Integrand size = 15, antiderivative size = 74 \[ \int \frac {x^{15}}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {a^3}{b^4 \sqrt [4]{a+b x^4}}+\frac {a^2 \left (a+b x^4\right )^{3/4}}{b^4}-\frac {3 a \left (a+b x^4\right )^{7/4}}{7 b^4}+\frac {\left (a+b x^4\right )^{11/4}}{11 b^4} \]
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Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {272, 45} \[ \int \frac {x^{15}}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {a^3}{b^4 \sqrt [4]{a+b x^4}}+\frac {a^2 \left (a+b x^4\right )^{3/4}}{b^4}-\frac {3 a \left (a+b x^4\right )^{7/4}}{7 b^4}+\frac {\left (a+b x^4\right )^{11/4}}{11 b^4} \]
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Rule 45
Rule 272
Rubi steps \begin{align*} \text {integral}& = \frac {1}{4} \text {Subst}\left (\int \frac {x^3}{(a+b x)^{5/4}} \, dx,x,x^4\right ) \\ & = \frac {1}{4} \text {Subst}\left (\int \left (-\frac {a^3}{b^3 (a+b x)^{5/4}}+\frac {3 a^2}{b^3 \sqrt [4]{a+b x}}-\frac {3 a (a+b x)^{3/4}}{b^3}+\frac {(a+b x)^{7/4}}{b^3}\right ) \, dx,x,x^4\right ) \\ & = \frac {a^3}{b^4 \sqrt [4]{a+b x^4}}+\frac {a^2 \left (a+b x^4\right )^{3/4}}{b^4}-\frac {3 a \left (a+b x^4\right )^{7/4}}{7 b^4}+\frac {\left (a+b x^4\right )^{11/4}}{11 b^4} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.68 \[ \int \frac {x^{15}}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {128 a^3+32 a^2 b x^4-12 a b^2 x^8+7 b^3 x^{12}}{77 b^4 \sqrt [4]{a+b x^4}} \]
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Time = 4.38 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.64
method | result | size |
gosper | \(\frac {7 b^{3} x^{12}-12 a \,b^{2} x^{8}+32 a^{2} b \,x^{4}+128 a^{3}}{77 \left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{4}}\) | \(47\) |
trager | \(\frac {7 b^{3} x^{12}-12 a \,b^{2} x^{8}+32 a^{2} b \,x^{4}+128 a^{3}}{77 \left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{4}}\) | \(47\) |
pseudoelliptic | \(\frac {7 b^{3} x^{12}-12 a \,b^{2} x^{8}+32 a^{2} b \,x^{4}+128 a^{3}}{77 \left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{4}}\) | \(47\) |
risch | \(\frac {\left (7 b^{2} x^{8}-19 a b \,x^{4}+51 a^{2}\right ) \left (b \,x^{4}+a \right )^{\frac {3}{4}}}{77 b^{4}}+\frac {a^{3}}{b^{4} \left (b \,x^{4}+a \right )^{\frac {1}{4}}}\) | \(53\) |
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Time = 0.28 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.78 \[ \int \frac {x^{15}}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {{\left (7 \, b^{3} x^{12} - 12 \, a b^{2} x^{8} + 32 \, a^{2} b x^{4} + 128 \, a^{3}\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{77 \, {\left (b^{5} x^{4} + a b^{4}\right )}} \]
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Time = 0.65 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.24 \[ \int \frac {x^{15}}{\left (a+b x^4\right )^{5/4}} \, dx=\begin {cases} \frac {128 a^{3}}{77 b^{4} \sqrt [4]{a + b x^{4}}} + \frac {32 a^{2} x^{4}}{77 b^{3} \sqrt [4]{a + b x^{4}}} - \frac {12 a x^{8}}{77 b^{2} \sqrt [4]{a + b x^{4}}} + \frac {x^{12}}{11 b \sqrt [4]{a + b x^{4}}} & \text {for}\: b \neq 0 \\\frac {x^{16}}{16 a^{\frac {5}{4}}} & \text {otherwise} \end {cases} \]
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Time = 0.19 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.84 \[ \int \frac {x^{15}}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {{\left (b x^{4} + a\right )}^{\frac {11}{4}}}{11 \, b^{4}} - \frac {3 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} a}{7 \, b^{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{2}}{b^{4}} + \frac {a^{3}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{4}} \]
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Time = 0.28 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.93 \[ \int \frac {x^{15}}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {a^{3}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{4}} + \frac {7 \, {\left (b x^{4} + a\right )}^{\frac {11}{4}} b^{40} - 33 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} a b^{40} + 77 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{2} b^{40}}{77 \, b^{44}} \]
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Time = 5.79 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.74 \[ \int \frac {x^{15}}{\left (a+b x^4\right )^{5/4}} \, dx=\frac {77\,a^2\,\left (b\,x^4+a\right )-33\,a\,{\left (b\,x^4+a\right )}^2+7\,{\left (b\,x^4+a\right )}^3+77\,a^3}{77\,b^4\,{\left (b\,x^4+a\right )}^{1/4}} \]
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