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

Optimal. Leaf size=63 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {x^3 (-a-b-c)+x^2 (a b+a c+b c)-a b c x+x^4}}{a-x}\right )}{\sqrt {d}} \]

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Rubi [F]  time = 4.09, 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 c+2 a (b+c) x-(3 a+b+c) x^2+2 x^3}{\sqrt {x (-a+x) (-b+x) (-c+x)} \left (a+(-1+b c d) x-(b+c) d x^2+d x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin {align*} \int \frac {-a b c+2 a (b+c) x-(3 a+b+c) x^2+2 x^3}{\sqrt {x (-a+x) (-b+x) (-c+x)} \left (a+(-1+b c d) x-(b+c) d x^2+d x^3\right )} \, dx &=\int \frac {-a b c+2 a (b+c) x-(3 a+b+c) x^2+2 x^3}{\sqrt {x (-a+x) (-b+x) (-c+x)} \left (a-(1-b c d) x-(b+c) d x^2+d x^3\right )} \, dx\\ &=\int \left (\frac {2}{d \sqrt {x (-a+x) (-b+x) (-c+x)}}-\frac {a (2+b c d)-2 (1-b c d+a (b+c) d) x+(3 a-b-c) d x^2}{d \sqrt {x (-a+x) (-b+x) (-c+x)} \left (a-(1-b c d) x-(b+c) d x^2+d x^3\right )}\right ) \, dx\\ &=-\frac {\int \frac {a (2+b c d)-2 (1-b c d+a (b+c) d) x+(3 a-b-c) d x^2}{\sqrt {x (-a+x) (-b+x) (-c+x)} \left (a-(1-b c d) x-(b+c) d x^2+d x^3\right )} \, dx}{d}+\frac {2 \int \frac {1}{\sqrt {x (-a+x) (-b+x) (-c+x)}} \, dx}{d}\\ &=-\frac {\int \left (\frac {a (2+b c d)}{\sqrt {x (-a+x) (-b+x) (-c+x)} \left (a-(1-b c d) x-(b+c) d x^2+d x^3\right )}+\frac {2 (-1+b c d-a (b+c) d) x}{\sqrt {x (-a+x) (-b+x) (-c+x)} \left (a-(1-b c d) x-(b+c) d x^2+d x^3\right )}+\frac {(3 a-b-c) d x^2}{\sqrt {x (-a+x) (-b+x) (-c+x)} \left (a-(1-b c d) x-(b+c) d x^2+d x^3\right )}\right ) \, dx}{d}+\frac {2 \int \frac {1}{\sqrt {x (-a+x) (-b+x) (-c+x)}} \, dx}{d}\\ &=-\left ((3 a-b-c) \int \frac {x^2}{\sqrt {x (-a+x) (-b+x) (-c+x)} \left (a-(1-b c d) x-(b+c) d x^2+d x^3\right )} \, dx\right )+\frac {2 \int \frac {1}{\sqrt {x (-a+x) (-b+x) (-c+x)}} \, dx}{d}-\frac {(a (2+b c d)) \int \frac {1}{\sqrt {x (-a+x) (-b+x) (-c+x)} \left (a-(1-b c d) x-(b+c) d x^2+d x^3\right )} \, dx}{d}+\frac {(2 (1-b c d+a (b+c) d)) \int \frac {x}{\sqrt {x (-a+x) (-b+x) (-c+x)} \left (a-(1-b c d) x-(b+c) d x^2+d x^3\right )} \, dx}{d}\\ \end {align*}

