Integrand size = 40, antiderivative size = 34 \[ \int \frac {1}{\left (1-c^2 x^2\right ) \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2} \, dx=\frac {1}{b c \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )} \]
[Out]
Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {2573, 6818} \[ \int \frac {1}{\left (1-c^2 x^2\right ) \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2} \, dx=\frac {1}{b c \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {c x+1}}\right )\right )} \]
[In]
[Out]
Rule 2573
Rule 6818
Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {1}{\left (1-c^2 x^2\right ) \left (a+b \log \left (\sqrt {\frac {1-c x}{1+c x}}\right )\right )^2} \, dx,\sqrt {\frac {1-c x}{1+c x}},\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right ) \\ & = \frac {1}{b c \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (1-c^2 x^2\right ) \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2} \, dx=\frac {1}{b c \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )} \]
[In]
[Out]
\[\int \frac {1}{\left (-x^{2} c^{2}+1\right ) \left (a +b \ln \left (\frac {\sqrt {-x c +1}}{\sqrt {x c +1}}\right )\right )^{2}}d x\]
[In]
[Out]
none
Time = 0.28 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\left (1-c^2 x^2\right ) \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2} \, dx=\frac {1}{b^{2} c \log \left (\frac {\sqrt {-c x + 1}}{\sqrt {c x + 1}}\right ) + a b c} \]
[In]
[Out]
Leaf count of result is larger than twice the leaf count of optimal. 61 vs. \(2 (26) = 52\).
Time = 74.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.79 \[ \int \frac {1}{\left (1-c^2 x^2\right ) \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2} \, dx=\begin {cases} \frac {x}{a^{2}} & \text {for}\: b = 0 \wedge c = 0 \\\frac {- \frac {\log {\left (x - \frac {1}{c} \right )}}{2 c} + \frac {\log {\left (x + \frac {1}{c} \right )}}{2 c}}{a^{2}} & \text {for}\: b = 0 \\\frac {x}{a^{2}} & \text {for}\: c = 0 \\\frac {1}{a b c + b^{2} c \log {\left (\frac {\sqrt {- c x + 1}}{\sqrt {c x + 1}} \right )}} & \text {otherwise} \end {cases} \]
[In]
[Out]
none
Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (1-c^2 x^2\right ) \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2} \, dx=-\frac {2}{b^{2} c \log \left (c x + 1\right ) - b^{2} c \log \left (-c x + 1\right ) - 2 \, a b c} \]
[In]
[Out]
none
Time = 0.35 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (1-c^2 x^2\right ) \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2} \, dx=-\frac {2}{b^{2} c \log \left (c x + 1\right ) - b^{2} c \log \left (-c x + 1\right ) - 2 \, a b c} \]
[In]
[Out]
Timed out. \[ \int \frac {1}{\left (1-c^2 x^2\right ) \left (a+b \log \left (\frac {\sqrt {1-c x}}{\sqrt {1+c x}}\right )\right )^2} \, dx=-\int \frac {1}{{\left (a+b\,\ln \left (\frac {\sqrt {1-c\,x}}{\sqrt {c\,x+1}}\right )\right )}^2\,\left (c^2\,x^2-1\right )} \,d x \]
[In]
[Out]