Integrand size = 63, antiderivative size = 622 \[ \int \frac {\left (1+x^2\right ) \left (-a-b x+a x^2\right )}{x^2 \left (-c+d x+c x^2\right ) \sqrt [3]{\frac {-1+x^2-x c_0}{-1+x^2-x c_1}}} \, dx=\frac {\sqrt {3} (b c+a d) \arctan \left (\frac {\sqrt [3]{d+c c_0}-2 \sqrt [3]{-d-c c_1} \sqrt [3]{\frac {1-x^2+x c_0}{1-x^2+x c_1}}}{\sqrt {3} \sqrt [3]{d+c c_0}}\right ) \sqrt [3]{-d-c c_1}}{c^2 \sqrt [3]{d+c c_0}}+\frac {\arctan \left (\frac {1+2 \sqrt [3]{\frac {-1+x^2-x c_0}{-1+x^2-x c_1}}}{\sqrt {3}}\right ) \left (3 \sqrt {3} b c+3 \sqrt {3} a d-\sqrt {3} a c c_0+\sqrt {3} a c c_1\right )}{3 c^2}+\frac {\left (\frac {-1+x^2-x c_0}{-1+x^2-x c_1}\right ){}^{2/3} \left (-a+a x^2-a x c_1\right )}{c x}+\frac {(3 b c+3 a d-a c c_0+a c c_1) \log \left (-1+\sqrt [3]{\frac {-1+x^2-x c_0}{-1+x^2-x c_1}}\right )}{3 c^2}+\frac {(-3 b c-3 a d+a c c_0-a c c_1) \log \left (1+\sqrt [3]{\frac {-1+x^2-x c_0}{-1+x^2-x c_1}}+\left (\frac {-1+x^2-x c_0}{-1+x^2-x c_1}\right ){}^{2/3}\right )}{6 c^2}+\frac {(b c+a d) \sqrt [3]{-d-c c_1} \log \left (\sqrt [3]{d+c c_0}+\sqrt [3]{-d-c c_1} \sqrt [3]{\frac {1-x^2+x c_0}{1-x^2+x c_1}}\right )}{c^2 \sqrt [3]{d+c c_0}}-\frac {(b c+a d) \sqrt [3]{-d-c c_1} \log \left ((d+c c_0){}^{2/3}-\sqrt [3]{d+c c_0} \sqrt [3]{-d-c c_1} \sqrt [3]{\frac {1-x^2+x c_0}{1-x^2+x c_1}}+(-d-c c_1){}^{2/3} \left (\frac {1-x^2+x c_0}{1-x^2+x c_1}\right ){}^{2/3}\right )}{2 c^2 \sqrt [3]{d+c c_0}} \]
3^(1/2)*(a*d+b*c)*arctan(1/3*((_C0*c+d)^(1/3)-2*(-_C1*c-d)^(1/3)*((_C0*x-x ^2+1)/(_C1*x-x^2+1))^(1/3))*3^(1/2)/(_C0*c+d)^(1/3))*(-_C1*c-d)^(1/3)/c^2/ (_C0*c+d)^(1/3)+1/3*arctan(1/3*(1+2*((-_C0*x+x^2-1)/(-_C1*x+x^2-1))^(1/3)) *3^(1/2))*(3*3^(1/2)*b*c+3*3^(1/2)*a*d-3^(1/2)*a*c*_C0+3^(1/2)*a*c*_C1)/c^ 2+((-_C0*x+x^2-1)/(-_C1*x+x^2-1))^(2/3)*(-_C1*a*x+a*x^2-a)/c/x+1/3*(-_C0*a *c+_C1*a*c+3*a*d+3*b*c)*ln(-1+((-_C0*x+x^2-1)/(-_C1*x+x^2-1))^(1/3))/c^2+1 /6*(_C0*a*c-_C1*a*c-3*a*d-3*b*c)*ln(1+((-_C0*x+x^2-1)/(-_C1*x+x^2-1))^(1/3 )+((-_C0*x+x^2-1)/(-_C1*x+x^2-1))^(2/3))/c^2+(a*d+b*c)*(-_C1*c-d)^(1/3)*ln ((_C0*c+d)^(1/3)+(-_C1*c-d)^(1/3)*((_C0*x-x^2+1)/(_C1*x-x^2+1))^(1/3))/c^2 /(_C0*c+d)^(1/3)-1/2*(a*d+b*c)*(-_C1*c-d)^(1/3)*ln((_C0*c+d)^(2/3)-(_C0*c+ d)^(1/3)*(-_C1*c-d)^(1/3)*((_C0*x-x^2+1)/(_C1*x-x^2+1))^(1/3)+(-_C1*c-d)^( 2/3)*((_C0*x-x^2+1)/(_C1*x-x^2+1))^(2/3))/c^2/(_C0*c+d)^(1/3)
