Integrand size = 24, antiderivative size = 309 \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{5/2}} \, dx=\frac {2 \left (c d^2-b d e+a e^2\right ) x}{9 d e^2 \left (d+e x^3\right )^{3/2}}-\frac {2 \left (11 c d^2-2 b d e-7 a e^2\right ) x}{27 d^2 e^2 \sqrt {d+e x^3}}+\frac {2 \sqrt {2+\sqrt {3}} \left (16 c d^2+e (2 b d+7 a e)\right ) \left (\sqrt [3]{d}+\sqrt [3]{e} x\right ) \sqrt {\frac {d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x}\right ),-7-4 \sqrt {3}\right )}{27 \sqrt [4]{3} d^2 e^{7/3} \sqrt {\frac {\sqrt [3]{d} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \sqrt {d+e x^3}} \]
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Time = 0.14 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1423, 393, 224} \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{5/2}} \, dx=\frac {2 \sqrt {2+\sqrt {3}} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right ) \sqrt {\frac {d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \left (e (7 a e+2 b d)+16 c d^2\right ) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{e} x+\left (1-\sqrt {3}\right ) \sqrt [3]{d}}{\sqrt [3]{e} x+\left (1+\sqrt {3}\right ) \sqrt [3]{d}}\right ),-7-4 \sqrt {3}\right )}{27 \sqrt [4]{3} d^2 e^{7/3} \sqrt {\frac {\sqrt [3]{d} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \sqrt {d+e x^3}}-\frac {2 x \left (-7 a e^2-2 b d e+11 c d^2\right )}{27 d^2 e^2 \sqrt {d+e x^3}}+\frac {2 x \left (a e^2-b d e+c d^2\right )}{9 d e^2 \left (d+e x^3\right )^{3/2}} \]
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Rule 224
Rule 393
Rule 1423
Rubi steps \begin{align*} \text {integral}& = \frac {2 \left (c d^2-b d e+a e^2\right ) x}{9 d e^2 \left (d+e x^3\right )^{3/2}}-\frac {2 \int \frac {\frac {1}{2} \left (2 c d^2-e (2 b d+7 a e)\right )-\frac {9}{2} c d e x^3}{\left (d+e x^3\right )^{3/2}} \, dx}{9 d e^2} \\ & = \frac {2 \left (c d^2-b d e+a e^2\right ) x}{9 d e^2 \left (d+e x^3\right )^{3/2}}-\frac {2 \left (11 c d^2-2 b d e-7 a e^2\right ) x}{27 d^2 e^2 \sqrt {d+e x^3}}-\frac {\left (4 \left (-\frac {9}{2} c d^2 e+\frac {1}{4} e \left (2 c d^2-e (2 b d+7 a e)\right )\right )\right ) \int \frac {1}{\sqrt {d+e x^3}} \, dx}{27 d^2 e^3} \\ & = \frac {2 \left (c d^2-b d e+a e^2\right ) x}{9 d e^2 \left (d+e x^3\right )^{3/2}}-\frac {2 \left (11 c d^2-2 b d e-7 a e^2\right ) x}{27 d^2 e^2 \sqrt {d+e x^3}}+\frac {2 \sqrt {2+\sqrt {3}} \left (16 c d^2+e (2 b d+7 a e)\right ) \left (\sqrt [3]{d}+\sqrt [3]{e} x\right ) \sqrt {\frac {d^{2/3}-\sqrt [3]{d} \sqrt [3]{e} x+e^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x}\right )|-7-4 \sqrt {3}\right )}{27 \sqrt [4]{3} d^2 e^{7/3} \sqrt {\frac {\sqrt [3]{d} \left (\sqrt [3]{d}+\sqrt [3]{e} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{d}+\sqrt [3]{e} x\right )^2}} \sqrt {d+e x^3}} \\ \end{align*}
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.09 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.42 \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{5/2}} \, dx=\frac {-2 x \left (c d^2 \left (8 d+11 e x^3\right )+e \left (b d \left (d-2 e x^3\right )-a e \left (10 d+7 e x^3\right )\right )\right )+\left (16 c d^2+e (2 b d+7 a e)\right ) x \left (d+e x^3\right ) \sqrt {1+\frac {e x^3}{d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{2},\frac {4}{3},-\frac {e x^3}{d}\right )}{27 d^2 e^2 \left (d+e x^3\right )^{3/2}} \]
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Time = 1.00 (sec) , antiderivative size = 401, normalized size of antiderivative = 1.30
method | result | size |
