Integrand size = 26, antiderivative size = 414 \[ \int \frac {x^{-1-\frac {n}{4}}}{a+b x^n+c x^{2 n}} \, dx=-\frac {4 x^{-n/4}}{a n}-\frac {2^{3/4} \left (b+\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x^{-n/4}}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{a^{5/4} \left (-b-\sqrt {b^2-4 a c}\right )^{3/4} n}-\frac {2^{3/4} \left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x^{-n/4}}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{a^{5/4} \left (-b+\sqrt {b^2-4 a c}\right )^{3/4} n}-\frac {2^{3/4} \left (b+\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x^{-n/4}}{\sqrt [4]{-b-\sqrt {b^2-4 a c}}}\right )}{a^{5/4} \left (-b-\sqrt {b^2-4 a c}\right )^{3/4} n}-\frac {2^{3/4} \left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x^{-n/4}}{\sqrt [4]{-b+\sqrt {b^2-4 a c}}}\right )}{a^{5/4} \left (-b+\sqrt {b^2-4 a c}\right )^{3/4} n} \]
-4/a/n/(x^(1/4*n))-2^(3/4)*arctan(2^(1/4)*a^(1/4)/(x^(1/4*n))/(-b-(-4*a*c+ b^2)^(1/2))^(1/4))*(b+(-2*a*c+b^2)/(-4*a*c+b^2)^(1/2))/a^(5/4)/n/(-b-(-4*a *c+b^2)^(1/2))^(3/4)-2^(3/4)*arctanh(2^(1/4)*a^(1/4)/(x^(1/4*n))/(-b-(-4*a *c+b^2)^(1/2))^(1/4))*(b+(-2*a*c+b^2)/(-4*a*c+b^2)^(1/2))/a^(5/4)/n/(-b-(- 4*a*c+b^2)^(1/2))^(3/4)-2^(3/4)*arctan(2^(1/4)*a^(1/4)/(x^(1/4*n))/(-b+(-4 *a*c+b^2)^(1/2))^(1/4))*(b+(2*a*c-b^2)/(-4*a*c+b^2)^(1/2))/a^(5/4)/n/(-b+( -4*a*c+b^2)^(1/2))^(3/4)-2^(3/4)*arctanh(2^(1/4)*a^(1/4)/(x^(1/4*n))/(-b+( -4*a*c+b^2)^(1/2))^(1/4))*(b+(2*a*c-b^2)/(-4*a*c+b^2)^(1/2))/a^(5/4)/n/(-b +(-4*a*c+b^2)^(1/2))^(3/4)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.17 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.31 \[ \int \frac {x^{-1-\frac {n}{4}}}{a+b x^n+c x^{2 n}} \, dx=\frac {8 c x^{-n/4} \left (\frac {\operatorname {Hypergeometric2F1}\left (-\frac {1}{4},1,\frac {3}{4},\frac {2 c x^n}{-b+\sqrt {b^2-4 a c}}\right )}{b^2-4 a c-b \sqrt {b^2-4 a c}}+\frac {\operatorname {Hypergeometric2F1}\left (-\frac {1}{4},1,\frac {3}{4},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{b^2-4 a c+b \sqrt {b^2-4 a c}}\right )}{n} \]
(8*c*(Hypergeometric2F1[-1/4, 1, 3/4, (2*c*x^n)/(-b + Sqrt[b^2 - 4*a*c])]/ (b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c]) + Hypergeometric2F1[-1/4, 1, 3/4, (-2* c*x^n)/(b + Sqrt[b^2 - 4*a*c])]/(b^2 - 4*a*c + b*Sqrt[b^2 - 4*a*c])))/(n*x ^(n/4))
Time = 0.62 (sec) , antiderivative size = 372, normalized size of antiderivative = 0.90, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {1717, 1679, 1703, 1752, 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.
