3.1236 \(\int \frac {x^{11}}{(a-b x^4)^{3/4}} \, dx\)

Optimal. Leaf size=60 \[ -\frac {a^2 \sqrt [4]{a-b x^4}}{b^3}-\frac {\left (a-b x^4\right )^{9/4}}{9 b^3}+\frac {2 a \left (a-b x^4\right )^{5/4}}{5 b^3} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {266, 43} \[ -\frac {a^2 \sqrt [4]{a-b x^4}}{b^3}-\frac {\left (a-b x^4\right )^{9/4}}{9 b^3}+\frac {2 a \left (a-b x^4\right )^{5/4}}{5 b^3} \]

Antiderivative was successfully verified.

[In]

Int[x^11/(a - b*x^4)^(3/4),x]

[Out]

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

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^{11}}{\left (a-b x^4\right )^{3/4}} \, dx &=\frac {1}{4} \operatorname {Subst}\left (\int \frac {x^2}{(a-b x)^{3/4}} \, dx,x,x^4\right )\\ &=\frac {1}{4} \operatorname {Subst}\left (\int \left (\frac {a^2}{b^2 (a-b x)^{3/4}}-\frac {2 a \sqrt [4]{a-b x}}{b^2}+\frac {(a-b x)^{5/4}}{b^2}\right ) \, dx,x,x^4\right )\\ &=-\frac {a^2 \sqrt [4]{a-b x^4}}{b^3}+\frac {2 a \left (a-b x^4\right )^{5/4}}{5 b^3}-\frac {\left (a-b x^4\right )^{9/4}}{9 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 40, normalized size = 0.67 \[ -\frac {\sqrt [4]{a-b x^4} \left (32 a^2+8 a b x^4+5 b^2 x^8\right )}{45 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^11/(a - b*x^4)^(3/4),x]

[Out]

-1/45*((a - b*x^4)^(1/4)*(32*a^2 + 8*a*b*x^4 + 5*b^2*x^8))/b^3

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fricas [A]  time = 0.69, size = 36, normalized size = 0.60 \[ -\frac {{\left (5 \, b^{2} x^{8} + 8 \, a b x^{4} + 32 \, a^{2}\right )} {\left (-b x^{4} + a\right )}^{\frac {1}{4}}}{45 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(-b*x^4+a)^(3/4),x, algorithm="fricas")

[Out]

-1/45*(5*b^2*x^8 + 8*a*b*x^4 + 32*a^2)*(-b*x^4 + a)^(1/4)/b^3

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giac [A]  time = 0.16, size = 61, normalized size = 1.02 \[ -\frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}} a^{2}}{b^{3}} - \frac {5 \, {\left (b x^{4} - a\right )}^{2} {\left (-b x^{4} + a\right )}^{\frac {1}{4}} - 18 \, {\left (-b x^{4} + a\right )}^{\frac {5}{4}} a}{45 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(-b*x^4+a)^(3/4),x, algorithm="giac")

[Out]

-(-b*x^4 + a)^(1/4)*a^2/b^3 - 1/45*(5*(b*x^4 - a)^2*(-b*x^4 + a)^(1/4) - 18*(-b*x^4 + a)^(5/4)*a)/b^3

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maple [A]  time = 0.01, size = 37, normalized size = 0.62 \[ -\frac {\left (-b \,x^{4}+a \right )^{\frac {1}{4}} \left (5 b^{2} x^{8}+8 a b \,x^{4}+32 a^{2}\right )}{45 b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11/(-b*x^4+a)^(3/4),x)

[Out]

-1/45*(-b*x^4+a)^(1/4)*(5*b^2*x^8+8*a*b*x^4+32*a^2)/b^3

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maxima [A]  time = 1.06, size = 50, normalized size = 0.83 \[ -\frac {{\left (-b x^{4} + a\right )}^{\frac {9}{4}}}{9 \, b^{3}} + \frac {2 \, {\left (-b x^{4} + a\right )}^{\frac {5}{4}} a}{5 \, b^{3}} - \frac {{\left (-b x^{4} + a\right )}^{\frac {1}{4}} a^{2}}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^11/(-b*x^4+a)^(3/4),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.15, size = 38, normalized size = 0.63 \[ -{\left (a-b\,x^4\right )}^{1/4}\,\left (\frac {32\,a^2}{45\,b^3}+\frac {x^8}{9\,b}+\frac {8\,a\,x^4}{45\,b^2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11/(a - b*x^4)^(3/4),x)

[Out]

-(a - b*x^4)^(1/4)*((32*a^2)/(45*b^3) + x^8/(9*b) + (8*a*x^4)/(45*b^2))

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sympy [A]  time = 5.67, size = 70, normalized size = 1.17 \[ \begin {cases} - \frac {32 a^{2} \sqrt [4]{a - b x^{4}}}{45 b^{3}} - \frac {8 a x^{4} \sqrt [4]{a - b x^{4}}}{45 b^{2}} - \frac {x^{8} \sqrt [4]{a - b x^{4}}}{9 b} & \text {for}\: b \neq 0 \\\frac {x^{12}}{12 a^{\frac {3}{4}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**11/(-b*x**4+a)**(3/4),x)

[Out]

Piecewise((-32*a**2*(a - b*x**4)**(1/4)/(45*b**3) - 8*a*x**4*(a - b*x**4)**(1/4)/(45*b**2) - x**8*(a - b*x**4)
**(1/4)/(9*b), Ne(b, 0)), (x**12/(12*a**(3/4)), True))

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