\(\int x^{11} (a+b x^4)^{3/4} \, dx\) [1019]

Optimal result

Integrand size = 15, antiderivative size = 59 \[ \int x^{11} \left (a+b x^4\right )^{3/4} \, dx=\frac {a^2 \left (a+b x^4\right )^{7/4}}{7 b^3}-\frac {2 a \left (a+b x^4\right )^{11/4}}{11 b^3}+\frac {\left (a+b x^4\right )^{15/4}}{15 b^3} \]

[Out]

1/7*a^2*(b*x^4+a)^(7/4)/b^3-2/11*a*(b*x^4+a)^(11/4)/b^3+1/15*(b*x^4+a)^(15/4)/b^3

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {272, 45} \[ \int x^{11} \left (a+b x^4\right )^{3/4} \, dx=\frac {a^2 \left (a+b x^4\right )^{7/4}}{7 b^3}+\frac {\left (a+b x^4\right )^{15/4}}{15 b^3}-\frac {2 a \left (a+b x^4\right )^{11/4}}{11 b^3} \]

[In]

Int[x^11*(a + b*x^4)^(3/4),x]

[Out]

(a^2*(a + b*x^4)^(7/4))/(7*b^3) - (2*a*(a + b*x^4)^(11/4))/(11*b^3) + (a + b*x^4)^(15/4)/(15*b^3)

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 x^2 (a+b x)^{3/4} \, dx,x,x^4\right ) \\ & = \frac {1}{4} \text {Subst}\left (\int \left (\frac {a^2 (a+b x)^{3/4}}{b^2}-\frac {2 a (a+b x)^{7/4}}{b^2}+\frac {(a+b x)^{11/4}}{b^2}\right ) \, dx,x,x^4\right ) \\ & = \frac {a^2 \left (a+b x^4\right )^{7/4}}{7 b^3}-\frac {2 a \left (a+b x^4\right )^{11/4}}{11 b^3}+\frac {\left (a+b x^4\right )^{15/4}}{15 b^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.66 \[ \int x^{11} \left (a+b x^4\right )^{3/4} \, dx=\frac {\left (a+b x^4\right )^{7/4} \left (32 a^2-56 a b x^4+77 b^2 x^8\right )}{1155 b^3} \]

[In]

Integrate[x^11*(a + b*x^4)^(3/4),x]

[Out]

((a + b*x^4)^(7/4)*(32*a^2 - 56*a*b*x^4 + 77*b^2*x^8))/(1155*b^3)

Maple [A] (verified)

Time = 4.18 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.61

method	result	size
gosper	\(\frac {\left (b \,x^{4}+a \right )^{\frac {7}{4}} \left (77 b^{2} x^{8}-56 a b \,x^{4}+32 a^{2}\right )}{1155 b^{3}}\)	\(36\)
pseudoelliptic	\(\frac {\left (b \,x^{4}+a \right )^{\frac {7}{4}} \left (77 b^{2} x^{8}-56 a b \,x^{4}+32 a^{2}\right )}{1155 b^{3}}\)	\(36\)
trager	\(\frac {\left (77 b^{3} x^{12}+21 a \,b^{2} x^{8}-24 a^{2} b \,x^{4}+32 a^{3}\right ) \left (b \,x^{4}+a \right )^{\frac {3}{4}}}{1155 b^{3}}\)	\(47\)
risch	\(\frac {\left (77 b^{3} x^{12}+21 a \,b^{2} x^{8}-24 a^{2} b \,x^{4}+32 a^{3}\right ) \left (b \,x^{4}+a \right )^{\frac {3}{4}}}{1155 b^{3}}\)	\(47\)

[In]

int(x^11*(b*x^4+a)^(3/4),x,method=_RETURNVERBOSE)

[Out]

1/1155*(b*x^4+a)^(7/4)*(77*b^2*x^8-56*a*b*x^4+32*a^2)/b^3

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.78 \[ \int x^{11} \left (a+b x^4\right )^{3/4} \, dx=\frac {{\left (77 \, b^{3} x^{12} + 21 \, a b^{2} x^{8} - 24 \, a^{2} b x^{4} + 32 \, a^{3}\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{1155 \, b^{3}} \]

