Integrand size = 90, antiderivative size = 250 \[ \int \frac {x \left (x^2 c_3-c_4\right )}{\sqrt {\frac {x c_0+x^2 c_3+c_4}{x c_1+x^2 c_3+c_4}} \left (x+3 x^2 c_3+3 c_4\right ) \left (-x^2+x^4 c_3{}^2+2 x^2 c_3 c_4+c_4{}^2\right )} \, dx=-\frac {\arctan \left (\frac {\sqrt {1-c_0} \sqrt {-1+c_1} \sqrt {\frac {x c_0+x^2 c_3+c_4}{x c_1+x^2 c_3+c_4}}}{-1+c_0}\right ) \sqrt {-1+c_1}}{2 \sqrt {1-c_0}}-\frac {\arctan \left (\frac {\sqrt {-1-c_0} \sqrt {1+c_1} \sqrt {\frac {x c_0+x^2 c_3+c_4}{x c_1+x^2 c_3+c_4}}}{1+c_0}\right ) \sqrt {1+c_1}}{4 \sqrt {-1-c_0}}+\frac {3 \arctan \left (\frac {\sqrt {1-3 c_0} \sqrt {-1+3 c_1} \sqrt {\frac {x c_0+x^2 c_3+c_4}{x c_1+x^2 c_3+c_4}}}{-1+3 c_0}\right ) \sqrt {-1+3 c_1}}{4 \sqrt {1-3 c_0}} \]
-1/2*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)-1/4*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)+3/4*arctan((1-3*_C0)^(1/2)*(-1+3*_C1)^(1/2 )*((_C3*x^2+_C0*x+_C4)/(_C3*x^2+_C1*x+_C4))^(1/2)/(-1+3*_C0))*(-1+3*_C1)^( 1/2)/(1-3*_C0)^(1/2)
Leaf count is larger than twice the leaf count of optimal. \(57413\) vs. \(2(250)=500\).
Time = 6.61 (sec) , antiderivative size = 57413, normalized size of antiderivative = 229.65 \[ \int \frac {x \left (x^2 c_3-c_4\right )}{\sqrt {\frac {x c_0+x^2 c_3+c_4}{x c_1+x^2 c_3+c_4}} \left (x+3 x^2 c_3+3 c_4\right ) \left (-x^2+x^4 c_3{}^2+2 x^2 c_3 c_4+c_4{}^2\right )} \, dx=\text {Result too large to show} \]
Integrate[(x*(x^2*C[3] - C[4]))/(Sqrt[(x*C[0] + x^2*C[3] + C[4])/(x*C[1] + x^2*C[3] + C[4])]*(x + 3*x^2*C[3] + 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.
| \(\displaystyle \int \frac {x \left (c_3 x^2-c_4\right )}{\sqrt {\frac {c_3 x^2+c_0 x+c_4}{c_3 x^2+c_1 x+c_4}} \left (3 c_3 x^2+x+3 c_4\right ) \left (c_3{}^2 x^4-x^2+2 c_3 c_4 x^2+c_4{}^2\right )} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {x \left (c_3 x^2-c_4\right )}{\sqrt {\frac {c_3 x^2+c_0 x+c_4}{c_3 x^2+c_1 x+c_4}} \left (3 c_3 x^2+x+3 c_4\right ) \left (c_3{}^2 x^4+(-1+2 c_3 c_4) x^2+c_4{}^2\right )}dx\) |
| \(\Big \downarrow \) 7270 |
