3.1141 \(\int \frac{x^{11}}{(a+b x^4)^{5/4}} \, dx\)

Optimal. Leaf size=57 \[ -\frac{a^2}{b^3 \sqrt [4]{a+b x^4}}-\frac{2 a \left (a+b x^4\right )^{3/4}}{3 b^3}+\frac{\left (a+b x^4\right )^{7/4}}{7 b^3} \]

[Out]

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

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Rubi [A]  time = 0.0334128, antiderivative size = 57, 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} \[ -\frac{a^2}{b^3 \sqrt [4]{a+b x^4}}-\frac{2 a \left (a+b x^4\right )^{3/4}}{3 b^3}+\frac{\left (a+b x^4\right )^{7/4}}{7 b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^11/(a + b*x^4)^(5/4),x]

[Out]

-(a^2/(b^3*(a + b*x^4)^(1/4))) - (2*a*(a + b*x^4)^(3/4))/(3*b^3) + (a + b*x^4)^(7/4)/(7*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 \frac{x^{11}}{\left (a+b x^4\right )^{5/4}} \, dx &=\frac{1}{4} \operatorname{Subst}\left (\int \frac{x^2}{(a+b x)^{5/4}} \, dx,x,x^4\right )\\ &=\frac{1}{4} \operatorname{Subst}\left (\int \left (\frac{a^2}{b^2 (a+b x)^{5/4}}-\frac{2 a}{b^2 \sqrt [4]{a+b x}}+\frac{(a+b x)^{3/4}}{b^2}\right ) \, dx,x,x^4\right )\\ &=-\frac{a^2}{b^3 \sqrt [4]{a+b x^4}}-\frac{2 a \left (a+b x^4\right )^{3/4}}{3 b^3}+\frac{\left (a+b x^4\right )^{7/4}}{7 b^3}\\ \end{align*}

Mathematica [A]  time = 0.0173317, size = 39, normalized size = 0.68 \[ \frac{-32 a^2-8 a b x^4+3 b^2 x^8}{21 b^3 \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^11/(a + b*x^4)^(5/4),x]

[Out]

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

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Maple [A]  time = 0.004, size = 36, normalized size = 0.6 \begin{align*} -{\frac{-3\,{b}^{2}{x}^{8}+8\,ab{x}^{4}+32\,{a}^{2}}{21\,{b}^{3}}{\frac{1}{\sqrt [4]{b{x}^{4}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.980901, size = 63, normalized size = 1.11 \begin{align*} \frac{{\left (b x^{4} + a\right )}^{\frac{7}{4}}}{7 \, b^{3}} - \frac{2 \,{\left (b x^{4} + a\right )}^{\frac{3}{4}} a}{3 \, b^{3}} - \frac{a^{2}}{{\left (b x^{4} + a\right )}^{\frac{1}{4}} 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]

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

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Fricas [A]  time = 1.48, size = 101, normalized size = 1.77 \begin{align*} \frac{{\left (3 \, b^{2} x^{8} - 8 \, a b x^{4} - 32 \, a^{2}\right )}{\left (b x^{4} + a\right )}^{\frac{3}{4}}}{21 \,{\left (b^{4} x^{4} + a b^{3}\right )}} \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]

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

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Sympy [A]  time = 3.00822, size = 68, normalized size = 1.19 \begin{align*} \begin{cases} - \frac{32 a^{2}}{21 b^{3} \sqrt [4]{a + b x^{4}}} - \frac{8 a x^{4}}{21 b^{2} \sqrt [4]{a + b x^{4}}} + \frac{x^{8}}{7 b \sqrt [4]{a + b x^{4}}} & \text{for}\: b \neq 0 \\\frac{x^{12}}{12 a^{\frac{5}{4}}} & \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**2/(21*b**3*(a + b*x**4)**(1/4)) - 8*a*x**4/(21*b**2*(a + b*x**4)**(1/4)) + x**8/(7*b*(a + b*
x**4)**(1/4)), Ne(b, 0)), (x**12/(12*a**(5/4)), True))

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Giac [A]  time = 1.0869, size = 58, normalized size = 1.02 \begin{align*} \frac{3 \,{\left (b x^{4} + a\right )}^{\frac{7}{4}} - 14 \,{\left (b x^{4} + a\right )}^{\frac{3}{4}} a - \frac{21 \, a^{2}}{{\left (b x^{4} + a\right )}^{\frac{1}{4}}}}{21 \, 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="giac")

[Out]

1/21*(3*(b*x^4 + a)^(7/4) - 14*(b*x^4 + a)^(3/4)*a - 21*a^2/(b*x^4 + a)^(1/4))/b^3