3.19.50 \(\int \frac {-((2 a-b) b^2)+(4 a-b) b x-(2 a+b) x^2+x^3}{((-a+x) (-b+x)^2)^{3/4} (a+b^2 d-(1+2 b d) x+d x^2)} \, dx\) [1850]

Optimal. Leaf size=127 \[ -\frac {2 \text {ArcTan}\left (\frac {\sqrt [4]{d} \left (-a b^2+\left (2 a b+b^2\right ) x+(-a-2 b) x^2+x^3\right )^{3/4}}{(b-x) (-a+x)}\right )}{d^{3/4}}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{d} \left (-a b^2+\left (2 a b+b^2\right ) x+(-a-2 b) x^2+x^3\right )^{3/4}}{(b-x) (-a+x)}\right )}{d^{3/4}} \]

[Out]

-2*arctan(d^(1/4)*(-a*b^2+(2*a*b+b^2)*x+(-a-2*b)*x^2+x^3)^(3/4)/(b-x)/(-a+x))/d^(3/4)+2*arctanh(d^(1/4)*(-a*b^
2+(2*a*b+b^2)*x+(-a-2*b)*x^2+x^3)^(3/4)/(b-x)/(-a+x))/d^(3/4)

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Rubi [C] Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
time = 1.81, antiderivative size = 283, normalized size of antiderivative = 2.23, number of steps used = 8, number of rules used = 5, integrand size = 77, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {6820, 6851, 6860, 142, 141} \begin {gather*} -\frac {4 \left (1-\sqrt {-4 a d+4 b d+1}\right ) \sqrt [4]{-\left ((a-x) (b-x)^2\right )} F_1\left (\frac {1}{4};-\frac {1}{2},1;\frac {5}{4};\frac {a-x}{a-b},-\frac {2 d (a-x)}{-2 a d+2 b d-\sqrt {-4 a d+4 b d+1}+1}\right )}{\left (-\sqrt {-4 a d+4 b d+1}-2 a d+2 b d+1\right ) \sqrt {-\frac {b-x}{a-b}}}-\frac {4 \left (\sqrt {-4 a d+4 b d+1}+1\right ) \sqrt [4]{-\left ((a-x) (b-x)^2\right )} F_1\left (\frac {1}{4};-\frac {1}{2},1;\frac {5}{4};\frac {a-x}{a-b},-\frac {2 d (a-x)}{-2 a d+2 b d+\sqrt {-4 a d+4 b d+1}+1}\right )}{\left (\sqrt {-4 a d+4 b d+1}-2 a d+2 b d+1\right ) \sqrt {-\frac {b-x}{a-b}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(1 - Sqrt[1 - 4*a*d + 4*b*d])*(-((a - x)*(b - x)^2))^(1/4)*AppellF1[1/4, -1/2, 1, 5/4, (a - x)/(a - b), (-
2*d*(a - x))/(1 - 2*a*d + 2*b*d - Sqrt[1 - 4*a*d + 4*b*d])])/((1 - 2*a*d + 2*b*d - Sqrt[1 - 4*a*d + 4*b*d])*Sq
rt[-((b - x)/(a - b))]) - (4*(1 + Sqrt[1 - 4*a*d + 4*b*d])*(-((a - x)*(b - x)^2))^(1/4)*AppellF1[1/4, -1/2, 1,
5/4, (a - x)/(a - b), (-2*d*(a - x))/(1 - 2*a*d + 2*b*d + Sqrt[1 - 4*a*d + 4*b*d])])/((1 - 2*a*d + 2*b*d + Sq
rt[1 - 4*a*d + 4*b*d])*Sqrt[-((b - x)/(a - b))])

Rule 141

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)/(b^(p + 1)*(m + 1)*(b/(b*c - a*d))^n))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(
b*c - a*d)), (-f)*((a + b*x)/(b*e - a*f))], 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 142

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 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6851

Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a*v^m*w^n)^FracPart[p]/(v^(m*Fr
acPart[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 6860

