3.1088 \(\int \frac {x^{13}}{\sqrt [4]{a+b x^4}} \, dx\)

Optimal. Leaf size=152 \[ \frac {8 a^{7/2} \sqrt [4]{\frac {b x^4}{a}+1} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{39 b^{7/2} \sqrt [4]{a+b x^4}}-\frac {8 a^3 x^2}{39 b^3 \sqrt [4]{a+b x^4}}+\frac {4 a^2 x^2 \left (a+b x^4\right )^{3/4}}{39 b^3}-\frac {10 a x^6 \left (a+b x^4\right )^{3/4}}{117 b^2}+\frac {x^{10} \left (a+b x^4\right )^{3/4}}{13 b} \]

[Out]

-8/39*a^3*x^2/b^3/(b*x^4+a)^(1/4)+4/39*a^2*x^2*(b*x^4+a)^(3/4)/b^3-10/117*a*x^6*(b*x^4+a)^(3/4)/b^2+1/13*x^10*
(b*x^4+a)^(3/4)/b+8/39*a^(7/2)*(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^(7/2)/(b*x^4+a)^(1/4)

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Rubi [A]  time = 0.10, antiderivative size = 152, 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 = {275, 321, 229, 227, 196} \[ \frac {4 a^2 x^2 \left (a+b x^4\right )^{3/4}}{39 b^3}-\frac {8 a^3 x^2}{39 b^3 \sqrt [4]{a+b x^4}}+\frac {8 a^{7/2} \sqrt [4]{\frac {b x^4}{a}+1} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{39 b^{7/2} \sqrt [4]{a+b x^4}}-\frac {10 a x^6 \left (a+b x^4\right )^{3/4}}{117 b^2}+\frac {x^{10} \left (a+b x^4\right )^{3/4}}{13 b} \]

Antiderivative was successfully verified.

[In]

Int[x^13/(a + b*x^4)^(1/4),x]

[Out]

(-8*a^3*x^2)/(39*b^3*(a + b*x^4)^(1/4)) + (4*a^2*x^2*(a + b*x^4)^(3/4))/(39*b^3) - (10*a*x^6*(a + b*x^4)^(3/4)
)/(117*b^2) + (x^10*(a + b*x^4)^(3/4))/(13*b) + (8*a^(7/2)*(1 + (b*x^4)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x^2
)/Sqrt[a]]/2, 2])/(39*b^(7/2)*(a + b*x^4)^(1/4))

Rule 196

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

Rule 227

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

Rule 229

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

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]

Rubi steps

\begin {align*} \int \frac {x^{13}}{\sqrt [4]{a+b x^4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^6}{\sqrt [4]{a+b x^2}} \, dx,x,x^2\right )\\ &=\frac {x^{10} \left (a+b x^4\right )^{3/4}}{13 b}-\frac {(5 a) \operatorname {Subst}\left (\int \frac {x^4}{\sqrt [4]{a+b x^2}} \, dx,x,x^2\right )}{13 b}\\ &=-\frac {10 a x^6 \left (a+b x^4\right )^{3/4}}{117 b^2}+\frac {x^{10} \left (a+b x^4\right )^{3/4}}{13 b}+\frac {\left (10 a^2\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt [4]{a+b x^2}} \, dx,x,x^2\right )}{39 b^2}\\ &=\frac {4 a^2 x^2 \left (a+b x^4\right )^{3/4}}{39 b^3}-\frac {10 a x^6 \left (a+b x^4\right )^{3/4}}{117 b^2}+\frac {x^{10} \left (a+b x^4\right )^{3/4}}{13 b}-\frac {\left (4 a^3\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{a+b x^2}} \, dx,x,x^2\right )}{39 b^3}\\ &=\frac {4 a^2 x^2 \left (a+b x^4\right )^{3/4}}{39 b^3}-\frac {10 a x^6 \left (a+b x^4\right )^{3/4}}{117 b^2}+\frac {x^{10} \left (a+b x^4\right )^{3/4}}{13 b}-\frac {\left (4 a^3 \sqrt [4]{1+\frac {b x^4}{a}}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{1+\frac {b x^2}{a}}} \, dx,x,x^2\right )}{39 b^3 \sqrt [4]{a+b x^4}}\\ &=-\frac {8 a^3 x^2}{39 b^3 \sqrt [4]{a+b x^4}}+\frac {4 a^2 x^2 \left (a+b x^4\right )^{3/4}}{39 b^3}-\frac {10 a x^6 \left (a+b x^4\right )^{3/4}}{117 b^2}+\frac {x^{10} \left (a+b x^4\right )^{3/4}}{13 b}+\frac {\left (4 a^3 \sqrt [4]{1+\frac {b x^4}{a}}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{5/4}} \, dx,x,x^2\right )}{39 b^3 \sqrt [4]{a+b x^4}}\\ &=-\frac {8 a^3 x^2}{39 b^3 \sqrt [4]{a+b x^4}}+\frac {4 a^2 x^2 \left (a+b x^4\right )^{3/4}}{39 b^3}-\frac {10 a x^6 \left (a+b x^4\right )^{3/4}}{117 b^2}+\frac {x^{10} \left (a+b x^4\right )^{3/4}}{13 b}+\frac {8 a^{7/2} \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 )}{39 b^{7/2} \sqrt [4]{a+b x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 91, normalized size = 0.60 \[ \frac {x^2 \left (-12 a^3 \sqrt [4]{\frac {b x^4}{a}+1} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};-\frac {b x^4}{a}\right )+12 a^3+2 a^2 b x^4-a b^2 x^8+9 b^3 x^{12}\right )}{117 b^3 \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^13/(a + b*x^4)^(1/4),x]

[Out]

(x^2*(12*a^3 + 2*a^2*b*x^4 - a*b^2*x^8 + 9*b^3*x^12 - 12*a^3*(1 + (b*x^4)/a)^(1/4)*Hypergeometric2F1[1/4, 1/2,
3/2, -((b*x^4)/a)]))/(117*b^3*(a + b*x^4)^(1/4))

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fricas [F]  time = 1.18, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{13}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^13/(b*x^4+a)^(1/4),x, algorithm="fricas")

[Out]

integral(x^13/(b*x^4 + a)^(1/4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{13}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^13/(b*x^4+a)^(1/4),x, algorithm="giac")

[Out]

integrate(x^13/(b*x^4 + a)^(1/4), x)

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \frac {x^{13}}{\left (b \,x^{4}+a \right )^{\frac {1}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^13/(b*x^4+a)^(1/4),x)

[Out]

int(x^13/(b*x^4+a)^(1/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{13}}{{\left (b x^{4} + a\right )}^{\frac {1}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^13/(b*x^4+a)^(1/4),x, algorithm="maxima")

[Out]

integrate(x^13/(b*x^4 + a)^(1/4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^{13}}{{\left (b\,x^4+a\right )}^{1/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^13/(a + b*x^4)^(1/4),x)

[Out]

int(x^13/(a + b*x^4)^(1/4), x)

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sympy [C]  time = 1.63, size = 27, normalized size = 0.18 \[ \frac {x^{14} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {7}{2} \\ \frac {9}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{14 \sqrt [4]{a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**13/(b*x**4+a)**(1/4),x)

[Out]

x**14*hyper((1/4, 7/2), (9/2,), b*x**4*exp_polar(I*pi)/a)/(14*a**(1/4))

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