\(\int \frac {1}{x^4 (a+b x) (c+d x)} \, dx\) [241]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.08 (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 = {84} \[ \int \frac {1}{x^4 (a+b x) (c+d x)} \, dx=\frac {b^4 \log (a+b x)}{a^4 (b c-a d)}+\frac {a d+b c}{2 a^2 c^2 x^2}-\frac {\log (x) (a d+b c) \left (a^2 d^2+b^2 c^2\right )}{a^4 c^4}-\frac {a^2 d^2+a b c d+b^2 c^2}{a^3 c^3 x}-\frac {d^4 \log (c+d x)}{c^4 (b c-a d)}-\frac {1}{3 a c x^3} \]

[In]

Int[1/(x^4*(a + b*x)*(c + d*x)),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.97 \[ \int \frac {1}{x^4 (a+b x) (c+d x)} \, dx=\frac {6 \left (b^4 c^4-a^4 d^4\right ) x^3 \log (x)-6 b^4 c^4 x^3 \log (a+b x)+a \left (2 a^2 b c^4-3 a b^2 c^4 x+6 b^3 c^4 x^2+a^3 c d \left (-2 c^2+3 c d x-6 d^2 x^2\right )+6 a^3 d^4 x^3 \log (c+d x)\right )}{6 a^4 c^4 (-b c+a d) x^3} \]

[In]

Integrate[1/(x^4*(a + b*x)*(c + d*x)),x]

[Out]

(6*(b^4*c^4 - a^4*d^4)*x^3*Log[x] - 6*b^4*c^4*x^3*Log[a + b*x] + a*(2*a^2*b*c^4 - 3*a*b^2*c^4*x + 6*b^3*c^4*x^
2 + a^3*c*d*(-2*c^2 + 3*c*d*x - 6*d^2*x^2) + 6*a^3*d^4*x^3*Log[c + d*x]))/(6*a^4*c^4*(-(b*c) + a*d)*x^3)

Maple [A] (verified)

Time = 1.06 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.03

method	result	size
parallelrisch	\(-\frac {6 \ln \left (x \right ) x^{3} a^{4} d^{4}-6 \ln \left (x \right ) x^{3} b^{4} c^{4}+6 b^{4} \ln \left (b x +a \right ) c^{4} x^{3}-6 d^{4} \ln \left (d x +c \right ) a^{4} x^{3}+6 x^{2} a^{4} c \,d^{3}-6 x^{2} a \,b^{3} c^{4}-3 x \,a^{4} c^{2} d^{2}+3 x \,a^{2} b^{2} c^{4}+2 a^{4} c^{3} d -2 a^{3} b \,c^{4}}{6 c^{4} a^{4} x^{3} \left (a d -b c \right )}\)	\(149\)
norman	\(\frac {-\frac {1}{3 a c}+\frac {\left (a d +b c \right ) x}{2 c^{2} a^{2}}-\frac {\left (a^{2} d^{2}+a b c d +b^{2} c^{2}\right ) x^{2}}{c^{3} a^{3}}}{x^{3}}+\frac {d^{4} \ln \left (d x +c \right )}{c^{4} \left (a d -b c \right )}-\frac {b^{4} \ln \left (b x +a \right )}{a^{4} \left (a d -b c \right )}-\frac {\left (a^{3} d^{3}+a^{2} b c \,d^{2}+a \,b^{2} c^{2} d +b^{3} c^{3}\right ) \ln \left (x \right )}{a^{4} c^{4}}\)	\(152\)
default	\(-\frac {1}{3 a c \,x^{3}}-\frac {-a d -b c}{2 c^{2} a^{2} x^{2}}-\frac {a^{2} d^{2}+a b c d +b^{2} c^{2}}{c^{3} a^{3} x}+\frac {\left (-a^{3} d^{3}-a^{2} b c \,d^{2}-a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \ln \left (x \right )}{c^{4} a^{4}}+\frac {d^{4} \ln \left (d x +c \right )}{c^{4} \left (a d -b c \right )}-\frac {b^{4} \ln \left (b x +a \right )}{a^{4} \left (a d -b c \right )}\)	\(157\)
risch	\(\frac {-\frac {1}{3 a c}+\frac {\left (a d +b c \right ) x}{2 c^{2} a^{2}}-\frac {\left (a^{2} d^{2}+a b c d +b^{2} c^{2}\right ) x^{2}}{c^{3} a^{3}}}{x^{3}}-\frac {\ln \left (-x \right ) d^{3}}{a \,c^{4}}-\frac {\ln \left (-x \right ) b \,d^{2}}{a^{2} c^{3}}-\frac {\ln \left (-x \right ) b^{2} d}{a^{3} c^{2}}-\frac {\ln \left (-x \right ) b^{3}}{a^{4} c}-\frac {b^{4} \ln \left (b x +a \right )}{a^{4} \left (a d -b c \right )}+\frac {d^{4} \ln \left (-d x -c \right )}{c^{4} \left (a d -b c \right )}\)	\(174\)

