3.1139 \(\int \frac{x^{19}}{(a+b x^4)^{5/4}} \, dx\)

Optimal. Leaf size=99 \[ -\frac{a^4}{b^5 \sqrt [4]{a+b x^4}}-\frac{4 a^3 \left (a+b x^4\right )^{3/4}}{3 b^5}+\frac{6 a^2 \left (a+b x^4\right )^{7/4}}{7 b^5}-\frac{4 a \left (a+b x^4\right )^{11/4}}{11 b^5}+\frac{\left (a+b x^4\right )^{15/4}}{15 b^5} \]

[Out]

-(a^4/(b^5*(a + b*x^4)^(1/4))) - (4*a^3*(a + b*x^4)^(3/4))/(3*b^5) + (6*a^2*(a + b*x^4)^(7/4))/(7*b^5) - (4*a*
(a + b*x^4)^(11/4))/(11*b^5) + (a + b*x^4)^(15/4)/(15*b^5)

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Rubi [A]  time = 0.0540024, antiderivative size = 99, 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} \[ -\frac{a^4}{b^5 \sqrt [4]{a+b x^4}}-\frac{4 a^3 \left (a+b x^4\right )^{3/4}}{3 b^5}+\frac{6 a^2 \left (a+b x^4\right )^{7/4}}{7 b^5}-\frac{4 a \left (a+b x^4\right )^{11/4}}{11 b^5}+\frac{\left (a+b x^4\right )^{15/4}}{15 b^5} \]

Antiderivative was successfully verified.

[In]

Int[x^19/(a + b*x^4)^(5/4),x]

[Out]

-(a^4/(b^5*(a + b*x^4)^(1/4))) - (4*a^3*(a + b*x^4)^(3/4))/(3*b^5) + (6*a^2*(a + b*x^4)^(7/4))/(7*b^5) - (4*a*
(a + b*x^4)^(11/4))/(11*b^5) + (a + b*x^4)^(15/4)/(15*b^5)

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^{19}}{\left (a+b x^4\right )^{5/4}} \, dx &=\frac{1}{4} \operatorname{Subst}\left (\int \frac{x^4}{(a+b x)^{5/4}} \, dx,x,x^4\right )\\ &=\frac{1}{4} \operatorname{Subst}\left (\int \left (\frac{a^4}{b^4 (a+b x)^{5/4}}-\frac{4 a^3}{b^4 \sqrt [4]{a+b x}}+\frac{6 a^2 (a+b x)^{3/4}}{b^4}-\frac{4 a (a+b x)^{7/4}}{b^4}+\frac{(a+b x)^{11/4}}{b^4}\right ) \, dx,x,x^4\right )\\ &=-\frac{a^4}{b^5 \sqrt [4]{a+b x^4}}-\frac{4 a^3 \left (a+b x^4\right )^{3/4}}{3 b^5}+\frac{6 a^2 \left (a+b x^4\right )^{7/4}}{7 b^5}-\frac{4 a \left (a+b x^4\right )^{11/4}}{11 b^5}+\frac{\left (a+b x^4\right )^{15/4}}{15 b^5}\\ \end{align*}

Mathematica [A]  time = 0.0266705, size = 61, normalized size = 0.62 \[ \frac{192 a^2 b^2 x^8-512 a^3 b x^4-2048 a^4-112 a b^3 x^{12}+77 b^4 x^{16}}{1155 b^5 \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^19/(a + b*x^4)^(5/4),x]

[Out]

(-2048*a^4 - 512*a^3*b*x^4 + 192*a^2*b^2*x^8 - 112*a*b^3*x^12 + 77*b^4*x^16)/(1155*b^5*(a + b*x^4)^(1/4))

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Maple [A]  time = 0.007, size = 58, normalized size = 0.6 \begin{align*} -{\frac{-77\,{x}^{16}{b}^{4}+112\,a{x}^{12}{b}^{3}-192\,{a}^{2}{x}^{8}{b}^{2}+512\,{a}^{3}{x}^{4}b+2048\,{a}^{4}}{1155\,{b}^{5}}{\frac{1}{\sqrt [4]{b{x}^{4}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^19/(b*x^4+a)^(5/4),x)

[Out]

-1/1155*(-77*b^4*x^16+112*a*b^3*x^12-192*a^2*b^2*x^8+512*a^3*b*x^4+2048*a^4)/(b*x^4+a)^(1/4)/b^5

