Integrand size = 99, antiderivative size = 145 \[ \int \frac {-3 a b^3+2 b^2 (3 a+b) x-3 b (a+b) x^2+x^4}{\sqrt [4]{x (-a+x) (-b+x)^3} \left (a b^3 d-b^2 (3 a+b) d x+3 b (a+b) d x^2-(1+a d+3 b d) x^3+d x^4\right )} \, dx=\frac {2 \arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{a b^3 x+\left (-3 a b^2-b^3\right ) x^2+\left (3 a b+3 b^2\right ) x^3+(-a-3 b) x^4+x^5}}{x}\right )}{d^{3/4}}-\frac {2 \text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{a b^3 x+\left (-3 a b^2-b^3\right ) x^2+\left (3 a b+3 b^2\right ) x^3+(-a-3 b) x^4+x^5}}{x}\right )}{d^{3/4}} \]
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\[ \int \frac {-3 a b^3+2 b^2 (3 a+b) x-3 b (a+b) x^2+x^4}{\sqrt [4]{x (-a+x) (-b+x)^3} \left (a b^3 d-b^2 (3 a+b) d x+3 b (a+b) d x^2-(1+a d+3 b d) x^3+d x^4\right )} \, dx=\int \frac {-3 a b^3+2 b^2 (3 a+b) x-3 b (a+b) x^2+x^4}{\sqrt [4]{x (-a+x) (-b+x)^3} \left (a b^3 d-b^2 (3 a+b) d x+3 b (a+b) d x^2-(1+a d+3 b d) x^3+d x^4\right )} \, dx \]
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Rubi steps \begin{align*} \text {integral}& = \frac {\left (\sqrt [4]{x} \sqrt [4]{-a+x} (-b+x)^{3/4}\right ) \int \frac {-3 a b^3+2 b^2 (3 a+b) x-3 b (a+b) x^2+x^4}{\sqrt [4]{x} \sqrt [4]{-a+x} (-b+x)^{3/4} \left (a b^3 d-b^2 (3 a+b) d x+3 b (a+b) d x^2-(1+a d+3 b d) x^3+d x^4\right )} \, dx}{\sqrt [4]{x (-a+x) (-b+x)^3}} \\ & = \frac {\left (\sqrt [4]{x} \sqrt [4]{-a+x} (-b+x)^{3/4}\right ) \int \frac {\sqrt [4]{-b+x} \left (3 a b^2+\left (-3 a b-2 b^2\right ) x+b x^2+x^3\right )}{\sqrt [4]{x} \sqrt [4]{-a+x} \left (a b^3 d-b^2 (3 a+b) d x+3 b (a+b) d x^2-(1+a d+3 b d) x^3+d x^4\right )} \, dx}{\sqrt [4]{x (-a+x) (-b+x)^3}} \\ & = \frac {\left (\sqrt [4]{x} \sqrt [4]{-a+x} (-b+x)^{3/4}\right ) \int \frac {(-b+x)^{5/4} \left (-3 a b+2 b x+x^2\right )}{\sqrt [4]{x} \sqrt [4]{-a+x} \left (a b^3 d-b^2 (3 a+b) d x+3 b (a+b) d x^2-(1+a d+3 b d) x^3+d x^4\right )} \, dx}{\sqrt [4]{x (-a+x) (-b+x)^3}} \\ & = \frac {\left (4 \sqrt [4]{x} \sqrt [4]{-a+x} (-b+x)^{3/4}\right ) \text {Subst}\left (\int \frac {x^2 \left (-b+x^4\right )^{5/4} \left (-3 a b+2 b x^4+x^8\right )}{\sqrt [4]{-a+x^4} \left (a b^3 d-b^2 (3 a+b) d x^4+3 b (a+b) d x^8-(1+a d+3 b d) x^{12}+d x^{16}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{x (-a+x) (-b+x)^3}} \\ & = \frac {\left (4 \sqrt [4]{x} \sqrt [4]{-a+x} (-b+x)^{3/4}\right ) \text {Subst}\left (\int \left (\frac {3 a b x^2 \left (-b+x^4\right )^{5/4}}{\sqrt [4]{-a+x^4} \left (-a b^3 d+3 a b^2 \left (1+\frac {b}{3 a}\right ) d x^4-3 