3.1169 \(\int \frac {1}{x^{14} (a+b x^4)^{5/4}} \, dx\)

Optimal. Leaf size=153 \[ -\frac {112 b^{7/2} x \sqrt [4]{\frac {a}{b x^4}+1} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{39 a^{9/2} \sqrt [4]{a+b x^4}}+\frac {56 b^3}{39 a^4 x \sqrt [4]{a+b x^4}}-\frac {28 b^2}{117 a^3 x^5 \sqrt [4]{a+b x^4}}+\frac {14 b}{117 a^2 x^9 \sqrt [4]{a+b x^4}}-\frac {1}{13 a x^{13} \sqrt [4]{a+b x^4}} \]

[Out]

-1/13/a/x^13/(b*x^4+a)^(1/4)+14/117*b/a^2/x^9/(b*x^4+a)^(1/4)-28/117*b^2/a^3/x^5/(b*x^4+a)^(1/4)+56/39*b^3/a^4
/x/(b*x^4+a)^(1/4)-112/39*b^(7/2)*(1+a/b/x^4)^(1/4)*x*(cos(1/2*arccot(x^2*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*a
rccot(x^2*b^(1/2)/a^(1/2)))*EllipticE(sin(1/2*arccot(x^2*b^(1/2)/a^(1/2))),2^(1/2))/a^(9/2)/(b*x^4+a)^(1/4)

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Rubi [A]  time = 0.08, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {283, 281, 335, 275, 196} \[ \frac {56 b^3}{39 a^4 x \sqrt [4]{a+b x^4}}-\frac {28 b^2}{117 a^3 x^5 \sqrt [4]{a+b x^4}}-\frac {112 b^{7/2} x \sqrt [4]{\frac {a}{b x^4}+1} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{39 a^{9/2} \sqrt [4]{a+b x^4}}+\frac {14 b}{117 a^2 x^9 \sqrt [4]{a+b x^4}}-\frac {1}{13 a x^{13} \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^14*(a + b*x^4)^(5/4)),x]

[Out]

-1/(13*a*x^13*(a + b*x^4)^(1/4)) + (14*b)/(117*a^2*x^9*(a + b*x^4)^(1/4)) - (28*b^2)/(117*a^3*x^5*(a + b*x^4)^
(1/4)) + (56*b^3)/(39*a^4*x*(a + b*x^4)^(1/4)) - (112*b^(7/2)*(1 + a/(b*x^4))^(1/4)*x*EllipticE[ArcCot[(Sqrt[b
]*x^2)/Sqrt[a]]/2, 2])/(39*a^(9/2)*(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 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 281

Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Dist[(x*(1 + a/(b*x^4))^(1/4))/(b*(a + b*x^4)^(1/4)), Int
[1/(x^3*(1 + a/(b*x^4))^(5/4)), x], x] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 283

