Integrand size = 25, antiderivative size = 216 \[ \int \frac {x^3 (e+f x)^n}{(a+b x) (c+d x)} \, dx=-\frac {e (e+f x)^{1+n}}{b d f^2 (1+n)}-\frac {(b c+a d) (e+f x)^{1+n}}{b^2 d^2 f (1+n)}+\frac {(e+f x)^{2+n}}{b d f^2 (2+n)}+\frac {a^3 (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {b (e+f x)}{b e-a f}\right )}{b^2 (b c-a d) (b e-a f) (1+n)}-\frac {c^3 (e+f x)^{1+n} \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {d (e+f x)}{d e-c f}\right )}{d^2 (b c-a d) (d e-c f) (1+n)} \]
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Time = 0.11 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {186, 45, 70} \[ \int \frac {x^3 (e+f x)^n}{(a+b x) (c+d x)} \, dx=\frac {a^3 (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {b (e+f x)}{b e-a f}\right )}{b^2 (n+1) (b c-a d) (b e-a f)}-\frac {(a d+b c) (e+f x)^{n+1}}{b^2 d^2 f (n+1)}-\frac {c^3 (e+f x)^{n+1} \operatorname {Hypergeometric2F1}\left (1,n+1,n+2,\frac {d (e+f x)}{d e-c f}\right )}{d^2 (n+1) (b c-a d) (d e-c f)}-\frac {e (e+f x)^{n+1}}{b d f^2 (n+1)}+\frac {(e+f x)^{n+2}}{b d f^2 (n+2)} \]
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Rule 45
Rule 70
Rule 186
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {(-b c-a d) (e+f x)^n}{b^2 d^2}+\frac {x (e+f x)^n}{b d}-\frac {a^3 (e+f x)^n}{b^2 (b c-a d) (a+b x)}-\frac {c^3 (e+f x)^n}{d^2 (-b c+a d) (c+d x)}\right ) \, dx \\ & = -\frac {(b c+a d) (e+f x)^{1+n}}{b^2 d^2 f (1+n)}+\frac {\int x (e+f x)^n \, dx}{b d}-\frac {a^3 \int \frac {(e+f x)^n}{a+b x} \, dx}{b^2 (b c-a d)}+\frac {c^3 \int \frac {(e+f x)^n}{c+d x} \, dx}{d^2 (b c-a d)} \\ & = -\frac {(b c+a d) (e+f x)^{1+n}}{b^2 d^2 f (1+n)}+\frac {a^3 (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {b (e+f x)}{b e-a f}\right )}{b^2 (b c-a d) (b e-a f) (1+n)}-\frac {c^3 (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {d (e+f x)}{d e-c f}\right )}{d^2 (b c-a d) (d e-c f) (1+n)}+\frac {\int \left (-\frac {e (e+f x)^n}{f}+\frac {(e+f x)^{1+n}}{f}\right ) \, dx}{b d} \\ & = -\frac {e (e+f x)^{1+n}}{b d f^2 (1+n)}-\frac {(b c+a d) (e+f x)^{1+n}}{b^2 d^2 f (1+n)}+\frac {(e+f x)^{2+n}}{b d f^2 (2+n)}+\frac {a^3 (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {b (e+f x)}{b e-a f}\right )}{b^2 (b c-a d) (b e-a f) (1+n)}-\frac {c^3 (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {d (e+f x)}{d e-c f}\right )}{d^2 (b c-a d) (d e-c f) (1+n)} \\ \end{align*}
Time = 0.45 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.81 \[ \int \frac {x^3 (e+f x)^n}{(a+b x) (c+d x)} \, dx=\frac {(e+f x)^{1+n} \left (\frac {a^3 \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {b (e+f x)}{b e-a f}\right )}{b e-a f}+\frac {(b c-a d) (-d e+c f) (b c f (2+n)+a d f (2+n)+b d (e-f (1+n) x))-b^2 c^3 f^2 (2+n) \operatorname {Hypergeometric2F1}\left (1,1+n,2+n,\frac {d (e+f x)}{d e-c f}\right )}{d^2 f^2 (d e-c f) (2+n)}\right )}{b^2 (b c-a d) (1+n)} \]
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\[\int \frac {x^{3} \left (f x +e \right )^{n}}{\left (b x +a \right ) \left (d x +c \right )}d x\]
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\[ \int \frac {x^3 (e+f x)^n}{(a+b x) (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{n} x^{3}}{{\left (b x + a\right )} {\left (d x + c\right )}} \,d x } \]
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\[ \int \frac {x^3 (e+f x)^n}{(a+b x) (c+d x)} \, dx=\int \frac {x^{3} \left (e + f x\right )^{n}}{\left (a + b x\right ) \left (c + d x\right )}\, dx \]
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\[ \int \frac {x^3 (e+f x)^n}{(a+b x) (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{n} x^{3}}{{\left (b x + a\right )} {\left (d x + c\right )}} \,d x } \]
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\[ \int \frac {x^3 (e+f x)^n}{(a+b x) (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{n} x^{3}}{{\left (b x + a\right )} {\left (d x + c\right )}} \,d x } \]
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Timed out. \[ \int \frac {x^3 (e+f x)^n}{(a+b x) (c+d x)} \, dx=\int \frac {x^3\,{\left (e+f\,x\right )}^n}{\left (a+b\,x\right )\,\left (c+d\,x\right )} \,d x \]
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