3.1.15 \(\int \frac {A+B x}{(a+b x+c x^2) (d+e x+f x^2)} \, dx\) [15]

3.1.15.1 Optimal result

Integrand size = 30, antiderivative size = 406 \[ \int \frac {A+B x}{\left (a+b x+c x^2\right ) \left (d+e x+f x^2\right )} \, dx=-\frac {\left (A b^2 f+2 c (A c d+a B e-a A f)-b (B c d+A c e+a B f)\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (c^2 d^2+f \left (b^2 d-a b e+a^2 f\right )-c \left (b d e-a \left (e^2-2 d f\right )\right )\right )}+\frac {\left (B (c d e-2 b d f+a e f)-A \left (c e^2-2 c d f-b e f+2 a f^2\right )\right ) \text {arctanh}\left (\frac {e+2 f x}{\sqrt {e^2-4 d f}}\right )}{\sqrt {e^2-4 d f} \left (c^2 d^2+f \left (b^2 d-a b e+a^2 f\right )-c \left (b d e-a \left (e^2-2 d f\right )\right )\right )}+\frac {(B c d-A c e+A b f-a B f) \log \left (a+b x+c x^2\right )}{2 \left (c^2 d^2+f \left (b^2 d-a b e+a^2 f\right )-c \left (b d e-a \left (e^2-2 d f\right )\right )\right )}-\frac {(B c d-A c e+A b f-a B f) \log \left (d+e x+f x^2\right )}{2 \left (c^2 d^2+f \left (b^2 d-a b e+a^2 f\right )-c \left (b d e-a \left (e^2-2 d f\right )\right )\right )} \]

output

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

3.1.15.2 Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 267, normalized size of antiderivative = 0.66 \[ \int \frac {A+B x}{\left (a+b x+c x^2\right ) \left (d+e x+f x^2\right )} \, dx=\frac {\frac {2 \left (A b^2 f+2 c (A c d+a B e-a A f)-b (B c d+A c e+a B f)\right ) \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{\sqrt {-b^2+4 a c}}-\frac {2 \left (B (c d e-2 b d f+a e f)+A \left (-c e^2+2 c d f+b e f-2 a f^2\right )\right ) \arctan \left (\frac {e+2 f x}{\sqrt {-e^2+4 d f}}\right )}{\sqrt {-e^2+4 d f}}+(B c d-A c e+A b f-a B f) \log (a+x (b+c x))+(-B c d+A c e-A b f+a B f) \log (d+x (e+f x))}{2 \left (c^2 d^2-b c d e+f \left (b^2 d-a b e+a^2 f\right )+a c \left (e^2-2 d f\right )\right )} \]

input

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

output

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

3.1.15.3 Rubi [A] (verified)

Time = 0.59 (sec) , antiderivative size = 314, normalized size of antiderivative = 0.77, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1355, 25, 1142, 1083, 219, 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.

