\(\int \frac {1}{x^2 (a+b x) (c+d x)^2} \, dx\) [249]

Optimal result

Integrand size = 18, antiderivative size = 110 \[ \int \frac {1}{x^2 (a+b x) (c+d x)^2} \, dx=-\frac {1}{a c^2 x}+\frac {d^2}{c^2 (b c-a d) (c+d x)}-\frac {(b c+2 a d) \log (x)}{a^2 c^3}+\frac {b^3 \log (a+b x)}{a^2 (b c-a d)^2}-\frac {d^2 (3 b c-2 a d) \log (c+d x)}{c^3 (b c-a d)^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 110, 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 = {90} \[ \int \frac {1}{x^2 (a+b x) (c+d x)^2} \, dx=\frac {b^3 \log (a+b x)}{a^2 (b c-a d)^2}-\frac {\log (x) (2 a d+b c)}{a^2 c^3}-\frac {d^2 (3 b c-2 a d) \log (c+d x)}{c^3 (b c-a d)^2}+\frac {d^2}{c^2 (c+d x) (b c-a d)}-\frac {1}{a c^2 x} \]

[In]

Int[1/(x^2*(a + b*x)*(c + d*x)^2),x]

[Out]

-(1/(a*c^2*x)) + d^2/(c^2*(b*c - a*d)*(c + d*x)) - ((b*c + 2*a*d)*Log[x])/(a^2*c^3) + (b^3*Log[a + b*x])/(a^2*
(b*c - a*d)^2) - (d^2*(3*b*c - 2*a*d)*Log[c + d*x])/(c^3*(b*c - a*d)^2)

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{a c^2 x^2}+\frac {-b c-2 a d}{a^2 c^3 x}+\frac {b^4}{a^2 (-b c+a d)^2 (a+b x)}-\frac {d^3}{c^2 (b c-a d) (c+d x)^2}-\frac {d^3 (3 b c-2 a d)}{c^3 (b c-a d)^2 (c+d x)}\right ) \, dx \\ & = -\frac {1}{a c^2 x}+\frac {d^2}{c^2 (b c-a d) (c+d x)}-\frac {(b c+2 a d) \log (x)}{a^2 c^3}+\frac {b^3 \log (a+b x)}{a^2 (b c-a d)^2}-\frac {d^2 (3 b c-2 a d) \log (c+d x)}{c^3 (b c-a d)^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.01 \[ \int \frac {1}{x^2 (a+b x) (c+d x)^2} \, dx=-\frac {1}{a c^2 x}+\frac {d^2}{c^2 (b c-a d) (c+d x)}+\frac {(-b c-2 a d) \log (x)}{a^2 c^3}+\frac {b^3 \log (a+b x)}{a^2 (-b c+a d)^2}+\frac {\left (-3 b c d^2+2 a d^3\right ) \log (c+d x)}{c^3 (b c-a d)^2} \]

[In]

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

[Out]

-(1/(a*c^2*x)) + d^2/(c^2*(b*c - a*d)*(c + d*x)) + ((-(b*c) - 2*a*d)*Log[x])/(a^2*c^3) + (b^3*Log[a + b*x])/(a
^2*(-(b*c) + a*d)^2) + ((-3*b*c*d^2 + 2*a*d^3)*Log[c + d*x])/(c^3*(b*c - a*d)^2)

Maple [A] (verified)

