\(\int \frac {1}{x^4 (a+b x^3)^3} \, dx\) [345]

Optimal result

Integrand size = 13, antiderivative size = 66 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^3} \, dx=-\frac {1}{3 a^3 x^3}-\frac {b}{6 a^2 \left (a+b x^3\right )^2}-\frac {2 b}{3 a^3 \left (a+b x^3\right )}-\frac {3 b \log (x)}{a^4}+\frac {b \log \left (a+b x^3\right )}{a^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {272, 46} \[ \int \frac {1}{x^4 \left (a+b x^3\right )^3} \, dx=\frac {b \log \left (a+b x^3\right )}{a^4}-\frac {3 b \log (x)}{a^4}-\frac {2 b}{3 a^3 \left (a+b x^3\right )}-\frac {1}{3 a^3 x^3}-\frac {b}{6 a^2 \left (a+b x^3\right )^2} \]

[In]

Int[1/(x^4*(a + b*x^3)^3),x]

[Out]

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

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])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^3} \, dx=-\frac {\frac {a \left (2 a^2+9 a b x^3+6 b^2 x^6\right )}{x^3 \left (a+b x^3\right )^2}+18 b \log (x)-6 b \log \left (a+b x^3\right )}{6 a^4} \]

[In]

Integrate[1/(x^4*(a + b*x^3)^3),x]

[Out]

-1/6*((a*(2*a^2 + 9*a*b*x^3 + 6*b^2*x^6))/(x^3*(a + b*x^3)^2) + 18*b*Log[x] - 6*b*Log[a + b*x^3])/a^4

Maple [A] (verified)

Time = 3.64 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.97

method	result	size
norman	\(\frac {-\frac {1}{3 a}+\frac {2 b^{2} x^{6}}{a^{3}}+\frac {3 b^{3} x^{9}}{2 a^{4}}}{x^{3} \left (b \,x^{3}+a \right )^{2}}+\frac {b \ln \left (b \,x^{3}+a \right )}{a^{4}}-\frac {3 b \ln \left (x \right )}{a^{4}}\)	\(64\)
risch	\(\frac {-\frac {b^{2} x^{6}}{a^{3}}-\frac {3 b \,x^{3}}{2 a^{2}}-\frac {1}{3 a}}{x^{3} \left (b \,x^{3}+a \right )^{2}}-\frac {3 b \ln \left (x \right )}{a^{4}}+\frac {b \ln \left (-b \,x^{3}-a \right )}{a^{4}}\)	\(65\)
default	\(-\frac {1}{3 a^{3} x^{3}}-\frac {3 b \ln \left (x \right )}{a^{4}}+\frac {b^{2} \left (\frac {3 \ln \left (b \,x^{3}+a \right )}{b}-\frac {a^{2}}{2 b \left (b \,x^{3}+a \right )^{2}}-\frac {2 a}{b \left (b \,x^{3}+a \right )}\right )}{3 a^{4}}\)	\(72\)
parallelrisch	\(-\frac {18 \ln \left (x \right ) x^{9} b^{3}-6 \ln \left (b \,x^{3}+a \right ) x^{9} b^{3}-9 b^{3} x^{9}+36 \ln \left (x \right ) x^{6} a \,b^{2}-12 \ln \left (b \,x^{3}+a \right ) x^{6} a \,b^{2}-12 a \,b^{2} x^{6}+18 a^{2} b \ln \left (x \right ) x^{3}-6 \ln \left (b \,x^{3}+a \right ) x^{3} a^{2} b +2 a^{3}}{6 a^{4} x^{3} \left (b \,x^{3}+a \right )^{2}}\)	\(123\)

[In]

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

[Out]

(-1/3/a+2*b^2/a^3*x^6+3/2*b^3/a^4*x^9)/x^3/(b*x^3+a)^2+b*ln(b*x^3+a)/a^4-3*b*ln(x)/a^4