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

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}{x^{14} \left (a+b x^4\right )^{5/4}} \, dx &=-\frac {1}{13 a x^{13} \sqrt [4]{a+b x^4}}-\frac {(14 b) \int \frac {1}{x^{10} \left (a+b x^4\right )^{5/4}} \, dx}{13 a}\\ &=-\frac {1}{13 a x^{13} \sqrt [4]{a+b x^4}}+\frac {14 b}{117 a^2 x^9 \sqrt [4]{a+b x^4}}+\frac {\left (140 b^2\right ) \int \frac {1}{x^6 \left (a+b x^4\right )^{5/4}} \, dx}{117 a^2}\\ &=-\frac {1}{13 a x^{13} \sqrt [4]{a+b x^4}}+\frac {14 b}{117 a^2 x^9 \sqrt [4]{a+b x^4}}-\frac {28 b^2}{117 a^3 x^5 \sqrt [4]{a+b x^4}}-\frac {\left (56 b^3\right ) \int \frac {1}{x^2 \left (a+b x^4\right )^{5/4}} \, dx}{39 a^3}\\ &=-\frac {1}{13 a x^{13} \sqrt [4]{a+b x^4}}+\frac {14 b}{117 a^2 x^9 \sqrt [4]{a+b x^4}}-\frac {28 b^2}{117 a^3 x^5 \sqrt [4]{a+b x^4}}+\frac {56 b^3}{39 a^4 x \sqrt [4]{a+b x^4}}+\frac {\left (112 b^4\right ) \int \frac {x^2}{\left (a+b x^4\right )^{5/4}} \, dx}{39 a^4}\\ &=-\frac {1}{13 a x^{13} \sqrt [4]{a+b x^4}}+\frac {14 b}{117 a^2 x^9 \sqrt [4]{a+b x^4}}-\frac {28 b^2}{117 a^3 x^5 \sqrt [4]{a+b x^4}}+\frac {56 b^3}{39 a^4 x \sqrt [4]{a+b x^4}}+\frac {\left (112 b^3 \sqrt [4]{1+\frac {a}{b x^4}} x\right ) \int \frac {1}{\left (1+\frac {a}{b x^4}\right )^{5/4} x^3} \, dx}{39 a^4 \sqrt [4]{a+b x^4}}\\ &=-\frac {1}{13 a x^{13} \sqrt [4]{a+b x^4}}+\frac {14 b}{117 a^2 x^9 \sqrt [4]{a+b x^4}}-\frac {28 b^2}{117 a^3 x^5 \sqrt [4]{a+b x^4}}+\frac {56 b^3}{39 a^4 x \sqrt [4]{a+b x^4}}-\frac {\left (112 b^3 \sqrt [4]{1+\frac {a}{b x^4}} x\right ) \operatorname {Subst}\left (\int \frac {x}{\left (1+\frac {a x^4}{b}\right )^{5/4}} \, dx,x,\frac {1}{x}\right )}{39 a^4 \sqrt [4]{a+b x^4}}\\ &=-\frac {1}{13 a x^{13} \sqrt [4]{a+b x^4}}+\frac {14 b}{117 a^2 x^9 \sqrt [4]{a+b x^4}}-\frac {28 b^2}{117 a^3 x^5 \sqrt [4]{a+b x^4}}+\frac {56 b^3}{39 a^4 x \sqrt [4]{a+b x^4}}-\frac {\left (56 b^3 \sqrt [4]{1+\frac {a}{b x^4}} x\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+\frac {a x^2}{b}\right )^{5/4}} \, dx,x,\frac {1}{x^2}\right )}{39 a^4 \sqrt [4]{a+b x^4}}\\ &=-\frac {1}{13 a x^{13} \sqrt [4]{a+b x^4}}+\frac {14 b}{117 a^2 x^9 \sqrt [4]{a+b x^4}}-\frac {28 b^2}{117 a^3 x^5 \sqrt [4]{a+b x^4}}+\frac {56 b^3}{39 a^4 x \sqrt [4]{a+b x^4}}-\frac {112 b^{7/2} \sqrt [4]{1+\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{39 a^{9/2} \sqrt [4]{a+b x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 54, normalized size = 0.35 \[ -\frac {\sqrt [4]{\frac {b x^4}{a}+1} \, _2F_1\left (-\frac {13}{4},\frac {5}{4};-\frac {9}{4};-\frac {b x^4}{a}\right )}{13 a x^{13} \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^14*(a + b*x^4)^(5/4)),x]

[Out]

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

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fricas [F]  time = 0.81, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{b^{2} x^{22} + 2 \, a b x^{18} + a^{2} x^{14}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^14/(b*x^4+a)^(5/4),x, algorithm="fricas")

[Out]

integral((b*x^4 + a)^(3/4)/(b^2*x^22 + 2*a*b*x^18 + a^2*x^14), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{14}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^14/(b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

integrate(1/((b*x^4 + a)^(5/4)*x^14), x)

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maple [F]  time = 0.17, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {5}{4}} x^{14}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^14/(b*x^4+a)^(5/4),x)

[Out]

int(1/x^14/(b*x^4+a)^(5/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{4} + a\right )}^{\frac {5}{4}} x^{14}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^14/(b*x^4+a)^(5/4),x, algorithm="maxima")

[Out]

integrate(1/((b*x^4 + a)^(5/4)*x^14), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^{14}\,{\left (b\,x^4+a\right )}^{5/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^14*(a + b*x^4)^(5/4)),x)

[Out]

int(1/(x^14*(a + b*x^4)^(5/4)), x)

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sympy [C]  time = 3.96, size = 44, normalized size = 0.29 \[ \frac {\Gamma \left (- \frac {13}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {13}{4}, \frac {5}{4} \\ - \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{4}} x^{13} \Gamma \left (- \frac {9}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**14/(b*x**4+a)**(5/4),x)

[Out]

gamma(-13/4)*hyper((-13/4, 5/4), (-9/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(5/4)*x**13*gamma(-9/4))

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