Integrand size = 80, antiderivative size = 107 \[ \int \frac {x^2 \left (3 a b^2-2 b (2 a+b) x+(a+2 b) x^2\right )}{\left (x (-a+x) (-b+x)^2\right )^{3/4} \left (a b^2-b (2 a+b) x+(a+2 b) x^2+(-1+d) x^3\right )} \, dx=-\frac {2 \arctan \left (\frac {\sqrt [4]{d} x}{\sqrt [4]{-a b^2 x+\left (2 a b+b^2\right ) x^2+(-a-2 b) x^3+x^4}}\right )}{d^{3/4}}+\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{d} x}{\sqrt [4]{-a b^2 x+\left (2 a b+b^2\right ) x^2+(-a-2 b) x^3+x^4}}\right )}{d^{3/4}} \]
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\[ \int \frac {x^2 \left (3 a b^2-2 b (2 a+b) x+(a+2 b) x^2\right )}{\left (x (-a+x) (-b+x)^2\right )^{3/4} \left (a b^2-b (2 a+b) x+(a+2 b) x^2+(-1+d) x^3\right )} \, dx=\int \frac {x^2 \left (3 a b^2-2 b (2 a+b) x+(a+2 b) x^2\right )}{\left (x (-a+x) (-b+x)^2\right )^{3/4} \left (a b^2-b (2 a+b) x+(a+2 b) x^2+(-1+d) x^3\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {\left (x^{3/4} (-a+x)^{3/4} (-b+x)^{3/2}\right ) \int \frac {x^{5/4} \left (3 a b^2-2 b (2 a+b) x+(a+2 b) x^2\right )}{(-a+x)^{3/4} (-b+x)^{3/2} \left (a b^2-b (2 a+b) x+(a+2 b) x^2+(-1+d) x^3\right )} \, dx}{\left (x (-a+x) (-b+x)^2\right )^{3/4}} \\ & = \frac {\left (x^{3/4} (-a+x)^{3/4} (-b+x)^{3/2}\right ) \int \frac {x^{5/4} (-3 a b+(a+2 b) x)}{(-a+x)^{3/4} \sqrt {-b+x} \left (a b^2-b (2 a+b) x+(a+2 b) x^2+(-1+d) x^3\right )} \, dx}{\left (x (-a+x) (-b+x)^2\right )^{3/4}} \\ & = \frac {\left (4 x^{3/4} (-a+x)^{3/4} (-b+x)^{3/2}\right ) \text {Subst}\left (\int \frac {x^8 \left (-3 a b+(a+2 b) x^4\right )}{\left (-a+x^4\right )^{3/4} \sqrt {-b+x^4} \left (a b^2-b (2 a+b) x^4+(a+2 b) x^8+(-1+d) x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{\left (x (-a+x) (-b+x)^2\right )^{3/4}} \\ & = \frac {\left (4 x^{3/4} (-a+x)^{3/4} (-b+x)^{3/2}\right ) \text {Subst}\left (\int \left (-\frac {a+2 b}{(1-d) \left (-a+x^4\right )^{3/4} \sqrt {-b+x^4}}-\frac {a b^2 (a+2 b)-b (2 a+b) (a+2 b) x^4+\left (a^2+4 b^2+a (b+3 b d)\right ) x^8}{(-1+d) \left (-a+x^4\right )^{3/4} \sqrt {-b+x^4} \left (a b^2-b (2 a+b) x^4+(a+2 b) x^8+(-1+d) x^{12}\right )}\right ) \, dx,x,\sqrt [4]{x}\right )}{\left (x (-a+x) (-b+x)^2\right )^{3/4}} \\ & = -\frac {\left (4 (a+2 b) x^{3/4} (-a+x)^{3/4} (-b+x)^{3/2}\right ) \text {Subst}\left (\int \frac {1}{\left (-a+x^4\right )^{3/4} \sqrt {-b+x^4}} \, dx,x,\sqrt [4]{x}\right )}{(1-d) \left (x (-a+x) (-b+x)^2\right )^{3/4}}-\frac {\left (4 x^{3/4} (-a+x)^{3/4} (-b+x)^{3/2}\right ) \text {Subst}\left (\int \frac {a b^2 (a+2 b)-b (2 a+b) (a+2 b) x^4+\left (a^2+4 b^2+a (b+3 b d)\right ) x^8}{\left (-a+x^4\right )^{3/4} \sqrt {-b+x^4} \left (a b^2-b (2 a+b) x^4+(a+2 b) x^8+(-1+d) x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{(-1+d) \left (x (-a+x) (-b+x)^2\right )^{3/4}} \\ & = \frac {4 (a+2 b) \left (\frac {b (a-x)}{a (b-x)}\right )^{3/4} (b-x)^2 x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {5}{4},-\frac {(a-b) x}{a (b-x)}\right )}{b (1-d) \left (-\left ((a-x) (b-x)^2 x\right )\right )^{3/4}}-\frac {\left (4 x^{3/4} (-a+x)^{3/4} (-b+x)^{3/2}\right ) \text {Subst}\left (\int \left (\frac {a b^2 (a+2 b)}{\left (-a+x^4\right )^{3/4} \sqrt {-b+x^4} \left (a b^2-2 a b \left (1+\frac {b}{2 a}\right ) x^4+a \left (1+\frac {2 b}{a}\right ) x^8-(1-d) x^{12}\right )}+\frac {(-a-2 b) b (2 a+b) x^4}{\left (-a+x^4\right )^{3/4} \sqrt {-b+x^4} \left (a b^2-2 a b \left (1+\frac {b}{2 a}\right ) x^4+a \left (1+\frac {2 b}{a}\right ) x^8-(1-d) x^{12}\right )}+\frac {\left (a^2+4 b^2+a (b+3 b d)\right ) x^8}{\left (-a+x^4\right )^{3/4} \sqrt {-b+x^4} \left (a b^2-2 a b \left (1+\frac {b}{2 a}\right ) x^4+a \left (1+\frac {2 b}{a}\right ) x^8-(1-d) x^{12}\right )}\right ) \, dx,x,\sqrt [4]{x}\right )}{(-1+d) \left (x (-a+x) (-b+x)^2\right )^{3/4}} \\ & = \frac {4 (a+2 b) \left (\frac {b (a-x)}{a (b-x)}\right )^{3/4} (b-x)^2 x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {5}{4},-\frac {(a-b) x}{a (b-x)}\right )}{b (1-d) \left (-\left ((a-x) (b-x)^2 x\right )\right )^{3/4}}-\frac {\left (4 a b^2 (a+2 b) x^{3/4} (-a+x)^{3/4} (-b+x)^{3/2}\right ) \text {Subst}\left (\int \frac {1}{\left (-a+x^4\right )^{3/4} \sqrt {-b+x^4} \left (a b^2-2 a b \left (1+\frac {b}{2 a}\right ) x^4+a \left (1+\frac {2 b}{a}\right ) x^8-(1-d) x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{(-1+d) \left (x (-a+x) (-b+x)^2\right )^{3/4}}+\frac {\left (4 b (2 a+b) (a+2 b) x^{3/4} (-a+x)^{3/4} (-b+x)^{3/2}\right ) \text {Subst}\left (\int \frac {x^4}{\left (-a+x^4\right )^{3/4} \sqrt {-b+x^4} \left (a b^2-2 a b \left (1+\frac {b}{2 a}\right ) x^4+a \left (1+\frac {2 b}{a}\right ) x^8-(1-d) x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{(-1+d) \left (x (-a+x) (-b+x)^2\right )^{3/4}}-\frac {\left (4 \left (a^2+4 b^2+a (b+3 b d)\right ) x^{3/4} (-a+x)^{3/4} (-b+x)^{3/2}\right ) \text {Subst}\left (\int \frac {x^8}{\left (-a+x^4\right )^{3/4} \sqrt {-b+x^4} \left (a b^2-2 a b \left (1+\frac {b}{2 a}\right ) x^4+a \left (1+\frac {2 b}{a}\right ) x^8-(1-d) x^{12}\right )} \, dx,x,\sqrt [4]{x}\right )}{(-1+d) \left (x (-a+x) (-b+x)^2\right )^{3/4}} \\ & = \frac {4 (a+2 b) \left (\frac {b (a-x)}{a (b-x)}\right )^{3/4} (b-x)^2 x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {5}{4},-\frac {(a-b) x}{a (b-x)}\right )}{b (1-d) \left (-\left ((a-x) (b-x)^2 x\right )\right )^{3/4}}-\frac {\left (4 a b^2 (a+2 b) x^{3/4} (-a+x)^{3/4} (-b+x)^{3/2}\right ) \text {Subst}\left (\int \frac {1}{\left (-a+x^4\right )^{3/4} \sqrt {-b+x^4} \left (-b^2 x^4+2 b x^8+(-1+d) x^{12}+a \left (b-x^4\right )^2\right )} \, dx,x,\sqrt [4]{x}\right )}{(-1+d) \left (x (-a+x) (-b+x)^2\right )^{3/4}}+\frac {\left (4 b (2 a+b) (a+2 b) x^{3/4} (-a+x)^{3/4} (-b+x)^{3/2}\right ) \text {Subst}\left (\int \frac {x^4}{\left (-a+x^4\right )^{3/4} \sqrt {-b+x^4} \left (-b^2 x^4+2 b x^8+(-1+d) x^{12}+a \left (b-x^4\right )^2\right )} \, dx,x,\sqrt [4]{x}\right )}{(-1+d) \left (x (-a+x) (-b+x)^2\right )^{3/4}}-\frac {\left (4 \left (a^2+4 b^2+a (b+3 b d)\right ) x^{3/4} (-a+x)^{3/4} (-b+x)^{3/2}\right ) \text {Subst}\left (\int \frac {x^8}{\left (-a+x^4\right )^{3/4} \sqrt {-b+x^4} \left (-b^2 x^4+2 b x^8+(-1+d) x^{12}+a \left (b-x^4\right )^2\right )} \, dx,x,\sqrt [4]{x}\right )}{(-1+d) \left (x (-a+x) (-b+x)^2\right )^{3/4}} \\ \end{align*}