a b \left (1+\frac {b}{a}\right ) d x^8+(1+(a+3 b) d) x^{12}-d x^{16}\right )}+\frac {2 b x^6 \left (-b+x^4\right )^{5/4}}{\sqrt [4]{-a+x^4} \left (a b^3 d-3 a b^2 \left (1+\frac {b}{3 a}\right ) d x^4+3 a b \left (1+\frac {b}{a}\right ) d x^8-(1+(a+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 b^3 d-3 a b^2 \left (1+\frac {b}{3 a}\right ) d x^4+3 a b \left (1+\frac {b}{a}\right ) d x^8-(1+(a+3 b) d) x^{12}+d x^{16}\right )}\right ) \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{x (-a+x) (-b+x)^3}} \\ & = \frac {\left (4 \sqrt [4]{x} \sqrt [4]{-a+x} (-b+x)^{3/4}\right ) \text {Subst}\left (\int \frac {x^{10} \left (-b+x^4\right )^{5/4}}{\sqrt [4]{-a+x^4} \left (a b^3 d-3 a b^2 \left (1+\frac {b}{3 a}\right ) d x^4+3 a b \left (1+\frac {b}{a}\right ) d x^8-(1+(a+3 b) d) x^{12}+d x^{16}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{x (-a+x) (-b+x)^3}}+\frac {\left (8 b \sqrt [4]{x} \sqrt [4]{-a+x} (-b+x)^{3/4}\right ) \text {Subst}\left (\int \frac {x^6 \left (-b+x^4\right )^{5/4}}{\sqrt [4]{-a+x^4} \left (a b^3 d-3 a b^2 \left (1+\frac {b}{3 a}\right ) d x^4+3 a b \left (1+\frac {b}{a}\right ) d x^8-(1+(a+3 b) d) x^{12}+d x^{16}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{x (-a+x) (-b+x)^3}}+\frac {\left (12 a b \sqrt [4]{x} \sqrt [4]{-a+x} (-b+x)^{3/4}\right ) \text {Subst}\left (\int \frac {x^2 \left (-b+x^4\right )^{5/4}}{\sqrt [4]{-a+x^4} \left (-a b^3 d+3 a b^2 \left (1+\frac {b}{3 a}\right ) d x^4-3 a b \left (1+\frac {b}{a}\right ) d x^8+(1+(a+3 b) d) x^{12}-d x^{16}\right )} \, dx,x,\sqrt [4]{x}\right )}{\sqrt [4]{x (-a+x) (-b+x)^3}} \\ \end{align*}
Time = 10.42 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.46 \[ \int \frac {-3 a b^3+2 b^2 (3 a+b) x-3 b (a+b) x^2+x^4}{\sqrt [4]{x (-a+x) (-b+x)^3} \left (a b^3 d-b^2 (3 a+b) d x+3 b (a+b) d x^2-(1+a d+3 b d) x^3+d x^4\right )} \, dx=\frac {2 \left (\arctan \left (\frac {\sqrt [4]{d} \sqrt [4]{x (-a+x) (-b+x)^3}}{x}\right )-\text {arctanh}\left (\frac {\sqrt [4]{d} \sqrt [4]{x (-a+x) (-b+x)^3}}{x}\right )\right )}{d^{3/4}} \]
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Time = 1.74 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.65
method | result | size |
pseudoelliptic | \(\frac {2 \arctan \left (\frac {\left (x \left (a -x \right ) \left (b -x \right )^{3}\right )^{\frac {1}{4}}}{x \left (\frac {1}{d}\right )^{\frac {1}{4}}}\right )-\ln \left (\frac {x \left (\frac {1}{d}\right )^{\frac {1}{4}}+\left (x \left (a -x \right ) \left (b -x \right )^{3}\right )^{\frac {1}{4}}}{-x \left (\frac {1}{d}\right )^{\frac {1}{4}}+\left (x \left (a -x \right ) \left (b -x \right )^{3}\right )^{\frac {1}{4}}}\right )}{\left (\frac {1}{d}\right )^{\frac {1}{4}} d}\) | \(94\) |