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 {-\left ((2 a-b) b^2\right )+(4 a-b) b x-(2 a+b) x^2+x^3}{\left ((-a+x) (-b+x)^2\right )^{3/4} \left (a+b^2 d-(1+2 b d) x+d x^2\right )} \, dx &=\int \frac {(2 a-b-x) \sqrt [4]{-\left ((a-x) (b-x)^2\right )}}{(a-x) \left (a+b^2 d-(1+2 b d) x+d x^2\right )} \, dx\\ &=\frac {\sqrt [4]{-\left ((a-x) (b-x)^2\right )} \int \frac {(2 a-b-x) \sqrt {b-x}}{(a-x)^{3/4} \left (a+b^2 d-(1+2 b d) x+d x^2\right )} \, dx}{\sqrt [4]{a-x} \sqrt {b-x}}\\ &=\frac {\sqrt [4]{-\left ((a-x) (b-x)^2\right )} \int \left (\frac {\left (-1-\sqrt {1-4 a d+4 b d}\right ) \sqrt {b-x}}{(a-x)^{3/4} \left (-1-2 b 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 {b-x}}{(a-x)^{3/4} \left (-1-2 b d+\sqrt {1-4 a d+4 b d}+2 d x\right )}\right ) \, dx}{\sqrt [4]{a-x} \sqrt {b-x}}\\ &=\frac {\left (\left (-1-\sqrt {1-4 a d+4 b d}\right ) \sqrt [4]{-\left ((a-x) (b-x)^2\right )}\right ) \int \frac {\sqrt {b-x}}{(a-x)^{3/4} \left (-1-2 b d-\sqrt {1-4 a d+4 b d}+2 d x\right )} \, dx}{\sqrt [4]{a-x} \sqrt {b-x}}+\frac {\left (\left (-1+\sqrt {1-4 a d+4 b d}\right ) \sqrt [4]{-\left ((a-x) (b-x)^2\right )}\right ) \int \frac {\sqrt {b-x}}{(a-x)^{3/4} \left (-1-2 b d+\sqrt {1-4 a d+4 b d}+2 d x\right )} \, dx}{\sqrt [4]{a-x} \sqrt {b-x}}\\ &=\frac {\left (\left (-1-\sqrt {1-4 a d+4 b d}\right ) \sqrt [4]{-\left ((a-x) (b-x)^2\right )}\right ) \int \frac {\sqrt {-\frac {b}{a-b}+\frac {x}{a-b}}}{(a-x)^{3/4} \left (-1-2 b d-\sqrt {1-4 a d+4 b d}+2 d x\right )} \, dx}{\sqrt [4]{a-x} \sqrt {-\frac {b-x}{a-b}}}+\frac {\left (\left (-1+\sqrt {1-4 a d+4 b d}\right ) \sqrt [4]{-\left ((a-x) (b-x)^2\right )}\right ) \int \frac {\sqrt {-\frac {b}{a-b}+\frac {x}{a-b}}}{(a-x)^{3/4} \left (-1-2 b d+\sqrt {1-4 a d+4 b d}+2 d x\right )} \, dx}{\sqrt [4]{a-x} \sqrt {-\frac {b-x}{a-b}}}\\ &=-\frac {4 \left (1-\sqrt {1-4 a d+4 b d}\right ) \sqrt [4]{-\left ((a-x) (b-x)^2\right )} F_1\left (\frac {1}{4};-\frac {1}{2},1;\frac {5}{4};\frac {a-x}{a-b},-\frac {2 d (a-x)}{1-2 a d+2 b d-\sqrt {1-4 a d+4 b d}}\right )}{\left (1-2 a d+2 b d-\sqrt {1-4 a d+4 b d}\right ) \sqrt {-\frac {b-x}{a-b}}}-\frac {4 \left (1+\sqrt {1-4 a d+4 b d}\right ) \sqrt [4]{-\left ((a-x) (b-x)^2\right )} F_1\left (\frac {1}{4};-\frac {1}{2},1;\frac {5}{4};\frac {a-x}{a-b},-\frac {2 d (a-x)}{1-2 a d+2 b d+\sqrt {1-4 a d+4 b d}}\right )}{\left (1-2 a d+2 b d+\sqrt {1-4 a d+4 b d}\right ) \sqrt {-\frac {b-x}{a-b}}}\\ \end {align*}

