\(\int \frac {A+B x+C x^2}{(3 a b+3 b^2 x+3 b c x^2+c^2 x^3)^2} \, dx\) [72]

Optimal result

Integrand size = 38, antiderivative size = 484 \[ \int \frac {A+B x+C x^2}{\left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )^2} \, dx=-\frac {C}{3 c^2 \left (b \left (3 a-\frac {b^2}{c}\right )+\frac {(b+c x)^3}{c}\right )}-\frac {(b+c x) \left (A c^2-b^2 C+c (B c-2 b C) x\right )}{3 b c^2 \left (b^2-3 a c\right ) \left (b \left (3 a-\frac {b^2}{c}\right )+\frac {(b+c x)^3}{c}\right )}-\frac {\left (2 b B c-2 A c^2+\sqrt [3]{b} B c \sqrt [3]{b^2-3 a c}-2 b^2 C-2 b^{4/3} \sqrt [3]{b^2-3 a c} C\right ) \arctan \left (\frac {\sqrt [3]{b}+\frac {2 (b+c x)}{\sqrt [3]{b^2-3 a c}}}{\sqrt {3} \sqrt [3]{b}}\right )}{3 \sqrt {3} b^{5/3} c \left (b^2-3 a c\right )^{5/3}}+\frac {\left (2 b B c-2 A c^2-\sqrt [3]{b} B c \sqrt [3]{b^2-3 a c}-2 b^2 C+2 b^{4/3} \sqrt [3]{b^2-3 a c} C\right ) \log \left (\sqrt [3]{b} \left (b^{2/3}-\sqrt [3]{b^2-3 a c}\right )+c x\right )}{9 b^{5/3} c \left (b^2-3 a c\right )^{5/3}}-\frac {\left (2 b B c-2 A c^2-\sqrt [3]{b} B c \sqrt [3]{b^2-3 a c}-2 b^2 C+2 b^{4/3} \sqrt [3]{b^2-3 a c} C\right ) \log \left (b^{2/3} \left (b^2-3 a c\right )^{2/3}+\sqrt [3]{b} \sqrt [3]{b^2-3 a c} (b+c x)+(b+c x)^2\right )}{18 b^{5/3} c \left (b^2-3 a c\right )^{5/3}} \] Output:

-1/3*C/c^2/(b*(3*a-b^2/c)+(c*x+b)^3/c)-1/3*(c*x+b)*(A*c^2-C*b^2+c*(B*c-2*C 
*b)*x)/b/c^2/(-3*a*c+b^2)/(b*(3*a-b^2/c)+(c*x+b)^3/c)-1/9*(2*B*b*c-2*A*c^2 
+b^(1/3)*B*c*(-3*a*c+b^2)^(1/3)-2*C*b^2-2*b^(4/3)*(-3*a*c+b^2)^(1/3)*C)*ar 
ctan(1/3*(b^(1/3)+2*(c*x+b)/(-3*a*c+b^2)^(1/3))*3^(1/2)/b^(1/3))*3^(1/2)/b 
^(5/3)/c/(-3*a*c+b^2)^(5/3)+1/9*(2*B*b*c-2*A*c^2-b^(1/3)*B*c*(-3*a*c+b^2)^ 
(1/3)-2*C*b^2+2*b^(4/3)*(-3*a*c+b^2)^(1/3)*C)*ln(b^(1/3)*(b^(2/3)-(-3*a*c+ 
b^2)^(1/3))+c*x)/b^(5/3)/c/(-3*a*c+b^2)^(5/3)-1/18*(2*B*b*c-2*A*c^2-b^(1/3 
)*B*c*(-3*a*c+b^2)^(1/3)-2*C*b^2+2*b^(4/3)*(-3*a*c+b^2)^(1/3)*C)*ln(b^(2/3 
)*(-3*a*c+b^2)^(2/3)+b^(1/3)*(-3*a*c+b^2)^(1/3)*(c*x+b)+(c*x+b)^2)/b^(5/3) 
/c/(-3*a*c+b^2)^(5/3)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 0.12 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.38 \[ \int \frac {A+B x+C x^2}{\left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )^2} \, dx=\frac {-\frac {3 \left (-3 a b C+A c (b+c x)+x \left (-3 b^2 C+B c^2 x+b c (B-2 C x)\right )\right )}{c \left (3 a b+x \left (3 b^2+3 b c x+c^2 x^2\right )\right )}+\text {RootSum}\left [3 a b+3 b^2 \text {$\#$1}+3 b c \text {$\#$1}^2+c^2 \text {$\#$1}^3\&,\frac {b B \log (x-\text {$\#$1})-2 A c \log (x-\text {$\#$1})-B c \log (x-\text {$\#$1}) \text {$\#$1}+2 b C \log (x-\text {$\#$1}) \text {$\#$1}}{b^2+2 b c \text {$\#$1}+c^2 \text {$\#$1}^2}\&\right ]}{9 \left (b^3-3 a b c\right )} \] Input:

Integrate[(A + B*x + C*x^2)/(3*a*b + 3*b^2*x + 3*b*c*x^2 + c^2*x^3)^2,x]
 

