Integrand size = 28, antiderivative size = 184 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{d+e x+f x^2} \, dx=-\frac {(B c e-b B f-A c f) x}{f^2}+\frac {B c x^2}{2 f}-\frac {\left (A f \left (c e^2-2 c d f-b e f+2 a f^2\right )+B \left (f \left (b e^2-2 b d f-a e f\right )-c \left (e^3-3 d e f\right )\right )\right ) \text {arctanh}\left (\frac {e+2 f x}{\sqrt {e^2-4 d f}}\right )}{f^3 \sqrt {e^2-4 d f}}-\frac {\left (A f (c e-b f)-B \left (c e^2-c d f-b e f+a f^2\right )\right ) \log \left (d+e x+f x^2\right )}{2 f^3} \]
-(-A*c*f-B*b*f+B*c*e)*x/f^2+1/2*B*c*x^2/f-1/2*(A*f*(-b*f+c*e)-B*(a*f^2-b*e *f-c*d*f+c*e^2))*ln(f*x^2+e*x+d)/f^3-(A*f*(2*a*f^2-b*e*f-2*c*d*f+c*e^2)+B* (f*(-a*e*f-2*b*d*f+b*e^2)-c*(-3*d*e*f+e^3)))*arctanh((2*f*x+e)/(-4*d*f+e^2 )^(1/2))/f^3/(-4*d*f+e^2)^(1/2)
Time = 0.10 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.95 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{d+e x+f x^2} \, dx=\frac {2 f (-B c e+b B f+A c f) x+B c f^2 x^2-\frac {2 \left (B f \left (-b e^2+2 b d f+a e f\right )+B c \left (e^3-3 d e f\right )+A f \left (-c e^2+2 c d f+b e f-2 a f^2\right )\right ) \arctan \left (\frac {e+2 f x}{\sqrt {-e^2+4 d f}}\right )}{\sqrt {-e^2+4 d f}}+\left (B f (-b e+a f)+A f (-c e+b f)+B c \left (e^2-d f\right )\right ) \log (d+x (e+f x))}{2 f^3} \]
(2*f*(-(B*c*e) + b*B*f + A*c*f)*x + B*c*f^2*x^2 - (2*(B*f*(-(b*e^2) + 2*b* d*f + a*e*f) + B*c*(e^3 - 3*d*e*f) + A*f*(-(c*e^2) + 2*c*d*f + b*e*f - 2*a *f^2))*ArcTan[(e + 2*f*x)/Sqrt[-e^2 + 4*d*f]])/Sqrt[-e^2 + 4*d*f] + (B*f*( -(b*e) + a*f) + A*f*(-(c*e) + b*f) + B*c*(e^2 - d*f))*Log[d + x*(e + f*x)] )/(2*f^3)
Time = 0.47 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.99, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2159, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {(A+B x) \left (a+b x+c x^2\right )}{d+e x+f x^2} \, dx\) |
\(\Big \downarrow \) 2159 |
\(\displaystyle \int \left (\frac {-x \left (B f (b e-a f)+A f (c e-b f)-B c \left (e^2-d f\right )\right )-A f (c d-a f)+B d (c e-b f)}{f^2 \left (d+e x+f x^2\right )}-\frac {-A c f-b B f+B c e}{f^2}+\frac {B c x}{f}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\text {arctanh}\left (\frac {e+2 f x}{\sqrt {e^2-4 d f}}\right ) \left (A f \left (2 a f^2-b e f-2 c d f+c e^2\right )+B f \left (-a e f-2 b d f+b e^2\right )-B c \left (e^3-3 d e f\right )\right )}{f^3 \sqrt {e^2-4 d f}}-\frac {\log \left (d+e x+f x^2\right ) \left (A f (c e-b f)-B \left (a f^2-b e f-c d f+c e^2\right )\right )}{2 f^3}-\frac {x (-A c f-b B f+B c e)}{f^2}+\frac {B c x^2}{2 f}\) |
-(((B*c*e - b*B*f - A*c*f)*x)/f^2) + (B*c*x^2)/(2*f) - ((B*f*(b*e^2 - 2*b* d*f - a*e*f) - B*c*(e^3 - 3*d*e*f) + A*f*(c*e^2 - 2*c*d*f - b*e*f + 2*a*f^ 2))*ArcTanh[(e + 2*f*x)/Sqrt[e^2 - 4*d*f]])/(f^3*Sqrt[e^2 - 4*d*f]) - ((A* f*(c*e - b*f) - B*(c*e^2 - c*d*f - b*e*f + a*f^2))*Log[d + e*x + f*x^2])/( 2*f^3)
3.1.13.3.1 Defintions of rubi rules used
Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*Pq*(a + b*x + c*x^2)^p, x ], x] /; FreeQ[{a, b, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
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 | \(\text {Expression too large to display}\) | \(8247\) |
