3.1009 \(\int x^{12} \sqrt [4]{a+b x^4} \, dx\)

Optimal. Leaf size=150 \[ \frac {3 a^{7/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{112 b^{5/2} \left (a+b x^4\right )^{3/4}}+\frac {3 a^3 x \sqrt [4]{a+b x^4}}{112 b^3}-\frac {3 a^2 x^5 \sqrt [4]{a+b x^4}}{280 b^2}+\frac {1}{14} x^{13} \sqrt [4]{a+b x^4}+\frac {a x^9 \sqrt [4]{a+b x^4}}{140 b} \]

[Out]

3/112*a^3*x*(b*x^4+a)^(1/4)/b^3-3/280*a^2*x^5*(b*x^4+a)^(1/4)/b^2+1/140*a*x^9*(b*x^4+a)^(1/4)/b+1/14*x^13*(b*x
^4+a)^(1/4)+3/112*a^(7/2)*(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 = 0.07, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {279, 321, 237, 335, 275, 231} \[ -\frac {3 a^2 x^5 \sqrt [4]{a+b x^4}}{280 b^2}+\frac {3 a^3 x \sqrt [4]{a+b x^4}}{112 b^3}+\frac {3 a^{7/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{112 b^{5/2} \left (a+b x^4\right )^{3/4}}+\frac {1}{14} x^{13} \sqrt [4]{a+b x^4}+\frac {a x^9 \sqrt [4]{a+b x^4}}{140 b} \]

Antiderivative was successfully verified.

[In]

Int[x^12*(a + b*x^4)^(1/4),x]

[Out]

(3*a^3*x*(a + b*x^4)^(1/4))/(112*b^3) - (3*a^2*x^5*(a + b*x^4)^(1/4))/(280*b^2) + (a*x^9*(a + b*x^4)^(1/4))/(1
40*b) + (x^13*(a + b*x^4)^(1/4))/14 + (3*a^(7/2)*(1 + a/(b*x^4))^(3/4)*x^3*EllipticF[ArcCot[(Sqrt[b]*x^2)/Sqrt
[a]]/2, 2])/(112*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 279

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

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 x^{12} \sqrt [4]{a+b x^4} \, dx &=\frac {1}{14} x^{13} \sqrt [4]{a+b x^4}+\frac {1}{14} a \int \frac {x^{12}}{\left (a+b x^4\right )^{3/4}} \, dx\\ &=\frac {a x^9 \sqrt [4]{a+b x^4}}{140 b}+\frac {1}{14} x^{13} \sqrt [4]{a+b x^4}-\frac {\left (9 a^2\right ) \int \frac {x^8}{\left (a+b x^4\right )^{3/4}} \, dx}{140 b}\\ &=-\frac {3 a^2 x^5 \sqrt [4]{a+b x^4}}{280 b^2}+\frac {a x^9 \sqrt [4]{a+b x^4}}{140 b}+\frac {1}{14} x^{13} \sqrt [4]{a+b x^4}+\frac {\left (3 a^3\right ) \int \frac {x^4}{\left (a+b x^4\right )^{3/4}} \, dx}{56 b^2}\\ &=\frac {3 a^3 x \sqrt [4]{a+b x^4}}{112 b^3}-\frac {3 a^2 x^5 \sqrt [4]{a+b x^4}}{280 b^2}+\frac {a x^9 \sqrt [4]{a+b x^4}}{140 b}+\frac {1}{14} x^{13} \sqrt [4]{a+b x^4}-\frac {\left (3 a^4\right ) \int \frac {1}{\left (a+b x^4\right )^{3/4}} \, dx}{112 b^3}\\ &=\frac {3 a^3 x \sqrt [4]{a+b x^4}}{112 b^3}-\frac {3 a^2 x^5 \sqrt [4]{a+b x^4}}{280 b^2}+\frac {a x^9 \sqrt [4]{a+b x^4}}{140 b}+\frac {1}{14} x^{13} \sqrt [4]{a+b x^4}-\frac {\left (3 a^4 \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}{112 b^3 \left (a+b x^4\right )^{3/4}}\\ &=\frac {3 a^3 x \sqrt [4]{a+b x^4}}{112 b^3}-\frac {3 a^2 x^5 \sqrt [4]{a+b x^4}}{280 b^2}+\frac {a x^9 \sqrt [4]{a+b x^4}}{140 b}+\frac {1}{14} x^{13} \sqrt [4]{a+b x^4}+\frac {\left (3 a^4 \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 )}{112 b^3 \left (a+b x^4\right )^{3/4}}\\ &=\frac {3 a^3 x \sqrt [4]{a+b x^4}}{112 b^3}-\frac {3 a^2 x^5 \sqrt [4]{a+b x^4}}{280 b^2}+\frac {a x^9 \sqrt [4]{a+b x^4}}{140 b}+\frac {1}{14} x^{13} \sqrt [4]{a+b x^4}+\frac {\left (3 a^4 \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 )}{224 b^3 \left (a+b x^4\right )^{3/4}}\\ &=\frac {3 a^3 x \sqrt [4]{a+b x^4}}{112 b^3}-\frac {3 a^2 x^5 \sqrt [4]{a+b x^4}}{280 b^2}+\frac {a x^9 \sqrt [4]{a+b x^4}}{140 b}+\frac {1}{14} x^{13} \sqrt [4]{a+b x^4}+\frac {3 a^{7/2} \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 )}{112 b^{5/2} \left (a+b x^4\right )^{3/4}}\\ \end {align*}

