Optimal. Leaf size=538 \[ \frac {(d+e x)^{1+m} \left (a B (2 c d-b e)-A \left (b c d-b^2 e+2 a c e\right )+c (b B d-2 A c d+A b e-2 a B e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )}+\frac {c \left (e (b B d-2 A c d+A b e-2 a B e) m-\frac {2 b \left (B c d^2+2 A c d e+a B e^2\right )-b^2 e (B d (2-m)+A e m)-4 c \left (A \left (c d^2+a e^2 (1-m)\right )+a B d e m\right )}{\sqrt {b^2-4 a c}}\right ) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )}{\left (b^2-4 a c\right ) \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) \left (c d^2-b d e+a e^2\right ) (1+m)}+\frac {c \left (e (b B d-2 A c d+A b e-2 a B e) m+\frac {2 b \left (B c d^2+2 A c d e+a B e^2\right )-b^2 e (B d (2-m)+A e m)-4 c \left (A \left (c d^2+a e^2 (1-m)\right )+a B d e m\right )}{\sqrt {b^2-4 a c}}\right ) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{\left (b^2-4 a c\right ) \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (c d^2-b d e+a e^2\right ) (1+m)} \]
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Rubi [A]
time = 2.85, antiderivative size = 536, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {836, 844, 70}
\begin {gather*} \frac {c (d+e x)^{m+1} \left (e m (-2 a B e+A b e-2 A c d+b B d)-\frac {2 b \left (a B e^2+2 A c d e+B c d^2\right )-4 c \left (a A e^2 (1-m)+a B d e m+A c d^2\right )+b^2 (-e) (A e m+B d (2-m))}{\sqrt {b^2-4 a c}}\right ) \, _2F_1\left (1,m+1;m+2;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )}{(m+1) \left (b^2-4 a c\right ) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right ) \left (a e^2-b d e+c d^2\right )}+\frac {c (d+e x)^{m+1} \left (\frac {2 b \left (a B e^2+2 A c d e+B c d^2\right )-4 c \left (a A e^2 (1-m)+a B d e m+A c d^2\right )+b^2 (-e) (A e m+B d (2-m))}{\sqrt {b^2-4 a c}}+e m (-2 a B e+A b e-2 A c d+b B d)\right ) \, _2F_1\left (1,m+1;m+2;\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{(m+1) \left (b^2-4 a c\right ) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right ) \left (a e^2-b d e+c d^2\right )}+\frac {(d+e x)^{m+1} \left (-A \left (2 a c e+b^2 (-e)+b c d\right )+c x (-2 a B e+A b e-2 A c d+b B d)+a B (2 c d-b e)\right )}{\left (b^2-4 a c\right ) \left (a+b x+c x^2\right ) \left (a e^2-b d e+c d^2\right )} \end {gather*}
Antiderivative was successfully verified.
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Rule 70
Rule 836
Rule 844
Rubi steps
\begin {align*} \int \frac {(A+B x) (d+e x)^m}{\left (a+b x+c x^2\right )^2} \, dx &=\frac {(d+e x)^{1+m} \left (a B (2 c d-b e)-A \left (b c d-b^2 e+2 a c e\right )+c (b B d-2 A c d+A b e-2 a B e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )}-\frac {\int \frac {(d+e x)^m \left (b^2 e (B d+A e m)+2 c \left (A c d^2+a A e^2 (1-m)+a B d e m\right )-b \left (B c d^2+a B e^2 (1+m)+A c d e (2+m)\right )+c e (b B d-2 A c d+A b e-2 a B e) m x\right )}{a+b x+c x^2} \, dx}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )}\\ &=\frac {(d+e x)^{1+m} \left (a B (2 c d-b e)-A \left (b c d-b^2 e+2 a c e\right )+c (b B d-2 A c d+A b e-2 a B e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )}-\frac {\int \left (\frac {\left (c e (b B d-2 A c d+A b e-2 a B e) m-\frac {c \left (2 b B c d^2-4 A c^2 d^2-2 b^2 B d e+4 A b c d e+2 a b B e^2-4 a A c e^2+b^2 B d e m-4 a B c d e m-A b^2 e^2 m+4 a A c e^2 m\right )}{\sqrt {b^2-4 a c}}\right ) (d+e x)^m}{b-\sqrt {b^2-4 a c}+2 c x}+\frac {\left (c e (b B d-2 A c d+A b e-2 a B e) m+\frac {c \left (2 b B c d^2-4 A c^2 d^2-2 b^2 B d e+4 A b c d e+2 a b B e^2-4 a A c e^2+b^2 B d e m-4 a B c d e m-A b^2 e^2 m+4 a A c e^2 m\right )}{\sqrt {b^2-4 a c}}\right ) (d+e x)^m}{b+\sqrt {b^2-4 a c}+2 c x}\right ) \, dx}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )}\\ &=\frac {(d+e x)^{1+m} \left (a B (2 c d-b e)-A \left (b c d-b^2 e+2 a c e\right )+c (b B d-2 A c d+A b e-2 a B e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )}-\frac {\left (c \left (e (b B d-2 A c d+A b e-2 a B e) m-\frac {2 b \left (B c d^2+2 A c d e+a B e^2\right )-b^2 e (B d (2-m)+A e m)-4 c \left (A c d^2+a A e^2 (1-m)+a B d e m\right )}{\sqrt {b^2-4 a c}}\right )\right ) \int \frac {(d+e x)^m}{b-\sqrt {b^2-4 a c}+2 c x} \, dx}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )}-\frac {\left (c \left (e (b B d-2 A c d+A b e-2 a B e) m+\frac {2 b \left (B c d^2+2 A c d e+a B e^2\right )-b^2 e (B d (2-m)+A e m)-4 c \left (A c d^2+a A e^2 (1-m)+a B d e m\right )}{\sqrt {b^2-4 a c}}\right )\right ) \int \frac {(d+e x)^m}{b+\sqrt {b^2-4 a c}+2 c x} \, dx}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )}\\ &=\frac {(d+e x)^{1+m} \left (a B (2 c d-b e)-A \left (b c d-b^2 e+2 a c e\right )+c (b B d-2 A c d+A b e-2 a B e) x\right )}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) \left (a+b x+c x^2\right )}+\frac {c \left (e (b B d-2 A c d+A b e-2 a B e) m-\frac {2 b \left (B c d^2+2 A c d e+a B e^2\right )-b^2 e (B d (2-m)+A e m)-4 c \left (A c d^2+a A e^2 (1-m)+a B d e m\right )}{\sqrt {b^2-4 a c}}\right ) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )}{\left (b^2-4 a c\right ) \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) \left (c d^2-b d e+a e^2\right ) (1+m)}+\frac {c \left (e (b B d-2 A c d+A b e-2 a B e) m+\frac {2 b \left (B c d^2+2 A c d e+a B e^2\right )-b^2 e (B d (2-m)+A e m)-4 c \left (A c d^2+a A e^2 (1-m)+a B d e m\right )}{\sqrt {b^2-4 a c}}\right ) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{\left (b^2-4 a c\right ) \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (c d^2-b d e+a e^2\right ) (1+m)}\\ \end {align*}
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Mathematica [A]
time = 4.04, size = 452, normalized size = 0.84 \begin {gather*} \frac {(d+e x)^{1+m} \left (\frac {A b^2 e+b B c d x-2 A c (a e+c d x)+A b c (-d+e x)+a B (-b e+2 c (d-e x))}{a+x (b+c x)}+\frac {c \left (e (b B d-2 A c d+A b e-2 a B e) m+\frac {-2 b \left (B c d^2+2 A c d e+a B e^2\right )+b^2 e (-B d (-2+m)+A e m)+4 c \left (A c d^2-a A e^2 (-1+m)+a B d e m\right )}{\sqrt {b^2-4 a c}}\right ) \, _2F_1\left (1,1+m;2+m;\frac {2 c (d+e x)}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )}{\left (2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e\right ) (1+m)}+\frac {c \left (e (b B d-2 A c d+A b e-2 a B e) m+\frac {2 b \left (B c d^2+2 A c d e+a B e^2\right )+b^2 e (B d (-2+m)-A e m)-4 c \left (A c d^2-a A e^2 (-1+m)+a B d e m\right )}{\sqrt {b^2-4 a c}}\right ) \, _2F_1\left (1,1+m;2+m;\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (1+m)}\right )}{\left (b^2-4 a c\right ) \left (c d^2+e (-b d+a e)\right )} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {\left (B x +A \right ) \left (e x +d \right )^{m}}{\left (c \,x^{2}+b x +a \right )^{2}}\, dx\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^m}{{\left (c\,x^2+b\,x+a\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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