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

\(\Big \downarrow \) 1355

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1142

\(\displaystyle \frac {\frac {1}{2} \left (-b (a B f+A c e+B c d)+2 c (-a A f+a B e+A c d)+A b^2 f\right ) \int \frac {1}{c x^2+b x+a}dx+\frac {1}{2} (-a B f+A b f-A c e+B c d) \int \frac {b+2 c x}{c x^2+b x+a}dx}{f \left (a^2 f-a b e+b^2 d\right )+a c \left (e^2-2 d f\right )-b c d e+c^2 d^2}-\frac {\frac {1}{2} \left (B (a e f-2 b d f+c d e)-A \left (2 a f^2-b e f-2 c d f+c e^2\right )\right ) \int \frac {1}{f x^2+e x+d}dx+\frac {1}{2} (-a B f+A b f-A c e+B c d) \int \frac {e+2 f x}{f x^2+e x+d}dx}{f \left (a^2 f-a b e+b^2 d\right )+a c \left (e^2-2 d f\right )-b c d e+c^2 d^2}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {\frac {1}{2} (-a B f+A b f-A c e+B c d) \int \frac {b+2 c x}{c x^2+b x+a}dx-\left (-b (a B f+A c e+B c d)+2 c (-a A f+a B e+A c d)+A b^2 f\right ) \int \frac {1}{b^2-(b+2 c x)^2-4 a c}d(b+2 c x)}{f \left (a^2 f-a b e+b^2 d\right )+a c \left (e^2-2 d f\right )-b c d e+c^2 d^2}-\frac {\frac {1}{2} (-a B f+A b f-A c e+B c d) \int \frac {e+2 f x}{f x^2+e x+d}dx-\left (B (a e f-2 b d f+c d e)-A \left (2 a f^2-b e f-2 c d f+c e^2\right )\right ) \int \frac {1}{e^2-(e+2 f x)^2-4 d f}d(e+2 f x)}{f \left (a^2 f-a b e+b^2 d\right )+a c \left (e^2-2 d f\right )-b c d e+c^2 d^2}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{2} (-a B f+A b f-A c e+B c d) \int \frac {b+2 c x}{c x^2+b x+a}dx-\frac {\text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \left (-b (a B f+A c e+B c d)+2 c (-a A f+a B e+A c d)+A b^2 f\right )}{\sqrt {b^2-4 a c}}}{f \left (a^2 f-a b e+b^2 d\right )+a c \left (e^2-2 d f\right )-b c d e+c^2 d^2}-\frac {\frac {1}{2} (-a B f+A b f-A c e+B c d) \int \frac {e+2 f x}{f x^2+e x+d}dx-\frac {\text {arctanh}\left (\frac {e+2 f x}{\sqrt {e^2-4 d f}}\right ) \left (B (a e f-2 b d f+c d e)-A \left (2 a f^2-b e f-2 c d f+c e^2\right )\right )}{\sqrt {e^2-4 d f}}}{f \left (a^2 f-a b e+b^2 d\right )+a c \left (e^2-2 d f\right )-b c d e+c^2 d^2}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {\frac {1}{2} \log \left (a+b x+c x^2\right ) (-a B f+A b f-A c e+B c d)-\frac {\text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ) \left (-b (a B f+A c e+B c d)+2 c (-a A f+a B e+A c d)+A b^2 f\right )}{\sqrt {b^2-4 a c}}}{f \left (a^2 f-a b e+b^2 d\right )+a c \left (e^2-2 d f\right )-b c d e+c^2 d^2}-\frac {\frac {1}{2} \log \left (d+e x+f x^2\right ) (-a B f+A b f-A c e+B c d)-\frac {\text {arctanh}\left (\frac {e+2 f x}{\sqrt {e^2-4 d f}}\right ) \left (B (a e f-2 b d f+c d e)-A \left (2 a f^2-b e f-2 c d f+c e^2\right )\right )}{\sqrt {e^2-4 d f}}}{f \left (a^2 f-a b e+b^2 d\right )+a c \left (e^2-2 d f\right )-b c d e+c^2 d^2}\)

input

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

output

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

3.1.15.3.1 Defintions of rubi rules used

rule 25

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

rule 219

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

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{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 1355

Int[((g_.) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_) + (e_.)* 
(x_) + (f_.)*(x_)^2)), x_Symbol] :> With[{q = Simplify[c^2*d^2 - b*c*d*e + 
a*c*e^2 + b^2*d*f - 2*a*c*d*f - a*b*e*f + a^2*f^2]}, Simp[1/q   Int[Simp[g* 
c^2*d - g*b*c*e + a*h*c*e + g*b^2*f - a*b*h*f - a*g*c*f + c*(h*c*d - g*c*e 
+ g*b*f - a*h*f)*x, x]/(a + b*x + c*x^2), x], x] + Simp[1/q   Int[Simp[(-h) 
*c*d*e + g*c*e^2 + b*h*d*f - g*c*d*f - g*b*e*f + a*g*f^2 - f*(h*c*d - g*c*e 
 + g*b*f - a*h*f)*x, x]/(d + e*x + f*x^2), x], x] /; NeQ[q, 0]] /; FreeQ[{a 
, b, c, d, e, f, g, h}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0]
 