[In]

integrate(x^11*(b*x^4+a)^(3/4),x, algorithm="fricas")

[Out]

1/1155*(77*b^3*x^12 + 21*a*b^2*x^8 - 24*a^2*b*x^4 + 32*a^3)*(b*x^4 + a)^(3/4)/b^3

Sympy [A] (verification not implemented)

Time = 0.65 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.47 \[ \int x^{11} \left (a+b x^4\right )^{3/4} \, dx=\begin {cases} \frac {32 a^{3} \left (a + b x^{4}\right )^{\frac {3}{4}}}{1155 b^{3}} - \frac {8 a^{2} x^{4} \left (a + b x^{4}\right )^{\frac {3}{4}}}{385 b^{2}} + \frac {a x^{8} \left (a + b x^{4}\right )^{\frac {3}{4}}}{55 b} + \frac {x^{12} \left (a + b x^{4}\right )^{\frac {3}{4}}}{15} & \text {for}\: b \neq 0 \\\frac {a^{\frac {3}{4}} x^{12}}{12} & \text {otherwise} \end {cases} \]

[In]

integrate(x**11*(b*x**4+a)**(3/4),x)

[Out]

Piecewise((32*a**3*(a + b*x**4)**(3/4)/(1155*b**3) - 8*a**2*x**4*(a + b*x**4)**(3/4)/(385*b**2) + a*x**8*(a +
b*x**4)**(3/4)/(55*b) + x**12*(a + b*x**4)**(3/4)/15, Ne(b, 0)), (a**(3/4)*x**12/12, True))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.80 \[ \int x^{11} \left (a+b x^4\right )^{3/4} \, dx=\frac {{\left (b x^{4} + a\right )}^{\frac {15}{4}}}{15 \, b^{3}} - \frac {2 \, {\left (b x^{4} + a\right )}^{\frac {11}{4}} a}{11 \, b^{3}} + \frac {{\left (b x^{4} + a\right )}^{\frac {7}{4}} a^{2}}{7 \, b^{3}} \]

[In]

integrate(x^11*(b*x^4+a)^(3/4),x, algorithm="maxima")

[Out]

1/15*(b*x^4 + a)^(15/4)/b^3 - 2/11*(b*x^4 + a)^(11/4)*a/b^3 + 1/7*(b*x^4 + a)^(7/4)*a^2/b^3

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 106 vs. \(2 (47) = 94\).

Time = 0.28 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.80 \[ \int x^{11} \left (a+b x^4\right )^{3/4} \, dx=\frac {\frac {5 \, {\left (21 \, {\left (b x^{4} + a\right )}^{\frac {11}{4}} - 66 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} a + 77 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{2}\right )} a}{b^{2}} + \frac {77 \, {\left (b x^{4} + a\right )}^{\frac {15}{4}} - 315 \, {\left (b x^{4} + a\right )}^{\frac {11}{4}} a + 495 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} a^{2} - 385 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{3}}{b^{2}}}{1155 \, b} \]

[In]

integrate(x^11*(b*x^4+a)^(3/4),x, algorithm="giac")

[Out]

1/1155*(5*(21*(b*x^4 + a)^(11/4) - 66*(b*x^4 + a)^(7/4)*a + 77*(b*x^4 + a)^(3/4)*a^2)*a/b^2 + (77*(b*x^4 + a)^
(15/4) - 315*(b*x^4 + a)^(11/4)*a + 495*(b*x^4 + a)^(7/4)*a^2 - 385*(b*x^4 + a)^(3/4)*a^3)/b^2)/b

Mupad [B] (verification not implemented)

Time = 5.60 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.75 \[ \int x^{11} \left (a+b x^4\right )^{3/4} \, dx={\left (b\,x^4+a\right )}^{3/4}\,\left (\frac {x^{12}}{15}+\frac {32\,a^3}{1155\,b^3}+\frac {a\,x^8}{55\,b}-\frac {8\,a^2\,x^4}{385\,b^2}\right ) \]

[In]

int(x^11*(a + b*x^4)^(3/4),x)

[Out]

(a + b*x^4)^(3/4)*(x^12/15 + (32*a^3)/(1155*b^3) + (a*x^8)/(55*b) - (8*a^2*x^4)/(385*b^2))