3.1163 \(\int \frac{x^{10}}{(a+b x^4)^{5/4}} \, dx\)

Optimal. Leaf size=105 \[ -\frac{7 a^{3/2} x \sqrt [4]{\frac{a}{b x^4}+1} E\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{4 b^{5/2} \sqrt [4]{a+b x^4}}-\frac{7 a x^3}{12 b^2 \sqrt [4]{a+b x^4}}+\frac{x^7}{6 b \sqrt [4]{a+b x^4}} \]

[Out]

(-7*a*x^3)/(12*b^2*(a + b*x^4)^(1/4)) + x^7/(6*b*(a + b*x^4)^(1/4)) - (7*a^(3/2)*(1 + a/(b*x^4))^(1/4)*x*Ellip
ticE[ArcCot[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(4*b^(5/2)*(a + b*x^4)^(1/4))

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Rubi [A]  time = 0.0493878, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {282, 281, 335, 275, 196} \[ -\frac{7 a^{3/2} x \sqrt [4]{\frac{a}{b x^4}+1} E\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{4 b^{5/2} \sqrt [4]{a+b x^4}}-\frac{7 a x^3}{12 b^2 \sqrt [4]{a+b x^4}}+\frac{x^7}{6 b \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Int[x^10/(a + b*x^4)^(5/4),x]

[Out]

(-7*a*x^3)/(12*b^2*(a + b*x^4)^(1/4)) + x^7/(6*b*(a + b*x^4)^(1/4)) - (7*a^(3/2)*(1 + a/(b*x^4))^(1/4)*x*Ellip
ticE[ArcCot[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(4*b^(5/2)*(a + b*x^4)^(1/4))

Rule 282

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Simp[x^(m - 3)/(b*(m - 4)*(a + b*x^4)^(1/4)), x] - Dis
t[(a*(m - 3))/(b*(m - 4)), Int[x^(m - 4)/(a + b*x^4)^(5/4), x], x] /; FreeQ[{a, b}, x] && PosQ[b/a] && IGtQ[(m
- 2)/4, 0]

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 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]

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 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{x^{10}}{\left (a+b x^4\right )^{5/4}} \, dx &=\frac{x^7}{6 b \sqrt [4]{a+b x^4}}-\frac{(7 a) \int \frac{x^6}{\left (a+b x^4\right )^{5/4}} \, dx}{6 b}\\ &=-\frac{7 a x^3}{12 b^2 \sqrt [4]{a+b x^4}}+\frac{x^7}{6 b \sqrt [4]{a+b x^4}}+\frac{\left (7 a^2\right ) \int \frac{x^2}{\left (a+b x^4\right )^{5/4}} \, dx}{4 b^2}\\ &=-\frac{7 a x^3}{12 b^2 \sqrt [4]{a+b x^4}}+\frac{x^7}{6 b \sqrt [4]{a+b x^4}}+\frac{\left (7 a^2 \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}{4 b^3 \sqrt [4]{a+b x^4}}\\ &=-\frac{7 a x^3}{12 b^2 \sqrt [4]{a+b x^4}}+\frac{x^7}{6 b \sqrt [4]{a+b x^4}}-\frac{\left (7 a^2 \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 )}{4 b^3 \sqrt [4]{a+b x^4}}\\ &=-\frac{7 a x^3}{12 b^2 \sqrt [4]{a+b x^4}}+\frac{x^7}{6 b \sqrt [4]{a+b x^4}}-\frac{\left (7 a^2 \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 )}{8 b^3 \sqrt [4]{a+b x^4}}\\ &=-\frac{7 a x^3}{12 b^2 \sqrt [4]{a+b x^4}}+\frac{x^7}{6 b \sqrt [4]{a+b x^4}}-\frac{7 a^{3/2} \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 )}{4 b^{5/2} \sqrt [4]{a+b x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0238079, size = 67, normalized size = 0.64 \[ \frac{x^3 \left (7 a \sqrt [4]{\frac{b x^4}{a}+1} \, _2F_1\left (\frac{3}{4},\frac{5}{4};\frac{7}{4};-\frac{b x^4}{a}\right )-7 a+2 b x^4\right )}{12 b^2 \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^10/(a + b*x^4)^(5/4),x]

[Out]

(x^3*(-7*a + 2*b*x^4 + 7*a*(1 + (b*x^4)/a)^(1/4)*Hypergeometric2F1[3/4, 5/4, 7/4, -((b*x^4)/a)]))/(12*b^2*(a +
b*x^4)^(1/4))

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Maple [F]  time = 0.046, size = 0, normalized size = 0. \begin{align*} \int{{x}^{10} \left ( b{x}^{4}+a \right ) ^{-{\frac{5}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^10/(b*x^4+a)^(5/4),x)

[Out]

int(x^10/(b*x^4+a)^(5/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{10}}{{\left (b x^{4} + a\right )}^{\frac{5}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(b*x^4+a)^(5/4),x, algorithm="maxima")

[Out]

integrate(x^10/(b*x^4 + a)^(5/4), 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}} x^{10}}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(b*x^4+a)^(5/4),x, algorithm="fricas")

[Out]

integral((b*x^4 + a)^(3/4)*x^10/(b^2*x^8 + 2*a*b*x^4 + a^2), x)

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Sympy [C]  time = 1.72837, size = 37, normalized size = 0.35 \begin{align*} \frac{x^{11} \Gamma \left (\frac{11}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{5}{4}, \frac{11}{4} \\ \frac{15}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac{5}{4}} \Gamma \left (\frac{15}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**10/(b*x**4+a)**(5/4),x)

[Out]

x**11*gamma(11/4)*hyper((5/4, 11/4), (15/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(5/4)*gamma(15/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{10}}{{\left (b x^{4} + a\right )}^{\frac{5}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

integrate(x^10/(b*x^4 + a)^(5/4), x)