3.12.64 \(\int \frac {x^6}{(a+b x^4)^{5/4}} \, dx\) [1164]

Optimal. Leaf size=83 \[ \frac {x^3}{2 b \sqrt [4]{a+b x^4}}+\frac {3 \sqrt {a} \sqrt [4]{1+\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{2 b^{3/2} \sqrt [4]{a+b x^4}} \]

[Out]

1/2*x^3/b/(b*x^4+a)^(1/4)+3/2*(1+a/b/x^4)^(1/4)*x*(cos(1/2*arccot(x^2*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arcco
t(x^2*b^(1/2)/a^(1/2)))*EllipticE(sin(1/2*arccot(x^2*b^(1/2)/a^(1/2))),2^(1/2))*a^(1/2)/b^(3/2)/(b*x^4+a)^(1/4
)

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Rubi [A]
time = 0.03, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {288, 287, 342, 281, 202} \begin {gather*} \frac {3 \sqrt {a} x \sqrt [4]{\frac {a}{b x^4}+1} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{2 b^{3/2} \sqrt [4]{a+b x^4}}+\frac {x^3}{2 b \sqrt [4]{a+b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^6/(a + b*x^4)^(5/4),x]

[Out]

x^3/(2*b*(a + b*x^4)^(1/4)) + (3*Sqrt[a]*(1 + a/(b*x^4))^(1/4)*x*EllipticE[ArcCot[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2]
)/(2*b^(3/2)*(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 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 287

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

Rule 288

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Simp[x^(m - 3)/(b*(m - 4)*(a + b*x^4)^(1/4)), x] - Dis
t[a*((m - 3)/(b*(m - 4))), Int[x^(m - 4)/(a + b*x^4)^(5/4), x], x] /; FreeQ[{a, b}, x] && PosQ[b/a] && IGtQ[(m
- 2)/4, 0]

Rule 342

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {x^6}{\left (a+b x^4\right )^{5/4}} \, dx &=\frac {x^3}{2 b \sqrt [4]{a+b x^4}}-\frac {(3 a) \int \frac {x^2}{\left (a+b x^4\right )^{5/4}} \, dx}{2 b}\\ &=\frac {x^3}{2 b \sqrt [4]{a+b x^4}}-\frac {\left (3 a \sqrt [4]{1+\frac {a}{b x^4}} x\right ) \int \frac {1}{\left (1+\frac {a}{b x^4}\right )^{5/4} x^3} \, dx}{2 b^2 \sqrt [4]{a+b x^4}}\\ &=\frac {x^3}{2 b \sqrt [4]{a+b x^4}}+\frac {\left (3 a \sqrt [4]{1+\frac {a}{b x^4}} x\right ) \text {Subst}\left (\int \frac {x}{\left (1+\frac {a x^4}{b}\right )^{5/4}} \, dx,x,\frac {1}{x}\right )}{2 b^2 \sqrt [4]{a+b x^4}}\\ &=\frac {x^3}{2 b \sqrt [4]{a+b x^4}}+\frac {\left (3 a \sqrt [4]{1+\frac {a}{b x^4}} x\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {a x^2}{b}\right )^{5/4}} \, dx,x,\frac {1}{x^2}\right )}{4 b^2 \sqrt [4]{a+b x^4}}\\ &=\frac {x^3}{2 b \sqrt [4]{a+b x^4}}+\frac {3 \sqrt {a} \sqrt [4]{1+\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{2 b^{3/2} \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 = 7.66, size = 60, normalized size = 0.72 \begin {gather*} \frac {x^3-x^3 \sqrt [4]{1+\frac {b x^4}{a}} \, _2F_1\left (\frac {3}{4},\frac {5}{4};\frac {7}{4};-\frac {b x^4}{a}\right )}{2 b \sqrt [4]{a+b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^6/(a + b*x^4)^(5/4),x]

[Out]

(x^3 - x^3*(1 + (b*x^4)/a)^(1/4)*Hypergeometric2F1[3/4, 5/4, 7/4, -((b*x^4)/a)])/(2*b*(a + b*x^4)^(1/4))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {x^{6}}{\left (b \,x^{4}+a \right )^{\frac {5}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(b*x^4+a)^(5/4),x)

[Out]

int(x^6/(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^6/(b*x^4+a)^(5/4),x, algorithm="maxima")

[Out]

integrate(x^6/(b*x^4 + a)^(5/4), x)

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Fricas [F]
time = 0.07, size = 35, normalized size = 0.42 \begin {gather*} {\rm integral}\left (\frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{6}}{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^6/(b*x^4+a)^(5/4),x, algorithm="fricas")

[Out]

integral((b*x^4 + a)^(3/4)*x^6/(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 = 0.44, size = 37, normalized size = 0.45 \begin {gather*} \frac {x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{4}} \Gamma \left (\frac {11}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(b*x**4+a)**(5/4),x)

[Out]

x**7*gamma(7/4)*hyper((5/4, 7/4), (11/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(5/4)*gamma(11/4))

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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^6/(b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

integrate(x^6/(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^6}{{\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^6/(a + b*x^4)^(5/4),x)

[Out]

int(x^6/(a + b*x^4)^(5/4), x)

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