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Mathematica [C]  time = 0.08, size = 105, normalized size = 0.70 \[ \frac {x \sqrt [4]{a+b x^4} \left (\sqrt [4]{\frac {b x^4}{a}+1} \left (15 a^3-3 a^2 b x^4+2 a b^2 x^8+20 b^3 x^{12}\right )-15 a^3 \, _2F_1\left (-\frac {1}{4},\frac {1}{4};\frac {5}{4};-\frac {b x^4}{a}\right )\right )}{280 b^3 \sqrt [4]{\frac {b x^4}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^12*(a + b*x^4)^(1/4),x]

[Out]

(x*(a + b*x^4)^(1/4)*((1 + (b*x^4)/a)^(1/4)*(15*a^3 - 3*a^2*b*x^4 + 2*a*b^2*x^8 + 20*b^3*x^12) - 15*a^3*Hyperg
eometric2F1[-1/4, 1/4, 5/4, -((b*x^4)/a)]))/(280*b^3*(1 + (b*x^4)/a)^(1/4))

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fricas [F]  time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} x^{12}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^12*(b*x^4+a)^(1/4),x, algorithm="fricas")

[Out]

integral((b*x^4 + a)^(1/4)*x^12, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{4} + a\right )}^{\frac {1}{4}} x^{12}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^12*(b*x^4+a)^(1/4),x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(1/4)*x^12, x)

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \left (b \,x^{4}+a \right )^{\frac {1}{4}} x^{12}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^12*(b*x^4+a)^(1/4),x)

[Out]

int(x^12*(b*x^4+a)^(1/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{4} + a\right )}^{\frac {1}{4}} x^{12}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^12*(b*x^4+a)^(1/4),x, algorithm="maxima")

[Out]

integrate((b*x^4 + a)^(1/4)*x^12, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^12*(a + b*x^4)^(1/4),x)

[Out]

int(x^12*(a + b*x^4)^(1/4), x)

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sympy [C]  time = 2.91, size = 39, normalized size = 0.26 \[ \frac {\sqrt [4]{a} x^{13} \Gamma \left (\frac {13}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {13}{4} \\ \frac {17}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {17}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**12*(b*x**4+a)**(1/4),x)

[Out]

a**(1/4)*x**13*gamma(13/4)*hyper((-1/4, 13/4), (17/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(17/4))

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