Integrand size = 29, antiderivative size = 86 \[ \int \frac {1}{(e+f x) \left (a c+(b c+a d) x+b d x^2\right )} \, dx=\frac {b \log (a+b x)}{(b c-a d) (b e-a f)}-\frac {d \log (c+d x)}{(b c-a d) (d e-c f)}+\frac {f \log (e+f x)}{(b e-a f) (d e-c f)} \] Output:
b*ln(b*x+a)/(-a*d+b*c)/(-a*f+b*e)-d*ln(d*x+c)/(-a*d+b*c)/(-c*f+d*e)+f*ln(f *x+e)/(-a*f+b*e)/(-c*f+d*e)
Time = 0.04 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.93 \[ \int \frac {1}{(e+f x) \left (a c+(b c+a d) x+b d x^2\right )} \, dx=\frac {b (-d e+c f) \log (a+b x)+d (b e-a f) \log (c+d x)+(-b c+a d) f \log (e+f x)}{(b c-a d) (b e-a f) (-d e+c f)} \] Input:
Integrate[1/((e + f*x)*(a*c + (b*c + a*d)*x + b*d*x^2)),x]
Output:
(b*(-(d*e) + c*f)*Log[a + b*x] + d*(b*e - a*f)*Log[c + d*x] + (-(b*c) + a* d)*f*Log[e + f*x])/((b*c - a*d)*(b*e - a*f)*(-(d*e) + c*f))
Time = 0.30 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.15, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {1141, 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.
\(\displaystyle \int \frac {1}{(e+f x) \left (x (a d+b c)+a c+b d x^2\right )} \, dx\) |
\(\Big \downarrow \) 1141 |
\(\displaystyle b d \int \left (\frac {f^2}{b d (b e-a f) (d e-c f) (e+f x)}+\frac {b}{d (b c-a d) (b e-a f) (a+b x)}-\frac {d}{b (b c-a d) (d e-c f) (c+d x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle b d \left (\frac {\log (a+b x)}{d (b c-a d) (b e-a f)}-\frac {\log (c+d x)}{b (b c-a d) (d e-c f)}+\frac {f \log (e+f x)}{b d (b e-a f) (d e-c f)}\right )\) |
Input:
Int[1/((e + f*x)*(a*c + (b*c + a*d)*x + b*d*x^2)),x]
Output:
b*d*(Log[a + b*x]/(d*(b*c - a*d)*(b*e - a*f)) - Log[c + d*x]/(b*(b*c - a*d )*(d*e - c*f)) + (f*Log[e + f*x])/(b*d*(b*e - a*f)*(d*e - c*f)))
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_ Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[1/c^p Int[ExpandIntegrand[ (d + e*x)^m*(b/2 - q/2 + c*x)^p*(b/2 + q/2 + c*x)^p, x], x], x] /; EqQ[p, - 1] || !FractionalPowerFactorQ[q]] /; FreeQ[{a, b, c, d, e}, x] && ILtQ[p, 0] && IntegerQ[m] && NiceSqrtQ[b^2 - 4*a*c]
Time = 1.14 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.01
method | result | size |
default | \(\frac {b \ln \left (b x +a \right )}{\left (a d -b c \right ) \left (a f -b e \right )}+\frac {f \ln \left (f x +e \right )}{\left (a f -b e \right ) \left (c f -d e \right )}-\frac {d \ln \left (d x +c \right )}{\left (a d -b c \right ) \left (c f -d e \right )}\) | \(87\) |
