\(\int \frac {A+B x+C x^2}{a d+(b d+a e) x+(c d+b e) x^2+c e x^3} \, dx\) [127]

Optimal result

Integrand size = 43, antiderivative size = 196 \[ \int \frac {A+B x+C x^2}{a d+(b d+a e) x+(c d+b e) x^2+c e x^3} \, dx=-\frac {\left (b^2 C d+2 c (A c d-a C d+a B e)-b (B c d+A c e+a C e)\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{c \sqrt {b^2-4 a c} \left (c d^2-b d e+a e^2\right )}+\frac {\left (C d^2-e (B d-A e)\right ) \log (d+e x)}{e \left (c d^2-b d e+a e^2\right )}+\frac {(B c d-b C d-A c e+a C e) \log \left (a+b x+c x^2\right )}{2 c \left (c d^2-b d e+a e^2\right )} \] Output:

-(b^2*C*d+2*c*(A*c*d+B*a*e-C*a*d)-b*(A*c*e+B*c*d+C*a*e))*arctanh((2*c*x+b) 
/(-4*a*c+b^2)^(1/2))/c/(-4*a*c+b^2)^(1/2)/(a*e^2-b*d*e+c*d^2)+(C*d^2-e*(-A 
*e+B*d))*ln(e*x+d)/e/(a*e^2-b*d*e+c*d^2)+1/2*(-A*c*e+B*c*d+C*a*e-C*b*d)*ln 
(c*x^2+b*x+a)/c/(a*e^2-b*d*e+c*d^2)
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x+C x^2}{a d+(b d+a e) x+(c d+b e) x^2+c e x^3} \, dx=\frac {-2 e \left (-b^2 C d-2 c (A c d-a C d+a B e)+b (B c d+A c e+a C e)\right ) \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )+2 c \sqrt {-b^2+4 a c} \left (C d^2+e (-B d+A e)\right ) \log (d+e x)+\sqrt {-b^2+4 a c} e (B c d-b C d-A c e+a C e) \log (a+x (b+c x))}{2 c \sqrt {-b^2+4 a c} e \left (c d^2+e (-b d+a e)\right )} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {2462, 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.

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

\(\Big \downarrow \) 2462

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \left (-b (a C e+A c e+B c d)+2 c (a B e-a C d+A c d)+b^2 C d\right )}{c \sqrt {b^2-4 a c} \left (a e^2-b d e+c d^2\right )}+\frac {\log \left (a+b x+c x^2\right ) (a C e-A c e-b C d+B c d)}{2 c \left (a e^2-b d e+c d^2\right )}+\frac {\log (d+e x) \left (C d^2-e (B d-A e)\right )}{e \left (a e^2-b d e+c d^2\right )}\)

Input:

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

Output:

-(((b^2*C*d + 2*c*(A*c*d - a*C*d + a*B*e) - b*(B*c*d + A*c*e + a*C*e))*Arc 
Tanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/(c*Sqrt[b^2 - 4*a*c]*(c*d^2 - b*d*e + 
 a*e^2))) + ((C*d^2 - e*(B*d - A*e))*Log[d + e*x])/(e*(c*d^2 - b*d*e + a*e 
^2)) + ((B*c*d - b*C*d - A*c*e + a*C*e)*Log[a + b*x + c*x^2])/(2*c*(c*d^2 
- b*d*e + a*e^2))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2462

Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr 
and[u*Qx^p, x], x] /;  !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && GtQ 
[Expon[Px, x], 2] &&  !BinomialQ[Px, x] &&  !TrinomialQ[Px, x] && ILtQ[p, 0 
] && RationalFunctionQ[u, x]
 

