\(\int x^{14} (a+b x^3)^3 \, dx\) [236]

Optimal result

Integrand size = 13, antiderivative size = 43 \[ \int x^{14} \left (a+b x^3\right )^3 \, dx=\frac {a^3 x^{15}}{15}+\frac {1}{6} a^2 b x^{18}+\frac {1}{7} a b^2 x^{21}+\frac {b^3 x^{24}}{24} \]

[Out]

1/15*a^3*x^15+1/6*a^2*b*x^18+1/7*a*b^2*x^21+1/24*b^3*x^24

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 43, 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} \[ \int x^{14} \left (a+b x^3\right )^3 \, dx=\frac {a^3 x^{15}}{15}+\frac {1}{6} a^2 b x^{18}+\frac {1}{7} a b^2 x^{21}+\frac {b^3 x^{24}}{24} \]

[In]

Int[x^14*(a + b*x^3)^3,x]

[Out]

(a^3*x^15)/15 + (a^2*b*x^18)/6 + (a*b^2*x^21)/7 + (b^3*x^24)/24

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 x^4 (a+b x)^3 \, dx,x,x^3\right ) \\ & = \frac {1}{3} \text {Subst}\left (\int \left (a^3 x^4+3 a^2 b x^5+3 a b^2 x^6+b^3 x^7\right ) \, dx,x,x^3\right ) \\ & = \frac {a^3 x^{15}}{15}+\frac {1}{6} a^2 b x^{18}+\frac {1}{7} a b^2 x^{21}+\frac {b^3 x^{24}}{24} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00 \[ \int x^{14} \left (a+b x^3\right )^3 \, dx=\frac {a^3 x^{15}}{15}+\frac {1}{6} a^2 b x^{18}+\frac {1}{7} a b^2 x^{21}+\frac {b^3 x^{24}}{24} \]

[In]

Integrate[x^14*(a + b*x^3)^3,x]

[Out]

(a^3*x^15)/15 + (a^2*b*x^18)/6 + (a*b^2*x^21)/7 + (b^3*x^24)/24

Maple [A] (verified)

Time = 3.65 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84

method	result	size
gosper	\(\frac {1}{15} a^{3} x^{15}+\frac {1}{6} a^{2} b \,x^{18}+\frac {1}{7} a \,b^{2} x^{21}+\frac {1}{24} b^{3} x^{24}\)	\(36\)
default	\(\frac {1}{15} a^{3} x^{15}+\frac {1}{6} a^{2} b \,x^{18}+\frac {1}{7} a \,b^{2} x^{21}+\frac {1}{24} b^{3} x^{24}\)	\(36\)
norman	\(\frac {1}{15} a^{3} x^{15}+\frac {1}{6} a^{2} b \,x^{18}+\frac {1}{7} a \,b^{2} x^{21}+\frac {1}{24} b^{3} x^{24}\)	\(36\)
risch	\(\frac {1}{15} a^{3} x^{15}+\frac {1}{6} a^{2} b \,x^{18}+\frac {1}{7} a \,b^{2} x^{21}+\frac {1}{24} b^{3} x^{24}\)	\(36\)
parallelrisch	\(\frac {1}{15} a^{3} x^{15}+\frac {1}{6} a^{2} b \,x^{18}+\frac {1}{7} a \,b^{2} x^{21}+\frac {1}{24} b^{3} x^{24}\)	\(36\)

[In]

int(x^14*(b*x^3+a)^3,x,method=_RETURNVERBOSE)

[Out]

1/15*a^3*x^15+1/6*a^2*b*x^18+1/7*a*b^2*x^21+1/24*b^3*x^24

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int x^{14} \left (a+b x^3\right )^3 \, dx=\frac {1}{24} \, b^{3} x^{24} + \frac {1}{7} \, a b^{2} x^{21} + \frac {1}{6} \, a^{2} b x^{18} + \frac {1}{15} \, a^{3} x^{15} \]

[In]

integrate(x^14*(b*x^3+a)^3,x, algorithm="fricas")

[Out]

1/24*b^3*x^24 + 1/7*a*b^2*x^21 + 1/6*a^2*b*x^18 + 1/15*a^3*x^15

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84 \[ \int x^{14} \left (a+b x^3\right )^3 \, dx=\frac {a^{3} x^{15}}{15} + \frac {a^{2} b x^{18}}{6} + \frac {a b^{2} x^{21}}{7} + \frac {b^{3} x^{24}}{24} \]

[In]

integrate(x**14*(b*x**3+a)**3,x)

[Out]

a**3*x**15/15 + a**2*b*x**18/6 + a*b**2*x**21/7 + b**3*x**24/24

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int x^{14} \left (a+b x^3\right )^3 \, dx=\frac {1}{24} \, b^{3} x^{24} + \frac {1}{7} \, a b^{2} x^{21} + \frac {1}{6} \, a^{2} b x^{18} + \frac {1}{15} \, a^{3} x^{15} \]

[In]

integrate(x^14*(b*x^3+a)^3,x, algorithm="maxima")

[Out]

1/24*b^3*x^24 + 1/7*a*b^2*x^21 + 1/6*a^2*b*x^18 + 1/15*a^3*x^15

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int x^{14} \left (a+b x^3\right )^3 \, dx=\frac {1}{24} \, b^{3} x^{24} + \frac {1}{7} \, a b^{2} x^{21} + \frac {1}{6} \, a^{2} b x^{18} + \frac {1}{15} \, a^{3} x^{15} \]

[In]

integrate(x^14*(b*x^3+a)^3,x, algorithm="giac")

[Out]

1/24*b^3*x^24 + 1/7*a*b^2*x^21 + 1/6*a^2*b*x^18 + 1/15*a^3*x^15

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int x^{14} \left (a+b x^3\right )^3 \, dx=\frac {a^3\,x^{15}}{15}+\frac {a^2\,b\,x^{18}}{6}+\frac {a\,b^2\,x^{21}}{7}+\frac {b^3\,x^{24}}{24} \]

[In]

int(x^14*(a + b*x^3)^3,x)

[Out]

(a^3*x^15)/15 + (b^3*x^24)/24 + (a^2*b*x^18)/6 + (a*b^2*x^21)/7