3.9.61 \(\int \frac {x^{12}}{(a+b x^4)^{3/2}} \, dx\) [861]

Optimal. Leaf size=151 \[ -\frac {x^9}{2 b \sqrt {a+b x^4}}-\frac {15 a x \sqrt {a+b x^4}}{14 b^3}+\frac {9 x^5 \sqrt {a+b x^4}}{14 b^2}+\frac {15 a^{7/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{28 b^{13/4} \sqrt {a+b x^4}} \]

[Out]

-1/2*x^9/b/(b*x^4+a)^(1/2)-15/14*a*x*(b*x^4+a)^(1/2)/b^3+9/14*x^5*(b*x^4+a)^(1/2)/b^2+15/28*a^(7/4)*(cos(2*arc
tan(b^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x/a^(1/4)))*EllipticF(sin(2*arctan(b^(1/4)*x/a^(1/4))),1
/2*2^(1/2))*(a^(1/2)+x^2*b^(1/2))*((b*x^4+a)/(a^(1/2)+x^2*b^(1/2))^2)^(1/2)/b^(13/4)/(b*x^4+a)^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {294, 327, 226} \begin {gather*} \frac {15 a^{7/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{28 b^{13/4} \sqrt {a+b x^4}}-\frac {15 a x \sqrt {a+b x^4}}{14 b^3}+\frac {9 x^5 \sqrt {a+b x^4}}{14 b^2}-\frac {x^9}{2 b \sqrt {a+b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^12/(a + b*x^4)^(3/2),x]

[Out]

-1/2*x^9/(b*Sqrt[a + b*x^4]) - (15*a*x*Sqrt[a + b*x^4])/(14*b^3) + (9*x^5*Sqrt[a + b*x^4])/(14*b^2) + (15*a^(7
/4)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)
], 1/2])/(28*b^(13/4)*Sqrt[a + b*x^4])

Rule 226

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rubi steps

\begin {align*} \int \frac {x^{12}}{\left (a+b x^4\right )^{3/2}} \, dx &=-\frac {x^9}{2 b \sqrt {a+b x^4}}+\frac {9 \int \frac {x^8}{\sqrt {a+b x^4}} \, dx}{2 b}\\ &=-\frac {x^9}{2 b \sqrt {a+b x^4}}+\frac {9 x^5 \sqrt {a+b x^4}}{14 b^2}-\frac {(45 a) \int \frac {x^4}{\sqrt {a+b x^4}} \, dx}{14 b^2}\\ &=-\frac {x^9}{2 b \sqrt {a+b x^4}}-\frac {15 a x \sqrt {a+b x^4}}{14 b^3}+\frac {9 x^5 \sqrt {a+b x^4}}{14 b^2}+\frac {\left (15 a^2\right ) \int \frac {1}{\sqrt {a+b x^4}} \, dx}{14 b^3}\\ &=-\frac {x^9}{2 b \sqrt {a+b x^4}}-\frac {15 a x \sqrt {a+b x^4}}{14 b^3}+\frac {9 x^5 \sqrt {a+b x^4}}{14 b^2}+\frac {15 a^{7/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{28 b^{13/4} \sqrt {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 = 10.03, size = 79, normalized size = 0.52 \begin {gather*} \frac {-15 a^2 x-6 a b x^5+2 b^2 x^9+15 a^2 x \sqrt {1+\frac {b x^4}{a}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\frac {b x^4}{a}\right )}{14 b^3 \sqrt {a+b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^12/(a + b*x^4)^(3/2),x]

[Out]

(-15*a^2*x - 6*a*b*x^5 + 2*b^2*x^9 + 15*a^2*x*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^4)/a
)])/(14*b^3*Sqrt[a + b*x^4])

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Maple [C] Result contains complex when optimal does not.
time = 0.17, size = 133, normalized size = 0.88

method	result	size
default	\(-\frac {a^{2} x}{2 b^{3} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {x^{5} \sqrt {b \,x^{4}+a}}{7 b^{2}}-\frac {4 a x \sqrt {b \,x^{4}+a}}{7 b^{3}}+\frac {15 a^{2} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{14 b^{3} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(133\)
elliptic	\(-\frac {a^{2} x}{2 b^{3} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {x^{5} \sqrt {b \,x^{4}+a}}{7 b^{2}}-\frac {4 a x \sqrt {b \,x^{4}+a}}{7 b^{3}}+\frac {15 a^{2} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{14 b^{3} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(133\)
risch	\(-\frac {x \left (-b \,x^{4}+4 a \right ) \sqrt {b \,x^{4}+a}}{7 b^{3}}+\frac {a^{2} \left (11 b \left (-\frac {x}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 b \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+4 a \left (\frac {x}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \EllipticF \left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\right )}{7 b^{3}}\)	\(228\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^12/(b*x^4+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/2/b^3*a^2*x/((x^4+a/b)*b)^(1/2)+1/7*x^5*(b*x^4+a)^(1/2)/b^2-4/7*a*x*(b*x^4+a)^(1/2)/b^3+15/14*a^2/b^3/(I/a^
(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)*x^2)^(1/2)*(1+I/a^(1/2)*b^(1/2)*x^2)^(1/2)/(b*x^4+a)^(1/2)*EllipticF
(x*(I/a^(1/2)*b^(1/2))^(1/2),I)

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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^12/(b*x^4+a)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^12/(b*x^4 + a)^(3/2), x)

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Fricas [A]
time = 0.08, size = 88, normalized size = 0.58 \begin {gather*} \frac {15 \, {\left (a b x^{4} + a^{2}\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + {\left (2 \, b^{2} x^{9} - 6 \, a b x^{5} - 15 \, a^{2} x\right )} \sqrt {b x^{4} + a}}{14 \, {\left (b^{4} x^{4} + a b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^12/(b*x^4+a)^(3/2),x, algorithm="fricas")

[Out]

1/14*(15*(a*b*x^4 + a^2)*sqrt(b)*(-a/b)^(3/4)*elliptic_f(arcsin((-a/b)^(1/4)/x), -1) + (2*b^2*x^9 - 6*a*b*x^5
- 15*a^2*x)*sqrt(b*x^4 + a))/(b^4*x^4 + a*b^3)

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Sympy [C] Result contains complex when optimal does not.
time = 0.66, size = 37, normalized size = 0.25 \begin {gather*} \frac {x^{13} \Gamma \left (\frac {13}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {13}{4} \\ \frac {17}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {17}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**12/(b*x**4+a)**(3/2),x)

[Out]

x**13*gamma(13/4)*hyper((3/2, 13/4), (17/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(3/2)*gamma(17/4))

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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^12/(b*x^4+a)^(3/2),x, algorithm="giac")

[Out]

integrate(x^12/(b*x^4 + a)^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{12}}{{\left (b\,x^4+a\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^12/(a + b*x^4)^(3/2),x)

[Out]

int(x^12/(a + b*x^4)^(3/2), x)

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