Integrand size = 15, antiderivative size = 97 \[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^2} \, dx=\frac {3 b x^3}{2 \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{x}+\frac {3 \sqrt {a} \sqrt {b} \sqrt [4]{1+\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{2 \sqrt [4]{a+b x^4}} \] Output:
3/2*b*x^3/(b*x^4+a)^(1/4)-(b*x^4+a)^(3/4)/x+3/2*a^(1/2)*b^(1/2)*(1+a/b/x^4 )^(1/4)*x*EllipticE(sin(1/2*arccot(b^(1/2)*x^2/a^(1/2))),2^(1/2))/(b*x^4+a )^(1/4)
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 \[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^2} \, dx=-\frac {\left (a+b x^4\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-\frac {1}{4},\frac {3}{4},-\frac {b x^4}{a}\right )}{x \left (1+\frac {b x^4}{a}\right )^{3/4}} \] Input:
Integrate[(a + b*x^4)^(3/4)/x^2,x]
Output:
-(((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)))
Time = 0.41 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {809, 839, 813, 858, 807, 212}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\left (a+b x^4\right )^{3/4}}{x^2} \, dx\) |
\(\Big \downarrow \) 809 |
\(\displaystyle 3 b \int \frac {x^2}{\sqrt [4]{b x^4+a}}dx-\frac {\left (a+b x^4\right )^{3/4}}{x}\) |
\(\Big \downarrow \) 839 |
\(\displaystyle 3 b \left (\frac {x^3}{2 \sqrt [4]{a+b x^4}}-\frac {1}{2} a \int \frac {x^2}{\left (b x^4+a\right )^{5/4}}dx\right )-\frac {\left (a+b x^4\right )^{3/4}}{x}\) |
\(\Big \downarrow \) 813 |
\(\displaystyle 3 b \left (\frac {x^3}{2 \sqrt [4]{a+b x^4}}-\frac {a x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{5/4} x^3}dx}{2 b \sqrt [4]{a+b x^4}}\right )-\frac {\left (a+b x^4\right )^{3/4}}{x}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle 3 b \left (\frac {a x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^4}+1\right )^{5/4} x}d\frac {1}{x}}{2 b \sqrt [4]{a+b x^4}}+\frac {x^3}{2 \sqrt [4]{a+b x^4}}\right )-\frac {\left (a+b x^4\right )^{3/4}}{x}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle 3 b \left (\frac {a x \sqrt [4]{\frac {a}{b x^4}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4}}d\frac {1}{x^2}}{4 b \sqrt [4]{a+b x^4}}+\frac {x^3}{2 \sqrt [4]{a+b x^4}}\right )-\frac {\left (a+b x^4\right )^{3/4}}{x}\) |
\(\Big \downarrow \) 212 |
\(\displaystyle 3 b \left (\frac {\sqrt {a} x \sqrt [4]{\frac {a}{b x^4}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x^2}\right )\right |2\right )}{2 \sqrt {b} \sqrt [4]{a+b x^4}}+\frac {x^3}{2 \sqrt [4]{a+b x^4}}\right )-\frac {\left (a+b x^4\right )^{3/4}}{x}\) |
Input:
Int[(a + b*x^4)^(3/4)/x^2,x]
Output:
-((a + b*x^4)^(3/4)/x) + 3*b*(x^3/(2*(a + b*x^4)^(1/4)) + (Sqrt[a]*(1 + a/ (b*x^4))^(1/4)*x*EllipticE[ArcTan[Sqrt[a]/(Sqrt[b]*x^2)]/2, 2])/(2*Sqrt[b] *(a + b*x^4)^(1/4)))
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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[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]
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] - Simp[b*n*(p/(c^n*(m + 1))) I nt[(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] && IntB inomialQ[a, b, c, n, m, p, x]
Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> Simp[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]
Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(1/4), x_Symbol] :> Simp[x^3/(2*(a + b*x^4 )^(1/4)), x] - Simp[a/2 Int[x^2/(a + b*x^4)^(5/4), x], x] /; FreeQ[{a, b} , x] && PosQ[b/a]
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] && Int egerQ[m]
\[\int \frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}}}{x^{2}}d x\]
Input:
int((b*x^4+a)^(3/4)/x^2,x)
Output:
int((b*x^4+a)^(3/4)/x^2,x)
\[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^2} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{x^{2}} \,d x } \] Input:
integrate((b*x^4+a)^(3/4)/x^2,x, algorithm="fricas")
Output:
integral((b*x^4 + a)^(3/4)/x^2, x)
Result contains complex when optimal does not.
Time = 0.53 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.42 \[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^2} \, dx=\frac {a^{\frac {3}{4}} \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x \Gamma \left (\frac {3}{4}\right )} \] Input:
integrate((b*x**4+a)**(3/4)/x**2,x)
Output:
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))
\[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^2} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{x^{2}} \,d x } \] Input:
integrate((b*x^4+a)^(3/4)/x^2,x, algorithm="maxima")
Output:
integrate((b*x^4 + a)^(3/4)/x^2, x)
\[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^2} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{x^{2}} \,d x } \] Input:
integrate((b*x^4+a)^(3/4)/x^2,x, algorithm="giac")
Output:
integrate((b*x^4 + a)^(3/4)/x^2, x)
Time = 0.56 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.41 \[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^2} \, dx=\frac {{\left (b\,x^4+a\right )}^{3/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},-\frac {1}{2};\ \frac {1}{2};\ -\frac {a}{b\,x^4}\right )}{2\,x\,{\left (\frac {a}{b\,x^4}+1\right )}^{3/4}} \] Input:
int((a + b*x^4)^(3/4)/x^2,x)
Output:
((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))
\[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^2} \, dx=\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}}+3 \left (\int \frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}}}{b \,x^{6}+a \,x^{2}}d x \right ) a x}{2 x} \] Input:
int((b*x^4+a)^(3/4)/x^2,x)
Output:
((a + b*x**4)**(3/4) + 3*int((a + b*x**4)**(3/4)/(a*x**2 + b*x**6),x)*a*x) /(2*x)