3.1094 \(\int \frac{1}{x^{11} \sqrt [4]{a+b x^4}} \, dx\)

Optimal. Leaf size=152 \[ \frac{7 b^3 x^2}{40 a^3 \sqrt [4]{a+b x^4}}-\frac{7 b^2 \left (a+b x^4\right )^{3/4}}{40 a^3 x^2}-\frac{7 b^{5/2} \sqrt [4]{\frac{b x^4}{a}+1} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{40 a^{5/2} \sqrt [4]{a+b x^4}}+\frac{7 b \left (a+b x^4\right )^{3/4}}{60 a^2 x^6}-\frac{\left (a+b x^4\right )^{3/4}}{10 a x^{10}} \]

[Out]

(7*b^3*x^2)/(40*a^3*(a + b*x^4)^(1/4)) - (a + b*x^4)^(3/4)/(10*a*x^10) + (7*b*(a + b*x^4)^(3/4))/(60*a^2*x^6)
- (7*b^2*(a + b*x^4)^(3/4))/(40*a^3*x^2) - (7*b^(5/2)*(1 + (b*x^4)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x^2)/Sqr
t[a]]/2, 2])/(40*a^(5/2)*(a + b*x^4)^(1/4))

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Rubi [A]  time = 0.0954142, antiderivative size = 152, normalized size of antiderivative = 1., 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, 325, 229, 227, 196} \[ \frac{7 b^3 x^2}{40 a^3 \sqrt [4]{a+b x^4}}-\frac{7 b^2 \left (a+b x^4\right )^{3/4}}{40 a^3 x^2}-\frac{7 b^{5/2} \sqrt [4]{\frac{b x^4}{a}+1} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{40 a^{5/2} \sqrt [4]{a+b x^4}}+\frac{7 b \left (a+b x^4\right )^{3/4}}{60 a^2 x^6}-\frac{\left (a+b x^4\right )^{3/4}}{10 a x^{10}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^11*(a + b*x^4)^(1/4)),x]

[Out]

(7*b^3*x^2)/(40*a^3*(a + b*x^4)^(1/4)) - (a + b*x^4)^(3/4)/(10*a*x^10) + (7*b*(a + b*x^4)^(3/4))/(60*a^2*x^6)
- (7*b^2*(a + b*x^4)^(3/4))/(40*a^3*x^2) - (7*b^(5/2)*(1 + (b*x^4)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x^2)/Sqr
t[a]]/2, 2])/(40*a^(5/2)*(a + b*x^4)^(1/4))

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 325

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

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 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 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]

Rubi steps

\begin{align*} \int \frac{1}{x^{11} \sqrt [4]{a+b x^4}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x^6 \sqrt [4]{a+b x^2}} \, dx,x,x^2\right )\\ &=-\frac{\left (a+b x^4\right )^{3/4}}{10 a x^{10}}-\frac{(7 b) \operatorname{Subst}\left (\int \frac{1}{x^4 \sqrt [4]{a+b x^2}} \, dx,x,x^2\right )}{20 a}\\ &=-\frac{\left (a+b x^4\right )^{3/4}}{10 a x^{10}}+\frac{7 b \left (a+b x^4\right )^{3/4}}{60 a^2 x^6}+\frac{\left (7 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt [4]{a+b x^2}} \, dx,x,x^2\right )}{40 a^2}\\ &=-\frac{\left (a+b x^4\right )^{3/4}}{10 a x^{10}}+\frac{7 b \left (a+b x^4\right )^{3/4}}{60 a^2 x^6}-\frac{7 b^2 \left (a+b x^4\right )^{3/4}}{40 a^3 x^2}+\frac{\left (7 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [4]{a+b x^2}} \, dx,x,x^2\right )}{80 a^3}\\ &=-\frac{\left (a+b x^4\right )^{3/4}}{10 a x^{10}}+\frac{7 b \left (a+b x^4\right )^{3/4}}{60 a^2 x^6}-\frac{7 b^2 \left (a+b x^4\right )^{3/4}}{40 a^3 x^2}+\frac{\left (7 b^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 )}{80 a^3 \sqrt [4]{a+b x^4}}\\ &=\frac{7 b^3 x^2}{40 a^3 \sqrt [4]{a+b x^4}}-\frac{\left (a+b x^4\right )^{3/4}}{10 a x^{10}}+\frac{7 b \left (a+b x^4\right )^{3/4}}{60 a^2 x^6}-\frac{7 b^2 \left (a+b x^4\right )^{3/4}}{40 a^3 x^2}-\frac{\left (7 b^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 )}{80 a^3 \sqrt [4]{a+b x^4}}\\ &=\frac{7 b^3 x^2}{40 a^3 \sqrt [4]{a+b x^4}}-\frac{\left (a+b x^4\right )^{3/4}}{10 a x^{10}}+\frac{7 b \left (a+b x^4\right )^{3/4}}{60 a^2 x^6}-\frac{7 b^2 \left (a+b x^4\right )^{3/4}}{40 a^3 x^2}-\frac{7 b^{5/2} \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 )}{40 a^{5/2} \sqrt [4]{a+b x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0107569, size = 51, normalized size = 0.34 \[ -\frac{\sqrt [4]{\frac{b x^4}{a}+1} \, _2F_1\left (-\frac{5}{2},\frac{1}{4};-\frac{3}{2};-\frac{b x^4}{a}\right )}{10 x^{10} \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^11*(a + b*x^4)^(1/4)),x]

[Out]

-((1 + (b*x^4)/a)^(1/4)*Hypergeometric2F1[-5/2, 1/4, -3/2, -((b*x^4)/a)])/(10*x^10*(a + b*x^4)^(1/4))

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Maple [F]  time = 0.033, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{11}}{\frac{1}{\sqrt [4]{b{x}^{4}+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{4} + a\right )}^{\frac{1}{4}} x^{11}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x^{4} + a\right )}^{\frac{3}{4}}}{b x^{15} + a x^{11}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^4 + a)^(3/4)/(b*x^15 + a*x^11), x)

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Sympy [C]  time = 2.10036, size = 32, normalized size = 0.21 \begin{align*} - \frac{{{}_{2}F_{1}\left (\begin{matrix} - \frac{5}{2}, \frac{1}{4} \\ - \frac{3}{2} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{10 \sqrt [4]{a} x^{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-hyper((-5/2, 1/4), (-3/2,), b*x**4*exp_polar(I*pi)/a)/(10*a**(1/4)*x**10)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{4} + a\right )}^{\frac{1}{4}} x^{11}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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