\(\int \frac {1}{x^2 (a+b x) (c+d x)} \, dx\) [239]

Optimal result

Integrand size = 18, antiderivative size = 76 \[ \int \frac {1}{x^2 (a+b x) (c+d x)} \, dx=-\frac {1}{a c x}-\frac {(b c+a d) \log (x)}{a^2 c^2}+\frac {b^2 \log (a+b x)}{a^2 (b c-a d)}-\frac {d^2 \log (c+d x)}{c^2 (b c-a d)} \]

[Out]

-1/a/c/x-(a*d+b*c)*ln(x)/a^2/c^2+b^2*ln(b*x+a)/a^2/(-a*d+b*c)-d^2*ln(d*x+c)/c^2/(-a*d+b*c)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {84} \[ \int \frac {1}{x^2 (a+b x) (c+d x)} \, dx=\frac {b^2 \log (a+b x)}{a^2 (b c-a d)}-\frac {\log (x) (a d+b c)}{a^2 c^2}-\frac {d^2 \log (c+d x)}{c^2 (b c-a d)}-\frac {1}{a c x} \]

[In]

Int[1/(x^2*(a + b*x)*(c + d*x)),x]

[Out]

-(1/(a*c*x)) - ((b*c + a*d)*Log[x])/(a^2*c^2) + (b^2*Log[a + b*x])/(a^2*(b*c - a*d)) - (d^2*Log[c + d*x])/(c^2
*(b*c - a*d))

Rule 84

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{a c x^2}+\frac {-b c-a d}{a^2 c^2 x}-\frac {b^3}{a^2 (-b c+a d) (a+b x)}-\frac {d^3}{c^2 (b c-a d) (c+d x)}\right ) \, dx \\ & = -\frac {1}{a c x}-\frac {(b c+a d) \log (x)}{a^2 c^2}+\frac {b^2 \log (a+b x)}{a^2 (b c-a d)}-\frac {d^2 \log (c+d x)}{c^2 (b c-a d)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.03 \[ \int \frac {1}{x^2 (a+b x) (c+d x)} \, dx=-\frac {1}{a c x}+\frac {(-b c-a d) \log (x)}{a^2 c^2}-\frac {b^2 \log (a+b x)}{a^2 (-b c+a d)}-\frac {d^2 \log (c+d x)}{c^2 (b c-a d)} \]

[In]

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

[Out]

-(1/(a*c*x)) + ((-(b*c) - a*d)*Log[x])/(a^2*c^2) - (b^2*Log[a + b*x])/(a^2*(-(b*c) + a*d)) - (d^2*Log[c + d*x]
)/(c^2*(b*c - a*d))

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.01

method	result	size
norman	\(-\frac {1}{a c x}+\frac {d^{2} \ln \left (d x +c \right )}{c^{2} \left (a d -b c \right )}-\frac {b^{2} \ln \left (b x +a \right )}{a^{2} \left (a d -b c \right )}-\frac {\left (a d +b c \right ) \ln \left (x \right )}{a^{2} c^{2}}\)	\(77\)
default	\(-\frac {1}{a c x}+\frac {\left (-a d -b c \right ) \ln \left (x \right )}{a^{2} c^{2}}+\frac {d^{2} \ln \left (d x +c \right )}{c^{2} \left (a d -b c \right )}-\frac {b^{2} \ln \left (b x +a \right )}{a^{2} \left (a d -b c \right )}\)	\(78\)
parallelrisch	\(-\frac {\ln \left (x \right ) x \,a^{2} d^{2}-\ln \left (x \right ) x \,b^{2} c^{2}+\ln \left (b x +a \right ) x \,b^{2} c^{2}-d^{2} \ln \left (d x +c \right ) a^{2} x +a^{2} c d -b \,c^{2} a}{a^{2} c^{2} x \left (a d -b c \right )}\)	\(86\)
risch	\(-\frac {1}{a c x}-\frac {\ln \left (-x \right ) d}{c^{2} a}-\frac {\ln \left (-x \right ) b}{c \,a^{2}}-\frac {b^{2} \ln \left (b x +a \right )}{a^{2} \left (a d -b c \right )}+\frac {d^{2} \ln \left (-d x -c \right )}{c^{2} \left (a d -b c \right )}\)	\(89\)