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.91

method	result	size
default	\(\frac {\frac {\left (-A c e +B c d +C a e -C b d \right ) \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {2 \left (-A b e +A c d +B a e -C a d -\frac {\left (-A c e +B c d +C a e -C b d \right ) b}{2 c}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{a \,e^{2}-b d e +c \,d^{2}}+\frac {\left (A \,e^{2}-B d e +C \,d^{2}\right ) \ln \left (e x +d \right )}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) e}\)	\(179\)
risch	\(\frac {e \ln \left (e x +d \right ) A}{a \,e^{2}-b d e +c \,d^{2}}-\frac {\ln \left (e x +d \right ) B d}{a \,e^{2}-b d e +c \,d^{2}}+\frac {\ln \left (e x +d \right ) C \,d^{2}}{\left (a \,e^{2}-b d e +c \,d^{2}\right ) e}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (4 a^{2} c^{2} e^{2}-a \,b^{2} c \,e^{2}-4 a b \,c^{2} d e +4 a \,c^{3} d^{2}+b^{3} c d e -b^{2} c^{2} d^{2}\right ) \textit {\_Z}^{2}+\left (4 A a \,c^{2} e -A \,b^{2} c e -4 B \,c^{2} d a +B \,b^{2} c d -4 C \,a^{2} c e +C a \,b^{2} e +4 C a b c d -C \,b^{3} d \right ) \textit {\_Z} +A^{2} c^{2}-A B b c -2 A C a c +A C \,b^{2}+B^{2} a c -B C a b +C^{2} a^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (6 a \,c^{2} e^{3}-2 b^{2} c \,e^{3}+2 d \,e^{2} b \,c^{2}-2 d^{2} e \,c^{3}\right ) \textit {\_R}^{2}+\left (3 A \,c^{2} e^{2}-B b c \,e^{2}-B \,c^{2} d e -5 C a c \,e^{2}+2 b^{2} C \,e^{2}-C b c d e +2 C \,c^{2} d^{2}\right ) \textit {\_R} -A C c e +B^{2} c e -B C b e -B C c d +C^{2} a e +C^{2} b d \right ) x +\left (-a b c \,e^{3}+8 a \,c^{2} d \,e^{2}-b^{2} c d \,e^{2}-d^{2} e b \,c^{2}\right ) \textit {\_R}^{2}+\left (A b c \,e^{2}+A \,c^{2} d e -B a c \,e^{2}-B b c d e +C a b \,e^{2}-5 C a d e c +b^{2} C d e +C b c \,d^{2}\right ) \textit {\_R} +A B c e -A C b e -A C c d +C^{2} a d \right )\right )\)	\(535\)

Input:

int((C*x^2+B*x+A)/(a*d+(a*e+b*d)*x+(b*e+c*d)*x^2+c*e*x^3),x,method=_RETURN 
VERBOSE)
 

Output:

1/(a*e^2-b*d*e+c*d^2)*(1/2*(-A*c*e+B*c*d+C*a*e-C*b*d)/c*ln(c*x^2+b*x+a)+2* 
(-A*b*e+A*c*d+B*a*e-C*a*d-1/2*(-A*c*e+B*c*d+C*a*e-C*b*d)*b/c)/(4*a*c-b^2)^ 
(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2)))+(A*e^2-B*d*e+C*d^2)/(a*e^2-b*d* 
e+c*d^2)/e*ln(e*x+d)
 

Fricas [A] (verification not implemented)

Time = 32.62 (sec) , antiderivative size = 613, normalized size of antiderivative = 3.13 \[ \int \frac {A+B x+C x^2}{a d+(b d+a e) x+(c d+b e) x^2+c e x^3} \, dx=\left [-\frac {{\left ({\left (C b^{2} + 2 \, A c^{2} - {\left (2 \, C a + B b\right )} c\right )} d e - {\left (C a b - {\left (2 \, B a - A b\right )} c\right )} e^{2}\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + {\left ({\left (C b^{3} + 4 \, B a c^{2} - {\left (4 \, C a b + B b^{2}\right )} c\right )} d e - {\left (C a b^{2} + 4 \, A a c^{2} - {\left (4 \, C a^{2} + A b^{2}\right )} c\right )} e^{2}\right )} \log \left (c x^{2} + b x + a\right ) - 2 \, {\left ({\left (C b^{2} c - 4 \, C a c^{2}\right )} d^{2} - {\left (B b^{2} c - 4 \, B a c^{2}\right )} d e + {\left (A b^{2} c - 4 \, A a c^{2}\right )} e^{2}\right )} \log \left (e x + d\right )}{2 \, {\left ({\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d^{2} e - {\left (b^{3} c - 4 \, a b c^{2}\right )} d e^{2} + {\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} e^{3}\right )}}, -\frac {2 \, {\left ({\left (C b^{2} + 2 \, A c^{2} - {\left (2 \, C a + B b\right )} c\right )} d e - {\left (C a b - {\left (2 \, B a - A b\right )} c\right )} e^{2}\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) + {\left ({\left (C b^{3} + 4 \, B a c^{2} - {\left (4 \, C a b + B b^{2}\right )} c\right )} d e - {\left (C a b^{2} + 4 \, A a c^{2} - {\left (4 \, C a^{2} + A b^{2}\right )} c\right )} e^{2}\right )} \log \left (c x^{2} + b x + a\right ) - 2 \, {\left ({\left (C b^{2} c - 4 \, C a c^{2}\right )} d^{2} - {\left (B b^{2} c - 4 \, B a c^{2}\right )} d e + {\left (A b^{2} c - 4 \, A a c^{2}\right )} e^{2}\right )} \log \left (e x + d\right )}{2 \, {\left ({\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d^{2} e - {\left (b^{3} c - 4 \, a b c^{2}\right )} d e^{2} + {\left (a b^{2} c - 4 \, a^{2} c^{2}\right )} e^{3}\right )}}\right ] \] Input:

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

Output:

[-1/2*(((C*b^2 + 2*A*c^2 - (2*C*a + B*b)*c)*d*e - (C*a*b - (2*B*a - A*b)*c 
)*e^2)*sqrt(b^2 - 4*a*c)*log((2*c^2*x^2 + 2*b*c*x + b^2 - 2*a*c + sqrt(b^2 
 - 4*a*c)*(2*c*x + b))/(c*x^2 + b*x + a)) + ((C*b^3 + 4*B*a*c^2 - (4*C*a*b 
 + B*b^2)*c)*d*e - (C*a*b^2 + 4*A*a*c^2 - (4*C*a^2 + A*b^2)*c)*e^2)*log(c* 
x^2 + b*x + a) - 2*((C*b^2*c - 4*C*a*c^2)*d^2 - (B*b^2*c - 4*B*a*c^2)*d*e 
+ (A*b^2*c - 4*A*a*c^2)*e^2)*log(e*x + d))/((b^2*c^2 - 4*a*c^3)*d^2*e - (b 
^3*c - 4*a*b*c^2)*d*e^2 + (a*b^2*c - 4*a^2*c^2)*e^3), -1/2*(2*((C*b^2 + 2* 
A*c^2 - (2*C*a + B*b)*c)*d*e - (C*a*b - (2*B*a - A*b)*c)*e^2)*sqrt(-b^2 + 
4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) + ((C*b^3 + 4 
*B*a*c^2 - (4*C*a*b + B*b^2)*c)*d*e - (C*a*b^2 + 4*A*a*c^2 - (4*C*a^2 + A* 
b^2)*c)*e^2)*log(c*x^2 + b*x + a) - 2*((C*b^2*c - 4*C*a*c^2)*d^2 - (B*b^2* 
c - 4*B*a*c^2)*d*e + (A*b^2*c - 4*A*a*c^2)*e^2)*log(e*x + d))/((b^2*c^2 - 
4*a*c^3)*d^2*e - (b^3*c - 4*a*b*c^2)*d*e^2 + (a*b^2*c - 4*a^2*c^2)*e^3)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B x+C x^2}{a d+(b d+a e) x+(c d+b e) x^2+c e x^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((C*x^2+B*x+A)/(a*d+(a*e+b*d)*x+(b*e+c*d)*x^2+c*e*x^3),x, algorit 
hm="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
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.02 \[ \int \frac {A+B x+C x^2}{a d+(b d+a e) x+(c d+b e) x^2+c e x^3} \, dx=-\frac {{\left (C b d - B c d - C a e + A c e\right )} \log \left (c x^{2} + b x + a\right )}{2 \, {\left (c^{2} d^{2} - b c d e + a c e^{2}\right )}} + \frac {{\left (C d^{2} - B d e + A e^{2}\right )} \log \left ({\left | e x + d \right |}\right )}{c d^{2} e - b d e^{2} + a e^{3}} + \frac {{\left (C b^{2} d - 2 \, C a c d - B b c d + 2 \, A c^{2} d - C a b e + 2 \, B a c e - A b c e\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (c^{2} d^{2} - b c d e + a c e^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} \] Input:

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

Output:

-1/2*(C*b*d - B*c*d - C*a*e + A*c*e)*log(c*x^2 + b*x + a)/(c^2*d^2 - b*c*d 
*e + a*c*e^2) + (C*d^2 - B*d*e + A*e^2)*log(abs(e*x + d))/(c*d^2*e - b*d*e 
^2 + a*e^3) + (C*b^2*d - 2*C*a*c*d - B*b*c*d + 2*A*c^2*d - C*a*b*e + 2*B*a 
*c*e - A*b*c*e)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/((c^2*d^2 - b*c*d*e 
 + a*c*e^2)*sqrt(-b^2 + 4*a*c))
 

