Integrand size = 60, antiderivative size = 708 \[ \int \sqrt [4]{\frac {1+a x-4 x^2-4 a x^3+6 x^4+6 a x^5-4 x^6-4 a x^7+x^8+a x^9}{-c+b x}} \, dx=\frac {7 b^2+96 a^2 b^2-6 a b c-45 a^2 c^2+3 a b^2 x+96 a^3 b^2 x-42 a^2 b c x-45 a^3 c^2 x-21 b^2 x^2-324 a^2 b^2 x^2+18 a b c x^2-36 a^3 b c x^2+135 a^2 c^2 x^2-9 a b^2 x^3-320 a^3 b^2 x^3+126 a^2 b c x^3+135 a^3 c^2 x^3+21 b^2 x^4+396 a^2 b^2 x^4-18 a b c x^4+108 a^3 b c x^4-135 a^2 c^2 x^4+9 a b^2 x^5+384 a^3 b^2 x^5-126 a^2 b c x^5-135 a^3 c^2 x^5-7 b^2 x^6-204 a^2 b^2 x^6+6 a b c x^6-108 a^3 b c x^6+45 a^2 c^2 x^6-3 a b^2 x^7-192 a^3 b^2 x^7+42 a^2 b c x^7+45 a^3 c^2 x^7+36 a^2 b^2 x^8+36 a^3 b c x^8+32 a^3 b^2 x^9}{96 a^2 b^3 \left (\frac {1+a x-4 x^2-4 a x^3+6 x^4+6 a x^5-4 x^6-4 a x^7+x^8+a x^9}{-c+b x}\right )^{3/4}}+\frac {\left (-7 b^3+32 a^2 b^3+3 a b^2 c+32 a^3 b^2 c-5 a^2 b c^2-15 a^3 c^3\right ) \arctan \left (\frac {\sqrt [4]{a} (-1+x) (1+x)}{\sqrt [4]{b} \sqrt [4]{\frac {1+a x-4 x^2-4 a x^3+6 x^4+6 a x^5-4 x^6-4 a x^7+x^8+a x^9}{-c+b x}}}\right )}{64 a^{11/4} b^{13/4}}+\frac {\left (7 b^3-32 a^2 b^3-3 a b^2 c-32 a^3 b^2 c+5 a^2 b c^2+15 a^3 c^3\right ) \text {arctanh}\left (\frac {\sqrt [4]{a} (-1+x) (1+x)}{\sqrt [4]{b} \sqrt [4]{\frac {1+a x-4 x^2-4 a x^3+6 x^4+6 a x^5-4 x^6-4 a x^7+x^8+a x^9}{-c+b x}}}\right )}{64 a^{11/4} b^{13/4}} \]
1/96*(32*a^3*b^2*x^9+36*a^3*b*c*x^8-192*a^3*b^2*x^7+45*a^3*c^2*x^7+36*a^2* b^2*x^8-108*a^3*b*c*x^6+42*a^2*b*c*x^7+384*a^3*b^2*x^5-135*a^3*c^2*x^5-204 *a^2*b^2*x^6+45*a^2*c^2*x^6-3*a*b^2*x^7+108*a^3*b*c*x^4-126*a^2*b*c*x^5+6* a*b*c*x^6-320*a^3*b^2*x^3+135*a^3*c^2*x^3+396*a^2*b^2*x^4-135*a^2*c^2*x^4+ 9*a*b^2*x^5-7*b^2*x^6-36*a^3*b*c*x^2+126*a^2*b*c*x^3-18*a*b*c*x^4+96*a^3*b ^2*x-45*a^3*c^2*x-324*a^2*b^2*x^2+135*a^2*c^2*x^2-9*a*b^2*x^3+21*b^2*x^4-4 2*a^2*b*c*x+18*a*b*c*x^2+96*a^2*b^2-45*a^2*c^2+3*a*b^2*x-21*b^2*x^2-6*a*b* c+7*b^2)/a^2/b^3/((a*x^9-4*a*x^7+x^8+6*a*x^5-4*x^6-4*a*x^3+6*x^4+a*x-4*x^2 +1)/(b*x-c))^(3/4)+1/64*(32*a^3*b^2*c-15*a^3*c^3+32*a^2*b^3-5*a^2*b*c^2+3* a*b^2*c-7*b^3)*arctan(a^(1/4)*(-1+x)*(1+x)/b^(1/4)/((a*x^9-4*a*x^7+x^8+6*a *x^5-4*x^6-4*a*x^3+6*x^4+a*x-4*x^2+1)/(b*x-c))^(1/4))/a^(11/4)/b^(13/4)+1/ 64*(-32*a^3*b^2*c+15*a^3*c^3-32*a^2*b^3+5*a^2*b*c^2-3*a*b^2*c+7*b^3)*arcta nh(a^(1/4)*(-1+x)*(1+x)/b^(1/4)/((a*x^9-4*a*x^7+x^8+6*a*x^5-4*x^6-4*a*x^3+ 6*x^4+a*x-4*x^2+1)/(b*x-c))^(1/4))/a^(11/4)/b^(13/4)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 10.17 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.24 \[ \int \sqrt [4]{\frac {1+a x-4 x^2-4 a x^3+6 x^4+6 a x^5-4 x^6-4 a x^7+x^8+a x^9}{-c+b x}} \, dx=\frac {4 (1+a x) \sqrt [4]{\frac {a (c-b x)}{b+a c}} \sqrt [4]{\frac {(1+a x) \left (-1+x^2\right )^4}{-c+b x}} \left ((b+a c)^2 \operatorname {Hypergeometric2F1}\left (-\frac {7}{4},\frac {5}{4},\frac {9}{4},\frac {b+a b x}{b+a c}\right )-a \left (2 c (b+a c) \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {5}{4},\frac {9}{4},\frac {b+a b x}{b+a c}\right )+a \left (b^2-c^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {5}{4},\frac {9}{4},\frac {b+a b x}{b+a c}\right )\right )\right )}{5 a^3 b^2 \left (-1+x^2\right )} \]
Integrate[((1 + a*x - 4*x^2 - 4*a*x^3 + 6*x^4 + 6*a*x^5 - 4*x^6 - 4*a*x^7 + x^8 + a*x^9)/(-c + b*x))^(1/4),x]
(4*(1 + a*x)*((a*(c - b*x))/(b + a*c))^(1/4)*(((1 + a*x)*(-1 + x^2)^4)/(-c + b*x))^(1/4)*((b + a*c)^2*Hypergeometric2F1[-7/4, 5/4, 9/4, (b + a*b*x)/ (b + a*c)] - a*(2*c*(b + a*c)*Hypergeometric2F1[-3/4, 5/4, 9/4, (b + a*b*x )/(b + a*c)] + a*(b^2 - c^2)*Hypergeometric2F1[1/4, 5/4, 9/4, (b + a*b*x)/ (b + a*c)])))/(5*a^3*b^2*(-1 + x^2))
Time = 0.58 (sec) , antiderivative size = 325, normalized size of antiderivative = 0.46, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {7239, 7269, 25, 651, 27, 90, 60, 73, 770, 756, 218, 221}
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.
| \(\displaystyle \int \sqrt [4]{\frac {a x^9-4 a x^7+6 a x^5-4 a x^3+a x+x^8-4 x^6+6 x^4-4 x^2+1}{b x-c}} \, dx\) |
| \(\Big \downarrow \) 7239 |
| \(\displaystyle \int \sqrt [4]{\frac {\left (x^2-1\right )^4 (a x+1)}{b x-c}}dx\) |
| \(\Big \downarrow \) 7269 |
| \(\displaystyle -\frac {\sqrt [4]{b x-c} \sqrt [4]{-\frac {\left (1-x^2\right )^4 (a x+1)}{c-b x}} \int -\frac {\sqrt [4]{a x+1} \left (1-x^2\right )}{\sqrt [4]{b x-c}}dx}{\left (1-x^2\right ) \sqrt [4]{a x+1}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\sqrt [4]{b x-c} \sqrt [4]{-\frac {\left (1-x^2\right )^4 (a x+1)}{c-b x}} \int \frac {\sqrt [4]{a x+1} \left (1-x^2\right )}{\sqrt [4]{b x-c}}dx}{\left (1-x^2\right ) \sqrt [4]{a x+1}}\) |
| \(\Big \downarrow \) 651 |
| \(\displaystyle \frac {\sqrt [4]{b x-c} \sqrt [4]{-\frac {\left (1-x^2\right )^4 (a x+1)}{c-b x}} \left (\frac {\int \frac {3 \sqrt [4]{a x+1} \left (4 b a^2-3 c a+(5 b-3 a c) x a+b\right )}{4 \sqrt [4]{b x-c}}dx}{3 a^2 b}-\frac {(a x+1)^{9/4} (b x-c)^{3/4}}{3 a^2 b}\right )}{\left (1-x^2\right ) \sqrt [4]{a x+1}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt [4]{b x-c} \sqrt [4]{-\frac {\left (1-x^2\right )^4 (a x+1)}{c-b x}} \left (\frac {\int \frac {\sqrt [4]{a x+1} \left (4 b a^2-3 c a+(5 b-3 a c) x a+b\right )}{\sqrt [4]{b x-c}}dx}{4 a^2 b}-\frac {(a x+1)^{9/4} (b x-c)^{3/4}}{3 a^2 b}\right )}{\left (1-x^2\right ) \sqrt [4]{a x+1}}\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {\sqrt [4]{b x-c} \sqrt [4]{-\frac {\left (1-x^2\right )^4 (a x+1)}{c-b x}} \left (\frac {\frac {(a x+1)^{5/4} (5 b-3 a c) (b x-c)^{3/4}}{2 b}-\frac {\left (\left (7-32 a^2\right ) b^2+15 a^2 c^2-10 a b c\right ) \int \frac {\sqrt [4]{a x+1}}{\sqrt [4]{b x-c}}dx}{8 b}}{4 a^2 b}-\frac {(a x+1)^{9/4} (b x-c)^{3/4}}{3 a^2 b}\right )}{\left (1-x^2\right ) \sqrt [4]{a x+1}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {\sqrt [4]{b x-c} \sqrt [4]{-\frac {\left (1-x^2\right )^4 (a x+1)}{c-b x}} \left (\frac {\frac {(a x+1)^{5/4} (5 b-3 a c) (b x-c)^{3/4}}{2 b}-\frac {\left (\left (7-32 a^2\right ) b^2+15 a^2 c^2-10 a b c\right ) \left (\frac {(a c+b) \int \frac {1}{(a x+1)^{3/4} \sqrt [4]{b x-c}}dx}{4 b}+\frac {\sqrt [4]{a x+1} (b x-c)^{3/4}}{b}\right )}{8 b}}{4 a^2 b}-\frac {(a x+1)^{9/4} (b x-c)^{3/4}}{3 a^2 b}\right )}{\left (1-x^2\right ) \sqrt [4]{a x+1}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {\sqrt [4]{b x-c} \sqrt [4]{-\frac {\left (1-x^2\right )^4 (a x+1)}{c-b x}} \left (\frac {\frac {(a x+1)^{5/4} (5 b-3 a c) (b x-c)^{3/4}}{2 b}-\frac {\left (\left (7-32 a^2\right ) b^2+15 a^2 c^2-10 a b c\right ) \left (\frac {(a c+b) \int \frac {1}{\sqrt [4]{\frac {b (a x+1)}{a}-\frac {b+a c}{a}}}d\sqrt [4]{a x+1}}{a b}+\frac {\sqrt [4]{a x+1} (b x-c)^{3/4}}{b}\right )}{8 b}}{4 a^2 b}-\frac {(a x+1)^{9/4} (b x-c)^{3/4}}{3 a^2 b}\right )}{\left (1-x^2\right ) \sqrt [4]{a x+1}}\) |
| \(\Big \downarrow \) 770 |
| \(\displaystyle \frac {\sqrt [4]{b x-c} \sqrt [4]{-\frac {\left (1-x^2\right )^4 (a x+1)}{c-b x}} \left (\frac {\frac {(a x+1)^{5/4} (5 b-3 a c) (b x-c)^{3/4}}{2 b}-\frac {\left (\left (7-32 a^2\right ) b^2+15 a^2 c^2-10 a b c\right ) \left (\frac {(a c+b) \int \frac {1}{1-\frac {b (a x+1)}{a}}d\frac {\sqrt [4]{a x+1}}{\sqrt [4]{\frac {b (a x+1)}{a}-\frac {b+a c}{a}}}}{a b}+\frac {\sqrt [4]{a x+1} (b x-c)^{3/4}}{b}\right )}{8 b}}{4 a^2 b}-\frac {(a x+1)^{9/4} (b x-c)^{3/4}}{3 a^2 b}\right )}{\left (1-x^2\right ) \sqrt [4]{a x+1}}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle \frac {\sqrt [4]{b x-c} \sqrt [4]{-\frac {\left (1-x^2\right )^4 (a x+1)}{c-b x}} \left (\frac {\frac {(a x+1)^{5/4} (5 b-3 a c) (b x-c)^{3/4}}{2 b}-\frac {\left (\left (7-32 a^2\right ) b^2+15 a^2 c^2-10 a b c\right ) \left (\frac {(a c+b) \left (\frac {1}{2} \sqrt {a} \int \frac {1}{\sqrt {a}-\sqrt {b} \sqrt {a x+1}}d\frac {\sqrt [4]{a x+1}}{\sqrt [4]{\frac {b (a x+1)}{a}-\frac {b+a c}{a}}}+\frac {1}{2} \sqrt {a} \int \frac {1}{\sqrt {a}+\sqrt {b} \sqrt {a x+1}}d\frac {\sqrt [4]{a x+1}}{\sqrt [4]{\frac {b (a x+1)}{a}-\frac {b+a c}{a}}}\right )}{a b}+\frac {\sqrt [4]{a x+1} (b x-c)^{3/4}}{b}\right )}{8 b}}{4 a^2 b}-\frac {(a x+1)^{9/4} (b x-c)^{3/4}}{3 