Integrand size = 33, antiderivative size = 169 \[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {(A b-a B) (d+e x)^{1+m}}{2 b (b d-a e) (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e (b (2 B d-A e (1-m))-a B e (1+m)) (a+b x) (d+e x)^{1+m} \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,\frac {b (d+e x)}{b d-a e}\right )}{2 b (b d-a e)^3 (1+m) \sqrt {a^2+2 a b x+b^2 x^2}} \]
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Time = 0.10 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {784, 79, 70} \[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {e (a+b x) (d+e x)^{m+1} (-a B e (m+1)-A b e (1-m)+2 b B d) \operatorname {Hypergeometric2F1}\left (2,m+1,m+2,\frac {b (d+e x)}{b d-a e}\right )}{2 b (m+1) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac {(A b-a B) (d+e x)^{m+1}}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)} \]
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Rule 70
Rule 79
Rule 784
Rubi steps \begin{align*} \text {integral}& = \frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac {(A+B x) (d+e x)^m}{\left (a b+b^2 x\right )^3} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = -\frac {(A b-a B) (d+e x)^{1+m}}{2 b (b d-a e) (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left ((2 b B d-A b e (1-m)-a B e (1+m)) \left (a b+b^2 x\right )\right ) \int \frac {(d+e x)^m}{\left (a b+b^2 x\right )^2} \, dx}{2 (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}} \\ & = -\frac {(A b-a B) (d+e x)^{1+m}}{2 b (b d-a e) (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e (2 b B d-A b e (1-m)-a B e (1+m)) (a+b x) (d+e x)^{1+m} \, _2F_1\left (2,1+m;2+m;\frac {b (d+e x)}{b d-a e}\right )}{2 b (b d-a e)^3 (1+m) \sqrt {a^2+2 a b x+b^2 x^2}} \\ \end{align*}
Time = 0.15 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.71 \[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {(a+b x) (d+e x)^{1+m} \left (-A b+a B+\frac {e (2 b B d+A b e (-1+m)-a B e (1+m)) (a+b x)^2 \operatorname {Hypergeometric2F1}\left (2,1+m,2+m,\frac {b (d+e x)}{b d-a e}\right )}{(b d-a e)^2 (1+m)}\right )}{2 b (b d-a e) \left ((a+b x)^2\right )^{3/2}} \]
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\[\int \frac {\left (B x +A \right ) \left (e x +d \right )^{m}}{\left (b^{2} x^{2}+2 a b x +a^{2}\right )^{\frac {3}{2}}}d x\]
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\[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int { \frac {{\left (B x + A\right )} {\left (e x + d\right )}^{m}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}}} \,d x } \]
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Exception generated. \[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\text {Exception raised: HeuristicGCDFailed} \]
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\[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int { \frac {{\left (B x + A\right )} {\left (e x + d\right )}^{m}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}}} \,d x } \]
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\[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int { \frac {{\left (B x + A\right )} {\left (e x + d\right )}^{m}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}}} \,d x } \]
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Timed out. \[ \int \frac {(A+B x) (d+e x)^m}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^m}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]
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