Optimal. Leaf size=59 \[ \frac {a^2 \left (a+b x^4\right )^{9/4}}{9 b^3}-\frac {2 a \left (a+b x^4\right )^{13/4}}{13 b^3}+\frac {\left (a+b x^4\right )^{17/4}}{17 b^3} \]
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Rubi [A]
time = 0.03, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps
used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {272, 45}
\begin {gather*} \frac {a^2 \left (a+b x^4\right )^{9/4}}{9 b^3}+\frac {\left (a+b x^4\right )^{17/4}}{17 b^3}-\frac {2 a \left (a+b x^4\right )^{13/4}}{13 b^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 45
Rule 272
Rubi steps
\begin {align*} \int x^{11} \left (a+b x^4\right )^{5/4} \, dx &=\frac {1}{4} \text {Subst}\left (\int x^2 (a+b x)^{5/4} \, dx,x,x^4\right )\\ &=\frac {1}{4} \text {Subst}\left (\int \left (\frac {a^2 (a+b x)^{5/4}}{b^2}-\frac {2 a (a+b x)^{9/4}}{b^2}+\frac {(a+b x)^{13/4}}{b^2}\right ) \, dx,x,x^4\right )\\ &=\frac {a^2 \left (a+b x^4\right )^{9/4}}{9 b^3}-\frac {2 a \left (a+b x^4\right )^{13/4}}{13 b^3}+\frac {\left (a+b x^4\right )^{17/4}}{17 b^3}\\ \end {align*}
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Mathematica [A]
time = 0.03, size = 39, normalized size = 0.66 \begin {gather*} \frac {\left (a+b x^4\right )^{9/4} \left (32 a^2-72 a b x^4+117 b^2 x^8\right )}{1989 b^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.16, size = 36, normalized size = 0.61
method | result | size |
gosper | \(\frac {\left (b \,x^{4}+a \right )^{\frac {9}{4}} \left (117 b^{2} x^{8}-72 a b \,x^{4}+32 a^{2}\right )}{1989 b^{3}}\) | \(36\) |
trager | \(\frac {\left (117 x^{16} b^{4}+162 a \,b^{3} x^{12}+5 a^{2} b^{2} x^{8}-8 a^{3} b \,x^{4}+32 a^{4}\right ) \left (b \,x^{4}+a \right )^{\frac {1}{4}}}{1989 b^{3}}\) | \(58\) |
risch | \(\frac {\left (117 x^{16} b^{4}+162 a \,b^{3} x^{12}+5 a^{2} b^{2} x^{8}-8 a^{3} b \,x^{4}+32 a^{4}\right ) \left (b \,x^{4}+a \right )^{\frac {1}{4}}}{1989 b^{3}}\) | \(58\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.30, size = 47, normalized size = 0.80 \begin {gather*} \frac {{\left (b x^{4} + a\right )}^{\frac {17}{4}}}{17 \, b^{3}} - \frac {2 \, {\left (b x^{4} + a\right )}^{\frac {13}{4}} a}{13 \, b^{3}} + \frac {{\left (b x^{4} + a\right )}^{\frac {9}{4}} a^{2}}{9 \, b^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 0.38, size = 57, normalized size = 0.97 \begin {gather*} \frac {{\left (117 \, b^{4} x^{16} + 162 \, a b^{3} x^{12} + 5 \, a^{2} b^{2} x^{8} - 8 \, a^{3} b x^{4} + 32 \, a^{4}\right )} {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{1989 \, b^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 110 vs.
\(2 (51) = 102\).
time = 1.04, size = 110, normalized size = 1.86 \begin {gather*} \begin {cases} \frac {32 a^{4} \sqrt [4]{a + b x^{4}}}{1989 b^{3}} - \frac {8 a^{3} x^{4} \sqrt [4]{a + b x^{4}}}{1989 b^{2}} + \frac {5 a^{2} x^{8} \sqrt [4]{a + b x^{4}}}{1989 b} + \frac {18 a x^{12} \sqrt [4]{a + b x^{4}}}{221} + \frac {b x^{16} \sqrt [4]{a + b x^{4}}}{17} & \text {for}\: b \neq 0 \\\frac {a^{\frac {5}{4}} x^{12}}{12} & \text {otherwise} \end {cases} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 1.02, size = 43, normalized size = 0.73 \begin {gather*} \frac {117 \, {\left (b x^{4} + a\right )}^{\frac {17}{4}} - 306 \, {\left (b x^{4} + a\right )}^{\frac {13}{4}} a + 221 \, {\left (b x^{4} + a\right )}^{\frac {9}{4}} a^{2}}{1989 \, b^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [B]
time = 1.14, size = 53, normalized size = 0.90 \begin {gather*} {\left (b\,x^4+a\right )}^{1/4}\,\left (\frac {18\,a\,x^{12}}{221}+\frac {b\,x^{16}}{17}+\frac {32\,a^4}{1989\,b^3}-\frac {8\,a^3\,x^4}{1989\,b^2}+\frac {5\,a^2\,x^8}{1989\,b}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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