Integrand size = 25, antiderivative size = 539 \[ \int \frac {f+g x}{(d+e x)^2 \left (a+b x+c x^2\right )^2} \, dx=-\frac {e^2 (e f-d g)}{\left (c d^2-b d e+a e^2\right )^2 (d+e x)}-\frac {b^3 e^2 f-b^2 e (2 c d f+a e g)+b c \left (c d^2 f-a e (3 e f-2 d g)\right )+2 a c \left (a e^2 g+c d (2 e f-d g)\right )+c \left (2 c^2 d^2 f+b e^2 (b f-a g)-c (2 a e (e f-2 d g)+b d (2 e f+d g))\right ) x}{\left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right )^2 \left (a+b x+c x^2\right )}+\frac {\left (4 c^4 d^4 f-b^3 e^3 (2 b e f-b d g-a e g)-6 c^2 \left (4 a b d e^3 f-b^2 d^3 e g+2 a^2 e^3 (e f-2 d g)\right )-b c e^2 \left (6 a^2 e^2 g-b^2 d (4 e f-3 d g)-6 a b e (2 e f-d g)\right )+2 c^3 d^2 (4 a e (3 e f-d g)-b d (4 e f+d g))\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c\right )^{3/2} \left (c d^2-b d e+a e^2\right )^3}+\frac {e^2 (c d (4 e f-3 d g)-e (2 b e f-b d g-a e g)) \log (d+e x)}{\left (c d^2-b d e+a e^2\right )^3}-\frac {e^2 (c d (4 e f-3 d g)-e (2 b e f-b d g-a e g)) \log \left (a+b x+c x^2\right )}{2 \left (c d^2-b d e+a e^2\right )^3} \] Output:
-e^2*(-d*g+e*f)/(a*e^2-b*d*e+c*d^2)^2/(e*x+d)-(b^3*e^2*f-b^2*e*(a*e*g+2*c* d*f)+b*c*(c*d^2*f-a*e*(-2*d*g+3*e*f))+2*a*c*(a*e^2*g+c*d*(-d*g+2*e*f))+c*( 2*c^2*d^2*f+b*e^2*(-a*g+b*f)-c*(2*a*e*(-2*d*g+e*f)+b*d*(d*g+2*e*f)))*x)/(- 4*a*c+b^2)/(a*e^2-b*d*e+c*d^2)^2/(c*x^2+b*x+a)+(4*c^4*d^4*f-b^3*e^3*(-a*e* g-b*d*g+2*b*e*f)-6*c^2*(4*a*b*d*e^3*f-b^2*d^3*e*g+2*a^2*e^3*(-2*d*g+e*f))- b*c*e^2*(6*a^2*e^2*g-b^2*d*(-3*d*g+4*e*f)-6*a*b*e*(-d*g+2*e*f))+2*c^3*d^2* (4*a*e*(-d*g+3*e*f)-b*d*(d*g+4*e*f)))*arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2) )/(-4*a*c+b^2)^(3/2)/(a*e^2-b*d*e+c*d^2)^3+e^2*(c*d*(-3*d*g+4*e*f)-e*(-a*e *g-b*d*g+2*b*e*f))*ln(e*x+d)/(a*e^2-b*d*e+c*d^2)^3-1/2*e^2*(c*d*(-3*d*g+4* e*f)-e*(-a*e*g-b*d*g+2*b*e*f))*ln(c*x^2+b*x+a)/(a*e^2-b*d*e+c*d^2)^3
