Integrand size = 35, antiderivative size = 154 \[ \int \frac {(a+b x)^{1-n} (c+d x)^{1+n}}{(b c+a d+2 b d x)^2} \, dx=-\frac {(b c-a d) (a+b x)^{1-n} (c+d x)^{-1+n} \operatorname {Hypergeometric2F1}\left (2,1-n,2-n,-\frac {d (a+b x)}{b (c+d x)}\right )}{4 b^3 d (1-n)}+\frac {(a+b x)^{-n} \left (-\frac {d (a+b x)}{b c-a d}\right )^n (c+d x)^{1+n} \operatorname {Hypergeometric2F1}\left (n,1+n,2+n,\frac {b (c+d x)}{b c-a d}\right )}{4 b d^2 (1+n)} \]
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Time = 0.07 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {130, 72, 71, 133} \[ \int \frac {(a+b x)^{1-n} (c+d x)^{1+n}}{(b c+a d+2 b d x)^2} \, dx=\frac {(a+b x)^{-n} (c+d x)^{n+1} \left (-\frac {d (a+b x)}{b c-a d}\right )^n \operatorname {Hypergeometric2F1}\left (n,n+1,n+2,\frac {b (c+d x)}{b c-a d}\right )}{4 b d^2 (n+1)}-\frac {(b c-a d) (a+b x)^{1-n} (c+d x)^{n-1} \operatorname {Hypergeometric2F1}\left (2,1-n,2-n,-\frac {d (a+b x)}{b (c+d x)}\right )}{4 b^3 d (1-n)} \]
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Rule 71
Rule 72
Rule 130
Rule 133
Rubi steps \begin{align*} \text {integral}& = \frac {\int (a+b x)^{-n} (c+d x)^n \, dx}{4 b d}-\frac {(b c-a d)^2 \int \frac {(a+b x)^{-n} (c+d x)^n}{(b c+a d+2 b d x)^2} \, dx}{4 b d} \\ & = -\frac {(b c-a d) (a+b x)^{1-n} (c+d x)^{-1+n} \, _2F_1\left (2,1-n;2-n;-\frac {d (a+b x)}{b (c+d x)}\right )}{4 b^3 d (1-n)}+\frac {\left ((a+b x)^{-n} \left (\frac {d (a+b x)}{-b c+a d}\right )^n\right ) \int (c+d x)^n \left (-\frac {a d}{b c-a d}-\frac {b d x}{b c-a d}\right )^{-n} \, dx}{4 b d} \\ & = -\frac {(b c-a d) (a+b x)^{1-n} (c+d x)^{-1+n} \, _2F_1\left (2,1-n;2-n;-\frac {d (a+b x)}{b (c+d x)}\right )}{4 b^3 d (1-n)}+\frac {(a+b x)^{-n} \left (-\frac {d (a+b x)}{b c-a d}\right )^n (c+d x)^{1+n} \, _2F_1\left (n,1+n;2+n;\frac {b (c+d x)}{b c-a d}\right )}{4 b d^2 (1+n)} \\ \end{align*}
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 1.28 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.52 \[ \int \frac {(a+b x)^{1-n} (c+d x)^{1+n}}{(b c+a d+2 b d x)^2} \, dx=-\frac {(a+b x)^{-n} (c+d x)^n \left (-(b c-a d)^2 (-1+n) \left (\frac {d (a+b x)}{a d+b (c+2 d x)}\right )^n \left (\frac {b (c+d x)}{a d+b (c+2 d x)}\right )^{-n} \operatorname {AppellF1}\left (1,-n,n,2,\frac {-b c+a d}{a d+b (c+2 d x)},\frac {b c-a d}{b c+a d+2 b d x}\right )+2 d (a+b x) \left (\frac {b (c+d x)}{b c-a d}\right )^{-n} (a d+b (c+2 d x)) \operatorname {Hypergeometric2F1}\left (1-n,-n,2-n,\frac {d (a+b x)}{-b c+a d}\right )\right )}{8 b^2 d^2 (-1+n) (a d+b (c+2 d x))} \]
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\[\int \frac {\left (b x +a \right )^{1-n} \left (d x +c \right )^{1+n}}{\left (2 b d x +a d +b c \right )^{2}}d x\]
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\[ \int \frac {(a+b x)^{1-n} (c+d x)^{1+n}}{(b c+a d+2 b d x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{-n + 1} {\left (d x + c\right )}^{n + 1}}{{\left (2 \, b d x + b c + a d\right )}^{2}} \,d x } \]
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Exception generated. \[ \int \frac {(a+b x)^{1-n} (c+d x)^{1+n}}{(b c+a d+2 b d x)^2} \, dx=\text {Exception raised: HeuristicGCDFailed} \]
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\[ \int \frac {(a+b x)^{1-n} (c+d x)^{1+n}}{(b c+a d+2 b d x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{-n + 1} {\left (d x + c\right )}^{n + 1}}{{\left (2 \, b d x + b c + a d\right )}^{2}} \,d x } \]
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\[ \int \frac {(a+b x)^{1-n} (c+d x)^{1+n}}{(b c+a d+2 b d x)^2} \, dx=\int { \frac {{\left (b x + a\right )}^{-n + 1} {\left (d x + c\right )}^{n + 1}}{{\left (2 \, b d x + b c + a d\right )}^{2}} \,d x } \]
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Timed out. \[ \int \frac {(a+b x)^{1-n} (c+d x)^{1+n}}{(b c+a d+2 b d x)^2} \, dx=\int \frac {{\left (a+b\,x\right )}^{1-n}\,{\left (c+d\,x\right )}^{n+1}}{{\left (a\,d+b\,c+2\,b\,d\,x\right )}^2} \,d x \]
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