Integrand size = 57, antiderivative size = 514 \[ \int \frac {\left (-3+x^2\right ) \left (1-2 x^2+x^4+x^6\right )}{x^{10} \sqrt [4]{\frac {-a+a x^2+b x^3}{-c+c x^2+d x^3}}} \, dx=\frac {\left (\frac {-a+a x^2+b x^3}{-c+c x^2+d x^3}\right )^{3/4} \left (32 a^2 c^3-96 a^2 c^3 x^2+36 a b c^3 x^3-36 a^2 c^2 d x^3+96 a^2 c^3 x^4-72 a b c^3 x^5+72 a^2 c^2 d x^5+64 a^2 c^3 x^6+45 b^2 c^3 x^6-42 a b c^2 d x^6-3 a^2 c d^2 x^6+36 a b c^3 x^7-36 a^2 c^2 d x^7-96 a^2 c^3 x^8-45 b^2 c^3 x^8+42 a b c^2 d x^8+3 a^2 c d^2 x^8-96 a^2 c^2 d x^9-45 b^2 c^2 d x^9+6 a b c d^2 x^9+7 a^2 d^3 x^9\right )}{96 a^3 c^2 x^9}+\frac {\left (-32 a^2 b c^3-15 b^3 c^3+32 a^3 c^2 d+5 a b^2 c^2 d+3 a^2 b c d^2+7 a^3 d^3\right ) \arctan \left (\frac {\sqrt [4]{c} \sqrt [4]{\frac {-a+a x^2+b x^3}{-c+c x^2+d x^3}}}{\sqrt [4]{a}}\right )}{64 a^{13/4} c^{11/4}}+\frac {\left (32 a^2 b c^3+15 b^3 c^3-32 a^3 c^2 d-5 a b^2 c^2 d-3 a^2 b c d^2-7 a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt [4]{c} \sqrt [4]{\frac {-a+a x^2+b x^3}{-c+c x^2+d x^3}}}{\sqrt [4]{a}}\right )}{64 a^{13/4} c^{11/4}} \]
1/96*((b*x^3+a*x^2-a)/(d*x^3+c*x^2-c))^(3/4)*(-96*a^2*c^2*d*x^9+7*a^2*d^3* x^9+6*a*b*c*d^2*x^9-45*b^2*c^2*d*x^9-96*a^2*c^3*x^8+3*a^2*c*d^2*x^8+42*a*b *c^2*d*x^8-45*b^2*c^3*x^8-36*a^2*c^2*d*x^7+36*a*b*c^3*x^7+64*a^2*c^3*x^6-3 *a^2*c*d^2*x^6-42*a*b*c^2*d*x^6+45*b^2*c^3*x^6+72*a^2*c^2*d*x^5-72*a*b*c^3 *x^5+96*a^2*c^3*x^4-36*a^2*c^2*d*x^3+36*a*b*c^3*x^3-96*a^2*c^3*x^2+32*a^2* c^3)/a^3/c^2/x^9+1/64*(32*a^3*c^2*d+7*a^3*d^3-32*a^2*b*c^3+3*a^2*b*c*d^2+5 *a*b^2*c^2*d-15*b^3*c^3)*arctan(c^(1/4)*((b*x^3+a*x^2-a)/(d*x^3+c*x^2-c))^ (1/4)/a^(1/4))/a^(13/4)/c^(11/4)+1/64*(-32*a^3*c^2*d-7*a^3*d^3+32*a^2*b*c^ 3-3*a^2*b*c*d^2-5*a*b^2*c^2*d+15*b^3*c^3)*arctanh(c^(1/4)*((b*x^3+a*x^2-a) /(d*x^3+c*x^2-c))^(1/4)/a^(1/4))/a^(13/4)/c^(11/4)
Time = 16.59 (sec) , antiderivative size = 298, normalized size of antiderivative = 0.58 \[ \int \frac {\left (-3+x^2\right ) \left (1-2 x^2+x^4+x^6\right )}{x^{10} \sqrt [4]{\frac {-a+a x^2+b x^3}{-c+c x^2+d x^3}}} \, dx=-\frac {\left (\frac {b x^3+a \left (-1+x^2\right )}{d x^3+c \left (-1+x^2\right )}\right )^{3/4} \left (d x^3+c \left (-1+x^2\right )\right ) \left (45 b^2 c^2 x^6-6 a b c x^3 \left (d x^3+6 c \left (-1+x^2\right )\right )+a^2 \left (-7 d^2 x^6+4 c d x^3 \left (-1+x^2\right )+32 c^2 \left (1-2 x^2+x^4+3 x^6\right )\right )\right )}{96 a^3 c^2 x^9}+\frac {\left (-15 b^3 c^3+5 a b^2 c^2 d+a^2 b \left (-32 c^3+3 c d^2\right )+a^3 \left (32 c^2 d+7 d^3\right )\right ) \left (\arctan \left (\frac {\sqrt [4]{c} \sqrt [4]{\frac {b x^3+a \left (-1+x^2\right )}{d x^3+c \left (-1+x^2\right )}}}{\sqrt [4]{a}}\right )-\text {arctanh}\left (\frac {\sqrt [4]{c} \sqrt [4]{\frac {b x^3+a \left (-1+x^2\right )}{d x^3+c \left (-1+x^2\right )}}}{\sqrt [4]{a}}\right )\right )}{64 a^{13/4} c^{11/4}} \]
Integrate[((-3 + x^2)*(1 - 2*x^2 + x^4 + x^6))/(x^10*((-a + a*x^2 + b*x^3) /(-c + c*x^2 + d*x^3))^(1/4)),x]
-1/96*(((b*x^3 + a*(-1 + x^2))/(d*x^3 + c*(-1 + x^2)))^(3/4)*(d*x^3 + c*(- 1 + x^2))*(45*b^2*c^2*x^6 - 6*a*b*c*x^3*(d*x^3 + 6*c*(-1 + x^2)) + a^2*(-7 *d^2*x^6 + 4*c*d*x^3*(-1 + x^2) + 32*c^2*(1 - 2*x^2 + x^4 + 3*x^6))))/(a^3 *c^2*x^9) + ((-15*b^3*c^3 + 5*a*b^2*c^2*d + a^2*b*(-32*c^3 + 3*c*d^2) + a^ 3*(32*c^2*d + 7*d^3))*(ArcTan[(c^(1/4)*((b*x^3 + a*(-1 + x^2))/(d*x^3 + c* (-1 + x^2)))^(1/4))/a^(1/4)] - ArcTanh[(c^(1/4)*((b*x^3 + a*(-1 + x^2))/(d *x^3 + c*(-1 + x^2)))^(1/4))/a^(1/4)]))/(64*a^(13/4)*c^(11/4))
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.
| \(\displaystyle \int \frac {\left (x^2-3\right ) \left (x^6+x^4-2 x^2+1\right )}{x^{10} \sqrt [4]{\frac {a x^2-a+b x^3}{c x^2-c+d x^3}}} \, dx\) |
| \(\Big \downarrow \) 7270 |
| \(\displaystyle \frac {\sqrt [4]{a x^2-a+b x^3} \int -\frac {\left (3-x^2\right ) \sqrt [4]{d x^3+c x^2-c} \left (x^6+x^4-2 x^2+1\right )}{x^{10} \sqrt [4]{b x^3+a x^2-a}}dx}{\sqrt [4]{c x^2-c+d x^3} \sqrt [4]{\frac {-a x^2+a-b x^3}{-c x^2+c-d x^3}}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\sqrt [4]{a x^2-a+b x^3} \int \frac {\left (3-x^2\right ) \sqrt [4]{d x^3+c x^2-c} \left (x^6+x^4-2 x^2+1\right )}{x^{10} \sqrt [4]{b x^3+a x^2-a}}dx}{\sqrt [4]{c x^2-c+d x^3} \sqrt [4]{\frac {-a x^2+a-b x^3}{-c x^2+c-d x^3}}}\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle -\frac {\sqrt [4]{a x^2-a+b x^3} \int \left (-\frac {\sqrt [4]{d x^3+c x^2-c}}{x^2 \sqrt [4]{b x^3+a x^2-a}}+\frac {2 \sqrt [4]{d x^3+c x^2-c}}{x^4 \sqrt [4]{b x^3+a x^2-a}}+\frac {5 \sqrt [4]{d x^3+c x^2-c}}{x^6 \sqrt [4]{b x^3+a x^2-a}}-\frac {7 \sqrt [4]{d x^3+c x^2-c}}{x^8 \sqrt [4]{b x^3+a x^2-a}}+\frac {3 \sqrt [4]{d x^3+c x^2-c}}{x^{10} \sqrt [4]{b x^3+a x^2-a}}\right )dx}{\sqrt [4]{c x^2-c+d x^3} \sqrt [4]{\frac {-a x^2+a-b x^3}{-c x^2+c-d x^3}}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\sqrt [4]{a x^2-a+b x^3} \left (-\int \frac {\sqrt [4]{d x^3+c x^2-c}}{x^2 \sqrt [4]{b x^3+a x^2-a}}dx+3 \int \frac {\sqrt [4]{d x^3+c x^2-c}}{x^{10} \sqrt [4]{b x^3+a x^2-a}}dx-7 \int \frac {\sqrt [4]{d x^3+c x^2-c}}{x^8 \sqrt [4]{b x^3+a x^2-a}}dx+5 \int \frac {\sqrt [4]{d x^3+c x^2-c}}{x^6 \sqrt [4]{b x^3+a x^2-a}}dx+2 \int \frac {\sqrt [4]{d x^3+c x^2-c}}{x^4 \sqrt [4]{b x^3+a x^2-a}}dx\right )}{\sqrt [4]{c x^2-c+d x^3} \sqrt [4]{\frac {-a x^2+a-b x^3}{-c x^2+c-d x^3}}}\) |