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Maxima [A]  time = 0.96882, size = 109, normalized size = 1.1 \begin{align*} \frac{{\left (b x^{4} + a\right )}^{\frac{15}{4}}}{15 \, b^{5}} - \frac{4 \,{\left (b x^{4} + a\right )}^{\frac{11}{4}} a}{11 \, b^{5}} + \frac{6 \,{\left (b x^{4} + a\right )}^{\frac{7}{4}} a^{2}}{7 \, b^{5}} - \frac{4 \,{\left (b x^{4} + a\right )}^{\frac{3}{4}} a^{3}}{3 \, b^{5}} - \frac{a^{4}}{{\left (b x^{4} + a\right )}^{\frac{1}{4}} b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^19/(b*x^4+a)^(5/4),x, algorithm="maxima")

[Out]

1/15*(b*x^4 + a)^(15/4)/b^5 - 4/11*(b*x^4 + a)^(11/4)*a/b^5 + 6/7*(b*x^4 + a)^(7/4)*a^2/b^5 - 4/3*(b*x^4 + a)^
(3/4)*a^3/b^5 - a^4/((b*x^4 + a)^(1/4)*b^5)

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Fricas [A]  time = 1.46559, size = 162, normalized size = 1.64 \begin{align*} \frac{{\left (77 \, b^{4} x^{16} - 112 \, a b^{3} x^{12} + 192 \, a^{2} b^{2} x^{8} - 512 \, a^{3} b x^{4} - 2048 \, a^{4}\right )}{\left (b x^{4} + a\right )}^{\frac{3}{4}}}{1155 \,{\left (b^{6} x^{4} + a b^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^19/(b*x^4+a)^(5/4),x, algorithm="fricas")

[Out]

1/1155*(77*b^4*x^16 - 112*a*b^3*x^12 + 192*a^2*b^2*x^8 - 512*a^3*b*x^4 - 2048*a^4)*(b*x^4 + a)^(3/4)/(b^6*x^4
+ a*b^5)

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Sympy [A]  time = 19.8584, size = 116, normalized size = 1.17 \begin{align*} \begin{cases} - \frac{2048 a^{4}}{1155 b^{5} \sqrt [4]{a + b x^{4}}} - \frac{512 a^{3} x^{4}}{1155 b^{4} \sqrt [4]{a + b x^{4}}} + \frac{64 a^{2} x^{8}}{385 b^{3} \sqrt [4]{a + b x^{4}}} - \frac{16 a x^{12}}{165 b^{2} \sqrt [4]{a + b x^{4}}} + \frac{x^{16}}{15 b \sqrt [4]{a + b x^{4}}} & \text{for}\: b \neq 0 \\\frac{x^{20}}{20 a^{\frac{5}{4}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**19/(b*x**4+a)**(5/4),x)

[Out]

Piecewise((-2048*a**4/(1155*b**5*(a + b*x**4)**(1/4)) - 512*a**3*x**4/(1155*b**4*(a + b*x**4)**(1/4)) + 64*a**
2*x**8/(385*b**3*(a + b*x**4)**(1/4)) - 16*a*x**12/(165*b**2*(a + b*x**4)**(1/4)) + x**16/(15*b*(a + b*x**4)**
(1/4)), Ne(b, 0)), (x**20/(20*a**(5/4)), True))

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Giac [A]  time = 1.09859, size = 96, normalized size = 0.97 \begin{align*} \frac{77 \,{\left (b x^{4} + a\right )}^{\frac{15}{4}} - 420 \,{\left (b x^{4} + a\right )}^{\frac{11}{4}} a + 990 \,{\left (b x^{4} + a\right )}^{\frac{7}{4}} a^{2} - 1540 \,{\left (b x^{4} + a\right )}^{\frac{3}{4}} a^{3} - \frac{1155 \, a^{4}}{{\left (b x^{4} + a\right )}^{\frac{1}{4}}}}{1155 \, b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^19/(b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

1/1155*(77*(b*x^4 + a)^(15/4) - 420*(b*x^4 + a)^(11/4)*a + 990*(b*x^4 + a)^(7/4)*a^2 - 1540*(b*x^4 + a)^(3/4)*
a^3 - 1155*a^4/(b*x^4 + a)^(1/4))/b^5