Integrand size = 24, antiderivative size = 362 \[ \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\frac {x \left (b^2 d-2 a c d-a b e+c (b d-2 a e) x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}-\frac {c \left (2 a \left (2 c d (1-2 n)+\sqrt {b^2-4 a c} e (1-n)\right )-b^2 (d-d n)-b \left (\sqrt {b^2-4 a c} d (1-n)-2 a e n\right )\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right ) \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) n}-\frac {c \left (2 a \left (c d (2-4 n)-\sqrt {b^2-4 a c} e (1-n)\right )-b^2 d (1-n)+b \left (\sqrt {b^2-4 a c} d (1-n)+2 a e n\right )\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right ) \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) n} \]
x*(b^2*d-2*a*c*d-a*b*e+c*(-2*a*e+b*d)*x^n)/a/(-4*a*c+b^2)/n/(a+b*x^n+c*x^( 2*n))-c*x*hypergeom([1, 1/n],[1+1/n],-2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))*(-b^ 2*d*(1-n)+b*(2*a*e*n+d*(1-n)*(-4*a*c+b^2)^(1/2))+2*a*(c*d*(2-4*n)-e*(1-n)* (-4*a*c+b^2)^(1/2)))/a/(-4*a*c+b^2)/n/(b^2-4*a*c+b*(-4*a*c+b^2)^(1/2))-c*x *hypergeom([1, 1/n],[1+1/n],-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))*(-b^2*(-d*n+d )-b*(-2*a*e*n+d*(1-n)*(-4*a*c+b^2)^(1/2))+2*a*(2*c*d*(1-2*n)+e*(1-n)*(-4*a *c+b^2)^(1/2)))/a/(-4*a*c+b^2)/n/(b^2-4*a*c-b*(-4*a*c+b^2)^(1/2))
Time = 4.18 (sec) , antiderivative size = 603, normalized size of antiderivative = 1.67 \[ \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\frac {c x \left (\frac {4 \left (b^2-4 a c\right ) \left (b^2 d (-1+n) x^n \left (b+c x^n\right )-2 a^2 c \left (2 d n+e x^n\right )+a \left (-2 c^2 d (-1+2 n) x^{2 n}+b c x^n \left (3 d-4 d n+e x^n\right )+b^2 \left (d n+e x^n\right )\right )\right )}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) \left (a+x^n \left (b+c x^n\right )\right )}+\frac {2^{-1/n} \left (4 a c \left (\sqrt {b^2-4 a c} d (1-2 n)+2 a e (-1+n)\right )+b^3 d (-1+n)+b^2 \left (\sqrt {b^2-4 a c} d-2 a e\right ) (-1+n)+2 a b \left (-2 c d (-1+n)+\sqrt {b^2-4 a c} e n\right )\right ) \left (\frac {c x^n}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )^{-1/n} \operatorname {Hypergeometric2F1}\left (-\frac {1}{n},-\frac {1}{n},\frac {-1+n}{n},\frac {b-\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )}{\sqrt {b^2-4 a c} \left (-b^2+4 a c+b \sqrt {b^2-4 a c}\right )}+\frac {2^{-1/n} \left (b \sqrt {b^2-4 a c} d (-1+n)-2 a \sqrt {b^2-4 a c} e (-1+n)-2 a b e n+4 a c d (-1+2 n)+b^2 (d-d n)\right ) \left (\frac {c x^n}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )^{-1/n} \operatorname {Hypergeometric2F1}\left (-\frac {1}{n},-\frac {1}{n},\frac {-1+n}{n},\frac {b+\sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right )}\right )}{a \left (-b^2+4 a c\right ) n} \]
(c*x*((4*(b^2 - 4*a*c)*(b^2*d*(-1 + n)*x^n*(b + c*x^n) - 2*a^2*c*(2*d*n + e*x^n) + a*(-2*c^2*d*(-1 + 2*n)*x^(2*n) + b*c*x^n*(3*d - 4*d*n + e*x^n) + b^2*(d*n + e*x^n))))/((b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c])*(b^2 - 4*a*c + b *Sqrt[b^2 - 4*a*c])*(a + x^n*(b + c*x^n))) + ((4*a*c*(Sqrt[b^2 - 4*a*c]*d* (1 - 2*n) + 2*a*e*(-1 + n)) + b^3*d*(-1 + n) + b^2*(Sqrt[b^2 - 4*a*c]*d - 2*a*e)*(-1 + n) + 2*a*b*(-2*c*d*(-1 + n) + Sqrt[b^2 - 4*a*c]*e*n))*Hyperge ometric2F1[-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)])/(2^n^(-1)*Sqrt[b^2 - 4*a*c]*(-b^2 + 4*a*c + b*S qrt[b^2 - 4*a*c])*((c*x^n)/(b - Sqrt[b^2 - 4*a*c] + 2*c*x^n))^n^(-1)) + (( b*Sqrt[b^2 - 4*a*c]*d*(-1 + n) - 2*a*Sqrt[b^2 - 4*a*c]*e*(-1 + n) - 2*a*b* e*n + 4*a*c*d*(-1 + 2*n) + b^2*(d - d*n))*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)] )/(2^n^(-1)*Sqrt[b^2 - 4*a*c]*(b + Sqrt[b^2 - 4*a*c])*((c*x^n)/(b + Sqrt[b ^2 - 4*a*c] + 2*c*x^n))^n^(-1))))/(a*(-b^2 + 4*a*c)*n)
Time = 1.17 (sec) , antiderivative size = 312, normalized size of antiderivative = 0.86, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1760, 25, 1752, 778}
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.
