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

Optimal. Leaf size=857 \[ \frac {(b-x)^{2/3} \sqrt [3]{x-a} \left (-\frac {\sqrt {3} (d-1) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{x-a}}{\sqrt [3]{x-a}-2 \sqrt [3]{d} \sqrt [3]{b-x}}\right ) a}{(a-b)^2 \sqrt [3]{d}}-\frac {(d-1) \log \left (\sqrt [3]{d} \sqrt [3]{b-x}+\sqrt [3]{x-a}\right ) a}{(a-b)^2 \sqrt [3]{d}}+\frac {(d-1) \log \left (d^{2/3} (b-x)^{2/3}-\sqrt [3]{d} \sqrt [3]{x-a} \sqrt [3]{b-x}+(x-a)^{2/3}\right ) a}{2 (a-b)^2 \sqrt [3]{d}}+\frac {3 \sqrt [3]{b-x} (x-a)^{2/3} a}{2 (a-b)^2 (x-b)}-\frac {\sqrt {3} c (a-b d) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{x-a}}{\sqrt [3]{x-a}-2 \sqrt [3]{d} \sqrt [3]{b-x}}\right )}{(a-b)^2 \sqrt [3]{d}}-\frac {\sqrt {3} (a-b d) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{x-a}}{\sqrt [3]{x-a}-2 \sqrt [3]{d} \sqrt [3]{b-x}}\right )}{(a-b)^2 \sqrt [3]{d}}-\frac {\sqrt {3} b c (d-1) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{x-a}}{\sqrt [3]{x-a}-2 \sqrt [3]{d} \sqrt [3]{b-x}}\right )}{(a-b)^2 \sqrt [3]{d}}-\frac {c (a-b d) \log \left (\sqrt [3]{d} \sqrt [3]{b-x}+\sqrt [3]{x-a}\right )}{(a-b)^2 \sqrt [3]{d}}+\frac {(b d-a) \log \left (\sqrt [3]{d} \sqrt [3]{b-x}+\sqrt [3]{x-a}\right )}{(a-b)^2 \sqrt [3]{d}}-\frac {b c (d-1) \log \left (\sqrt [3]{d} \sqrt [3]{b-x}+\sqrt [3]{x-a}\right )}{(a-b)^2 \sqrt [3]{d}}+\frac {c (a-b d) \log \left (d^{2/3} (b-x)^{2/3}-\sqrt [3]{d} \sqrt [3]{x-a} \sqrt [3]{b-x}+(x-a)^{2/3}\right )}{2 (a-b)^2 \sqrt [3]{d}}+\frac {(a-b d) \log \left (d^{2/3} (b-x)^{2/3}-\sqrt [3]{d} \sqrt [3]{x-a} \sqrt [3]{b-x}+(x-a)^{2/3}\right )}{2 (a-b)^2 \sqrt [3]{d}}+\frac {b c (d-1) \log \left (d^{2/3} (b-x)^{2/3}-\sqrt [3]{d} \sqrt [3]{x-a} \sqrt [3]{b-x}+(x-a)^{2/3}\right )}{2 (a-b)^2 \sqrt [3]{d}}-\frac {3 b \sqrt [3]{b-x} (x-a)^{2/3}}{2 (a-b)^2 (x-b)}\right )}{\sqrt [3]{(b-x)^2 (x-a)}} \]