| \(\displaystyle \int \frac {x^{-\frac {n}{4}-1}}{a+b x^n+c x^{2 n}} \, dx\) |
| \(\Big \downarrow \) 1717 |
| \(\displaystyle -\frac {4 \int \frac {1}{b x^n+c x^{2 n}+a}dx^{-n/4}}{n}\) |
| \(\Big \downarrow \) 1679 |
| \(\displaystyle -\frac {4 \int \frac {x^{-2 n}}{a x^{-2 n}+b x^{-n}+c}dx^{-n/4}}{n}\) |
| \(\Big \downarrow \) 1703 |
| \(\displaystyle -\frac {4 \left (\frac {x^{-n/4}}{a}-\frac {\int \frac {b x^{-n}+c}{a x^{-2 n}+b x^{-n}+c}dx^{-n/4}}{a}\right )}{n}\) |
| \(\Big \downarrow \) 1752 |
| \(\displaystyle -\frac {4 \left (\frac {x^{-n/4}}{a}-\frac {\frac {1}{2} \left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \int \frac {1}{a x^{-n}+\frac {1}{2} \left (b-\sqrt {b^2-4 a c}\right )}dx^{-n/4}+\frac {1}{2} \left (\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}+b\right ) \int \frac {1}{a x^{-n}+\frac {1}{2} \left (b+\sqrt {b^2-4 a c}\right )}dx^{-n/4}}{a}\right )}{n}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle -\frac {4 \left (\frac {x^{-n/4}}{a}-\frac {\frac {1}{2} \left (\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}+b\right ) \left (-\frac {\int \frac {1}{\sqrt {-b-\sqrt {b^2-4 a c}}-\sqrt {2} \sqrt {a} x^{-n/2}}dx^{-n/4}}{\sqrt {-\sqrt {b^2-4 a c}-b}}-\frac {\int \frac {1}{\sqrt {2} \sqrt {a} x^{-n/2}+\sqrt {-b-\sqrt {b^2-4 a c}}}dx^{-n/4}}{\sqrt {-\sqrt {b^2-4 a c}-b}}\right )+\frac {1}{2} \left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \left (-\frac {\int \frac {1}{\sqrt {\sqrt {b^2-4 a c}-b}-\sqrt {2} \sqrt {a} x^{-n/2}}dx^{-n/4}}{\sqrt {\sqrt {b^2-4 a c}-b}}-\frac {\int \frac {1}{\sqrt {2} \sqrt {a} x^{-n/2}+\sqrt {\sqrt {b^2-4 a c}-b}}dx^{-n/4}}{\sqrt {\sqrt {b^2-4 a c}-b}}\right )}{a}\right )}{n}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle -\frac {4 \left (\frac {x^{-n/4}}{a}-\frac {\frac {1}{2} \left (\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}+b\right ) \left (-\frac {\int \frac {1}{\sqrt {-b-\sqrt {b^2-4 a c}}-\sqrt {2} \sqrt {a} x^{-n/2}}dx^{-n/4}}{\sqrt {-\sqrt {b^2-4 a c}-b}}-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x^{-n/4}}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{a} \left (-\sqrt {b^2-4 a c}-b\right )^{3/4}}\right )+\frac {1}{2} \left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \left (-\frac {\int \frac {1}{\sqrt {\sqrt {b^2-4 a c}-b}-\sqrt {2} \sqrt {a} x^{-n/2}}dx^{-n/4}}{\sqrt {\sqrt {b^2-4 a c}-b}}-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x^{-n/4}}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{a} \left (\sqrt {b^2-4 a c}-b\right )^{3/4}}\right )}{a}\right )}{n}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {4 \left (\frac {x^{-n/4}}{a}-\frac {\frac {1}{2} \left (\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}+b\right ) \left (-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x^{-n/4}}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{a} \left (-\sqrt {b^2-4 a c}-b\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x^{-n/4}}{\sqrt [4]{-\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{a} \left (-\sqrt {b^2-4 a c}-b\right )^{3/4}}\right )+\frac {1}{2} \left (b-\frac {b^2-2 a c}{\sqrt {b^2-4 a c}}\right ) \left (-\frac {\arctan \left (\frac {\sqrt [4]{2} \sqrt [4]{a} x^{-n/4}}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{a} \left (\sqrt {b^2-4 a c}-b\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt [4]{a} x^{-n/4}}{\sqrt [4]{\sqrt {b^2-4 a c}-b}}\right )}{\sqrt [4]{2} \sqrt [4]{a} \left (\sqrt {b^2-4 a c}-b\right )^{3/4}}\right )}{a}\right )}{n}\) |
(-4*(1/(a*x^(n/4)) - (((b + (b^2 - 2*a*c)/Sqrt[b^2 - 4*a*c])*(-(ArcTan[(2^ (1/4)*a^(1/4))/((-b - Sqrt[b^2 - 4*a*c])^(1/4)*x^(n/4))]/(2^(1/4)*a^(1/4)* (-b - Sqrt[b^2 - 4*a*c])^(3/4))) - ArcTanh[(2^(1/4)*a^(1/4))/((-b - Sqrt[b ^2 - 4*a*c])^(1/4)*x^(n/4))]/(2^(1/4)*a^(1/4)*(-b - Sqrt[b^2 - 4*a*c])^(3/ 4))))/2 + ((b - (b^2 - 2*a*c)/Sqrt[b^2 - 4*a*c])*(-(ArcTan[(2^(1/4)*a^(1/4 ))/((-b + Sqrt[b^2 - 4*a*c])^(1/4)*x^(n/4))]/(2^(1/4)*a^(1/4)*(-b + Sqrt[b ^2 - 4*a*c])^(3/4))) - ArcTanh[(2^(1/4)*a^(1/4))/((-b + Sqrt[b^2 - 4*a*c]) ^(1/4)*x^(n/4))]/(2^(1/4)*a^(1/4)*(-b + Sqrt[b^2 - 4*a*c])^(3/4))))/2)/a)) /n
3.6.61.3.1 Defintions of rubi rules used
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[((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_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^( 2*n*p)*(c + b/x^n + a/x^(2*n))^p, x] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && LtQ[n, 0] && IntegerQ[p]
Int[((d_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x _Symbol] :> Simp[d^(2*n - 1)*(d*x)^(m - 2*n + 1)*((a + b*x^n + c*x^(2*n))^( p + 1)/(c*(m + 2*n*p + 1))), x] - Simp[d^(2*n)/(c*(m + 2*n*p + 1)) Int[(d *x)^(m - 2*n)*Simp[a*(m - 2*n + 1) + b*(m + n*(p - 1) + 1)*x^n, x]*(a + b*x ^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && EqQ[n2, 2*n] && N eQ[b^2 - 4*a*c, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1] && NeQ[m + 2*n*p + 1, 0 ] && IntegerQ[p]
Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/(m + 1) Subst[Int[(a + b*x^Simplify[n/(m + 1)] + c*x^Simplify[ 2*(n/(m + 1))])^p, x], x, x^(m + 1)], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && IntegerQ[Simplify[n/(m + 1)]] && ! IntegerQ[n]
Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x _Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) Int[1/(b/2 - q/2 + c*x^n), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q)) I nt[1/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2 , 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && (PosQ[b^2 - 4*a*c] || !IGtQ[n/2, 0])
