Integrand size = 90, antiderivative size = 195 \[ \int \frac {\left (x^2 c_3-c_4\right ) \left (x+3 x^2 c_3+3 c_4\right )}{x \sqrt {\frac {x c_0+x^2 c_3+c_4}{x c_1+x^2 c_3+c_4}} \left (-x^2+x^4 c_3{}^2+2 x^2 c_3 c_4+c_4{}^2\right )} \, dx=6 \text {arctanh}\left (\sqrt {\frac {x c_0+x^2 c_3+c_4}{x c_1+x^2 c_3+c_4}}\right )-\frac {2 \arctan \left (\frac {\sqrt {1-c_0} \sqrt {-1+c_1} \sqrt {\frac {x c_0+x^2 c_3+c_4}{x c_1+x^2 c_3+c_4}}}{-1+c_0}\right ) \sqrt {-1+c_1}}{\sqrt {1-c_0}}-\frac {4 \arctan \left (\frac {\sqrt {-1-c_0} \sqrt {1+c_1} \sqrt {\frac {x c_0+x^2 c_3+c_4}{x c_1+x^2 c_3+c_4}}}{1+c_0}\right ) \sqrt {1+c_1}}{\sqrt {-1-c_0}} \]
6*arctanh(((_C3*x^2+_C0*x+_C4)/(_C3*x^2+_C1*x+_C4))^(1/2))-2*arctan((1-_C0 )^(1/2)*(-1+_C1)^(1/2)*((_C3*x^2+_C0*x+_C4)/(_C3*x^2+_C1*x+_C4))^(1/2)/(-1 +_C0))*(-1+_C1)^(1/2)/(1-_C0)^(1/2)-4*arctan((-1-_C0)^(1/2)*(1+_C1)^(1/2)* ((_C3*x^2+_C0*x+_C4)/(_C3*x^2+_C1*x+_C4))^(1/2)/(1+_C0))*(1+_C1)^(1/2)/(-1 -_C0)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(43491\) vs. \(2(195)=390\).
Time = 6.51 (sec) , antiderivative size = 43491, normalized size of antiderivative = 223.03 \[ \int \frac {\left (x^2 c_3-c_4\right ) \left (x+3 x^2 c_3+3 c_4\right )}{x \sqrt {\frac {x c_0+x^2 c_3+c_4}{x c_1+x^2 c_3+c_4}} \left (-x^2+x^4 c_3{}^2+2 x^2 c_3 c_4+c_4{}^2\right )} \, dx=\text {Result too large to show} \]
Integrate[((x^2*C[3] - C[4])*(x + 3*x^2*C[3] + 3*C[4]))/(x*Sqrt[(x*C[0] + x^2*C[3] + C[4])/(x*C[1] + x^2*C[3] + C[4])]*(-x^2 + x^4*C[3]^2 + 2*x^2*C[ 3]*C[4] + C[4]^2)),x]
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.