Mupad [B] (verification not implemented)

Time = 18.29 (sec) , antiderivative size = 2467, normalized size of antiderivative = 12.59 \[ \int \frac {A+B x+C x^2}{a d+(b d+a e) x+(c d+b e) x^2+c e x^3} \, dx=\text {Too large to display} \] Input:

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

Output:

(log(d + e*x)*(A*e^2 + C*d^2 - B*d*e))/(a*e^3 - b*d*e^2 + c*d^2*e) - (log( 
6*A*a^2*c*e^4 - 2*A*a*b^2*e^4 - 2*C*a^3*e^4 + B*a^2*b*e^4 - 4*C*a*c^2*d^4 
+ C*b^2*c*d^4 + C*b^3*d^3*e - 2*A*b^3*e^4*x - B*a^2*e^4*(b^2 - 4*a*c)^(1/2 
) + B*a*b^2*e^4*x - 2*B*a^2*c*e^4*x - C*a^2*b*e^4*x + 2*A*c^3*d^3*e*x + B* 
b^3*d*e^3*x + A*c^2*d^3*e*(b^2 - 4*a*c)^(1/2) + 3*C*a^2*d*e^3*(b^2 - 4*a*c 
)^(1/2) + C*b^2*d^3*e*(b^2 - 4*a*c)^(1/2) + 2*A*b^2*e^4*x*(b^2 - 4*a*c)^(1 
/2) + C*a^2*e^4*x*(b^2 - 4*a*c)^(1/2) + 2*C*c^2*d^4*x*(b^2 - 4*a*c)^(1/2) 
- 10*A*a*c^2*d^2*e^2 + A*b^2*c*d^2*e^2 - 4*C*a*b^2*d^2*e^2 + 10*C*a^2*c*d^ 
2*e^2 - C*b^3*d^2*e^2*x + 2*A*a*b*e^4*(b^2 - 4*a*c)^(1/2) + C*b*c*d^4*(b^2 
 - 4*a*c)^(1/2) + B*a*b^2*d*e^3 + A*b*c^2*d^3*e + 6*B*a*c^2*d^3*e - 10*B*a 
^2*c*d*e^3 + 3*C*a^2*b*d*e^3 - 2*B*b^2*c*d^3*e + 7*A*a*b*c*e^4*x + 5*A*c^2 
*d^2*e^2*x*(b^2 - 4*a*c)^(1/2) + C*b^2*d^2*e^2*x*(b^2 - 4*a*c)^(1/2) - 7*A 
*a*c*d*e^3*(b^2 - 4*a*c)^(1/2) - B*a*b*d*e^3*(b^2 - 4*a*c)^(1/2) - 2*B*b*c 
*d^3*e*(b^2 - 4*a*c)^(1/2) - 5*C*a*c*d^3*e*(b^2 - 4*a*c)^(1/2) - 3*A*a*c*e 
^4*x*(b^2 - 4*a*c)^(1/2) - B*a*b*e^4*x*(b^2 - 4*a*c)^(1/2) + 3*B*a*b*c*d^2 
*e^2 - 14*A*a*c^2*d*e^3*x + 5*A*b^2*c*d*e^3*x - B*b*c^2*d^3*e*x - 10*C*a*c 
^2*d^3*e*x + 6*C*a^2*c*d*e^3*x + 3*C*b^2*c*d^3*e*x + A*b*c*d^2*e^2*(b^2 - 
4*a*c)^(1/2) + 7*B*a*c*d^2*e^2*(b^2 - 4*a*c)^(1/2) - 2*C*a*b*d^2*e^2*(b^2 
- 4*a*c)^(1/2) - B*b^2*d*e^3*x*(b^2 - 4*a*c)^(1/2) - 3*B*c^2*d^3*e*x*(b^2 
- 4*a*c)^(1/2) - 3*A*b*c^2*d^2*e^2*x + 14*B*a*c^2*d^2*e^2*x - 2*B*b^2*c...
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.05 \[ \int \frac {A+B x+C x^2}{a d+(b d+a e) x+(c d+b e) x^2+c e x^3} \, dx=\frac {\mathrm {log}\left (e x +d \right )}{e} \] Input:

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

Output:

log(d + e*x)/e