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

Optimal. Leaf size=153 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a b^3 x+x^3 \left (3 a b+3 b^2\right )+x^2 \left (-3 a b^2-b^3\right )+x^4 (-a-3 b)+x^5}}{a-x}\right )}{d^{3/4}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a b^3 x+x^3 \left (3 a b+3 b^2\right )+x^2 \left (-3 a b^2-b^3\right )+x^4 (-a-3 b)+x^5}}{a-x}\right )}{d^{3/4}} \]

________________________________________________________________________________________

Rubi [F]  time = 19.22, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {a b^3-2 (3 a-b) b^2 x+3 (3 a-b) b x^2-4 a x^3+x^4}{\sqrt [4]{x (-a+x) (-b+x)^3} \left (a^3-\left (3 a^2+b^3 d\right ) x+3 \left (a+b^2 d\right ) x^2-(1+3 b d) x^3+d x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

(4*a*b*(a - x)^(1/4)*(b - x)^(3/4)*x^(1/4)*Defer[Subst][Defer[Int][(x^2*(b - x^4)^(5/4))/((a - x^4)^(1/4)*(a^3
- 3*a^2*(1 + (b^3*d)/(3*a^2))*x^4 + 3*a*(1 + (b^2*d)/a)*x^8 - (1 + 3*b*d)*x^12 + d*x^16)), x], x, x^(1/4)])/(
(a - x)*(b - x)^3*x)^(1/4) - (8*(2*a - b)*(a - x)^(1/4)*(b - x)^(3/4)*x^(1/4)*Defer[Subst][Defer[Int][(x^6*(b
- x^4)^(5/4))/((a - x^4)^(1/4)*(a^3 - 3*a^2*(1 + (b^3*d)/(3*a^2))*x^4 + 3*a*(1 + (b^2*d)/a)*x^8 - (1 + 3*b*d)*
x^12 + d*x^16)), x], x, x^(1/4)])/((a - x)*(b - x)^3*x)^(1/4) + (4*(a - x)^(1/4)*(b - x)^(3/4)*x^(1/4)*Defer[S
ubst][Defer[Int][(x^10*(b - x^4)^(5/4))/((a - x^4)^(1/4)*(a^3 - 3*a^2*(1 + (b^3*d)/(3*a^2))*x^4 + 3*a*(1 + (b^
2*d)/a)*x^8 - (1 + 3*b*d)*x^12 + d*x^16)), x], x, x^(1/4)])/((a - x)*(b - x)^3*x)^(1/4)

