Integrand size = 15, antiderivative size = 31 \[ \int (c x)^m \left (b x^n\right )^{5/2} \, dx=\frac {2 (c x)^{1+m} \left (b x^n\right )^{5/2}}{c (2+2 m+5 n)} \]
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Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {15, 20, 30} \[ \int (c x)^m \left (b x^n\right )^{5/2} \, dx=\frac {2 b^2 x^{2 n+1} \sqrt {b x^n} (c x)^m}{2 m+5 n+2} \]
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Rule 15
Rule 20
Rule 30
Rubi steps \begin{align*} \text {integral}& = \left (b^2 x^{-n/2} \sqrt {b x^n}\right ) \int x^{5 n/2} (c x)^m \, dx \\ & = \left (b^2 x^{-m-\frac {n}{2}} (c x)^m \sqrt {b x^n}\right ) \int x^{m+\frac {5 n}{2}} \, dx \\ & = \frac {2 b^2 x^{1+2 n} (c x)^m \sqrt {b x^n}}{2+2 m+5 n} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84 \[ \int (c x)^m \left (b x^n\right )^{5/2} \, dx=\frac {x (c x)^m \left (b x^n\right )^{5/2}}{1+m+\frac {5 n}{2}} \]
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Time = 0.10 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.84
method | result | size |
gosper | \(\frac {2 x \left (c x \right )^{m} \left (b \,x^{n}\right )^{\frac {5}{2}}}{2+2 m +5 n}\) | \(26\) |
risch | \(\frac {2 b^{3} x \,x^{m} c^{m} {\mathrm e}^{\frac {i \operatorname {csgn}\left (i c x \right ) \pi m \left (\operatorname {csgn}\left (i c x \right )-\operatorname {csgn}\left (i x \right )\right ) \left (-\operatorname {csgn}\left (i c x \right )+\operatorname {csgn}\left (i c \right )\right )}{2}} x^{3 n}}{\left (2+2 m +5 n \right ) \sqrt {b \,x^{n}}}\) | \(75\) |
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Exception generated. \[ \int (c x)^m \left (b x^n\right )^{5/2} \, dx=\text {Exception raised: TypeError} \]
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Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (26) = 52\).
Time = 138.73 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.81 \[ \int (c x)^m \left (b x^n\right )^{5/2} \, dx=\begin {cases} \frac {2 x \left (b x^{n}\right )^{\frac {5}{2}} \left (c x\right )^{m}}{2 m + 5 n + 2} & \text {for}\: m \neq - \frac {5 n}{2} - 1 \\x \left (b x^{n}\right )^{\frac {5}{2}} \left (c x\right )^{- \frac {5 n}{2} - 1} \log {\left (x \right )} & \text {otherwise} \end {cases} \]
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none
Time = 0.21 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87 \[ \int (c x)^m \left (b x^n\right )^{5/2} \, dx=\frac {2 \, b^{\frac {5}{2}} c^{m} x x^{m} {\left (x^{n}\right )}^{\frac {5}{2}}}{2 \, m + 5 \, n + 2} \]
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\[ \int (c x)^m \left (b x^n\right )^{5/2} \, dx=\int { \left (b x^{n}\right )^{\frac {5}{2}} \left (c x\right )^{m} \,d x } \]
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Time = 5.57 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.10 \[ \int (c x)^m \left (b x^n\right )^{5/2} \, dx=\frac {2\,b^2\,x^{2\,n+1}\,\sqrt {b\,x^n}\,{\left (c\,x\right )}^m}{2\,m+5\,n+2} \]
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