Integrand size = 22, antiderivative size = 144 \[ \int \frac {1}{x^4 \left (1+x^2\right ) \left (1+2 x^2\right )^{3/4}} \, dx=-\frac {\sqrt [4]{1+2 x^2}}{3 x^3}+\frac {8 \sqrt [4]{1+2 x^2}}{3 x}-\frac {\arctan \left (\frac {\sqrt {2} x \sqrt [4]{1+2 x^2}}{1+\sqrt {1+2 x^2}}\right )}{\sqrt {2}}-\frac {\text {arctanh}\left (\frac {1-\sqrt {1+2 x^2}}{\sqrt {2} x \sqrt [4]{1+2 x^2}}\right )}{\sqrt {2}}+\frac {11}{3} \sqrt {2} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\sqrt {2} x\right ),2\right ) \] Output:
-1/3*(2*x^2+1)^(1/4)/x^3+8/3*(2*x^2+1)^(1/4)/x-1/2*arctan(2^(1/2)*x*(2*x^2 +1)^(1/4)/(1+(2*x^2+1)^(1/2)))*2^(1/2)-1/2*arctanh(1/2*(1-(2*x^2+1)^(1/2)) *2^(1/2)/x/(2*x^2+1)^(1/4))*2^(1/2)+11/3*2^(1/2)*InverseJacobiAM(1/2*arcta n(x*2^(1/2)),2^(1/2))
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 11.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.19 \[ \int \frac {1}{x^4 \left (1+x^2\right ) \left (1+2 x^2\right )^{3/4}} \, dx=-\frac {\operatorname {AppellF1}\left (-\frac {3}{2},\frac {3}{4},1,-\frac {1}{2},-2 x^2,-x^2\right )}{3 x^3} \] Input:
Integrate[1/(x^4*(1 + x^2)*(1 + 2*x^2)^(3/4)),x]
Output:
-1/3*AppellF1[-3/2, 3/4, 1, -1/2, -2*x^2, -x^2]/x^3
Time = 0.27 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.99, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {352, 2009}
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.
\(\displaystyle \int \frac {1}{x^4 \left (x^2+1\right ) \left (2 x^2+1\right )^{3/4}} \, dx\) |
\(\Big \downarrow \) 352 |
\(\displaystyle \int \left (\frac {1}{\left (x^2+1\right ) \left (2 x^2+1\right )^{3/4}}-\frac {1}{x^2 \left (2 x^2+1\right )^{3/4}}+\frac {1}{x^4 \left (2 x^2+1\right )^{3/4}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\arctan \left (\frac {\sqrt {2 x^2+1}+1}{\sqrt {2} x \sqrt [4]{2 x^2+1}}\right )}{\sqrt {2}}+\frac {11}{3} \sqrt {2} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\sqrt {2} x\right ),2\right )-\frac {\text {arctanh}\left (\frac {1-\sqrt {2 x^2+1}}{\sqrt {2} x \sqrt [4]{2 x^2+1}}\right )}{\sqrt {2}}+\frac {8 \sqrt [4]{2 x^2+1}}{3 x}-\frac {\sqrt [4]{2 x^2+1}}{3 x^3}\) |
Input:
Int[1/(x^4*(1 + x^2)*(1 + 2*x^2)^(3/4)),x]
Output:
-1/3*(1 + 2*x^2)^(1/4)/x^3 + (8*(1 + 2*x^2)^(1/4))/(3*x) + ArcTan[(1 + Sqr t[1 + 2*x^2])/(Sqrt[2]*x*(1 + 2*x^2)^(1/4))]/Sqrt[2] - ArcTanh[(1 - Sqrt[1 + 2*x^2])/(Sqrt[2]*x*(1 + 2*x^2)^(1/4))]/Sqrt[2] + (11*Sqrt[2]*EllipticF[ ArcTan[Sqrt[2]*x]/2, 2])/3
Int[(x_)^(m_)/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol ] :> Int[ExpandIntegrand[x^m/((a + b*x^2)^(3/4)*(c + d*x^2)), x], x] /; Fre eQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0] && IntegerQ[m] && (PosQ[a] || In tegerQ[m/2])
\[\int \frac {1}{x^{4} \left (x^{2}+1\right ) \left (2 x^{2}+1\right )^{\frac {3}{4}}}d x\]
Input:
int(1/x^4/(x^2+1)/(2*x^2+1)^(3/4),x)
Output:
int(1/x^4/(x^2+1)/(2*x^2+1)^(3/4),x)
\[ \int \frac {1}{x^4 \left (1+x^2\right ) \left (1+2 x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (2 \, x^{2} + 1\right )}^{\frac {3}{4}} {\left (x^{2} + 1\right )} x^{4}} \,d x } \] Input:
integrate(1/x^4/(x^2+1)/(2*x^2+1)^(3/4),x, algorithm="fricas")
Output:
integral((2*x^2 + 1)^(1/4)/(2*x^8 + 3*x^6 + x^4), x)
\[ \int \frac {1}{x^4 \left (1+x^2\right ) \left (1+2 x^2\right )^{3/4}} \, dx=\int \frac {1}{x^{4} \left (x^{2} + 1\right ) \left (2 x^{2} + 1\right )^{\frac {3}{4}}}\, dx \] Input:
integrate(1/x**4/(x**2+1)/(2*x**2+1)**(3/4),x)
Output:
Integral(1/(x**4*(x**2 + 1)*(2*x**2 + 1)**(3/4)), x)
\[ \int \frac {1}{x^4 \left (1+x^2\right ) \left (1+2 x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (2 \, x^{2} + 1\right )}^{\frac {3}{4}} {\left (x^{2} + 1\right )} x^{4}} \,d x } \] Input:
integrate(1/x^4/(x^2+1)/(2*x^2+1)^(3/4),x, algorithm="maxima")
Output:
integrate(1/((2*x^2 + 1)^(3/4)*(x^2 + 1)*x^4), x)
\[ \int \frac {1}{x^4 \left (1+x^2\right ) \left (1+2 x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (2 \, x^{2} + 1\right )}^{\frac {3}{4}} {\left (x^{2} + 1\right )} x^{4}} \,d x } \] Input:
integrate(1/x^4/(x^2+1)/(2*x^2+1)^(3/4),x, algorithm="giac")
Output:
integrate(1/((2*x^2 + 1)^(3/4)*(x^2 + 1)*x^4), x)
Timed out. \[ \int \frac {1}{x^4 \left (1+x^2\right ) \left (1+2 x^2\right )^{3/4}} \, dx=\int \frac {1}{x^4\,\left (x^2+1\right )\,{\left (2\,x^2+1\right )}^{3/4}} \,d x \] Input:
int(1/(x^4*(x^2 + 1)*(2*x^2 + 1)^(3/4)),x)
Output:
int(1/(x^4*(x^2 + 1)*(2*x^2 + 1)^(3/4)), x)
\[ \int \frac {1}{x^4 \left (1+x^2\right ) \left (1+2 x^2\right )^{3/4}} \, dx=\int \frac {1}{\left (2 x^{2}+1\right )^{\frac {3}{4}} x^{6}+\left (2 x^{2}+1\right )^{\frac {3}{4}} x^{4}}d x \] Input:
int(1/x^4/(x^2+1)/(2*x^2+1)^(3/4),x)
Output:
int(1/((2*x**2 + 1)**(3/4)*x**6 + (2*x**2 + 1)**(3/4)*x**4),x)