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.54 (sec) , antiderivative size = 630, normalized size of antiderivative = 1.52
| method | result | size |
| risch | \(-\frac {4 x^{-\frac {n}{4}}}{a n}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (256 a^{9} c^{4} n^{8}-256 a^{8} b^{2} c^{3} n^{8}+96 a^{7} b^{4} c^{2} n^{8}-16 a^{6} b^{6} c \,n^{8}+a^{5} b^{8} n^{8}\right ) \textit {\_Z}^{8}+\left (80 a^{4} b \,c^{4} n^{4}-120 a^{3} b^{3} c^{3} n^{4}+61 a^{2} b^{5} c^{2} n^{4}-13 a \,b^{7} c \,n^{4}+b^{9} n^{4}\right ) \textit {\_Z}^{4}+c^{5}\right )}{\sum }\textit {\_R} \ln \left (x^{\frac {n}{4}}+\left (-\frac {128 n^{7} a^{10} c^{5}}{a^{2} c^{6}-3 a \,b^{2} c^{5}+b^{4} c^{4}}+\frac {352 n^{7} b^{2} a^{9} c^{4}}{a^{2} c^{6}-3 a \,b^{2} c^{5}+b^{4} c^{4}}-\frac {280 n^{7} b^{4} a^{8} c^{3}}{a^{2} c^{6}-3 a \,b^{2} c^{5}+b^{4} c^{4}}+\frac {98 n^{7} b^{6} a^{7} c^{2}}{a^{2} c^{6}-3 a \,b^{2} c^{5}+b^{4} c^{4}}-\frac {16 n^{7} b^{8} a^{6} c}{a^{2} c^{6}-3 a \,b^{2} c^{5}+b^{4} c^{4}}+\frac {n^{7} b^{10} a^{5}}{a^{2} c^{6}-3 a \,b^{2} c^{5}+b^{4} c^{4}}\right ) \textit {\_R}^{7}+\left (-\frac {36 n^{3} b \,a^{5} c^{5}}{a^{2} c^{6}-3 a \,b^{2} c^{5}+b^{4} c^{4}}+\frac {129 n^{3} b^{3} a^{4} c^{4}}{a^{2} c^{6}-3 a \,b^{2} c^{5}+b^{4} c^{4}}-\frac {138 n^{3} b^{5} a^{3} c^{3}}{a^{2} c^{6}-3 a \,b^{2} c^{5}+b^{4} c^{4}}+\frac {63 n^{3} b^{7} a^{2} c^{2}}{a^{2} c^{6}-3 a \,b^{2} c^{5}+b^{4} c^{4}}-\frac {13 n^{3} b^{9} a c}{a^{2} c^{6}-3 a \,b^{2} c^{5}+b^{4} c^{4}}+\frac {n^{3} b^{11}}{a^{2} c^{6}-3 a \,b^{2} c^{5}+b^{4} c^{4}}\right ) \textit {\_R}^{3}\right )\right )\) | \(630\) |
-4/a/n/(x^(1/4*n))+sum(_R*ln(x^(1/4*n)+(-128/(a^2*c^6-3*a*b^2*c^5+b^4*c^4) *n^7*a^10*c^5+352/(a^2*c^6-3*a*b^2*c^5+b^4*c^4)*n^7*b^2*a^9*c^4-280/(a^2*c ^6-3*a*b^2*c^5+b^4*c^4)*n^7*b^4*a^8*c^3+98/(a^2*c^6-3*a*b^2*c^5+b^4*c^4)*n ^7*b^6*a^7*c^2-16/(a^2*c^6-3*a*b^2*c^5+b^4*c^4)*n^7*b^8*a^6*c+1/(a^2*c^6-3 *a*b^2*c^5+b^4*c^4)*n^7*b^10*a^5)*_R^7+(-36/(a^2*c^6-3*a*b^2*c^5+b^4*c^4)* n^3*b*a^5*c^5+129/(a^2*c^6-3*a*b^2*c^5+b^4*c^4)*n^3*b^3*a^4*c^4-138/(a^2*c ^6-3*a*b^2*c^5+b^4*c^4)*n^3*b^5*a^3*c^3+63/(a^2*c^6-3*a*b^2*c^5+b^4*c^4)*n ^3*b^7*a^2*c^2-13/(a^2*c^6-3*a*b^2*c^5+b^4*c^4)*n^3*b^9*a*c+1/(a^2*c^6-3*a *b^2*c^5+b^4*c^4)*n^3*b^11)*_R^3),_R=RootOf((256*a^9*c^4*n^8-256*a^8*b^2*c ^3*n^8+96*a^7*b^4*c^2*n^8-16*a^6*b^6*c*n^8+a^5*b^8*n^8)*_Z^8+(80*a^4*b*c^4 *n^4-120*a^3*b^3*c^3*n^4+61*a^2*b^5*c^2*n^4-13*a*b^7*c*n^4+b^9*n^4)*_Z^4+c ^5))
Leaf count of result is larger than twice the leaf count of optimal. 4375 vs. \(2 (342) = 684\).