Output:

((-3*(-3*a*b*C + A*c*(b + c*x) + x*(-3*b^2*C + B*c^2*x + b*c*(B - 2*C*x))) 
)/(c*(3*a*b + x*(3*b^2 + 3*b*c*x + c^2*x^2))) + RootSum[3*a*b + 3*b^2*#1 + 
 3*b*c*#1^2 + c^2*#1^3 & , (b*B*Log[x - #1] - 2*A*c*Log[x - #1] - B*c*Log[ 
x - #1]*#1 + 2*b*C*Log[x - #1]*#1)/(b^2 + 2*b*c*#1 + c^2*#1^2) & ])/(9*(b^ 
3 - 3*a*b*c))
 

Rubi [A] (verified)

Time = 2.40 (sec) , antiderivative size = 481, normalized size of antiderivative = 0.99, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {2459, 2393, 25, 2400, 16, 1142, 27, 1082, 217, 1103}

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.

\(\displaystyle \int \frac {A+B x+C x^2}{\left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )^2} \, dx\)

\(\Big \downarrow \) 2459

\(\displaystyle \int \frac {A-\frac {b (B c-b C)}{c^2}+\left (\frac {b}{c}+x\right ) \left (B-\frac {2 b C}{c}\right )+C \left (\frac {b}{c}+x\right )^2}{\left (b \left (3 a-\frac {b^2}{c}\right )+c^2 \left (\frac {b}{c}+x\right )^3\right )^2}d\left (\frac {b}{c}+x\right )\)

\(\Big \downarrow \) 2393

\(\displaystyle \frac {c \int -\frac {2 \left (A-\frac {b (B c-b C)}{c^2}\right )+\left (B-\frac {2 b C}{c}\right ) \left (\frac {b}{c}+x\right )}{c^2 \left (\frac {b}{c}+x\right )^3+b \left (3 a-\frac {b^2}{c}\right )}d\left (\frac {b}{c}+x\right )}{3 b \left (b^2-3 a c\right )}+\frac {b C \left (3 a-\frac {b^2}{c}\right )-c^2 \left (\frac {b}{c}+x\right ) \left (A-\frac {b (B c-b C)}{c^2}+\left (\frac {b}{c}+x\right ) \left (B-\frac {2 b C}{c}\right )\right )}{3 b c \left (b^2-3 a c\right ) \left (b \left (3 a-\frac {b^2}{c}\right )+c^2 \left (\frac {b}{c}+x\right )^3\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {b C \left (3 a-\frac {b^2}{c}\right )-c^2 \left (\frac {b}{c}+x\right ) \left (A-\frac {b (B c-b C)}{c^2}+\left (\frac {b}{c}+x\right ) \left (B-\frac {2 b C}{c}\right )\right )}{3 b c \left (b^2-3 a c\right ) \left (b \left (3 a-\frac {b^2}{c}\right )+c^2 \left (\frac {b}{c}+x\right )^3\right )}-\frac {c \int \frac {2 \left (A-\frac {b (B c-b C)}{c^2}\right )+\left (B-\frac {2 b C}{c}\right ) \left (\frac {b}{c}+x\right )}{c^2 \left (\frac {b}{c}+x\right )^3+b \left (3 a-\frac {b^2}{c}\right )}d\left (\frac {b}{c}+x\right )}{3 b \left (b^2-3 a c\right )}\)

