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]

(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)

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Rubi [A]  time = 0.0320943, antiderivative size = 59, normalized size of antiderivative = 1., 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 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]]

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])

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*}

Mathematica [A]  time = 0.0195229, 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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Maple [A]  time = 0.004, size = 36, normalized size = 0.6 \begin{align*}{\frac{117\,{b}^{2}{x}^{8}-72\,ab{x}^{4}+32\,{a}^{2}}{1989\,{b}^{3}} \left ( b{x}^{4}+a \right ) ^{{\frac{9}{4}}}} \end{align*}

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 = 0.960163, size = 63, normalized size = 1.07 \begin{align*} \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}} \end{align*}

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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Fricas [A]  time = 1.70786, size = 136, normalized size = 2.31 \begin{align*} \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}} \end{align*}

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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Sympy [A]  time = 27.5379, size = 110, normalized size = 1.86 \begin{align*} \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} \end{align*}

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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Giac [B]  time = 1.11559, size = 144, normalized size = 2.44 \begin{align*} \frac{\frac{17 \,{\left (45 \,{\left (b x^{4} + a\right )}^{\frac{13}{4}} - 130 \,{\left (b x^{4} + a\right )}^{\frac{9}{4}} a + 117 \,{\left (b x^{4} + a\right )}^{\frac{5}{4}} a^{2}\right )} a}{b^{2}} + \frac{3 \,{\left (195 \,{\left (b x^{4} + a\right )}^{\frac{17}{4}} - 765 \,{\left (b x^{4} + a\right )}^{\frac{13}{4}} a + 1105 \,{\left (b x^{4} + a\right )}^{\frac{9}{4}} a^{2} - 663 \,{\left (b x^{4} + a\right )}^{\frac{5}{4}} a^{3}\right )}}{b^{2}}}{9945 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9945*(17*(45*(b*x^4 + a)^(13/4) - 130*(b*x^4 + a)^(9/4)*a + 117*(b*x^4 + a)^(5/4)*a^2)*a/b^2 + 3*(195*(b*x^4
+ a)^(17/4) - 765*(b*x^4 + a)^(13/4)*a + 1105*(b*x^4 + a)^(9/4)*a^2 - 663*(b*x^4 + a)^(5/4)*a^3)/b^2)/b