Integrand size = 28, antiderivative size = 184 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{d+e x+f x^2} \, dx=-\frac {(B c e-b B f-A c f) x}{f^2}+\frac {B c x^2}{2 f}-\frac {\left (A f \left (c e^2-2 c d f-b e f+2 a f^2\right )+B \left (f \left (b e^2-2 b d f-a e f\right )-c \left (e^3-3 d e f\right )\right )\right ) \text {arctanh}\left (\frac {e+2 f x}{\sqrt {e^2-4 d f}}\right )}{f^3 \sqrt {e^2-4 d f}}-\frac {\left (A f (c e-b f)-B \left (c e^2-c d f-b e f+a f^2\right )\right ) \log \left (d+e x+f x^2\right )}{2 f^3} \]
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Time = 0.22 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {1642, 648, 632, 212, 642} \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{d+e x+f x^2} \, dx=-\frac {\text {arctanh}\left (\frac {e+2 f x}{\sqrt {e^2-4 d f}}\right ) \left (A f \left (2 a f^2-b e f-2 c d f+c e^2\right )+B f \left (-a e f-2 b d f+b e^2\right )-B c \left (e^3-3 d e f\right )\right )}{f^3 \sqrt {e^2-4 d f}}-\frac {\log \left (d+e x+f x^2\right ) \left (B f (b e-a f)+A f (c e-b f)-B c \left (e^2-d f\right )\right )}{2 f^3}-\frac {x (-A c f-b B f+B c e)}{f^2}+\frac {B c x^2}{2 f} \]
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Rule 212
Rule 632
Rule 642
Rule 648
Rule 1642
Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {B c e-b B f-A c f}{f^2}+\frac {B c x}{f}+\frac {-A f (c d-a f)+B d (c e-b f)-\left (B f (b e-a f)+A f (c e-b f)-B c \left (e^2-d f\right )\right ) x}{f^2 \left (d+e x+f x^2\right )}\right ) \, dx \\ & = -\frac {(B c e-b B f-A c f) x}{f^2}+\frac {B c x^2}{2 f}+\frac {\int \frac {-A f (c d-a f)+B d (c e-b f)-\left (B f (b e-a f)+A f (c e-b f)-B c \left (e^2-d f\right )\right ) x}{d+e x+f x^2} \, dx}{f^2} \\ & = -\frac {(B c e-b B f-A c f) x}{f^2}+\frac {B c x^2}{2 f}+\frac {\left (-B f (b e-a f)-A f (c e-b f)+B c \left (e^2-d f\right )\right ) \int \frac {e+2 f x}{d+e x+f x^2} \, dx}{2 f^3}+\frac {\left (B f \left (b e^2-2 b d f-a e f\right )-B c \left (e^3-3 d e f\right )+A f \left (c e^2-2 c d f-b e f+2 a f^2\right )\right ) \int \frac {1}{d+e x+f x^2} \, dx}{2 f^3} \\ & = -\frac {(B c e-b B f-A c f) x}{f^2}+\frac {B c x^2}{2 f}-\frac {\left (B f (b e-a f)+A f (c e-b f)-B c \left (e^2-d f\right )\right ) \log \left (d+e x+f x^2\right )}{2 f^3}-\frac {\left (B f \left (b e^2-2 b d f-a e f\right )-B c \left (e^3-3 d e f\right )+A f \left (c e^2-2 c d f-b e f+2 a f^2\right )\right ) \text {Subst}\left (\int \frac {1}{e^2-4 d f-x^2} \, dx,x,e+2 f x\right )}{f^3} \\ & = -\frac {(B c e-b B f-A c f) x}{f^2}+\frac {B c x^2}{2 f}-\frac {\left (B f \left (b e^2-2 b d f-a e f\right )-B c \left (e^3-3 d e f\right )+A f \left (c e^2-2 c d f-b e f+2 a f^2\right )\right ) \tanh ^{-1}\left (\frac {e+2 f x}{\sqrt {e^2-4 d f}}\right )}{f^3 \sqrt {e^2-4 d f}}-\frac {\left (B f (b e-a f)+A f (c e-b f)-B c \left (e^2-d f\right )\right ) \log \left (d+e x+f x^2\right )}{2 f^3} \\ \end{align*}
Time = 0.10 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.95 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{d+e x+f x^2} \, dx=\frac {2 f (-B c e+b B f+A c f) x+B c f^2 x^2-\frac {2 \left (B f \left (-b e^2+2 b d f+a e f\right )+B c \left (e^3-3 d e f\right )+A f \left (-c e^2+2 c d f+b e f-2 a f^2\right )\right ) \arctan \left (\frac {e+2 f x}{\sqrt {-e^2+4 d f}}\right )}{\sqrt {-e^2+4 d f}}+\left (B f (-b e+a f)+A f (-c e+b f)+B c \left (e^2-d f\right )\right ) \log (d+x (e+f x))}{2 f^3} \]