Time = 1.28 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.01

method	result	size
default	\(-\frac {1}{a \,c^{2} x}+\frac {\left (-2 a d -b c \right ) \ln \left (x \right )}{a^{2} c^{3}}-\frac {d^{2}}{c^{2} \left (a d -b c \right ) \left (d x +c \right )}+\frac {d^{2} \left (2 a d -3 b c \right ) \ln \left (d x +c \right )}{c^{3} \left (a d -b c \right )^{2}}+\frac {b^{3} \ln \left (b x +a \right )}{a^{2} \left (a d -b c \right )^{2}}\)	\(111\)
norman	\(\frac {\frac {\left (-2 a \,d^{3}+b c \,d^{2}\right ) x}{c^{2} d a \left (a d -b c \right )}-\frac {1}{a c}}{\left (d x +c \right ) x}+\frac {b^{3} \ln \left (b x +a \right )}{a^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {d^{2} \left (2 a d -3 b c \right ) \ln \left (d x +c \right )}{c^{3} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}-\frac {\left (2 a d +b c \right ) \ln \left (x \right )}{a^{2} c^{3}}\)	\(155\)
risch	\(\frac {-\frac {d \left (2 a d -b c \right ) x}{a \,c^{2} \left (a d -b c \right )}-\frac {1}{a c}}{\left (d x +c \right ) x}-\frac {2 \ln \left (-x \right ) d}{a \,c^{3}}-\frac {\ln \left (-x \right ) b}{a^{2} c^{2}}+\frac {b^{3} \ln \left (b x +a \right )}{a^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {2 d^{3} \ln \left (-d x -c \right ) a}{c^{3} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}-\frac {3 d^{2} \ln \left (-d x -c \right ) b}{c^{2} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}\)	\(195\)
parallelrisch	\(-\frac {2 \ln \left (x \right ) x^{2} a^{3} d^{5}-3 \ln \left (x \right ) x^{2} a^{2} b c \,d^{4}+\ln \left (x \right ) x^{2} b^{3} c^{3} d^{2}-\ln \left (b x +a \right ) x^{2} b^{3} c^{3} d^{2}-2 \ln \left (d x +c \right ) x^{2} a^{3} d^{5}+3 \ln \left (d x +c \right ) x^{2} a^{2} b c \,d^{4}+2 \ln \left (x \right ) x \,a^{3} c \,d^{4}-3 \ln \left (x \right ) x \,a^{2} b \,c^{2} d^{3}+\ln \left (x \right ) x \,b^{3} c^{4} d -\ln \left (b x +a \right ) x \,b^{3} c^{4} d -2 \ln \left (d x +c \right ) x \,a^{3} c \,d^{4}+3 \ln \left (d x +c \right ) x \,a^{2} b \,c^{2} d^{3}+2 x \,a^{3} c \,d^{4}-3 x \,a^{2} b \,c^{2} d^{3}+x a \,b^{2} c^{3} d^{2}+a^{3} c^{2} d^{3}-2 a^{2} b \,c^{3} d^{2}+a \,b^{2} c^{4} d}{\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \left (d x +c \right ) x \,a^{2} c^{3} d}\)	\(300\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (110) = 220\).

Time = 2.21 (sec) , antiderivative size = 287, normalized size of antiderivative = 2.61 \[ \int \frac {1}{x^2 (a+b x) (c+d x)^2} \, dx=-\frac {a b^{2} c^{4} - 2 \, a^{2} b c^{3} d + a^{3} c^{2} d^{2} + {\left (a b^{2} c^{3} d - 3 \, a^{2} b c^{2} d^{2} + 2 \, a^{3} c d^{3}\right )} x - {\left (b^{3} c^{3} d x^{2} + b^{3} c^{4} x\right )} \log \left (b x + a\right ) + {\left ({\left (3 \, a^{2} b c d^{3} - 2 \, a^{3} d^{4}\right )} x^{2} + {\left (3 \, a^{2} b c^{2} d^{2} - 2 \, a^{3} c d^{3}\right )} x\right )} \log \left (d x + c\right ) + {\left ({\left (b^{3} c^{3} d - 3 \, a^{2} b c d^{3} + 2 \, a^{3} d^{4}\right )} x^{2} + {\left (b^{3} c^{4} - 3 \, a^{2} b c^{2} d^{2} + 2 \, a^{3} c d^{3}\right )} x\right )} \log \left (x\right )}{{\left (a^{2} b^{2} c^{5} d - 2 \, a^{3} b c^{4} d^{2} + a^{4} c^{3} d^{3}\right )} x^{2} + {\left (a^{2} b^{2} c^{6} - 2 \, a^{3} b c^{5} d + a^{4} c^{4} d^{2}\right )} x} \]

[In]

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

[Out]

