Optimal. Leaf size=113 \[ \frac {2 \tan ^{-1}\left (\frac {\sqrt [4]{d} \left (x \left (2 a b+b^2\right )-a b^2+x^2 (-a-2 b)+x^3\right )^{3/4}}{(b-x)^2}\right )}{d^{3/4}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt [4]{d} \left (x \left (2 a b+b^2\right )-a b^2+x^2 (-a-2 b)+x^3\right )^{3/4}}{(b-x)^2}\right )}{d^{3/4}} \]
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Rubi [F] time = 9.25, 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 {\left (a^2-2 a x+x^2\right ) \left (-((2 a-3 b) b)+2 (a-2 b) x+x^2\right )}{\left ((-a+x) (-b+x)^2\right )^{3/4} \left (-b^2-a^3 d+\left (2 b+3 a^2 d\right ) x-(1+3 a d) x^2+d x^3\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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\begin {align*} \int \frac {\left (a^2-2 a x+x^2\right ) \left (-((2 a-3 b) b)+2 (a-2 b) x+x^2\right )}{\left ((-a+x) (-b+x)^2\right )^{3/4} \left (-b^2-a^3 d+\left (2 b+3 a^2 d\right ) x-(1+3 a d) x^2+d x^3\right )} \, dx &=\int \frac {(-a+x)^2 \left (-((2 a-3 b) b)+2 (a-2 b) x+x^2\right )}{\left ((-a+x) (-b+x)^2\right )^{3/4} \left (-b^2-a^3 d+\left (2 b+3 a^2 d\right ) x-(1+3 a d) x^2+d x^3\right )} \, dx\\ &=\frac {\left ((-a+x)^{3/4} (-b+x)^{3/2}\right ) \int \frac {(-a+x)^{5/4} \left (-((2 a-3 b) b)+2 (a-2 b) x+x^2\right )}{(-b+x)^{3/2} \left (-b^2-a^3 d+\left (2 b+3 a^2 d\right ) x-(1+3 a d) x^2+d x^3\right )} \, dx}{\left ((-a+x) (-b+x)^2\right )^{3/4}}\\ &=\frac {\left ((-a+x)^{3/4} (-b+x)^{3/2}\right ) \int \frac {(-a+x)^{5/4} (2 a-3 b+x)}{\sqrt {-b+x} \left (-b^2-a^3 d+\left (2 b+3 a^2 d\right ) x-(1+3 a d) x^2+d x^3\right )} \, dx}{\left ((-a+x) (-b+x)^2\right )^{3/4}}\\ &=\frac {\left ((-a+x)^{3/4} (-b+x)^{3/2}\right ) \int \left (\frac {3 \left (1-\frac {2 a}{3 b}\right ) b (-a+x)^{5/4}}{\sqrt {-b+x} \left (b^2+a^3 d-\left (2 b+3 a^2 d\right ) x+(1+3 a d) x^2-d x^3\right )}+\frac {x (-a+x)^{5/4}}{\sqrt {-b+x} \left (-b^2-a^3 d+\left (2 b+3 a^2 d\right ) x-(1+3 a d) x^2+d x^3\right )}\right ) \, dx}{\left ((-a+x) (-b+x)^2\right )^{3/4}}\\ &=\frac {\left ((-a+x)^{3/4} (-b+x)^{3/2}\right ) \int \frac {x (-a+x)^{5/4}}{\sqrt {-b+x} \left (-b^2-a^3 d+\left (2 b+3 a^2 d\right ) x-(1+3 a d) x^2+d x^3\right )} \, dx}{\left ((-a+x) (-b+x)^2\right )^{3/4}}+\frac {\left ((-2 a+3 b) (-a+x)^{3/4} (-b+x)^{3/2}\right ) \int \frac {(-a+x)^{5/4}}{\sqrt {-b+x} \left (b^2+a^3 d-\left (2 b+3 a^2 d\right ) x+(1+3 a d) x^2-d x^3\right )} \, dx}{\left ((-a+x) (-b+x)^2\right )^{3/4}}\\ &=-\frac {\left (4 (-a+x)^{3/4} (-b+x)^{3/2}\right ) \operatorname {Subst}\left (\int \frac {x^8 \left (a+x^4\right )}{\sqrt {a-b+x^4} \left (a^2-2 a b+b^2+2 a x^4-2 b x^4+x^8-d x^{12}\right )} \, dx,x,\sqrt [4]{-a+x}\right )}{\left ((-a+x) (-b+x)^2\right )^{3/4}}+\frac {\left (4 (-2 a+3 b) (-a+x)^{3/4} (-b+x)^{3/2}\right ) \operatorname {Subst}\left (\int \frac {x^8}{\sqrt {a-b+x^4} \left (a^2-2 a b+b^2+2 a x^4-2 b x^4+x^8-d x^{12}\right )} \, dx,x,\sqrt [4]{-a+x}\right )}{\left ((-a+x) (-b+x)^2\right )^{3/4}}\\ &=-\frac {\left (4 (-a+x)^{3/4} (-b+x)^{3/2}\right ) \operatorname {Subst}\left (\int \frac {x^8 \left (a+x^4\right )}{\sqrt {a-b+x^4} \left (a^2-2 a b+b^2+(2 a-2 b) x^4+x^8-d x^{12}\right )} \, dx,x,\sqrt [4]{-a+x}\right )}{\left ((-a+x) (-b+x)^2\right )^{3/4}}+\frac {\left (4 (-2 a+3 b) (-a+x)^{3/4} (-b+x)^{3/2}\right ) \operatorname {Subst}\left (\int \frac {x^8}{\sqrt {a-b+x^4} \left (a^2-2 a b+b^2+(2 a-2 b) x^4+x^8-d x^{12}\right )} \, dx,x,\sqrt [4]{-a+x}\right )}{\left ((-a+x) (-b+x)^2\right )^{3/4}}\\ &=-\frac {\left (4 (-a+x)^{3/4} (-b+x)^{3/2}\right ) \operatorname {Subst}\left (\int \frac {x^8 \left (a+x^4\right )}{\sqrt {a-b+x^4} \left (a^2 \left (1+\frac {b (-2 a+b)}{a^2}\right )+(2 a-2 b) x^4+x^8-d x^{12}\right )} \, dx,x,\sqrt [4]{-a+x}\right )}{\left ((-a+x) (-b+x)^2\right )^{3/4}}+\frac {\left (4 (-2 a+3 b) (-a+x)^{3/4} (-b+x)^{3/2}\right ) \operatorname {Subst}\left (\int \frac {x^8}{\sqrt {a-b+x^4} \left (a^2 \left (1+\frac {b (-2 a+b)}{a^2}\right )+(2 a-2 b) x^4+x^8-d x^{12}\right )} \, dx,x,\sqrt [4]{-a+x}\right )}{\left ((-a+x) (-b+x)^2\right )^{3/4}}\\ &=-\frac {\left (4 (-a+x)^{3/4} (-b+x)^{3/2}\right ) \operatorname {Subst}\left (\int \left (-\frac {1}{d \sqrt {a-b+x^4}}+\frac {(a-b)^2+2 (a-b) x^4+(1+a d) x^8}{d \sqrt {a-b+x^4} \left (a^2 \left (1+\frac {b (-2 a+b)}{a^2}\right )+(2 a-2 b) x^4+x^8-d x^{12}\right )}\right ) \, dx,x,\sqrt [4]{-a+x}\right )}{\left ((-a+x) (-b+x)^2\right )^{3/4}}+\frac {\left (4 (-2 a+3 b) (-a+x)^{3/4} (-b+x)^{3/2}\right ) \operatorname {Subst}\left (\int \frac {x^8}{\sqrt {a-b+x^4} \left (a^2 \left (1+\frac {b (-2 a+b)}{a^2}\right )+(2 a-2 b) x^4+x^8-d x^{12}\right )} \, dx,x,\sqrt [4]{-a+x}\right )}{\left ((-a+x) (-b+x)^2\right )^{3/4}}\\ &=\frac {\left (4 (-2 a+3 b) (-a+x)^{3/4} (-b+x)^{3/2}\right ) \operatorname {Subst}\left (\int \frac {x^8}{\sqrt {a-b+x^4} \left (a^2 \left (1+\frac {b (-2 a+b)}{a^2}\right )+(2 a-2 b) x^4+x^8-d x^{12}\right )} \, dx,x,\sqrt [4]{-a+x}\right )}{\left ((-a+x) (-b+x)^2\right )^{3/4}}+\frac {\left (4 (-a+x)^{3/4} (-b+x)^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a-b+x^4}} \, dx,x,\sqrt [4]{-a+x}\right )}{d \left ((-a+x) (-b+x)^2\right )^{3/4}}-\frac {\left (4 (-a+x)^{3/4} (-b+x)^{3/2}\right ) \operatorname {Subst}\left (\int \frac {(a-b)^2+2 (a-b) x^4+(1+a d) x^8}{\sqrt {a-b+x^4} \left (a^2 \left (1+\frac {b (-2 a+b)}{a^2}\right )+(2 a-2 b) x^4+x^8-d x^{12}\right )} \, dx,x,\sqrt [4]{-a+x}\right )}{d \left ((-a+x) (-b+x)^2\right )^{3/4}}\\ &=-\frac {2 \sqrt [4]{a-b} (b-x) (-a+x)^{3/4} \sqrt {-\frac {b-x}{(a-b) \left (1+\frac {\sqrt {-a+x}}{\sqrt {a-b}}\right )^2}} \left (1+\frac {\sqrt {-a+x}}{\sqrt {a-b}}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{-a+x}}{\sqrt [4]{a-b}}\right )|\frac {1}{2}\right )}{d \left (-\left ((a-x) (b-x)^2\right )\right )^{3/4}}+\frac {\left (4 (-2 a+3 b) (-a+x)^{3/4} (-b+x)^{3/2}\right ) \operatorname {Subst}\left (\int \frac {x^8}{\sqrt {a-b+x^4} \left (a^2 \left (1+\frac {b (-2 a+b)}{a^2}\right )+(2 a-2 b) x^4+x^8-d x^{12}\right )} \, dx,x,\sqrt [4]{-a+x}\right )}{\left ((-a+x) (-b+x)^2\right )^{3/4}}-\frac {\left (4 (-a+x)^{3/4} (-b+x)^{3/2}\right ) \operatorname {Subst}\left (\int \left (\frac {(a-b)^2}{\sqrt {a-b+x^4} \left (a^2 \left (1+\frac {b (-2 a+b)}{a^2}\right )+2 a \left (1-\frac {b}{a}\right ) x^4+x^8-d x^{12}\right )}+\frac {2 (a-b) x^4}{\sqrt {a-b+x^4} \left (a^2 \left (1+\frac {b (-2 a+b)}{a^2}\right )+2 a \left (1-\frac {b}{a}\right ) x^4+x^8-d x^{12}\right )}+\frac {(1+a d) x^8}{\sqrt {a-b+x^4} \left (a^2 \left (1+\frac {b (-2 a+b)}{a^2}\right )+2 a \left (1-\frac {b}{a}\right ) x^4+x^8-d x^{12}\right )}\right ) \, dx,x,\sqrt [4]{-a+x}\right )}{d \left ((-a+x) (-b+x)^2\right )^{3/4}}\\ &=-\frac {2 \sqrt [4]{a-b} (b-x) (-a+x)^{3/4} \sqrt {-\frac {b-x}{(a-b) \left (1+\frac {\sqrt {-a+x}}{\sqrt {a-b}}\right )^2}} \left (1+\frac {\sqrt {-a+x}}{\sqrt {a-b}}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{-a+x}}{\sqrt [4]{a-b}}\right )|\frac {1}{2}\right )}{d \left (-\left ((a-x) (b-x)^2\right )\right )^{3/4}}+\frac {\left (4 (-2 a+3 b) (-a+x)^{3/4} (-b+x)^{3/2}\right ) \operatorname {Subst}\left (\int \frac {x^8}{\sqrt {a-b+x^4} \left (a^2 \left (1+\frac {b (-2 a+b)}{a^2}\right )+(2 a-2 b) x^4+x^8-d x^{12}\right )} \, dx,x,\sqrt [4]{-a+x}\right )}{\left ((-a+x) (-b+x)^2\right )^{3/4}}-\frac {\left (8 (a-b) (-a+x)^{3/4} (-b+x)^{3/2}\right ) \operatorname {Subst}\left (\int \frac {x^4}{\sqrt {a-b+x^4} \left (a^2 \left (1+\frac {b (-2 a+b)}{a^2}\right )+2 a \left (1-\frac {b}{a}\right ) x^4+x^8-d x^{12}\right )} \, dx,x,\sqrt [4]{-a+x}\right )}{d \left ((-a+x) (-b+x)^2\right )^{3/4}}-\frac {\left (4 (a-b)^2 (-a+x)^{3/4} (-b+x)^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a-b+x^4} \left (a^2 \left (1+\frac {b (-2 a+b)}{a^2}\right )+2 a \left (1-\frac {b}{a}\right ) x^4+x^8-d x^{12}\right )} \, dx,x,\sqrt [4]{-a+x}\right )}{d \left ((-a+x) (-b+x)^2\right )^{3/4}}-\frac {\left (4 (1+a d) (-a+x)^{3/4} (-b+x)^{3/2}\right ) \operatorname {Subst}\left (\int \frac {x^8}{\sqrt {a-b+x^4} \left (a^2 \left (1+\frac {b (-2 a+b)}{a^2}\right )+2 a \left (1-\frac {b}{a}\right ) x^4+x^8-d x^{12}\right )} \, dx,x,\sqrt [4]{-a+x}\right )}{d \left ((-a+x) (-b+x)^2\right )^{3/4}}\\ \end {align*}