| \(\displaystyle \frac {\sqrt {c_3 x^2+c_0 x+c_4} \int \frac {x \left (x^2 c_3-c_4\right ) \sqrt {c_3 x^2+c_1 x+c_4}}{\sqrt {c_3 x^2+c_0 x+c_4} \left (3 c_3 x^2+x+3 c_4\right ) \left (c_3{}^2 x^4-(1-2 c_3 c_4) x^2+c_4{}^2\right )}dx}{\sqrt {\frac {c_3 x^2+c_0 x+c_4}{c_3 x^2+c_1 x+c_4}} \sqrt {c_3 x^2+c_1 x+c_4}}\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \frac {\sqrt {c_3 x^2+c_0 x+c_4} \int \left (\frac {\sqrt {c_3 x^2+c_1 x+c_4} (2 x c_3-1)}{8 \left (c_3 x^2-x+c_4\right ) \sqrt {c_3 x^2+c_0 x+c_4}}+\frac {(2 x c_3+1) \sqrt {c_3 x^2+c_1 x+c_4}}{4 \left (c_3 x^2+x+c_4\right ) \sqrt {c_3 x^2+c_0 x+c_4}}-\frac {3 (6 x c_3+1) \sqrt {c_3 x^2+c_1 x+c_4}}{8 \sqrt {c_3 x^2+c_0 x+c_4} \left (3 c_3 x^2+x+3 c_4\right )}\right )dx}{\sqrt {\frac {c_3 x^2+c_0 x+c_4}{c_3 x^2+c_1 x+c_4}} \sqrt {c_3 x^2+c_1 x+c_4}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {c_3 x^2+c_0 x+c_4} \left (-\frac {9}{4} c_3 \int \frac {\sqrt {c_3 x^2+c_1 x+c_4}}{\sqrt {c_3 x^2+c_0 x+c_4} \left (6 x c_3-\sqrt {1-36 c_3 c_4}+1\right )}dx-\frac {9}{4} c_3 \int \frac {\sqrt {c_3 x^2+c_1 x+c_4}}{\sqrt {c_3 x^2+c_0 x+c_4} \left (6 x c_3+\sqrt {1-36 c_3 c_4}+1\right )}dx+\frac {1}{4} c_3 \int \frac {\sqrt {c_3 x^2+c_1 x+c_4}}{\sqrt {c_3 x^2+c_0 x+c_4} \left (2 x c_3-\sqrt {1-4 c_3 c_4}-1\right )}dx+\frac {1}{2} c_3 \int \frac {\sqrt {c_3 x^2+c_1 x+c_4}}{\sqrt {c_3 x^2+c_0 x+c_4} \left (2 x c_3-\sqrt {1-4 c_3 c_4}+1\right )}dx+\frac {1}{4} c_3 \int \frac {\sqrt {c_3 x^2+c_1 x+c_4}}{\sqrt {c_3 x^2+c_0 x+c_4} \left (2 x c_3+\sqrt {1-4 c_3 c_4}-1\right )}dx+\frac {1}{2} c_3 \int \frac {\sqrt {c_3 x^2+c_1 x+c_4}}{\sqrt {c_3 x^2+c_0 x+c_4} \left (2 x c_3+\sqrt {1-4 c_3 c_4}+1\right )}dx\right )}{\sqrt {\frac {c_3 x^2+c_0 x+c_4}{c_3 x^2+c_1 x+c_4}} \sqrt {c_3 x^2+c_1 x+c_4}}\) |
Int[(x*(x^2*C[3] - C[4]))/(Sqrt[(x*C[0] + x^2*C[3] + C[4])/(x*C[1] + x^2*C [3] + C[4])]*(x + 3*x^2*C[3] + 3*C[4])*(-x^2 + x^4*C[3]^2 + 2*x^2*C[3]*C[4 ] + C[4]^2)),x]
3.28.28.3.1 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]
Leaf count of result is larger than twice the leaf count of optimal. \(942\) vs. \(2(181)=362\).
