Integrand size = 32, antiderivative size = 545 \[ \int \frac {d+e x+f x^2+g x^3}{a+b x^n+c x^{2 n}} \, dx=-\frac {2 c d x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{b^2-4 a c-b \sqrt {b^2-4 a c}}-\frac {2 c d x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{b^2-4 a c+b \sqrt {b^2-4 a c}}-\frac {c e x^2 \operatorname {Hypergeometric2F1}\left (1,\frac {2}{n},\frac {2+n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{b^2-4 a c-b \sqrt {b^2-4 a c}}-\frac {c e x^2 \operatorname {Hypergeometric2F1}\left (1,\frac {2}{n},\frac {2+n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{b^2-4 a c+b \sqrt {b^2-4 a c}}-\frac {2 c f x^3 \operatorname {Hypergeometric2F1}\left (1,\frac {3}{n},\frac {3+n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{3 \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right )}-\frac {2 c f x^3 \operatorname {Hypergeometric2F1}\left (1,\frac {3}{n},\frac {3+n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{3 \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right )}-\frac {c g x^4 \operatorname {Hypergeometric2F1}\left (1,\frac {4}{n},\frac {4+n}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{2 \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right )}-\frac {c g x^4 \operatorname {Hypergeometric2F1}\left (1,\frac {4}{n},\frac {4+n}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{2 \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right )} \]
-2*c*d*x*hypergeom([1, 1/n],[1+1/n],-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))/(b^2- 4*a*c-b*(-4*a*c+b^2)^(1/2))-c*e*x^2*hypergeom([1, 2/n],[(2+n)/n],-2*c*x^n/ (b-(-4*a*c+b^2)^(1/2)))/(b^2-4*a*c-b*(-4*a*c+b^2)^(1/2))-2/3*c*f*x^3*hyper geom([1, 3/n],[(3+n)/n],-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))/(b^2-4*a*c-b*(-4* a*c+b^2)^(1/2))-1/2*c*g*x^4*hypergeom([1, 4/n],[(4+n)/n],-2*c*x^n/(b-(-4*a *c+b^2)^(1/2)))/(b^2-4*a*c-b*(-4*a*c+b^2)^(1/2))-2*c*d*x*hypergeom([1, 1/n ],[1+1/n],-2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))/(b^2-4*a*c+b*(-4*a*c+b^2)^(1/2) )-c*e*x^2*hypergeom([1, 2/n],[(2+n)/n],-2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))/(b ^2-4*a*c+b*(-4*a*c+b^2)^(1/2))-2/3*c*f*x^3*hypergeom([1, 3/n],[(3+n)/n],-2 *c*x^n/(b+(-4*a*c+b^2)^(1/2)))/(b^2-4*a*c+b*(-4*a*c+b^2)^(1/2))-1/2*c*g*x^ 4*hypergeom([1, 4/n],[(4+n)/n],-2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))/(b^2-4*a*c +b*(-4*a*c+b^2)^(1/2))
Leaf count is larger than twice the leaf count of optimal. \(1093\) vs. \(2(545)=1090\).