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Mathematica [A]
time = 2.70, size = 155, normalized size = 1.22 \begin {gather*} \frac {\sqrt {2} \sqrt [4]{(b-x)^2 (-a+x)} \left (\text {ArcTan}\left (\frac {-b \sqrt {d}+\sqrt {a-x}+\sqrt {d} x}{\sqrt {2} \sqrt [4]{d} \sqrt [4]{a-x} \sqrt {b-x}}\right )+\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{d} \sqrt [4]{a-x} \sqrt {b-x}}{\sqrt {a-x}+\sqrt {d} (b-x)}\right )\right )}{d^{3/4} \sqrt [4]{a-x} \sqrt {b-x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {-\left (2 a -b \right ) b^{2}+\left (4 a -b \right ) b x -\left (2 a +b \right ) x^{2}+x^{3}}{\left (\left (-a +x \right ) \left (-b +x \right )^{2}\right )^{\frac {3}{4}} \left (a +b^{2} d -\left (2 b d +1\right ) x +d \,x^{2}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-(2*a-b)*b^2+(4*a-b)*b*x-(2*a+b)*x^2+x^3)/((-a+x)*(-b+x)^2)^(3/4)/(a+b^2*d-(2*b*d+1)*x+d*x^2),x)

[Out]

int((-(2*a-b)*b^2+(4*a-b)*b*x-(2*a+b)*x^2+x^3)/((-a+x)*(-b+x)^2)^(3/4)/(a+b^2*d-(2*b*d+1)*x+d*x^2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-(2*a-b)*b^2+(4*a-b)*b*x-(2*a+b)*x^2+x^3)/((-a+x)*(-b+x)^2)^(3/4)/(a+b^2*d-(2*b*d+1)*x+d*x^2),x, al
gorithm="maxima")

[Out]

-integrate(((2*a - b)*b^2 - (4*a - b)*b*x + (2*a + b)*x^2 - x^3)/((-(a - x)*(b - x)^2)^(3/4)*(b^2*d + d*x^2 -
(2*b*d + 1)*x + a)), x)

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Fricas [F(-1)] Timed out
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((-(2*a-b)*b^2+(4*a-b)*b*x-(2*a+b)*x^2+x^3)/((-a+x)*(-b+x)^2)^(3/4)/(a+b^2*d-(2*b*d+1)*x+d*x^2),x, al
gorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)] Timed out
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((-(2*a-b)*b**2+(4*a-b)*b*x-(2*a+b)*x**2+x**3)/((-a+x)*(-b+x)**2)**(3/4)/(a+b**2*d-(2*b*d+1)*x+d*x**2
),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-(2*a-b)*b^2+(4*a-b)*b*x-(2*a+b)*x^2+x^3)/((-a+x)*(-b+x)^2)^(3/4)/(a+b^2*d-(2*b*d+1)*x+d*x^2),x, al
gorithm="giac")

[Out]

integrate(-((2*a - b)*b^2 - (4*a - b)*b*x + (2*a + b)*x^2 - x^3)/((-(a - x)*(b - x)^2)^(3/4)*(b^2*d + d*x^2 -
(2*b*d + 1)*x + a)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int -\frac {b^2\,\left (2\,a-b\right )+x^2\,\left (2\,a+b\right )-x^3-b\,x\,\left (4\,a-b\right )}{{\left (-\left (a-x\right )\,{\left (b-x\right )}^2\right )}^{3/4}\,\left (a-x\,\left (2\,b\,d+1\right )+b^2\,d+d\,x^2\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(b^2*(2*a - b) + x^2*(2*a + b) - x^3 - b*x*(4*a - b))/((-(a - x)*(b - x)^2)^(3/4)*(a - x*(2*b*d + 1) + b^
2*d + d*x^2)),x)

[Out]

int(-(b^2*(2*a - b) + x^2*(2*a + b) - x^3 - b*x*(4*a - b))/((-(a - x)*(b - x)^2)^(3/4)*(a - x*(2*b*d + 1) + b^
2*d + d*x^2)), x)

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