Time = 8.22 (sec) , antiderivative size = 943, normalized size of antiderivative = 3.77
int(x*(_C3*x^2-_C4)/((_C3*x^2+_C0*x+_C4)/(_C3*x^2+_C1*x+_C4))^(1/2)/(3*_C3 *x^2+3*_C4+x)/(_C3^2*x^4+2*_C3*_C4*x^2+_C4^2-x^2),x,method=_RETURNVERBOSE)
-1/8*(_C3*x^2+_C0*x+_C4)*(((-1+3*_C1)*(-1+3*_C0))^(1/2)*ln((x^2*_C0*_C3+x^ 2*_C1*_C3+2*x^2*_C3+2*x*_C0*_C1+2*((1+_C1)*(1+_C0))^(1/2)*((_C3*x^2+_C1*x+ _C4)*(_C3*x^2+_C0*x+_C4))^(1/2)+x*_C0+x*_C1+_C0*_C4+_C1*_C4+2*_C4)*_C3/(_C 3*x^2+_C4-x))*((1+_C1)*(1+_C0))^(1/2)*_C0+2*((-1+3*_C1)*(-1+3*_C0))^(1/2)* ln((x^2*_C0*_C3+x^2*_C1*_C3-2*x^2*_C3+2*x*_C0*_C1+2*((-1+_C1)*(-1+_C0))^(1 /2)*((_C3*x^2+_C1*x+_C4)*(_C3*x^2+_C0*x+_C4))^(1/2)-x*_C0-x*_C1+_C0*_C4+_C 1*_C4-2*_C4)*_C3/(_C3*x^2+_C4+x))*((-1+_C1)*(-1+_C0))^(1/2)*_C0-9*ln((3*x^ 2*_C0*_C3+3*x^2*_C1*_C3-2*x^2*_C3+6*x*_C0*_C1+2*((-1+3*_C1)*(-1+3*_C0))^(1 /2)*((_C3*x^2+_C1*x+_C4)*(_C3*x^2+_C0*x+_C4))^(1/2)-x*_C0-x*_C1+3*_C0*_C4+ 3*_C1*_C4-2*_C4)*_C3/(3*_C3*x^2+3*_C4+x))*_C0^2*_C1-((-1+3*_C1)*(-1+3*_C0) )^(1/2)*ln((x^2*_C0*_C3+x^2*_C1*_C3+2*x^2*_C3+2*x*_C0*_C1+2*((1+_C1)*(1+_C 0))^(1/2)*((_C3*x^2+_C1*x+_C4)*(_C3*x^2+_C0*x+_C4))^(1/2)+x*_C0+x*_C1+_C0* _C4+_C1*_C4+2*_C4)*_C3/(_C3*x^2+_C4-x))*((1+_C1)*(1+_C0))^(1/2)+2*((-1+3*_ C1)*(-1+3*_C0))^(1/2)*ln((x^2*_C0*_C3+x^2*_C1*_C3-2*x^2*_C3+2*x*_C0*_C1+2* ((-1+_C1)*(-1+_C0))^(1/2)*((_C3*x^2+_C1*x+_C4)*(_C3*x^2+_C0*x+_C4))^(1/2)- x*_C0-x*_C1+_C0*_C4+_C1*_C4-2*_C4)*_C3/(_C3*x^2+_C4+x))*((-1+_C1)*(-1+_C0) )^(1/2)+3*ln((3*x^2*_C0*_C3+3*x^2*_C1*_C3-2*x^2*_C3+6*x*_C0*_C1+2*((-1+3*_ C1)*(-1+3*_C0))^(1/2)*((_C3*x^2+_C1*x+_C4)*(_C3*x^2+_C0*x+_C4))^(1/2)-x*_C 0-x*_C1+3*_C0*_C4+3*_C1*_C4-2*_C4)*_C3/(3*_C3*x^2+3*_C4+x))*_C0^2+9*ln((3* x^2*_C0*_C3+3*x^2*_C1*_C3-2*x^2*_C3+6*x*_C0*_C1+2*((-1+3*_C1)*(-1+3*_C0...