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Mathematica [F] time = 3.55, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a^2-2 a x+x^2\right ) \left (-((2 a-3 b) b)+2 (a-2 b) x+x^2\right )}{\left ((-a+x) (-b+x)^2\right )^{3/4} \left (-b^2-a^3 d+\left (2 b+3 a^2 d\right ) x-(1+3 a d) x^2+d x^3\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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IntegrateAlgebraic [A] time = 3.64, size = 113, normalized size = 1.00 \begin {gather*} \frac {2 \tan ^{-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)^2}\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)^2}\right )}{d^{3/4}} \end {gather*}
Antiderivative was successfully verified.
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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.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (a^{2} - 2 \, a x + x^{2}\right )} {\left ({\left (2 \, a - 3 \, b\right )} b - 2 \, {\left (a - 2 \, b\right )} x - x^{2}\right )}}{{\left (a^{3} d - d x^{3} + {\left (3 \, a d + 1\right )} x^{2} + b^{2} - {\left (3 \, a^{2} d + 2 \, b\right )} x\right )} \left (-{\left (a - x\right )} {\left (b - x\right )}^{2}\right )^{\frac {3}{4}}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (a^{2}-2 a x +x^{2}\right ) \left (-\left (2 a -3 b \right ) b +2 \left (a -2 b \right ) x +x^{2}\right )}{\left (\left (-a +x \right ) \left (-b +x \right )^{2}\right )^{\frac {3}{4}} \left (-b^{2}-a^{3} d +\left (3 a^{2} d +2 b \right ) x -\left (3 a d +1\right ) x^{2}+d \,x^{3}\right )}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (a^{2} - 2 \, a x + x^{2}\right )} {\left ({\left (2 \, a - 3 \, b\right )} b - 2 \, {\left (a - 2 \, b\right )} x - x^{2}\right )}}{{\left (a^{3} d - d x^{3} + {\left (3 \, a d + 1\right )} x^{2} + b^{2} - {\left (3 \, a^{2} d + 2 \, b\right )} x\right )} \left (-{\left (a - x\right )} {\left (b - x\right )}^{2}\right )^{\frac {3}{4}}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int -\frac {\left (a^2-2\,a\,x+x^2\right )\,\left (2\,x\,\left (a-2\,b\right )-b\,\left (2\,a-3\,b\right )+x^2\right )}{{\left (-\left (a-x\right )\,{\left (b-x\right )}^2\right )}^{3/4}\,\left (a^3\,d-x\,\left (3\,d\,a^2+2\,b\right )-d\,x^3+x^2\,\left (3\,a\,d+1\right )+b^2\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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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.
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