\(\int \frac {1}{x^4 (a+b x)^3} \, dx\) [191]

Optimal result

Integrand size = 11, antiderivative size = 89 \[ \int \frac {1}{x^4 (a+b x)^3} \, dx=-\frac {1}{3 a^3 x^3}+\frac {3 b}{2 a^4 x^2}-\frac {6 b^2}{a^5 x}-\frac {b^3}{2 a^4 (a+b x)^2}-\frac {4 b^3}{a^5 (a+b x)}-\frac {10 b^3 \log (x)}{a^6}+\frac {10 b^3 \log (a+b x)}{a^6} \]

[Out]

-1/3/a^3/x^3+3/2*b/a^4/x^2-6*b^2/a^5/x-1/2*b^3/a^4/(b*x+a)^2-4*b^3/a^5/(b*x+a)-10*b^3*ln(x)/a^6+10*b^3*ln(b*x+
a)/a^6

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {46} \[ \int \frac {1}{x^4 (a+b x)^3} \, dx=-\frac {10 b^3 \log (x)}{a^6}+\frac {10 b^3 \log (a+b x)}{a^6}-\frac {4 b^3}{a^5 (a+b x)}-\frac {6 b^2}{a^5 x}-\frac {b^3}{2 a^4 (a+b x)^2}+\frac {3 b}{2 a^4 x^2}-\frac {1}{3 a^3 x^3} \]

[In]

Int[1/(x^4*(a + b*x)^3),x]

[Out]

-1/3*1/(a^3*x^3) + (3*b)/(2*a^4*x^2) - (6*b^2)/(a^5*x) - b^3/(2*a^4*(a + b*x)^2) - (4*b^3)/(a^5*(a + b*x)) - (
10*b^3*Log[x])/a^6 + (10*b^3*Log[a + b*x])/a^6

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{a^3 x^4}-\frac {3 b}{a^4 x^3}+\frac {6 b^2}{a^5 x^2}-\frac {10 b^3}{a^6 x}+\frac {b^4}{a^4 (a+b x)^3}+\frac {4 b^4}{a^5 (a+b x)^2}+\frac {10 b^4}{a^6 (a+b x)}\right ) \, dx \\ & = -\frac {1}{3 a^3 x^3}+\frac {3 b}{2 a^4 x^2}-\frac {6 b^2}{a^5 x}-\frac {b^3}{2 a^4 (a+b x)^2}-\frac {4 b^3}{a^5 (a+b x)}-\frac {10 b^3 \log (x)}{a^6}+\frac {10 b^3 \log (a+b x)}{a^6} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x^4 (a+b x)^3} \, dx=-\frac {\frac {a \left (2 a^4-5 a^3 b x+20 a^2 b^2 x^2+90 a b^3 x^3+60 b^4 x^4\right )}{x^3 (a+b x)^2}+60 b^3 \log (x)-60 b^3 \log (a+b x)}{6 a^6} \]

[In]

Integrate[1/(x^4*(a + b*x)^3),x]

[Out]

-1/6*((a*(2*a^4 - 5*a^3*b*x + 20*a^2*b^2*x^2 + 90*a*b^3*x^3 + 60*b^4*x^4))/(x^3*(a + b*x)^2) + 60*b^3*Log[x] -
60*b^3*Log[a + b*x])/a^6

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.93

method	result	size
norman	\(\frac {\frac {15 b^{5} x^{5}}{a^{6}}-\frac {1}{3 a}+\frac {5 b x}{6 a^{2}}-\frac {10 b^{2} x^{2}}{3 a^{3}}+\frac {20 b^{4} x^{4}}{a^{5}}}{x^{3} \left (b x +a \right )^{2}}-\frac {10 b^{3} \ln \left (x \right )}{a^{6}}+\frac {10 b^{3} \ln \left (b x +a \right )}{a^{6}}\)	\(83\)
default	\(-\frac {1}{3 a^{3} x^{3}}+\frac {3 b}{2 a^{4} x^{2}}-\frac {6 b^{2}}{a^{5} x}-\frac {b^{3}}{2 a^{4} \left (b x +a \right )^{2}}-\frac {4 b^{3}}{a^{5} \left (b x +a \right )}-\frac {10 b^{3} \ln \left (x \right )}{a^{6}}+\frac {10 b^{3} \ln \left (b x +a \right )}{a^{6}}\)	\(84\)
risch	\(\frac {-\frac {10 b^{4} x^{4}}{a^{5}}-\frac {15 b^{3} x^{3}}{a^{4}}-\frac {10 b^{2} x^{2}}{3 a^{3}}+\frac {5 b x}{6 a^{2}}-\frac {1}{3 a}}{x^{3} \left (b x +a \right )^{2}}-\frac {10 b^{3} \ln \left (x \right )}{a^{6}}+\frac {10 b^{3} \ln \left (-b x -a \right )}{a^{6}}\)	\(86\)
parallelrisch	\(-\frac {60 b^{5} \ln \left (x \right ) x^{5}-60 \ln \left (b x +a \right ) x^{5} b^{5}+120 a \,b^{4} \ln \left (x \right ) x^{4}-120 \ln \left (b x +a \right ) x^{4} a \,b^{4}-90 b^{5} x^{5}+60 a^{2} b^{3} \ln \left (x \right ) x^{3}-60 \ln \left (b x +a \right ) x^{3} a^{2} b^{3}-120 a \,b^{4} x^{4}+20 a^{3} b^{2} x^{2}-5 a^{4} b x +2 a^{5}}{6 a^{6} x^{3} \left (b x +a \right )^{2}}\)	\(137\)

[In]

int(1/x^4/(b*x+a)^3,x,method=_RETURNVERBOSE)

[Out]