\(\Big \downarrow \) 2400

\(\displaystyle \frac {b C \left (3 a-\frac {b^2}{c}\right )-c^2 \left (\frac {b}{c}+x\right ) \left (A-\frac {b (B c-b C)}{c^2}+\left (\frac {b}{c}+x\right ) \left (B-\frac {2 b C}{c}\right )\right )}{3 b c \left (b^2-3 a c\right ) \left (b \left (3 a-\frac {b^2}{c}\right )+c^2 \left (\frac {b}{c}+x\right )^3\right )}-\frac {c \left (\frac {\left (-\sqrt [3]{b} B c \sqrt [3]{b^2-3 a c}+2 b^{4/3} C \sqrt [3]{b^2-3 a c}-2 A c^2-2 b^2 C+2 b B c\right ) \int \frac {1}{\sqrt [3]{b} \sqrt [3]{b^2-3 a c}-c \left (\frac {b}{c}+x\right )}d\left (\frac {b}{c}+x\right )}{3 b^{2/3} c \left (b^2-3 a c\right )^{2/3}}+\frac {\int \frac {\sqrt [3]{b} \sqrt [3]{b^2-3 a c} \left (\sqrt [3]{b} \sqrt [3]{b^2-3 a c} \left (B-\frac {2 b C}{c}\right )-4 c \left (A-\frac {b (B c-b C)}{c^2}\right )\right )-c \left (2 A c+\sqrt [3]{b} \sqrt [3]{b^2-3 a c} \left (B-\frac {2 b C}{c}\right )-\frac {2 b (B c-b C)}{c}\right ) \left (\frac {b}{c}+x\right )}{c^2 \left (\frac {b}{c}+x\right )^2+\sqrt [3]{b} c \sqrt [3]{b^2-3 a c} \left (\frac {b}{c}+x\right )+b^{2/3} \left (b^2-3 a c\right )^{2/3}}d\left (\frac {b}{c}+x\right )}{3 b^{2/3} \left (b^2-3 a c\right )^{2/3}}\right )}{3 b \left (b^2-3 a c\right )}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {b C \left (3 a-\frac {b^2}{c}\right )-c^2 \left (\frac {b}{c}+x\right ) \left (A-\frac {b (B c-b C)}{c^2}+\left (\frac {b}{c}+x\right ) \left (B-\frac {2 b C}{c}\right )\right )}{3 b c \left (b^2-3 a c\right ) \left (b \left (3 a-\frac {b^2}{c}\right )+c^2 \left (\frac {b}{c}+x\right )^3\right )}-\frac {c \left (\frac {\int \frac {\sqrt [3]{b} \sqrt [3]{b^2-3 a c} \left (\sqrt [3]{b} \sqrt [3]{b^2-3 a c} \left (B-\frac {2 b C}{c}\right )-4 c \left (A-\frac {b (B c-b C)}{c^2}\right )\right )-c \left (2 A c+\sqrt [3]{b} \sqrt [3]{b^2-3 a c} \left (B-\frac {2 b C}{c}\right )-\frac {2 b (B c-b C)}{c}\right ) \left (\frac {b}{c}+x\right )}{c^2 \left (\frac {b}{c}+x\right )^2+\sqrt [3]{b} c \sqrt [3]{b^2-3 a c} \left (\frac {b}{c}+x\right )+b^{2/3} \left (b^2-3 a c\right )^{2/3}}d\left (\frac {b}{c}+x\right )}{3 b^{2/3} \left (b^2-3 a c\right )^{2/3}}-\frac {\log \left (\sqrt [3]{b} \sqrt [3]{b^2-3 a c}-c \left (\frac {b}{c}+x\right )\right ) \left (-\sqrt [3]{b} B c \sqrt [3]{b^2-3 a c}+2 b^{4/3} C \sqrt [3]{b^2-3 a c}-2 A c^2-2 b^2 C+2 b B c\right )}{3 b^{2/3} c^2 \left (b^2-3 a c\right )^{2/3}}\right )}{3 b \left (b^2-3 a c\right )}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {b C \left (3 a-\frac {b^2}{c}\right )-c^2 \left (\frac {b}{c}+x\right ) \left (A-\frac {b (B c-b C)}{c^2}+\left (\frac {b}{c}+x\right ) \left (B-\frac {2 b C}{c}\right )\right )}{3 b c \left (b^2-3 a c\right ) \left (b \left (3 a-\frac {b^2}{c}\right )+c^2 \left (\frac {b}{c}+x\right )^3\right )}-\frac {c \left (\frac {\frac {3 \sqrt [3]{b} \sqrt [3]{b^2-3 a c} \left (\sqrt [3]{b} B c \sqrt [3]{b^2-3 a c}-2 b^{4/3} C \sqrt [3]{b^2-3 a c}-2 A c^2-2 b^2 C+2 b B c\right ) \int \frac {1}{c^2 \left (\frac {b}{c}+x\right )^2+\sqrt [3]{b} c \sqrt [3]{b^2-3 a c} \left (\frac {b}{c}+x\right )+b^{2/3} \left (b^2-3 a c\right )^{2/3}}d\left (\frac {b}{c}+x\right )}{2 c}-\frac {\left (\sqrt [3]{b} \sqrt [3]{b^2-3 a c} \left (B-\frac {2 b C}{c}\right )+2 A c-\frac {2 b (B c-b C)}{c}\right ) \int \frac {c \left (2 c \left (\frac {b}{c}+x\right )+\sqrt [3]{b} \sqrt [3]{b^2-3 a c}\right )}{c^2 \left (\frac {b}{c}+x\right )^2+\sqrt [3]{b} c \sqrt [3]{b^2-3 a c} \left (\frac {b}{c}+x\right )+b^{2/3} \left (b^2-3 a c\right )^{2/3}}d\left (\frac {b}{c}+x\right )}{2 c}}{3 b^{2/3} \left (b^2-3 a c\right )^{2/3}}-\frac {\log \left (\sqrt [3]{b} \sqrt [3]{b^2-3 a c}-c \left (\frac {b}{c}+x\right )\right ) \left (-\sqrt [3]{b} B c \sqrt [3]{b^2-3 a c}+2 b^{4/3} C \sqrt [3]{b^2-3 a c}-2 A c^2-2 b^2 C+2 b B c\right )}{3 b^{2/3} c^2 \left (b^2-3 a c\right )^{2/3}}\right )}{3 b \left (b^2-3 a c\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {b C \left (3 a-\frac {b^2}{c}\right )-c^2 \left (\frac {b}{c}+x\right ) \left (A-\frac {b (B c-b C)}{c^2}+\left (\frac {b}{c}+x\right ) \left (B-\frac {2 b C}{c}\right )\right )}{3 b c \left (b^2-3 a c\right ) \left (b \left (3 a-\frac {b^2}{c}\right )+c^2 \left (\frac {b}{c}+x\right )^3\right )}-\frac {c \left (\frac {\frac {3 \sqrt [3]{b} \sqrt [3]{b^2-3 a c} \left (\sqrt [3]{b} B c \sqrt [3]{b^2-3 a c}-2 b^{4/3} C \sqrt [3]{b^2-3 a c}-2 A c^2-2 b^2 C+2 b B c\right ) \int \frac {1}{c^2 \left (\frac {b}{c}+x\right )^2+\sqrt [3]{b} c \sqrt [3]{b^2-3 a c} \left (\frac {b}{c}+x\right )+b^{2/3} \left (b^2-3 a c\right )^{2/3}}d\left (\frac {b}{c}+x\right )}{2 c}-\frac {1}{2} \left (\sqrt [3]{b} \sqrt [3]{b^2-3 a c} \left (B-\frac {2 