\(\int \frac {1}{(2-3 x) (5+x) (a+b x+c x^2)} \, dx\) [89]

Optimal result

Integrand size = 25, antiderivative size = 159 \[ \int \frac {1}{(2-3 x) (5+x) \left (a+b x+c x^2\right )} \, dx=\frac {\left (3 b^2-6 a c-13 b c-20 c^2\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{(9 a+6 b+4 c) (a-5 b+25 c) \sqrt {b^2-4 a c}}-\frac {9 \log (2-3 x)}{17 (9 a+6 b+4 c)}+\frac {\log (5+x)}{17 (a-5 b+25 c)}-\frac {(3 b-13 c) \log \left (a+b x+c x^2\right )}{2 (9 a+6 b+4 c) (a-5 b+25 c)} \] Output:

(-6*a*c+3*b^2-13*b*c-20*c^2)*arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2))/(9*a+6* 
b+4*c)/(a-5*b+25*c)/(-4*a*c+b^2)^(1/2)-9*ln(2-3*x)/(153*a+102*b+68*c)+ln(5 
+x)/(17*a-85*b+425*c)-1/2*(3*b-13*c)*ln(c*x^2+b*x+a)/(9*a+6*b+4*c)/(a-5*b+ 
25*c)
 

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.03 \[ \int \frac {1}{(2-3 x) (5+x) \left (a+b x+c x^2\right )} \, dx=-\frac {\left (3 b^2-13 b c-2 c (3 a+10 c)\right ) \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )}{(9 a+6 b+4 c) (a-5 b+25 c) \sqrt {-b^2+4 a c}}-\frac {9 \log (2-3 x)}{17 (9 a+6 b+4 c)}+\frac {\log (5+x)}{17 (a-5 b+25 c)}-\frac {(3 b-13 c) \log (a+x (b+c x))}{2 (9 a+6 b+4 c) (a-5 b+25 c)} \] Input:

Integrate[1/((2 - 3*x)*(5 + x)*(a + b*x + c*x^2)),x]
 

Output:

-(((3*b^2 - 13*b*c - 2*c*(3*a + 10*c))*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a* 
c]])/((9*a + 6*b + 4*c)*(a - 5*b + 25*c)*Sqrt[-b^2 + 4*a*c])) - (9*Log[2 - 
 3*x])/(17*(9*a + 6*b + 4*c)) + Log[5 + x]/(17*(a - 5*b + 25*c)) - ((3*b - 
 13*c)*Log[a + x*(b + c*x)])/(2*(9*a + 6*b + 4*c)*(a - 5*b + 25*c))
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {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.

\(\displaystyle \int \frac {1}{(2-3 x) (x+5) \left (a+b x+c x^2\right )} \, dx\)

\(\Big \downarrow \) 1200

\(\displaystyle \int \left (\frac {c (3 a+10 c)-3 b^2-c x (3 b-13 c)+13 b c}{(9 a+6 b+4 c) (a-5 b+25 c) \left (a+b x+c x^2\right )}+\frac {1}{17 (x+5) (a-5 b+25 c)}-\frac {27}{17 (3 x-2) (9 a+6 b+4 c)}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\left (-6 a c+3 b^2-13 b c-20 c^2\right ) \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{(9 a+6 b+4 c) (a-5 b+25 c) \sqrt {b^2-4 a c}}-\frac {(3 b-13 c) \log \left (a+b x+c x^2\right )}{2 (9 a+6 b+4 c) (a-5 b+25 c)}-\frac {9 \log (2-3 x)}{17 (9 a+6 b+4 c)}+\frac {\log (x+5)}{17 (a-5 b+25 c)}\)

Input:

Int[1/((2 - 3*x)*(5 + x)*(a + b*x + c*x^2)),x]
 

Output:

