\(\int \frac {(a+b x^3)^3}{x^4} \, dx\) [242]

Optimal result

Integrand size = 13, antiderivative size = 37 \[ \int \frac {\left (a+b x^3\right )^3}{x^4} \, dx=-\frac {a^3}{3 x^3}+a b^2 x^3+\frac {b^3 x^6}{6}+3 a^2 b \log (x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, 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, 45} \[ \int \frac {\left (a+b x^3\right )^3}{x^4} \, dx=-\frac {a^3}{3 x^3}+3 a^2 b \log (x)+a b^2 x^3+\frac {b^3 x^6}{6} \]

[In]

Int[(a + b*x^3)^3/x^4,x]

[Out]

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

Rule 45

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[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 {(a+b x)^3}{x^2} \, dx,x,x^3\right ) \\ & = \frac {1}{3} \text {Subst}\left (\int \left (3 a b^2+\frac {a^3}{x^2}+\frac {3 a^2 b}{x}+b^3 x\right ) \, dx,x,x^3\right ) \\ & = -\frac {a^3}{3 x^3}+a b^2 x^3+\frac {b^3 x^6}{6}+3 a^2 b \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^3\right )^3}{x^4} \, dx=-\frac {a^3}{3 x^3}+a b^2 x^3+\frac {b^3 x^6}{6}+3 a^2 b \log (x) \]

[In]

Integrate[(a + b*x^3)^3/x^4,x]

[Out]

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

Maple [A] (verified)

Time = 3.66 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.92

method	result	size
default	\(-\frac {a^{3}}{3 x^{3}}+a \,b^{2} x^{3}+\frac {b^{3} x^{6}}{6}+3 a^{2} b \ln \left (x \right )\)	\(34\)
norman	\(\frac {a \,b^{2} x^{6}-\frac {1}{3} a^{3}+\frac {1}{6} b^{3} x^{9}}{x^{3}}+3 a^{2} b \ln \left (x \right )\)	\(36\)
parallelrisch	\(\frac {b^{3} x^{9}+6 a \,b^{2} x^{6}+18 a^{2} b \ln \left (x \right ) x^{3}-2 a^{3}}{6 x^{3}}\)	\(39\)
risch	\(\frac {b^{3} x^{6}}{6}+a \,b^{2} x^{3}+\frac {3 a^{2} b}{2}-\frac {a^{3}}{3 x^{3}}+3 a^{2} b \ln \left (x \right )\)	\(40\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a+b x^3\right )^3}{x^4} \, dx=\frac {b^{3} x^{9} + 6 \, a b^{2} x^{6} + 18 \, a^{2} b x^{3} \log \left (x\right ) - 2 \, a^{3}}{6 \, x^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a+b x^3\right )^3}{x^4} \, dx=- \frac {a^{3}}{3 x^{3}} + 3 a^{2} b \log {\left (x \right )} + a b^{2} x^{3} + \frac {b^{3} x^{6}}{6} \]

[In]

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

[Out]

-a**3/(3*x**3) + 3*a**2*b*log(x) + a*b**2*x**3 + b**3*x**6/6

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a+b x^3\right )^3}{x^4} \, dx=\frac {1}{6} \, b^{3} x^{6} + a b^{2} x^{3} + a^{2} b \log \left (x^{3}\right ) - \frac {a^{3}}{3 \, x^{3}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.19 \[ \int \frac {\left (a+b x^3\right )^3}{x^4} \, dx=\frac {1}{6} \, b^{3} x^{6} + a b^{2} x^{3} + 3 \, a^{2} b \log \left ({\left | x \right |}\right ) - \frac {3 \, a^{2} b x^{3} + a^{3}}{3 \, x^{3}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.89 \[ \int \frac {\left (a+b x^3\right )^3}{x^4} \, dx=\frac {b^3\,x^6}{6}-\frac {a^3}{3\,x^3}+a\,b^2\,x^3+3\,a^2\,b\,\ln \left (x\right ) \]

[In]

int((a + b*x^3)^3/x^4,x)

[Out]

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