b C}{c}\right )+2 A c-\frac {2 b (B c-b C)}{c}\right ) \int \frac {2 c \left (\frac {b}{c}+x\right )+\sqrt [3]{b} \sqrt [3]{b^2-3 a c}}{c^2 \left (\frac {b}{c}+x\right )^2+\sqrt [3]{b} c \sqrt [3]{b^2-3 a c} \left (\frac {b}{c}+x\right )+b^{2/3} \left (b^2-3 a c\right )^{2/3}}d\left (\frac {b}{c}+x\right )}{3 b^{2/3} \left (b^2-3 a c\right )^{2/3}}-\frac {\log \left (\sqrt [3]{b} \sqrt [3]{b^2-3 a c}-c \left (\frac {b}{c}+x\right )\right ) \left (-\sqrt [3]{b} B c \sqrt [3]{b^2-3 a c}+2 b^{4/3} C \sqrt [3]{b^2-3 a c}-2 A c^2-2 b^2 C+2 b B c\right )}{3 b^{2/3} c^2 \left (b^2-3 a c\right )^{2/3}}\right )}{3 b \left (b^2-3 a c\right )}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {b C \left (3 a-\frac {b^2}{c}\right )-c^2 \left (\frac {b}{c}+x\right ) \left (A-\frac {b (B c-b C)}{c^2}+\left (\frac {b}{c}+x\right ) \left (B-\frac {2 b C}{c}\right )\right )}{3 b c \left (b^2-3 a c\right ) \left (b \left (3 a-\frac {b^2}{c}\right )+c^2 \left (\frac {b}{c}+x\right )^3\right )}-\frac {c \left (\frac {-\frac {1}{2} \left (\sqrt [3]{b} \sqrt [3]{b^2-3 a c} \left (B-\frac {2 b C}{c}\right )+2 A c-\frac {2 b (B c-b C)}{c}\right ) \int \frac {2 c \left (\frac {b}{c}+x\right )+\sqrt [3]{b} \sqrt [3]{b^2-3 a c}}{c^2 \left (\frac {b}{c}+x\right )^2+\sqrt [3]{b} c \sqrt [3]{b^2-3 a c} \left (\frac {b}{c}+x\right )+b^{2/3} \left (b^2-3 a c\right )^{2/3}}d\left (\frac {b}{c}+x\right )-\frac {3 \left (\sqrt [3]{b} B c \sqrt [3]{b^2-3 a c}-2 b^{4/3} C \sqrt [3]{b^2-3 a c}-2 A c^2-2 b^2 C+2 b B c\right ) \int \frac {1}{-\left (\frac {2 c \left (\frac {b}{c}+x\right )}{\sqrt [3]{b} \sqrt [3]{b^2-3 a c}}+1\right )^2-3}d\left (\frac {2 c \left (\frac {b}{c}+x\right )}{\sqrt [3]{b} \sqrt [3]{b^2-3 a c}}+1\right )}{c^2}}{3 b^{2/3} \left (b^2-3 a c\right )^{2/3}}-\frac {\log \left (\sqrt [3]{b} \sqrt [3]{b^2-3 a c}-c \left (\frac {b}{c}+x\right )\right ) \left (-\sqrt [3]{b} B c \sqrt [3]{b^2-3 a c}+2 b^{4/3} C \sqrt [3]{b^2-3 a c}-2 A c^2-2 b^2 C+2 b B c\right )}{3 b^{2/3} c^2 \left (b^2-3 a c\right )^{2/3}}\right )}{3 b \left (b^2-3 a c\right )}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {b C \left (3 a-\frac {b^2}{c}\right )-c^2 \left (\frac {b}{c}+x\right ) \left (A-\frac {b (B c-b C)}{c^2}+\left (\frac {b}{c}+x\right ) \left (B-\frac {2 b C}{c}\right )\right )}{3 b c \left (b^2-3 a c\right ) \left (b \left (3 a-\frac {b^2}{c}\right )+c^2 \left (\frac {b}{c}+x\right )^3\right )}-\frac {c \left (\frac {\frac {\sqrt {3} \arctan \left (\frac {\frac {2 c \left (\frac {b}{c}+x\right )}{\sqrt [3]{b} \sqrt [3]{b^2-3 a c}}+1}{\sqrt {3}}\right ) \left (\sqrt [3]{b} B c \sqrt [3]{b^2-3 a c}-2 b^{4/3} C \sqrt [3]{b^2-3 a c}-2 A c^2-2 b^2 C+2 b B c\right )}{c^2}-\frac {1}{2} \left (\sqrt [3]{b} \sqrt [3]{b^2-3 a c} \left (B-\frac {2 b C}{c}\right )+2 A c-\frac {2 b (B c-b C)}{c}\right ) \int \frac {2 c \left (\frac {b}{c}+x\right )+\sqrt [3]{b} \sqrt [3]{b^2-3 a c}}{c^2 \left (\frac {b}{c}+x\right )^2+\sqrt [3]{b} c \sqrt [3]{b^2-3 a c} \left (\frac {b}{c}+x\right )+b^{2/3} \left (b^2-3 a c\right )^{2/3}}d\left (\frac {b}{c}+x\right )}{3 b^{2/3} \left (b^2-3 a c\right )^{2/3}}-\frac {\log \left (\sqrt [3]{b} \sqrt [3]{b^2-3 a c}-c \left (\frac {b}{c}+x\right )\right ) \left (-\sqrt [3]{b} B c \sqrt [3]{b^2-3 a c}+2 b^{4/3} C \sqrt [3]{b^2-3 a c}-2 A c^2-2 b^2 C+2 b B c\right )}{3 b^{2/3} c^2 \left (b^2-3 a c\right )^{2/3}}\right )}{3 b \left (b^2-3 a c\right )}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {b C \left (3 a-\frac {b^2}{c}\right )-c^2 \left (\frac {b}{c}+x\right ) \left (A-\frac {b (B c-b C)}{c^2}+\left (\frac {b}{c}+x\right ) \left (B-\frac {2 b C}{c}\right )\right )}{3 b c \left (b^2-3 a c\right ) \left (b \left (3 a-\frac {b^2}{c}\right )+c^2 \left (\frac {b}{c}+x\right )^3\right )}-\frac {c \left (\frac {\frac {\sqrt {3} \arctan \left (\frac {\frac {2 c \left (\frac {b}{c}+x\right )}{\sqrt [3]{b} \sqrt [3]{b^2-3 a c}}+1}{\sqrt {3}}\right ) \left (\sqrt [3]{b} B c \sqrt [3]{b^2-3 a c}-2 b^{4/3} C \sqrt [3]{b^2-3 a c}-2 A c^2-2 b^2 C+2 b B c\right )}{c^2}-\frac {\log \left (\sqrt [3]{b} c \sqrt [3]{b^2-3 a c} \left (\frac {b}{c}+x\right )+b^{2/3} \left (b^2-3 a c\right )^{2/3}+c^2 \left (\frac {b}{c}+x\right )^2\right ) \left (\sqrt [3]{b} \sqrt [3]{b^2-3 a c} \left (B-\frac {2 b C}{c}\right )+2 A c-\frac {2 b (B c-b C)}{c}\right )}{2 c}}{3 b^{2/3} \left (b^2-3 a c\right )^{2/3}}-\frac {\log \left (\sqrt [3]{b} \sqrt [3]{b^2-3 a c}-c \left (\frac {b}{c}+x\right )\right ) \left (-\sqrt [3]{b} B c \sqrt [3]{b^2-3 a c}+2 b^{4/3} C \sqrt [3]{b^2-3 a c}-2 A c^2-2 b^2 C+2 b B c\right )}{3 b^{2/3} c^2 \left (b^2-3 a c\right )^{2/3}}\right )}{3 b \left (b^2-3 a c\right )}\)