-(a*b^2*c^4 - 2*a^2*b*c^3*d + a^3*c^2*d^2 + (a*b^2*c^3*d - 3*a^2*b*c^2*d^2 + 2*a^3*c*d^3)*x - (b^3*c^3*d*x^2 +
b^3*c^4*x)*log(b*x + a) + ((3*a^2*b*c*d^3 - 2*a^3*d^4)*x^2 + (3*a^2*b*c^2*d^2 - 2*a^3*c*d^3)*x)*log(d*x + c)
+ ((b^3*c^3*d - 3*a^2*b*c*d^3 + 2*a^3*d^4)*x^2 + (b^3*c^4 - 3*a^2*b*c^2*d^2 + 2*a^3*c*d^3)*x)*log(x))/((a^2*b^
2*c^5*d - 2*a^3*b*c^4*d^2 + a^4*c^3*d^3)*x^2 + (a^2*b^2*c^6 - 2*a^3*b*c^5*d + a^4*c^4*d^2)*x)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.61 \[ \int \frac {1}{x^2 (a+b x) (c+d x)^2} \, dx=\frac {b^{3} \log \left (b x + a\right )}{a^{2} b^{2} c^{2} - 2 \, a^{3} b c d + a^{4} d^{2}} - \frac {{\left (3 \, b c d^{2} - 2 \, a d^{3}\right )} \log \left (d x + c\right )}{b^{2} c^{5} - 2 \, a b c^{4} d + a^{2} c^{3} d^{2}} - \frac {b c^{2} - a c d + {\left (b c d - 2 \, a d^{2}\right )} x}{{\left (a b c^{3} d - a^{2} c^{2} d^{2}\right )} x^{2} + {\left (a b c^{4} - a^{2} c^{3} d\right )} x} - \frac {{\left (b c + 2 \, a d\right )} \log \left (x\right )}{a^{2} c^{3}} \]

[In]

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

[Out]

b^3*log(b*x + a)/(a^2*b^2*c^2 - 2*a^3*b*c*d + a^4*d^2) - (3*b*c*d^2 - 2*a*d^3)*log(d*x + c)/(b^2*c^5 - 2*a*b*c
^4*d + a^2*c^3*d^2) - (b*c^2 - a*c*d + (b*c*d - 2*a*d^2)*x)/((a*b*c^3*d - a^2*c^2*d^2)*x^2 + (a*b*c^4 - a^2*c^
3*d)*x) - (b*c + 2*a*d)*log(x)/(a^2*c^3)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.36 \[ \int \frac {1}{x^2 (a+b x) (c+d x)^2} \, dx=\frac {b^{3} d \log \left ({\left | b - \frac {b c}{d x + c} + \frac {a d}{d x + c} \right |}\right )}{a^{2} b^{2} c^{2} d - 2 \, a^{3} b c d^{2} + a^{4} d^{3}} + \frac {d^{5}}{{\left (b c^{3} d^{3} - a c^{2} d^{4}\right )} {\left (d x + c\right )}} + \frac {d}{a c^{3} {\left (\frac {c}{d x + c} - 1\right )}} - \frac {{\left (b c d + 2 \, a d^{2}\right )} \log \left ({\left | -\frac {c}{d x + c} + 1 \right |}\right )}{a^{2} c^{3} d} \]

[In]

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

[Out]

b^3*d*log(abs(b - b*c/(d*x + c) + a*d/(d*x + c)))/(a^2*b^2*c^2*d - 2*a^3*b*c*d^2 + a^4*d^3) + d^5/((b*c^3*d^3
- a*c^2*d^4)*(d*x + c)) + d/(a*c^3*(c/(d*x + c) - 1)) - (b*c*d + 2*a*d^2)*log(abs(-c/(d*x + c) + 1))/(a^2*c^3*
d)

Mupad [B] (verification not implemented)

Time = 0.78 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.43 \[ \int \frac {1}{x^2 (a+b x) (c+d x)^2} \, dx=\frac {b^3\,\ln \left (a+b\,x\right )}{a^4\,d^2-2\,a^3\,b\,c\,d+a^2\,b^2\,c^2}-\frac {\frac {1}{a\,c}+\frac {x\,\left (2\,a\,d^2-b\,c\,d\right )}{a\,c^2\,\left (a\,d-b\,c\right )}}{d\,x^2+c\,x}+\frac {\ln \left (c+d\,x\right )\,\left (2\,a\,d^3-3\,b\,c\,d^2\right )}{a^2\,c^3\,d^2-2\,a\,b\,c^4\,d+b^2\,c^5}-\frac {\ln \left (x\right )\,\left (2\,a\,d+b\,c\right )}{a^2\,c^3} \]

[In]

int(1/(x^2*(a + b*x)*(c + d*x)^2),x)

[Out]

(b^3*log(a + b*x))/(a^4*d^2 + a^2*b^2*c^2 - 2*a^3*b*c*d) - (1/(a*c) + (x*(2*a*d^2 - b*c*d))/(a*c^2*(a*d - b*c)
))/(c*x + d*x^2) + (log(c + d*x)*(2*a*d^3 - 3*b*c*d^2))/(b^2*c^5 + a^2*c^3*d^2 - 2*a*b*c^4*d) - (log(x)*(2*a*d
+ b*c))/(a^2*c^3)