Time = 1.28 (sec) , antiderivative size = 1093, normalized size of antiderivative = 2.01 \[ \int \frac {d+e x+f x^2+g x^3}{a+b x^n+c x^{2 n}} \, dx=\frac {x \left (3 g x^3 \left (\left (-b^2+4 a c-b \sqrt {b^2-4 a c}\right ) \left (1-\left (\frac {x^n}{-\frac {-b+\sqrt {b^2-4 a c}}{2 c}+x^n}\right )^{-4/n} \operatorname {Hypergeometric2F1}\left (-\frac {4}{n},-\frac {4}{n},\frac {-4+n}{n},\frac {b-\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )\right )+\left (-b^2+4 a c+b \sqrt {b^2-4 a c}\right ) \left (1-2^{-4/n} \left (\frac {c x^n}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )^{-4/n} \operatorname {Hypergeometric2F1}\left (-\frac {4}{n},-\frac {4}{n},\frac {-4+n}{n},\frac {b+\sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )\right )\right )+4 f x^2 \left (\left (-b^2+4 a c-b \sqrt {b^2-4 a c}\right ) \left (1-\left (\frac {x^n}{-\frac {-b+\sqrt {b^2-4 a c}}{2 c}+x^n}\right )^{-3/n} \operatorname {Hypergeometric2F1}\left (-\frac {3}{n},-\frac {3}{n},\frac {-3+n}{n},\frac {b-\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )\right )+\left (-b^2+4 a c+b \sqrt {b^2-4 a c}\right ) \left (1-8^{-1/n} \left (\frac {c x^n}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )^{-3/n} \operatorname {Hypergeometric2F1}\left (-\frac {3}{n},-\frac {3}{n},\frac {-3+n}{n},\frac {b+\sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )\right )\right )+6 e x \left (\left (-b^2+4 a c-b \sqrt {b^2-4 a c}\right ) \left (1-\left (\frac {x^n}{-\frac {-b+\sqrt {b^2-4 a c}}{2 c}+x^n}\right )^{-2/n} \operatorname {Hypergeometric2F1}\left (-\frac {2}{n},-\frac {2}{n},\frac {-2+n}{n},\frac {b-\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )\right )+\left (-b^2+4 a c+b \sqrt {b^2-4 a c}\right ) \left (1-4^{-1/n} \left (\frac {c x^n}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )^{-2/n} \operatorname {Hypergeometric2F1}\left (-\frac {2}{n},-\frac {2}{n},\frac {-2+n}{n},\frac {b+\sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )\right )\right )+12 d \left (\left (-b^2+4 a c-b \sqrt {b^2-4 a c}\right ) \left (1-\left (\frac {x^n}{-\frac {-b+\sqrt {b^2-4 a c}}{2 c}+x^n}\right )^{-1/n} \operatorname {Hypergeometric2F1}\left (-\frac {1}{n},-\frac {1}{n},\frac {-1+n}{n},\frac {b-\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )\right )-2^{-1/n} \sqrt {b^2-4 a c} \left (-b+\sqrt {b^2-4 a c}\right ) \left (\frac {c x^n}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )^{-1/n} \left (2^{\frac {1}{n}} \left (\frac {c x^n}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )^{\frac {1}{n}}-\operatorname {Hypergeometric2F1}\left (-\frac {1}{n},-\frac {1}{n},\frac {-1+n}{n},\frac {b+\sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )\right )\right )\right )}{24 a \left (-b^2+4 a c\right )} \]
