3.3.46 \(\int \frac {(a+b x^3)^3}{x^{16}} \, dx\) [246]

Optimal. Leaf size=40 \[ -\frac {\left (a+b x^3\right )^4}{15 a x^{15}}+\frac {b \left (a+b x^3\right )^4}{60 a^2 x^{12}} \]

[Out]

-1/15*(b*x^3+a)^4/a/x^15+1/60*b*(b*x^3+a)^4/a^2/x^12

________________________________________________________________________________________

Rubi [A]
time = 0.01, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {272, 47, 37} \begin {gather*} \frac {b \left (a+b x^3\right )^4}{60 a^2 x^{12}}-\frac {\left (a+b x^3\right )^4}{15 a x^{15}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^3/x^16,x]

[Out]

-1/15*(a + b*x^3)^4/(a*x^15) + (b*(a + b*x^3)^4)/(60*a^2*x^12)

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*(Simplify[m + n + 2]/((b*c - a*d)*(m + 1))), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

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*} \int \frac {\left (a+b x^3\right )^3}{x^{16}} \, dx &=\frac {1}{3} \text {Subst}\left (\int \frac {(a+b x)^3}{x^6} \, dx,x,x^3\right )\\ &=-\frac {\left (a+b x^3\right )^4}{15 a x^{15}}-\frac {b \text {Subst}\left (\int \frac {(a+b x)^3}{x^5} \, dx,x,x^3\right )}{15 a}\\ &=-\frac {\left (a+b x^3\right )^4}{15 a x^{15}}+\frac {b \left (a+b x^3\right )^4}{60 a^2 x^{12}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.00, size = 43, normalized size = 1.08 \begin {gather*} -\frac {a^3}{15 x^{15}}-\frac {a^2 b}{4 x^{12}}-\frac {a b^2}{3 x^9}-\frac {b^3}{6 x^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^3)^3/x^16,x]

[Out]

-1/15*a^3/x^15 - (a^2*b)/(4*x^12) - (a*b^2)/(3*x^9) - b^3/(6*x^6)

________________________________________________________________________________________

Maple [A]
time = 0.12, size = 36, normalized size = 0.90

method	result	size
default	\(-\frac {a^{3}}{15 x^{15}}-\frac {a \,b^{2}}{3 x^{9}}-\frac {a^{2} b}{4 x^{12}}-\frac {b^{3}}{6 x^{6}}\)	\(36\)
norman	\(\frac {-\frac {1}{15} a^{3}-\frac {1}{4} a^{2} b \,x^{3}-\frac {1}{3} a \,b^{2} x^{6}-\frac {1}{6} b^{3} x^{9}}{x^{15}}\)	\(37\)
risch	\(\frac {-\frac {1}{15} a^{3}-\frac {1}{4} a^{2} b \,x^{3}-\frac {1}{3} a \,b^{2} x^{6}-\frac {1}{6} b^{3} x^{9}}{x^{15}}\)	\(37\)
gosper	\(-\frac {10 b^{3} x^{9}+20 a \,b^{2} x^{6}+15 a^{2} b \,x^{3}+4 a^{3}}{60 x^{15}}\)	\(38\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)^3/x^16,x,method=_RETURNVERBOSE)

[Out]

-1/15*a^3/x^15-1/3*a*b^2/x^9-1/4*a^2*b/x^12-1/6*b^3/x^6

________________________________________________________________________________________

Maxima [A]
time = 0.30, size = 37, normalized size = 0.92 \begin {gather*} -\frac {10 \, b^{3} x^{9} + 20 \, a b^{2} x^{6} + 15 \, a^{2} b x^{3} + 4 \, a^{3}}{60 \, x^{15}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^3/x^16,x, algorithm="maxima")

[Out]

-1/60*(10*b^3*x^9 + 20*a*b^2*x^6 + 15*a^2*b*x^3 + 4*a^3)/x^15

________________________________________________________________________________________

Fricas [A]
time = 0.33, size = 37, normalized size = 0.92 \begin {gather*} -\frac {10 \, b^{3} x^{9} + 20 \, a b^{2} x^{6} + 15 \, a^{2} b x^{3} + 4 \, a^{3}}{60 \, x^{15}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^3/x^16,x, algorithm="fricas")

[Out]

-1/60*(10*b^3*x^9 + 20*a*b^2*x^6 + 15*a^2*b*x^3 + 4*a^3)/x^15

________________________________________________________________________________________

Sympy [A]
time = 0.15, size = 39, normalized size = 0.98 \begin {gather*} \frac {- 4 a^{3} - 15 a^{2} b x^{3} - 20 a b^{2} x^{6} - 10 b^{3} x^{9}}{60 x^{15}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)**3/x**16,x)

[Out]

(-4*a**3 - 15*a**2*b*x**3 - 20*a*b**2*x**6 - 10*b**3*x**9)/(60*x**15)

________________________________________________________________________________________

Giac [A]
time = 1.67, size = 37, normalized size = 0.92 \begin {gather*} -\frac {10 \, b^{3} x^{9} + 20 \, a b^{2} x^{6} + 15 \, a^{2} b x^{3} + 4 \, a^{3}}{60 \, x^{15}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)^3/x^16,x, algorithm="giac")

[Out]

-1/60*(10*b^3*x^9 + 20*a*b^2*x^6 + 15*a^2*b*x^3 + 4*a^3)/x^15

________________________________________________________________________________________

Mupad [B]
time = 0.03, size = 37, normalized size = 0.92 \begin {gather*} -\frac {\frac {a^3}{15}+\frac {a^2\,b\,x^3}{4}+\frac {a\,b^2\,x^6}{3}+\frac {b^3\,x^9}{6}}{x^{15}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^3)^3/x^16,x)

[Out]

-(a^3/15 + (b^3*x^9)/6 + (a^2*b*x^3)/4 + (a*b^2*x^6)/3)/x^15

________________________________________________________________________________________