1/f^2*(1/2*B*c*x^2*f+A*c*f*x+B*b*f*x-B*c*e*x)+1/f^2*(1/2*(A*b*f^2-A*c*e*f+ B*a*f^2-B*b*e*f-B*c*d*f+B*c*e^2)/f*ln(f*x^2+e*x+d)+2*(A*a*f^2-A*c*d*f-B*b* d*f+B*c*d*e-1/2*(A*b*f^2-A*c*e*f+B*a*f^2-B*b*e*f-B*c*d*f+B*c*e^2)*e/f)/(4* d*f-e^2)^(1/2)*arctan((2*f*x+e)/(4*d*f-e^2)^(1/2)))
Time = 0.31 (sec) , antiderivative size = 583, normalized size of antiderivative = 3.17 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{d+e x+f x^2} \, dx=\left [\frac {{\left (B c e^{2} f^{2} - 4 \, B c d f^{3}\right )} x^{2} - {\left (B c e^{3} - 2 \, A a f^{3} + {\left (2 \, {\left (B b + A c\right )} d + {\left (B a + A b\right )} e\right )} f^{2} - {\left (3 \, B c d e + {\left (B b + A c\right )} e^{2}\right )} f\right )} \sqrt {e^{2} - 4 \, d f} \log \left (\frac {2 \, f^{2} x^{2} + 2 \, e f x + e^{2} - 2 \, d f - \sqrt {e^{2} - 4 \, d f} {\left (2 \, f x + e\right )}}{f x^{2} + e x + d}\right ) - 2 \, {\left (B c e^{3} f + 4 \, {\left (B b + A c\right )} d f^{3} - {\left (4 \, B c d e + {\left (B b + A c\right )} e^{2}\right )} f^{2}\right )} x + {\left (B c e^{4} - 4 \, {\left (B a + A b\right )} d f^{3} + {\left (4 \, B c d^{2} + 4 \, {\left (B b + A c\right )} d e + {\left (B a + A b\right )} e^{2}\right )} f^{2} - {\left (5 \, B c d e^{2} + {\left (B b + A c\right )} e^{3}\right )} f\right )} \log \left (f x^{2} + e x + d\right )}{2 \, {\left (e^{2} f^{3} - 4 \, d f^{4}\right )}}, \frac {{\left (B c e^{2} f^{2} - 4 \, B c d f^{3}\right )} x^{2} + 2 \, {\left (B c e^{3} - 2 \, A a f^{3} + {\left (2 \, {\left (B b + A c\right )} d + {\left (B a + A b\right )} e\right )} f^{2} - {\left (3 \, B c d e + {\left (B b + A c\right )} e^{2}\right )} f\right )} \sqrt {-e^{2} + 4 \, d f} \arctan \left (-\frac {\sqrt {-e^{2} + 4 \, d f} {\left (2 \, f x + e\right )}}{e^{2} - 4 \, d f}\right ) - 2 \, {\left (B c e^{3} f + 4 \, {\left (B b + A c\right )} d f^{3} - {\left (4 \, B c d e + {\left (B b + A c\right )} e^{2}\right )} f^{2}\right )} x + {\left (B c e^{4} - 4 \, {\left (B a + A b\right )} d f^{3} + {\left (4 \, B c d^{2} + 4 \, {\left (B b + A c\right )} d e + {\left (B a + A b\right )} e^{2}\right )} f^{2} - {\left (5 \, B c d e^{2} + {\left (B b + A c\right )} e^{3}\right )} f\right )} \log \left (f x^{2} + e x + d\right )}{2 \, {\left (e^{2} f^{3} - 4 \, d f^{4}\right )}}\right ] \]
[1/2*((B*c*e^2*f^2 - 4*B*c*d*f^3)*x^2 - (B*c*e^3 - 2*A*a*f^3 + (2*(B*b + A *c)*d + (B*a + A*b)*e)*f^2 - (3*B*c*d*e + (B*b + A*c)*e^2)*f)*sqrt(e^2 - 4 *d*f)*log((2*f^2*x^2 + 2*e*f*x + e^2 - 2*d*f - sqrt(e^2 - 4*d*f)*(2*f*x + e))/(f*x^2 + e*x + d)) - 2*(B*c*e^3*f + 4*(B*b + A*c)*d*f^3 - (4*B*c*d*e + (B*b + A*c)*e^2)*f^2)*x + (B*c*e^4 - 4*(B*a + A*b)*d*f^3 + (4*B*c*d^2 + 4 *(B*b + A*c)*d*e + (B*a + A*b)*e^2)*f^2 - (5*B*c*d*e^2 + (B*b + A*c)*e^3)* f)*log(f*x^2 + e*x + d))/(e^2*f^3 - 4*d*f^4), 1/2*((B*c*e^2*f^2 - 4*B*c*d* f^3)*x^2 + 2*(B*c*e^3 - 2*A*a*f^3 + (2*(B*b + A*c)*d + (B*a + A*b)*e)*f^2 - (3*B*c*d*e + (B*b + A*c)*e^2)*f)*sqrt(-e^2 + 4*d*f)*arctan(-sqrt(-e^2 + 4*d*f)*(2*f*x + e)/(e^2 - 4*d*f)) - 2*(B*c*e^3*f + 4*(B*b + A*c)*d*f^3 - ( 4*B*c*d*e + (B*b + A*c)*e^2)*f^2)*x + (B*c*e^4 - 4*(B*a + A*b)*d*f^3 + (4* B*c*d^2 + 4*(B*b + A*c)*d*e + (B*a + A*b)*e^2)*f^2 - (5*B*c*d*e^2 + (B*b + A*c)*e^3)*f)*log(f*x^2 + e*x + d))/(e^2*f^3 - 4*d*f^4)]
