Integrand size = 20, antiderivative size = 119 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{(d+e x)^2} \, dx=\frac {2 \sqrt {a} \sqrt {b} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b d^2+a e^2}-\frac {2 b d p \log (d+e x)}{e \left (b d^2+a e^2\right )}+\frac {b d p \log \left (a+b x^2\right )}{e \left (b d^2+a e^2\right )}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{e (d+e x)} \]
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Time = 0.06 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2513, 815, 649, 211, 266} \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{(d+e x)^2} \, dx=\frac {2 \sqrt {a} \sqrt {b} p \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{a e^2+b d^2}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{e (d+e x)}+\frac {b d p \log \left (a+b x^2\right )}{e \left (a e^2+b d^2\right )}-\frac {2 b d p \log (d+e x)}{e \left (a e^2+b d^2\right )} \]
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Rule 211
Rule 266
Rule 649
Rule 815
Rule 2513
Rubi steps \begin{align*} \text {integral}& = -\frac {\log \left (c \left (a+b x^2\right )^p\right )}{e (d+e x)}+\frac {(2 b p) \int \frac {x}{(d+e x) \left (a+b x^2\right )} \, dx}{e} \\ & = -\frac {\log \left (c \left (a+b x^2\right )^p\right )}{e (d+e x)}+\frac {(2 b p) \int \left (-\frac {d e}{\left (b d^2+a e^2\right ) (d+e x)}+\frac {a e+b d x}{\left (b d^2+a e^2\right ) \left (a+b x^2\right )}\right ) \, dx}{e} \\ & = -\frac {2 b d p \log (d+e x)}{e \left (b d^2+a e^2\right )}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{e (d+e x)}+\frac {(2 b p) \int \frac {a e+b d x}{a+b x^2} \, dx}{e \left (b d^2+a e^2\right )} \\ & = -\frac {2 b d p \log (d+e x)}{e \left (b d^2+a e^2\right )}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{e (d+e x)}+\frac {(2 a b p) \int \frac {1}{a+b x^2} \, dx}{b d^2+a e^2}+\frac {\left (2 b^2 d p\right ) \int \frac {x}{a+b x^2} \, dx}{e \left (b d^2+a e^2\right )} \\ & = \frac {2 \sqrt {a} \sqrt {b} p \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b d^2+a e^2}-\frac {2 b d p \log (d+e x)}{e \left (b d^2+a e^2\right )}+\frac {b d p \log \left (a+b x^2\right )}{e \left (b d^2+a e^2\right )}-\frac {\log \left (c \left (a+b x^2\right )^p\right )}{e (d+e x)} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.15 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{(d+e x)^2} \, dx=\frac {2 \sqrt {a} \sqrt {b} e p (d+e x) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )-2 b d p (d+e x) \log (d+e x)+b d^2 p \log \left (a+b x^2\right )+b d e p x \log \left (a+b x^2\right )-b d^2 \log \left (c \left (a+b x^2\right )^p\right )-a e^2 \log \left (c \left (a+b x^2\right )^p\right )}{e \left (b d^2+a e^2\right ) (d+e x)} \]
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Time = 1.29 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.83
method | result | size |
parts | \(-\frac {\ln \left (c \left (b \,x^{2}+a \right )^{p}\right )}{e \left (e x +d \right )}+\frac {2 p b \left (-\frac {d \ln \left (e x +d \right )}{a \,e^{2}+b \,d^{2}}+\frac {\frac {d \ln \left (b \,x^{2}+a \right )}{2}+\frac {a e \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}}}{a \,e^{2}+b \,d^{2}}\right )}{e}\) | \(99\) |
risch | \(\text {Expression too large to display}\) | \(1233\) |