a^2 b}\right )}{\left (1-x^2\right ) \sqrt [4]{a x+1}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\sqrt [4]{b x-c} \sqrt [4]{-\frac {\left (1-x^2\right )^4 (a x+1)}{c-b x}} \left (\frac {\frac {(a x+1)^{5/4} (5 b-3 a c) (b x-c)^{3/4}}{2 b}-\frac {\left (\left (7-32 a^2\right ) b^2+15 a^2 c^2-10 a b c\right ) \left (\frac {(a c+b) \left (\frac {1}{2} \sqrt {a} \int \frac {1}{\sqrt {a}-\sqrt {b} \sqrt {a x+1}}d\frac {\sqrt [4]{a x+1}}{\sqrt [4]{\frac {b (a x+1)}{a}-\frac {b+a c}{a}}}+\frac {\sqrt [4]{a} \arctan \left (\frac {\sqrt [4]{b} \sqrt [4]{a x+1}}{\sqrt [4]{a} \sqrt [4]{\frac {b (a x+1)}{a}-\frac {a c+b}{a}}}\right )}{2 \sqrt [4]{b}}\right )}{a b}+\frac {\sqrt [4]{a x+1} (b x-c)^{3/4}}{b}\right )}{8 b}}{4 a^2 b}-\frac {(a x+1)^{9/4} (b x-c)^{3/4}}{3 a^2 b}\right )}{\left (1-x^2\right ) \sqrt [4]{a x+1}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\sqrt [4]{b x-c} \sqrt [4]{-\frac {\left (1-x^2\right )^4 (a x+1)}{c-b x}} \left (\frac {\frac {(a x+1)^{5/4} (5 b-3 a c) (b x-c)^{3/4}}{2 b}-\frac {\left (\left (7-32 a^2\right ) b^2+15 a^2 c^2-10 a b c\right ) \left (\frac {(a c+b) \left (\frac {\sqrt [4]{a} \arctan \left (\frac {\sqrt [4]{b} \sqrt [4]{a x+1}}{\sqrt [4]{a} \sqrt [4]{\frac {b (a x+1)}{a}-\frac {a c+b}{a}}}\right )}{2 \sqrt [4]{b}}+\frac {\sqrt [4]{a} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt [4]{a x+1}}{\sqrt [4]{a} \sqrt [4]{\frac {b (a x+1)}{a}-\frac {a c+b}{a}}}\right )}{2 \sqrt [4]{b}}\right )}{a b}+\frac {\sqrt [4]{a x+1} (b x-c)^{3/4}}{b}\right )}{8 b}}{4 a^2 b}-\frac {(a x+1)^{9/4} (b x-c)^{3/4}}{3 a^2 b}\right )}{\left (1-x^2\right ) \sqrt [4]{a x+1}}\) |
Int[((1 + a*x - 4*x^2 - 4*a*x^3 + 6*x^4 + 6*a*x^5 - 4*x^6 - 4*a*x^7 + x^8 + a*x^9)/(-c + b*x))^(1/4),x]
((-c + b*x)^(1/4)*(-(((1 + a*x)*(1 - x^2)^4)/(c - b*x)))^(1/4)*(-1/3*((1 + a*x)^(9/4)*(-c + b*x)^(3/4))/(a^2*b) + (((5*b - 3*a*c)*(1 + a*x)^(5/4)*(- c + b*x)^(3/4))/(2*b) - (((7 - 32*a^2)*b^2 - 10*a*b*c + 15*a^2*c^2)*(((1 + a*x)^(1/4)*(-c + b*x)^(3/4))/b + ((b + a*c)*((a^(1/4)*ArcTan[(b^(1/4)*(1 + a*x)^(1/4))/(a^(1/4)*(-((b + a*c)/a) + (b*(1 + a*x))/a)^(1/4))])/(2*b^(1 /4)) + (a^(1/4)*ArcTanh[(b^(1/4)*(1 + a*x)^(1/4))/(a^(1/4)*(-((b + a*c)/a) + (b*(1 + a*x))/a)^(1/4))])/(2*b^(1/4))))/(a*b)))/(8*b))/(4*a^2*b)))/((1 + a*x)^(1/4)*(1 - x^2))
3.32.22.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^ 2)^(p_.), x_Symbol] :> Simp[c^p*(d + e*x)^(m + 2*p)*((f + g*x)^(n + 1)/(g*e ^(2*p)*(m + n + 2*p + 1))), x] + Simp[1/(g*e^(2*p)*(m + n + 2*p + 1)) Int [(d + e*x)^m*(f + g*x)^n*ExpandToSum[g*(m + n + 2*p + 1)*(e^(2*p)*(a + c*x^ 2)^p - c^p*(d + e*x)^(2*p)) - c^p*(e*f - d*g)*(m + 2*p)*(d + e*x)^(2*p - 1) , x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && IGtQ[p, 0] && !IntegerQ[m] && !IntegerQ[n] && NeQ[m + n + 2*p + 1, 0]