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Time = 0.85 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.03
method | result | size |
default | \(\frac {\frac {1}{2} B c \,x^{2} f +A c f x +B b f x -B c e x}{f^{2}}+\frac {\frac {\left (A b \,f^{2}-A c e f +B a \,f^{2}-B b e f -B c d f +B c \,e^{2}\right ) \ln \left (f \,x^{2}+e x +d \right )}{2 f}+\frac {2 \left (A a \,f^{2}-A c d f -B b d f +B c d e -\frac {\left (A b \,f^{2}-A c e f +B a \,f^{2}-B b e f -B c d f +B c \,e^{2}\right ) e}{2 f}\right ) \arctan \left (\frac {2 f x +e}{\sqrt {4 d f -e^{2}}}\right )}{\sqrt {4 d f -e^{2}}}}{f^{2}}\) | \(190\) |
risch | \(\text {Expression too large to display}\) | \(8247\) |
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Time = 0.31 (sec) , antiderivative size = 583, normalized size of antiderivative = 3.17 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{d+e x+f x^2} \, dx=\left [\frac {{\left (B c e^{2} f^{2} - 4 \, B c d f^{3}\right )} x^{2} - {\left (B c e^{3} - 2 \, A a f^{3} + {\left (2 \, {\left (B b + A c\right )} d + {\left (B a + A b\right )} e\right )} f^{2} - {\left (3 \, B c d e + {\left (B b + A c\right )} e^{2}\right )} f\right )} \sqrt {e^{2} - 4 \, d f} \log \left (\frac {2 \, f^{2} x^{2} + 2 \, e f x + e^{2} - 2 \, d f - \sqrt {e^{2} - 4 \, d f} {\left (2 \, f x + e\right )}}{f x^{2} + e x + d}\right ) - 2 \, {\left (B c e^{3} f + 4 \, {\left (B b + A c\right )} d f^{3} - {\left (4 \, B c d e + {\left (B b + A c\right )} e^{2}\right )} f^{2}\right )} x + {\left (B c e^{4} - 4 \, {\left (B a + A b\right )} d f^{3} + {\left (4 \, B c d^{2} + 4 \, {\left (B b + A c\right )} d e + {\left (B a + A b\right )} e^{2}\right )} f^{2} - {\left (5 \, B c d e^{2} + {\left (B b + A c\right )} e^{3}\right )} f\right )} \log \left (f x^{2} + e x + d\right )}{2 \, {\left (e^{2} f^{3} - 4 \, d f^{4}\right )}}, \frac {{\left (B c e^{2} f^{2} - 4 \, B c d f^{3}\right )} x^{2} + 2 \, {\left (B c e^{3} - 2 \, A a f^{3} + {\left (2 \, {\left (B b + A c\right )} d + {\left (B a + A b\right )} e\right )} f^{2} - {\left (3 \, B c d e + {\left (B b + A c\right )} e^{2}\right )} f\right )} \sqrt {-e^{2} + 4 \, d f} \arctan \left (-\frac {\sqrt {-e^{2} + 4 \, d f} {\left (2 \, f x + e\right )}}{e^{2} - 4 \, d f}\right ) - 2 \, {\left (B c e^{3} f + 4 \, {\left (B b + A c\right )} d f^{3} - {\left (4 \, B c d e + {\left (B b + A c\right )} e^{2}\right )} f^{2}\right )} x + {\left (B c e^{4} - 4 \, {\left (B a + A b\right )} d f^{3} + {\left (4 \, B c d^{2} + 4 \, {\left (B b + A c\right )} d e + {\left (B a + A b\right )} e^{2}\right )} f^{2} - {\left (5 \, B c d e^{2} + {\left (B b + A c\right )} e^{3}\right )} f\right )} \log \left (f x^{2} + e x + d\right )}{2 \, {\left (e^{2} f^{3} - 4 \, d f^{4}\right )}}\right ] \]
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Leaf count of result is larger than twice the leaf count of optimal. 1260 vs. \(2 (175) = 350\).
