3.1147 \(\int \frac{x^{13}}{(a+b x^4)^{5/4}} \, dx\)

Optimal. Leaf size=128 \[ \frac{4 a^2 x^2}{3 b^3 \sqrt [4]{a+b x^4}}-\frac{8 a^{5/2} \sqrt [4]{\frac{b x^4}{a}+1} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{3 b^{7/2} \sqrt [4]{a+b x^4}}-\frac{2 a x^6}{9 b^2 \sqrt [4]{a+b x^4}}+\frac{x^{10}}{9 b \sqrt [4]{a+b x^4}} \]

[Out]

(4*a^2*x^2)/(3*b^3*(a + b*x^4)^(1/4)) - (2*a*x^6)/(9*b^2*(a + b*x^4)^(1/4)) + x^10/(9*b*(a + b*x^4)^(1/4)) - (
8*a^(5/2)*(1 + (b*x^4)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(3*b^(7/2)*(a + b*x^4)^(1/4))

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Rubi [A]  time = 0.0851698, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {275, 285, 197, 196} \[ \frac{4 a^2 x^2}{3 b^3 \sqrt [4]{a+b x^4}}-\frac{8 a^{5/2} \sqrt [4]{\frac{b x^4}{a}+1} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{3 b^{7/2} \sqrt [4]{a+b x^4}}-\frac{2 a x^6}{9 b^2 \sqrt [4]{a+b x^4}}+\frac{x^{10}}{9 b \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Int[x^13/(a + b*x^4)^(5/4),x]

[Out]

(4*a^2*x^2)/(3*b^3*(a + b*x^4)^(1/4)) - (2*a*x^6)/(9*b^2*(a + b*x^4)^(1/4)) + x^10/(9*b*(a + b*x^4)^(1/4)) - (
8*a^(5/2)*(1 + (b*x^4)/a)^(1/4)*EllipticE[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(3*b^(7/2)*(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 285

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

Rule 197

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Dist[(1 + (b*x^2)/a)^(1/4)/(a*(a + b*x^2)^(1/4)), Int[1/(1 + (b
*x^2)/a)^(5/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a] && 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{x^{13}}{\left (a+b x^4\right )^{5/4}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^6}{\left (a+b x^2\right )^{5/4}} \, dx,x,x^2\right )\\ &=\frac{x^{10}}{9 b \sqrt [4]{a+b x^4}}-\frac{(5 a) \operatorname{Subst}\left (\int \frac{x^4}{\left (a+b x^2\right )^{5/4}} \, dx,x,x^2\right )}{9 b}\\ &=-\frac{2 a x^6}{9 b^2 \sqrt [4]{a+b x^4}}+\frac{x^{10}}{9 b \sqrt [4]{a+b x^4}}+\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{\left (a+b x^2\right )^{5/4}} \, dx,x,x^2\right )}{3 b^2}\\ &=\frac{4 a^2 x^2}{3 b^3 \sqrt [4]{a+b x^4}}-\frac{2 a x^6}{9 b^2 \sqrt [4]{a+b x^4}}+\frac{x^{10}}{9 b \sqrt [4]{a+b x^4}}-\frac{\left (4 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^2\right )^{5/4}} \, dx,x,x^2\right )}{3 b^3}\\ &=\frac{4 a^2 x^2}{3 b^3 \sqrt [4]{a+b x^4}}-\frac{2 a x^6}{9 b^2 \sqrt [4]{a+b x^4}}+\frac{x^{10}}{9 b \sqrt [4]{a+b x^4}}-\frac{\left (4 a^2 \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 )}{3 b^3 \sqrt [4]{a+b x^4}}\\ &=\frac{4 a^2 x^2}{3 b^3 \sqrt [4]{a+b x^4}}-\frac{2 a x^6}{9 b^2 \sqrt [4]{a+b x^4}}+\frac{x^{10}}{9 b \sqrt [4]{a+b x^4}}-\frac{8 a^{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 )}{3 b^{7/2} \sqrt [4]{a+b x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0297703, size = 79, normalized size = 0.62 \[ \frac{x^2 \left (12 a^2 \sqrt [4]{\frac{b x^4}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{3}{2};-\frac{b x^4}{a}\right )-12 a^2-2 a b x^4+b^2 x^8\right )}{9 b^3 \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^13/(a + b*x^4)^(5/4),x]

[Out]

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

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Maple [F]  time = 0.047, size = 0, normalized size = 0. \begin{align*} \int{{x}^{13} \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^13/(b*x^4+a)^(5/4),x)

[Out]

int(x^13/(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^{13}}{{\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^13/(b*x^4+a)^(5/4),x, algorithm="maxima")

[Out]

integrate(x^13/(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^{13}}{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^13/(b*x^4+a)^(5/4),x, algorithm="fricas")

[Out]

integral((b*x^4 + a)^(3/4)*x^13/(b^2*x^8 + 2*a*b*x^4 + a^2), x)

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Sympy [C]  time = 2.58765, size = 27, normalized size = 0.21 \begin{align*} \frac{x^{14}{{}_{2}F_{1}\left (\begin{matrix} \frac{5}{4}, \frac{7}{2} \\ \frac{9}{2} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{14 a^{\frac{5}{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**13/(b*x**4+a)**(5/4),x)

[Out]

x**14*hyper((5/4, 7/2), (9/2,), b*x**4*exp_polar(I*pi)/a)/(14*a**(5/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{13}}{{\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^13/(b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

integrate(x^13/(b*x^4 + a)^(5/4), x)