3.1.15.4 Maple [A] (verified)

Time = 1.57 (sec) , antiderivative size = 384, normalized size of antiderivative = 0.95

method	result	size
default	\(\frac {\frac {\left (-A b \,f^{2}+A c e f +B a \,f^{2}-B c d f \right ) \ln \left (f \,x^{2}+e x +d \right )}{2 f}+\frac {2 \left (A a \,f^{2}-A b e f -A c d f +A c \,e^{2}+B b d f -B c d e -\frac {\left (-A b \,f^{2}+A c e f +B a \,f^{2}-B c d f \right ) e}{2 f}\right ) \arctan \left (\frac {2 f x +e}{\sqrt {4 d f -e^{2}}}\right )}{\sqrt {4 d f -e^{2}}}}{a^{2} f^{2}-a b e f -2 a c d f +a c \,e^{2}+b^{2} d f -b c d e +c^{2} d^{2}}+\frac {\frac {\left (A b c f -A \,c^{2} e -B a c f +B \,c^{2} d \right ) \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {2 \left (-A a c f +A \,b^{2} f -A b c e +A \,c^{2} d -B a b f +B a c e -\frac {\left (A b c f -A \,c^{2} e -B a c f +B \,c^{2} d \right ) b}{2 c}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{a^{2} f^{2}-a b e f -2 a c d f +a c \,e^{2}+b^{2} d f -b c d e +c^{2} d^{2}}\)	\(384\)
risch	\(\text {Expression too large to display}\)	\(2635514\)

input

int((B*x+A)/(c*x^2+b*x+a)/(f*x^2+e*x+d),x,method=_RETURNVERBOSE)
 

output

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

3.1.15.5 Fricas [F(-1)]

Timed out. \[ \int \frac {A+B x}{\left (a+b x+c x^2\right ) \left (d+e x+f x^2\right )} \, dx=\text {Timed out} \]

input

integrate((B*x+A)/(c*x^2+b*x+a)/(f*x^2+e*x+d),x, algorithm="fricas")
 

output

Timed out
 

3.1.15.6 Sympy [F(-1)]

Timed out. \[ \int \frac {A+B x}{\left (a+b x+c x^2\right ) \left (d+e x+f x^2\right )} \, dx=\text {Timed out} \]

input

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

output

Timed out
 

3.1.15.7 Maxima [F(-2)]

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

input

integrate((B*x+A)/(c*x^2+b*x+a)/(f*x^2+e*x+d),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*d*f-e^2>0)', see `assume?` for 
 more deta
 

3.1.15.8 Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 405, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x}{\left (a+b x+c x^2\right ) \left (d+e x+f x^2\right )} \, dx=\frac {{\left (B c d - A c e - B a f + A b f\right )} \log \left (c x^{2} + b x + a\right )}{2 \, {\left (c^{2} d^{2} - b c d e + a c e^{2} + b^{2} d f - 2 \, a c d f - a b e f + a^{2} f^{2}\right )}} - \frac {{\left (B c d - A c e - B a f + A b f\right )} \log \left (f x^{2} + e x + d\right )}{2 \, {\left (c^{2} d^{2} - b c d e + a c e^{2} + b^{2} d f - 2 \, a c d f - a b e f + a^{2} f^{2}\right )}} - \frac {{\left (B b c d - 2 \, A c^{2} d - 2 \, B a c e + A b c e + B a b f - A b^{2} f + 2 \, A a c f\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (c^{2} d^{2} - b c d e + a c e^{2} + b^{2} d f - 2 \, a c d f - a b e f + a^{2} f^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {{\left (B c d e - A c e^{2} - 2 \, B b d f + 2 \, A c d f + B a e f + A b e f - 2 \, A a f^{2}\right )} \arctan \left (\frac {2 \, f x + e}{\sqrt {-e^{2} + 4 \, d f}}\right )}{{\left (c^{2} d^{2} - b c d e + a c e^{2} + b^{2} d f - 2 \, a c d f - a b e f + a^{2} f^{2}\right )} \sqrt {-e^{2} + 4 \, d f}} \]

input

integrate((B*x+A)/(c*x^2+b*x+a)/(f*x^2+e*x+d),x, algorithm="giac")
 

output

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

3.1.15.9 Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x}{\left (a+b x+c x^2\right ) \left (d+e x+f x^2\right )} \, dx=\text {Hanged} \]

input

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

output

\text{Hanged}