norman | \(\frac {f \ln \left (f x +e \right )}{a c \,f^{2}-a d e f -b c e f +b d \,e^{2}}+\frac {b \ln \left (b x +a \right )}{\left (a d -b c \right ) \left (a f -b e \right )}-\frac {d \ln \left (d x +c \right )}{\left (a d -b c \right ) \left (c f -d e \right )}\) | \(94\) |
parallelrisch | \(\frac {\ln \left (b x +a \right ) b c f -\ln \left (b x +a \right ) b d e -\ln \left (d x +c \right ) a d f +\ln \left (d x +c \right ) b d e +\ln \left (f x +e \right ) a d f -\ln \left (f x +e \right ) b c f}{\left (a c \,f^{2}-a d e f -b c e f +b d \,e^{2}\right ) \left (a d -b c \right )}\) | \(103\) |
risch | \(\frac {b \ln \left (b x +a \right )}{a^{2} d f -a b c f -a b d e +c e \,b^{2}}+\frac {f \ln \left (-f x -e \right )}{a c \,f^{2}-a d e f -b c e f +b d \,e^{2}}-\frac {d \ln \left (d x +c \right )}{a c d f -a \,d^{2} e -b \,c^{2} f +b c d e}\) | \(111\) |
Input:
int(1/(f*x+e)/(a*c+(a*d+b*c)*x+b*d*x^2),x,method=_RETURNVERBOSE)
Output:
b/(a*d-b*c)/(a*f-b*e)*ln(b*x+a)+f/(a*f-b*e)/(c*f-d*e)*ln(f*x+e)-d/(a*d-b*c )/(c*f-d*e)*ln(d*x+c)
Time = 2.88 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.30 \[ \int \frac {1}{(e+f x) \left (a c+(b c+a d) x+b d x^2\right )} \, dx=\frac {{\left (b c - a d\right )} f \log \left (f x + e\right ) + {\left (b d e - b c f\right )} \log \left (b x + a\right ) - {\left (b d e - a d f\right )} \log \left (d x + c\right )}{{\left (b^{2} c d - a b d^{2}\right )} e^{2} - {\left (b^{2} c^{2} - a^{2} d^{2}\right )} e f + {\left (a b c^{2} - a^{2} c d\right )} f^{2}} \] Input:
integrate(1/(f*x+e)/(a*c+(a*d+b*c)*x+b*d*x^2),x, algorithm="fricas")
Output:
((b*c - a*d)*f*log(f*x + e) + (b*d*e - b*c*f)*log(b*x + a) - (b*d*e - a*d* f)*log(d*x + c))/((b^2*c*d - a*b*d^2)*e^2 - (b^2*c^2 - a^2*d^2)*e*f + (a*b *c^2 - a^2*c*d)*f^2)
Timed out. \[ \int \frac {1}{(e+f x) \left (a c+(b c+a d) x+b d x^2\right )} \, dx=\text {Timed out} \] Input:
integrate(1/(f*x+e)/(a*c+(a*d+b*c)*x+b*d*x**2),x)
Output:
Timed out
Time = 0.05 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.30 \[ \int \frac {1}{(e+f x) \left (a c+(b c+a d) x+b d x^2\right )} \, dx=\frac {b \log \left (b x + a\right )}{{\left (b^{2} c - a b d\right )} e - {\left (a b c - a^{2} d\right )} f} - \frac {d \log \left (d x + c\right )}{{\left (b c d - a d^{2}\right )} e - {\left (b c^{2} - a c d\right )} f} + \frac {f \log \left (f x + e\right )}{b d e^{2} + a c f^{2} - {\left (b c + a d\right )} e f} \] Input:
integrate(1/(f*x+e)/(a*c+(a*d+b*c)*x+b*d*x^2),x, algorithm="maxima")
Output:
b*log(b*x + a)/((b^2*c - a*b*d)*e - (a*b*c - a^2*d)*f) - d*log(d*x + c)/(( b*c*d - a*d^2)*e - (b*c^2 - a*c*d)*f) + f*log(f*x + e)/(b*d*e^2 + a*c*f^2 - (b*c + a*d)*e*f)
