\(\int \frac {1}{x^4 (a+b x^3)} \, dx\) [322]

Optimal result

Integrand size = 13, antiderivative size = 35 \[ \int \frac {1}{x^4 \left (a+b x^3\right )} \, dx=-\frac {1}{3 a x^3}-\frac {b \log (x)}{a^2}+\frac {b \log \left (a+b x^3\right )}{3 a^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 35, 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} \[ \int \frac {1}{x^4 \left (a+b x^3\right )} \, dx=\frac {b \log \left (a+b x^3\right )}{3 a^2}-\frac {b \log (x)}{a^2}-\frac {1}{3 a x^3} \]

[In]

Int[1/(x^4*(a + b*x^3)),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^4 \left (a+b x^3\right )} \, dx=-\frac {1}{3 a x^3}-\frac {b \log (x)}{a^2}+\frac {b \log \left (a+b x^3\right )}{3 a^2} \]

[In]

Integrate[1/(x^4*(a + b*x^3)),x]

[Out]

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

Maple [A] (verified)

Time = 3.72 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.91

method	result	size
default	\(-\frac {1}{3 a \,x^{3}}-\frac {b \ln \left (x \right )}{a^{2}}+\frac {b \ln \left (b \,x^{3}+a \right )}{3 a^{2}}\)	\(32\)
norman	\(-\frac {1}{3 a \,x^{3}}-\frac {b \ln \left (x \right )}{a^{2}}+\frac {b \ln \left (b \,x^{3}+a \right )}{3 a^{2}}\)	\(32\)
parallelrisch	\(-\frac {3 b \ln \left (x \right ) x^{3}-b \ln \left (b \,x^{3}+a \right ) x^{3}+a}{3 a^{2} x^{3}}\)	\(33\)
risch	\(-\frac {1}{3 a \,x^{3}}-\frac {b \ln \left (x \right )}{a^{2}}+\frac {b \ln \left (-b \,x^{3}-a \right )}{3 a^{2}}\)	\(35\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x^4 \left (a+b x^3\right )} \, dx=\frac {b x^{3} \log \left (b x^{3} + a\right ) - 3 \, b x^{3} \log \left (x\right ) - a}{3 \, a^{2} x^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x^4 \left (a+b x^3\right )} \, dx=- \frac {1}{3 a x^{3}} - \frac {b \log {\left (x \right )}}{a^{2}} + \frac {b \log {\left (\frac {a}{b} + x^{3} \right )}}{3 a^{2}} \]

[In]

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

[Out]

-1/(3*a*x**3) - b*log(x)/a**2 + b*log(a/b + x**3)/(3*a**2)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.94 \[ \int \frac {1}{x^4 \left (a+b x^3\right )} \, dx=\frac {b \log \left (b x^{3} + a\right )}{3 \, a^{2}} - \frac {b \log \left (x^{3}\right )}{3 \, a^{2}} - \frac {1}{3 \, a x^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.20 \[ \int \frac {1}{x^4 \left (a+b x^3\right )} \, dx=\frac {b \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, a^{2}} - \frac {b \log \left ({\left | x \right |}\right )}{a^{2}} + \frac {b x^{3} - a}{3 \, a^{2} x^{3}} \]

[In]

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

[Out]

1/3*b*log(abs(b*x^3 + a))/a^2 - b*log(abs(x))/a^2 + 1/3*(b*x^3 - a)/(a^2*x^3)

Mupad [B] (verification not implemented)

Time = 5.45 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.89 \[ \int \frac {1}{x^4 \left (a+b x^3\right )} \, dx=\frac {b\,\ln \left (b\,x^3+a\right )}{3\,a^2}-\frac {1}{3\,a\,x^3}-\frac {b\,\ln \left (x\right )}{a^2} \]

[In]

int(1/(x^4*(a + b*x^3)),x)

[Out]

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