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

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Rubi [A]  time = 0.04, antiderivative size = 74, normalized size of antiderivative = 1.00, 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 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])

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]]

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*}

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Mathematica [A]  time = 0.02, size = 50, normalized size = 0.68 \[ \frac {128 a^3+32 a^2 b x^4-12 a b^2 x^8+7 b^3 x^{12}}{77 b^4 \sqrt [4]{a+b x^4}} \]

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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fricas [A]  time = 1.04, size = 58, normalized size = 0.78 \[ \frac {{\left (7 \, b^{3} x^{12} - 12 \, a b^{2} x^{8} + 32 \, a^{2} b x^{4} + 128 \, a^{3}\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{77 \, {\left (b^{5} x^{4} + a b^{4}\right )}} \]

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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giac [A]  time = 0.16, size = 69, normalized size = 0.93 \[ \frac {a^{3}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{4}} + \frac {7 \, {\left (b x^{4} + a\right )}^{\frac {11}{4}} b^{40} - 33 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} a b^{40} + 77 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{2} b^{40}}{77 \, b^{44}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^3/((b*x^4 + a)^(1/4)*b^4) + 1/77*(7*(b*x^4 + a)^(11/4)*b^40 - 33*(b*x^4 + a)^(7/4)*a*b^40 + 77*(b*x^4 + a)^(
3/4)*a^2*b^40)/b^44

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maple [A]  time = 0.01, size = 47, normalized size = 0.64 \[ \frac {7 b^{3} x^{12}-12 a \,b^{2} x^{8}+32 a^{2} b \,x^{4}+128 a^{3}}{77 \left (b \,x^{4}+a \right )^{\frac {1}{4}} b^{4}} \]

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 = 1.36, size = 62, normalized size = 0.84 \[ \frac {{\left (b x^{4} + a\right )}^{\frac {11}{4}}}{11 \, b^{4}} - \frac {3 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} a}{7 \, b^{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{2}}{b^{4}} + \frac {a^{3}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}} b^{4}} \]

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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mupad [B]  time = 1.23, size = 55, normalized size = 0.74 \[ \frac {77\,a^2\,\left (b\,x^4+a\right )-33\,a\,{\left (b\,x^4+a\right )}^2+7\,{\left (b\,x^4+a\right )}^3+77\,a^3}{77\,b^4\,{\left (b\,x^4+a\right )}^{1/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^15/(a + b*x^4)^(5/4),x)

[Out]

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

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sympy [A]  time = 12.75, size = 92, normalized size = 1.24 \[ \begin {cases} \frac {128 a^{3}}{77 b^{4} \sqrt [4]{a + b x^{4}}} + \frac {32 a^{2} x^{4}}{77 b^{3} \sqrt [4]{a + b x^{4}}} - \frac {12 a x^{8}}{77 b^{2} \sqrt [4]{a + b x^{4}}} + \frac {x^{12}}{11 b \sqrt [4]{a + b x^{4}}} & \text {for}\: b \neq 0 \\\frac {x^{16}}{16 a^{\frac {5}{4}}} & \text {otherwise} \end {cases} \]

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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