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n) Subst[In t[1/(1 - b*x^n)^(p + 1/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p + 1 /n]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Int[(u_.)*((a_.)*(v_)^(m_.)*(w_)^(n_.)*(z_)^(q_.))^(p_), x_Symbol] :> Simp[ a^IntPart[p]*((a*v^m*w^n*z^q)^FracPart[p]/(v^(m*FracPart[p])*w^(n*FracPart[ p])*z^(q*FracPart[p]))) Int[u*v^(m*p)*w^(n*p)*z^(p*q), x], x] /; FreeQ[{a , m, n, p, q}, x] && !IntegerQ[p] && !FreeQ[v, x] && !FreeQ[w, x] && !F reeQ[z, x]
\[\int \left (\frac {a \,x^{9}-4 a \,x^{7}+x^{8}+6 a \,x^{5}-4 x^{6}-4 a \,x^{3}+6 x^{4}+a x -4 x^{2}+1}{b x -c}\right )^{\frac {1}{4}}d x\]
Result contains complex when optimal does not.
Time = 0.35 (sec) , antiderivative size = 3408, normalized size of antiderivative = 4.81 \[ \int \sqrt [4]{\frac {1+a x-4 x^2-4 a x^3+6 x^4+6 a x^5-4 x^6-4 a x^7+x^8+a x^9}{-c+b x}} \, dx=\text {Too large to display} \]
integrate(((a*x^9-4*a*x^7+x^8+6*a*x^5-4*x^6-4*a*x^3+6*x^4+a*x-4*x^2+1)/(b* x-c))^(1/4),x, algorithm="fricas")
1/384*(3*(a^2*b^3*x^2 - a^2*b^3)*((50625*a^12*c^12 + 67500*a^11*b*c^11 + ( 1048576*a^8 - 917504*a^6 + 301056*a^4 - 43904*a^2 + 2401)*b^12 + 4*(104857 6*a^9 - 589824*a^7 + 86016*a^5 + 3136*a^3 - 1029*a)*b^11*c + 2*(3145728*a^ 10 - 1114112*a^8 + 135168*a^6 - 30912*a^4 + 4753*a^2)*b^10*c^2 + 4*(104857 6*a^11 - 917504*a^9 + 454656*a^7 - 79104*a^5 + 2751*a^3)*b^9*c^3 + (104857 6*a^12 - 7471104*a^10 + 2297856*a^8 - 40704*a^6 - 15249*a^4)*b^8*c^4 - 200 *(32768*a^11 - 6144*a^9 + 1344*a^7 - 243*a^5)*b^7*c^5 - 20*(98304*a^12 - 1 41312*a^10 + 43712*a^8 - 1579*a^6)*b^6*c^6 + 200*(18432*a^11 - 1664*a^9 - 93*a^7)*b^5*c^7 + 25*(55296*a^12 - 12672*a^10 + 3751*a^8)*b^4*c^8 - 1500*( 576*a^11 - 41*a^9)*b^3*c^9 - 6750*(64*a^12 + a^10)*b^2*c^10)/(a^11*b^13))^ (1/4)*log(((15*a^3*c^3 + 5*a^2*b*c^2 - (32*a^2 - 7)*b^3 - (32*a^3 + 3*a)*b ^2*c)*((a*x^9 - 4*a*x^7 + x^8 + 6*a*x^5 - 4*x^6 - 4*a*x^3 + 6*x^4 + a*x - 4*x^2 + 1)/(b*x - c))^(1/4) + (a^3*b^3*x^2 - a^3*b^3)*((50625*a^12*c^12 + 67500*a^11*b*c^11 + (1048576*a^8 - 917504*a^6 + 301056*a^4 - 43904*a^2 + 2 401)*b^12 + 4*(1048576*a^9 - 589824*a^7 + 86016*a^5 + 3136*a^3 - 1029*a)*b ^11*c + 2*(3145728*a^10 - 1114112*a^8 + 135168*a^6 - 30912*a^4 + 4753*a^2) *b^10*c^2 + 4*(1048576*a^11 - 917504*a^9 + 454656*a^7 - 79104*a^5 + 2751*a ^3)*b^9*c^3 + (1048576*a^12 - 7471104*a^10 + 2297856*a^8 - 40704*a^6 - 152 49*a^4)*b^8*c^4 - 200*(32768*a^11 - 6144*a^9 + 1344*a^7 - 243*a^5)*b^7*c^5 - 20*(98304*a^12 - 141312*a^10 + 43712*a^8 - 1579*a^6)*b^6*c^6 + 200*(...