| \(\displaystyle \int \frac {\left (c_3 x^2-c_4\right ) \left (3 c_3 x^2+x+3 c_4\right )}{x \sqrt {\frac {c_3 x^2+c_0 x+c_4}{c_3 x^2+c_1 x+c_4}} \left (c_3{}^2 x^4-x^2+2 c_3 c_4 x^2+c_4{}^2\right )} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {\left (c_3 x^2-c_4\right ) \left (3 c_3 x^2+x+3 c_4\right )}{x \sqrt {\frac {c_3 x^2+c_0 x+c_4}{c_3 x^2+c_1 x+c_4}} \left (c_3{}^2 x^4+(-1+2 c_3 c_4) x^2+c_4{}^2\right )}dx\) |
| \(\Big \downarrow \) 7270 |
| \(\displaystyle \frac {\sqrt {c_3 x^2+c_0 x+c_4} \int \frac {\left (x^2 c_3-c_4\right ) \sqrt {c_3 x^2+c_1 x+c_4} \left (3 c_3 x^2+x+3 c_4\right )}{x \sqrt {c_3 x^2+c_0 x+c_4} \left (c_3{}^2 x^4-(1-2 c_3 c_4) x^2+c_4{}^2\right )}dx}{\sqrt {\frac {c_3 x^2+c_0 x+c_4}{c_3 x^2+c_1 x+c_4}} \sqrt {c_3 x^2+c_1 x+c_4}}\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \frac {\sqrt {c_3 x^2+c_0 x+c_4} \int \left (\frac {2 \sqrt {c_3 x^2+c_1 x+c_4} (2 x c_3-1)}{\left (c_3 x^2-x+c_4\right ) \sqrt {c_3 x^2+c_0 x+c_4}}-\frac {3 \sqrt {c_3 x^2+c_1 x+c_4}}{x \sqrt {c_3 x^2+c_0 x+c_4}}+\frac {(2 x c_3+1) \sqrt {c_3 x^2+c_1 x+c_4}}{\left (c_3 x^2+x+c_4\right ) \sqrt {c_3 x^2+c_0 x+c_4}}\right )dx}{\sqrt {\frac {c_3 x^2+c_0 x+c_4}{c_3 x^2+c_1 x+c_4}} \sqrt {c_3 x^2+c_1 x+c_4}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {c_3 x^2+c_0 x+c_4} \left (-3 \int \frac {\sqrt {c_3 x^2+c_1 x+c_4}}{x \sqrt {c_3 x^2+c_0 x+c_4}}dx+4 c_3 \int \frac {\sqrt {c_3 x^2+c_1 x+c_4}}{\sqrt {c_3 x^2+c_0 x+c_4} \left (2 x c_3-\sqrt {1-4 c_3 c_4}-1\right )}dx+2 c_3 \int \frac {\sqrt {c_3 x^2+c_1 x+c_4}}{\sqrt {c_3 x^2+c_0 x+c_4} \left (2 x c_3-\sqrt {1-4 c_3 c_4}+1\right )}dx+4 c_3 \int \frac {\sqrt {c_3 x^2+c_1 x+c_4}}{\sqrt {c_3 x^2+c_0 x+c_4} \left (2 x c_3+\sqrt {1-4 c_3 c_4}-1\right )}dx+2 c_3 \int \frac {\sqrt {c_3 x^2+c_1 x+c_4}}{\sqrt {c_3 x^2+c_0 x+c_4} \left (2 x c_3+\sqrt {1-4 c_3 c_4}+1\right )}dx\right )}{\sqrt {\frac {c_3 x^2+c_0 x+c_4}{c_3 x^2+c_1 x+c_4}} \sqrt {c_3 x^2+c_1 x+c_4}}\) |
Int[((x^2*C[3] - C[4])*(x + 3*x^2*C[3] + 3*C[4]))/(x*Sqrt[(x*C[0] + x^2*C[ 3] + C[4])/(x*C[1] + x^2*C[3] + C[4])]*(-x^2 + x^4*C[3]^2 + 2*x^2*C[3]*C[4 ] + C[4]^2)),x]
3.25.20.3.1 Defintions of rubi rules used
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.))^(p_), x_Symbol] :> Simp[a^IntPart[p ]*((a*v^m*w^n)^FracPart[p]/(v^(m*FracPart[p])*w^(n*FracPart[p]))) Int[u*v ^(m*p)*w^(n*p), x], x] /; FreeQ[{a, m, n, p}, x] && !IntegerQ[p] && !Free Q[v, x] && !FreeQ[w, x]