\[ \int \frac {x^2 \left (3 a b^2-2 b (2 a+b) x+(a+2 b) x^2\right )}{\left (x (-a+x) (-b+x)^2\right )^{3/4} \left (a b^2-b (2 a+b) x+(a+2 b) x^2+(-1+d) x^3\right )} \, dx=\int \frac {x^2 \left (3 a b^2-2 b (2 a+b) x+(a+2 b) x^2\right )}{\left (x (-a+x) (-b+x)^2\right )^{3/4} \left (a b^2-b (2 a+b) x+(a+2 b) x^2+(-1+d) x^3\right )} \, dx \]
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Time = 0.92 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.79
method | result | size |
pseudoelliptic | \(\frac {\ln \left (\frac {d^{\frac {1}{4}} x +\left (-x \left (a -x \right ) \left (b -x \right )^{2}\right )^{\frac {1}{4}}}{-d^{\frac {1}{4}} x +\left (-x \left (a -x \right ) \left (b -x \right )^{2}\right )^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {\left (-x \left (a -x \right ) \left (b -x \right )^{2}\right )^{\frac {1}{4}}}{x \,d^{\frac {1}{4}}}\right )}{d^{\frac {3}{4}}}\) | \(84\) |
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Timed out. \[ \int \frac {x^2 \left (3 a b^2-2 b (2 a+b) x+(a+2 b) x^2\right )}{\left (x (-a+x) (-b+x)^2\right )^{3/4} \left (a b^2-b (2 a+b) x+(a+2 b) x^2+(-1+d) x^3\right )} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {x^2 \left (3 a b^2-2 b (2 a+b) x+(a+2 b) x^2\right )}{\left (x (-a+x) (-b+x)^2\right )^{3/4} \left (a b^2-b (2 a+b) x+(a+2 b) x^2+(-1+d) x^3\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {x^2 \left (3 a b^2-2 b (2 a+b) x+(a+2 b) x^2\right )}{\left (x (-a+x) (-b+x)^2\right )^{3/4} \left (a b^2-b (2 a+b) x+(a+2 b) x^2+(-1+d) x^3\right )} \, dx=\int { \frac {{\left (3 \, a b^{2} - 2 \, {\left (2 \, a + b\right )} b x + {\left (a + 2 \, b\right )} x^{2}\right )} x^{2}}{\left (-{\left (a - x\right )} {\left (b - x\right )}^{2} x\right )^{\frac {3}{4}} {\left ({\left (d - 1\right )} x^{3} + a b^{2} - {\left (2 \, a + b\right )} b x + {\left (a + 2 \, b\right )} x^{2}\right )}} \,d x } \]
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\[ \int \frac {x^2 \left (3 a b^2-2 b (2 a+b) x+(a+2 b) x^2\right )}{\left (x (-a+x) (-b+x)^2\right )^{3/4} \left (a b^2-b (2 a+b) x+(a+2 b) x^2+(-1+d) x^3\right )} \, dx=\int { \frac {{\left (3 \, a b^{2} - 2 \, {\left (2 \, a + b\right )} b x + {\left (a + 2 \, b\right )} x^{2}\right )} x^{2}}{\left (-{\left (a - x\right )} {\left (b - x\right )}^{2} x\right )^{\frac {3}{4}} {\left ({\left (d - 1\right )} x^{3} + a b^{2} - {\left (2 \, a + b\right )} b x + {\left (a + 2 \, b\right )} x^{2}\right )}} \,d x } \]
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Timed out. \[ \int \frac {x^2 \left (3 a b^2-2 b (2 a+b) x+(a+2 b) x^2\right )}{\left (x (-a+x) (-b+x)^2\right )^{3/4} \left (a b^2-b (2 a+b) x+(a+2 b) x^2+(-1+d) x^3\right )} \, dx=\int \frac {x^2\,\left (3\,a\,b^2+x^2\,\left (a+2\,b\right )-2\,b\,x\,\left (2\,a+b\right )\right )}{{\left (-x\,\left (a-x\right )\,{\left (b-x\right )}^2\right )}^{3/4}\,\left (a\,b^2+x^2\,\left (a+2\,b\right )+x^3\,\left (d-1\right )-b\,x\,\left (2\,a+b\right )\right )} \,d x \]
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