3.1019 \(\int x^{11} (a+b x^4)^{3/4} \, dx\)

Optimal. Leaf size=59 \[ \frac{a^2 \left (a+b x^4\right )^{7/4}}{7 b^3}+\frac{\left (a+b x^4\right )^{15/4}}{15 b^3}-\frac{2 a \left (a+b x^4\right )^{11/4}}{11 b^3} \]

[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)

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Rubi [A]  time = 0.0340583, 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} \[ \frac{a^2 \left (a+b x^4\right )^{7/4}}{7 b^3}+\frac{\left (a+b x^4\right )^{15/4}}{15 b^3}-\frac{2 a \left (a+b x^4\right )^{11/4}}{11 b^3} \]

Antiderivative was successfully verified.

[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 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 )^{3/4} \, dx &=\frac{1}{4} \operatorname{Subst}\left (\int x^2 (a+b x)^{3/4} \, dx,x,x^4\right )\\ &=\frac{1}{4} \operatorname{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]  time = 0.0178629, size = 39, normalized size = 0.66 \[ \frac{\left (a+b x^4\right )^{7/4} \left (32 a^2-56 a b x^4+77 b^2 x^8\right )}{1155 b^3} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.005, size = 36, normalized size = 0.6 \begin{align*}{\frac{77\,{b}^{2}{x}^{8}-56\,ab{x}^{4}+32\,{a}^{2}}{1155\,{b}^{3}} \left ( b{x}^{4}+a \right ) ^{{\frac{7}{4}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.992419, size = 63, normalized size = 1.07 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 1.63973, size = 112, normalized size = 1.9 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [A]  time = 13.9351, size = 87, normalized size = 1.47 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Giac [A]  time = 1.36871, size = 58, normalized size = 0.98 \begin{align*} \frac{77 \,{\left (b x^{4} + a\right )}^{\frac{15}{4}} - 210 \,{\left (b x^{4} + a\right )}^{\frac{11}{4}} a + 165 \,{\left (b x^{4} + a\right )}^{\frac{7}{4}} a^{2}}{1155 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1155*(77*(b*x^4 + a)^(15/4) - 210*(b*x^4 + a)^(11/4)*a + 165*(b*x^4 + a)^(7/4)*a^2)/b^3