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

Optimal. Leaf size=100 \[ \frac{24 a^{3/2} \sqrt [4]{\frac{b x^2}{a}+1} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{5 b^{5/2} \sqrt [4]{a+b x^2}}-\frac{12 a x}{5 b^2 \sqrt [4]{a+b x^2}}+\frac{2 x^3}{5 b \sqrt [4]{a+b x^2}} \]

[Out]

(-12*a*x)/(5*b^2*(a + b*x^2)^(1/4)) + (2*x^3)/(5*b*(a + b*x^2)^(1/4)) + (24*a^(3/2)*(1 + (b*x^2)/a)^(1/4)*Elli
pticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(5*b^(5/2)*(a + b*x^2)^(1/4))

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Rubi [A]  time = 0.0314686, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {285, 197, 196} \[ \frac{24 a^{3/2} \sqrt [4]{\frac{b x^2}{a}+1} E\left (\left .\frac{1}{2} \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right |2\right )}{5 b^{5/2} \sqrt [4]{a+b x^2}}-\frac{12 a x}{5 b^2 \sqrt [4]{a+b x^2}}+\frac{2 x^3}{5 b \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^4/(a + b*x^2)^(5/4),x]

[Out]

(-12*a*x)/(5*b^2*(a + b*x^2)^(1/4)) + (2*x^3)/(5*b*(a + b*x^2)^(1/4)) + (24*a^(3/2)*(1 + (b*x^2)/a)^(1/4)*Elli
pticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(5*b^(5/2)*(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^4}{\left (a+b x^2\right )^{5/4}} \, dx &=\frac{2 x^3}{5 b \sqrt [4]{a+b x^2}}-\frac{(6 a) \int \frac{x^2}{\left (a+b x^2\right )^{5/4}} \, dx}{5 b}\\ &=-\frac{12 a x}{5 b^2 \sqrt [4]{a+b x^2}}+\frac{2 x^3}{5 b \sqrt [4]{a+b x^2}}+\frac{\left (12 a^2\right ) \int \frac{1}{\left (a+b x^2\right )^{5/4}} \, dx}{5 b^2}\\ &=-\frac{12 a x}{5 b^2 \sqrt [4]{a+b x^2}}+\frac{2 x^3}{5 b \sqrt [4]{a+b x^2}}+\frac{\left (12 a \sqrt [4]{1+\frac{b x^2}{a}}\right ) \int \frac{1}{\left (1+\frac{b x^2}{a}\right )^{5/4}} \, dx}{5 b^2 \sqrt [4]{a+b x^2}}\\ &=-\frac{12 a x}{5 b^2 \sqrt [4]{a+b x^2}}+\frac{2 x^3}{5 b \sqrt [4]{a+b x^2}}+\frac{24 a^{3/2} \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 )}{5 b^{5/2} \sqrt [4]{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0203067, size = 65, normalized size = 0.65 \[ \frac{2 \left (-6 a x \sqrt [4]{\frac{b x^2}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{3}{2};-\frac{b x^2}{a}\right )+6 a x+b x^3\right )}{5 b^2 \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/(a + b*x^2)^(5/4),x]

[Out]

(2*(6*a*x + b*x^3 - 6*a*x*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, -((b*x^2)/a)]))/(5*b^2*(a + b
*x^2)^(1/4))

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Maple [F]  time = 0.048, size = 0, normalized size = 0. \begin{align*} \int{{x}^{4} \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^4/(b*x^2+a)^(5/4),x)

[Out]

int(x^4/(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^{4}}{{\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^4/(b*x^2+a)^(5/4),x, algorithm="maxima")

[Out]

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

[Out]

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

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Sympy [C]  time = 0.819233, size = 27, normalized size = 0.27 \begin{align*} \frac{x^{5}{{}_{2}F_{1}\left (\begin{matrix} \frac{5}{4}, \frac{5}{2} \\ \frac{7}{2} \end{matrix}\middle |{\frac{b x^{2} e^{i \pi }}{a}} \right )}}{5 a^{\frac{5}{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{{\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^4/(b*x^2+a)^(5/4),x, algorithm="giac")

[Out]

integrate(x^4/(b*x^2 + a)^(5/4), x)