Time = 0.44 (sec) , antiderivative size = 4375, normalized size of antiderivative = 10.57 \[ \int \frac {x^{-1-\frac {n}{4}}}{a+b x^n+c x^{2 n}} \, dx=\text {Too large to display} \]
1/2*(sqrt(2)*a*n*sqrt(sqrt(2)*sqrt(-((a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)* n^4*sqrt((b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)/((a^ 10*b^6 - 12*a^11*b^4*c + 48*a^12*b^2*c^2 - 64*a^13*c^3)*n^8)) + b^5 - 5*a* b^3*c + 5*a^2*b*c^2)/((a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)*n^4)))*log((4*( b^4*c - 3*a*b^2*c^2 + a^2*c^3)*x*x^(-1/4*n - 1) + sqrt(2)*((a^5*b^5 - 8*a^ 6*b^3*c + 16*a^7*b*c^2)*n^5*sqrt((b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3 *b^2*c^3 + a^4*c^4)/((a^10*b^6 - 12*a^11*b^4*c + 48*a^12*b^2*c^2 - 64*a^13 *c^3)*n^8)) - (b^6 - 7*a*b^4*c + 13*a^2*b^2*c^2 - 4*a^3*c^3)*n)*sqrt(sqrt( 2)*sqrt(-((a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)*n^4*sqrt((b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)/((a^10*b^6 - 12*a^11*b^4*c + 48 *a^12*b^2*c^2 - 64*a^13*c^3)*n^8)) + b^5 - 5*a*b^3*c + 5*a^2*b*c^2)/((a^5* b^4 - 8*a^6*b^2*c + 16*a^7*c^2)*n^4))))/x) - sqrt(2)*a*n*sqrt(sqrt(2)*sqrt (-((a^5*b^4 - 8*a^6*b^2*c + 16*a^7*c^2)*n^4*sqrt((b^8 - 6*a*b^6*c + 11*a^2 *b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)/((a^10*b^6 - 12*a^11*b^4*c + 48*a^12*b ^2*c^2 - 64*a^13*c^3)*n^8)) + b^5 - 5*a*b^3*c + 5*a^2*b*c^2)/((a^5*b^4 - 8 *a^6*b^2*c + 16*a^7*c^2)*n^4)))*log((4*(b^4*c - 3*a*b^2*c^2 + a^2*c^3)*x*x ^(-1/4*n - 1) - sqrt(2)*((a^5*b^5 - 8*a^6*b^3*c + 16*a^7*b*c^2)*n^5*sqrt(( b^8 - 6*a*b^6*c + 11*a^2*b^4*c^2 - 6*a^3*b^2*c^3 + a^4*c^4)/((a^10*b^6 - 1 2*a^11*b^4*c + 48*a^12*b^2*c^2 - 64*a^13*c^3)*n^8)) - (b^6 - 7*a*b^4*c + 1 3*a^2*b^2*c^2 - 4*a^3*c^3)*n)*sqrt(sqrt(2)*sqrt(-((a^5*b^4 - 8*a^6*b^2*...
\[ \int \frac {x^{-1-\frac {n}{4}}}{a+b x^n+c x^{2 n}} \, dx=\int \frac {x^{- \frac {n}{4} - 1}}{a + b x^{n} + c x^{2 n}}\, dx \]
\[ \int \frac {x^{-1-\frac {n}{4}}}{a+b x^n+c x^{2 n}} \, dx=\int { \frac {x^{-\frac {1}{4} \, n - 1}}{c x^{2 \, n} + b x^{n} + a} \,d x } \]
-4/(a*n*x^(1/4*n)) - integrate((c*x^(7/4*n) + b*x^(3/4*n))/(a*c*x*x^(2*n) + a*b*x*x^n + a^2*x), x)
\[ \int \frac {x^{-1-\frac {n}{4}}}{a+b x^n+c x^{2 n}} \, dx=\int { \frac {x^{-\frac {1}{4} \, n - 1}}{c x^{2 \, n} + b x^{n} + a} \,d x } \]
Timed out. \[ \int \frac {x^{-1-\frac {n}{4}}}{a+b x^n+c x^{2 n}} \, dx=\int \frac {1}{x^{\frac {n}{4}+1}\,\left (a+b\,x^n+c\,x^{2\,n}\right )} \,d x \]