3.12.12 \(\int \frac {x^{11}}{(a+b x^4)^{3/4}} \, dx\) [1112]

Optimal. Leaf size=56 \[ \frac {a^2 \sqrt [4]{a+b x^4}}{b^3}-\frac {2 a \left (a+b x^4\right )^{5/4}}{5 b^3}+\frac {\left (a+b x^4\right )^{9/4}}{9 b^3} \]

[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 = 56, 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 = {272, 45} \begin {gather*} \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} \end {gather*}

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 45

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 272

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} \text {Subst}\left (\int \frac {x^2}{(a+b x)^{3/4}} \, dx,x,x^4\right )\\ &=\frac {1}{4} \text {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 = 0.03, size = 39, normalized size = 0.70 \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^11/(a + b*x^4)^(3/4),x]

[Out]

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

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Maple [A]
time = 0.18, size = 36, normalized size = 0.64

method	result	size
gosper	\(\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (5 b^{2} x^{8}-8 a b \,x^{4}+32 a^{2}\right )}{45 b^{3}}\)	\(36\)
trager	\(\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (5 b^{2} x^{8}-8 a b \,x^{4}+32 a^{2}\right )}{45 b^{3}}\)	\(36\)
risch	\(\frac {\left (b \,x^{4}+a \right )^{\frac {1}{4}} \left (5 b^{2} x^{8}-8 a b \,x^{4}+32 a^{2}\right )}{45 b^{3}}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^11/(b*x^4+a)^(3/4),x,method=_RETURNVERBOSE)

[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 = 0.29, size = 46, normalized size = 0.82 \begin {gather*} \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}} \end {gather*}

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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Fricas [A]
time = 0.37, size = 35, normalized size = 0.62 \begin {gather*} \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}} \end {gather*}

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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Sympy [A]
time = 0.48, size = 68, normalized size = 1.21 \begin {gather*} \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} \end {gather*}

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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Giac [A]
time = 0.86, size = 46, normalized size = 0.82 \begin {gather*} \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{2}}{b^{3}} + \frac {5 \, {\left (b x^{4} + a\right )}^{\frac {9}{4}} - 18 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} a}{45 \, b^{3}} \end {gather*}

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)^(9/4) - 18*(b*x^4 + a)^(5/4)*a)/b^3

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Mupad [B]
time = 1.14, size = 36, normalized size = 0.64 \begin {gather*} {\left (b\,x^4+a\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 ) \end {gather*}

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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