Time = 1.26 (sec) , antiderivative size = 522, normalized size of antiderivative = 0.97 \[ \int \frac {f+g x}{(d+e x)^2 \left (a+b x+c x^2\right )^2} \, dx=\frac {e^2 (-e f+d g)}{\left (c d^2+e (-b d+a e)\right )^2 (d+e x)}+\frac {-b^3 e^2 f+b^2 e (a e g+c f (2 d-e x))+b c (c d (-d f+2 e f x+d g x)+a e (3 e f-2 d g+e g x))-2 c \left (a^2 e^2 g+c^2 d^2 f x-a c \left (d^2 g+e^2 f x-2 d e (f+g x)\right )\right )}{\left (b^2-4 a c\right ) \left (c d^2+e (-b d+a e)\right )^2 (a+x (b+c x))}-\frac {\left (4 c^4 d^4 f+b^3 e^3 (-2 b e f+b d g+a e g)+6 c^2 \left (-4 a b d e^3 f+b^2 d^3 e g-2 a^2 e^3 (e f-2 d g)\right )+b c e^2 \left (-6 a^2 e^2 g+b^2 d (4 e f-3 d g)+6 a b e (2 e f-d g)\right )-2 c^3 d^2 (4 a e (-3 e f+d g)+b d (4 e f+d g))\right ) \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\left (-b^2+4 a c\right )^{3/2} \left (-c d^2+e (b d-a e)\right )^3}-\frac {e^2 (c d (-4 e f+3 d g)+e (2 b e f-b d g-a e g)) \log (d+e x)}{\left (c d^2+e (-b d+a e)\right )^3}+\frac {e^2 (c d (-4 e f+3 d g)+e (2 b e f-b d g-a e g)) \log (a+x (b+c x))}{2 \left (c d^2+e (-b d+a e)\right )^3} \] Input:
Integrate[(f + g*x)/((d + e*x)^2*(a + b*x + c*x^2)^2),x]
Output:
(e^2*(-(e*f) + d*g))/((c*d^2 + e*(-(b*d) + a*e))^2*(d + e*x)) + (-(b^3*e^2 *f) + b^2*e*(a*e*g + c*f*(2*d - e*x)) + b*c*(c*d*(-(d*f) + 2*e*f*x + d*g*x ) + a*e*(3*e*f - 2*d*g + e*g*x)) - 2*c*(a^2*e^2*g + c^2*d^2*f*x - a*c*(d^2 *g + e^2*f*x - 2*d*e*(f + g*x))))/((b^2 - 4*a*c)*(c*d^2 + e*(-(b*d) + a*e) )^2*(a + x*(b + c*x))) - ((4*c^4*d^4*f + b^3*e^3*(-2*b*e*f + b*d*g + a*e*g ) + 6*c^2*(-4*a*b*d*e^3*f + b^2*d^3*e*g - 2*a^2*e^3*(e*f - 2*d*g)) + b*c*e ^2*(-6*a^2*e^2*g + b^2*d*(4*e*f - 3*d*g) + 6*a*b*e*(2*e*f - d*g)) - 2*c^3* d^2*(4*a*e*(-3*e*f + d*g) + b*d*(4*e*f + d*g)))*ArcTan[(b + 2*c*x)/Sqrt[-b ^2 + 4*a*c]])/((-b^2 + 4*a*c)^(3/2)*(-(c*d^2) + e*(b*d - a*e))^3) - (e^2*( c*d*(-4*e*f + 3*d*g) + e*(2*b*e*f - b*d*g - a*e*g))*Log[d + e*x])/(c*d^2 + e*(-(b*d) + a*e))^3 + (e^2*(c*d*(-4*e*f + 3*d*g) + e*(2*b*e*f - b*d*g - a *e*g))*Log[a + x*(b + c*x)])/(2*(c*d^2 + e*(-(b*d) + a*e))^3)
Time = 2.40 (sec) , antiderivative size = 570, normalized size of antiderivative = 1.06, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {1235, 1200, 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.
