3.843 \(\int \frac{x^6}{(a+b x^2)^{5/4}} \, dx\)

Optimal. Leaf size=124 \[ \frac{8 a^2 x}{3 b^3 \sqrt [4]{a+b x^2}}-\frac{16 a^{5/2} \sqrt [4]{\frac{b x^2}{a}+1} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{3 b^{7/2} \sqrt [4]{a+b x^2}}-\frac{4 a x^3}{9 b^2 \sqrt [4]{a+b x^2}}+\frac{2 x^5}{9 b \sqrt [4]{a+b x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0434474, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {285, 197, 196} \[ \frac{8 a^2 x}{3 b^3 \sqrt [4]{a+b x^2}}-\frac{16 a^{5/2} \sqrt [4]{\frac{b x^2}{a}+1} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{3 b^{7/2} \sqrt [4]{a+b x^2}}-\frac{4 a x^3}{9 b^2 \sqrt [4]{a+b x^2}}+\frac{2 x^5}{9 b \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^6/(a + b*x^2)^(5/4),x]

[Out]

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

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^6}{\left (a+b x^2\right )^{5/4}} \, dx &=\frac{2 x^5}{9 b \sqrt [4]{a+b x^2}}-\frac{(10 a) \int \frac{x^4}{\left (a+b x^2\right )^{5/4}} \, dx}{9 b}\\ &=-\frac{4 a x^3}{9 b^2 \sqrt [4]{a+b x^2}}+\frac{2 x^5}{9 b \sqrt [4]{a+b x^2}}+\frac{\left (4 a^2\right ) \int \frac{x^2}{\left (a+b x^2\right )^{5/4}} \, dx}{3 b^2}\\ &=\frac{8 a^2 x}{3 b^3 \sqrt [4]{a+b x^2}}-\frac{4 a x^3}{9 b^2 \sqrt [4]{a+b x^2}}+\frac{2 x^5}{9 b \sqrt [4]{a+b x^2}}-\frac{\left (8 a^3\right ) \int \frac{1}{\left (a+b x^2\right )^{5/4}} \, dx}{3 b^3}\\ &=\frac{8 a^2 x}{3 b^3 \sqrt [4]{a+b x^2}}-\frac{4 a x^3}{9 b^2 \sqrt [4]{a+b x^2}}+\frac{2 x^5}{9 b \sqrt [4]{a+b x^2}}-\frac{\left (8 a^2 \sqrt [4]{1+\frac{b x^2}{a}}\right ) \int \frac{1}{\left (1+\frac{b x^2}{a}\right )^{5/4}} \, dx}{3 b^3 \sqrt [4]{a+b x^2}}\\ &=\frac{8 a^2 x}{3 b^3 \sqrt [4]{a+b x^2}}-\frac{4 a x^3}{9 b^2 \sqrt [4]{a+b x^2}}+\frac{2 x^5}{9 b \sqrt [4]{a+b x^2}}-\frac{16 a^{5/2} \sqrt [4]{1+\frac{b x^2}{a}} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{3 b^{7/2} \sqrt [4]{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0284226, size = 78, normalized size = 0.63 \[ \frac{2 \left (12 a^2 x \sqrt [4]{\frac{b x^2}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{3}{2};-\frac{b x^2}{a}\right )-12 a^2 x-2 a b x^3+b^2 x^5\right )}{9 b^3 \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/(a + b*x^2)^(5/4),x]

[Out]

(2*(-12*a^2*x - 2*a*b*x^3 + b^2*x^5 + 12*a^2*x*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, -((b*x^2
)/a)]))/(9*b^3*(a + b*x^2)^(1/4))

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Maple [F]  time = 0.045, size = 0, normalized size = 0. \begin{align*} \int{{x}^{6} \left ( b{x}^{2}+a \right ) ^{-{\frac{5}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(b*x^2+a)^(5/4),x)

[Out]

int(x^6/(b*x^2+a)^(5/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{6}}{{\left (b x^{2} + a\right )}^{\frac{5}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^2+a)^(5/4),x, algorithm="maxima")

[Out]

integrate(x^6/(b*x^2 + 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^{2} + a\right )}^{\frac{3}{4}} x^{6}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^2+a)^(5/4),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(3/4)*x^6/(b^2*x^4 + 2*a*b*x^2 + a^2), x)

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Sympy [C]  time = 0.902676, size = 27, normalized size = 0.22 \begin{align*} \frac{x^{7}{{}_{2}F_{1}\left (\begin{matrix} \frac{5}{4}, \frac{7}{2} \\ \frac{9}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{7 a^{\frac{5}{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(b*x**2+a)**(5/4),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{6}}{{\left (b x^{2} + a\right )}^{\frac{5}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^2+a)^(5/4),x, algorithm="giac")

[Out]

integrate(x^6/(b*x^2 + a)^(5/4), x)