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Time = 0.30 (sec) , antiderivative size = 261, normalized size of antiderivative = 2.19 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{(d+e x)^2} \, dx=\left [\frac {{\left (e^{2} p x + d e p\right )} \sqrt {-a b} \log \left (\frac {b x^{2} + 2 \, \sqrt {-a b} x - a}{b x^{2} + a}\right ) + {\left (b d e p x - a e^{2} p\right )} \log \left (b x^{2} + a\right ) - 2 \, {\left (b d e p x + b d^{2} p\right )} \log \left (e x + d\right ) - {\left (b d^{2} + a e^{2}\right )} \log \left (c\right )}{b d^{3} e + a d e^{3} + {\left (b d^{2} e^{2} + a e^{4}\right )} x}, \frac {2 \, {\left (e^{2} p x + d e p\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b} x}{a}\right ) + {\left (b d e p x - a e^{2} p\right )} \log \left (b x^{2} + a\right ) - 2 \, {\left (b d e p x + b d^{2} p\right )} \log \left (e x + d\right ) - {\left (b d^{2} + a e^{2}\right )} \log \left (c\right )}{b d^{3} e + a d e^{3} + {\left (b d^{2} e^{2} + a e^{4}\right )} x}\right ] \]
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Timed out. \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{(d+e x)^2} \, dx=\text {Timed out} \]
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Time = 0.37 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.91 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{(d+e x)^2} \, dx=\frac {{\left (\frac {2 \, a e \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (b d^{2} + a e^{2}\right )} \sqrt {a b}} + \frac {d \log \left (b x^{2} + a\right )}{b d^{2} + a e^{2}} - \frac {2 \, d \log \left (e x + d\right )}{b d^{2} + a e^{2}}\right )} b p}{e} - \frac {\log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{{\left (e x + d\right )} e} \]
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Time = 0.33 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.02 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{(d+e x)^2} \, dx=\frac {b d p \log \left (b x^{2} + a\right )}{b d^{2} e + a e^{3}} - \frac {2 \, b d p \log \left (e x + d\right )}{b d^{2} e + a e^{3}} + \frac {2 \, a b p \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{{\left (b d^{2} + a e^{2}\right )} \sqrt {a b}} - \frac {p \log \left (b x^{2} + a\right )}{e^{2} x + d e} - \frac {\log \left (c\right )}{e^{2} x + d e} \]
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Time = 2.31 (sec) , antiderivative size = 337, normalized size of antiderivative = 2.83 \[ \int \frac {\log \left (c \left (a+b x^2\right )^p\right )}{(d+e x)^2} \, dx=\frac {\ln \left (\frac {4\,b^3\,p^2\,x}{e}-\frac {p\,\left (b\,d+e\,\sqrt {-a\,b}\right )\,\left (2\,a\,b^2\,e\,p+2\,b^3\,d\,p\,x-\frac {2\,b^2\,e\,p\,\left (b\,d+e\,\sqrt {-a\,b}\right )\,\left (-b\,x\,d^2+4\,a\,d\,e+3\,a\,x\,e^2\right )}{b\,d^2\,e+a\,e^3}\right )}{b\,d^2\,e+a\,e^3}\right )\,\left (b\,d\,p+e\,p\,\sqrt {-a\,b}\right )}{b\,d^2\,e+a\,e^3}-\frac {\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}{e\,\left (d+e\,x\right )}+\frac {\ln \left (\frac {4\,b^3\,p^2\,x}{e}-\frac {p\,\left (b\,d-e\,\sqrt {-a\,b}\right )\,\left (2\,a\,b^2\,e\,p+2\,b^3\,d\,p\,x-\frac {2\,b^2\,e\,p\,\left (b\,d-e\,\sqrt {-a\,b}\right )\,\left (-b\,x\,d^2+4\,a\,d\,e+3\,a\,x\,e^2\right )}{b\,d^2\,e+a\,e^3}\right )}{b\,d^2\,e+a\,e^3}\right )\,\left (b\,d\,p-e\,p\,\sqrt {-a\,b}\right )}{b\,d^2\,e+a\,e^3}-\frac {2\,b\,d\,p\,\ln \left (d+e\,x\right )}{b\,d^2\,e+a\,e^3} \]
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