Integrand size = 98, antiderivative size = 259 \[ \int \frac {x^3 \left (x^3 c_3-2 c_4\right ) \sqrt {\frac {x^2 c_0+x^3 c_3+c_4}{x^2 c_1+x^3 c_3+c_4}}}{\left (x^2+2 x^3 c_3+2 c_4\right ) \left (-x^4+x^6 c_3{}^2+2 x^3 c_3 c_4+c_4{}^2\right )} \, dx=\frac {\arctan \left (\frac {\sqrt {1-c_0} \sqrt {-1+c_1} \sqrt {\frac {x^2 c_0+x^3 c_3+c_4}{x^2 c_1+x^3 c_3+c_4}}}{-1+c_0}\right ) \sqrt {1-c_0}}{\sqrt {-1+c_1}}+\frac {\arctan \left (\frac {\sqrt {-1-c_0} \sqrt {1+c_1} \sqrt {\frac {x^2 c_0+x^3 c_3+c_4}{x^2 c_1+x^3 c_3+c_4}}}{1+c_0}\right ) \sqrt {-1-c_0}}{3 \sqrt {1+c_1}}-\frac {4 \arctan \left (\frac {\sqrt {1-2 c_0} \sqrt {-1+2 c_1} \sqrt {\frac {x^2 c_0+x^3 c_3+c_4}{x^2 c_1+x^3 c_3+c_4}}}{-1+2 c_0}\right ) \sqrt {1-2 c_0}}{3 \sqrt {-1+2 c_1}} \]
arctan((1-_C0)^(1/2)*(-1+_C1)^(1/2)*((_C3*x^3+_C0*x^2+_C4)/(_C3*x^3+_C1*x^ 2+_C4))^(1/2)/(-1+_C0))*(1-_C0)^(1/2)/(-1+_C1)^(1/2)+1/3*arctan((-1-_C0)^( 1/2)*(1+_C1)^(1/2)*((_C3*x^3+_C0*x^2+_C4)/(_C3*x^3+_C1*x^2+_C4))^(1/2)/(1+ _C0))*(-1-_C0)^(1/2)/(1+_C1)^(1/2)-4/3*arctan((1-2*_C0)^(1/2)*(-1+2*_C1)^( 1/2)*((_C3*x^3+_C0*x^2+_C4)/(_C3*x^3+_C1*x^2+_C4))^(1/2)/(-1+2*_C0))*(1-2* _C0)^(1/2)/(-1+2*_C1)^(1/2)
\[ \int \frac {x^3 \left (x^3 c_3-2 c_4\right ) \sqrt {\frac {x^2 c_0+x^3 c_3+c_4}{x^2 c_1+x^3 c_3+c_4}}}{\left (x^2+2 x^3 c_3+2 c_4\right ) \left (-x^4+x^6 c_3{}^2+2 x^3 c_3 c_4+c_4{}^2\right )} \, dx=\int \frac {x^3 \left (x^3 c_3-2 c_4\right ) \sqrt {\frac {x^2 c_0+x^3 c_3+c_4}{x^2 c_1+x^3 c_3+c_4}}}{\left (x^2+2 x^3 c_3+2 c_4\right ) \left (-x^4+x^6 c_3{}^2+2 x^3 c_3 c_4+c_4{}^2\right )} \, dx \]
Integrate[(x^3*(x^3*C[3] - 2*C[4])*Sqrt[(x^2*C[0] + x^3*C[3] + C[4])/(x^2* C[1] + x^3*C[3] + C[4])])/((x^2 + 2*x^3*C[3] + 2*C[4])*(-x^4 + x^6*C[3]^2 + 2*x^3*C[3]*C[4] + C[4]^2)),x]
Integrate[(x^3*(x^3*C[3] - 2*C[4])*Sqrt[(x^2*C[0] + x^3*C[3] + C[4])/(x^2* C[1] + x^3*C[3] + C[4])])/((x^2 + 2*x^3*C[3] + 2*C[4])*(-x^4 + x^6*C[3]^2 + 2*x^3*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.