\[ \int \frac {\left (1+x^2\right ) \left (-a-b x+a x^2\right )}{x^2 \left (-c+d x+c x^2\right ) \sqrt [3]{\frac {-1+x^2-x c_0}{-1+x^2-x c_1}}} \, dx=\int \frac {\left (1+x^2\right ) \left (-a-b x+a x^2\right )}{x^2 \left (-c+d x+c x^2\right ) \sqrt [3]{\frac {-1+x^2-x c_0}{-1+x^2-x c_1}}} \, dx \]
Integrate[((1 + x^2)*(-a - b*x + a*x^2))/(x^2*(-c + d*x + c*x^2)*((-1 + x^ 2 - x*C[0])/(-1 + x^2 - x*C[1]))^(1/3)),x]
Integrate[((1 + x^2)*(-a - b*x + a*x^2))/(x^2*(-c + d*x + c*x^2)*((-1 + x^ 2 - x*C[0])/(-1 + x^2 - x*C[1]))^(1/3)), 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.
\(\displaystyle \int \frac {\left (x^2+1\right ) \left (a x^2-a-b x\right )}{x^2 \sqrt [3]{\frac {x^2-c_0 x-1}{x^2-c_1 x-1}} \left (c x^2-c+d x\right )} \, dx\) |
\(\Big \downarrow \) 7270 |
\(\displaystyle \frac {\sqrt [3]{x^2-c_0 x-1} \int \frac {\left (x^2+1\right ) \left (-a x^2+b x+a\right ) \sqrt [3]{x^2-c_1 x-1}}{x^2 \left (-c x^2-d x+c\right ) \sqrt [3]{x^2-c_0 x-1}}dx}{\sqrt [3]{x^2-c_1 x-1} \sqrt [3]{\frac {-x^2+c_0 x+1}{-x^2+c_1 x+1}}}\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \frac {\sqrt [3]{x^2-c_0 x-1} \int \left (\frac {\sqrt [3]{x^2-c_1 x-1} a}{c \sqrt [3]{x^2-c_0 x-1}}+\frac {\sqrt [3]{x^2-c_1 x-1} a}{c x^2 \sqrt [3]{x^2-c_0 x-1}}+\frac {(b c+a d) \sqrt [3]{x^2-c_1 x-1}}{c^2 x \sqrt [3]{x^2-c_0 x-1}}-\frac {(b c+a d) (d+2 c x) \sqrt [3]{x^2-c_1 x-1}}{c^2 \left (c x^2+d x-c\right ) \sqrt [3]{x^2-c_0 x-1}}\right )dx}{\sqrt [3]{x^2-c_1 x-1} \sqrt [3]{\frac {-x^2+c_0 x+1}{-x^2+c_1 x+1}}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt [3]{x^2-c_0 x-1} \left (-\frac {2 (a d+b c) \int \frac {\sqrt [3]{x^2-c_1 x-1}}{\left (d+2 c x-\sqrt {4 c^2+d^2}\right ) \sqrt [3]{x^2-c_0 x-1}}dx}{c}-\frac {2 (a d+b c) \int \frac {\sqrt [3]{x^2-c_1 x-1}}{\left (d+2 c x+\sqrt {4 c^2+d^2}\right ) \sqrt [3]{x^2-c_0 x-1}}dx}{c}+\frac {(a d+b c) \int \frac {\sqrt [3]{x^2-c_1 x-1}}{x \sqrt [3]{x^2-c_0 x-1}}dx}{c^2}+\frac {a \int \frac {\sqrt [3]{x^2-c_1 x-1}}{\sqrt [3]{x^2-c_0 x-1}}dx}{c}+\frac {a \int \frac {\sqrt [3]{x^2-c_1 x-1}}{x^2 \sqrt [3]{x^2-c_0 x-1}}dx}{c}\right )}{\sqrt [3]{x^2-c_1 x-1} \sqrt [3]{\frac {-x^2+c_0 x+1}{-x^2+c_1 x+1}}}\) |