\(\displaystyle \int \frac {f+g x}{(d+e x)^2 \left (a+b x+c x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 1235 |
\(\displaystyle -\frac {\int \frac {2 c^2 f d^2-b c g d^2+2 a c e (3 e f-2 d g)-b e (2 b e f-b d g-a e g)+2 c e (2 c d f+2 a e g-b (e f+d g)) x}{(d+e x)^2 \left (c x^2+b x+a\right )}dx}{\left (b^2-4 a c\right ) \left (a e^2-b d e+c d^2\right )}-\frac {c x (2 a e g-b (d g+e f)+2 c d f)+a b e g-2 a c d g+2 a c e f+b^2 (-e) f+b c d f}{\left (b^2-4 a c\right ) (d+e x) \left (a+b x+c x^2\right ) \left (a e^2-b d e+c d^2\right )}\) |
\(\Big \downarrow \) 1200 |
\(\displaystyle -\frac {\int \left (\frac {\left (b^2-4 a c\right ) (e (2 b e f-b d g-a e g)-c d (4 e f-3 d g)) e^3}{\left (c d^2-b e d+a e^2\right )^2 (d+e x)}+\frac {\left (-2 c^2 f d^2+b c (2 e f+d g) d+2 a c e (3 e f-4 d g)-b e (2 b e f-b d g-a e g)\right ) e^2}{\left (c d^2-b e d+a e^2\right ) (d+e x)^2}+\frac {2 c^4 f d^4+c^3 (4 a e (3 e f-d g)-b d (4 e f+d g)) d^2-b^3 e^3 (2 b e f-b d g-a e g)+c^2 e \left (3 b^2 g d^3-2 a b e (10 e f-3 d g) d-6 a^2 e^2 (e f-2 d g)\right )-b c e^2 \left (-d (4 e f-3 d g) b^2-5 a e (2 e f-d g) b+5 a^2 e^2 g\right )+c \left (b^2-4 a c\right ) e^2 (c d (4 e f-3 d g)-e (2 b e f-b d g-a e g)) x}{\left (c d^2-b e d+a e^2\right )^2 \left (c x^2+b x+a\right )}\right )dx}{\left (b^2-4 a c\right ) \left (a e^2-b d e+c d^2\right )}-\frac {c x (2 a e g-b (d g+e f)+2 c d f)+a b e g-2 a c d g+2 a c e f+b^2 (-e) f+b c d f}{\left (b^2-4 a c\right ) (d+e x) \left (a+b x+c x^2\right ) \left (a e^2-b d e+c d^2\right )}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {\text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \left (-6 c^2 \left (2 a^2 e^3 (e f-2 d g)+4 a b d e^3 f-b^2 d^3 e g\right )-b c e^2 \left (6 a^2 e^2 g-6 a b e (2 e f-d g)+b^2 (-d) (4 e f-3 d g)\right )-b^3 e^3 (-a e g-b d g+2 b e f)+2 c^3 d^2 (4 a e (3 e f-d g)-b d (d g+4 e f))+4 c^4 d^4 f\right )}{\sqrt {b^2-4 a c} \left (a e^2-b d e+c d^2\right )^2}+\frac {e^2 \left (b^2-4 a c\right ) \log \left (a+b x+c x^2\right ) (c d (4 e f-3 d g)-e (-a e g-b d g+2 b e f))}{2 \left (a e^2-b d e+c d^2\right )^2}-\frac {e^2 \left (b^2-4 a c\right ) \log (d+e x) (c d (4 e f-3 d g)-e (-a e g-b d g+2 b e f))}{\left (a e^2-b d e+c d^2\right )^2}+\frac {e \left (b e (-a e g-b d g+2 b e f)-2 a c e (3 e f-4 d g)-b c d (d g+2 e f)+2 c^2 d^2 f\right )}{(d+e x) \left (a e^2-b d e+c d^2\right )}}{\left (b^2-4 a c\right ) \left (a e^2-b d e+c d^2\right )}-\frac {c x (2 a e g-b (d g+e f)+2 c d f)+a b e g-2 a c d g+2 a c e f+b^2 (-e) f+b c d f}{\left (b^2-4 a c\right ) (d+e x) \left (a+b x+c x^2\right ) \left (a e^2-b d e+c d^2\right )}\) |
