3.23.17 \(\int \frac {-((2 a-b) b^2)+(4 a-b) b x-(2 a+b) x^2+x^3}{(-a+x) \sqrt [4]{(-a+x) (-b+x)^2} (b^2+a d-(2 b+d) x+x^2)} \, dx\)

Optimal. Leaf size=165 \[ -2 \sqrt [4]{d} \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{x \left (2 a b+b^2\right )-a b^2+x^2 (-a-2 b)+x^3}}{b-x}\right )+2 \sqrt [4]{d} \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{x \left (2 a b+b^2\right )-a b^2+x^2 (-a-2 b)+x^3}}{b-x}\right )-\frac {4 \left (-a b^2+2 a b x-a x^2+b^2 x-2 b x^2+x^3\right )^{3/4}}{(x-a) (b-x)} \]

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Rubi [C]  time = 3.34, antiderivative size = 325, normalized size of antiderivative = 1.97, number of steps used = 8, number of rules used = 5, integrand size = 81, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {6688, 6719, 6728, 137, 136} \begin {gather*} \frac {4 (a-b) (b-x) \left (\sqrt {-4 a+4 b+d}+\sqrt {d}\right ) F_1\left (-\frac {1}{4};-\frac {3}{2},1;\frac {3}{4};\frac {a-x}{a-b},\frac {2 (a-x)}{2 a-2 b-d-\sqrt {d} \sqrt {-4 a+4 b+d}}\right )}{\sqrt {d} \left (-\sqrt {d} \sqrt {-4 a+4 b+d}+2 a-2 b-d\right ) \sqrt {-\frac {b-x}{a-b}} \sqrt [4]{-\left ((a-x) (b-x)^2\right )}}+\frac {4 (a-b) (b-x) \left (1-\frac {\sqrt {-4 a+4 b+d}}{\sqrt {d}}\right ) F_1\left (-\frac {1}{4};-\frac {3}{2},1;\frac {3}{4};\frac {a-x}{a-b},\frac {2 (a-x)}{2 a-2 b-d+\sqrt {d} \sqrt {-4 a+4 b+d}}\right )}{\left (\sqrt {d} \sqrt {-4 a+4 b+d}+2 a-2 b-d\right ) \sqrt {-\frac {b-x}{a-b}} \sqrt [4]{-\left ((a-x) (b-x)^2\right )}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(4*(a - b)*(Sqrt[d] + Sqrt[-4*a + 4*b + d])*(b - x)*AppellF1[-1/4, -3/2, 1, 3/4, (a - x)/(a - b), (2*(a - x))/
(2*a - 2*b - d - Sqrt[d]*Sqrt[-4*a + 4*b + d])])/(Sqrt[d]*(2*a - 2*b - d - Sqrt[d]*Sqrt[-4*a + 4*b + d])*Sqrt[
-((b - x)/(a - b))]*(-((a - x)*(b - x)^2))^(1/4)) + (4*(a - b)*(1 - Sqrt[-4*a + 4*b + d]/Sqrt[d])*(b - x)*Appe
llF1[-1/4, -3/2, 1, 3/4, (a - x)/(a - b), (2*(a - x))/(2*a - 2*b - d + Sqrt[d]*Sqrt[-4*a + 4*b + d])])/((2*a -
2*b - d + Sqrt[d]*Sqrt[-4*a + 4*b + d])*Sqrt[-((b - x)/(a - b))]*(-((a - x)*(b - x)^2))^(1/4))