((3*b^2 - 6*a*c - 13*b*c - 20*c^2)*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]]) 
/((9*a + 6*b + 4*c)*(a - 5*b + 25*c)*Sqrt[b^2 - 4*a*c]) - (9*Log[2 - 3*x]) 
/(17*(9*a + 6*b + 4*c)) + Log[5 + x]/(17*(a - 5*b + 25*c)) - ((3*b - 13*c) 
*Log[a + b*x + c*x^2])/(2*(9*a + 6*b + 4*c)*(a - 5*b + 25*c))
 

Defintions of rubi rules used

rule 1200

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]
 

rule 2009

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

Maple [A] (verified)

Time = 1.94 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.99

method	result	size
default	\(\frac {\ln \left (x +5\right )}{17 a -85 b +425 c}-\frac {9 \ln \left (-2+3 x \right )}{153 a +102 b +68 c}+\frac {\frac {\left (-3 b c +13 c^{2}\right ) \ln \left (c \,x^{2}+b x +a \right )}{2 c}+\frac {2 \left (3 a c -3 b^{2}+13 b c +10 c^{2}-\frac {\left (-3 b c +13 c^{2}\right ) b}{2 c}\right ) \arctan \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{\left (a -5 b +25 c \right ) \left (9 a +6 b +4 c \right )}\)	\(158\)
risch	\(\text {Expression too large to display}\)	\(21328\)

Input:

int(1/(2-3*x)/(x+5)/(c*x^2+b*x+a),x,method=_RETURNVERBOSE)
 

Output:

ln(x+5)/(17*a-85*b+425*c)-9/(153*a+102*b+68*c)*ln(-2+3*x)+1/(a-5*b+25*c)/( 
9*a+6*b+4*c)*(1/2*(-3*b*c+13*c^2)/c*ln(c*x^2+b*x+a)+2*(3*a*c-3*b^2+13*b*c+ 
10*c^2-1/2*(-3*b*c+13*c^2)*b/c)/(4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c- 
b^2)^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 1.81 (sec) , antiderivative size = 553, normalized size of antiderivative = 3.48 \[ \int \frac {1}{(2-3 x) (5+x) \left (a+b x+c x^2\right )} \, dx=\left [-\frac {17 \, {\left (3 \, b^{2} - {\left (6 \, a + 13 \, b\right )} c - 20 \, c^{2}\right )} \sqrt {b^{2} - 4 \, a c} \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c - \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) + 17 \, {\left (3 \, b^{3} + 52 \, a c^{2} - {\left (12 \, a b + 13 \, b^{2}\right )} c\right )} \log \left (c x^{2} + b x + a\right ) + 18 \, {\left (a b^{2} - 5 \, b^{3} - 100 \, a c^{2} - {\left (4 \, a^{2} - 20 \, a b - 25 \, b^{2}\right )} c\right )} \log \left (3 \, x - 2\right ) - 2 \, {\left (9 \, a b^{2} + 6 \, b^{3} - 16 \, a c^{2} - 4 \, {\left (9 \, a^{2} + 6 \, a b - b^{2}\right )} c\right )} \log \left (x + 5\right )}{34 \, {\left (9 \, a^{2} b^{2} - 39 \, a b^{3} - 30 \, b^{4} - 400 \, a c^{3} - 4 \, {\left (229 \, a^{2} + 130 \, a b - 25 \, b^{2}\right )} c^{2} - {\left (36 \, a^{3} - 156 \, a^{2} b - 349 \, a b^{2} - 130 \, b^{3}\right )} c\right )}}, \frac {34 \, {\left (3 \, b^{2} - {\left (6 \, a + 13 \, b\right )} c - 20 \, c^{2}\right )} \sqrt {-b^{2} + 4 \, a c} \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) - 17 \, {\left (3 \, b^{3} + 52 \, a c^{2} - {\left (12 \, a b + 13 \, b^{2}\right )} c\right )} \log \left (c x^{2} + b x + a\right ) - 18 \, {\left (a b^{2} - 5 \, b^{3} - 100 \, a c^{2} - {\left (4 \, a^{2} - 20 \, a b - 25 \, b^{2}\right )} c\right )} \log \left (3 \, x - 2\right ) + 2 \, {\left (9 \, a b^{2} + 6 \, b^{3} - 16 \, a c^{2} - 4 \, {\left (9 \, a^{2} + 6 \, a b - b^{2}\right )} c\right )} \log \left (x + 5\right )}{34 \, {\left (9 \, a^{2} b^{2} - 39 \, a b^{3} - 30 \, b^{4} - 400 \, a c^{3} - 4 \, {\left (229 \, a^{2} + 130 \, a b - 25 \, b^{2}\right )} c^{2} - {\left (36 \, a^{3} - 156 \, a^{2} b - 349 \, a b^{2} - 130 \, b^{3}\right )} c\right )}}\right ] \] Input:

