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

Optimal. Leaf size=1387 \[ \frac {\left (1+i \sqrt {3}\right ) \sqrt [3]{a-x} (x-b)^{2/3} \left (\sqrt [3]{d} \sqrt [3]{a-x}+\sqrt [3]{x-b}\right ) \left (d^{2/3} (a-x)^{2/3}-\sqrt [3]{d} \sqrt [3]{x-b} \sqrt [3]{a-x}+(x-b)^{2/3}\right ) \left (\frac {3 \left (a \sqrt [3]{x-b}-i \sqrt {3} a \sqrt [3]{x-b}\right )}{2 (a-b)^2 \sqrt [3]{a-x}}+\frac {3 i \left (\sqrt {3} \sqrt [3]{x-b} b+i \sqrt [3]{x-b} b\right )}{2 (a-b)^2 \sqrt [3]{a-x}}+\frac {3 \left (a c \sqrt [3]{x-b}-i \sqrt {3} a c \sqrt [3]{x-b}\right )}{2 (a-b)^2 \sqrt [3]{a-x}}+\frac {3 i \left (\sqrt {3} a \sqrt [3]{x-b} c+i a \sqrt [3]{x-b} c\right )}{2 (a-b)^2 \sqrt [3]{a-x}}+\frac {\left (-\sqrt {3} b+3 i b+\sqrt {3} a d-3 i a d\right ) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{x-b}}{\sqrt [3]{x-b}-2 \sqrt [3]{d} \sqrt [3]{a-x}}\right )}{2 (a-b)^2 d^{2/3}}+\frac {\left (-\sqrt {3} d b+3 i d b+\sqrt {3} b-3 i b\right ) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{x-b}}{\sqrt [3]{x-b}-2 \sqrt [3]{d} \sqrt [3]{a-x}}\right )}{2 (a-b)^2 d^{2/3}}+\frac {\left (\sqrt {3} a c-3 i a c-\sqrt {3} a d c+3 i a d c\right ) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{x-b}}{\sqrt [3]{x-b}-2 \sqrt [3]{d} \sqrt [3]{a-x}}\right )}{2 (a-b)^2 d^{2/3}}+\frac {\left (-\sqrt {3} b c+3 i b c+\sqrt {3} a d c-3 i a d c\right ) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{x-b}}{\sqrt [3]{x-b}-2 \sqrt [3]{d} \sqrt [3]{a-x}}\right )}{2 (a-b)^2 d^{2/3}}+\frac {\left (-i \sqrt {3} b+b+i \sqrt {3} a d-a d\right ) \log \left (\sqrt [3]{d} \sqrt [3]{a-x}+\sqrt [3]{x-b}\right )}{2 (a-b)^2 d^{2/3}}+\frac {\left (-i \sqrt {3} d b+d b+i \sqrt {3} b-b\right ) \log \left (\sqrt [3]{d} \sqrt [3]{a-x}+\sqrt [3]{x-b}\right )}{2 (a-b)^2 d^{2/3}}+\frac {\left (i \sqrt {3} a c-a c-i \sqrt {3} a d c+a d c\right ) \log \left (\sqrt [3]{d} \sqrt [3]{a-x}+\sqrt [3]{x-b}\right )}{2 (a-b)^2 d^{2/3}}+\frac {\left (-i \sqrt {3} b c+b c+i \sqrt {3} a d c-a d c\right ) \log \left (\sqrt [3]{d} \sqrt [3]{a-x}+\sqrt [3]{x-b}\right )}{2 (a-b)^2 d^{2/3}}+\frac {\left (i \sqrt {3} b-b-i \sqrt {3} a d+a d\right ) \log \left (d^{2/3} (a-x)^{2/3}-\sqrt [3]{d} \sqrt [3]{x-b} \sqrt [3]{a-x}+(x-b)^{2/3}\right )}{4 (a-b)^2 d^{2/3}}+\frac {\left (i \sqrt {3} d b-d b-i \sqrt {3} b+b\right ) \log \left (d^{2/3} (a-x)^{2/3}-\sqrt [3]{d} \sqrt [3]{x-b} \sqrt [3]{a-x}+(x-b)^{2/3}\right )}{4 (a-b)^2 d^{2/3}}+\frac {\left (i \sqrt {3} b c-b c-i \sqrt {3} a d c+a d c\right ) \log \left (d^{2/3} (a-x)^{2/3}-\sqrt [3]{d} \sqrt [3]{x-b} \sqrt [3]{a-x}+(x-b)^{2/3}\right )}{4 (a-b)^2 d^{2/3}}+\frac {\left (-i \sqrt {3} a c+a c+i \sqrt {3} a d c-a d c\right ) \log \left (d^{2/3} (a-x)^{2/3}-\sqrt [3]{d} \sqrt [3]{x-b} \sqrt [3]{a-x}+(x-b)^{2/3}\right )}{4 (a-b)^2 d^{2/3}}\right )}{2 \sqrt [3]{-\left ((a-x) (b-x)^2\right )} (-b+a d-d x+x)} \]