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 6688

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

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

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Mathematica [C]  time = 18.73, size = 736, normalized size = 4.46 \begin {gather*} -\frac {4 (b-x)}{\sqrt [4]{(x-a) (b-x)^2}}-\frac {i \sqrt {2} d (a-x)^{11/4} (b-x)^3 \sqrt {\frac {b}{a-x}-\frac {a}{a-x}+1} \left (\left (\sqrt {d (-4 a+4 b+d)}+4 a-4 b-d\right ) \sqrt {-\frac {\sqrt {d (-4 a+4 b+d)}-2 a+2 b+d}{(a-b)^2}} \Pi \left (-\frac {\sqrt {2}}{\sqrt {a-b} \sqrt {\frac {2 a-2 b-d+\sqrt {d (-4 a+4 b+d)}}{(a-b)^2}}};\left .i \sinh ^{-1}\left (\frac {\sqrt {-\sqrt {a-b}}}{\sqrt [4]{a-x}}\right )\right |-1\right )-\left (\sqrt {d (-4 a+4 b+d)}+4 a-4 b-d\right ) \sqrt {-\frac {\sqrt {d (-4 a+4 b+d)}-2 a+2 b+d}{(a-b)^2}} \Pi \left (\frac {\sqrt {2}}{\sqrt {a-b} \sqrt {\frac {2 a-2 b-d+\sqrt {d (-4 a+4 b+d)}}{(a-b)^2}}};\left .i \sinh ^{-1}\left (\frac {\sqrt {-\sqrt {a-b}}}{\sqrt [4]{a-x}}\right )\right |-1\right )-\left (-\sqrt {d (-4 a+4 b+d)}+4 a-4 b-d\right ) \sqrt {\frac {\sqrt {d (-4 a+4 b+d)}+2 a-2 b-d}{(a-b)^2}} \left (\Pi \left (-\frac {\sqrt {2}}{\sqrt {a-b} \sqrt {-\frac {-2 a+2 b+d+\sqrt {d (-4 a+4 b+d)}}{(a-b)^2}}};\left .i \sinh ^{-1}\left (\frac {\sqrt {-\sqrt {a-b}}}{\sqrt [4]{a-x}}\right )\right |-1\right )-\Pi \left (\frac {\sqrt {2}}{\sqrt {a-b} \sqrt {-\frac {-2 a+2 b+d+\sqrt {d (-4 a+4 b+d)}}{(a-b)^2}}};\left .i \sinh ^{-1}\left (\frac {\sqrt {-\sqrt {a-b}}}{\sqrt [4]{a-x}}\right )\right |-1\right )\right )\right )}{\sqrt {-\sqrt {a-b}} (a-b) (x-a) (x-b) \sqrt {d (-4 a+4 b+d)} \sqrt {\frac {\sqrt {d (-4 a+4 b+d)}+2 a-2 b-d}{(a-b)^2}} \sqrt {-\frac {\sqrt {d (-4 a+4 b+d)}-2 a+2 b+d}{(a-b)^2}} \left (-\left ((a-x) (b-x)^2\right )\right )^{5/4}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-4*(b - x))/((b - x)^2*(-a + x))^(1/4) - (I*Sqrt[2]*d*Sqrt[1 - a/(a - x) + b/(a - x)]*(a - x)^(11/4)*(b - x)^
3*((4*a - 4*b - d + Sqrt[d*(-4*a + 4*b + d)])*Sqrt[-((-2*a + 2*b + d + Sqrt[d*(-4*a + 4*b + d)])/(a - b)^2)]*E
llipticPi[-(Sqrt[2]/(Sqrt[a - b]*Sqrt[(2*a - 2*b - d + Sqrt[d*(-4*a + 4*b + d)])/(a - b)^2])), I*ArcSinh[Sqrt[
-Sqrt[a - b]]/(a - x)^(1/4)], -1] - (4*a - 4*b - d + Sqrt[d*(-4*a + 4*b + d)])*Sqrt[-((-2*a + 2*b + d + Sqrt[d
*(-4*a + 4*b + d)])/(a - b)^2)]*EllipticPi[Sqrt[2]/(Sqrt[a - b]*Sqrt[(2*a - 2*b - d + Sqrt[d*(-4*a + 4*b + d)]
)/(a - b)^2]), I*ArcSinh[Sqrt[-Sqrt[a - b]]/(a - x)^(1/4)], -1] - (4*a - 4*b - d - Sqrt[d*(-4*a + 4*b + d)])*S
qrt[(2*a - 2*b - d + Sqrt[d*(-4*a + 4*b + d)])/(a - b)^2]*(EllipticPi[-(Sqrt[2]/(Sqrt[a - b]*Sqrt[-((-2*a + 2*
b + d + Sqrt[d*(-4*a + 4*b + d)])/(a - b)^2)])), I*ArcSinh[Sqrt[-Sqrt[a - b]]/(a - x)^(1/4)], -1] - EllipticPi
[Sqrt[2]/(Sqrt[a - b]*Sqrt[-((-2*a + 2*b + d + Sqrt[d*(-4*a + 4*b + d)])/(a - b)^2)]), I*ArcSinh[Sqrt[-Sqrt[a
- b]]/(a - x)^(1/4)], -1])))/(Sqrt[-Sqrt[a - b]]*(a - b)*Sqrt[d*(-4*a + 4*b + d)]*Sqrt[(2*a - 2*b - d + Sqrt[d
*(-4*a + 4*b + d)])/(a - b)^2]*Sqrt[-((-2*a + 2*b + d + Sqrt[d*(-4*a + 4*b + d)])/(a - b)^2)]*(-((a - x)*(b -
x)^2))^(5/4)*(-a + x)*(-b + x))