integrate(1/(2-3*x)/(5+x)/(c*x^2+b*x+a),x, algorithm="fricas")
 

Output:

[-1/34*(17*(3*b^2 - (6*a + 13*b)*c - 20*c^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)) + 17*(3*b^3 + 52*a*c^2 - (12*a*b + 13*b^2)*c)*log(c*x^2 + b*x + a) + 
 18*(a*b^2 - 5*b^3 - 100*a*c^2 - (4*a^2 - 20*a*b - 25*b^2)*c)*log(3*x - 2) 
 - 2*(9*a*b^2 + 6*b^3 - 16*a*c^2 - 4*(9*a^2 + 6*a*b - b^2)*c)*log(x + 5))/ 
(9*a^2*b^2 - 39*a*b^3 - 30*b^4 - 400*a*c^3 - 4*(229*a^2 + 130*a*b - 25*b^2 
)*c^2 - (36*a^3 - 156*a^2*b - 349*a*b^2 - 130*b^3)*c), 1/34*(34*(3*b^2 - ( 
6*a + 13*b)*c - 20*c^2)*sqrt(-b^2 + 4*a*c)*arctan(-sqrt(-b^2 + 4*a*c)*(2*c 
*x + b)/(b^2 - 4*a*c)) - 17*(3*b^3 + 52*a*c^2 - (12*a*b + 13*b^2)*c)*log(c 
*x^2 + b*x + a) - 18*(a*b^2 - 5*b^3 - 100*a*c^2 - (4*a^2 - 20*a*b - 25*b^2 
)*c)*log(3*x - 2) + 2*(9*a*b^2 + 6*b^3 - 16*a*c^2 - 4*(9*a^2 + 6*a*b - b^2 
)*c)*log(x + 5))/(9*a^2*b^2 - 39*a*b^3 - 30*b^4 - 400*a*c^3 - 4*(229*a^2 + 
 130*a*b - 25*b^2)*c^2 - (36*a^3 - 156*a^2*b - 349*a*b^2 - 130*b^3)*c)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(2-3 x) (5+x) \left (a+b x+c x^2\right )} \, dx=\text {Timed out} \] Input:

integrate(1/(2-3*x)/(5+x)/(c*x**2+b*x+a),x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(2-3 x) (5+x) \left (a+b x+c x^2\right )} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(2-3*x)/(5+x)/(c*x^2+b*x+a),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
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.08 \[ \int \frac {1}{(2-3 x) (5+x) \left (a+b x+c x^2\right )} \, dx=-\frac {{\left (3 \, b - 13 \, c\right )} \log \left (c x^{2} + b x + a\right )}{2 \, {\left (9 \, a^{2} - 39 \, a b - 30 \, b^{2} + 229 \, a c + 130 \, b c + 100 \, c^{2}\right )}} - \frac {{\left (3 \, b^{2} - 6 \, a c - 13 \, b c - 20 \, c^{2}\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{{\left (9 \, a^{2} - 39 \, a b - 30 \, b^{2} + 229 \, a c + 130 \, b c + 100 \, c^{2}\right )} \sqrt {-b^{2} + 4 \, a c}} - \frac {9 \, \log \left ({\left | 3 \, x - 2 \right |}\right )}{17 \, {\left (9 \, a + 6 \, b + 4 \, c\right )}} + \frac {\log \left ({\left | x + 5 \right |}\right )}{17 \, {\left (a - 5 \, b + 25 \, c\right )}} \] Input:

integrate(1/(2-3*x)/(5+x)/(c*x^2+b*x+a),x, algorithm="giac")
 

