\(\int \frac {1}{x^3 (a+b x) (c+d x)^2} \, dx\) [250]

Optimal result

Integrand size = 18, antiderivative size = 144 \[ \int \frac {1}{x^3 (a+b x) (c+d x)^2} \, dx=-\frac {1}{2 a c^2 x^2}+\frac {b c+2 a d}{a^2 c^3 x}-\frac {d^3}{c^3 (b c-a d) (c+d x)}+\frac {\left (b^2 c^2+2 a b c d+3 a^2 d^2\right ) \log (x)}{a^3 c^4}-\frac {b^4 \log (a+b x)}{a^3 (b c-a d)^2}+\frac {d^3 (4 b c-3 a d) \log (c+d x)}{c^4 (b c-a d)^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 144, 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} \[ \int \frac {1}{x^3 (a+b x) (c+d x)^2} \, dx=-\frac {b^4 \log (a+b x)}{a^3 (b c-a d)^2}+\frac {2 a d+b c}{a^2 c^3 x}+\frac {\log (x) \left (3 a^2 d^2+2 a b c d+b^2 c^2\right )}{a^3 c^4}+\frac {d^3 (4 b c-3 a d) \log (c+d x)}{c^4 (b c-a d)^2}-\frac {d^3}{c^3 (c+d x) (b c-a d)}-\frac {1}{2 a c^2 x^2} \]

[In]

Int[1/(x^3*(a + b*x)*(c + d*x)^2),x]

[Out]

-1/2*1/(a*c^2*x^2) + (b*c + 2*a*d)/(a^2*c^3*x) - d^3/(c^3*(b*c - a*d)*(c + d*x)) + ((b^2*c^2 + 2*a*b*c*d + 3*a
^2*d^2)*Log[x])/(a^3*c^4) - (b^4*Log[a + b*x])/(a^3*(b*c - a*d)^2) + (d^3*(4*b*c - 3*a*d)*Log[c + d*x])/(c^4*(
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^3}+\frac {-b c-2 a d}{a^2 c^3 x^2}+\frac {b^2 c^2+2 a b c d+3 a^2 d^2}{a^3 c^4 x}-\frac {b^5}{a^3 (-b c+a d)^2 (a+b x)}+\frac {d^4}{c^3 (b c-a d) (c+d x)^2}+\frac {d^4 (4 b c-3 a d)}{c^4 (b c-a d)^2 (c+d x)}\right ) \, dx \\ & = -\frac {1}{2 a c^2 x^2}+\frac {b c+2 a d}{a^2 c^3 x}-\frac {d^3}{c^3 (b c-a d) (c+d x)}+\frac {\left (b^2 c^2+2 a b c d+3 a^2 d^2\right ) \log (x)}{a^3 c^4}-\frac {b^4 \log (a+b x)}{a^3 (b c-a d)^2}+\frac {d^3 (4 b c-3 a d) \log (c+d x)}{c^4 (b c-a d)^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.99 \[ \int \frac {1}{x^3 (a+b x) (c+d x)^2} \, dx=\frac {\left (b^2 c^2+2 a b c d+3 a^2 d^2\right ) \log (x)}{a^3 c^4}-\frac {b^4 \log (a+b x)}{a^3 (b c-a d)^2}+\frac {c \left (\frac {2 b c}{a^2 x}-\frac {c-4 d x}{a x^2}+\frac {2 d^3}{(-b c+a d) (c+d x)}\right )+\frac {2 d^3 (4 b c-3 a d) \log (c+d x)}{(b c-a d)^2}}{2 c^4} \]

[In]

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

[Out]

((b^2*c^2 + 2*a*b*c*d + 3*a^2*d^2)*Log[x])/(a^3*c^4) - (b^4*Log[a + b*x])/(a^3*(b*c - a*d)^2) + (c*((2*b*c)/(a
^2*x) - (c - 4*d*x)/(a*x^2) + (2*d^3)/((-(b*c) + a*d)*(c + d*x))) + (2*d^3*(4*b*c - 3*a*d)*Log[c + d*x])/(b*c
- a*d)^2)/(2*c^4)

Maple [A] (verified)