(15*b^5/a^6*x^5-1/3/a+5/6*b/a^2*x-10/3*b^2/a^3*x^2+20*b^4/a^5*x^4)/x^3/(b*x+a)^2-10*b^3*ln(x)/a^6+10*b^3*ln(b*
x+a)/a^6

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.58 \[ \int \frac {1}{x^4 (a+b x)^3} \, dx=-\frac {60 \, a b^{4} x^{4} + 90 \, a^{2} b^{3} x^{3} + 20 \, a^{3} b^{2} x^{2} - 5 \, a^{4} b x + 2 \, a^{5} - 60 \, {\left (b^{5} x^{5} + 2 \, a b^{4} x^{4} + a^{2} b^{3} x^{3}\right )} \log \left (b x + a\right ) + 60 \, {\left (b^{5} x^{5} + 2 \, a b^{4} x^{4} + a^{2} b^{3} x^{3}\right )} \log \left (x\right )}{6 \, {\left (a^{6} b^{2} x^{5} + 2 \, a^{7} b x^{4} + a^{8} x^{3}\right )}} \]

[In]

integrate(1/x^4/(b*x+a)^3,x, algorithm="fricas")

[Out]

-1/6*(60*a*b^4*x^4 + 90*a^2*b^3*x^3 + 20*a^3*b^2*x^2 - 5*a^4*b*x + 2*a^5 - 60*(b^5*x^5 + 2*a*b^4*x^4 + a^2*b^3
*x^3)*log(b*x + a) + 60*(b^5*x^5 + 2*a*b^4*x^4 + a^2*b^3*x^3)*log(x))/(a^6*b^2*x^5 + 2*a^7*b*x^4 + a^8*x^3)

Sympy [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.03 \[ \int \frac {1}{x^4 (a+b x)^3} \, dx=\frac {- 2 a^{4} + 5 a^{3} b x - 20 a^{2} b^{2} x^{2} - 90 a b^{3} x^{3} - 60 b^{4} x^{4}}{6 a^{7} x^{3} + 12 a^{6} b x^{4} + 6 a^{5} b^{2} x^{5}} + \frac {10 b^{3} \left (- \log {\left (x \right )} + \log {\left (\frac {a}{b} + x \right )}\right )}{a^{6}} \]

[In]

integrate(1/x**4/(b*x+a)**3,x)

[Out]

(-2*a**4 + 5*a**3*b*x - 20*a**2*b**2*x**2 - 90*a*b**3*x**3 - 60*b**4*x**4)/(6*a**7*x**3 + 12*a**6*b*x**4 + 6*a
**5*b**2*x**5) + 10*b**3*(-log(x) + log(a/b + x))/a**6

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.09 \[ \int \frac {1}{x^4 (a+b x)^3} \, dx=-\frac {60 \, b^{4} x^{4} + 90 \, a b^{3} x^{3} + 20 \, a^{2} b^{2} x^{2} - 5 \, a^{3} b x + 2 \, a^{4}}{6 \, {\left (a^{5} b^{2} x^{5} + 2 \, a^{6} b x^{4} + a^{7} x^{3}\right )}} + \frac {10 \, b^{3} \log \left (b x + a\right )}{a^{6}} - \frac {10 \, b^{3} \log \left (x\right )}{a^{6}} \]

[In]

integrate(1/x^4/(b*x+a)^3,x, algorithm="maxima")

[Out]

-1/6*(60*b^4*x^4 + 90*a*b^3*x^3 + 20*a^2*b^2*x^2 - 5*a^3*b*x + 2*a^4)/(a^5*b^2*x^5 + 2*a^6*b*x^4 + a^7*x^3) +
10*b^3*log(b*x + a)/a^6 - 10*b^3*log(x)/a^6

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.97 \[ \int \frac {1}{x^4 (a+b x)^3} \, dx=\frac {10 \, b^{3} \log \left ({\left | b x + a \right |}\right )}{a^{6}} - \frac {10 \, b^{3} \log \left ({\left | x \right |}\right )}{a^{6}} - \frac {60 \, a b^{4} x^{4} + 90 \, a^{2} b^{3} x^{3} + 20 \, a^{3} b^{2} x^{2} - 5 \, a^{4} b x + 2 \, a^{5}}{6 \, {\left (b x + a\right )}^{2} a^{6} x^{3}} \]

[In]

integrate(1/x^4/(b*x+a)^3,x, algorithm="giac")

[Out]

10*b^3*log(abs(b*x + a))/a^6 - 10*b^3*log(abs(x))/a^6 - 1/6*(60*a*b^4*x^4 + 90*a^2*b^3*x^3 + 20*a^3*b^2*x^2 -
5*a^4*b*x + 2*a^5)/((b*x + a)^2*a^6*x^3)

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.02 \[ \int \frac {1}{x^4 (a+b x)^3} \, dx=\frac {20\,b^3\,\mathrm {atanh}\left (\frac {2\,b\,x}{a}+1\right )}{a^6}-\frac {\frac {1}{3\,a}+\frac {10\,b^2\,x^2}{3\,a^3}+\frac {15\,b^3\,x^3}{a^4}+\frac {10\,b^4\,x^4}{a^5}-\frac {5\,b\,x}{6\,a^2}}{a^2\,x^3+2\,a\,b\,x^4+b^2\,x^5} \]

[In]

int(1/(x^4*(a + b*x)^3),x)

[Out]

(20*b^3*atanh((2*b*x)/a + 1))/a^6 - (1/(3*a) + (10*b^2*x^2)/(3*a^3) + (15*b^3*x^3)/a^4 + (10*b^4*x^4)/a^5 - (5
*b*x)/(6*a^2))/(a^2*x^3 + b^2*x^5 + 2*a*b*x^4)