Output:

-1/2*(3*b - 13*c)*log(c*x^2 + b*x + a)/(9*a^2 - 39*a*b - 30*b^2 + 229*a*c 
+ 130*b*c + 100*c^2) - (3*b^2 - 6*a*c - 13*b*c - 20*c^2)*arctan((2*c*x + b 
)/sqrt(-b^2 + 4*a*c))/((9*a^2 - 39*a*b - 30*b^2 + 229*a*c + 130*b*c + 100* 
c^2)*sqrt(-b^2 + 4*a*c)) - 9/17*log(abs(3*x - 2))/(9*a + 6*b + 4*c) + 1/17 
*log(abs(x + 5))/(a - 5*b + 25*c)
 

Mupad [B] (verification not implemented)

Time = 20.08 (sec) , antiderivative size = 2331, normalized size of antiderivative = 14.66 \[ \int \frac {1}{(2-3 x) (5+x) \left (a+b x+c x^2\right )} \, dx=\text {Too large to display} \] Input:

int(-1/((3*x - 2)*(x + 5)*(a + b*x + c*x^2)),x)
 

Output:

log(x + 5)/(17*a - 85*b + 425*c) - (9*log(x - 2/3))/(17*(9*a + 6*b + 4*c)) 
 - (log(3757000*a*c^5 - 16416*a*b^5 + 289000*b*c^5 + 2700*b^5*c - 16416*b^ 
6*x + 578000*c^6*x - 289000*c^5*(b^2 - 4*a*c)^(1/2) - 1053*a^2*b^4 + 243*a 
^3*b^3 - 1461798*a^2*c^4 + 552474*a^3*c^3 + 8424*a^4*c^2 - 375700*b^2*c^4 
+ 77700*b^3*c^3 - 11700*b^4*c^2 - 16416*a*b^4*(b^2 - 4*a*c)^(1/2) + 312277 
0*a*c^4*(b^2 - 4*a*c)^(1/2) - 972*a^4*c*(b^2 - 4*a*c)^(1/2) + 375700*b*c^4 
*(b^2 - 4*a*c)^(1/2) - 2700*b^4*c*(b^2 - 4*a*c)^(1/2) - 16416*b^5*x*(b^2 - 
 4*a*c)^(1/2) + 1878500*c^5*x*(b^2 - 4*a*c)^(1/2) + 391586*a*b^2*c^3 - 808 
218*a*b^3*c^2 + 2828133*a^2*b*c^3 + 95769*a^2*b^3*c - 128223*a^3*b*c^2 + 2 
106*a^3*b^2*c + 243*a^2*b^4*x - 1708782*a^2*c^4*x + 71550*a^3*c^3*x + 1944 
*a^4*c^2*x + 2163950*b^2*c^4*x + 107666*b^3*c^3*x - 756198*b^4*c^2*x - 105 
3*a^2*b^3*(b^2 - 4*a*c)^(1/2) + 243*a^3*b^2*(b^2 - 4*a*c)^(1/2) + 854391*a 
^2*c^3*(b^2 - 4*a*c)^(1/2) - 35775*a^3*c^2*(b^2 - 4*a*c)^(1/2) - 77700*b^2 
*c^3*(b^2 - 4*a*c)^(1/2) + 11700*b^3*c^2*(b^2 - 4*a*c)^(1/2) - 955071*a^2* 
b^2*c^2 + 894330*a*b*c^4 + 216918*a*b^4*c - 972*a^4*b*c - 1053*a*b^5*x - 6 
245540*a*c^5*x + 1127100*b*c^5*x + 209898*b^5*c*x - 196101*a^2*b^2*c^2*x - 
 471991*a*b*c^4*x + 109242*a*b^4*c*x - 129454*a*b*c^3*(b^2 - 4*a*c)^(1/2) 
+ 202878*a*b^3*c*(b^2 - 4*a*c)^(1/2) + 4212*a^3*b*c*(b^2 - 4*a*c)^(1/2) - 
1053*a*b^4*x*(b^2 - 4*a*c)^(1/2) - 730899*a*c^4*x*(b^2 - 4*a*c)^(1/2) + 20 
08550*b*c^4*x*(b^2 - 4*a*c)^(1/2) + 209898*b^4*c*x*(b^2 - 4*a*c)^(1/2) ...
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 419, normalized size of antiderivative = 2.64 \[ \int \frac {1}{(2-3 x) (5+x) \left (a+b x+c x^2\right )} \, dx=\frac {204 \sqrt {4 a c -b^{2}}\, \mathit {atan} \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right ) a c -102 \sqrt {4 a c -b^{2}}\, \mathit {atan} \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right ) b^{2}+442 \sqrt {4 a c -b^{2}}\, \mathit {atan} \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right ) b c +680 \sqrt {4 a c -b^{2}}\, \mathit {atan} \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right ) c^{2}-72 \,\mathrm {log}\left (3 x -2\right ) a^{2} c +18 \,\mathrm {log}\left (3 x -2\right ) a \,b^{2}+360 \,\mathrm {log}\left (3 x -2\right ) a b c -1800 \,\mathrm {log}\left (3 x -2\right ) a \,c^{2}-90 \,\mathrm {log}\left (3 x -2\right ) b^{3}+450 \,\mathrm {log}\left (3 x -2\right ) b^{2} c -204 \,\mathrm {log}\left (c \,x^{2}+b x +a \right ) a b c +884 \,\mathrm {log}\left (c \,x^{2}+b x +a \right ) a \,c^{2}+51 \,\mathrm {log}\left (c \,x^{2}+b x +a \right ) b^{3}-221 \,\mathrm {log}\left (c \,x^{2}+b x +a \right ) b^{2} c +72 \,\mathrm {log}\left (x +5\right ) a^{2} c -18 \,\mathrm {log}\left (x +5\right ) a \,b^{2}+48 \,\mathrm {log}\left (x +5\right ) a b c +32 \,\mathrm {log}\left (x +5\right ) a \,c^{2}-12 \,\mathrm {log}\left (x +5\right ) b^{3}-8 \,\mathrm {log}\left (x +5\right ) b^{2} c}{1224 a^{3} c -306 a^{2} b^{2}-5304 a^{2} b c +31144 a^{2} c^{2}+1326 a \,b^{3}-11866 a \,b^{2} c +17680 a b \,c^{2}+13600 a \,c^{3}+1020 b^{4}-4420 b^{3} c -3400 b^{2} c^{2}} \] Input:

int(1/(2-3*x)/(5+x)/(c*x^2+b*x+a),x)
 

Output:

(204*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*a*c - 102*sqr 
t(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b**2 + 442*sqrt(4*a*c 
 - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*b*c + 680*sqrt(4*a*c - b**2) 
*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*c**2 - 72*log(3*x - 2)*a**2*c + 18*l 
og(3*x - 2)*a*b**2 + 360*log(3*x - 2)*a*b*c - 1800*log(3*x - 2)*a*c**2 - 9 
0*log(3*x - 2)*b**3 + 450*log(3*x - 2)*b**2*c - 204*log(a + b*x + c*x**2)* 
a*b*c + 884*log(a + b*x + c*x**2)*a*c**2 + 51*log(a + b*x + c*x**2)*b**3 - 
 221*log(a + b*x + c*x**2)*b**2*c + 72*log(x + 5)*a**2*c - 18*log(x + 5)*a 
*b**2 + 48*log(x + 5)*a*b*c + 32*log(x + 5)*a*c**2 - 12*log(x + 5)*b**3 - 
8*log(x + 5)*b**2*c)/(34*(36*a**3*c - 9*a**2*b**2 - 156*a**2*b*c + 916*a** 
2*c**2 + 39*a*b**3 - 349*a*b**2*c + 520*a*b*c**2 + 400*a*c**3 + 30*b**4 - 
130*b**3*c - 100*b**2*c**2))