Int[((-3 + x^2)*(1 - 2*x^2 + x^4 + x^6))/(x^10*((-a + a*x^2 + b*x^3)/(-c + c*x^2 + d*x^3))^(1/4)),x]
3.31.80.3.1 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]
\[\int \frac {\left (x^{2}-3\right ) \left (x^{6}+x^{4}-2 x^{2}+1\right )}{x^{10} \left (\frac {b \,x^{3}+a \,x^{2}-a}{d \,x^{3}+c \,x^{2}-c}\right )^{\frac {1}{4}}}d x\]
Timed out. \[ \int \frac {\left (-3+x^2\right ) \left (1-2 x^2+x^4+x^6\right )}{x^{10} \sqrt [4]{\frac {-a+a x^2+b x^3}{-c+c x^2+d x^3}}} \, dx=\text {Timed out} \]
integrate((x^2-3)*(x^6+x^4-2*x^2+1)/x^10/((b*x^3+a*x^2-a)/(d*x^3+c*x^2-c)) ^(1/4),x, algorithm="fricas")
Timed out. \[ \int \frac {\left (-3+x^2\right ) \left (1-2 x^2+x^4+x^6\right )}{x^{10} \sqrt [4]{\frac {-a+a x^2+b x^3}{-c+c x^2+d x^3}}} \, dx=\text {Timed out} \]
integrate((x**2-3)*(x**6+x**4-2*x**2+1)/x**10/((b*x**3+a*x**2-a)/(d*x**3+c *x**2-c))**(1/4),x)
\[ \int \frac {\left (-3+x^2\right ) \left (1-2 x^2+x^4+x^6\right )}{x^{10} \sqrt [4]{\frac {-a+a x^2+b x^3}{-c+c x^2+d x^3}}} \, dx=\int { \frac {{\left (x^{6} + x^{4} - 2 \, x^{2} + 1\right )} {\left (x^{2} - 3\right )}}{x^{10} \left (\frac {b x^{3} + a x^{2} - a}{d x^{3} + c x^{2} - c}\right )^{\frac {1}{4}}} \,d x } \]
integrate((x^2-3)*(x^6+x^4-2*x^2+1)/x^10/((b*x^3+a*x^2-a)/(d*x^3+c*x^2-c)) ^(1/4),x, algorithm="maxima")
integrate((x^6 + x^4 - 2*x^2 + 1)*(x^2 - 3)/(x^10*((b*x^3 + a*x^2 - a)/(d* x^3 + c*x^2 - c))^(1/4)), x)
\[ \int \frac {\left (-3+x^2\right ) \left (1-2 x^2+x^4+x^6\right )}{x^{10} \sqrt [4]{\frac {-a+a x^2+b x^3}{-c+c x^2+d x^3}}} \, dx=\int { \frac {{\left (x^{6} + x^{4} - 2 \, x^{2} + 1\right )} {\left (x^{2} - 3\right )}}{x^{10} \left (\frac {b x^{3} + a x^{2} - a}{d x^{3} + c x^{2} - c}\right )^{\frac {1}{4}}} \,d x } \]
integrate((x^2-3)*(x^6+x^4-2*x^2+1)/x^10/((b*x^3+a*x^2-a)/(d*x^3+c*x^2-c)) ^(1/4),x, algorithm="giac")
integrate((x^6 + x^4 - 2*x^2 + 1)*(x^2 - 3)/(x^10*((b*x^3 + a*x^2 - a)/(d* x^3 + c*x^2 - c))^(1/4)), x)
Timed out. \[ \int \frac {\left (-3+x^2\right ) \left (1-2 x^2+x^4+x^6\right )}{x^{10} \sqrt [4]{\frac {-a+a x^2+b x^3}{-c+c x^2+d x^3}}} \, dx=\text {Hanged} \]