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 )} \]
B*c*x**2/(2*f) + x*(A*c/f + B*b/f - B*c*e/f**2) + (-sqrt(-4*d*f + e**2)*(- 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)/(2*f**3*(4*d*f - e**2)) + (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))*log( x + (-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*(-sqrt(-4*d*f + e**2)*(-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)/(2*f**3*(4*d*f - e**2)) + (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)) + e**2*f**2*(-sqrt (-4*d*f + e**2)*(-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)/(2*f**3*(4* d*f - e**2)) + (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)))/(-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)) + (sqrt(-4* d*f + e**2)*(-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)/(2*f**3*(4*d*f - e**2)) + (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))*log(x + (-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*(sqrt(-4*d*f + e**2)*(-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...
Exception generated. \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{d+e x+f x^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*d*f-e^2>0)', see `assume?` for more deta
Time = 0.26 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.03 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{d+e x+f x^2} \, dx=\frac {B c f x^{2} - 2 \, B c e x + 2 \, B b f x + 2 \, A c f x}{2 \, f^{2}} + \frac {{\left (B c e^{2} - B c d f - B b e f - A c e f + B a f^{2} + A b f^{2}\right )} \log \left (f x^{2} + e x + d\right )}{2 \, f^{3}} - \frac {{\left (B c e^{3} - 3 \, B c d e f - B b e^{2} f - A c e^{2} f + 2 \, B b d f^{2} + 2 \, A c d f^{2} + B a e f^{2} + A b e f^{2} - 2 \, A a f^{3}\right )} \arctan \left (\frac {2 \, f x + e}{\sqrt {-e^{2} + 4 \, d f}}\right )}{\sqrt {-e^{2} + 4 \, d f} f^{3}} \]
1/2*(B*c*f*x^2 - 2*B*c*e*x + 2*B*b*f*x + 2*A*c*f*x)/f^2 + 1/2*(B*c*e^2 - B *c*d*f - B*b*e*f - A*c*e*f + B*a*f^2 + A*b*f^2)*log(f*x^2 + e*x + d)/f^3 - (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)*arctan((2*f*x + e)/sqrt(-e^2 + 4*d* f))/(sqrt(-e^2 + 4*d*f)*f^3)
Time = 13.66 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.48 \[ \int \frac {(A+B x) \left (a+b x+c x^2\right )}{d+e x+f x^2} \, dx=x\,\left (\frac {A\,c+B\,b}{f}-\frac {B\,c\,e}{f^2}\right )-\frac {\ln \left (f\,x^2+e\,x+d\right )\,\left (B\,c\,e^4-4\,A\,b\,d\,f^3-4\,B\,a\,d\,f^3-A\,c\,e^3\,f-B\,b\,e^3\,f+A\,b\,e^2\,f^2+B\,a\,e^2\,f^2+4\,B\,c\,d^2\,f^2+4\,A\,c\,d\,e\,f^2+4\,B\,b\,d\,e\,f^2-5\,B\,c\,d\,e^2\,f\right )}{2\,\left (4\,d\,f^4-e^2\,f^3\right )}-\frac {\mathrm {atan}\left (\frac {e}{\sqrt {4\,d\,f-e^2}}+\frac {2\,f\,x}{\sqrt {4\,d\,f-e^2}}\right )\,\left (B\,c\,e^3-2\,A\,a\,f^3+A\,b\,e\,f^2+2\,A\,c\,d\,f^2+B\,a\,e\,f^2+2\,B\,b\,d\,f^2-A\,c\,e^2\,f-B\,b\,e^2\,f-3\,B\,c\,d\,e\,f\right )}{f^3\,\sqrt {4\,d\,f-e^2}}+\frac {B\,c\,x^2}{2\,f} \]
x*((A*c + B*b)/f - (B*c*e)/f^2) - (log(d + e*x + f*x^2)*(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))/(2*(4*d*f^4 - e^2*f^3)) - (atan(e/(4*d*f - e^2)^(1/2) + (2*f*x)/(4*d*f - e^2)^(1/2))* (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))/(f^3*(4*d*f - e^2)^(1/2)) + (B*c*x^ 2)/(2*f)