3.1132 \(\int \frac {x^{12}}{(a+b x^4)^{3/4}} \, dx\)

Optimal. Leaf size=129 \[ \frac {3 a^{5/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{8 b^{5/2} \left (a+b x^4\right )^{3/4}}+\frac {3 a^2 x \sqrt [4]{a+b x^4}}{8 b^3}-\frac {3 a x^5 \sqrt [4]{a+b x^4}}{20 b^2}+\frac {x^9 \sqrt [4]{a+b x^4}}{10 b} \]

[Out]

3/8*a^2*x*(b*x^4+a)^(1/4)/b^3-3/20*a*x^5*(b*x^4+a)^(1/4)/b^2+1/10*x^9*(b*x^4+a)^(1/4)/b+3/8*a^(5/2)*(1+a/b/x^4
)^(3/4)*x^3*(cos(1/2*arccot(x^2*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arccot(x^2*b^(1/2)/a^(1/2)))*EllipticF(sin(
1/2*arccot(x^2*b^(1/2)/a^(1/2))),2^(1/2))/b^(5/2)/(b*x^4+a)^(3/4)

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Rubi [A]  time = 0.06, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {321, 237, 335, 275, 231} \[ \frac {3 a^2 x \sqrt [4]{a+b x^4}}{8 b^3}+\frac {3 a^{5/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{8 b^{5/2} \left (a+b x^4\right )^{3/4}}-\frac {3 a x^5 \sqrt [4]{a+b x^4}}{20 b^2}+\frac {x^9 \sqrt [4]{a+b x^4}}{10 b} \]

Antiderivative was successfully verified.

[In]

Int[x^12/(a + b*x^4)^(3/4),x]

[Out]

(3*a^2*x*(a + b*x^4)^(1/4))/(8*b^3) - (3*a*x^5*(a + b*x^4)^(1/4))/(20*b^2) + (x^9*(a + b*x^4)^(1/4))/(10*b) +
(3*a^(5/2)*(1 + a/(b*x^4))^(3/4)*x^3*EllipticF[ArcCot[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(8*b^(5/2)*(a + b*x^4)^(3/
4))

Rule 231

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(3/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 237

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

Rule 275

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 321

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]

Rule 335

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^{12}}{\left (a+b x^4\right )^{3/4}} \, dx &=\frac {x^9 \sqrt [4]{a+b x^4}}{10 b}-\frac {(9 a) \int \frac {x^8}{\left (a+b x^4\right )^{3/4}} \, dx}{10 b}\\ &=-\frac {3 a x^5 \sqrt [4]{a+b x^4}}{20 b^2}+\frac {x^9 \sqrt [4]{a+b x^4}}{10 b}+\frac {\left (3 a^2\right ) \int \frac {x^4}{\left (a+b x^4\right )^{3/4}} \, dx}{4 b^2}\\ &=\frac {3 a^2 x \sqrt [4]{a+b x^4}}{8 b^3}-\frac {3 a x^5 \sqrt [4]{a+b x^4}}{20 b^2}+\frac {x^9 \sqrt [4]{a+b x^4}}{10 b}-\frac {\left (3 a^3\right ) \int \frac {1}{\left (a+b x^4\right )^{3/4}} \, dx}{8 b^3}\\ &=\frac {3 a^2 x \sqrt [4]{a+b x^4}}{8 b^3}-\frac {3 a x^5 \sqrt [4]{a+b x^4}}{20 b^2}+\frac {x^9 \sqrt [4]{a+b x^4}}{10 b}-\frac {\left (3 a^3 \left (1+\frac {a}{b x^4}\right )^{3/4} x^3\right ) \int \frac {1}{\left (1+\frac {a}{b x^4}\right )^{3/4} x^3} \, dx}{8 b^3 \left (a+b x^4\right )^{3/4}}\\ &=\frac {3 a^2 x \sqrt [4]{a+b x^4}}{8 b^3}-\frac {3 a x^5 \sqrt [4]{a+b x^4}}{20 b^2}+\frac {x^9 \sqrt [4]{a+b x^4}}{10 b}+\frac {\left (3 a^3 \left (1+\frac {a}{b x^4}\right )^{3/4} x^3\right ) \operatorname {Subst}\left (\int \frac {x}{\left (1+\frac {a x^4}{b}\right )^{3/4}} \, dx,x,\frac {1}{x}\right )}{8 b^3 \left (a+b x^4\right )^{3/4}}\\ &=\frac {3 a^2 x \sqrt [4]{a+b x^4}}{8 b^3}-\frac {3 a x^5 \sqrt [4]{a+b x^4}}{20 b^2}+\frac {x^9 \sqrt [4]{a+b x^4}}{10 b}+\frac {\left (3 a^3 \left (1+\frac {a}{b x^4}\right )^{3/4} x^3\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+\frac {a x^2}{b}\right )^{3/4}} \, dx,x,\frac {1}{x^2}\right )}{16 b^3 \left (a+b x^4\right )^{3/4}}\\ &=\frac {3 a^2 x \sqrt [4]{a+b x^4}}{8 b^3}-\frac {3 a x^5 \sqrt [4]{a+b x^4}}{20 b^2}+\frac {x^9 \sqrt [4]{a+b x^4}}{10 b}+\frac {3 a^{5/2} \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{8 b^{5/2} \left (a+b x^4\right )^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 90, normalized size = 0.70 \[ \frac {-15 a^3 x \left (\frac {b x^4}{a}+1\right )^{3/4} \, _2F_1\left (\frac {1}{4},\frac {3}{4};\frac {5}{4};-\frac {b x^4}{a}\right )+15 a^3 x+9 a^2 b x^5-2 a b^2 x^9+4 b^3 x^{13}}{40 b^3 \left (a+b x^4\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^12/(a + b*x^4)^(3/4),x]

[Out]

(15*a^3*x + 9*a^2*b*x^5 - 2*a*b^2*x^9 + 4*b^3*x^13 - 15*a^3*x*(1 + (b*x^4)/a)^(3/4)*Hypergeometric2F1[1/4, 3/4
, 5/4, -((b*x^4)/a)])/(40*b^3*(a + b*x^4)^(3/4))

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fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{12}}{{\left (b x^{4} + a\right )}^{\frac {3}{4}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^12/(b*x^4 + a)^(3/4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{12}}{{\left (b x^{4} + a\right )}^{\frac {3}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^12/(b*x^4 + a)^(3/4), x)

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \frac {x^{12}}{\left (b \,x^{4}+a \right )^{\frac {3}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^12/(b*x^4+a)^(3/4),x)

[Out]

int(x^12/(b*x^4+a)^(3/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{12}}{{\left (b x^{4} + a\right )}^{\frac {3}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^12/(b*x^4 + a)^(3/4), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^12/(a + b*x^4)^(3/4),x)

[Out]

int(x^12/(a + b*x^4)^(3/4), x)

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sympy [C]  time = 2.14, size = 37, normalized size = 0.29 \[ \frac {x^{13} \Gamma \left (\frac {13}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {13}{4} \\ \frac {17}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{4}} \Gamma \left (\frac {17}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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