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.80 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^3} \, dx=-\frac {6 \, a b^{2} x^{6} + 9 \, a^{2} b x^{3} + 2 \, a^{3} - 6 \, {\left (b^{3} x^{9} + 2 \, a b^{2} x^{6} + a^{2} b x^{3}\right )} \log \left (b x^{3} + a\right ) + 18 \, {\left (b^{3} x^{9} + 2 \, a b^{2} x^{6} + a^{2} b x^{3}\right )} \log \left (x\right )}{6 \, {\left (a^{4} b^{2} x^{9} + 2 \, a^{5} b x^{6} + a^{6} x^{3}\right )}} \]

[In]

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

[Out]

-1/6*(6*a*b^2*x^6 + 9*a^2*b*x^3 + 2*a^3 - 6*(b^3*x^9 + 2*a*b^2*x^6 + a^2*b*x^3)*log(b*x^3 + a) + 18*(b^3*x^9 +
2*a*b^2*x^6 + a^2*b*x^3)*log(x))/(a^4*b^2*x^9 + 2*a^5*b*x^6 + a^6*x^3)

Sympy [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.15 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^3} \, dx=\frac {- 2 a^{2} - 9 a b x^{3} - 6 b^{2} x^{6}}{6 a^{5} x^{3} + 12 a^{4} b x^{6} + 6 a^{3} b^{2} x^{9}} - \frac {3 b \log {\left (x \right )}}{a^{4}} + \frac {b \log {\left (\frac {a}{b} + x^{3} \right )}}{a^{4}} \]

[In]

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

[Out]

(-2*a**2 - 9*a*b*x**3 - 6*b**2*x**6)/(6*a**5*x**3 + 12*a**4*b*x**6 + 6*a**3*b**2*x**9) - 3*b*log(x)/a**4 + b*l
og(a/b + x**3)/a**4

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.15 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^3} \, dx=-\frac {6 \, b^{2} x^{6} + 9 \, a b x^{3} + 2 \, a^{2}}{6 \, {\left (a^{3} b^{2} x^{9} + 2 \, a^{4} b x^{6} + a^{5} x^{3}\right )}} + \frac {b \log \left (b x^{3} + a\right )}{a^{4}} - \frac {b \log \left (x^{3}\right )}{a^{4}} \]

[In]

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

[Out]

-1/6*(6*b^2*x^6 + 9*a*b*x^3 + 2*a^2)/(a^3*b^2*x^9 + 2*a^4*b*x^6 + a^5*x^3) + b*log(b*x^3 + a)/a^4 - b*log(x^3)
/a^4

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.21 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^3} \, dx=\frac {b \log \left ({\left | b x^{3} + a \right |}\right )}{a^{4}} - \frac {3 \, b \log \left ({\left | x \right |}\right )}{a^{4}} - \frac {9 \, b^{3} x^{6} + 22 \, a b^{2} x^{3} + 14 \, a^{2} b}{6 \, {\left (b x^{3} + a\right )}^{2} a^{4}} + \frac {3 \, b x^{3} - a}{3 \, a^{4} x^{3}} \]

[In]

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

[Out]

b*log(abs(b*x^3 + a))/a^4 - 3*b*log(abs(x))/a^4 - 1/6*(9*b^3*x^6 + 22*a*b^2*x^3 + 14*a^2*b)/((b*x^3 + a)^2*a^4
) + 1/3*(3*b*x^3 - a)/(a^4*x^3)

Mupad [B] (verification not implemented)

Time = 5.61 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.11 \[ \int \frac {1}{x^4 \left (a+b x^3\right )^3} \, dx=\frac {b\,\ln \left (b\,x^3+a\right )}{a^4}-\frac {\frac {1}{3\,a}+\frac {3\,b\,x^3}{2\,a^2}+\frac {b^2\,x^6}{a^3}}{a^2\,x^3+2\,a\,b\,x^6+b^2\,x^9}-\frac {3\,b\,\ln \left (x\right )}{a^4} \]

[In]

int(1/(x^4*(a + b*x^3)^3),x)

[Out]

(b*log(a + b*x^3))/a^4 - (1/(3*a) + (3*b*x^3)/(2*a^2) + (b^2*x^6)/a^3)/(a^2*x^3 + b^2*x^9 + 2*a*b*x^6) - (3*b*
log(x))/a^4