Input:

Int[(A + B*x + C*x^2)/(3*a*b + 3*b^2*x + 3*b*c*x^2 + c^2*x^3)^2,x]
 

Output:

(b*(3*a - b^2/c)*C - c^2*(b/c + x)*(A - (b*(B*c - b*C))/c^2 + (B - (2*b*C) 
/c)*(b/c + x)))/(3*b*c*(b^2 - 3*a*c)*(b*(3*a - b^2/c) + c^2*(b/c + x)^3)) 
- (c*(-1/3*((2*b*B*c - 2*A*c^2 - b^(1/3)*B*c*(b^2 - 3*a*c)^(1/3) - 2*b^2*C 
 + 2*b^(4/3)*(b^2 - 3*a*c)^(1/3)*C)*Log[b^(1/3)*(b^2 - 3*a*c)^(1/3) - c*(b 
/c + x)])/(b^(2/3)*c^2*(b^2 - 3*a*c)^(2/3)) + ((Sqrt[3]*(2*b*B*c - 2*A*c^2 
 + b^(1/3)*B*c*(b^2 - 3*a*c)^(1/3) - 2*b^2*C - 2*b^(4/3)*(b^2 - 3*a*c)^(1/ 
3)*C)*ArcTan[(1 + (2*c*(b/c + x))/(b^(1/3)*(b^2 - 3*a*c)^(1/3)))/Sqrt[3]]) 
/c^2 - ((2*A*c - (2*b*(B*c - b*C))/c + b^(1/3)*(b^2 - 3*a*c)^(1/3)*(B - (2 
*b*C)/c))*Log[b^(2/3)*(b^2 - 3*a*c)^(2/3) + b^(1/3)*c*(b^2 - 3*a*c)^(1/3)* 
(b/c + x) + c^2*(b/c + x)^2])/(2*c))/(3*b^(2/3)*(b^2 - 3*a*c)^(2/3))))/(3* 
b*(b^2 - 3*a*c))
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1142