\(\displaystyle \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx\) |
\(\Big \downarrow \) 1760 |
\(\displaystyle \frac {x \left (c x^n (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{a n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )}-\frac {\int -\frac {-c (b d-2 a e) (1-n) x^n+a b e+2 a c d (1-2 n)-b^2 (d-d n)}{b x^n+c x^{2 n}+a}dx}{a n \left (b^2-4 a c\right )}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {-c (b d-2 a e) (1-n) x^n+a b e+2 a c d (1-2 n)-b^2 d (1-n)}{b x^n+c x^{2 n}+a}dx}{a n \left (b^2-4 a c\right )}+\frac {x \left (c x^n (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{a n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )}\) |
\(\Big \downarrow \) 1752 |
\(\displaystyle \frac {\frac {c \left (-(1-n) \sqrt {b^2-4 a c} (b d-2 a e)+2 a b e n+2 a c d (2-4 n)+b^2 (-d) (1-n)\right ) \int \frac {1}{c x^n+\frac {1}{2} \left (b-\sqrt {b^2-4 a c}\right )}dx}{2 \sqrt {b^2-4 a c}}-\frac {1}{2} c \left (\frac {2 a b e n+4 a c d (1-2 n)+b^2 (-d) (1-n)}{\sqrt {b^2-4 a c}}+(1-n) (b d-2 a e)\right ) \int \frac {1}{c x^n+\frac {1}{2} \left (b+\sqrt {b^2-4 a c}\right )}dx}{a n \left (b^2-4 a c\right )}+\frac {x \left (c x^n (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{a n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )}\) |
\(\Big \downarrow \) 778 |
\(\displaystyle \frac {\frac {c x \left (-(1-n) \sqrt {b^2-4 a c} (b d-2 a e)+2 a b e n+2 a c d (2-4 n)+b^2 (-d) (1-n)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}\right )}-\frac {c x \left (\frac {2 a b e n+4 a c d (1-2 n)+b^2 (-d) (1-n)}{\sqrt {b^2-4 a c}}+(1-n) (b d-2 a e)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}+b}}{a n \left (b^2-4 a c\right )}+\frac {x \left (c x^n (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{a n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )}\) |
(x*(b^2*d - 2*a*c*d - a*b*e + c*(b*d - 2*a*e)*x^n))/(a*(b^2 - 4*a*c)*n*(a + b*x^n + c*x^(2*n))) + ((c*(2*a*c*d*(2 - 4*n) - b^2*d*(1 - n) - Sqrt[b^2 - 4*a*c]*(b*d - 2*a*e)*(1 - n) + 2*a*b*e*n)*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(Sqrt[b^2 - 4*a*c]*(b - Sqrt[b^2 - 4*a*c])) - (c*((b*d - 2*a*e)*(1 - n) + (4*a*c*d*(1 - 2*n) - b^2 *d*(1 - n) + 2*a*b*e*n)/Sqrt[b^2 - 4*a*c])*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]))/ (a*(b^2 - 4*a*c)*n)
3.1.77.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F 1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, n, p}, x] && !IGtQ[p , 0] && !IntegerQ[1/n] && !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p] || GtQ[a, 0])
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])
Int[((d_) + (e_.)*(x_)^(n_))*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p _), x_Symbol] :> Simp[(-x)*(d*b^2 - a*b*e - 2*a*c*d + (b*d - 2*a*e)*c*x^n)* ((a + b*x^n + c*x^(2*n))^(p + 1)/(a*n*(p + 1)*(b^2 - 4*a*c))), x] + Simp[1/ (a*n*(p + 1)*(b^2 - 4*a*c)) Int[Simp[(n*p + n + 1)*d*b^2 - a*b*e - 2*a*c* d*(2*n*p + 2*n + 1) + (2*n*p + 3*n + 1)*(d*b - 2*a*e)*c*x^n, x]*(a + b*x^n + c*x^(2*n))^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2, 2*n ] && NeQ[b^2 - 4*a*c, 0] && ILtQ[p, -1]
\[\int \frac {d +e \,x^{n}}{\left (a +b \,x^{n}+c \,x^{2 n}\right )^{2}}d x\]
\[ \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int { \frac {e x^{n} + d}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}} \,d x } \]
integral((e*x^n + d)/(c^2*x^(4*n) + b^2*x^(2*n) + 2*a*b*x^n + a^2 + 2*(b*c *x^n + a*c)*x^(2*n)), x)
Timed out. \[ \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int { \frac {e x^{n} + d}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}} \,d x } \]
((b*c*d - 2*a*c*e)*x*x^n + (b^2*d - (2*c*d + b*e)*a)*x)/(a^2*b^2*n - 4*a^3 *c*n + (a*b^2*c*n - 4*a^2*c^2*n)*x^(2*n) + (a*b^3*n - 4*a^2*b*c*n)*x^n) + integrate((b^2*d*(n - 1) - (2*c*d*(2*n - 1) - b*e)*a + (b*c*d*(n - 1) - 2* a*c*e*(n - 1))*x^n)/(a^2*b^2*n - 4*a^3*c*n + (a*b^2*c*n - 4*a^2*c^2*n)*x^( 2*n) + (a*b^3*n - 4*a^2*b*c*n)*x^n), x)
\[ \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int { \frac {e x^{n} + d}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int \frac {d+e\,x^n}{{\left (a+b\,x^n+c\,x^{2\,n}\right )}^2} \,d x \]