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Timed out. \[ \int \frac {-3 a b^3+2 b^2 (3 a+b) x-3 b (a+b) x^2+x^4}{\sqrt [4]{x (-a+x) (-b+x)^3} \left (a b^3 d-b^2 (3 a+b) d x+3 b (a+b) d x^2-(1+a d+3 b d) x^3+d x^4\right )} \, dx=\text {Timed out} \]
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Timed out. \[ \int \frac {-3 a b^3+2 b^2 (3 a+b) x-3 b (a+b) x^2+x^4}{\sqrt [4]{x (-a+x) (-b+x)^3} \left (a b^3 d-b^2 (3 a+b) d x+3 b (a+b) d x^2-(1+a d+3 b d) x^3+d x^4\right )} \, dx=\text {Timed out} \]
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\[ \int \frac {-3 a b^3+2 b^2 (3 a+b) x-3 b (a+b) x^2+x^4}{\sqrt [4]{x (-a+x) (-b+x)^3} \left (a b^3 d-b^2 (3 a+b) d x+3 b (a+b) d x^2-(1+a d+3 b d) x^3+d x^4\right )} \, dx=\int { -\frac {3 \, a b^{3} - 2 \, {\left (3 \, a + b\right )} b^{2} x + 3 \, {\left (a + b\right )} b x^{2} - x^{4}}{{\left (a b^{3} d - {\left (3 \, a + b\right )} b^{2} d x + 3 \, {\left (a + b\right )} b d x^{2} + d x^{4} - {\left (a d + 3 \, b d + 1\right )} x^{3}\right )} \left ({\left (a - x\right )} {\left (b - x\right )}^{3} x\right )^{\frac {1}{4}}} \,d x } \]
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\[ \int \frac {-3 a b^3+2 b^2 (3 a+b) x-3 b (a+b) x^2+x^4}{\sqrt [4]{x (-a+x) (-b+x)^3} \left (a b^3 d-b^2 (3 a+b) d x+3 b (a+b) d x^2-(1+a d+3 b d) x^3+d x^4\right )} \, dx=\int { -\frac {3 \, a b^{3} - 2 \, {\left (3 \, a + b\right )} b^{2} x + 3 \, {\left (a + b\right )} b x^{2} - x^{4}}{{\left (a b^{3} d - {\left (3 \, a + b\right )} b^{2} d x + 3 \, {\left (a + b\right )} b d x^{2} + d x^{4} - {\left (a d + 3 \, b d + 1\right )} x^{3}\right )} \left ({\left (a - x\right )} {\left (b - x\right )}^{3} x\right )^{\frac {1}{4}}} \,d x } \]
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Timed out. \[ \int \frac {-3 a b^3+2 b^2 (3 a+b) x-3 b (a+b) x^2+x^4}{\sqrt [4]{x (-a+x) (-b+x)^3} \left (a b^3 d-b^2 (3 a+b) d x+3 b (a+b) d x^2-(1+a d+3 b d) x^3+d x^4\right )} \, dx=\int -\frac {3\,a\,b^3-x^4-2\,b^2\,x\,\left (3\,a+b\right )+3\,b\,x^2\,\left (a+b\right )}{{\left (x\,\left (a-x\right )\,{\left (b-x\right )}^3\right )}^{1/4}\,\left (d\,x^4-x^3\,\left (a\,d+3\,b\,d+1\right )+a\,b^3\,d+3\,b\,d\,x^2\,\left (a+b\right )-b^2\,d\,x\,\left (3\,a+b\right )\right )} \,d x \]
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