Time = 34.26 (sec) , antiderivative size = 6425, normalized size of antiderivative = 25.70 \[ \int \frac {x \left (x^2 c_3-c_4\right )}{\sqrt {\frac {x c_0+x^2 c_3+c_4}{x c_1+x^2 c_3+c_4}} \left (x+3 x^2 c_3+3 c_4\right ) \left (-x^2+x^4 c_3{}^2+2 x^2 c_3 c_4+c_4{}^2\right )} \, dx=\text {Too large to display} \]
integrate(x*(_C3*x^2-_C4)/((_C3*x^2+_C0*x+_C4)/(_C3*x^2+_C1*x+_C4))^(1/2)/ (3*_C3*x^2+3*_C4+x)/(_C3^2*x^4+2*_C3*_C4*x^2+_C4^2-x^2),x, algorithm="fric as")
Timed out. \[ \int \frac {x \left (x^2 c_3-c_4\right )}{\sqrt {\frac {x c_0+x^2 c_3+c_4}{x c_1+x^2 c_3+c_4}} \left (x+3 x^2 c_3+3 c_4\right ) \left (-x^2+x^4 c_3{}^2+2 x^2 c_3 c_4+c_4{}^2\right )} \, dx=\text {Timed out} \]
integrate(x*(_C3*x**2-_C4)/((_C3*x**2+_C0*x+_C4)/(_C3*x**2+_C1*x+_C4))**(1 /2)/(3*_C3*x**2+3*_C4+x)/(_C3**2*x**4+2*_C3*_C4*x**2+_C4**2-x**2),x)
\[ \text {Unable to display latex} \]
integrate(x*(_C3*x^2-_C4)/((_C3*x^2+_C0*x+_C4)/(_C3*x^2+_C1*x+_C4))^(1/2)/ (3*_C3*x^2+3*_C4+x)/(_C3^2*x^4+2*_C3*_C4*x^2+_C4^2-x^2),x, algorithm="maxi ma")
integrate((_C3*x^2 - _C4)*x/((_C3^2*x^4 + 2*_C3*_C4*x^2 + _C4^2 - x^2)*(3* _C3*x^2 + 3*_C4 + x)*sqrt((_C3*x^2 + _C0*x + _C4)/(_C3*x^2 + _C1*x + _C4)) ), x)
\[ \text {Unable to display latex} \]
integrate(x*(_C3*x^2-_C4)/((_C3*x^2+_C0*x+_C4)/(_C3*x^2+_C1*x+_C4))^(1/2)/ (3*_C3*x^2+3*_C4+x)/(_C3^2*x^4+2*_C3*_C4*x^2+_C4^2-x^2),x, algorithm="giac ")
integrate((_C3*x^2 - _C4)*x/((_C3^2*x^4 + 2*_C3*_C4*x^2 + _C4^2 - x^2)*(3* _C3*x^2 + 3*_C4 + x)*sqrt((_C3*x^2 + _C0*x + _C4)/(_C3*x^2 + _C1*x + _C4)) ), x)
Timed out. \[ \int \frac {x \left (x^2 c_3-c_4\right )}{\sqrt {\frac {x c_0+x^2 c_3+c_4}{x c_1+x^2 c_3+c_4}} \left (x+3 x^2 c_3+3 c_4\right ) \left (-x^2+x^4 c_3{}^2+2 x^2 c_3 c_4+c_4{}^2\right )} \, dx=-\int \frac {x\,\left (_{\mathrm {C4}}-_{\mathrm {C3}}\,x^2\right )}{\sqrt {\frac {_{\mathrm {C3}}\,x^2+_{\mathrm {C0}}\,x+_{\mathrm {C4}}}{_{\mathrm {C3}}\,x^2+_{\mathrm {C1}}\,x+_{\mathrm {C4}}}}\,\left (3\,_{\mathrm {C3}}\,x^2+x+3\,_{\mathrm {C4}}\right )\,\left ({_{\mathrm {C3}}}^2\,x^4+2\,_{\mathrm {C3}}\,_{\mathrm {C4}}\,x^2+{_{\mathrm {C4}}}^2-x^2\right )} \,d x \]
int(-(x*(_C4 - _C3*x^2))/(((_C4 + _C0*x + _C3*x^2)/(_C4 + _C1*x + _C3*x^2) )^(1/2)*(3*_C4 + x + 3*_C3*x^2)*(_C4^2 - x^2 + _C3^2*x^4 + 2*_C3*_C4*x^2)) ,x)