3.1140 \(\int \frac{x^{15}}{(a+b x^4)^{5/4}} \, dx\)

Optimal. Leaf size=74 \[ \frac{a^3}{b^4 \sqrt [4]{a+b x^4}}+\frac{a^2 \left (a+b x^4\right )^{3/4}}{b^4}-\frac{3 a \left (a+b x^4\right )^{7/4}}{7 b^4}+\frac{\left (a+b x^4\right )^{11/4}}{11 b^4} \]

[Out]

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

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Rubi [A]  time = 0.0439708, antiderivative size = 74, 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} \[ \frac{a^3}{b^4 \sqrt [4]{a+b x^4}}+\frac{a^2 \left (a+b x^4\right )^{3/4}}{b^4}-\frac{3 a \left (a+b x^4\right )^{7/4}}{7 b^4}+\frac{\left (a+b x^4\right )^{11/4}}{11 b^4} \]

Antiderivative was successfully verified.

[In]

Int[x^15/(a + b*x^4)^(5/4),x]

[Out]

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

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^{15}}{\left (a+b x^4\right )^{5/4}} \, dx &=\frac{1}{4} \operatorname{Subst}\left (\int \frac{x^3}{(a+b x)^{5/4}} \, dx,x,x^4\right )\\ &=\frac{1}{4} \operatorname{Subst}\left (\int \left (-\frac{a^3}{b^3 (a+b x)^{5/4}}+\frac{3 a^2}{b^3 \sqrt [4]{a+b x}}-\frac{3 a (a+b x)^{3/4}}{b^3}+\frac{(a+b x)^{7/4}}{b^3}\right ) \, dx,x,x^4\right )\\ &=\frac{a^3}{b^4 \sqrt [4]{a+b x^4}}+\frac{a^2 \left (a+b x^4\right )^{3/4}}{b^4}-\frac{3 a \left (a+b x^4\right )^{7/4}}{7 b^4}+\frac{\left (a+b x^4\right )^{11/4}}{11 b^4}\\ \end{align*}

Mathematica [A]  time = 0.0213592, size = 50, normalized size = 0.68 \[ \frac{32 a^2 b x^4+128 a^3-12 a b^2 x^8+7 b^3 x^{12}}{77 b^4 \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^15/(a + b*x^4)^(5/4),x]

[Out]

(128*a^3 + 32*a^2*b*x^4 - 12*a*b^2*x^8 + 7*b^3*x^12)/(77*b^4*(a + b*x^4)^(1/4))

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Maple [A]  time = 0.005, size = 47, normalized size = 0.6 \begin{align*}{\frac{7\,{b}^{3}{x}^{12}-12\,a{b}^{2}{x}^{8}+32\,{a}^{2}b{x}^{4}+128\,{a}^{3}}{77\,{b}^{4}}{\frac{1}{\sqrt [4]{b{x}^{4}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^15/(b*x^4+a)^(5/4),x)

[Out]

1/77*(7*b^3*x^12-12*a*b^2*x^8+32*a^2*b*x^4+128*a^3)/(b*x^4+a)^(1/4)/b^4

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Maxima [A]  time = 1.00307, size = 84, normalized size = 1.14 \begin{align*} \frac{{\left (b x^{4} + a\right )}^{\frac{11}{4}}}{11 \, b^{4}} - \frac{3 \,{\left (b x^{4} + a\right )}^{\frac{7}{4}} a}{7 \, b^{4}} + \frac{{\left (b x^{4} + a\right )}^{\frac{3}{4}} a^{2}}{b^{4}} + \frac{a^{3}}{{\left (b x^{4} + a\right )}^{\frac{1}{4}} b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^15/(b*x^4+a)^(5/4),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.42926, size = 128, normalized size = 1.73 \begin{align*} \frac{{\left (7 \, b^{3} x^{12} - 12 \, a b^{2} x^{8} + 32 \, a^{2} b x^{4} + 128 \, a^{3}\right )}{\left (b x^{4} + a\right )}^{\frac{3}{4}}}{77 \,{\left (b^{5} x^{4} + a b^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^15/(b*x^4+a)^(5/4),x, algorithm="fricas")

[Out]

1/77*(7*b^3*x^12 - 12*a*b^2*x^8 + 32*a^2*b*x^4 + 128*a^3)*(b*x^4 + a)^(3/4)/(b^5*x^4 + a*b^4)

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Sympy [A]  time = 8.62131, size = 92, normalized size = 1.24 \begin{align*} \begin{cases} \frac{128 a^{3}}{77 b^{4} \sqrt [4]{a + b x^{4}}} + \frac{32 a^{2} x^{4}}{77 b^{3} \sqrt [4]{a + b x^{4}}} - \frac{12 a x^{8}}{77 b^{2} \sqrt [4]{a + b x^{4}}} + \frac{x^{12}}{11 b \sqrt [4]{a + b x^{4}}} & \text{for}\: b \neq 0 \\\frac{x^{16}}{16 a^{\frac{5}{4}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**15/(b*x**4+a)**(5/4),x)

[Out]

Piecewise((128*a**3/(77*b**4*(a + b*x**4)**(1/4)) + 32*a**2*x**4/(77*b**3*(a + b*x**4)**(1/4)) - 12*a*x**8/(77
*b**2*(a + b*x**4)**(1/4)) + x**12/(11*b*(a + b*x**4)**(1/4)), Ne(b, 0)), (x**16/(16*a**(5/4)), True))

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Giac [A]  time = 1.09573, size = 77, normalized size = 1.04 \begin{align*} \frac{7 \,{\left (b x^{4} + a\right )}^{\frac{11}{4}} - 33 \,{\left (b x^{4} + a\right )}^{\frac{7}{4}} a + 77 \,{\left (b x^{4} + a\right )}^{\frac{3}{4}} a^{2} + \frac{77 \, a^{3}}{{\left (b x^{4} + a\right )}^{\frac{1}{4}}}}{77 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^15/(b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

1/77*(7*(b*x^4 + a)^(11/4) - 33*(b*x^4 + a)^(7/4)*a + 77*(b*x^4 + a)^(3/4)*a^2 + 77*a^3/(b*x^4 + a)^(1/4))/b^4