Int[((1 + x^2)*(-a - b*x + a*x^2))/(x^2*(-c + d*x + c*x^2)*((-1 + x^2 - x* C[0])/(-1 + x^2 - x*C[1]))^(1/3)),x]
3.32.10.3.1 Defintions of rubi rules used
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]
\[\int \frac {\left (x^{2}+1\right ) \left (a \,x^{2}-b x -a \right )}{x^{2} \left (c \,x^{2}+d x -c \right ) \left (\frac {-\textit {\_C0} x +x^{2}-1}{-\textit {\_C1} x +x^{2}-1}\right )^{\frac {1}{3}}}d x\]
Timed out. \[ \int \frac {\left (1+x^2\right ) \left (-a-b x+a x^2\right )}{x^2 \left (-c+d x+c x^2\right ) \sqrt [3]{\frac {-1+x^2-x c_0}{-1+x^2-x c_1}}} \, dx=\text {Timed out} \]
integrate((x^2+1)*(a*x^2-b*x-a)/x^2/(c*x^2+d*x-c)/((-_C0*x+x^2-1)/(-_C1*x+ x^2-1))^(1/3),x, algorithm="fricas")
Timed out. \[ \int \frac {\left (1+x^2\right ) \left (-a-b x+a x^2\right )}{x^2 \left (-c+d x+c x^2\right ) \sqrt [3]{\frac {-1+x^2-x c_0}{-1+x^2-x c_1}}} \, dx=\text {Timed out} \]
integrate((x**2+1)*(a*x**2-b*x-a)/x**2/(c*x**2+d*x-c)/((-_C0*x+x**2-1)/(-_ C1*x+x**2-1))**(1/3),x)
\[ \text {Unable to display latex} \]
integrate((x^2+1)*(a*x^2-b*x-a)/x^2/(c*x^2+d*x-c)/((-_C0*x+x^2-1)/(-_C1*x+ x^2-1))^(1/3),x, algorithm="maxima")
integrate((a*x^2 - b*x - a)*(x^2 + 1)/((c*x^2 + d*x - c)*x^2*((_C0*x - x^2 + 1)/(_C1*x - x^2 + 1))^(1/3)), x)
\[ \text {Unable to display latex} \]
integrate((x^2+1)*(a*x^2-b*x-a)/x^2/(c*x^2+d*x-c)/((-_C0*x+x^2-1)/(-_C1*x+ x^2-1))^(1/3),x, algorithm="giac")
integrate((a*x^2 - b*x - a)*(x^2 + 1)/((c*x^2 + d*x - c)*x^2*((_C0*x - x^2 + 1)/(_C1*x - x^2 + 1))^(1/3)), x)
Timed out. \[ \int \frac {\left (1+x^2\right ) \left (-a-b x+a x^2\right )}{x^2 \left (-c+d x+c x^2\right ) \sqrt [3]{\frac {-1+x^2-x c_0}{-1+x^2-x c_1}}} \, dx=\int -\frac {\left (x^2+1\right )\,\left (-a\,x^2+b\,x+a\right )}{x^2\,{\left (\frac {-x^2+_{\mathrm {C0}}\,x+1}{-x^2+_{\mathrm {C1}}\,x+1}\right )}^{1/3}\,\left (c\,x^2+d\,x-c\right )} \,d x \]
int(-((x^2 + 1)*(a + b*x - a*x^2))/(x^2*((_C0*x - x^2 + 1)/(_C1*x - x^2 + 1))^(1/3)*(d*x - c + c*x^2)),x)