Timed out. \[ \int \sqrt [4]{\frac {1+a x-4 x^2-4 a x^3+6 x^4+6 a x^5-4 x^6-4 a x^7+x^8+a x^9}{-c+b x}} \, dx=\text {Timed out} \]
integrate(((a*x**9-4*a*x**7+x**8+6*a*x**5-4*x**6-4*a*x**3+6*x**4+a*x-4*x** 2+1)/(b*x-c))**(1/4),x)
\[ \int \sqrt [4]{\frac {1+a x-4 x^2-4 a x^3+6 x^4+6 a x^5-4 x^6-4 a x^7+x^8+a x^9}{-c+b x}} \, dx=\int { \left (\frac {a x^{9} - 4 \, a x^{7} + x^{8} + 6 \, a x^{5} - 4 \, x^{6} - 4 \, a x^{3} + 6 \, x^{4} + a x - 4 \, x^{2} + 1}{b x - c}\right )^{\frac {1}{4}} \,d x } \]
integrate(((a*x^9-4*a*x^7+x^8+6*a*x^5-4*x^6-4*a*x^3+6*x^4+a*x-4*x^2+1)/(b* x-c))^(1/4),x, algorithm="maxima")
integrate(((a*x^9 - 4*a*x^7 + x^8 + 6*a*x^5 - 4*x^6 - 4*a*x^3 + 6*x^4 + a* x - 4*x^2 + 1)/(b*x - c))^(1/4), x)
\[ \int \sqrt [4]{\frac {1+a x-4 x^2-4 a x^3+6 x^4+6 a x^5-4 x^6-4 a x^7+x^8+a x^9}{-c+b x}} \, dx=\int { \left (\frac {a x^{9} - 4 \, a x^{7} + x^{8} + 6 \, a x^{5} - 4 \, x^{6} - 4 \, a x^{3} + 6 \, x^{4} + a x - 4 \, x^{2} + 1}{b x - c}\right )^{\frac {1}{4}} \,d x } \]
integrate(((a*x^9-4*a*x^7+x^8+6*a*x^5-4*x^6-4*a*x^3+6*x^4+a*x-4*x^2+1)/(b* x-c))^(1/4),x, algorithm="giac")
integrate(((a*x^9 - 4*a*x^7 + x^8 + 6*a*x^5 - 4*x^6 - 4*a*x^3 + 6*x^4 + a* x - 4*x^2 + 1)/(b*x - c))^(1/4), x)
Timed out. \[ \int \sqrt [4]{\frac {1+a x-4 x^2-4 a x^3+6 x^4+6 a x^5-4 x^6-4 a x^7+x^8+a x^9}{-c+b x}} \, dx=\int {\left (-\frac {a\,x^9+x^8-4\,a\,x^7-4\,x^6+6\,a\,x^5+6\,x^4-4\,a\,x^3-4\,x^2+a\,x+1}{c-b\,x}\right )}^{1/4} \,d x \]
int((-(a*x - 4*a*x^3 + 6*a*x^5 - 4*a*x^7 + a*x^9 - 4*x^2 + 6*x^4 - 4*x^6 + x^8 + 1)/(c - b*x))^(1/4),x)