\(\int \frac {(a+b x^4)^3}{x^5} \, dx\) [643]

Optimal result

Integrand size = 13, antiderivative size = 40 \[ \int \frac {\left (a+b x^4\right )^3}{x^5} \, dx=-\frac {a^3}{4 x^4}+\frac {3}{4} a b^2 x^4+\frac {b^3 x^8}{8}+3 a^2 b \log (x) \]

[Out]

-1/4*a^3/x^4+3/4*a*b^2*x^4+1/8*b^3*x^8+3*a^2*b*ln(x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 40, 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} \[ \int \frac {\left (a+b x^4\right )^3}{x^5} \, dx=-\frac {a^3}{4 x^4}+3 a^2 b \log (x)+\frac {3}{4} a b^2 x^4+\frac {b^3 x^8}{8} \]

[In]

Int[(a + b*x^4)^3/x^5,x]

[Out]

-1/4*a^3/x^4 + (3*a*b^2*x^4)/4 + (b^3*x^8)/8 + 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}{4} \text {Subst}\left (\int \frac {(a+b x)^3}{x^2} \, dx,x,x^4\right ) \\ & = \frac {1}{4} \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^4\right ) \\ & = -\frac {a^3}{4 x^4}+\frac {3}{4} a b^2 x^4+\frac {b^3 x^8}{8}+3 a^2 b \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00 \[ \int \frac {\left (a+b x^4\right )^3}{x^5} \, dx=-\frac {a^3}{4 x^4}+\frac {3}{4} a b^2 x^4+\frac {b^3 x^8}{8}+3 a^2 b \log (x) \]

[In]

Integrate[(a + b*x^4)^3/x^5,x]

[Out]

-1/4*a^3/x^4 + (3*a*b^2*x^4)/4 + (b^3*x^8)/8 + 3*a^2*b*Log[x]

Maple [A] (verified)

Time = 3.83 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.88

method	result	size
default	\(-\frac {a^{3}}{4 x^{4}}+\frac {3 a \,b^{2} x^{4}}{4}+\frac {b^{3} x^{8}}{8}+3 a^{2} b \ln \left (x \right )\)	\(35\)
norman	\(\frac {-\frac {1}{4} a^{3}+\frac {1}{8} b^{3} x^{12}+\frac {3}{4} a \,b^{2} x^{8}}{x^{4}}+3 a^{2} b \ln \left (x \right )\)	\(37\)
parallelrisch	\(\frac {b^{3} x^{12}+6 a \,b^{2} x^{8}+24 a^{2} b \ln \left (x \right ) x^{4}-2 a^{3}}{8 x^{4}}\)	\(39\)
risch	\(\frac {b^{3} x^{8}}{8}+\frac {3 a \,b^{2} x^{4}}{4}+\frac {9 a^{2} b}{8}-\frac {a^{3}}{4 x^{4}}+3 a^{2} b \ln \left (x \right )\)	\(41\)

[In]

int((b*x^4+a)^3/x^5,x,method=_RETURNVERBOSE)

[Out]

-1/4*a^3/x^4+3/4*a*b^2*x^4+1/8*b^3*x^8+3*a^2*b*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.95 \[ \int \frac {\left (a+b x^4\right )^3}{x^5} \, dx=\frac {b^{3} x^{12} + 6 \, a b^{2} x^{8} + 24 \, a^{2} b x^{4} \log \left (x\right ) - 2 \, a^{3}}{8 \, x^{4}} \]

[In]

integrate((b*x^4+a)^3/x^5,x, algorithm="fricas")

[Out]

1/8*(b^3*x^12 + 6*a*b^2*x^8 + 24*a^2*b*x^4*log(x) - 2*a^3)/x^4

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a+b x^4\right )^3}{x^5} \, dx=- \frac {a^{3}}{4 x^{4}} + 3 a^{2} b \log {\left (x \right )} + \frac {3 a b^{2} x^{4}}{4} + \frac {b^{3} x^{8}}{8} \]

[In]

integrate((b*x**4+a)**3/x**5,x)

[Out]

-a**3/(4*x**4) + 3*a**2*b*log(x) + 3*a*b**2*x**4/4 + b**3*x**8/8

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.90 \[ \int \frac {\left (a+b x^4\right )^3}{x^5} \, dx=\frac {1}{8} \, b^{3} x^{8} + \frac {3}{4} \, a b^{2} x^{4} + \frac {3}{4} \, a^{2} b \log \left (x^{4}\right ) - \frac {a^{3}}{4 \, x^{4}} \]

[In]

integrate((b*x^4+a)^3/x^5,x, algorithm="maxima")

[Out]

1/8*b^3*x^8 + 3/4*a*b^2*x^4 + 3/4*a^2*b*log(x^4) - 1/4*a^3/x^4

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.15 \[ \int \frac {\left (a+b x^4\right )^3}{x^5} \, dx=\frac {1}{8} \, b^{3} x^{8} + \frac {3}{4} \, a b^{2} x^{4} + \frac {3}{4} \, a^{2} b \log \left (x^{4}\right ) - \frac {3 \, a^{2} b x^{4} + a^{3}}{4 \, x^{4}} \]

[In]

integrate((b*x^4+a)^3/x^5,x, algorithm="giac")

[Out]

1/8*b^3*x^8 + 3/4*a*b^2*x^4 + 3/4*a^2*b*log(x^4) - 1/4*(3*a^2*b*x^4 + a^3)/x^4

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.85 \[ \int \frac {\left (a+b x^4\right )^3}{x^5} \, dx=\frac {b^3\,x^8}{8}-\frac {a^3}{4\,x^4}+\frac {3\,a\,b^2\,x^4}{4}+3\,a^2\,b\,\ln \left (x\right ) \]

[In]

int((a + b*x^4)^3/x^5,x)

[Out]

(b^3*x^8)/8 - a^3/(4*x^4) + (3*a*b^2*x^4)/4 + 3*a^2*b*log(x)