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IntegrateAlgebraic [A]  time = 0.92, size = 165, normalized size = 1.00 \begin {gather*} -\frac {4 \left (-a b^2+2 a b x+b^2 x-a x^2-2 b x^2+x^3\right )^{3/4}}{(b-x) (-a+x)}-2 \sqrt [4]{d} \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{-a b^2+\left (2 a b+b^2\right ) x+(-a-2 b) x^2+x^3}}{b-x}\right )+2 \sqrt [4]{d} \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{-a b^2+\left (2 a b+b^2\right ) x+(-a-2 b) x^2+x^3}}{b-x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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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((-(2*a-b)*b^2+(4*a-b)*b*x-(2*a+b)*x^2+x^3)/(-a+x)/((-a+x)*(-b+x)^2)^(1/4)/(b^2+a*d-(2*b+d)*x+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 (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 {1}{4}} {\left (b^{2} + a d - {\left (2 \, b + d\right )} x + x^{2}\right )} {\left (a - x\right )}}\,{d x} \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)/((-a+x)*(-b+x)^2)^(1/4)/(b^2+a*d-(2*b+d)*x+x^2),x,
algorithm="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)^(1/4)*(b^2 + a*d - (2*b
+ d)*x + x^2)*(a - x)), x)

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maple [F]  time = 0.13, size = 0, normalized size = 0.00 \[\int \frac {-\left (2 a -b \right ) b^{2}+\left (4 a -b \right ) b x -\left (2 a +b \right ) x^{2}+x^{3}}{\left (-a +x \right ) \left (\left (-a +x \right ) \left (-b +x \right )^{2}\right )^{\frac {1}{4}} \left (b^{2}+a d -\left (2 b +d \right ) x +x^{2}\right )}\, dx\]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {1}{4}} {\left (b^{2} + a d - {\left (2 \, b + d\right )} x + x^{2}\right )} {\left (a - x\right )}}\,{d x} \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)/((-a+x)*(-b+x)^2)^(1/4)/(b^2+a*d-(2*b+d)*x+x^2),x,
algorithm="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)^(1/4)*(b^2 + a*d - (2*b
+ d)*x + x^2)*(a - x)), 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 (a-x\right )\,{\left (-\left (a-x\right )\,{\left (b-x\right )}^2\right )}^{1/4}\,\left (a\,d-x\,\left (2\,b+d\right )+b^2+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)*(-(a - x)*(b - x)^2)^(1/4)*(a*d - x*(2*b +
d) + b^2 + x^2)),x)

[Out]

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

[Out]

Timed out

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