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Rubi [A]  time = 1.44, antiderivative size = 283, normalized size of antiderivative = 0.33, 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-b d-(1-d) x)}{2 \sqrt [3]{d} (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 (\frac {\sqrt [3]{x-a}}{\sqrt [3]{d}}-\sqrt [3]{x-b}\right )}{2 \sqrt [3]{d} (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]{x-a}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{x-b}}+\frac {1}{\sqrt {3}}\right )}{\sqrt [3]{d} (a-b) \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}+\frac {3 (a-x)}{2 (a-b) \sqrt [3]{-\left ((a-x) (b-x)^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(a - x))/(2*(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*(-a + x)^(1/3))/(Sqrt[3]*d^(1/3)*(-b + x)^(1/3))])/((a - b)*d^(1/3)*(-((a - x)*(b - x)^2))^(1/3)
) + ((c + d)*(-a + x)^(1/3)*(-b + x)^(2/3)*Log[a - b*d - (1 - d)*x])/(2*(a - b)*d^(1/3)*(-((a - x)*(b - x)^2))
^(1/3)) - (3*(c + d)*(-a + x)^(1/3)*(-b + x)^(2/3)*Log[(-a + x)^(1/3)/d^(1/3) - (-b + x)^(1/3)])/(2*(a - b)*d^
(1/3)*(-((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 {-a-b c+(1+c) x}{(-b+x) \sqrt [3]{(-a+x) (-b+x)^2} (a-b d+(-1+d) x)} \, dx &=\frac {\left (\sqrt [3]{-a+x} (-b+x)^{2/3}\right ) \int \frac {-a-b c+(1+c) x}{\sqrt [3]{-a+x} (-b+x)^{5/3} (a-b d+(-1+d) x)} \, dx}{\sqrt [3]{(-a+x) (-b+x)^2}}\\ &=\frac {3 (a-x)}{2 (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 {2 (a-b)^2 (c+d)}{3 \sqrt [3]{-a+x} (-b+x)^{2/3} (a-b d+(-1+d) x)} \, dx}{2 (a-b)^2 \sqrt [3]{(-a+x) (-b+x)^2}}\\ &=\frac {3 (a-x)}{2 (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} (a-b d+(-1+d) x)} \, dx}{\sqrt [3]{(-a+x) (-b+x)^2}}\\ &=\frac {3 (a-x)}{2 (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]{-a+x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{-b+x}}\right )}{(a-b) \sqrt [3]{d} \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}+\frac {(c+d) \sqrt [3]{-a+x} (-b+x)^{2/3} \log (a-b d-(1-d) x)}{2 (a-b) \sqrt [3]{d} \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}-\frac {3 (c+d) \sqrt [3]{-a+x} (-b+x)^{2/3} \log \left (\frac {\sqrt [3]{-a+x}}{\sqrt [3]{d}}-\sqrt [3]{-b+x}\right )}{2 (a-b) \sqrt [3]{d} \sqrt [3]{-\left ((a-x) (b-x)^2\right )}}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 65, normalized size = 0.08 \begin {gather*} \frac {3 \left (2 (x-b) (c+d) \, _2F_1\left (\frac {1}{3},1;\frac {4}{3};\frac {d (b-x)}{a-x}\right )+a-x\right )}{2 (a-b) \sqrt [3]{(x-a) (b-x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 11.66, size = 242, normalized size = 0.28 \begin {gather*} \frac {(b-x)^{2/3} \sqrt [3]{-a+x} \left (-\frac {3 (-a+x)^{2/3}}{2 (a-b) (b-x)^{2/3}}+\frac {\sqrt {3} (c+d) \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) \sqrt [3]{d}}+\frac {(c+d) \log \left (1+\frac {d^{2/3} (b-x)^{2/3}}{(-a+x)^{2/3}}-\frac {\sqrt [3]{d} \sqrt [3]{b-x}}{\sqrt [3]{-a+x}}\right )}{2 (a-b) \sqrt [3]{d}}+\frac {(-c-d) \log \left (1+\frac {\sqrt [3]{d} \sqrt [3]{b-x}}{\sqrt [3]{-a+x}}\right )}{(a-b) \sqrt [3]{d}}\right )}{\sqrt [3]{(b-x)^2 (-a+x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.70, size = 976, normalized size = 1.14

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 {1}{3}} {\left (b d - {\left (d - 1\right )} x - a\right )} {\left (b - x\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

Timed out

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