| \(\displaystyle \int \frac {x^3 \left (c_3 x^3-2 c_4\right ) \sqrt {\frac {c_3 x^3+c_0 x^2+c_4}{c_3 x^3+c_1 x^2+c_4}}}{\left (2 c_3 x^3+x^2+2 c_4\right ) \left (c_3{}^2 x^6-x^4+2 c_3 c_4 x^3+c_4{}^2\right )} \, dx\) |
| \(\Big \downarrow \) 2463 |
| \(\displaystyle \int \left (\frac {x^3 (1-c_3 x) \left (c_3 x^3-2 c_4\right ) \sqrt {\frac {c_3 x^3+c_0 x^2+c_4}{c_3 x^3+c_1 x^2+c_4}}}{2 c_4 \left (c_3 x^3-x^2+c_4\right ) \left (2 c_3 x^3+x^2+2 c_4\right )}+\frac {x^3 (1+c_3 x) \left (c_3 x^3-2 c_4\right ) \sqrt {\frac {c_3 x^3+c_0 x^2+c_4}{c_3 x^3+c_1 x^2+c_4}}}{2 c_4 \left (c_3 x^3+x^2+c_4\right ) \left (2 c_3 x^3+x^2+2 c_4\right )}\right )dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \frac {x^3 \sqrt {\frac {c_3 x^3+c_0 x^2+c_4}{c_3 x^3+c_1 x^2+c_4}} \left (2 c_4-c_3 x^3\right )}{\left (-c_3 x^3+x^2-c_4\right ) \left (c_3 x^3+x^2+c_4\right ) \left (2 c_3 x^3+x^2+2 c_4\right )}dx\) |
| \(\Big \downarrow \) 7270 |
| \(\displaystyle \frac {\sqrt {\frac {c_3 x^3+c_0 x^2+c_4}{c_3 x^3+c_1 x^2+c_4}} \sqrt {c_3 x^3+c_1 x^2+c_4} \int -\frac {x^3 \left (x^3 c_3-2 c_4\right ) \sqrt {c_3 x^3+c_0 x^2+c_4}}{\left (-c_3 x^3+x^2-c_4\right ) \left (c_3 x^3+x^2+c_4\right ) \sqrt {c_3 x^3+c_1 x^2+c_4} \left (2 c_3 x^3+x^2+2 c_4\right )}dx}{\sqrt {c_3 x^3+c_0 x^2+c_4}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\sqrt {\frac {c_3 x^3+c_0 x^2+c_4}{c_3 x^3+c_1 x^2+c_4}} \sqrt {c_3 x^3+c_1 x^2+c_4} \int \frac {x^3 \left (x^3 c_3-2 c_4\right ) \sqrt {c_3 x^3+c_0 x^2+c_4}}{\left (-c_3 x^3+x^2-c_4\right ) \left (c_3 x^3+x^2+c_4\right ) \sqrt {c_3 x^3+c_1 x^2+c_4} \left (2 c_3 x^3+x^2+2 c_4\right )}dx}{\sqrt {c_3 x^3+c_0 x^2+c_4}}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {\sqrt {\frac {c_3 x^3+c_0 x^2+c_4}{c_3 x^3+c_1 x^2+c_4}} \sqrt {c_3 x^3+c_1 x^2+c_4} \int \left (-\frac {x \sqrt {c_3 x^3+c_0 x^2+c_4} (3 x c_3-2)}{6 \left (c_3 x^3-x^2+c_4\right ) \sqrt {c_3 x^3+c_1 x^2+c_4}}-\frac {x (3 x c_3+2) \sqrt {c_3 x^3+c_0 x^2+c_4}}{2 \left (c_3 x^3+x^2+c_4\right ) \sqrt {c_3 x^3+c_1 x^2+c_4}}+\frac {4 x (3 x c_3+1) \sqrt {c_3 x^3+c_0 x^2+c_4}}{3 \sqrt {c_3 x^3+c_1 x^2+c_4} \left (2 c_3 x^3+x^2+2 c_4\right )}\right )dx}{\sqrt {c_3 x^3+c_0 x^2+c_4}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\sqrt {\frac {c_3 x^3+c_0 x^2+c_4}{c_3 x^3+c_1 x^2+c_4}} \sqrt {c_3 x^3+c_1 x^2+c_4} \left (\frac {1}{3} \int \frac {x \sqrt {c_3 x^3+c_0 x^2+c_4}}{\left (c_3 x^3-x^2+c_4\right ) \sqrt {c_3 x^3+c_1 x^2+c_4}}dx-\frac {1}{2} c_3 \int \frac {x^2 \sqrt {c_3 x^3+c_0 x^2+c_4}}{\left (c_3 x^3-x^2+c_4\right ) \sqrt {c_3 x^3+c_1 x^2+c_4}}dx-\int \frac {x \sqrt {c_3 x^3+c_0 x^2+c_4}}{\left (c_3 x^3+x^2+c_4\right ) \sqrt {c_3 x^3+c_1 x^2+c_4}}dx-\frac {3}{2} c_3 \int \frac {x^2 \sqrt {c_3 x^3+c_0 x^2+c_4}}{\left (c_3 x^3+x^2+c_4\right ) \sqrt {c_3 x^3+c_1 x^2+c_4}}dx+\frac {4}{3} \int \frac {x \sqrt {c_3 x^3+c_0 x^2+c_4}}{\sqrt {c_3 x^3+c_1 x^2+c_4} \left (2 c_3 x^3+x^2+2 c_4\right )}dx+4 c_3 \int \frac {x^2 \sqrt {c_3 x^3+c_0 x^2+c_4}}{\sqrt {c_3 x^3+c_1 x^2+c_4} \left (2 c_3 x^3+x^2+2 c_4\right )}dx\right )}{\sqrt {c_3 x^3+c_0 x^2+c_4}}\) |
Int[(x^3*(x^3*C[3] - 2*C[4])*Sqrt[(x^2*C[0] + x^3*C[3] + C[4])/(x^2*C[1] + x^3*C[3] + C[4])])/((x^2 + 2*x^3*C[3] + 2*C[4])*(-x^4 + x^6*C[3]^2 + 2*x^ 3*C[3]*C[4] + C[4]^2)),x]
3.28.59.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, 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]
\[\int \frac {x^{3} \left (\textit {\_C3} \,x^{3}-2 \textit {\_C4} \right ) \sqrt {\frac {\textit {\_C3} \,x^{3}+\textit {\_C0} \,x^{2}+\textit {\_C4}}{\textit {\_C3} \,x^{3}+\textit {\_C1} \,x^{2}+\textit {\_C4}}}}{\left (2 \textit {\_C3} \,x^{3}+x^{2}+2 \textit {\_C4} \right ) \left (\textit {\_C3}^{2} x^{6}+2 \textit {\_C3} \textit {\_C4} \,x^{3}-x^{4}+\textit {\_C4}^{2}\right )}d x\]
int(x^3*(_C3*x^3-2*_C4)*((_C3*x^3+_C0*x^2+_C4)/(_C3*x^3+_C1*x^2+_C4))^(1/2 )/(2*_C3*x^3+x^2+2*_C4)/(_C3^2*x^6+2*_C3*_C4*x^3-x^4+_C4^2),x)
int(x^3*(_C3*x^3-2*_C4)*((_C3*x^3+_C0*x^2+_C4)/(_C3*x^3+_C1*x^2+_C4))^(1/2 )/(2*_C3*x^3+x^2+2*_C4)/(_C3^2*x^6+2*_C3*_C4*x^3-x^4+_C4^2),x)
