3.1043 \(\int \frac{(a+b x^4)^{3/4}}{x^2} \, dx\)

Optimal. Leaf size=97 \[ \frac{3 b x^3}{2 \sqrt [4]{a+b x^4}}-\frac{\left (a+b x^4\right )^{3/4}}{x}+\frac{3 \sqrt{a} \sqrt{b} x \sqrt [4]{\frac{a}{b x^4}+1} E\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{2 \sqrt [4]{a+b x^4}} \]

[Out]

(3*b*x^3)/(2*(a + b*x^4)^(1/4)) - (a + b*x^4)^(3/4)/x + (3*Sqrt[a]*Sqrt[b]*(1 + a/(b*x^4))^(1/4)*x*EllipticE[A
rcCot[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(2*(a + b*x^4)^(1/4))

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Rubi [A]  time = 0.0465699, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {277, 310, 281, 335, 275, 196} \[ \frac{3 b x^3}{2 \sqrt [4]{a+b x^4}}-\frac{\left (a+b x^4\right )^{3/4}}{x}+\frac{3 \sqrt{a} \sqrt{b} x \sqrt [4]{\frac{a}{b x^4}+1} E\left (\left .\frac{1}{2} \cot ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{2 \sqrt [4]{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^(3/4)/x^2,x]

[Out]

(3*b*x^3)/(2*(a + b*x^4)^(1/4)) - (a + b*x^4)^(3/4)/x + (3*Sqrt[a]*Sqrt[b]*(1 + a/(b*x^4))^(1/4)*x*EllipticE[A
rcCot[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(2*(a + b*x^4)^(1/4))

Rule 277

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

Rule 310

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

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

Mathematica [C]  time = 0.0102763, size = 49, normalized size = 0.51 \[ -\frac{\left (a+b x^4\right )^{3/4} \, _2F_1\left (-\frac{3}{4},-\frac{1}{4};\frac{3}{4};-\frac{b x^4}{a}\right )}{x \left (\frac{b x^4}{a}+1\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^(3/4)/x^2,x]

[Out]

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

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Maple [F]  time = 0.029, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{2}} \left ( b{x}^{4}+a \right ) ^{{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{4} + a\right )}^{\frac{3}{4}}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(3/4)/x^2,x, algorithm="maxima")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(3/4)/x^2,x, algorithm="fricas")

[Out]

integral((b*x^4 + a)^(3/4)/x^2, x)

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Sympy [C]  time = 0.972202, size = 41, normalized size = 0.42 \begin{align*} \frac{a^{\frac{3}{4}} \Gamma \left (- \frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{4}, - \frac{1}{4} \\ \frac{3}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 x \Gamma \left (\frac{3}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{4} + a\right )}^{\frac{3}{4}}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(3/4)/x^2,x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(3/4)/x^2, x)