Time = 1.29 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.01

method	result	size
default	\(-\frac {1}{2 a \,c^{2} x^{2}}-\frac {-2 a d -b c}{x \,a^{2} c^{3}}+\frac {\left (3 a^{2} d^{2}+2 a b c d +b^{2} c^{2}\right ) \ln \left (x \right )}{a^{3} c^{4}}+\frac {d^{3}}{c^{3} \left (a d -b c \right ) \left (d x +c \right )}-\frac {d^{3} \left (3 a d -4 b c \right ) \ln \left (d x +c \right )}{c^{4} \left (a d -b c \right )^{2}}-\frac {b^{4} \ln \left (b x +a \right )}{a^{3} \left (a d -b c \right )^{2}}\)	\(145\)
norman	\(\frac {\frac {\left (-3 a^{2} d^{3}+a b c \,d^{2}+b^{2} c^{2} d \right ) d \,x^{3}}{c^{4} a^{2} \left (a d -b c \right )}-\frac {1}{2 a c}+\frac {\left (3 a d +2 b c \right ) x}{2 c^{2} a^{2}}}{\left (d x +c \right ) x^{2}}+\frac {\left (3 a^{2} d^{2}+2 a b c d +b^{2} c^{2}\right ) \ln \left (x \right )}{a^{3} c^{4}}-\frac {b^{4} \ln \left (b x +a \right )}{a^{3} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}-\frac {d^{3} \left (3 a d -4 b c \right ) \ln \left (d x +c \right )}{c^{4} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}\)	\(199\)
risch	\(\frac {\frac {d \left (3 a^{2} d^{2}-a b c d -b^{2} c^{2}\right ) x^{2}}{a^{2} c^{3} \left (a d -b c \right )}+\frac {\left (3 a d +2 b c \right ) x}{2 c^{2} a^{2}}-\frac {1}{2 a c}}{\left (d x +c \right ) x^{2}}-\frac {b^{4} \ln \left (b x +a \right )}{a^{3} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}-\frac {3 d^{4} \ln \left (-d x -c \right ) a}{c^{4} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {4 d^{3} \ln \left (-d x -c \right ) b}{c^{3} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}+\frac {3 \ln \left (-x \right ) d^{2}}{a \,c^{4}}+\frac {2 \ln \left (-x \right ) b d}{a^{2} c^{3}}+\frac {\ln \left (-x \right ) b^{2}}{a^{3} c^{2}}\)	\(246\)
parallelrisch	\(\frac {2 a^{3} b \,c^{4} d^{2}+3 x \,a^{4} c^{2} d^{4}+6 \ln \left (x \right ) x^{3} a^{4} d^{6}-6 \ln \left (d x +c \right ) x^{3} a^{4} d^{6}-a^{4} c^{3} d^{3}-8 x^{2} a^{3} b \,c^{2} d^{4}+2 x^{2} a \,b^{3} c^{4} d^{2}-4 x \,a^{3} b \,c^{3} d^{3}-x \,a^{2} b^{2} c^{4} d^{2}+2 x a \,b^{3} c^{5} d +2 \ln \left (x \right ) x^{3} b^{4} c^{4} d^{2}-2 \ln \left (b x +a \right ) x^{3} b^{4} c^{4} d^{2}+6 \ln \left (x \right ) x^{2} a^{4} c \,d^{5}+2 \ln \left (x \right ) x^{2} b^{4} c^{5} d -2 \ln \left (b x +a \right ) x^{2} b^{4} c^{5} d -6 \ln \left (d x +c \right ) x^{2} a^{4} c \,d^{5}-a^{2} b^{2} c^{5} d +6 x^{2} a^{4} c \,d^{5}-8 \ln \left (x \right ) x^{3} a^{3} b c \,d^{5}+8 \ln \left (d x +c \right ) x^{3} a^{3} b c \,d^{5}-8 \ln \left (x \right ) x^{2} a^{3} b \,c^{2} d^{4}+8 \ln \left (d x +c \right ) x^{2} a^{3} b \,c^{2} d^{4}}{2 \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \left (d x +c \right ) x^{2} a^{3} c^{4} d}\)	\(376\)

[In]

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

[Out]

-1/2/a/c^2/x^2-(-2*a*d-b*c)/x/a^2/c^3+(3*a^2*d^2+2*a*b*c*d+b^2*c^2)*ln(x)/a^3/c^4+d^3/c^3/(a*d-b*c)/(d*x+c)-d^
3*(3*a*d-4*b*c)/c^4/(a*d-b*c)^2*ln(d*x+c)-b^4/a^3/(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. 350 vs. \(2 (142) = 284\).