Input:
Int[(f + g*x)/((d + e*x)^2*(a + b*x + c*x^2)^2),x]
Output:
-((b*c*d*f - b^2*e*f + 2*a*c*e*f - 2*a*c*d*g + a*b*e*g + c*(2*c*d*f + 2*a* e*g - b*(e*f + d*g))*x)/((b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)*(d + e*x)*( a + b*x + c*x^2))) - ((e*(2*c^2*d^2*f - 2*a*c*e*(3*e*f - 4*d*g) - b*c*d*(2 *e*f + d*g) + b*e*(2*b*e*f - b*d*g - a*e*g)))/((c*d^2 - b*d*e + a*e^2)*(d + e*x)) - ((4*c^4*d^4*f - b^3*e^3*(2*b*e*f - b*d*g - a*e*g) - 6*c^2*(4*a*b *d*e^3*f - b^2*d^3*e*g + 2*a^2*e^3*(e*f - 2*d*g)) - b*c*e^2*(6*a^2*e^2*g - b^2*d*(4*e*f - 3*d*g) - 6*a*b*e*(2*e*f - d*g)) + 2*c^3*d^2*(4*a*e*(3*e*f - d*g) - b*d*(4*e*f + d*g)))*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(Sqrt [b^2 - 4*a*c]*(c*d^2 - b*d*e + a*e^2)^2) - ((b^2 - 4*a*c)*e^2*(c*d*(4*e*f - 3*d*g) - e*(2*b*e*f - b*d*g - a*e*g))*Log[d + e*x])/(c*d^2 - b*d*e + a*e ^2)^2 + ((b^2 - 4*a*c)*e^2*(c*d*(4*e*f - 3*d*g) - e*(2*b*e*f - b*d*g - a*e *g))*Log[a + b*x + c*x^2])/(2*(c*d^2 - b*d*e + a*e^2)^2))/((b^2 - 4*a*c)*( c*d^2 - b*d*e + a*e^2))
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* (x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In tegersQ[n]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2 *a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^m *(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d* m + b*e*m) - b*d*(3*c*d - b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p] )
Time = 2.17 (sec) , antiderivative size = 998, normalized size of antiderivative = 1.85
method | result | size |
default | \(\frac {e^{2} \left (a \,e^{2} g +b d e g -2 b \,e^{2} f -3 c \,d^{2} g +4 c d e f \right ) \ln \left (e x +d \right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right )^{3}}+\frac {\left (d g -e f \right ) e^{2}}{\left (a \,e^{2}-b d e +c \,d^{2}\right )^{2} \left (e x +d \right )}-\frac {\frac {\frac {c \left (b g \,e^{4} a^{2}-4 a^{2} c d \,e^{3} g +2 a^{2} c \,e^{4} f -a \,b^{2} d \,e^{3} g -a \,b^{2} e^{4} f +6 a b c \,d^{2} e^{2} g -4 a \,c^{2} d^{3} e g +b^{3} d \,e^{3} f -b^{2} c \,d^{3} e g -3 b^{2} c \,d^{2} e^{2} f +b \,c^{2} d^{4} g +4 b \,c^{2} d^{3} e f -2 d^{4} f \,c^{3}\right ) x}{4 a c -b^{2}}-\frac {2 a^{3} c \,e^{4} g -a^{2} b^{2} e^{4} g -3 a^{2} b c \,e^{4} f +4 a^{2} c^{2} d \,e^{3} f +a \,b^{3} d \,e^{3} g +a \,b^{3} e^{4} f -3 a \,b^{2} c \,d^{2} e^{2} g +a \,b^{2} c d \,e^{3} f +4 a b \,c^{2} d^{3} e g -6 a b \,c^{2} d^{2} e^{2} f -2 a \,c^{3} d^{4} g +4 a \,c^{3} d^{3} e f -b^{4} d \,e^{3} f +3 b^{3} c \,d^{2} e^{2} f -3 b^{2} c^{2} d^{3} e f +b \,c^{3} d^{4} f}{4 a c -b^{2}}}{c \,x^{2}+b x +a}+\frac {\frac {\left (4 a^{2} c^{2} e^{4} g -a \,b^{2} c \,e^{4} g +4 a b \,c^{2} d \,e^{3} g -8 a b \,c^{2} e^{4} f -12 a \,c^{3} d^{2} e^{2} g +16 a \,c^{3} d \,e^{3} f -b^{3} c d \,e^{3} g +2 b^{3} c \,e^{4} f +3 b^{2} c^{2} d^{2} e^{2} g -4 b^{2} c^{2} d \,e^{3} f \right ) \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {2 \left (5 a^{2} b c \,e^{4} g -12 a^{2} c^{2} d \,e^{3} g +6 a^{2} c^{2} e^{4} f -a \,b^{3} e^{4} g +5 a \,b^{2} c d \,e^{3} g -10 a \,b^{2} c \,e^{4} f -6 a b \,c^{2} d^{2} e^{2} g +20 a b \,c^{2} d \,e^{3} f +4 a \,c^{3} d^{3} e g -12 a \,c^{3} d^{2} e^{2} f -b^{4} d \,e^{3} g +2 b^{4} e^{4} f +3 b^{3} c \,d^{2} e^{2} g -4 b^{3} c d \,e^{3} f -3 b^{2} c^{2} d^{3} e g +b \,c^{3} d^{4} g +4 b \,c^{3} d^{3} e f -2 c^{4} d^{4} f -\frac {\left (4 a^{2} c^{2} e^{4} g -a \,b^{2} c \,e^{4} g +4 a b \,c^{2} d \,e^{3} g -8 a b \,c^{2} e^{4} f -12 a \,c^{3} d^{2} e^{2} g +16 a \,c^{3} d \,e^{3} f -b^{3} c d \,e^{3} g +2 b^{3} c \,e^{4} f +3 b^{2} c^{2} d^{2} e^{2} g -4 b^{2} c^{2} d \,e^{3} f \right ) b}{2 c}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{4 a c -b^{2}}}{\left (a \,e^{2}-b d e +c \,d^{2}\right )^{3}}\) | \(998\) |
risch | \(\text {Expression too large to display}\) | \(6254\) |
Input:
int((g*x+f)/(e*x+d)^2/(c*x^2+b*x+a)^2,x,method=_RETURNVERBOSE)
Output:
e^2*(a*e^2*g+b*d*e*g-2*b*e^2*f-3*c*d^2*g+4*c*d*e*f)/(a*e^2-b*d*e+c*d^2)^3* ln(e*x+d)+(d*g-e*f)*e^2/(a*e^2-b*d*e+c*d^2)^2/(e*x+d)-1/(a*e^2-b*d*e+c*d^2 )^3*((c*(a^2*b*e^4*g-4*a^2*c*d*e^3*g+2*a^2*c*e^4*f-a*b^2*d*e^3*g-a*b^2*e^4 *f+6*a*b*c*d^2*e^2*g-4*a*c^2*d^3*e*g+b^3*d*e^3*f-b^2*c*d^3*e*g-3*b^2*c*d^2 *e^2*f+b*c^2*d^4*g+4*b*c^2*d^3*e*f-2*c^3*d^4*f)/(4*a*c-b^2)*x-(2*a^3*c*e^4 *g-a^2*b^2*e^4*g-3*a^2*b*c*e^4*f+4*a^2*c^2*d*e^3*f+a*b^3*d*e^3*g+a*b^3*e^4 *f-3*a*b^2*c*d^2*e^2*g+a*b^2*c*d*e^3*f+4*a*b*c^2*d^3*e*g-6*a*b*c^2*d^2*e^2 *f-2*a*c^3*d^4*g+4*a*c^3*d^3*e*f-b^4*d*e^3*f+3*b^3*c*d^2*e^2*f-3*b^2*c^2*d ^3*e*f+b*c^3*d^4*f)/(4*a*c-b^2))/(c*x^2+b*x+a)+1/(4*a*c-b^2)*(1/2*(4*a^2*c ^2*e^4*g-a*b^2*c*e^4*g+4*a*b*c^2*d*e^3*g-8*a*b*c^2*e^4*f-12*a*c^3*d^2*e^2* g+16*a*c^3*d*e^3*f-b^3*c*d*e^3*g+2*b^3*c*e^4*f+3*b^2*c^2*d^2*e^2*g-4*b^2*c ^2*d*e^3*f)/c*ln(c*x^2+b*x+a)+2*(5*a^2*b*c*e^4*g-12*a^2*c^2*d*e^3*g+6*a^2* c^2*e^4*f-a*b^3*e^4*g+5*a*b^2*c*d*e^3*g-10*a*b^2*c*e^4*f-6*a*b*c^2*d^2*e^2 *g+20*a*b*c^2*d*e^3*f+4*a*c^3*d^3*e*g-12*a*c^3*d^2*e^2*f-b^4*d*e^3*g+2*b^4 *e^4*f+3*b^3*c*d^2*e^2*g-4*b^3*c*d*e^3*f-3*b^2*c^2*d^3*e*g+b*c^3*d^4*g+4*b *c^3*d^3*e*f-2*c^4*d^4*f-1/2*(4*a^2*c^2*e^4*g-a*b^2*c*e^4*g+4*a*b*c^2*d*e^ 3*g-8*a*b*c^2*e^4*f-12*a*c^3*d^2*e^2*g+16*a*c^3*d*e^3*f-b^3*c*d*e^3*g+2*b^ 3*c*e^4*f+3*b^2*c^2*d^2*e^2*g-4*b^2*c^2*d*e^3*f)*b/c)/(4*a*c-b^2)^(1/2)*ar ctan((2*c*x+b)/(4*a*c-b^2)^(1/2))))
Timed out. \[ \int \frac {f+g x}{(d+e x)^2 \left (a+b x+c x^2\right )^2} \, dx=\text {Timed out} \] Input:
integrate((g*x+f)/(e*x+d)^2/(c*x^2+b*x+a)^2,x, algorithm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {f+g x}{(d+e x)^2 \left (a+b x+c x^2\right )^2} \, dx=\text {Timed out} \] Input:
integrate((g*x+f)/(e*x+d)**2/(c*x**2+b*x+a)**2,x)
Output:
Timed out
Exception generated. \[ \int \frac {f+g x}{(d+e x)^2 \left (a+b x+c x^2\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((g*x+f)/(e*x+d)^2/(c*x^2+b*x+a)^2,x, algorithm="maxima")
Output:
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*a*c-b^2>0)', see `assume?` for more deta
Leaf count of result is larger than twice the leaf count of optimal. 1249 vs. \(2 (532) = 1064\).
Time = 0.29 (sec) , antiderivative size = 1249, normalized size of antiderivative = 2.32 \[ \int \frac {f+g x}{(d+e x)^2 \left (a+b x+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:
integrate((g*x+f)/(e*x+d)^2/(c*x^2+b*x+a)^2,x, algorithm="giac")
Output:
-1/2*(4*c*d*e^3*f - 2*b*e^4*f - 3*c*d^2*e^2*g + b*d*e^3*g + a*e^4*g)*log(c - 2*c*d/(e*x + d) + c*d^2/(e*x + d)^2 + b*e/(e*x + d) - b*d*e/(e*x + d)^2 + a*e^2/(e*x + d)^2)/(c^3*d^6 - 3*b*c^2*d^5*e + 3*b^2*c*d^4*e^2 + 3*a*c^2 *d^4*e^2 - b^3*d^3*e^3 - 6*a*b*c*d^3*e^3 + 3*a*b^2*d^2*e^4 + 3*a^2*c*d^2*e ^4 - 3*a^2*b*d*e^5 + a^3*e^6) - (e^7*f/(e*x + d) - d*e^6*g/(e*x + d))/(c^2 *d^4*e^4 - 2*b*c*d^3*e^5 + b^2*d^2*e^6 + 2*a*c*d^2*e^6 - 2*a*b*d*e^7 + a^2 *e^8) - (4*c^4*d^4*e^2*f - 8*b*c^3*d^3*e^3*f + 24*a*c^3*d^2*e^4*f + 4*b^3* c*d*e^5*f - 24*a*b*c^2*d*e^5*f - 2*b^4*e^6*f + 12*a*b^2*c*e^6*f - 12*a^2*c ^2*e^6*f - 2*b*c^3*d^4*e^2*g + 6*b^2*c^2*d^3*e^3*g - 8*a*c^3*d^3*e^3*g - 3 *b^3*c*d^2*e^4*g + b^4*d*e^5*g - 6*a*b^2*c*d*e^5*g + 24*a^2*c^2*d*e^5*g + a*b^3*e^6*g - 6*a^2*b*c*e^6*g)*arctan((2*c*d - 2*c*d^2/(e*x + d) - b*e + 2 *b*d*e/(e*x + d) - 2*a*e^2/(e*x + d))/(sqrt(-b^2 + 4*a*c)*e))/((b^2*c^3*d^ 6 - 4*a*c^4*d^6 - 3*b^3*c^2*d^5*e + 12*a*b*c^3*d^5*e + 3*b^4*c*d^4*e^2 - 9 *a*b^2*c^2*d^4*e^2 - 12*a^2*c^3*d^4*e^2 - b^5*d^3*e^3 - 2*a*b^3*c*d^3*e^3 + 24*a^2*b*c^2*d^3*e^3 + 3*a*b^4*d^2*e^4 - 9*a^2*b^2*c*d^2*e^4 - 12*a^3*c^ 2*d^2*e^4 - 3*a^2*b^3*d*e^5 + 12*a^3*b*c*d*e^5 + a^3*b^2*e^6 - 4*a^4*c*e^6 )*sqrt(-b^2 + 4*a*c)*e^2) - ((2*c^4*d^3*e*f - 3*b*c^3*d^2*e^2*f + 3*b^2*c^ 2*d*e^3*f - 6*a*c^3*d*e^3*f - b^3*c*e^4*f + 3*a*b*c^2*e^4*f - b*c^3*d^3*e* g + 6*a*c^3*d^2*e^2*g - 3*a*b*c^2*d*e^3*g + a*b^2*c*e^4*g - 2*a^2*c^2*e^4* g)/(c*d^2 - b*d*e + a*e^2) - (2*c^4*d^4*e^2*f - 4*b*c^3*d^3*e^3*f + 6*b...
Time = 20.14 (sec) , antiderivative size = 16955, normalized size of antiderivative = 31.46 \[ \int \frac {f+g x}{(d+e x)^2 \left (a+b x+c x^2\right )^2} \, dx=\text {Too large to display} \] Input:
int((f + g*x)/((d + e*x)^2*(a + b*x + c*x^2)^2),x)
Output:
symsum(log((4*b^5*c^2*e^7*f^2 + 4*c^7*d^5*e^2*f^2 + a^2*b^3*c^2*e^7*g^2 - 32*a^2*c^5*d^3*e^4*g^2 + 4*b^2*c^5*d^3*e^4*f^2 + 4*b^3*c^4*d^2*e^5*f^2 + b ^2*c^5*d^5*e^2*g^2 - b^3*c^4*d^4*e^3*g^2 - b^4*c^3*d^3*e^4*g^2 + b^5*c^2*d ^2*e^5*g^2 - 24*a^3*c^4*e^7*f*g - 28*a*b^3*c^3*e^7*f^2 + 48*a^2*b*c^4*e^7* f^2 - 4*a^3*b*c^3*e^7*g^2 + 8*a*c^6*d^3*e^4*f^2 - 60*a^2*c^5*d*e^6*f^2 - 8 *b*c^6*d^4*e^3*f^2 + 32*a^3*c^4*d*e^6*g^2 - 8*b^4*c^3*d*e^6*f^2 - 24*a*b*c ^5*d^2*e^5*f^2 + 48*a*b^2*c^4*d*e^6*f^2 - 4*a*b*c^5*d^4*e^3*g^2 + 2*a*b^4* c^2*d*e^6*g^2 + 22*a^2*b^2*c^3*e^7*f*g + 112*a^2*c^5*d^2*e^5*f*g + 6*b^2*c ^5*d^4*e^3*f*g - 4*b^3*c^4*d^3*e^4*f*g + 6*b^4*c^3*d^2*e^5*f*g + 18*a*b^2* c^4*d^3*e^4*g^2 - 12*a*b^3*c^3*d^2*e^5*g^2 + 24*a^2*b*c^4*d^2*e^5*g^2 - 15 *a^2*b^2*c^3*d*e^6*g^2 - 4*a*b^4*c^2*e^7*f*g + 8*a*c^6*d^4*e^3*f*g - 4*b*c ^6*d^5*e^2*f*g - 4*b^5*c^2*d*e^6*f*g - 8*a*b*c^5*d^3*e^4*f*g + 40*a*b^3*c^ 3*d*e^6*f*g - 100*a^2*b*c^4*d*e^6*f*g - 