Optimal. Leaf size=402 \[ \frac {(a+b x)^{m+1} (c+d x)^n (e+f x)^{-m-n-1} \left (a^2 d^2 f^2 \left (m^2+3 m+2\right )+2 a b d f (m+1) (c f (n+1)-d e (m+n+3))-\left (b^2 \left (-c^2 f^2 \left (n^2+3 n+2\right )+2 c d e f (n+1) (m+n+3)-d^2 e^2 \left (m^2+m (2 n+5)+n^2+5 n+6\right )\right )\right )\right ) \left (\frac {(c+d x) (b e-a f)}{(e+f x) (b c-a d)}\right )^{-n} \, _2F_1\left (m+1,-n;m+2;-\frac {(d e-c f) (a+b x)}{(b c-a d) (e+f x)}\right )}{(m+1) (m+n+2) (m+n+3) (b e-a f)^3 (d e-c f)^2}-\frac {f (a+b x)^{m+1} (c+d x)^{n+1} (e+f x)^{-m-n-3}}{(m+n+3) (b e-a f) (d e-c f)}+\frac {f (a+b x)^{m+1} (c+d x)^{n+1} (e+f x)^{-m-n-2} (a d f (m+2)+b (c f (n+2)-d e (m+n+4)))}{(m+n+2) (m+n+3) (b e-a f)^2 (d e-c f)^2} \]
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Rubi [A] time = 0.54, antiderivative size = 401, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {129, 155, 12, 132} \[ \frac {(a+b x)^{m+1} (c+d x)^n (e+f x)^{-m-n-1} \left (a^2 d^2 f^2 \left (m^2+3 m+2\right )+2 a b d f (m+1) (c f (n+1)-d e (m+n+3))+b^2 \left (-\left (-c^2 f^2 \left (n^2+3 n+2\right )+2 c d e f (n+1) (m+n+3)-d^2 e^2 \left (m^2+m (2 n+5)+n^2+5 n+6\right )\right )\right )\right ) \left (\frac {(c+d x) (b e-a f)}{(e+f x) (b c-a d)}\right )^{-n} \, _2F_1\left (m+1,-n;m+2;-\frac {(d e-c f) (a+b x)}{(b c-a d) (e+f x)}\right )}{(m+1) (m+n+2) (m+n+3) (b e-a f)^3 (d e-c f)^2}-\frac {f (a+b x)^{m+1} (c+d x)^{n+1} (e+f x)^{-m-n-3}}{(m+n+3) (b e-a f) (d e-c f)}+\frac {f (a+b x)^{m+1} (c+d x)^{n+1} (e+f x)^{-m-n-2} (a d f (m+2)+b c f (n+2)-b d e (m+n+4))}{(m+n+2) (m+n+3) (b e-a f)^2 (d e-c f)^2} \]
Antiderivative was successfully verified.
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Rule 12
Rule 129
Rule 132
Rule 155
Rubi steps
\begin {align*} \int (a+b x)^m (c+d x)^n (e+f x)^{-4-m-n} \, dx &=-\frac {f (a+b x)^{1+m} (c+d x)^{1+n} (e+f x)^{-3-m-n}}{(b e-a f) (d e-c f) (3+m+n)}-\frac {\int (a+b x)^m (c+d x)^n (e+f x)^{-3-m-n} (a d f (2+m)+b c f (2+n)-b d e (3+m+n)+b d f x) \, dx}{(b e-a f) (d e-c f) (3+m+n)}\\ &=-\frac {f (a+b x)^{1+m} (c+d x)^{1+n} (e+f x)^{-3-m-n}}{(b e-a f) (d e-c f) (3+m+n)}+\frac {f (a d f (2+m)+b c f (2+n)-b d e (4+m+n)) (a+b x)^{1+m} (c+d x)^{1+n} (e+f x)^{-2-m-n}}{(b e-a f)^2 (d e-c f)^2 (2+m+n) (3+m+n)}+\frac {\int \left (-f (b c (1+m)+a d (1+n)) (a d f (2+m)+b c f (2+n)-b d e (4+m+n))-(2+m+n) \left (a b c d f^2+b d e (a d f (2+m)+b c f (2+n)-b d e (3+m+n))-(b c+a d) f (a d f (2+m)+b c f (2+n)-b d e (3+m+n))\right )\right ) (a+b x)^m (c+d x)^n (e+f x)^{-2-m-n} \, dx}{(b e-a f)^2 (d e-c f)^2 (2+m+n) (3+m+n)}\\ &=-\frac {f (a+b x)^{1+m} (c+d x)^{1+n} (e+f x)^{-3-m-n}}{(b e-a f) (d e-c f) (3+m+n)}+\frac {f (a d f (2+m)+b c f (2+n)-b d e (4+m+n)) (a+b x)^{1+m} (c+d x)^{1+n} (e+f x)^{-2-m-n}}{(b e-a f)^2 (d e-c f)^2 (2+m+n) (3+m+n)}-\frac {\left (f (b c (1+m)+a d (1+n)) (a d f (2+m)+b c f (2+n)-b d e (4+m+n))+(2+m+n) \left (a b c d f^2+b d e (a d f (2+m)+b c f (2+n)-b d e (3+m+n))-(b c+a d) f (a d f (2+m)+b c f (2+n)-b d e (3+m+n))\right )\right ) \int (a+b x)^m (c+d x)^n (e+f x)^{-2-m-n} \, dx}{(b e-a f)^2 (d e-c f)^2 (2+m+n) (3+m+n)}\\ &=-\frac {f (a+b x)^{1+m} (c+d x)^{1+n} (e+f x)^{-3-m-n}}{(b e-a f) (d e-c f) (3+m+n)}+\frac {f (a d f (2+m)+b c f (2+n)-b d e (4+m+n)) (a+b x)^{1+m} (c+d x)^{1+n} (e+f x)^{-2-m-n}}{(b e-a f)^2 (d e-c f)^2 (2+m+n) (3+m+n)}-\frac {\left (f (b c (1+m)+a d (1+n)) (a d f (2+m)+b c f (2+n)-b d e (4+m+n))+(2+m+n) \left (a b c d f^2+b d e (a d f (2+m)+b c f (2+n)-b d e (3+m+n))-(b c+a d) f (a d f (2+m)+b c f (2+n)-b d e (3+m+n))\right )\right ) (a+b x)^{1+m} (c+d x)^n \left (\frac {(b e-a f) (c+d x)}{(b c-a d) (e+f x)}\right )^{-n} (e+f x)^{-1-m-n} \, _2F_1\left (1+m,-n;2+m;-\frac {(d e-c f) (a+b x)}{(b c-a d) (e+f x)}\right )}{(b e-a f)^3 (d e-c f)^2 (1+m) (2+m+n) (3+m+n)}\\ \end {align*}
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Mathematica [A] time = 1.07, size = 346, normalized size = 0.86 \[ -\frac {(a+b x)^{m+1} (c+d x)^n (e+f x)^{-m-n-3} \left (-\frac {(e+f x)^2 \left (a^2 d^2 f^2 \left (m^2+3 m+2\right )-2 a b d f (m+1) (d e (m+n+3)-c f (n+1))+b^2 \left (c^2 f^2 \left (n^2+3 n+2\right )-2 c d e f (n+1) (m+n+3)+d^2 e^2 \left (m^2+m (2 n+5)+n^2+5 n+6\right )\right )\right ) \left (\frac {(c+d x) (b e-a f)}{(e+f x) (b c-a d)}\right )^{-n} \, _2F_1\left (m+1,-n;m+2;\frac {(c f-d e) (a+b x)}{(b c-a d) (e+f x)}\right )}{(m+1) (m+n+2) (b e-a f)^2 (d e-c f)}+\frac {f (c+d x) (e+f x) (-a d f (m+2)-b c f (n+2)+b d e (m+n+4))}{(m+n+2) (b e-a f) (d e-c f)}+f (c+d x)\right )}{(m+n+3) (b e-a f) (d e-c f)} \]
Antiderivative was successfully verified.
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fricas [F] time = 2.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b x + a\right )}^{m} {\left (d x + c\right )}^{n} {\left (f x + e\right )}^{-m - n - 4}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{n} {\left (f x + e\right )}^{-m - n - 4}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.24, size = 0, normalized size = 0.00 \[ \int \left (b x +a \right )^{m} \left (d x +c \right )^{n} \left (f x +e \right )^{-m -n -4}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{n} {\left (f x + e\right )}^{-m - n - 4}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^n}{{\left (e+f\,x\right )}^{m+n+4}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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