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 14.98 (sec) , antiderivative size = 35837869, normalized size of antiderivative = 183783.94
int((_C3*x^2-_C4)*(3*_C3*x^2+3*_C4+x)/x/((_C3*x^2+_C0*x+_C4)/(_C3*x^2+_C1* x+_C4))^(1/2)/(_C3^2*x^4+2*_C3*_C4*x^2+_C4^2-x^2),x,method=_RETURNVERBOSE)
Timed out. \[ \int \frac {\left (x^2 c_3-c_4\right ) \left (x+3 x^2 c_3+3 c_4\right )}{x \sqrt {\frac {x c_0+x^2 c_3+c_4}{x c_1+x^2 c_3+c_4}} \left (-x^2+x^4 c_3{}^2+2 x^2 c_3 c_4+c_4{}^2\right )} \, dx=\text {Timed out} \]
integrate((_C3*x^2-_C4)*(3*_C3*x^2+3*_C4+x)/x/((_C3*x^2+_C0*x+_C4)/(_C3*x^ 2+_C1*x+_C4))^(1/2)/(_C3^2*x^4+2*_C3*_C4*x^2+_C4^2-x^2),x, algorithm="fric as")
Timed out. \[ \int \frac {\left (x^2 c_3-c_4\right ) \left (x+3 x^2 c_3+3 c_4\right )}{x \sqrt {\frac {x c_0+x^2 c_3+c_4}{x c_1+x^2 c_3+c_4}} \left (-x^2+x^4 c_3{}^2+2 x^2 c_3 c_4+c_4{}^2\right )} \, dx=\text {Timed out} \]
integrate((_C3*x**2-_C4)*(3*_C3*x**2+3*_C4+x)/x/((_C3*x**2+_C0*x+_C4)/(_C3 *x**2+_C1*x+_C4))**(1/2)/(_C3**2*x**4+2*_C3*_C4*x**2+_C4**2-x**2),x)
\[ \text {Unable to display latex} \]
integrate((_C3*x^2-_C4)*(3*_C3*x^2+3*_C4+x)/x/((_C3*x^2+_C0*x+_C4)/(_C3*x^ 2+_C1*x+_C4))^(1/2)/(_C3^2*x^4+2*_C3*_C4*x^2+_C4^2-x^2),x, algorithm="maxi ma")
integrate((3*_C3*x^2 + 3*_C4 + x)*(_C3*x^2 - _C4)/((_C3^2*x^4 + 2*_C3*_C4* x^2 + _C4^2 - x^2)*x*sqrt((_C3*x^2 + _C0*x + _C4)/(_C3*x^2 + _C1*x + _C4)) ), x)
\[ \text {Unable to display latex} \]
integrate((_C3*x^2-_C4)*(3*_C3*x^2+3*_C4+x)/x/((_C3*x^2+_C0*x+_C4)/(_C3*x^ 2+_C1*x+_C4))^(1/2)/(_C3^2*x^4+2*_C3*_C4*x^2+_C4^2-x^2),x, algorithm="giac ")
integrate((3*_C3*x^2 + 3*_C4 + x)*(_C3*x^2 - _C4)/((_C3^2*x^4 + 2*_C3*_C4* x^2 + _C4^2 - x^2)*x*sqrt((_C3*x^2 + _C0*x + _C4)/(_C3*x^2 + _C1*x + _C4)) ), x)
Timed out. \[ \int \frac {\left (x^2 c_3-c_4\right ) \left (x+3 x^2 c_3+3 c_4\right )}{x \sqrt {\frac {x c_0+x^2 c_3+c_4}{x c_1+x^2 c_3+c_4}} \left (-x^2+x^4 c_3{}^2+2 x^2 c_3 c_4+c_4{}^2\right )} \, dx=\int -\frac {\left (_{\mathrm {C4}}-_{\mathrm {C3}}\,x^2\right )\,\left (3\,_{\mathrm {C3}}\,x^2+x+3\,_{\mathrm {C4}}\right )}{x\,\sqrt {\frac {_{\mathrm {C3}}\,x^2+_{\mathrm {C0}}\,x+_{\mathrm {C4}}}{_{\mathrm {C3}}\,x^2+_{\mathrm {C1}}\,x+_{\mathrm {C4}}}}\,\left ({_{\mathrm {C3}}}^2\,x^4+2\,_{\mathrm {C3}}\,_{\mathrm {C4}}\,x^2+{_{\mathrm {C4}}}^2-x^2\right )} \,d x \]
int(-((_C4 - _C3*x^2)*(3*_C4 + x + 3*_C3*x^2))/(x*((_C4 + _C0*x + _C3*x^2) /(_C4 + _C1*x + _C3*x^2))^(1/2)*(_C4^2 - x^2 + _C3^2*x^4 + 2*_C3*_C4*x^2)) ,x)