Timed out. \[ \int \frac {x^3 \left (x^3 c_3-2 c_4\right ) \sqrt {\frac {x^2 c_0+x^3 c_3+c_4}{x^2 c_1+x^3 c_3+c_4}}}{\left (x^2+2 x^3 c_3+2 c_4\right ) \left (-x^4+x^6 c_3{}^2+2 x^3 c_3 c_4+c_4{}^2\right )} \, dx=\text {Timed out} \]
integrate(x^3*(_C3*x^3-2*_C4)*((_C3*x^3+_C0*x^2+_C4)/(_C3*x^3+_C1*x^2+_C4) )^(1/2)/(2*_C3*x^3+x^2+2*_C4)/(_C3^2*x^6+2*_C3*_C4*x^3-x^4+_C4^2),x, algor ithm="fricas")
Timed out. \[ \int \frac {x^3 \left (x^3 c_3-2 c_4\right ) \sqrt {\frac {x^2 c_0+x^3 c_3+c_4}{x^2 c_1+x^3 c_3+c_4}}}{\left (x^2+2 x^3 c_3+2 c_4\right ) \left (-x^4+x^6 c_3{}^2+2 x^3 c_3 c_4+c_4{}^2\right )} \, dx=\text {Timed out} \]
integrate(x**3*(_C3*x**3-2*_C4)*((_C3*x**3+_C0*x**2+_C4)/(_C3*x**3+_C1*x** 2+_C4))**(1/2)/(2*_C3*x**3+x**2+2*_C4)/(_C3**2*x**6+2*_C3*_C4*x**3-x**4+_C 4**2),x)
\[ \text {Unable to display latex} \]
integrate(x^3*(_C3*x^3-2*_C4)*((_C3*x^3+_C0*x^2+_C4)/(_C3*x^3+_C1*x^2+_C4) )^(1/2)/(2*_C3*x^3+x^2+2*_C4)/(_C3^2*x^6+2*_C3*_C4*x^3-x^4+_C4^2),x, algor ithm="maxima")
integrate((_C3*x^3 - 2*_C4)*x^3*sqrt((_C3*x^3 + _C0*x^2 + _C4)/(_C3*x^3 + _C1*x^2 + _C4))/((_C3^2*x^6 + 2*_C3*_C4*x^3 - x^4 + _C4^2)*(2*_C3*x^3 + x^ 2 + 2*_C4)), x)
\[ \text {Unable to display latex} \]
integrate(x^3*(_C3*x^3-2*_C4)*((_C3*x^3+_C0*x^2+_C4)/(_C3*x^3+_C1*x^2+_C4) )^(1/2)/(2*_C3*x^3+x^2+2*_C4)/(_C3^2*x^6+2*_C3*_C4*x^3-x^4+_C4^2),x, algor ithm="giac")
Timed out. \[ \int \frac {x^3 \left (x^3 c_3-2 c_4\right ) \sqrt {\frac {x^2 c_0+x^3 c_3+c_4}{x^2 c_1+x^3 c_3+c_4}}}{\left (x^2+2 x^3 c_3+2 c_4\right ) \left (-x^4+x^6 c_3{}^2+2 x^3 c_3 c_4+c_4{}^2\right )} \, dx=\int -\frac {x^3\,\sqrt {\frac {_{\mathrm {C3}}\,x^3+_{\mathrm {C0}}\,x^2+_{\mathrm {C4}}}{_{\mathrm {C3}}\,x^3+_{\mathrm {C1}}\,x^2+_{\mathrm {C4}}}}\,\left (2\,_{\mathrm {C4}}-_{\mathrm {C3}}\,x^3\right )}{\left (2\,_{\mathrm {C3}}\,x^3+x^2+2\,_{\mathrm {C4}}\right )\,\left ({_{\mathrm {C3}}}^2\,x^6+2\,_{\mathrm {C3}}\,_{\mathrm {C4}}\,x^3+{_{\mathrm {C4}}}^2-x^4\right )} \,d x \]
int(-(x^3*((_C4 + _C0*x^2 + _C3*x^3)/(_C4 + _C1*x^2 + _C3*x^3))^(1/2)*(2*_ C4 - _C3*x^3))/((2*_C4 + 2*_C3*x^3 + x^2)*(_C4^2 - x^4 + _C3^2*x^6 + 2*_C3 *_C4*x^3)),x)