\(\int x^4 (a+b x^4)^3 \, dx\) [634]

Optimal result

Integrand size = 13, antiderivative size = 43 \[ \int x^4 \left (a+b x^4\right )^3 \, dx=\frac {a^3 x^5}{5}+\frac {1}{3} a^2 b x^9+\frac {3}{13} a b^2 x^{13}+\frac {b^3 x^{17}}{17} \]

[Out]

1/5*a^3*x^5+1/3*a^2*b*x^9+3/13*a*b^2*x^13+1/17*b^3*x^17

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {276} \[ \int x^4 \left (a+b x^4\right )^3 \, dx=\frac {a^3 x^5}{5}+\frac {1}{3} a^2 b x^9+\frac {3}{13} a b^2 x^{13}+\frac {b^3 x^{17}}{17} \]

[In]

Int[x^4*(a + b*x^4)^3,x]

[Out]

(a^3*x^5)/5 + (a^2*b*x^9)/3 + (3*a*b^2*x^13)/13 + (b^3*x^17)/17

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (a^3 x^4+3 a^2 b x^8+3 a b^2 x^{12}+b^3 x^{16}\right ) \, dx \\ & = \frac {a^3 x^5}{5}+\frac {1}{3} a^2 b x^9+\frac {3}{13} a b^2 x^{13}+\frac {b^3 x^{17}}{17} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00 \[ \int x^4 \left (a+b x^4\right )^3 \, dx=\frac {a^3 x^5}{5}+\frac {1}{3} a^2 b x^9+\frac {3}{13} a b^2 x^{13}+\frac {b^3 x^{17}}{17} \]

[In]

Integrate[x^4*(a + b*x^4)^3,x]

[Out]

(a^3*x^5)/5 + (a^2*b*x^9)/3 + (3*a*b^2*x^13)/13 + (b^3*x^17)/17

Maple [A] (verified)

Time = 3.89 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.84

method	result	size
gosper	\(\frac {1}{5} a^{3} x^{5}+\frac {1}{3} a^{2} b \,x^{9}+\frac {3}{13} a \,b^{2} x^{13}+\frac {1}{17} b^{3} x^{17}\)	\(36\)
default	\(\frac {1}{5} a^{3} x^{5}+\frac {1}{3} a^{2} b \,x^{9}+\frac {3}{13} a \,b^{2} x^{13}+\frac {1}{17} b^{3} x^{17}\)	\(36\)
norman	\(\frac {1}{5} a^{3} x^{5}+\frac {1}{3} a^{2} b \,x^{9}+\frac {3}{13} a \,b^{2} x^{13}+\frac {1}{17} b^{3} x^{17}\)	\(36\)
risch	\(\frac {1}{5} a^{3} x^{5}+\frac {1}{3} a^{2} b \,x^{9}+\frac {3}{13} a \,b^{2} x^{13}+\frac {1}{17} b^{3} x^{17}\)	\(36\)
parallelrisch	\(\frac {1}{5} a^{3} x^{5}+\frac {1}{3} a^{2} b \,x^{9}+\frac {3}{13} a \,b^{2} x^{13}+\frac {1}{17} b^{3} x^{17}\)	\(36\)

[In]

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

[Out]

1/5*a^3*x^5+1/3*a^2*b*x^9+3/13*a*b^2*x^13+1/17*b^3*x^17

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int x^4 \left (a+b x^4\right )^3 \, dx=\frac {1}{17} \, b^{3} x^{17} + \frac {3}{13} \, a b^{2} x^{13} + \frac {1}{3} \, a^{2} b x^{9} + \frac {1}{5} \, a^{3} x^{5} \]

[In]

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

[Out]

1/17*b^3*x^17 + 3/13*a*b^2*x^13 + 1/3*a^2*b*x^9 + 1/5*a^3*x^5

Sympy [A] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.86 \[ \int x^4 \left (a+b x^4\right )^3 \, dx=\frac {a^{3} x^{5}}{5} + \frac {a^{2} b x^{9}}{3} + \frac {3 a b^{2} x^{13}}{13} + \frac {b^{3} x^{17}}{17} \]

[In]

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

[Out]

a**3*x**5/5 + a**2*b*x**9/3 + 3*a*b**2*x**13/13 + b**3*x**17/17

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int x^4 \left (a+b x^4\right )^3 \, dx=\frac {1}{17} \, b^{3} x^{17} + \frac {3}{13} \, a b^{2} x^{13} + \frac {1}{3} \, a^{2} b x^{9} + \frac {1}{5} \, a^{3} x^{5} \]

[In]

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

[Out]

1/17*b^3*x^17 + 3/13*a*b^2*x^13 + 1/3*a^2*b*x^9 + 1/5*a^3*x^5

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int x^4 \left (a+b x^4\right )^3 \, dx=\frac {1}{17} \, b^{3} x^{17} + \frac {3}{13} \, a b^{2} x^{13} + \frac {1}{3} \, a^{2} b x^{9} + \frac {1}{5} \, a^{3} x^{5} \]

[In]

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

[Out]

1/17*b^3*x^17 + 3/13*a*b^2*x^13 + 1/3*a^2*b*x^9 + 1/5*a^3*x^5

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int x^4 \left (a+b x^4\right )^3 \, dx=\frac {a^3\,x^5}{5}+\frac {a^2\,b\,x^9}{3}+\frac {3\,a\,b^2\,x^{13}}{13}+\frac {b^3\,x^{17}}{17} \]

[In]

int(x^4*(a + b*x^4)^3,x)

[Out]

(a^3*x^5)/5 + (b^3*x^17)/17 + (a^2*b*x^9)/3 + (3*a*b^2*x^13)/13