Optimal. Leaf size=68 \[ -\frac{32 b^2 \sqrt [4]{a+b x^4}}{45 a^3 x}+\frac{8 b \sqrt [4]{a+b x^4}}{45 a^2 x^5}-\frac{\sqrt [4]{a+b x^4}}{9 a x^9} \]
[Out]
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Rubi [A] time = 0.0648762, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133 \[ -\frac{32 b^2 \sqrt [4]{a+b x^4}}{45 a^3 x}+\frac{8 b \sqrt [4]{a+b x^4}}{45 a^2 x^5}-\frac{\sqrt [4]{a+b x^4}}{9 a x^9} \]
Antiderivative was successfully verified.
[In] Int[1/(x^10*(a + b*x^4)^(3/4)),x]
[Out]
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Rubi in Sympy [A] time = 6.66349, size = 60, normalized size = 0.88 \[ - \frac{\sqrt [4]{a + b x^{4}}}{9 a x^{9}} + \frac{8 b \sqrt [4]{a + b x^{4}}}{45 a^{2} x^{5}} - \frac{32 b^{2} \sqrt [4]{a + b x^{4}}}{45 a^{3} x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] rubi_integrate(1/x**10/(b*x**4+a)**(3/4),x)
[Out]
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Mathematica [A] time = 0.0329346, size = 42, normalized size = 0.62 \[ -\frac{\sqrt [4]{a+b x^4} \left (5 a^2-8 a b x^4+32 b^2 x^8\right )}{45 a^3 x^9} \]
Antiderivative was successfully verified.
[In] Integrate[1/(x^10*(a + b*x^4)^(3/4)),x]
[Out]
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Maple [A] time = 0.009, size = 39, normalized size = 0.6 \[ -{\frac{32\,{b}^{2}{x}^{8}-8\,ab{x}^{4}+5\,{a}^{2}}{45\,{a}^{3}{x}^{9}}\sqrt [4]{b{x}^{4}+a}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] int(1/x^10/(b*x^4+a)^(3/4),x)
[Out]
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Maxima [A] time = 1.43906, size = 70, normalized size = 1.03 \[ -\frac{\frac{45 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}} b^{2}}{x} - \frac{18 \,{\left (b x^{4} + a\right )}^{\frac{5}{4}} b}{x^{5}} + \frac{5 \,{\left (b x^{4} + a\right )}^{\frac{9}{4}}}{x^{9}}}{45 \, a^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate(1/((b*x^4 + a)^(3/4)*x^10),x, algorithm="maxima")
[Out]
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Fricas [A] time = 0.233882, size = 51, normalized size = 0.75 \[ -\frac{{\left (32 \, b^{2} x^{8} - 8 \, a b x^{4} + 5 \, a^{2}\right )}{\left (b x^{4} + a\right )}^{\frac{1}{4}}}{45 \, a^{3} x^{9}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate(1/((b*x^4 + a)^(3/4)*x^10),x, algorithm="fricas")
[Out]
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Sympy [A] time = 10.4999, size = 406, normalized size = 5.97 \[ \frac{5 a^{4} b^{\frac{17}{4}} \sqrt [4]{\frac{a}{b x^{4}} + 1} \Gamma \left (- \frac{9}{4}\right )}{64 a^{5} b^{4} x^{8} \Gamma \left (\frac{3}{4}\right ) + 128 a^{4} b^{5} x^{12} \Gamma \left (\frac{3}{4}\right ) + 64 a^{3} b^{6} x^{16} \Gamma \left (\frac{3}{4}\right )} + \frac{2 a^{3} b^{\frac{21}{4}} x^{4} \sqrt [4]{\frac{a}{b x^{4}} + 1} \Gamma \left (- \frac{9}{4}\right )}{64 a^{5} b^{4} x^{8} \Gamma \left (\frac{3}{4}\right ) + 128 a^{4} b^{5} x^{12} \Gamma \left (\frac{3}{4}\right ) + 64 a^{3} b^{6} x^{16} \Gamma \left (\frac{3}{4}\right )} + \frac{21 a^{2} b^{\frac{25}{4}} x^{8} \sqrt [4]{\frac{a}{b x^{4}} + 1} \Gamma \left (- \frac{9}{4}\right )}{64 a^{5} b^{4} x^{8} \Gamma \left (\frac{3}{4}\right ) + 128 a^{4} b^{5} x^{12} \Gamma \left (\frac{3}{4}\right ) + 64 a^{3} b^{6} x^{16} \Gamma \left (\frac{3}{4}\right )} + \frac{56 a b^{\frac{29}{4}} x^{12} \sqrt [4]{\frac{a}{b x^{4}} + 1} \Gamma \left (- \frac{9}{4}\right )}{64 a^{5} b^{4} x^{8} \Gamma \left (\frac{3}{4}\right ) + 128 a^{4} b^{5} x^{12} \Gamma \left (\frac{3}{4}\right ) + 64 a^{3} b^{6} x^{16} \Gamma \left (\frac{3}{4}\right )} + \frac{32 b^{\frac{33}{4}} x^{16} \sqrt [4]{\frac{a}{b x^{4}} + 1} \Gamma \left (- \frac{9}{4}\right )}{64 a^{5} b^{4} x^{8} \Gamma \left (\frac{3}{4}\right ) + 128 a^{4} b^{5} x^{12} \Gamma \left (\frac{3}{4}\right ) + 64 a^{3} b^{6} x^{16} \Gamma \left (\frac{3}{4}\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate(1/x**10/(b*x**4+a)**(3/4),x)
[Out]
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GIAC/XCAS [F] time = 0., size = 0, normalized size = 0. \[ \int \frac{1}{{\left (b x^{4} + a\right )}^{\frac{3}{4}} x^{10}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In] integrate(1/((b*x^4 + a)^(3/4)*x^10),x, algorithm="giac")
[Out]