3.12.48 \(\int \frac {x^9}{(a+b x^4)^{5/4}} \, dx\) [1148]

Optimal. Leaf size=104 \[ -\frac {6 a x^2}{5 b^2 \sqrt [4]{a+b x^4}}+\frac {x^6}{5 b \sqrt [4]{a+b x^4}}+\frac {12 a^{3/2} \sqrt [4]{1+\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{5 b^{5/2} \sqrt [4]{a+b x^4}} \]

[Out]

-6/5*a*x^2/b^2/(b*x^4+a)^(1/4)+1/5*x^6/b/(b*x^4+a)^(1/4)+12/5*a^(3/2)*(1+b*x^4/a)^(1/4)*(cos(1/2*arctan(x^2*b^
(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arctan(x^2*b^(1/2)/a^(1/2)))*EllipticE(sin(1/2*arctan(x^2*b^(1/2)/a^(1/2))),2
^(1/2))/b^(5/2)/(b*x^4+a)^(1/4)

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Rubi [A]
time = 0.05, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {281, 291, 203, 202} \begin {gather*} \frac {12 a^{3/2} \sqrt [4]{\frac {b x^4}{a}+1} E\left (\left .\frac {1}{2} \text {ArcTan}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{5 b^{5/2} \sqrt [4]{a+b x^4}}-\frac {6 a x^2}{5 b^2 \sqrt [4]{a+b x^4}}+\frac {x^6}{5 b \sqrt [4]{a+b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^9/(a + b*x^4)^(5/4),x]

[Out]

(-6*a*x^2)/(5*b^2*(a + b*x^4)^(1/4)) + x^6/(5*b*(a + b*x^4)^(1/4)) + (12*a^(3/2)*(1 + (b*x^4)/a)^(1/4)*Ellipti
cE[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(5*b^(5/2)*(a + b*x^4)^(1/4))

Rule 202

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]))*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Dist[(1 + b*(x^2/a))^(1/4)/(a*(a + b*x^2)^(1/4)), Int[1/(1 + b*
(x^2/a))^(5/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a] && PosQ[b/a]

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 291

Int[((c_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Simp[2*c*((c*x)^(m - 1)/(b*(2*m - 3)*(a + b*x^
2)^(1/4))), x] - Dist[2*a*c^2*((m - 1)/(b*(2*m - 3))), Int[(c*x)^(m - 2)/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a
, b, c}, x] && PosQ[b/a] && IntegerQ[2*m] && GtQ[m, 3/2]

Rubi steps

\begin {align*} \int \frac {x^9}{\left (a+b x^4\right )^{5/4}} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {x^4}{\left (a+b x^2\right )^{5/4}} \, dx,x,x^2\right )\\ &=\frac {x^6}{5 b \sqrt [4]{a+b x^4}}-\frac {(3 a) \text {Subst}\left (\int \frac {x^2}{\left (a+b x^2\right )^{5/4}} \, dx,x,x^2\right )}{5 b}\\ &=-\frac {6 a x^2}{5 b^2 \sqrt [4]{a+b x^4}}+\frac {x^6}{5 b \sqrt [4]{a+b x^4}}+\frac {\left (6 a^2\right ) \text {Subst}\left (\int \frac {1}{\left (a+b x^2\right )^{5/4}} \, dx,x,x^2\right )}{5 b^2}\\ &=-\frac {6 a x^2}{5 b^2 \sqrt [4]{a+b x^4}}+\frac {x^6}{5 b \sqrt [4]{a+b x^4}}+\frac {\left (6 a \sqrt [4]{1+\frac {b x^4}{a}}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{5/4}} \, dx,x,x^2\right )}{5 b^2 \sqrt [4]{a+b x^4}}\\ &=-\frac {6 a x^2}{5 b^2 \sqrt [4]{a+b x^4}}+\frac {x^6}{5 b \sqrt [4]{a+b x^4}}+\frac {12 a^{3/2} \sqrt [4]{1+\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{5 b^{5/2} \sqrt [4]{a+b x^4}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 8.55, size = 66, normalized size = 0.63 \begin {gather*} \frac {x^2 \left (6 a+b x^4-6 a \sqrt [4]{1+\frac {b x^4}{a}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};-\frac {b x^4}{a}\right )\right )}{5 b^2 \sqrt [4]{a+b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^9/(a + b*x^4)^(5/4),x]

[Out]

(x^2*(6*a + b*x^4 - 6*a*(1 + (b*x^4)/a)^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, -((b*x^4)/a)]))/(5*b^2*(a + b*x
^4)^(1/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {x^{9}}{\left (b \,x^{4}+a \right )^{\frac {5}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(b*x^4+a)^(5/4),x)

[Out]

int(x^9/(b*x^4+a)^(5/4),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^4+a)^(5/4),x, algorithm="maxima")

[Out]

integrate(x^9/(b*x^4 + a)^(5/4), x)

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Fricas [F]
time = 0.08, size = 35, normalized size = 0.34 \begin {gather*} {\rm integral}\left (\frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{9}}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^4+a)^(5/4),x, algorithm="fricas")

[Out]

integral((b*x^4 + a)^(3/4)*x^9/(b^2*x^8 + 2*a*b*x^4 + a^2), x)

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Sympy [C] Result contains complex when optimal does not.
time = 0.53, size = 27, normalized size = 0.26 \begin {gather*} \frac {x^{10} {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{10 a^{\frac {5}{4}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9/(b*x**4+a)**(5/4),x)

[Out]

x**10*hyper((5/4, 5/2), (7/2,), b*x**4*exp_polar(I*pi)/a)/(10*a**(5/4))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

integrate(x^9/(b*x^4 + a)^(5/4), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^9}{{\left (b\,x^4+a\right )}^{5/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(a + b*x^4)^(5/4),x)

[Out]

int(x^9/(a + b*x^4)^(5/4), x)

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