Integrand size = 29, antiderivative size = 224 \[ \int \frac {(a+b x)^m}{(c+d x) (e+f x) (g+h x)} \, dx=\frac {d^2 (a+b x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {d (a+b x)}{b c-a d}\right )}{(b c-a d) (d e-c f) (d g-c h) (1+m)}-\frac {f^2 (a+b x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {f (a+b x)}{b e-a f}\right )}{(b e-a f) (d e-c f) (f g-e h) (1+m)}+\frac {h^2 (a+b x)^{1+m} \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {h (a+b x)}{b g-a h}\right )}{(b g-a h) (d g-c h) (f g-e h) (1+m)} \]
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Time = 0.15 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {186, 70} \[ \int \frac {(a+b x)^m}{(c+d x) (e+f x) (g+h x)} \, dx=\frac {d^2 (a+b x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {d (a+b x)}{b c-a d}\right )}{(m+1) (b c-a d) (d e-c f) (d g-c h)}-\frac {f^2 (a+b x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {f (a+b x)}{b e-a f}\right )}{(m+1) (b e-a f) (d e-c f) (f g-e h)}+\frac {h^2 (a+b x)^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {h (a+b x)}{b g-a h}\right )}{(m+1) (b g-a h) (d g-c h) (f g-e h)} \]
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Rule 70
Rule 186
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {d^2 (a+b x)^m}{(d e-c f) (d g-c h) (c+d x)}+\frac {f^2 (a+b x)^m}{(d e-c f) (-f g+e h) (e+f x)}+\frac {h^2 (a+b x)^m}{(d g-c h) (f g-e h) (g+h x)}\right ) \, dx \\ & = \frac {d^2 \int \frac {(a+b x)^m}{c+d x} \, dx}{(d e-c f) (d g-c h)}-\frac {f^2 \int \frac {(a+b x)^m}{e+f x} \, dx}{(d e-c f) (f g-e h)}+\frac {h^2 \int \frac {(a+b x)^m}{g+h x} \, dx}{(d g-c h) (f g-e h)} \\ & = \frac {d^2 (a+b x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {d (a+b x)}{b c-a d}\right )}{(b c-a d) (d e-c f) (d g-c h) (1+m)}-\frac {f^2 (a+b x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {f (a+b x)}{b e-a f}\right )}{(b e-a f) (d e-c f) (f g-e h) (1+m)}+\frac {h^2 (a+b x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {h (a+b x)}{b g-a h}\right )}{(b g-a h) (d g-c h) (f g-e h) (1+m)} \\ \end{align*}
Time = 0.41 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.86 \[ \int \frac {(a+b x)^m}{(c+d x) (e+f x) (g+h x)} \, dx=\frac {(a+b x)^{1+m} \left (\frac {d^2 \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {d (a+b x)}{-b c+a d}\right )}{(b c-a d) (-d e+c f) (-d g+c h)}+\frac {f^2 \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {f (a+b x)}{-b e+a f}\right )}{(b e-a f) (d e-c f) (-f g+e h)}+\frac {h^2 \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {h (a+b x)}{-b g+a h}\right )}{(b g-a h) (d g-c h) (f g-e h)}\right )}{1+m} \]
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\[\int \frac {\left (b x +a \right )^{m}}{\left (d x +c \right ) \left (f x +e \right ) \left (h x +g \right )}d x\]
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\[ \int \frac {(a+b x)^m}{(c+d x) (e+f x) (g+h x)} \, dx=\int { \frac {{\left (b x + a\right )}^{m}}{{\left (d x + c\right )} {\left (f x + e\right )} {\left (h x + g\right )}} \,d x } \]
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Exception generated. \[ \int \frac {(a+b x)^m}{(c+d x) (e+f x) (g+h x)} \, dx=\text {Exception raised: HeuristicGCDFailed} \]
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\[ \int \frac {(a+b x)^m}{(c+d x) (e+f x) (g+h x)} \, dx=\int { \frac {{\left (b x + a\right )}^{m}}{{\left (d x + c\right )} {\left (f x + e\right )} {\left (h x + g\right )}} \,d x } \]
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\[ \int \frac {(a+b x)^m}{(c+d x) (e+f x) (g+h x)} \, dx=\int { \frac {{\left (b x + a\right )}^{m}}{{\left (d x + c\right )} {\left (f x + e\right )} {\left (h x + g\right )}} \,d x } \]
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Timed out. \[ \int \frac {(a+b x)^m}{(c+d x) (e+f x) (g+h x)} \, dx=\int \frac {{\left (a+b\,x\right )}^m}{\left (e+f\,x\right )\,\left (g+h\,x\right )\,\left (c+d\,x\right )} \,d x \]
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