Integrand size = 30, antiderivative size = 302 \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {-c-d x-e x^2-f x^3}{2 b \sqrt {a+b x^4}}+\frac {3 f x \sqrt {a+b x^4}}{2 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {e \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{2 b^{3/2}}-\frac {3 \sqrt [4]{a} f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 b^{7/4} \sqrt {a+b x^4}}+\frac {\left (\sqrt {b} d+3 \sqrt {a} f\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{4 \sqrt [4]{a} b^{7/4} \sqrt {a+b x^4}} \]
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Time = 0.14 (sec) , antiderivative size = 297, normalized size of antiderivative = 0.98, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {1837, 1899, 281, 223, 212, 1212, 226, 1210} \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {\left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \left (3 \sqrt {a} f+\sqrt {b} d\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{4 \sqrt [4]{a} b^{7/4} \sqrt {a+b x^4}}-\frac {3 \sqrt [4]{a} f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 b^{7/4} \sqrt {a+b x^4}}+\frac {e \text {arctanh}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{2 b^{3/2}}+\frac {3 f x \sqrt {a+b x^4}}{2 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {c+d x+e x^2+f x^3}{2 b \sqrt {a+b x^4}} \]
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Rule 212
Rule 223
Rule 226
Rule 281
Rule 1210
Rule 1212
Rule 1837
Rule 1899
Rubi steps \begin{align*} \text {integral}& = -\frac {c+d x+e x^2+f x^3}{2 b \sqrt {a+b x^4}}+\frac {\int \frac {d+2 e x+3 f x^2}{\sqrt {a+b x^4}} \, dx}{2 b} \\ & = -\frac {c+d x+e x^2+f x^3}{2 b \sqrt {a+b x^4}}+\frac {\int \left (\frac {2 e x}{\sqrt {a+b x^4}}+\frac {d+3 f x^2}{\sqrt {a+b x^4}}\right ) \, dx}{2 b} \\ & = -\frac {c+d x+e x^2+f x^3}{2 b \sqrt {a+b x^4}}+\frac {\int \frac {d+3 f x^2}{\sqrt {a+b x^4}} \, dx}{2 b}+\frac {e \int \frac {x}{\sqrt {a+b x^4}} \, dx}{b} \\ & = -\frac {c+d x+e x^2+f x^3}{2 b \sqrt {a+b x^4}}+\frac {e \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,x^2\right )}{2 b}-\frac {\left (3 \sqrt {a} f\right ) \int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a+b x^4}} \, dx}{2 b^{3/2}}+\frac {\left (d+\frac {3 \sqrt {a} f}{\sqrt {b}}\right ) \int \frac {1}{\sqrt {a+b x^4}} \, dx}{2 b} \\ & = -\frac {c+d x+e x^2+f x^3}{2 b \sqrt {a+b x^4}}+\frac {3 f x \sqrt {a+b x^4}}{2 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {3 \sqrt [4]{a} f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 b^{7/4} \sqrt {a+b x^4}}+\frac {\left (\sqrt {b} d+3 \sqrt {a} f\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{a} b^{7/4} \sqrt {a+b x^4}}+\frac {e \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^2}{\sqrt {a+b x^4}}\right )}{2 b} \\ & = -\frac {c+d x+e x^2+f x^3}{2 b \sqrt {a+b x^4}}+\frac {3 f x \sqrt {a+b x^4}}{2 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a+b x^4}}\right )}{2 b^{3/2}}-\frac {3 \sqrt [4]{a} f \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 b^{7/4} \sqrt {a+b x^4}}+\frac {\left (\sqrt {b} d+3 \sqrt {a} f\right ) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{4 \sqrt [4]{a} b^{7/4} \sqrt {a+b x^4}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.13 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.60 \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {-\sqrt {b} c-\sqrt {b} d x-\sqrt {b} e x^2+2 \sqrt {b} f x^3+\sqrt {a} e \sqrt {1+\frac {b x^4}{a}} \text {arcsinh}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )+\sqrt {b} d x \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^4}{a}\right )-2 \sqrt {b} f x^3 \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\frac {b x^4}{a}\right )}{2 b^{3/2} \sqrt {a+b x^4}} \]
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Result contains complex when optimal does not.
Time = 1.80 (sec) , antiderivative size = 248, normalized size of antiderivative = 0.82
method | result | size |
elliptic | \(-\frac {2 b \left (\frac {f \,x^{3}}{4 b^{2}}+\frac {e \,x^{2}}{4 b^{2}}+\frac {d x}{4 b^{2}}+\frac {c}{4 b^{2}}\right )}{\sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {d \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}+\frac {e \ln \left (2 x^{2} \sqrt {b}+2 \sqrt {b \,x^{4}+a}\right )}{2 b^{\frac {3}{2}}}+\frac {3 i f \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) | \(248\) |
default | \(f \left (-\frac {x^{3}}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {3 i \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+e \left (-\frac {x^{2}}{2 b \sqrt {b \,x^{4}+a}}+\frac {\ln \left (x^{2} \sqrt {b}+\sqrt {b \,x^{4}+a}\right )}{2 b^{\frac {3}{2}}}\right )+d \left (-\frac {x}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )-\frac {c}{2 b \sqrt {b \,x^{4}+a}}\) | \(275\) |
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Time = 0.13 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.71 \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {6 \, {\left (a b f x^{5} + a^{2} f x\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + 2 \, {\left ({\left (b^{2} d - 3 \, a b f\right )} x^{5} + {\left (a b d - 3 \, a^{2} f\right )} x\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + {\left (a b e x^{5} + a^{2} e x\right )} \sqrt {b} \log \left (-2 \, b x^{4} - 2 \, \sqrt {b x^{4} + a} \sqrt {b} x^{2} - a\right ) + 2 \, {\left (2 \, a b f x^{4} - a b e x^{3} - a b d x^{2} - a b c x + 3 \, a^{2} f\right )} \sqrt {b x^{4} + a}}{4 \, {\left (a b^{3} x^{5} + a^{2} b^{2} x\right )}} \]
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Time = 6.39 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.52 \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=c \left (\begin {cases} - \frac {1}{2 b \sqrt {a + b x^{4}}} & \text {for}\: b \neq 0 \\\frac {x^{4}}{4 a^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) + e \left (\frac {\operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{2 b^{\frac {3}{2}}} - \frac {x^{2}}{2 \sqrt {a} b \sqrt {1 + \frac {b x^{4}}{a}}}\right ) + \frac {d x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {3}{2} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {9}{4}\right )} + \frac {f x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {11}{4}\right )} \]
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\[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {{\left (f x^{3} + e x^{2} + d x + c\right )} x^{3}}{{\left (b x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \]
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\[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {{\left (f x^{3} + e x^{2} + d x + c\right )} x^{3}}{{\left (b x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \]
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Timed out. \[ \int \frac {x^3 \left (c+d x+e x^2+f x^3\right )}{\left (a+b x^4\right )^{3/2}} \, dx=\int \frac {x^3\,\left (f\,x^3+e\,x^2+d\,x+c\right )}{{\left (b\,x^4+a\right )}^{3/2}} \,d x \]
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