3.1172 \(\int \frac {1}{(a+b x^4)^{11/4}} \, dx\)

Optimal. Leaf size=102 \[ -\frac {4 \sqrt {b} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{7 a^{5/2} \left (a+b x^4\right )^{3/4}}+\frac {2 x}{7 a^2 \left (a+b x^4\right )^{3/4}}+\frac {x}{7 a \left (a+b x^4\right )^{7/4}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used = {199, 237, 335, 275, 231} \[ \frac {2 x}{7 a^2 \left (a+b x^4\right )^{3/4}}-\frac {4 \sqrt {b} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{7 a^{5/2} \left (a+b x^4\right )^{3/4}}+\frac {x}{7 a \left (a+b x^4\right )^{7/4}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^(-11/4),x]

[Out]

x/(7*a*(a + b*x^4)^(7/4)) + (2*x)/(7*a^2*(a + b*x^4)^(3/4)) - (4*Sqrt[b]*(1 + a/(b*x^4))^(3/4)*x^3*EllipticF[A
rcCot[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(7*a^(5/2)*(a + b*x^4)^(3/4))

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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 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 {1}{\left (a+b x^4\right )^{11/4}} \, dx &=\frac {x}{7 a \left (a+b x^4\right )^{7/4}}+\frac {6 \int \frac {1}{\left (a+b x^4\right )^{7/4}} \, dx}{7 a}\\ &=\frac {x}{7 a \left (a+b x^4\right )^{7/4}}+\frac {2 x}{7 a^2 \left (a+b x^4\right )^{3/4}}+\frac {4 \int \frac {1}{\left (a+b x^4\right )^{3/4}} \, dx}{7 a^2}\\ &=\frac {x}{7 a \left (a+b x^4\right )^{7/4}}+\frac {2 x}{7 a^2 \left (a+b x^4\right )^{3/4}}+\frac {\left (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}{7 a^2 \left (a+b x^4\right )^{3/4}}\\ &=\frac {x}{7 a \left (a+b x^4\right )^{7/4}}+\frac {2 x}{7 a^2 \left (a+b x^4\right )^{3/4}}-\frac {\left (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 )}{7 a^2 \left (a+b x^4\right )^{3/4}}\\ &=\frac {x}{7 a \left (a+b x^4\right )^{7/4}}+\frac {2 x}{7 a^2 \left (a+b x^4\right )^{3/4}}-\frac {\left (2 \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 )}{7 a^2 \left (a+b x^4\right )^{3/4}}\\ &=\frac {x}{7 a \left (a+b x^4\right )^{7/4}}+\frac {2 x}{7 a^2 \left (a+b x^4\right )^{3/4}}-\frac {4 \sqrt {b} \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 )}{7 a^{5/2} \left (a+b x^4\right )^{3/4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.05, size = 72, normalized size = 0.71 \[ \frac {4 x \left (a+b x^4\right ) \left (\frac {b x^4}{a}+1\right )^{3/4} \, _2F_1\left (\frac {1}{4},\frac {3}{4};\frac {5}{4};-\frac {b x^4}{a}\right )+3 a x+2 b x^5}{7 a^2 \left (a+b x^4\right )^{7/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^(-11/4),x]

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.83, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{3} x^{12} + 3 \, a b^{2} x^{8} + 3 \, a^{2} b x^{4} + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^4+a)^(11/4),x, algorithm="fricas")

[Out]

integral((b*x^4 + a)^(1/4)/(b^3*x^12 + 3*a*b^2*x^8 + 3*a^2*b*x^4 + a^3), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{4} + a\right )}^{\frac {11}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^4+a)^(11/4),x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(-11/4), x)

________________________________________________________________________________________

maple [F]  time = 0.18, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {11}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^4+a)^(11/4),x)

[Out]

int(1/(b*x^4+a)^(11/4),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{4} + a\right )}^{\frac {11}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^4+a)^(11/4),x, algorithm="maxima")

[Out]

integrate((b*x^4 + a)^(-11/4), x)

________________________________________________________________________________________

mupad [B]  time = 1.13, size = 37, normalized size = 0.36 \[ \frac {x\,{\left (\frac {b\,x^4}{a}+1\right )}^{11/4}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},\frac {11}{4};\ \frac {5}{4};\ -\frac {b\,x^4}{a}\right )}{{\left (b\,x^4+a\right )}^{11/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*x^4)^(11/4),x)

[Out]

(x*((b*x^4)/a + 1)^(11/4)*hypergeom([1/4, 11/4], 5/4, -(b*x^4)/a))/(a + b*x^4)^(11/4)

________________________________________________________________________________________

sympy [C]  time = 3.53, size = 36, normalized size = 0.35 \[ \frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {11}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {11}{4}} \Gamma \left (\frac {5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**4+a)**(11/4),x)

[Out]

x*gamma(1/4)*hyper((1/4, 11/4), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(11/4)*gamma(5/4))

________________________________________________________________________________________