Integrand size = 17, antiderivative size = 45 \[ \int f^{a+b x^n} x^{-1+2 n} \, dx=-\frac {f^{a+b x^n}}{b^2 n \log ^2(f)}+\frac {f^{a+b x^n} x^n}{b n \log (f)} \]
[Out]
Time = 0.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2244, 2240} \[ \int f^{a+b x^n} x^{-1+2 n} \, dx=\frac {x^n f^{a+b x^n}}{b n \log (f)}-\frac {f^{a+b x^n}}{b^2 n \log ^2(f)} \]
[In]
[Out]
Rule 2240
Rule 2244
Rubi steps \begin{align*} \text {integral}& = \frac {f^{a+b x^n} x^n}{b n \log (f)}-\frac {\int f^{a+b x^n} x^{-1+n} \, dx}{b \log (f)} \\ & = -\frac {f^{a+b x^n}}{b^2 n \log ^2(f)}+\frac {f^{a+b x^n} x^n}{b n \log (f)} \\ \end{align*}
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.56 \[ \int f^{a+b x^n} x^{-1+2 n} \, dx=-\frac {f^a \Gamma \left (2,-b x^n \log (f)\right )}{b^2 n \log ^2(f)} \]
[In]
[Out]
Time = 0.08 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.67
method | result | size |
risch | \(\frac {\left (b \,x^{n} \ln \left (f \right )-1\right ) f^{a +b \,x^{n}}}{\ln \left (f \right )^{2} b^{2} n}\) | \(30\) |
meijerg | \(\frac {f^{a} \left (1-\frac {\left (2-2 b \,x^{n} \ln \left (f \right )\right ) {\mathrm e}^{b \,x^{n} \ln \left (f \right )}}{2}\right )}{\ln \left (f \right )^{2} b^{2} n}\) | \(37\) |
norman | \(\frac {{\mathrm e}^{n \ln \left (x \right )} {\mathrm e}^{\left (a +b \,{\mathrm e}^{n \ln \left (x \right )}\right ) \ln \left (f \right )}}{\ln \left (f \right ) b n}-\frac {{\mathrm e}^{\left (a +b \,{\mathrm e}^{n \ln \left (x \right )}\right ) \ln \left (f \right )}}{\ln \left (f \right )^{2} b^{2} n}\) | \(56\) |
[In]
[Out]
none
Time = 0.28 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.73 \[ \int f^{a+b x^n} x^{-1+2 n} \, dx=\frac {{\left (b x^{n} \log \left (f\right ) - 1\right )} e^{\left (b x^{n} \log \left (f\right ) + a \log \left (f\right )\right )}}{b^{2} n \log \left (f\right )^{2}} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 73 vs. \(2 (36) = 72\).
Time = 1.53 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.62 \[ \int f^{a+b x^n} x^{-1+2 n} \, dx=\begin {cases} \log {\left (x \right )} & \text {for}\: b = 0 \wedge f = 1 \wedge n = 0 \\\frac {f^{a} x x^{2 n - 1}}{2 n} & \text {for}\: b = 0 \\f^{a + b} \log {\left (x \right )} & \text {for}\: n = 0 \\\frac {x x^{2 n - 1}}{2 n} & \text {for}\: f = 1 \\\frac {f^{a + b x^{n}} x^{n}}{b n \log {\left (f \right )}} - \frac {f^{a + b x^{n}}}{b^{2} n \log {\left (f \right )}^{2}} & \text {otherwise} \end {cases} \]
[In]
[Out]
none
Time = 0.20 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.76 \[ \int f^{a+b x^n} x^{-1+2 n} \, dx=\frac {{\left (b f^{a} x^{n} \log \left (f\right ) - f^{a}\right )} f^{b x^{n}}}{b^{2} n \log \left (f\right )^{2}} \]
[In]
[Out]
\[ \int f^{a+b x^n} x^{-1+2 n} \, dx=\int { f^{b x^{n} + a} x^{2 \, n - 1} \,d x } \]
[In]
[Out]
Timed out. \[ \int f^{a+b x^n} x^{-1+2 n} \, dx=\int f^{a+b\,x^n}\,x^{2\,n-1} \,d x \]
[In]
[Out]