Integrand size = 31, antiderivative size = 126 \[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {(A b-a B) (d+e x)^{1+m}}{3 b (b d-a e) (a+b x)^3}-\frac {e^2 (b (3 B d-A e (2-m))-a B e (1+m)) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (3,1+m,2+m,\frac {b (d+e x)}{b d-a e}\right )}{3 b (b d-a e)^4 (1+m)} \]
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Time = 0.05 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {27, 79, 70} \[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=-\frac {e^2 (d+e x)^{m+1} (-a B e (m+1)-A b e (2-m)+3 b B d) \operatorname {Hypergeometric2F1}\left (3,m+1,m+2,\frac {b (d+e x)}{b d-a e}\right )}{3 b (m+1) (b d-a e)^4}-\frac {(A b-a B) (d+e x)^{m+1}}{3 b (a+b x)^3 (b d-a e)} \]
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Rule 27
Rule 70
Rule 79
Rubi steps \begin{align*} \text {integral}& = \int \frac {(A+B x) (d+e x)^m}{(a+b x)^4} \, dx \\ & = -\frac {(A b-a B) (d+e x)^{1+m}}{3 b (b d-a e) (a+b x)^3}+\frac {(3 b B d-A b e (2-m)-a B e (1+m)) \int \frac {(d+e x)^m}{(a+b x)^3} \, dx}{3 b (b d-a e)} \\ & = -\frac {(A b-a B) (d+e x)^{1+m}}{3 b (b d-a e) (a+b x)^3}-\frac {e^2 (3 b B d-A b e (2-m)-a B e (1+m)) (d+e x)^{1+m} \, _2F_1\left (3,1+m;2+m;\frac {b (d+e x)}{b d-a e}\right )}{3 b (b d-a e)^4 (1+m)} \\ \end{align*}
Time = 0.12 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.87 \[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\frac {(d+e x)^{1+m} \left (\frac {-A b+a B}{(a+b x)^3}-\frac {e^2 (3 b B d+A b e (-2+m)-a B e (1+m)) \operatorname {Hypergeometric2F1}\left (3,1+m,2+m,\frac {b (d+e x)}{b d-a e}\right )}{(b d-a e)^3 (1+m)}\right )}{3 b (b d-a e)} \]
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\[\int \frac {\left (B x +A \right ) \left (e x +d \right )^{m}}{\left (b^{2} x^{2}+2 a b x +a^{2}\right )^{2}}d x\]
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\[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\int { \frac {{\left (B x + A\right )} {\left (e x + d\right )}^{m}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{2}} \,d x } \]
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\[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\int \frac {\left (A + B x\right ) \left (d + e x\right )^{m}}{\left (a + b x\right )^{4}}\, dx \]
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\[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\int { \frac {{\left (B x + A\right )} {\left (e x + d\right )}^{m}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{2}} \,d x } \]
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\[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\int { \frac {{\left (B x + A\right )} {\left (e x + d\right )}^{m}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^m}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^2} \,d x \]
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