Time = 3.16 (sec) , antiderivative size = 350, normalized size of antiderivative = 2.43 \[ \int \frac {1}{x^3 (a+b x) (c+d x)^2} \, dx=-\frac {a^{2} b^{2} c^{5} - 2 \, a^{3} b c^{4} d + a^{4} c^{3} d^{2} - 2 \, {\left (a b^{3} c^{4} d - 4 \, a^{3} b c^{2} d^{3} + 3 \, a^{4} c d^{4}\right )} x^{2} - {\left (2 \, a b^{3} c^{5} - a^{2} b^{2} c^{4} d - 4 \, a^{3} b c^{3} d^{2} + 3 \, a^{4} c^{2} d^{3}\right )} x + 2 \, {\left (b^{4} c^{4} d x^{3} + b^{4} c^{5} x^{2}\right )} \log \left (b x + a\right ) - 2 \, {\left ({\left (4 \, a^{3} b c d^{4} - 3 \, a^{4} d^{5}\right )} x^{3} + {\left (4 \, a^{3} b c^{2} d^{3} - 3 \, a^{4} c d^{4}\right )} x^{2}\right )} \log \left (d x + c\right ) - 2 \, {\left ({\left (b^{4} c^{4} d - 4 \, a^{3} b c d^{4} + 3 \, a^{4} d^{5}\right )} x^{3} + {\left (b^{4} c^{5} - 4 \, a^{3} b c^{2} d^{3} + 3 \, a^{4} c d^{4}\right )} x^{2}\right )} \log \left (x\right )}{2 \, {\left ({\left (a^{3} b^{2} c^{6} d - 2 \, a^{4} b c^{5} d^{2} + a^{5} c^{4} d^{3}\right )} x^{3} + {\left (a^{3} b^{2} c^{7} - 2 \, a^{4} b c^{6} d + a^{5} c^{5} d^{2}\right )} x^{2}\right )}} \]

[In]

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

[Out]

-1/2*(a^2*b^2*c^5 - 2*a^3*b*c^4*d + a^4*c^3*d^2 - 2*(a*b^3*c^4*d - 4*a^3*b*c^2*d^3 + 3*a^4*c*d^4)*x^2 - (2*a*b
^3*c^5 - a^2*b^2*c^4*d - 4*a^3*b*c^3*d^2 + 3*a^4*c^2*d^3)*x + 2*(b^4*c^4*d*x^3 + b^4*c^5*x^2)*log(b*x + a) - 2
*((4*a^3*b*c*d^4 - 3*a^4*d^5)*x^3 + (4*a^3*b*c^2*d^3 - 3*a^4*c*d^4)*x^2)*log(d*x + c) - 2*((b^4*c^4*d - 4*a^3*
b*c*d^4 + 3*a^4*d^5)*x^3 + (b^4*c^5 - 4*a^3*b*c^2*d^3 + 3*a^4*c*d^4)*x^2)*log(x))/((a^3*b^2*c^6*d - 2*a^4*b*c^
5*d^2 + a^5*c^4*d^3)*x^3 + (a^3*b^2*c^7 - 2*a^4*b*c^6*d + a^5*c^5*d^2)*x^2)

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{x^3 (a+b x) (c+d x)^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.70 \[ \int \frac {1}{x^3 (a+b x) (c+d x)^2} \, dx=-\frac {b^{4} \log \left (b x + a\right )}{a^{3} b^{2} c^{2} - 2 \, a^{4} b c d + a^{5} d^{2}} + \frac {{\left (4 \, b c d^{3} - 3 \, a d^{4}\right )} \log \left (d x + c\right )}{b^{2} c^{6} - 2 \, a b c^{5} d + a^{2} c^{4} d^{2}} - \frac {a b c^{3} - a^{2} c^{2} d - 2 \, {\left (b^{2} c^{2} d + a b c d^{2} - 3 \, a^{2} d^{3}\right )} x^{2} - {\left (2 \, b^{2} c^{3} + a b c^{2} d - 3 \, a^{2} c d^{2}\right )} x}{2 \, {\left ({\left (a^{2} b c^{4} d - a^{3} c^{3} d^{2}\right )} x^{3} + {\left (a^{2} b c^{5} - a^{3} c^{4} d\right )} x^{2}\right )}} + \frac {{\left (b^{2} c^{2} + 2 \, a b c d + 3 \, a^{2} d^{2}\right )} \log \left (x\right )}{a^{3} c^{4}} \]

[In]

integrate(1/x^3/(b*x+a)/(d*x+c)^2,x, algorithm="maxima")

[Out]

