3.1111 \(\int \frac{x^{15}}{(a+b x^4)^{3/4}} \, dx\)

Optimal. Leaf size=78 \[ \frac{3 a^2 \left (a+b x^4\right )^{5/4}}{5 b^4}-\frac{a^3 \sqrt [4]{a+b x^4}}{b^4}+\frac{\left (a+b x^4\right )^{13/4}}{13 b^4}-\frac{a \left (a+b x^4\right )^{9/4}}{3 b^4} \]

[Out]

-((a^3*(a + b*x^4)^(1/4))/b^4) + (3*a^2*(a + b*x^4)^(5/4))/(5*b^4) - (a*(a + b*x^4)^(9/4))/(3*b^4) + (a + b*x^
4)^(13/4)/(13*b^4)

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Rubi [A]  time = 0.0449799, antiderivative size = 78, 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} \[ \frac{3 a^2 \left (a+b x^4\right )^{5/4}}{5 b^4}-\frac{a^3 \sqrt [4]{a+b x^4}}{b^4}+\frac{\left (a+b x^4\right )^{13/4}}{13 b^4}-\frac{a \left (a+b x^4\right )^{9/4}}{3 b^4} \]

Antiderivative was successfully verified.

[In]

Int[x^15/(a + b*x^4)^(3/4),x]

[Out]

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

Mathematica [A]  time = 0.0222493, size = 50, normalized size = 0.64 \[ \frac{\sqrt [4]{a+b x^4} \left (32 a^2 b x^4-128 a^3-20 a b^2 x^8+15 b^3 x^{12}\right )}{195 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^15/(a + b*x^4)^(3/4),x]

[Out]

((a + b*x^4)^(1/4)*(-128*a^3 + 32*a^2*b*x^4 - 20*a*b^2*x^8 + 15*b^3*x^12))/(195*b^4)

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Maple [A]  time = 0.006, size = 47, normalized size = 0.6 \begin{align*} -{\frac{-15\,{b}^{3}{x}^{12}+20\,a{b}^{2}{x}^{8}-32\,{a}^{2}b{x}^{4}+128\,{a}^{3}}{195\,{b}^{4}}\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)^(3/4),x)

[Out]

-1/195*(b*x^4+a)^(1/4)*(-15*b^3*x^12+20*a*b^2*x^8-32*a^2*b*x^4+128*a^3)/b^4

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Maxima [A]  time = 1.0279, size = 86, normalized size = 1.1 \begin{align*} \frac{{\left (b x^{4} + a\right )}^{\frac{13}{4}}}{13 \, b^{4}} - \frac{{\left (b x^{4} + a\right )}^{\frac{9}{4}} a}{3 \, b^{4}} + \frac{3 \,{\left (b x^{4} + a\right )}^{\frac{5}{4}} a^{2}}{5 \, b^{4}} - \frac{{\left (b x^{4} + a\right )}^{\frac{1}{4}} a^{3}}{b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^15/(b*x^4+a)^(3/4),x, algorithm="maxima")

[Out]

1/13*(b*x^4 + a)^(13/4)/b^4 - 1/3*(b*x^4 + a)^(9/4)*a/b^4 + 3/5*(b*x^4 + a)^(5/4)*a^2/b^4 - (b*x^4 + a)^(1/4)*
a^3/b^4

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Fricas [A]  time = 1.46871, size = 112, normalized size = 1.44 \begin{align*} \frac{{\left (15 \, b^{3} x^{12} - 20 \, a b^{2} x^{8} + 32 \, a^{2} b x^{4} - 128 \, a^{3}\right )}{\left (b x^{4} + a\right )}^{\frac{1}{4}}}{195 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^15/(b*x^4+a)^(3/4),x, algorithm="fricas")

[Out]

1/195*(15*b^3*x^12 - 20*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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Sympy [A]  time = 9.41193, size = 92, normalized size = 1.18 \begin{align*} \begin{cases} - \frac{128 a^{3} \sqrt [4]{a + b x^{4}}}{195 b^{4}} + \frac{32 a^{2} x^{4} \sqrt [4]{a + b x^{4}}}{195 b^{3}} - \frac{4 a x^{8} \sqrt [4]{a + b x^{4}}}{39 b^{2}} + \frac{x^{12} \sqrt [4]{a + b x^{4}}}{13 b} & \text{for}\: b \neq 0 \\\frac{x^{16}}{16 a^{\frac{3}{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)**(3/4),x)

[Out]

Piecewise((-128*a**3*(a + b*x**4)**(1/4)/(195*b**4) + 32*a**2*x**4*(a + b*x**4)**(1/4)/(195*b**3) - 4*a*x**8*(
a + b*x**4)**(1/4)/(39*b**2) + x**12*(a + b*x**4)**(1/4)/(13*b), Ne(b, 0)), (x**16/(16*a**(3/4)), True))

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Giac [A]  time = 1.10276, size = 77, normalized size = 0.99 \begin{align*} \frac{15 \,{\left (b x^{4} + a\right )}^{\frac{13}{4}} - 65 \,{\left (b x^{4} + a\right )}^{\frac{9}{4}} a + 117 \,{\left (b x^{4} + a\right )}^{\frac{5}{4}} a^{2} - 195 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}} a^{3}}{195 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^15/(b*x^4+a)^(3/4),x, algorithm="giac")

[Out]

1/195*(15*(b*x^4 + a)^(13/4) - 65*(b*x^4 + a)^(9/4)*a + 117*(b*x^4 + a)^(5/4)*a^2 - 195*(b*x^4 + a)^(1/4)*a^3)
/b^4