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Mathematica [C]  time = 13.32, size = 8822, normalized size = 140.03 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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IntegrateAlgebraic [A]  time = 0.54, size = 63, normalized size = 1.00 \begin {gather*} \frac {2 \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {x^3 (-a-b-c)+x^2 (a b+a c+b c)-a b c x+x^4}}{a-x}\right )}{\sqrt {d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.10, size = 502, normalized size = 7.97 \begin {gather*} -\frac {4 a \sqrt {\frac {\left (a -c \right ) x}{a \left (-c +x \right )}}\, \left (-c +x \right )^{2} \sqrt {\frac {c \left (-b +x \right )}{b \left (-c +x \right )}}\, \sqrt {\frac {c \left (-a +x \right )}{a \left (-c +x \right )}}\, \EllipticF \left (\sqrt {\frac {\left (a -c \right ) x}{a \left (-c +x \right )}}, \sqrt {\frac {\left (-b +c \right ) a}{b \left (c -a \right )}}\right )}{d \left (a -c \right ) c \sqrt {x \left (-a +x \right ) \left (-b +x \right ) \left (-c +x \right )}}-\frac {2 a \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\RootOf \left (d \,\textit {\_Z}^{3}+\left (-b d -c d \right ) \textit {\_Z}^{2}+\left (b c d -1\right ) \textit {\_Z} +a \right )}{\sum }\frac {\left (-3 \underline {\hspace {1.25 ex}}\alpha ^{2} a d +\underline {\hspace {1.25 ex}}\alpha ^{2} b d +\underline {\hspace {1.25 ex}}\alpha ^{2} c d +2 \underline {\hspace {1.25 ex}}\alpha a b d +2 \underline {\hspace {1.25 ex}}\alpha a c d -2 \underline {\hspace {1.25 ex}}\alpha b c d -a b c d +2 \underline {\hspace {1.25 ex}}\alpha -2 a \right ) \left (-c +x \right )^{2} \left (-d \,\underline {\hspace {1.25 ex}}\alpha ^{2}+\underline {\hspace {1.25 ex}}\alpha b d +1\right ) \sqrt {\frac {\left (a -c \right ) x}{a \left (-c +x \right )}}\, \sqrt {\frac {c \left (-b +x \right )}{b \left (-c +x \right )}}\, \sqrt {\frac {c \left (-a +x \right )}{a \left (-c +x \right )}}\, \left (\EllipticF \left (\sqrt {\frac {\left (a -c \right ) x}{a \left (-c +x \right )}}, \sqrt {\frac {\left (-b +c \right ) a}{b \left (c -a \right )}}\right )+\frac {c \left (d \,\underline {\hspace {1.25 ex}}\alpha ^{2}-\underline {\hspace {1.25 ex}}\alpha b d -d \underline {\hspace {1.25 ex}}\alpha c +b c d -1\right ) \EllipticPi \left (\sqrt {\frac {\left (a -c \right ) x}{a \left (-c +x \right )}}, \frac {\underline {\hspace {1.25 ex}}\alpha ^{2} c d -\underline {\hspace {1.25 ex}}\alpha b c d -\underline {\hspace {1.25 ex}}\alpha \,c^{2} d +b \,c^{2} d +a -c}{a -c}, \sqrt {\frac {\left (-b +c \right ) a}{b \left (c -a \right )}}\right )}{a}\right )}{\left (-3 d \,\underline {\hspace {1.25 ex}}\alpha ^{2}+2 \underline {\hspace {1.25 ex}}\alpha b d +2 d \underline {\hspace {1.25 ex}}\alpha c -b c d +1\right ) \left (a -c \right )^{2} \sqrt {x \left (-a +x \right ) \left (-b +x \right ) \left (-c +x \right )}}\right )}{d c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/d*a*((a-c)*x/a/(-c+x))^(1/2)*(-c+x)^2*(c*(-b+x)/b/(-c+x))^(1/2)*(c*(-a+x)/a/(-c+x))^(1/2)/(a-c)/c/(x*(-a+x)
*(-b+x)*(-c+x))^(1/2)*EllipticF(((a-c)*x/a/(-c+x))^(1/2),((-b+c)*a/b/(c-a))^(1/2))-2/d*a/c*sum((-3*_alpha^2*a*
d+_alpha^2*b*d+_alpha^2*c*d+2*_alpha*a*b*d+2*_alpha*a*c*d-2*_alpha*b*c*d-a*b*c*d+2*_alpha-2*a)/(-3*_alpha^2*d+
2*_alpha*b*d+2*_alpha*c*d-b*c*d+1)*(-c+x)^2/(a-c)^2*(-_alpha^2*d+_alpha*b*d+1)*((a-c)*x/a/(-c+x))^(1/2)*(c*(-b
+x)/b/(-c+x))^(1/2)*(c*(-a+x)/a/(-c+x))^(1/2)/(x*(-a+x)*(-b+x)*(-c+x))^(1/2)*(EllipticF(((a-c)*x/a/(-c+x))^(1/
2),((-b+c)*a/b/(c-a))^(1/2))+c*(_alpha^2*d-_alpha*b*d-_alpha*c*d+b*c*d-1)/a*EllipticPi(((a-c)*x/a/(-c+x))^(1/2
),(_alpha^2*c*d-_alpha*b*c*d-_alpha*c^2*d+b*c^2*d+a-c)/(a-c),((-b+c)*a/b/(c-a))^(1/2))),_alpha=RootOf(d*_Z^3+(
-b*d-c*d)*_Z^2+(b*c*d-1)*_Z+a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} -\int \frac {-2\,x^3+\left (3\,a+b+c\right )\,x^2-2\,a\,\left (b+c\right )\,x+a\,b\,c}{\sqrt {-x\,\left (a-x\right )\,\left (b-x\right )\,\left (c-x\right )}\,\left (d\,x^3-d\,\left (b+c\right )\,x^2+\left (b\,c\,d-1\right )\,x+a\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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