Time = 6.33 (sec) , antiderivative size = 1260, normalized size of antiderivative = 6.85 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{d+e x+f x^2} \, dx=\frac {B c x^{2}}{2 f} + x \left (\frac {A c}{f} + \frac {B b}{f} - \frac {B c e}{f^{2}}\right ) + \left (- \frac {\sqrt {- 4 d f + e^{2}} \left (- 2 A a f^{3} + A b e f^{2} + 2 A c d f^{2} - A c e^{2} f + B a e f^{2} + 2 B b d f^{2} - B b e^{2} f - 3 B c d e f + B c e^{3}\right )}{2 f^{3} \cdot \left (4 d f - e^{2}\right )} + \frac {A b f^{2} - A c e f + B a f^{2} - B b e f - B c d f + B c e^{2}}{2 f^{3}}\right ) \log {\left (x + \frac {- A a e f^{2} + 2 A b d f^{2} - A c d e f + 2 B a d f^{2} - B b d e f - 2 B c d^{2} f + B c d e^{2} - 4 d f^{3} \left (- \frac {\sqrt {- 4 d f + e^{2}} \left (- 2 A a f^{3} + A b e f^{2} + 2 A c d f^{2} - A c e^{2} f + B a e f^{2} + 2 B b d f^{2} - B b e^{2} f - 3 B c d e f + B c e^{3}\right )}{2 f^{3} \cdot \left (4 d f - e^{2}\right )} + \frac {A b f^{2} - A c e f + B a f^{2} - B b e f - B c d f + B c e^{2}}{2 f^{3}}\right ) + e^{2} f^{2} \left (- \frac {\sqrt {- 4 d f + e^{2}} \left (- 2 A a f^{3} + A b e f^{2} + 2 A c d f^{2} - A c e^{2} f + B a e f^{2} + 2 B b d f^{2} - B b e^{2} f - 3 B c d e f + B c e^{3}\right )}{2 f^{3} \cdot \left (4 d f - e^{2}\right )} + \frac {A b f^{2} - A c e f + B a f^{2} - B b e f - B c d f + B c e^{2}}{2 f^{3}}\right )}{- 2 A a f^{3} + A b e f^{2} + 2 A c d f^{2} - A c e^{2} f + B a e f^{2} + 2 B b d f^{2} - B b e^{2} f - 3 B c d e f + B c e^{3}} \right )} + \left (\frac {\sqrt {- 4 d f + e^{2}} \left (- 2 A a f^{3} + A b e f^{2} + 2 A c d f^{2} - A c e^{2} f + B a e f^{2} + 2 B b d f^{2} - B b e^{2} f - 3 B c d e f + B c e^{3}\right )}{2 f^{3} \cdot \left (4 d f - e^{2}\right )} + \frac {A b f^{2} - A c e f + B a f^{2} - B b e f - B c d f + B c e^{2}}{2 f^{3}}\right ) \log {\left (x + \frac {- A a e f^{2} + 2 A b d f^{2} - A c d e f + 2 B a d f^{2} - B b d e f - 2 B c d^{2} f + B c d e^{2} - 4 d f^{3} \left (\frac {\sqrt {- 4 d f + e^{2}} \left (- 2 A a f^{3} + A b e f^{2} + 2 A c d f^{2} - A c e^{2} f + B a e f^{2} + 2 B b d f^{2} - B b e^{2} f - 3 B c d e f + B c e^{3}\right )}{2 f^{3} \cdot \left (4 d f - e^{2}\right )} + \frac {A b f^{2} - A c e f + B a f^{2} - B b e f - B c d f + B c e^{2}}{2 f^{3}}\right ) + e^{2} f^{2} \left (\frac {\sqrt {- 4 d f + e^{2}} \left (- 2 A a f^{3} + A b e f^{2} + 2 A c d f^{2} - A c e^{2} f + B a e f^{2} + 2 B b d f^{2} - B b e^{2} f - 3 B c d e f + B c e^{3}\right )}{2 f^{3} \cdot \left (4 d f - e^{2}\right )} + \frac {A b f^{2} - A c e f + B a f^{2} - B b e f - B c d f + B c e^{2}}{2 f^{3}}\right )}{- 2 A a f^{3} + A b e f^{2} + 2 A c d f^{2} - A c e^{2} f + B a e f^{2} + 2 B b d f^{2} - B b e^{2} f - 3 B c d e f + B c e^{3}} \right )} \]
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Exception generated. \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{d+e x+f x^2} \, dx=\text {Exception raised: ValueError} \]
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Time = 0.26 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.03 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{d+e x+f x^2} \, dx=\frac {B c f x^{2} - 2 \, B c e x + 2 \, B b f x + 2 \, A c f x}{2 \, f^{2}} + \frac {{\left (B c e^{2} - B c d f - B b e f - A c e f + B a f^{2} + A b f^{2}\right )} \log \left (f x^{2} + e x + d\right )}{2 \, f^{3}} - \frac {{\left (B c e^{3} - 3 \, B c d e f - B b e^{2} f - A c e^{2} f + 2 \, B b d f^{2} + 2 \, A c d f^{2} + B a e f^{2} + A b e f^{2} - 2 \, A a f^{3}\right )} \arctan \left (\frac {2 \, f x + e}{\sqrt {-e^{2} + 4 \, d f}}\right )}{\sqrt {-e^{2} + 4 \, d f} f^{3}} \]
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Time = 13.66 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.48 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{d+e x+f x^2} \, dx=x\,\left (\frac {A\,c+B\,b}{f}-\frac {B\,c\,e}{f^2}\right )-\frac {\ln \left (f\,x^2+e\,x+d\right )\,\left (B\,c\,e^4-4\,A\,b\,d\,f^3-4\,B\,a\,d\,f^3-A\,c\,e^3\,f-B\,b\,e^3\,f+A\,b\,e^2\,f^2+B\,a\,e^2\,f^2+4\,B\,c\,d^2\,f^2+4\,A\,c\,d\,e\,f^2+4\,B\,b\,d\,e\,f^2-5\,B\,c\,d\,e^2\,f\right )}{2\,\left (4\,d\,f^4-e^2\,f^3\right )}-\frac {\mathrm {atan}\left (\frac {e}{\sqrt {4\,d\,f-e^2}}+\frac {2\,f\,x}{\sqrt {4\,d\,f-e^2}}\right )\,\left (B\,c\,e^3-2\,A\,a\,f^3+A\,b\,e\,f^2+2\,A\,c\,d\,f^2+B\,a\,e\,f^2+2\,B\,b\,d\,f^2-A\,c\,e^2\,f-B\,b\,e^2\,f-3\,B\,c\,d\,e\,f\right )}{f^3\,\sqrt {4\,d\,f-e^2}}+\frac {B\,c\,x^2}{2\,f} \]
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