elliptic | \(\frac {2 x \left (a \,e^{2}-b d e +c \,d^{2}\right ) \sqrt {e \,x^{3}+d}}{9 d \,e^{4} \left (x^{3}+\frac {d}{e}\right )^{2}}+\frac {2 x \left (7 a \,e^{2}+2 b d e -11 c \,d^{2}\right )}{27 e^{2} d^{2} \sqrt {\left (x^{3}+\frac {d}{e}\right ) e}}-\frac {2 i \left (\frac {c}{e^{2}}+\frac {7 a \,e^{2}+2 b d e -11 c \,d^{2}}{27 e^{2} d^{2}}\right ) \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (x +\frac {\left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}-\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}\right ) \sqrt {3}\, e}{\left (-d \,e^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {x -\frac {\left (-d \,e^{2}\right )^{\frac {1}{3}}}{e}}{-\frac {3 \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}+\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}}}\, \sqrt {-\frac {i \left (x +\frac {\left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}+\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}\right ) \sqrt {3}\, e}{\left (-d \,e^{2}\right )^{\frac {1}{3}}}}\, F\left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}-\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}\right ) \sqrt {3}\, e}{\left (-d \,e^{2}\right )^{\frac {1}{3}}}}}{3}, \sqrt {\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{e \left (-\frac {3 \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}+\frac {i \sqrt {3}\, \left (-d \,e^{2}\right )^{\frac {1}{3}}}{2 e}\right )}}\right )}{3 e \sqrt {e \,x^{3}+d}}\) | \(401\) |
default | \(\text {Expression too large to display}\) | \(1005\) |
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.11 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.61 \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{5/2}} \, dx=\frac {2 \, {\left ({\left ({\left (16 \, c d^{2} e^{2} + 2 \, b d e^{3} + 7 \, a e^{4}\right )} x^{6} + 16 \, c d^{4} + 2 \, b d^{3} e + 7 \, a d^{2} e^{2} + 2 \, {\left (16 \, c d^{3} e + 2 \, b d^{2} e^{2} + 7 \, a d e^{3}\right )} x^{3}\right )} \sqrt {e} {\rm weierstrassPInverse}\left (0, -\frac {4 \, d}{e}, x\right ) - {\left ({\left (11 \, c d^{2} e^{2} - 2 \, b d e^{3} - 7 \, a e^{4}\right )} x^{4} + {\left (8 \, c d^{3} e + b d^{2} e^{2} - 10 \, a d e^{3}\right )} x\right )} \sqrt {e x^{3} + d}\right )}}{27 \, {\left (d^{2} e^{5} x^{6} + 2 \, d^{3} e^{4} x^{3} + d^{4} e^{3}\right )}} \]
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Time = 34.98 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.39 \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{5/2}} \, dx=\frac {a x \Gamma \left (\frac {1}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{3}, \frac {5}{2} \\ \frac {4}{3} \end {matrix}\middle | {\frac {e x^{3} e^{i \pi }}{d}} \right )}}{3 d^{\frac {5}{2}} \Gamma \left (\frac {4}{3}\right )} + \frac {b x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {4}{3}, \frac {5}{2} \\ \frac {7}{3} \end {matrix}\middle | {\frac {e x^{3} e^{i \pi }}{d}} \right )}}{3 d^{\frac {5}{2}} \Gamma \left (\frac {7}{3}\right )} + \frac {c x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{3}, \frac {5}{2} \\ \frac {10}{3} \end {matrix}\middle | {\frac {e x^{3} e^{i \pi }}{d}} \right )}}{3 d^{\frac {5}{2}} \Gamma \left (\frac {10}{3}\right )} \]
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\[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{5/2}} \, dx=\int { \frac {c x^{6} + b x^{3} + a}{{\left (e x^{3} + d\right )}^{\frac {5}{2}}} \,d x } \]
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\[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{5/2}} \, dx=\int { \frac {c x^{6} + b x^{3} + a}{{\left (e x^{3} + d\right )}^{\frac {5}{2}}} \,d x } \]
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Timed out. \[ \int \frac {a+b x^3+c x^6}{\left (d+e x^3\right )^{5/2}} \, dx=\int \frac {c\,x^6+b\,x^3+a}{{\left (e\,x^3+d\right )}^{5/2}} \,d x \]
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