{2/3}}+\frac {\sqrt {3} (a-b d) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{x-a}}{\sqrt [3]{x-a}-2 \sqrt [3]{d} \sqrt [3]{b-x}}\right )}{(a-b)^2 d^{2/3}}+\frac {\sqrt {3} b c (d-1) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{x-a}}{\sqrt [3]{x-a}-2 \sqrt [3]{d} \sqrt [3]{b-x}}\right )}{(a-b)^2 d^{2/3}}-\frac {c (a-b d) \log \left (\sqrt [3]{d} \sqrt [3]{b-x}+\sqrt [3]{x-a}\right )}{(a-b)^2 d^{2/3}}+\frac {(b d-a) \log \left (\sqrt [3]{d} \sqrt [3]{b-x}+\sqrt [3]{x-a}\right )}{(a-b)^2 d^{2/3}}-\frac {b c (d-1) \log \left (\sqrt [3]{d} \sqrt [3]{b-x}+\sqrt [3]{x-a}\right )}{(a-b)^2 d^{2/3}}+\frac {c (a-b d) \log \left (d^{2/3} (b-x)^{2/3}-\sqrt [3]{d} \sqrt [3]{x-a} \sqrt [3]{b-x}+(x-a)^{2/3}\right )}{2 (a-b)^2 d^{2/3}}+\frac {(a-b d) \log \left (d^{2/3} (b-x)^{2/3}-\sqrt [3]{d} \sqrt [3]{x-a} \sqrt [3]{b-x}+(x-a)^{2/3}\right )}{2 (a-b)^2 d^{2/3}}+\frac {b c (d-1) \log \left (d^{2/3} (b-x)^{2/3}-\sqrt [3]{d} \sqrt [3]{x-a} \sqrt [3]{b-x}+(x-a)^{2/3}\right )}{2 (a-b)^2 d^{2/3}}+\frac {3 b (b-x)^{2/3} \sqrt [3]{x-a}}{(a-b)^2 (x-b)}\right )}{\left ((b-x)^2 (x-a)\right )^{2/3}} \]

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Rubi [A]  time = 0.88, antiderivative size = 286, normalized size of antiderivative = 0.34, number of steps used = 4, number of rules used = 4, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {6719, 155, 12, 91} \begin {gather*} \frac {(x-a)^{2/3} (x-b)^{4/3} (c+d) \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} (c+d) \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} (c+d) \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}}-\frac {3 (a-x) (b-x)}{(a-b) \left (-\left ((a-x) (b-x)^2\right )\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(a - x)*(b - x))/((a - b)*(-((a - x)*(b - x)^2))^(2/3)) - (Sqrt[3]*(c + d)*(-a + x)^(2/3)*(-b + x)^(4/3)*A
rcTan[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)) + ((c + d)*(-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*(c + d)*(-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 12

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

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 155

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m + n + p + 2, 0] && NeQ[m, -1] && (Sum
SimplerQ[m, 1] || ( !(NeQ[n, -1] && SumSimplerQ[n, 1]) &&  !(NeQ[p, -1] && SumSimplerQ[p, 1])))

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 {-a-b c+(1+c) 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 {-a-b c+(1+c) x}{(-a+x)^{2/3} (-b+x)^{4/3} (a-b d+(-1+d) x)} \, dx}{\left ((-a+x) (-b+x)^2\right )^{2/3}}\\ &=-\frac {3 (a-x) (b-x)}{(a-b) \left (-\left ((a-x) (b-x)^2\right )\right )^{2/3}}+\frac {\left (3 (-a+x)^{2/3} (-b+x)^{4/3}\right ) \int \frac {(a-b)^2 (c+d)}{3 (-a+x)^{2/3} \sqrt [3]{-b+x} (a-b d+(-1+d) x)} \, dx}{(a-b)^2 \left ((-a+x) (-b+x)^2\right )^{2/3}}\\ &=-\frac {3 (a-x) (b-x)}{(a-b) \left (-\left ((a-x) (b-x)^2\right )\right )^{2/3}}+\frac {\left ((c+d) (-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 {3 (a-x) (b-x)}{(a-b) \left (-\left ((a-x) (b-x)^2\right )\right )^{2/3}}-\frac {\sqrt {3} (c+d) (-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 {(c+d) (-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 (c+d) (-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 = 0.07, size = 72, normalized size = 0.08 \begin {gather*} \frac {3 (x-b) \left ((x-b) (c+d) \, _2F_1\left (\frac {2}{3},1;\frac {5}{3};\frac {d (b-x)}{a-x}\right )+2 (a-x)\right )}{2 (a-b) \left ((x-a) (b-x)^2\right )^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(-b + x)*(2*(a - x) + (c + d)*(-b + x)*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 [F]  time = 180.04, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

$Aborted

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fricas [A]  time = 0.65, size = 389, normalized size = 0.46 \begin {gather*} -\frac {2 \, \sqrt {3} {\left (b c d + b d^{2} - {\left (c d + d^{2}\right )} x\right )} {\left (d^{2}\right )}^{\frac {1}{6}} \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 (b c + b d - {\left (c + d\right )} x\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 (b c + b d - {\left (c + d\right )} x\right )} {\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 ) + 6 \, {\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^{2}}{2 \, {\left ({\left (a - b\right )} d^{2} x - {\left (a b - b^{2}\right )} d^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {-a -b c +\left (1+c \right ) x}{\left (\left (-a +x \right ) \left (-b +x \right )^{2}\right )^{\frac {2}{3}} \left (a -b d +\left (-1+d \right ) x \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((-a-b*c+(1+c)*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 c - {\left (c + 1\right )} x + a}{\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((-a-b*c+(1+c)*x)/((-a+x)*(-b+x)^2)^(2/3)/(a-b*d+(-1+d)*x),x, algorithm="maxima")

[Out]

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

[Out]

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

[Out]

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

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