Rubi steps

\begin {align*} \int \frac {a b^3-2 (3 a-b) b^2 x+3 (3 a-b) b x^2-4 a x^3+x^4}{\sqrt [4]{x (-a+x) (-b+x)^3} \left (a^3-\left (3 a^2+b^3 d\right ) x+3 \left (a+b^2 d\right ) x^2-(1+3 b d) x^3+d x^4\right )} \, dx &=\int \frac {(b-x)^2 \left (a b-2 (2 a-b) x+x^2\right )}{\sqrt [4]{(a-x) (b-x)^3 x} \left (a^3-\left (3 a^2+b^3 d\right ) x+3 \left (a+b^2 d\right ) x^2-(1+3 b d) x^3+d x^4\right )} \, dx\\ &=\frac {\left (\sqrt [4]{a-x} (b-x)^{3/4} \sqrt [4]{x}\right ) \int \frac {(b-x)^{5/4} \left (a b-2 (2 a-b) x+x^2\right )}{\sqrt [4]{a-x} \sqrt [4]{x} \left (a^3-\left (3 a^2+b^3 d\right ) x+3 \left (a+b^2 d\right ) x^2-(1+3 b d) x^3+d x^4\right )} \, dx}{\sqrt [4]{(a-x) (b-x)^3 x}}\\ &=\frac {\left (4 \sqrt [4]{a-x} (b-x)^{3/4} \sqrt [4]{x}\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (b-x^4\right )^{5/4} \left (a b-2 (2 a-b) x^4+x^8\right )}{\sqrt [4]{a-x^4} \left (a^3-\left (3 a^2+b^3 d\right ) x^4+3 \left (a+b^2 d\right ) x^8-(1+3 b d) x^{12}+d x^{16}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(a-x) (b-x)^3 x}}\\ &=\frac {\left (4 \sqrt [4]{a-x} (b-x)^{3/4} \sqrt [4]{x}\right ) \operatorname {Subst}\left (\int \left (\frac {a b x^2 \left (b-x^4\right )^{5/4}}{\sqrt [4]{a-x^4} \left (a^3-3 a^2 \left (1+\frac {b^3 d}{3 a^2}\right ) x^4+3 a \left (1+\frac {b^2 d}{a}\right ) x^8-(1+3 b d) x^{12}+d x^{16}\right )}+\frac {2 (-2 a+b) x^6 \left (b-x^4\right )^{5/4}}{\sqrt [4]{a-x^4} \left (a^3-3 a^2 \left (1+\frac {b^3 d}{3 a^2}\right ) x^4+3 a \left (1+\frac {b^2 d}{a}\right ) x^8-(1+3 b d) x^{12}+d x^{16}\right )}+\frac {x^{10} \left (b-x^4\right )^{5/4}}{\sqrt [4]{a-x^4} \left (a^3-3 a^2 \left (1+\frac {b^3 d}{3 a^2}\right ) x^4+3 a \left (1+\frac {b^2 d}{a}\right ) x^8-(1+3 b d) x^{12}+d x^{16}\right )}\right ) \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(a-x) (b-x)^3 x}}\\ &=\frac {\left (4 \sqrt [4]{a-x} (b-x)^{3/4} \sqrt [4]{x}\right ) \operatorname {Subst}\left (\int \frac {x^{10} \left (b-x^4\right )^{5/4}}{\sqrt [4]{a-x^4} \left (a^3-3 a^2 \left (1+\frac {b^3 d}{3 a^2}\right ) x^4+3 a \left (1+\frac {b^2 d}{a}\right ) x^8-(1+3 b d) x^{12}+d x^{16}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(a-x) (b-x)^3 x}}-\frac {\left (8 (2 a-b) \sqrt [4]{a-x} (b-x)^{3/4} \sqrt [4]{x}\right ) \operatorname {Subst}\left (\int \frac {x^6 \left (b-x^4\right )^{5/4}}{\sqrt [4]{a-x^4} \left (a^3-3 a^2 \left (1+\frac {b^3 d}{3 a^2}\right ) x^4+3 a \left (1+\frac {b^2 d}{a}\right ) x^8-(1+3 b d) x^{12}+d x^{16}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(a-x) (b-x)^3 x}}+\frac {\left (4 a b \sqrt [4]{a-x} (b-x)^{3/4} \sqrt [4]{x}\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (b-x^4\right )^{5/4}}{\sqrt [4]{a-x^4} \left (a^3-3 a^2 \left (1+\frac {b^3 d}{3 a^2}\right ) x^4+3 a \left (1+\frac {b^2 d}{a}\right ) x^8-(1+3 b d) x^{12}+d x^{16}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{(a-x) (b-x)^3 x}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [F]  time = 2.61, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a b^3-2 (3 a-b) b^2 x+3 (3 a-b) b x^2-4 a x^3+x^4}{\sqrt [4]{x (-a+x) (-b+x)^3} \left (a^3-\left (3 a^2+b^3 d\right ) x+3 \left (a+b^2 d\right ) x^2-(1+3 b d) x^3+d x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.50, size = 153, normalized size = 1.00 \begin {gather*} -\frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a b^3 x+\left (-3 a b^2-b^3\right ) x^2+\left (3 a b+3 b^2\right ) x^3+(-a-3 b) x^4+x^5}}{a-x}\right )}{d^{3/4}}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{d} \sqrt [4]{a b^3 x+\left (-3 a b^2-b^3\right ) x^2+\left (3 a b+3 b^2\right ) x^3+(-a-3 b) x^4+x^5}}{a-x}\right )}{d^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*ArcTan[(d^(1/4)*(a*b^3*x + (-3*a*b^2 - b^3)*x^2 + (3*a*b + 3*b^2)*x^3 + (-a - 3*b)*x^4 + x^5)^(1/4))/(a -
x)])/d^(3/4) + (2*ArcTanh[(d^(1/4)*(a*b^3*x + (-3*a*b^2 - b^3)*x^2 + (3*a*b + 3*b^2)*x^3 + (-a - 3*b)*x^4 + x^
5)^(1/4))/(a - x)])/d^(3/4)

________________________________________________________________________________________

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((a*b^3-2*(3*a-b)*b^2*x+3*(3*a-b)*b*x^2-4*a*x^3+x^4)/(x*(-a+x)*(-b+x)^3)^(1/4)/(a^3-(b^3*d+3*a^2)*x+3
*(b^2*d+a)*x^2-(3*b*d+1)*x^3+d*x^4),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a b^{3} - 2 \, {\left (3 \, a - b\right )} b^{2} x + 3 \, {\left (3 \, a - b\right )} b x^{2} - 4 \, a x^{3} + x^{4}}{\left ({\left (a - x\right )} {\left (b - x\right )}^{3} x\right )^{\frac {1}{4}} {\left (d x^{4} - {\left (3 \, b d + 1\right )} x^{3} + a^{3} + 3 \, {\left (b^{2} d + a\right )} x^{2} - {\left (b^{3} d + 3 \, a^{2}\right )} x\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {a \,b^{3}-2 \left (3 a -b \right ) b^{2} x +3 \left (3 a -b \right ) b \,x^{2}-4 a \,x^{3}+x^{4}}{\left (x \left (-a +x \right ) \left (-b +x \right )^{3}\right )^{\frac {1}{4}} \left (a^{3}-\left (b^{3} d +3 a^{2}\right ) x +3 \left (b^{2} d +a \right ) x^{2}-\left (3 b d +1\right ) x^{3}+d \,x^{4}\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a b^{3} - 2 \, {\left (3 \, a - b\right )} b^{2} x + 3 \, {\left (3 \, a - b\right )} b x^{2} - 4 \, a x^{3} + x^{4}}{\left ({\left (a - x\right )} {\left (b - x\right )}^{3} x\right )^{\frac {1}{4}} {\left (d x^{4} - {\left (3 \, b d + 1\right )} x^{3} + a^{3} + 3 \, {\left (b^{2} d + a\right )} x^{2} - {\left (b^{3} d + 3 \, a^{2}\right )} x\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________