44*a*b^2*c^4*d^2*e^5*f*g)/(16*a^2* c^6*d^8 + a^4*b^4*e^8 + 16*a^6*c^2*e^8 + b^4*c^4*d^8 + b^8*d^4*e^4 - 8*a*b ^2*c^5*d^8 - 8*a^5*b^2*c*e^8 - 4*a*b^7*d^3*e^5 - 4*a^3*b^5*d*e^7 - 4*b^5*c ^3*d^7*e - 4*b^7*c*d^5*e^3 + 6*a^2*b^6*d^2*e^6 + 64*a^3*c^5*d^6*e^2 + 96*a ^4*c^4*d^4*e^4 + 64*a^5*c^3*d^2*e^6 + 6*b^6*c^2*d^6*e^2 + 64*a^2*b^2*c^4*d ^6*e^2 + 32*a^2*b^3*c^3*d^5*e^3 - 74*a^2*b^4*c^2*d^4*e^4 + 144*a^3*b^2*c^3 *d^4*e^4 + 32*a^3*b^3*c^2*d^3*e^5 + 64*a^4*b^2*c^2*d^2*e^6 + 32*a*b^3*c^4* d^7*e + 4*a*b^6*c*d^4*e^4 - 64*a^2*b*c^5*d^7*e + 32*a^4*b^3*c*d*e^7 - 6...
Time = 0.23 (sec) , antiderivative size = 13839, normalized size of antiderivative = 25.68 \[ \int \frac {f+g x}{(d+e x)^2 \left (a+b x+c x^2\right )^2} \, dx =\text {Too large to display} \] Input:
int((g*x+f)/(e*x+d)^2/(c*x^2+b*x+a)^2,x)
Output:
( - 12*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**3*b**2*c *d*e**5*g - 12*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a** 3*b**2*c*e**6*g*x + 36*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b* *2))*a**3*b*c**2*d**2*e**4*g - 24*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt (4*a*c - b**2))*a**3*b*c**2*d*e**5*f + 36*sqrt(4*a*c - b**2)*atan((b + 2*c *x)/sqrt(4*a*c - b**2))*a**3*b*c**2*d*e**5*g*x - 24*sqrt(4*a*c - b**2)*ata n((b + 2*c*x)/sqrt(4*a*c - b**2))*a**3*b*c**2*e**6*f*x + 48*sqrt(4*a*c - b **2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**3*c**3*d**3*e**3*g - 24*sqrt( 4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**3*c**3*d**2*e**4*f + 48*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**3*c**3*d**2 *e**4*g*x - 24*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a** 3*c**3*d*e**5*f*x + 2*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b** 2))*a**2*b**4*d*e**5*g + 2*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**2*b**4*e**6*g*x - 10*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt( 4*a*c - b**2))*a**2*b**3*c*d**2*e**4*g + 24*sqrt(4*a*c - b**2)*atan((b + 2 *c*x)/sqrt(4*a*c - b**2))*a**2*b**3*c*d*e**5*f - 22*sqrt(4*a*c - b**2)*ata n((b + 2*c*x)/sqrt(4*a*c - b**2))*a**2*b**3*c*d*e**5*g*x + 24*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**2*b**3*c*e**6*f*x - 12*sqrt (4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**2*b**3*c*e**6*g*x** 2 - 12*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a**2*b**...