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 2393

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, 
 x], i}, Simp[(a*Coeff[Pq, x, q] - b*x*ExpandToSum[Pq - Coeff[Pq, x, q]*x^q 
, x])*((a + b*x^n)^(p + 1)/(a*b*n*(p + 1))), x] + Simp[1/(a*n*(p + 1))   In 
t[Sum[(n*(p + 1) + i + 1)*Coeff[Pq, x, i]*x^i, {i, 0, q - 1}]*(a + b*x^n)^( 
p + 1), x], x] /; q == n - 1] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[n 
, 0] && LtQ[p, -1]
 

rule 2400

Int[((A_) + (B_.)*(x_))/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{r = Numer 
ator[Rt[-a/b, 3]], s = Denominator[Rt[-a/b, 3]]}, Simp[r*((B*r + A*s)/(3*a* 
s))   Int[1/(r - s*x), x], x] - Simp[r/(3*a*s)   Int[(r*(B*r - 2*A*s) - s*( 
B*r + A*s)*x)/(r^2 + r*s*x + s^2*x^2), x], x]] /; FreeQ[{a, b, A, B}, x] && 
 NeQ[a*B^3 - b*A^3, 0] && NegQ[a/b]
 

rule 2459

Int[(Pn_)^(p_.)*(Qx_), x_Symbol] :> With[{S = Coeff[Pn, x, Expon[Pn, x] - 1 
]/(Expon[Pn, x]*Coeff[Pn, x, Expon[Pn, x]])}, Subst[Int[ExpandToSum[Pn /. x 
 -> x - S, x]^p*ExpandToSum[Qx /. x -> x - S, x], x], x, x + S] /; Binomial 
Q[Pn /. x -> x - S, x] || (IntegerQ[Expon[Pn, x]/2] && TrinomialQ[Pn /. x - 
> x - S, x])] /; FreeQ[p, x] && PolyQ[Pn, x] && GtQ[Expon[Pn, x], 2] && NeQ 
[Coeff[Pn, x, Expon[Pn, x] - 1], 0] && PolyQ[Qx, x] &&  !(MonomialQ[Qx, x] 
&& IGtQ[p, 0])
 

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.13 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.43

method	result	size
default	\(\frac {\frac {\left (B c -2 b C \right ) x^{2}}{9 b \left (3 a c -b^{2}\right )}+\frac {\left (A \,c^{2}+b B c -3 b^{2} C \right ) x}{9 b c \left (3 a c -b^{2}\right )}+\frac {A c -3 C a}{9 c \left (3 a c -b^{2}\right )}}{\frac {1}{3} c^{2} x^{3}+b c \,x^{2}+b^{2} x +a b}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3} c^{2}+3 \textit {\_Z}^{2} b c +3 b^{2} \textit {\_Z} +3 a b \right )}{\sum }\frac {\left (B \textit {\_R} c -2 C \textit {\_R} b +2 A c -B b \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2} c^{2}+2 \textit {\_R} b c +b^{2}}}{9 b \left (3 a c -b^{2}\right )}\)	\(208\)
risch	\(\frac {\frac {\left (B c -2 b C \right ) x^{2}}{9 b \left (3 a c -b^{2}\right )}+\frac {\left (A \,c^{2}+b B c -3 b^{2} C \right ) x}{9 b c \left (3 a c -b^{2}\right )}+\frac {A c -3 C a}{9 c \left (3 a c -b^{2}\right )}}{\frac {1}{3} c^{2} x^{3}+b c \,x^{2}+b^{2} x +a b}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3} c^{2}+3 \textit {\_Z}^{2} b c +3 b^{2} \textit {\_Z} +3 a b \right )}{\sum }\frac {\left (\frac {\left (B c -2 b C \right ) \textit {\_R}}{3 a c -b^{2}}+\frac {2 A c -B b}{3 a c -b^{2}}\right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2} c^{2}+2 \textit {\_R} b c +b^{2}}}{9 b}\)	\(223\)

Input:

int((C*x^2+B*x+A)/(c^2*x^3+3*b*c*x^2+3*b^2*x+3*a*b)^2,x,method=_RETURNVERB 
OSE)
 

Output:

(1/9*(B*c-2*C*b)/b/(3*a*c-b^2)*x^2+1/9*(A*c^2+B*b*c-3*C*b^2)/b/c/(3*a*c-b^ 
2)*x+1/9/c*(A*c-3*C*a)/(3*a*c-b^2))/(1/3*c^2*x^3+b*c*x^2+b^2*x+a*b)+1/9/b/ 
(3*a*c-b^2)*sum((B*_R*c-2*C*_R*b+2*A*c-B*b)/(_R^2*c^2+2*_R*b*c+b^2)*ln(x-_ 
R),_R=RootOf(_Z^3*c^2+3*_Z^2*b*c+3*_Z*b^2+3*a*b))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 11.54 (sec) , antiderivative size = 15642, normalized size of antiderivative = 32.32 \[ \int \frac {A+B x+C x^2}{\left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )^2} \, dx=\text {Too large to display} \] Input:

integrate((C*x^2+B*x+A)/(c^2*x^3+3*b*c*x^2+3*b^2*x+3*a*b)^2,x, algorithm=" 
fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B x+C x^2}{\left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )^2} \, dx=\text {Timed out} \] Input:

integrate((C*x**2+B*x+A)/(c**2*x**3+3*b*c*x**2+3*b**2*x+3*a*b)**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {A+B x+C x^2}{\left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )^2} \, dx=\int { \frac {C x^{2} + B x + A}{{\left (c^{2} x^{3} + 3 \, b c x^{2} + 3 \, b^{2} x + 3 \, a b\right )}^{2}} \,d x } \] Input:

integrate((C*x^2+B*x+A)/(c^2*x^3+3*b*c*x^2+3*b^2*x+3*a*b)^2,x, algorithm=" 
maxima")
 

Output:

1/3*(3*C*a*b - A*b*c + (2*C*b*c - B*c^2)*x^2 + (3*C*b^2 - B*b*c - A*c^2)*x 
)/(3*a*b^4*c - 9*a^2*b^2*c^2 + (b^3*c^3 - 3*a*b*c^4)*x^3 + 3*(b^4*c^2 - 3* 
a*b^2*c^3)*x^2 + 3*(b^5*c - 3*a*b^3*c^2)*x) - 1/3*integrate(-(B*b - 2*A*c 
+ (2*C*b - B*c)*x)/(c^2*x^3 + 3*b*c*x^2 + 3*b^2*x + 3*a*b), x)/(b^3 - 3*a* 
b*c)
 

Giac [F]

\[ \int \frac {A+B x+C x^2}{\left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )^2} \, dx=\int { \frac {C x^{2} + B x + A}{{\left (c^{2} x^{3} + 3 \, b c x^{2} + 3 \, b^{2} x + 3 \, a b\right )}^{2}} \,d x } \] Input:

integrate((C*x^2+B*x+A)/(c^2*x^3+3*b*c*x^2+3*b^2*x+3*a*b)^2,x, algorithm=" 
giac")
 

Output:

integrate((C*x^2 + B*x + A)/(c^2*x^3 + 3*b*c*x^2 + 3*b^2*x + 3*a*b)^2, x)
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 13.19 (sec) , antiderivative size = 1692, normalized size of antiderivative = 3.50 \[ \int \frac {A+B x+C x^2}{\left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )^2} \, dx=\text {Too large to display} \] Input:

int((A + B*x + C*x^2)/(3*a*b + 3*b^2*x + c^2*x^3 + 3*b*c*x^2)^2,x)
 

Output:

((A*c - 3*C*a)/(3*c*(3*a*c - b^2)) + (x^2*(B*c - 2*C*b))/(3*b*(3*a*c - b^2 
)) + (x*(A*c^2 - 3*C*b^2 + B*b*c))/(3*b*c*(3*a*c - b^2)))/(3*a*b + 3*b^2*x 
 + c^2*x^3 + 3*b*c*x^2) + symsum(log((x*(B^2*c^4 + 4*C^2*b^2*c^2 - 4*B*C*b 
*c^3))/(9*(b^6 + 9*a^2*b^2*c^2 - 6*a*b^4*c)) - root(295245*a^4*b^7*c^6*z^3 
 - 196830*a^3*b^9*c^5*z^3 - 177147*a^5*b^5*c^7*z^3 + 65610*a^2*b^11*c^4*z^ 
3 - 10935*a*b^13*c^3*z^3 + 729*b^15*c^2*z^3 - 1458*B*C*a^2*b^4*c^3*z + 972 
*A*C*a^2*b^3*c^4*z - 486*A*B*a^2*b^2*c^5*z + 972*B*C*a*b^6*c^2*z - 648*A*C 
*a*b^5*c^3*z + 324*A*B*a*b^4*c^4*z + 108*A*C*b^7*c^2*z - 54*A*B*b^6*c^3*z 
+ 972*C^2*a^2*b^5*c^2*z + 486*B^2*a^2*b^3*c^4*z - 162*B*C*b^8*c*z - 324*B^ 
2*a*b^5*c^3*z - 648*C^2*a*b^7*c*z + 54*B^2*b^7*c^2*z + 108*C^2*b^9*z - 36* 
B*C^2*a*b^3*c - 48*A*B*C*b^3*c^2 + 18*B^2*C*a*b^2*c^2 + 24*A^2*C*b^2*c^3 + 
 24*A*B^2*b^2*c^3 - 3*B^3*a*b*c^3 + 18*B^2*C*b^4*c + 24*A*C^2*b^4*c - 24*A 
^2*B*b*c^4 - 7*B^3*b^3*c^2 + 24*C^3*a*b^4 - 12*B*C^2*b^5 + 8*A^3*c^5, z, k 
)*((18*A*b^4*c^4 - 18*B*b^5*c^3 + 18*C*b^6*c^2 - 54*A*a*b^2*c^5 + 54*B*a*b 
^3*c^4 - 54*C*a*b^4*c^3)/(9*(b^6 + 9*a^2*b^2*c^2 - 6*a*b^4*c)) + (root(295 
245*a^4*b^7*c^6*z^3 - 196830*a^3*b^9*c^5*z^3 - 177147*a^5*b^5*c^7*z^3 + 65 
610*a^2*b^11*c^4*z^3 - 10935*a*b^13*c^3*z^3 + 729*b^15*c^2*z^3 - 1458*B*C* 
a^2*b^4*c^3*z + 972*A*C*a^2*b^3*c^4*z - 486*A*B*a^2*b^2*c^5*z + 972*B*C*a* 
b^6*c^2*z - 648*A*C*a*b^5*c^3*z + 324*A*B*a*b^4*c^4*z + 108*A*C*b^7*c^2*z 
- 54*A*B*b^6*c^3*z + 972*C^2*a^2*b^5*c^2*z + 486*B^2*a^2*b^3*c^4*z - 16...
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 2426, normalized size of antiderivative = 5.01 \[ \int \frac {A+B x+C x^2}{\left (3 a b+3 b^2 x+3 b c x^2+c^2 x^3\right )^2} \, dx =\text {Too large to display} \] Input:

int((C*x^2+B*x+A)/(c^2*x^3+3*b*c*x^2+3*b^2*x+3*a*b)^2,x)
 

Output:

( - 12*b**(2/3)*(3*a*c - b**2)**(2/3)*sqrt(3)*atan((b**(1/3)*(3*a*c - b**2 
)**(1/3) - 2*b - 2*c*x)/(b**(1/3)*(3*a*c - b**2)**(1/3)*sqrt(3)))*a**2*b*c 
 - 12*b**(2/3)*(3*a*c - b**2)**(2/3)*sqrt(3)*atan((b**(1/3)*(3*a*c - b**2) 
**(1/3) - 2*b - 2*c*x)/(b**(1/3)*(3*a*c - b**2)**(1/3)*sqrt(3)))*a*b**2*c* 
x - 12*b**(2/3)*(3*a*c - b**2)**(2/3)*sqrt(3)*atan((b**(1/3)*(3*a*c - b**2 
)**(1/3) - 2*b - 2*c*x)/(b**(1/3)*(3*a*c - b**2)**(1/3)*sqrt(3)))*a*b*c**2 
*x**2 - 4*b**(2/3)*(3*a*c - b**2)**(2/3)*sqrt(3)*atan((b**(1/3)*(3*a*c - b 
**2)**(1/3) - 2*b - 2*c*x)/(b**(1/3)*(3*a*c - b**2)**(1/3)*sqrt(3)))*a*c** 
3*x**3 + 18*sqrt(3)*atan((b**(1/3)*(3*a*c - b**2)**(1/3) - 2*b - 2*c*x)/(b 
**(1/3)*(3*a*c - b**2)**(1/3)*sqrt(3)))*a**2*b**3*c - 6*sqrt(3)*atan((b**( 
1/3)*(3*a*c - b**2)**(1/3) - 2*b - 2*c*x)/(b**(1/3)*(3*a*c - b**2)**(1/3)* 
sqrt(3)))*a*b**5 + 18*sqrt(3)*atan((b**(1/3)*(3*a*c - b**2)**(1/3) - 2*b - 
 2*c*x)/(b**(1/3)*(3*a*c - b**2)**(1/3)*sqrt(3)))*a*b**4*c*x + 18*sqrt(3)* 
atan((b**(1/3)*(3*a*c - b**2)**(1/3) - 2*b - 2*c*x)/(b**(1/3)*(3*a*c - b** 
2)**(1/3)*sqrt(3)))*a*b**3*c**2*x**2 + 6*sqrt(3)*atan((b**(1/3)*(3*a*c - b 
**2)**(1/3) - 2*b - 2*c*x)/(b**(1/3)*(3*a*c - b**2)**(1/3)*sqrt(3)))*a*b** 
2*c**3*x**3 - 6*sqrt(3)*atan((b**(1/3)*(3*a*c - b**2)**(1/3) - 2*b - 2*c*x 
)/(b**(1/3)*(3*a*c - b**2)**(1/3)*sqrt(3)))*b**6*x - 6*sqrt(3)*atan((b**(1 
/3)*(3*a*c - b**2)**(1/3) - 2*b - 2*c*x)/(b**(1/3)*(3*a*c - b**2)**(1/3)*s 
qrt(3)))*b**5*c*x**2 - 2*sqrt(3)*atan((b**(1/3)*(3*a*c - b**2)**(1/3) -...