\(\int \frac {(a+b x^4)^{3/4}}{x^2} \, dx\) [1043]

Optimal result

Integrand size = 15, antiderivative size = 97 \[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^2} \, dx=\frac {3 b x^3}{2 \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{x}+\frac {3 \sqrt {a} \sqrt {b} \sqrt [4]{1+\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{2 \sqrt [4]{a+b x^4}} \]

[Out]

3/2*b*x^3/(b*x^4+a)^(1/4)-(b*x^4+a)^(3/4)/x+3/2*(1+a/b/x^4)^(1/4)*x*(cos(1/2*arccot(x^2*b^(1/2)/a^(1/2)))^2)^(
1/2)/cos(1/2*arccot(x^2*b^(1/2)/a^(1/2)))*EllipticE(sin(1/2*arccot(x^2*b^(1/2)/a^(1/2))),2^(1/2))*a^(1/2)*b^(1
/2)/(b*x^4+a)^(1/4)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {283, 316, 287, 342, 281, 202} \[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^2} \, dx=-\frac {\left (a+b x^4\right )^{3/4}}{x}+\frac {3 b x^3}{2 \sqrt [4]{a+b x^4}}+\frac {3 \sqrt {a} \sqrt {b} x \sqrt [4]{\frac {a}{b x^4}+1} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{2 \sqrt [4]{a+b x^4}} \]

[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 202

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]))*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 281

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 283

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 287

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 316

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 342

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]

Rubi steps \begin{align*} \text {integral}& = -\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 ) \text {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 ) \text {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] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.01 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.51 \[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^2} \, dx=-\frac {\left (a+b x^4\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-\frac {1}{4},\frac {3}{4},-\frac {b x^4}{a}\right )}{x \left (1+\frac {b x^4}{a}\right )^{3/4}} \]

[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)))

Maple [F]

\[\int \frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}}}{x^{2}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^2} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{x^{2}} \,d x } \]

[In]

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

[Out]

integral((b*x^4 + a)^(3/4)/x^2, x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.57 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.42 \[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^2} \, dx=\frac {a^{\frac {3}{4}} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} \]

[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))

Maxima [F]

\[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^2} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{x^{2}} \,d x } \]

[In]

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

[Out]

integrate((b*x^4 + a)^(3/4)/x^2, x)

Giac [F]

\[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^2} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{x^{2}} \,d x } \]

[In]

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

[Out]

integrate((b*x^4 + a)^(3/4)/x^2, x)

Mupad [B] (verification not implemented)

Time = 5.89 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.41 \[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^2} \, dx=\frac {{\left (b\,x^4+a\right )}^{3/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},-\frac {1}{2};\ \frac {1}{2};\ -\frac {a}{b\,x^4}\right )}{2\,x\,{\left (\frac {a}{b\,x^4}+1\right )}^{3/4}} \]

[In]

int((a + b*x^4)^(3/4)/x^2,x)

[Out]

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