-b^4*log(b*x + a)/(a^3*b^2*c^2 - 2*a^4*b*c*d + a^5*d^2) + (4*b*c*d^3 - 3*a*d^4)*log(d*x + c)/(b^2*c^6 - 2*a*b*
c^5*d + a^2*c^4*d^2) - 1/2*(a*b*c^3 - a^2*c^2*d - 2*(b^2*c^2*d + a*b*c*d^2 - 3*a^2*d^3)*x^2 - (2*b^2*c^3 + a*b
*c^2*d - 3*a^2*c*d^2)*x)/((a^2*b*c^4*d - a^3*c^3*d^2)*x^3 + (a^2*b*c^5 - a^3*c^4*d)*x^2) + (b^2*c^2 + 2*a*b*c*
d + 3*a^2*d^2)*log(x)/(a^3*c^4)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.47 \[ \int \frac {1}{x^3 (a+b x) (c+d x)^2} \, dx=-\frac {d^{7}}{{\left (b c^{4} d^{4} - a c^{3} d^{5}\right )} {\left (d x + c\right )}} - \frac {b^{4} d \log \left ({\left | b - \frac {b c}{d x + c} + \frac {a d}{d x + c} \right |}\right )}{a^{3} b^{2} c^{2} d - 2 \, a^{4} b c d^{2} + a^{5} d^{3}} + \frac {{\left (b^{2} c^{2} d + 2 \, a b c d^{2} + 3 \, a^{2} d^{3}\right )} \log \left ({\left | -\frac {c}{d x + c} + 1 \right |}\right )}{a^{3} c^{4} d} + \frac {2 \, a b c d + 5 \, a^{2} d^{2} - \frac {2 \, {\left (a b c^{2} d^{2} + 3 \, a^{2} c d^{3}\right )}}{{\left (d x + c\right )} d}}{2 \, a^{3} c^{4} {\left (\frac {c}{d x + c} - 1\right )}^{2}} \]

[In]

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

[Out]

-d^7/((b*c^4*d^4 - a*c^3*d^5)*(d*x + c)) - b^4*d*log(abs(b - b*c/(d*x + c) + a*d/(d*x + c)))/(a^3*b^2*c^2*d -
2*a^4*b*c*d^2 + a^5*d^3) + (b^2*c^2*d + 2*a*b*c*d^2 + 3*a^2*d^3)*log(abs(-c/(d*x + c) + 1))/(a^3*c^4*d) + 1/2*
(2*a*b*c*d + 5*a^2*d^2 - 2*(a*b*c^2*d^2 + 3*a^2*c*d^3)/((d*x + c)*d))/(a^3*c^4*(c/(d*x + c) - 1)^2)

Mupad [B] (verification not implemented)

Time = 0.72 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.44 \[ \int \frac {1}{x^3 (a+b x) (c+d x)^2} \, dx=\frac {\ln \left (x\right )\,\left (3\,a^2\,d^2+2\,a\,b\,c\,d+b^2\,c^2\right )}{a^3\,c^4}-\frac {b^4\,\ln \left (a+b\,x\right )}{a^5\,d^2-2\,a^4\,b\,c\,d+a^3\,b^2\,c^2}-\frac {\ln \left (c+d\,x\right )\,\left (3\,a\,d^4-4\,b\,c\,d^3\right )}{a^2\,c^4\,d^2-2\,a\,b\,c^5\,d+b^2\,c^6}-\frac {\frac {1}{2\,a\,c}-\frac {x\,\left (3\,a\,d+2\,b\,c\right )}{2\,a^2\,c^2}+\frac {x^2\,\left (-3\,a^2\,d^3+a\,b\,c\,d^2+b^2\,c^2\,d\right )}{a^2\,c^3\,\left (a\,d-b\,c\right )}}{d\,x^3+c\,x^2} \]

[In]

int(1/(x^3*(a + b*x)*(c + d*x)^2),x)

[Out]

(log(x)*(3*a^2*d^2 + b^2*c^2 + 2*a*b*c*d))/(a^3*c^4) - (b^4*log(a + b*x))/(a^5*d^2 + a^3*b^2*c^2 - 2*a^4*b*c*d
) - (log(c + d*x)*(3*a*d^4 - 4*b*c*d^3))/(b^2*c^6 + a^2*c^4*d^2 - 2*a*b*c^5*d) - (1/(2*a*c) - (x*(3*a*d + 2*b*
c))/(2*a^2*c^2) + (x^2*(b^2*c^2*d - 3*a^2*d^3 + a*b*c*d^2))/(a^2*c^3*(a*d - b*c)))/(c*x^2 + d*x^3)