[In]

int(1/x^4/(b*x+a)/(d*x+c),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 1.89 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.03 \[ \int \frac {1}{x^4 (a+b x) (c+d x)} \, dx=\frac {6 \, b^{4} c^{4} x^{3} \log \left (b x + a\right ) - 6 \, a^{4} d^{4} x^{3} \log \left (d x + c\right ) - 2 \, a^{3} b c^{4} + 2 \, a^{4} c^{3} d - 6 \, {\left (b^{4} c^{4} - a^{4} d^{4}\right )} x^{3} \log \left (x\right ) - 6 \, {\left (a b^{3} c^{4} - a^{4} c d^{3}\right )} x^{2} + 3 \, {\left (a^{2} b^{2} c^{4} - a^{4} c^{2} d^{2}\right )} x}{6 \, {\left (a^{4} b c^{5} - a^{5} c^{4} d\right )} x^{3}} \]

[In]

integrate(1/x^4/(b*x+a)/(d*x+c),x, algorithm="fricas")

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{x^4 (a+b x) (c+d x)} \, dx=\text {Timed out} \]

[In]

integrate(1/x**4/(b*x+a)/(d*x+c),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.08 \[ \int \frac {1}{x^4 (a+b x) (c+d x)} \, dx=\frac {b^{4} \log \left (b x + a\right )}{a^{4} b c - a^{5} d} - \frac {d^{4} \log \left (d x + c\right )}{b c^{5} - a c^{4} d} - \frac {{\left (b^{3} c^{3} + a b^{2} c^{2} d + a^{2} b c d^{2} + a^{3} d^{3}\right )} \log \left (x\right )}{a^{4} c^{4}} - \frac {2 \, a^{2} c^{2} + 6 \, {\left (b^{2} c^{2} + a b c d + a^{2} d^{2}\right )} x^{2} - 3 \, {\left (a b c^{2} + a^{2} c d\right )} x}{6 \, a^{3} c^{3} x^{3}} \]

[In]

integrate(1/x^4/(b*x+a)/(d*x+c),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.22 \[ \int \frac {1}{x^4 (a+b x) (c+d x)} \, dx=\frac {b^{5} \log \left ({\left | b x + a \right |}\right )}{a^{4} b^{2} c - a^{5} b d} - \frac {d^{5} \log \left ({\left | d x + c \right |}\right )}{b c^{5} d - a c^{4} d^{2}} - \frac {{\left (b^{3} c^{3} + a b^{2} c^{2} d + a^{2} b c d^{2} + a^{3} d^{3}\right )} \log \left ({\left | x \right |}\right )}{a^{4} c^{4}} - \frac {2 \, a^{3} c^{3} + 6 \, {\left (a b^{2} c^{3} + a^{2} b c^{2} d + a^{3} c d^{2}\right )} x^{2} - 3 \, {\left (a^{2} b c^{3} + a^{3} c^{2} d\right )} x}{6 \, a^{4} c^{4} x^{3}} \]

[In]

integrate(1/x^4/(b*x+a)/(d*x+c),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.54 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.06 \[ \int \frac {1}{x^4 (a+b x) (c+d x)} \, dx=\frac {d^4\,\ln \left (c+d\,x\right )}{c^4\,\left (a\,d-b\,c\right )}-\frac {b^4\,\ln \left (a+b\,x\right )}{a^5\,d-a^4\,b\,c}-\frac {\ln \left (x\right )\,\left (a^3\,d^3+a^2\,b\,c\,d^2+a\,b^2\,c^2\,d+b^3\,c^3\right )}{a^4\,c^4}-\frac {\frac {1}{3\,a\,c}-\frac {x\,\left (a\,d+b\,c\right )}{2\,a^2\,c^2}+\frac {x^2\,\left (a^2\,d^2+a\,b\,c\,d+b^2\,c^2\right )}{a^3\,c^3}}{x^3} \]

[In]

int(1/(x^4*(a + b*x)*(c + d*x)),x)

[Out]

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