Integrand size = 15, antiderivative size = 124 \[ \int \frac {1}{x^4 \sqrt [4]{a+b x^2}} \, dx=-\frac {b^2 x}{2 a^2 \sqrt [4]{a+b x^2}}-\frac {\left (a+b x^2\right )^{3/4}}{3 a x^3}+\frac {b \left (a+b x^2\right )^{3/4}}{2 a^2 x}+\frac {b^{3/2} \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{2 a^{3/2} \sqrt [4]{a+b x^2}} \] Output:
-1/2*b^2*x/a^2/(b*x^2+a)^(1/4)-1/3*(b*x^2+a)^(3/4)/a/x^3+1/2*b*(b*x^2+a)^( 3/4)/a^2/x+1/2*b^(3/2)*(1+b*x^2/a)^(1/4)*EllipticE(sin(1/2*arctan(b^(1/2)* x/a^(1/2))),2^(1/2))/a^(3/2)/(b*x^2+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 = 0.41 \[ \int \frac {1}{x^4 \sqrt [4]{a+b x^2}} \, dx=-\frac {\sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{4},-\frac {1}{2},-\frac {b x^2}{a}\right )}{3 x^3 \sqrt [4]{a+b x^2}} \] Input:
Integrate[1/(x^4*(a + b*x^2)^(1/4)),x]
Output:
-1/3*((1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[-3/2, 1/4, -1/2, -((b*x^2)/a )])/(x^3*(a + b*x^2)^(1/4))
Time = 0.21 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.06, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {264, 264, 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.
\(\displaystyle \int \frac {1}{x^4 \sqrt [4]{a+b x^2}} \, dx\) |
\(\Big \downarrow \) 264 |
\(\displaystyle -\frac {b \int \frac {1}{x^2 \sqrt [4]{b x^2+a}}dx}{2 a}-\frac {\left (a+b x^2\right )^{3/4}}{3 a x^3}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle -\frac {b \left (\frac {b \int \frac {1}{\sqrt [4]{b x^2+a}}dx}{2 a}-\frac {\left (a+b x^2\right )^{3/4}}{a x}\right )}{2 a}-\frac {\left (a+b x^2\right )^{3/4}}{3 a x^3}\) |
\(\Big \downarrow \) 227 |
\(\displaystyle -\frac {b \left (\frac {b \sqrt [4]{\frac {b x^2}{a}+1} \int \frac {1}{\sqrt [4]{\frac {b x^2}{a}+1}}dx}{2 a \sqrt [4]{a+b x^2}}-\frac {\left (a+b x^2\right )^{3/4}}{a x}\right )}{2 a}-\frac {\left (a+b x^2\right )^{3/4}}{3 a x^3}\) |
\(\Big \downarrow \) 225 |
\(\displaystyle -\frac {b \left (\frac {b \sqrt [4]{\frac {b x^2}{a}+1} \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{5/4}}dx\right )}{2 a \sqrt [4]{a+b x^2}}-\frac {\left (a+b x^2\right )^{3/4}}{a x}\right )}{2 a}-\frac {\left (a+b x^2\right )^{3/4}}{3 a x^3}\) |
\(\Big \downarrow \) 212 |
\(\displaystyle -\frac {b \left (\frac {b \sqrt [4]{\frac {b x^2}{a}+1} \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\frac {2 \sqrt {a} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b}}\right )}{2 a \sqrt [4]{a+b x^2}}-\frac {\left (a+b x^2\right )^{3/4}}{a x}\right )}{2 a}-\frac {\left (a+b x^2\right )^{3/4}}{3 a x^3}\) |
Input:
Int[1/(x^4*(a + b*x^2)^(1/4)),x]
Output:
-1/3*(a + b*x^2)^(3/4)/(a*x^3) - (b*(-((a + b*x^2)^(3/4)/(a*x)) + (b*(1 + (b*x^2)/a)^(1/4)*((2*x)/(1 + (b*x^2)/a)^(1/4) - (2*Sqrt[a]*EllipticE[ArcTa n[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/Sqrt[b]))/(2*a*(a + b*x^2)^(1/4))))/(2*a)
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 + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
\[\int \frac {1}{x^{4} \left (b \,x^{2}+a \right )^{\frac {1}{4}}}d x\]
Input:
int(1/x^4/(b*x^2+a)^(1/4),x)
Output:
int(1/x^4/(b*x^2+a)^(1/4),x)
\[ \int \frac {1}{x^4 \sqrt [4]{a+b x^2}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} x^{4}} \,d x } \] Input:
integrate(1/x^4/(b*x^2+a)^(1/4),x, algorithm="fricas")
Output:
integral((b*x^2 + a)^(3/4)/(b*x^6 + a*x^4), x)
Result contains complex when optimal does not.
Time = 0.54 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.26 \[ \int \frac {1}{x^4 \sqrt [4]{a+b x^2}} \, dx=- \frac {{{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {1}{4} \\ - \frac {1}{2} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{3 \sqrt [4]{a} x^{3}} \] Input:
integrate(1/x**4/(b*x**2+a)**(1/4),x)
Output:
-hyper((-3/2, 1/4), (-1/2,), b*x**2*exp_polar(I*pi)/a)/(3*a**(1/4)*x**3)
\[ \int \frac {1}{x^4 \sqrt [4]{a+b x^2}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} x^{4}} \,d x } \] Input:
integrate(1/x^4/(b*x^2+a)^(1/4),x, algorithm="maxima")
Output:
integrate(1/((b*x^2 + a)^(1/4)*x^4), x)
\[ \int \frac {1}{x^4 \sqrt [4]{a+b x^2}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} x^{4}} \,d x } \] Input:
integrate(1/x^4/(b*x^2+a)^(1/4),x, algorithm="giac")
Output:
integrate(1/((b*x^2 + a)^(1/4)*x^4), x)
Timed out. \[ \int \frac {1}{x^4 \sqrt [4]{a+b x^2}} \, dx=\int \frac {1}{x^4\,{\left (b\,x^2+a\right )}^{1/4}} \,d x \] Input:
int(1/(x^4*(a + b*x^2)^(1/4)),x)
Output:
int(1/(x^4*(a + b*x^2)^(1/4)), x)
\[ \int \frac {1}{x^4 \sqrt [4]{a+b x^2}} \, dx=\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} x^{4}}d x \] Input:
int(1/x^4/(b*x^2+a)^(1/4),x)
Output:
int(1/((a + b*x**2)**(1/4)*x**4),x)