[In]

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

[Out]

-1/a/c/x+d^2/c^2/(a*d-b*c)*ln(d*x+c)-b^2/a^2/(a*d-b*c)*ln(b*x+a)-(a*d+b*c)*ln(x)/a^2/c^2

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.16 \[ \int \frac {1}{x^2 (a+b x) (c+d x)} \, dx=\frac {b^{2} c^{2} x \log \left (b x + a\right ) - a^{2} d^{2} x \log \left (d x + c\right ) - a b c^{2} + a^{2} c d - {\left (b^{2} c^{2} - a^{2} d^{2}\right )} x \log \left (x\right )}{{\left (a^{2} b c^{3} - a^{3} c^{2} d\right )} x} \]

[In]

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

[Out]

(b^2*c^2*x*log(b*x + a) - a^2*d^2*x*log(d*x + c) - a*b*c^2 + a^2*c*d - (b^2*c^2 - a^2*d^2)*x*log(x))/((a^2*b*c
^3 - a^3*c^2*d)*x)

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{x^2 (a+b x) (c+d x)} \, dx=\text {Timed out} \]

[In]

integrate(1/x**2/(b*x+a)/(d*x+c),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.05 \[ \int \frac {1}{x^2 (a+b x) (c+d x)} \, dx=\frac {b^{2} \log \left (b x + a\right )}{a^{2} b c - a^{3} d} - \frac {d^{2} \log \left (d x + c\right )}{b c^{3} - a c^{2} d} - \frac {{\left (b c + a d\right )} \log \left (x\right )}{a^{2} c^{2}} - \frac {1}{a c x} \]

[In]

integrate(1/x^2/(b*x+a)/(d*x+c),x, algorithm="maxima")

[Out]

b^2*log(b*x + a)/(a^2*b*c - a^3*d) - d^2*log(d*x + c)/(b*c^3 - a*c^2*d) - (b*c + a*d)*log(x)/(a^2*c^2) - 1/(a*
c*x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.17 \[ \int \frac {1}{x^2 (a+b x) (c+d x)} \, dx=\frac {b^{3} \log \left ({\left | b x + a \right |}\right )}{a^{2} b^{2} c - a^{3} b d} - \frac {d^{3} \log \left ({\left | d x + c \right |}\right )}{b c^{3} d - a c^{2} d^{2}} - \frac {{\left (b c + a d\right )} \log \left ({\left | x \right |}\right )}{a^{2} c^{2}} - \frac {1}{a c x} \]

[In]

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

[Out]

b^3*log(abs(b*x + a))/(a^2*b^2*c - a^3*b*d) - d^3*log(abs(d*x + c))/(b*c^3*d - a*c^2*d^2) - (b*c + a*d)*log(ab
s(x))/(a^2*c^2) - 1/(a*c*x)

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.03 \[ \int \frac {1}{x^2 (a+b x) (c+d x)} \, dx=\frac {d^2\,\ln \left (c+d\,x\right )}{c^2\,\left (a\,d-b\,c\right )}-\frac {1}{a\,c\,x}-\frac {b^2\,\ln \left (a+b\,x\right )}{a^3\,d-a^2\,b\,c}-\frac {\ln \left (x\right )\,\left (a\,d+b\,c\right )}{a^2\,c^2} \]

[In]

int(1/(x^2*(a + b*x)*(c + d*x)),x)

[Out]

(d^2*log(c + d*x))/(c^2*(a*d - b*c)) - 1/(a*c*x) - (b^2*log(a + b*x))/(a^3*d - a^2*b*c) - (log(x)*(a*d + b*c))
/(a^2*c^2)