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Rubi [A]  time = 1.47, antiderivative size = 280, normalized size of antiderivative = 0.20, number of steps used = 4, number of rules used = 4, integrand size = 51, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.078, Rules used = {6719, 155, 12, 91} \begin {gather*} -\frac {\sqrt [3]{x-a} (x-b)^{2/3} (c+d) \log (-a d+b-(1-d) x)}{2 d^{2/3} (a-b) \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}+\frac {3 \sqrt [3]{x-a} (x-b)^{2/3} (c+d) \log \left (\sqrt [3]{d} \sqrt [3]{x-a}-\sqrt [3]{x-b}\right )}{2 d^{2/3} (a-b) \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}+\frac {\sqrt {3} \sqrt [3]{x-a} (x-b)^{2/3} (c+d) \tan ^{-1}\left (\frac {2 \sqrt [3]{d} \sqrt [3]{x-a}}{\sqrt {3} \sqrt [3]{x-b}}+\frac {1}{\sqrt {3}}\right )}{d^{2/3} (a-b) \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}-\frac {3 (b-x)}{(a-b) \sqrt [3]{-\left ((a-x) (b-x)^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(b - x))/((a - b)*(-((a - x)*(b - x)^2))^(1/3)) + (Sqrt[3]*(c + d)*(-a + x)^(1/3)*(-b + x)^(2/3)*ArcTan[1/
Sqrt[3] + (2*d^(1/3)*(-a + x)^(1/3))/(Sqrt[3]*(-b + x)^(1/3))])/((a - b)*d^(2/3)*(-((a - x)*(b - x)^2))^(1/3))
- ((c + d)*(-a + x)^(1/3)*(-b + x)^(2/3)*Log[b - a*d - (1 - d)*x])/(2*(a - b)*d^(2/3)*(-((a - x)*(b - x)^2))^
(1/3)) + (3*(c + d)*(-a + x)^(1/3)*(-b + x)^(2/3)*Log[d^(1/3)*(-a + x)^(1/3) - (-b + x)^(1/3)])/(2*(a - b)*d^(
2/3)*(-((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 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 {-b-a c+(1+c) x}{(-a+x) \sqrt [3]{(-a+x) (-b+x)^2} (b-a d+(-1+d) x)} \, dx &=\frac {\left (\sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \frac {-b-a c+(1+c) x}{(-a+x)^{4/3} (-b+x)^{2/3} (b-a d+(-1+d) x)} \, dx}{\sqrt [3]{(-a+x) (-b+x)^2}}\\ &=-\frac {3 (b-x)}{(a-b) \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}+\frac {\left (3 \sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \frac {(a-b)^2 (c+d)}{3 \sqrt [3]{-a+x} (-b+x)^{2/3} (b-a d+(-1+d) x)} \, dx}{(a-b)^2 \sqrt [3]{(-a+x) (-b+x)^2}}\\ &=-\frac {3 (b-x)}{(a-b) \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}+\frac {\left ((c+d) \sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \frac {1}{\sqrt [3]{-a+x} (-b+x)^{2/3} (b-a d+(-1+d) x)} \, dx}{\sqrt [3]{(-a+x) (-b+x)^2}}\\ &=-\frac {3 (b-x)}{(a-b) \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}+\frac {\sqrt {3} (c+d) \sqrt [3]{-a+x} (-b+x)^{2/3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{d} \sqrt [3]{-a+x}}{\sqrt {3} \sqrt [3]{-b+x}}\right )}{(a-b) d^{2/3} \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}-\frac {(c+d) \sqrt [3]{-a+x} (-b+x)^{2/3} \log (b-a d-(1-d) x)}{2 (a-b) d^{2/3} \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}+\frac {3 (c+d) \sqrt [3]{-a+x} (-b+x)^{2/3} \log \left (\sqrt [3]{d} \sqrt [3]{-a+x}-\sqrt [3]{-b+x}\right )}{2 (a-b) d^{2/3} \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}\\ \end {align*}

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Mathematica [C]  time = 0.08, size = 66, normalized size = 0.05 \begin {gather*} -\frac {3 (x-b) \left ((c+d) \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};\frac {b-x}{a d-d x}\right )-d\right )}{d (a-b) \sqrt [3]{(x-a) (b-x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(-b + x)*(-d + (c + d)*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 [F]  time = 180.57, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

$Aborted

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fricas [A]  time = 0.65, size = 473, normalized size = 0.34 \begin {gather*} \frac {2 \, \sqrt {3} {\left (a b c d + a b d^{2} + {\left (c d + d^{2}\right )} x^{2} - {\left ({\left (a + b\right )} c d + {\left (a + b\right )} d^{2}\right )} x\right )} \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 ) + 6 \, {\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 c + a b d + {\left (c + d\right )} x^{2} - {\left ({\left (a + b\right )} c + {\left (a + b\right )} 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 {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 (a b c + a b d + {\left (c + d\right )} x^{2} - {\left ({\left (a + b\right )} c + {\left (a + b\right )} d\right )} x\right )} \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 ({\left (a - b\right )} d^{2} x^{2} - {\left (a^{2} - b^{2}\right )} d^{2} x + {\left (a^{2} b - a b^{2}\right )} d^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {-b -a c +\left (1+c \right ) x}{\left (-a +x \right ) \left (\left (-a +x \right ) \left (-b +x \right )^{2}\right )^{\frac {1}{3}} \left (b -a d +\left (-1+d \right ) x \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

Timed out

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