Integrand size = 20, antiderivative size = 89 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^2} \, dx=-\frac {c (2 B d-A e) x}{e^3}+\frac {B c x^2}{2 e^2}+\frac {(B d-A e) \left (c d^2+a e^2\right )}{e^4 (d+e x)}+\frac {\left (3 B c d^2-2 A c d e+a B e^2\right ) \log (d+e x)}{e^4} \]
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Time = 0.05 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {786} \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^2} \, dx=\frac {\left (a e^2+c d^2\right ) (B d-A e)}{e^4 (d+e x)}+\frac {\log (d+e x) \left (a B e^2-2 A c d e+3 B c d^2\right )}{e^4}-\frac {c x (2 B d-A e)}{e^3}+\frac {B c x^2}{2 e^2} \]
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Rule 786
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {c (-2 B d+A e)}{e^3}+\frac {B c x}{e^2}+\frac {(-B d+A e) \left (c d^2+a e^2\right )}{e^3 (d+e x)^2}+\frac {3 B c d^2-2 A c d e+a B e^2}{e^3 (d+e x)}\right ) \, dx \\ & = -\frac {c (2 B d-A e) x}{e^3}+\frac {B c x^2}{2 e^2}+\frac {(B d-A e) \left (c d^2+a e^2\right )}{e^4 (d+e x)}+\frac {\left (3 B c d^2-2 A c d e+a B e^2\right ) \log (d+e x)}{e^4} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.97 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^2} \, dx=\frac {2 c e (-2 B d+A e) x+B c e^2 x^2+\frac {2 (B d-A e) \left (c d^2+a e^2\right )}{d+e x}+2 \left (3 B c d^2-2 A c d e+a B e^2\right ) \log (d+e x)}{2 e^4} \]
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Time = 0.24 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.07
method | result | size |
default | \(\frac {c \left (\frac {1}{2} B e \,x^{2}+A e x -2 B d x \right )}{e^{3}}-\frac {A a \,e^{3}+A c \,d^{2} e -B a d \,e^{2}-B c \,d^{3}}{e^{4} \left (e x +d \right )}+\frac {\left (-2 A c d e +B a \,e^{2}+3 B c \,d^{2}\right ) \ln \left (e x +d \right )}{e^{4}}\) | \(95\) |
norman | \(\frac {-\frac {A a \,e^{3}+2 A c \,d^{2} e -B a d \,e^{2}-3 B c \,d^{3}}{e^{4}}+\frac {B c \,x^{3}}{2 e}+\frac {c \left (2 A e -3 B d \right ) x^{2}}{2 e^{2}}}{e x +d}-\frac {\left (2 A c d e -B a \,e^{2}-3 B c \,d^{2}\right ) \ln \left (e x +d \right )}{e^{4}}\) | \(106\) |
risch | \(\frac {B c \,x^{2}}{2 e^{2}}+\frac {A c x}{e^{2}}-\frac {2 B c d x}{e^{3}}-\frac {A a}{e \left (e x +d \right )}-\frac {d^{2} A c}{e^{3} \left (e x +d \right )}+\frac {B a d}{e^{2} \left (e x +d \right )}+\frac {d^{3} B c}{e^{4} \left (e x +d \right )}-\frac {2 \ln \left (e x +d \right ) A c d}{e^{3}}+\frac {\ln \left (e x +d \right ) B a}{e^{2}}+\frac {3 \ln \left (e x +d \right ) B c \,d^{2}}{e^{4}}\) | \(131\) |
parallelrisch | \(-\frac {-B c \,x^{3} e^{3}+4 A \ln \left (e x +d \right ) x c d \,e^{2}-2 A c \,e^{3} x^{2}-2 B \ln \left (e x +d \right ) x a \,e^{3}-6 B \ln \left (e x +d \right ) x c \,d^{2} e +3 B \,x^{2} c d \,e^{2}+4 A \ln \left (e x +d \right ) c \,d^{2} e -2 B \ln \left (e x +d \right ) a d \,e^{2}-6 B \ln \left (e x +d \right ) c \,d^{3}+2 A a \,e^{3}+4 A c \,d^{2} e -2 B a d \,e^{2}-6 B c \,d^{3}}{2 e^{4} \left (e x +d \right )}\) | \(160\) |
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Time = 0.30 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.71 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^2} \, dx=\frac {B c e^{3} x^{3} + 2 \, B c d^{3} - 2 \, A c d^{2} e + 2 \, B a d e^{2} - 2 \, A a e^{3} - {\left (3 \, B c d e^{2} - 2 \, A c e^{3}\right )} x^{2} - 2 \, {\left (2 \, B c d^{2} e - A c d e^{2}\right )} x + 2 \, {\left (3 \, B c d^{3} - 2 \, A c d^{2} e + B a d e^{2} + {\left (3 \, B c d^{2} e - 2 \, A c d e^{2} + B a e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (e^{5} x + d e^{4}\right )}} \]
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Time = 0.29 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.17 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^2} \, dx=\frac {B c x^{2}}{2 e^{2}} + x \left (\frac {A c}{e^{2}} - \frac {2 B c d}{e^{3}}\right ) + \frac {- A a e^{3} - A c d^{2} e + B a d e^{2} + B c d^{3}}{d e^{4} + e^{5} x} + \frac {\left (- 2 A c d e + B a e^{2} + 3 B c d^{2}\right ) \log {\left (d + e x \right )}}{e^{4}} \]
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Time = 0.20 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.13 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^2} \, dx=\frac {B c d^{3} - A c d^{2} e + B a d e^{2} - A a e^{3}}{e^{5} x + d e^{4}} + \frac {B c e x^{2} - 2 \, {\left (2 \, B c d - A c e\right )} x}{2 \, e^{3}} + \frac {{\left (3 \, B c d^{2} - 2 \, A c d e + B a e^{2}\right )} \log \left (e x + d\right )}{e^{4}} \]
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Time = 0.27 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.72 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^2} \, dx=\frac {{\left (B c - \frac {2 \, {\left (3 \, B c d e - A c e^{2}\right )}}{{\left (e x + d\right )} e}\right )} {\left (e x + d\right )}^{2}}{2 \, e^{4}} - \frac {{\left (3 \, B c d^{2} - 2 \, A c d e + B a e^{2}\right )} \log \left (\frac {{\left | e x + d \right |}}{{\left (e x + d\right )}^{2} {\left | e \right |}}\right )}{e^{4}} + \frac {\frac {B c d^{3} e^{2}}{e x + d} - \frac {A c d^{2} e^{3}}{e x + d} + \frac {B a d e^{4}}{e x + d} - \frac {A a e^{5}}{e x + d}}{e^{6}} \]
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Time = 0.08 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.18 \[ \int \frac {(A+B x) \left (a+c x^2\right )}{(d+e x)^2} \, dx=x\,\left (\frac {A\,c}{e^2}-\frac {2\,B\,c\,d}{e^3}\right )-\frac {-B\,c\,d^3+A\,c\,d^2\,e-B\,a\,d\,e^2+A\,a\,e^3}{e\,\left (x\,e^4+d\,e^3\right )}+\frac {\ln \left (d+e\,x\right )\,\left (3\,B\,c\,d^2-2\,A\,c\,d\,e+B\,a\,e^2\right )}{e^4}+\frac {B\,c\,x^2}{2\,e^2} \]
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