Time = 0.41 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.52 \[ \int \frac {1}{(e+f x) \left (a c+(b c+a d) x+b d x^2\right )} \, dx=\frac {b^{2} \log \left ({\left | b x + a \right |}\right )}{b^{3} c e - a b^{2} d e - a b^{2} c f + a^{2} b d f} - \frac {d^{2} \log \left ({\left | d x + c \right |}\right )}{b c d^{2} e - a d^{3} e - b c^{2} d f + a c d^{2} f} + \frac {f^{2} \log \left ({\left | f x + e \right |}\right )}{b d e^{2} f - b c e f^{2} - a d e f^{2} + a c f^{3}} \] Input:
integrate(1/(f*x+e)/(a*c+(a*d+b*c)*x+b*d*x^2),x, algorithm="giac")
Output:
b^2*log(abs(b*x + a))/(b^3*c*e - a*b^2*d*e - a*b^2*c*f + a^2*b*d*f) - d^2* log(abs(d*x + c))/(b*c*d^2*e - a*d^3*e - b*c^2*d*f + a*c*d^2*f) + f^2*log( abs(f*x + e))/(b*d*e^2*f - b*c*e*f^2 - a*d*e*f^2 + a*c*f^3)
Time = 6.17 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.99 \[ \int \frac {1}{(e+f x) \left (a c+(b c+a d) x+b d x^2\right )} \, dx=\frac {f\,\ln \left (\frac {{\left (e+f\,x\right )}^2}{\left (a+b\,x\right )\,\left (c+d\,x\right )}\right )}{2\,a\,c\,f^2-2\,e\,f\,\left (a\,d+b\,c\right )+2\,b\,d\,e^2}-\frac {\ln \left (\frac {a\,d+b\,c-\sqrt {{\left (a\,d-b\,c\right )}^2}+2\,b\,d\,x}{a\,d+b\,c+\sqrt {{\left (a\,d-b\,c\right )}^2}+2\,b\,d\,x}\right )\,\left (f\,\left (a\,d+b\,c\right )-2\,b\,d\,e\right )}{\sqrt {{\left (a\,d+b\,c\right )}^2-4\,a\,b\,c\,d}\,\left (2\,a\,c\,f^2-2\,e\,f\,\left (a\,d+b\,c\right )+2\,b\,d\,e^2\right )} \] Input:
int(1/((e + f*x)*(a*c + x*(a*d + b*c) + b*d*x^2)),x)
Output:
(f*log((e + f*x)^2/((a + b*x)*(c + d*x))))/(2*a*c*f^2 - 2*e*f*(a*d + b*c) + 2*b*d*e^2) - (log((a*d + b*c - ((a*d - b*c)^2)^(1/2) + 2*b*d*x)/(a*d + b *c + ((a*d - b*c)^2)^(1/2) + 2*b*d*x))*(f*(a*d + b*c) - 2*b*d*e))/(((a*d + b*c)^2 - 4*a*b*c*d)^(1/2)*(2*a*c*f^2 - 2*e*f*(a*d + b*c) + 2*b*d*e^2))
Time = 0.22 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.45 \[ \int \frac {1}{(e+f x) \left (a c+(b c+a d) x+b d x^2\right )} \, dx=\frac {\mathrm {log}\left (b x +a \right ) b c f -\mathrm {log}\left (b x +a \right ) b d e -\mathrm {log}\left (d x +c \right ) a d f +\mathrm {log}\left (d x +c \right ) b d e +\mathrm {log}\left (f x +e \right ) a d f -\mathrm {log}\left (f x +e \right ) b c f}{a^{2} c d \,f^{2}-a^{2} d^{2} e f -a b \,c^{2} f^{2}+a b \,d^{2} e^{2}+b^{2} c^{2} e f -b^{2} c d \,e^{2}} \] Input:
int(1/(f*x+e)/(a*c+(a*d+b*c)*x+b*d*x^2),x)
Output:
(log(a + b*x)*b*c*f - log(a + b*x)*b*d*e - log(c + d*x)*a*d*f + log(c + d* x)*b*d*e + log(e + f*x)*a*d*f - log(e + f*x)*b*c*f)/(a**2*c*d*f**2 - a**2* d**2*e*f - a*b*c**2*f**2 + a*b*d**2*e**2 + b**2*c**2*e*f - b**2*c*d*e**2)