(x*(3*g*x^3*((-b^2 + 4*a*c - b*Sqrt[b^2 - 4*a*c])*(1 - Hypergeometric2F1[- 4/n, -4/n, (-4 + n)/n, (b - Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c] + 2* c*x^n)]/(x^n/(-1/2*(-b + Sqrt[b^2 - 4*a*c])/c + x^n))^(4/n)) + (-b^2 + 4*a *c + b*Sqrt[b^2 - 4*a*c])*(1 - Hypergeometric2F1[-4/n, -4/n, (-4 + n)/n, ( b + Sqrt[b^2 - 4*a*c])/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)]/(2^(4/n)*((c*x^n )/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n))^(4/n)))) + 4*f*x^2*((-b^2 + 4*a*c - b *Sqrt[b^2 - 4*a*c])*(1 - Hypergeometric2F1[-3/n, -3/n, (-3 + n)/n, (b - Sq rt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c] + 2*c*x^n)]/(x^n/(-1/2*(-b + Sqrt[ b^2 - 4*a*c])/c + x^n))^(3/n)) + (-b^2 + 4*a*c + b*Sqrt[b^2 - 4*a*c])*(1 - Hypergeometric2F1[-3/n, -3/n, (-3 + n)/n, (b + Sqrt[b^2 - 4*a*c])/(b + Sq rt[b^2 - 4*a*c] + 2*c*x^n)]/(8^n^(-1)*((c*x^n)/(b + Sqrt[b^2 - 4*a*c] + 2* c*x^n))^(3/n)))) + 6*e*x*((-b^2 + 4*a*c - b*Sqrt[b^2 - 4*a*c])*(1 - Hyperg eometric2F1[-2/n, -2/n, (-2 + n)/n, (b - Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c] + 2*c*x^n)]/(x^n/(-1/2*(-b + Sqrt[b^2 - 4*a*c])/c + x^n))^(2/n)) + (-b^2 + 4*a*c + b*Sqrt[b^2 - 4*a*c])*(1 - Hypergeometric2F1[-2/n, -2/n, (-2 + n)/n, (b + Sqrt[b^2 - 4*a*c])/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n)]/(4^ n^(-1)*((c*x^n)/(b + Sqrt[b^2 - 4*a*c] + 2*c*x^n))^(2/n)))) + 12*d*((-b^2 + 4*a*c - b*Sqrt[b^2 - 4*a*c])*(1 - Hypergeometric2F1[-n^(-1), -n^(-1), (- 1 + n)/n, (b - Sqrt[b^2 - 4*a*c])/(b - Sqrt[b^2 - 4*a*c] + 2*c*x^n)]/(x^n/ (-1/2*(-b + Sqrt[b^2 - 4*a*c])/c + x^n))^n^(-1)) - (Sqrt[b^2 - 4*a*c]*(...
Time = 0.71 (sec) , antiderivative size = 511, normalized size of antiderivative = 0.94, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {2325, 2432, 2009}
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.
| \(\displaystyle \int \frac {d+e x+f x^2+g x^3}{a+b x^n+c x^{2 n}} \, dx\) |
| \(\Big \downarrow \) 2325 |
| \(\displaystyle \frac {2 c \int \frac {g x^3+f x^2+e x+d}{2 c x^n+b-\sqrt {b^2-4 a c}}dx}{\sqrt {b^2-4 a c}}-\frac {2 c \int \frac {g x^3+f x^2+e x+d}{2 c x^n+b+\sqrt {b^2-4 a c}}dx}{\sqrt {b^2-4 a c}}\) |
| \(\Big \downarrow \) 2432 |
| \(\displaystyle \frac {2 c \int \left (-\frac {g x^3}{-2 c x^n-b+\sqrt {b^2-4 a c}}-\frac {f x^2}{-2 c x^n-b+\sqrt {b^2-4 a c}}-\frac {e x}{-2 c x^n-b+\sqrt {b^2-4 a c}}-\frac {d}{-2 c x^n-b+\sqrt {b^2-4 a c}}\right )dx}{\sqrt {b^2-4 a c}}-\frac {2 c \int \left (\frac {g x^3}{2 c x^n+b+\sqrt {b^2-4 a c}}+\frac {f x^2}{2 c x^n+b+\sqrt {b^2-4 a c}}+\frac {e x}{2 c x^n+b+\sqrt {b^2-4 a c}}+\frac {d}{2 c x^n+b+\sqrt {b^2-4 a c}}\right )dx}{\sqrt {b^2-4 a c}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 c \left (\frac {d x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{b-\sqrt {b^2-4 a c}}+\frac {e x^2 \operatorname {Hypergeometric2F1}\left (1,\frac {2}{n},\frac {n+2}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{2 \left (b-\sqrt {b^2-4 a c}\right )}+\frac {f x^3 \operatorname {Hypergeometric2F1}\left (1,\frac {3}{n},\frac {n+3}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{3 \left (b-\sqrt {b^2-4 a c}\right )}+\frac {g x^4 \operatorname {Hypergeometric2F1}\left (1,\frac {4}{n},\frac {n+4}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{4 \left (b-\sqrt {b^2-4 a c}\right )}\right )}{\sqrt {b^2-4 a c}}-\frac {2 c \left (\frac {d x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}+b}+\frac {e x^2 \operatorname {Hypergeometric2F1}\left (1,\frac {2}{n},\frac {n+2}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{2 \left (\sqrt {b^2-4 a c}+b\right )}+\frac {f x^3 \operatorname {Hypergeometric2F1}\left (1,\frac {3}{n},\frac {n+3}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{3 \left (\sqrt {b^2-4 a c}+b\right )}+\frac {g x^4 \operatorname {Hypergeometric2F1}\left (1,\frac {4}{n},\frac {n+4}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{4 \left (\sqrt {b^2-4 a c}+b\right )}\right )}{\sqrt {b^2-4 a c}}\) |
(2*c*((d*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b - Sqrt[b ^2 - 4*a*c])])/(b - Sqrt[b^2 - 4*a*c]) + (e*x^2*Hypergeometric2F1[1, 2/n, (2 + n)/n, (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(2*(b - Sqrt[b^2 - 4*a*c]) ) + (f*x^3*Hypergeometric2F1[1, 3/n, (3 + n)/n, (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(3*(b - Sqrt[b^2 - 4*a*c])) + (g*x^4*Hypergeometric2F1[1, 4/n, (4 + n)/n, (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(4*(b - Sqrt[b^2 - 4*a*c]) )))/Sqrt[b^2 - 4*a*c] - (2*c*((d*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1) , (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(b + Sqrt[b^2 - 4*a*c]) + (e*x^2*Hy pergeometric2F1[1, 2/n, (2 + n)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(2 *(b + Sqrt[b^2 - 4*a*c])) + (f*x^3*Hypergeometric2F1[1, 3/n, (3 + n)/n, (- 2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(3*(b + Sqrt[b^2 - 4*a*c])) + (g*x^4*Hy pergeometric2F1[1, 4/n, (4 + n)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(4 *(b + Sqrt[b^2 - 4*a*c]))))/Sqrt[b^2 - 4*a*c]
3.1.5.3.1 Defintions of rubi rules used
Int[(Pq_)/((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {q = Rt[b^2 - 4*a*c, 2]}, Simp[2*(c/q) Int[Pq/(b - q + 2*c*x^n), x], x] - Simp[2*(c/q) Int[Pq/(b + q + 2*c*x^n), x], x]] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0]
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[ Pq*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, n, p}, x] && (PolyQ[Pq, x] || Poly Q[Pq, x^n])
\[\int \frac {g \,x^{3}+f \,x^{2}+e x +d}{a +b \,x^{n}+c \,x^{2 n}}d x\]
\[ \int \frac {d+e x+f x^2+g x^3}{a+b x^n+c x^{2 n}} \, dx=\int { \frac {g x^{3} + f x^{2} + e x + d}{c x^{2 \, n} + b x^{n} + a} \,d x } \]
Timed out. \[ \int \frac {d+e x+f x^2+g x^3}{a+b x^n+c x^{2 n}} \, dx=\text {Timed out} \]
\[ \int \frac {d+e x+f x^2+g x^3}{a+b x^n+c x^{2 n}} \, dx=\int { \frac {g x^{3} + f x^{2} + e x + d}{c x^{2 \, n} + b x^{n} + a} \,d x } \]
\[ \int \frac {d+e x+f x^2+g x^3}{a+b x^n+c x^{2 n}} \, dx=\int { \frac {g x^{3} + f x^{2} + e x + d}{c x^{2 \, n} + b x^{n} + a} \,d x } \]
Timed out. \[ \int \frac {d+e x+f x^2+g x^3}{a+b x^n+c x^{2 n}} \, dx=\int \frac {g\,x^3+f\,x^2+e\,x+d}{a+b\,x^n+c\,x^{2\,n}} \,d x \]