Integrand size = 15, antiderivative size = 98 \[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^3} \, dx=\frac {3 b x^2}{2 \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{2 x^2}-\frac {3 \sqrt {a} \sqrt {b} \sqrt [4]{1+\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{2 \sqrt [4]{a+b x^4}} \] Output:
3/2*b*x^2/(b*x^4+a)^(1/4)-1/2*(b*x^4+a)^(3/4)/x^2-3/2*a^(1/2)*b^(1/2)*(1+b *x^4/a)^(1/4)*EllipticE(sin(1/2*arctan(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 = 51, normalized size of antiderivative = 0.52 \[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^3} \, dx=-\frac {\left (a+b x^4\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-\frac {1}{2},\frac {1}{2},-\frac {b x^4}{a}\right )}{2 x^2 \left (1+\frac {b x^4}{a}\right )^{3/4}} \] Input:
Integrate[(a + b*x^4)^(3/4)/x^3,x]
Output:
-1/2*((a + b*x^4)^(3/4)*Hypergeometric2F1[-3/4, -1/2, 1/2, -((b*x^4)/a)])/ (x^2*(1 + (b*x^4)/a)^(3/4))
Time = 0.33 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {807, 247, 227, 225, 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.
\(\displaystyle \int \frac {\left (a+b x^4\right )^{3/4}}{x^3} \, dx\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{2} \int \frac {\left (b x^4+a\right )^{3/4}}{x^4}dx^2\) |
\(\Big \downarrow \) 247 |
\(\displaystyle \frac {1}{2} \left (\frac {3}{2} b \int \frac {1}{\sqrt [4]{b x^4+a}}dx^2-\frac {\left (a+b x^4\right )^{3/4}}{x^2}\right )\) |
\(\Big \downarrow \) 227 |
\(\displaystyle \frac {1}{2} \left (\frac {3 b \sqrt [4]{\frac {b x^4}{a}+1} \int \frac {1}{\sqrt [4]{\frac {b x^4}{a}+1}}dx^2}{2 \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{x^2}\right )\) |
\(\Big \downarrow \) 225 |
\(\displaystyle \frac {1}{2} \left (\frac {3 b \sqrt [4]{\frac {b x^4}{a}+1} \left (\frac {2 x^2}{\sqrt [4]{\frac {b x^4}{a}+1}}-\int \frac {1}{\left (\frac {b x^4}{a}+1\right )^{5/4}}dx^2\right )}{2 \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{x^2}\right )\) |
\(\Big \downarrow \) 212 |
\(\displaystyle \frac {1}{2} \left (\frac {3 b \sqrt [4]{\frac {b x^4}{a}+1} \left (\frac {2 x^2}{\sqrt [4]{\frac {b x^4}{a}+1}}-\frac {2 \sqrt {a} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b}}\right )}{2 \sqrt [4]{a+b x^4}}-\frac {\left (a+b x^4\right )^{3/4}}{x^2}\right )\) |
Input:
Int[(a + b*x^4)^(3/4)/x^3,x]
Output:
(-((a + b*x^4)^(3/4)/x^2) + (3*b*(1 + (b*x^4)/a)^(1/4)*((2*x^2)/(1 + (b*x^ 4)/a)^(1/4) - (2*Sqrt[a]*EllipticE[ArcTan[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/Sq rt[b]))/(2*(a + b*x^4)^(1/4)))/2
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[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[2*(x/(a + b*x^2)^(1/4)) , x] - Simp[a Int[1/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b}, x] && GtQ[ a, 0] && PosQ[b/a]
Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(1 + b*(x^2/a))^(1/4)/( a + b*x^2)^(1/4) Int[1/(1 + b*(x^2/a))^(1/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1))) Int[ (c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
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]
\[\int \frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}}}{x^{3}}d x\]
Input:
int((b*x^4+a)^(3/4)/x^3,x)
Output:
int((b*x^4+a)^(3/4)/x^3,x)
\[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^3} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{x^{3}} \,d x } \] Input:
integrate((b*x^4+a)^(3/4)/x^3,x, algorithm="fricas")
Output:
integral((b*x^4 + a)^(3/4)/x^3, x)
Result contains complex when optimal does not.
Time = 0.57 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.33 \[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^3} \, dx=- \frac {a^{\frac {3}{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{2} \\ \frac {1}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{2 x^{2}} \] Input:
integrate((b*x**4+a)**(3/4)/x**3,x)
Output:
-a**(3/4)*hyper((-3/4, -1/2), (1/2,), b*x**4*exp_polar(I*pi)/a)/(2*x**2)
\[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^3} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{x^{3}} \,d x } \] Input:
integrate((b*x^4+a)^(3/4)/x^3,x, algorithm="maxima")
Output:
integrate((b*x^4 + a)^(3/4)/x^3, x)
\[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^3} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{\frac {3}{4}}}{x^{3}} \,d x } \] Input:
integrate((b*x^4+a)^(3/4)/x^3,x, algorithm="giac")
Output:
integrate((b*x^4 + a)^(3/4)/x^3, x)
Timed out. \[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^3} \, dx=\int \frac {{\left (b\,x^4+a\right )}^{3/4}}{x^3} \,d x \] Input:
int((a + b*x^4)^(3/4)/x^3,x)
Output:
int((a + b*x^4)^(3/4)/x^3, x)
\[ \int \frac {\left (a+b x^4\right )^{3/4}}{x^3} \, dx=\frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}}+3 \left (\int \frac {\left (b \,x^{4}+a \right )^{\frac {3}{4}}}{b \,x^{7}+a \,x^{3}}d x \right ) a \,x^{2}}{x^{2}} \] Input:
int((b*x^4+a)^(3/4)/x^3,x)
Output:
((a + b*x**4)**(3/4) + 3*int((a + b*x**4)**(3/4)/(a*x**3 + b*x**7),x)*a*x* *2)/x**2