3.1049 \(\int x^{11} (a+b x^4)^{5/4} \, dx\)

Optimal. Leaf size=59 \[ \frac {a^2 \left (a+b x^4\right )^{9/4}}{9 b^3}+\frac {\left (a+b x^4\right )^{17/4}}{17 b^3}-\frac {2 a \left (a+b x^4\right )^{13/4}}{13 b^3} \]

[Out]

1/9*a^2*(b*x^4+a)^(9/4)/b^3-2/13*a*(b*x^4+a)^(13/4)/b^3+1/17*(b*x^4+a)^(17/4)/b^3

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Rubi [A]  time = 0.03, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {266, 43} \[ \frac {a^2 \left (a+b x^4\right )^{9/4}}{9 b^3}+\frac {\left (a+b x^4\right )^{17/4}}{17 b^3}-\frac {2 a \left (a+b x^4\right )^{13/4}}{13 b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^11*(a + b*x^4)^(5/4),x]

[Out]

(a^2*(a + b*x^4)^(9/4))/(9*b^3) - (2*a*(a + b*x^4)^(13/4))/(13*b^3) + (a + b*x^4)^(17/4)/(17*b^3)

Rule 43

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 266

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 x^{11} \left (a+b x^4\right )^{5/4} \, dx &=\frac {1}{4} \operatorname {Subst}\left (\int x^2 (a+b x)^{5/4} \, dx,x,x^4\right )\\ &=\frac {1}{4} \operatorname {Subst}\left (\int \left (\frac {a^2 (a+b x)^{5/4}}{b^2}-\frac {2 a (a+b x)^{9/4}}{b^2}+\frac {(a+b x)^{13/4}}{b^2}\right ) \, dx,x,x^4\right )\\ &=\frac {a^2 \left (a+b x^4\right )^{9/4}}{9 b^3}-\frac {2 a \left (a+b x^4\right )^{13/4}}{13 b^3}+\frac {\left (a+b x^4\right )^{17/4}}{17 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 39, normalized size = 0.66 \[ \frac {\left (a+b x^4\right )^{9/4} \left (32 a^2-72 a b x^4+117 b^2 x^8\right )}{1989 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^11*(a + b*x^4)^(5/4),x]

[Out]

((a + b*x^4)^(9/4)*(32*a^2 - 72*a*b*x^4 + 117*b^2*x^8))/(1989*b^3)

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fricas [A]  time = 0.78, size = 57, normalized size = 0.97 \[ \frac {{\left (117 \, b^{4} x^{16} + 162 \, a b^{3} x^{12} + 5 \, a^{2} b^{2} x^{8} - 8 \, a^{3} b x^{4} + 32 \, a^{4}\right )} {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{1989 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(b*x^4+a)^(5/4),x, algorithm="fricas")

[Out]

1/1989*(117*b^4*x^16 + 162*a*b^3*x^12 + 5*a^2*b^2*x^8 - 8*a^3*b*x^4 + 32*a^4)*(b*x^4 + a)^(1/4)/b^3

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giac [A]  time = 0.16, size = 43, normalized size = 0.73 \[ \frac {117 \, {\left (b x^{4} + a\right )}^{\frac {17}{4}} - 306 \, {\left (b x^{4} + a\right )}^{\frac {13}{4}} a + 221 \, {\left (b x^{4} + a\right )}^{\frac {9}{4}} a^{2}}{1989 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

1/1989*(117*(b*x^4 + a)^(17/4) - 306*(b*x^4 + a)^(13/4)*a + 221*(b*x^4 + a)^(9/4)*a^2)/b^3

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maple [A]  time = 0.00, size = 36, normalized size = 0.61 \[ \frac {\left (b \,x^{4}+a \right )^{\frac {9}{4}} \left (117 b^{2} x^{8}-72 a b \,x^{4}+32 a^{2}\right )}{1989 b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11*(b*x^4+a)^(5/4),x)

[Out]

1/1989*(b*x^4+a)^(9/4)*(117*b^2*x^8-72*a*b*x^4+32*a^2)/b^3

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maxima [A]  time = 1.29, size = 47, normalized size = 0.80 \[ \frac {{\left (b x^{4} + a\right )}^{\frac {17}{4}}}{17 \, b^{3}} - \frac {2 \, {\left (b x^{4} + a\right )}^{\frac {13}{4}} a}{13 \, b^{3}} + \frac {{\left (b x^{4} + a\right )}^{\frac {9}{4}} a^{2}}{9 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11*(b*x^4+a)^(5/4),x, algorithm="maxima")

[Out]

1/17*(b*x^4 + a)^(17/4)/b^3 - 2/13*(b*x^4 + a)^(13/4)*a/b^3 + 1/9*(b*x^4 + a)^(9/4)*a^2/b^3

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mupad [B]  time = 1.14, size = 53, normalized size = 0.90 \[ {\left (b\,x^4+a\right )}^{1/4}\,\left (\frac {18\,a\,x^{12}}{221}+\frac {b\,x^{16}}{17}+\frac {32\,a^4}{1989\,b^3}-\frac {8\,a^3\,x^4}{1989\,b^2}+\frac {5\,a^2\,x^8}{1989\,b}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11*(a + b*x^4)^(5/4),x)

[Out]

(a + b*x^4)^(1/4)*((18*a*x^12)/221 + (b*x^16)/17 + (32*a^4)/(1989*b^3) - (8*a^3*x^4)/(1989*b^2) + (5*a^2*x^8)/
(1989*b))

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sympy [A]  time = 25.29, size = 110, normalized size = 1.86 \[ \begin {cases} \frac {32 a^{4} \sqrt [4]{a + b x^{4}}}{1989 b^{3}} - \frac {8 a^{3} x^{4} \sqrt [4]{a + b x^{4}}}{1989 b^{2}} + \frac {5 a^{2} x^{8} \sqrt [4]{a + b x^{4}}}{1989 b} + \frac {18 a x^{12} \sqrt [4]{a + b x^{4}}}{221} + \frac {b x^{16} \sqrt [4]{a + b x^{4}}}{17} & \text {for}\: b \neq 0 \\\frac {a^{\frac {5}{4}} x^{12}}{12} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**11*(b*x**4+a)**(5/4),x)

[Out]

Piecewise((32*a**4*(a + b*x**4)**(1/4)/(1989*b**3) - 8*a**3*x**4*(a + b*x**4)**(1/4)/(1989*b**2) + 5*a**2*x**8
*(a + b*x**4)**(1/4)/(1989*b) + 18*a*x**12*(a + b*x**4)**(1/4)/221 + b*x**16*(a + b*x**4)**(1/4)/17, Ne(b, 0))
, (a**(5/4)*x**12/12, True))

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