3.1.8 \(\int \frac {d+e x+f x^2}{(a+b x^n+c x^{2 n})^2} \, dx\) [8]

Optimal. Leaf size=1194 \[ \frac {d x \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {e x^2 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {f x^3 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}-\frac {c d \left (4 a c (1-2 n)-b^2 (1-n)-b \sqrt {b^2-4 a c} (1-n)\right ) x \, _2F_1\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 d \left (4 a c (1-2 n)-b^2 (1-n)+b \sqrt {b^2-4 a c} (1-n)\right ) x \, _2F_1\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 e \left (4 a c (1-n)-b^2 (2-n)\right ) x^2 \, _2F_1\left (1,\frac {2}{n};\frac {2+n}{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 e \left (4 a c (1-n)-b^2 (2-n)\right ) x^2 \, _2F_1\left (1,\frac {2}{n};\frac {2+n}{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 {2 c f \left (2 a c (3-2 n)-b^2 (3-n)\right ) x^3 \, _2F_1\left (1,\frac {3}{n};\frac {3+n}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{3 a \left (b^2-4 a c\right ) \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) n}-\frac {2 c f \left (2 a c (3-2 n)-b^2 (3-n)\right ) x^3 \, _2F_1\left (1,\frac {3}{n};\frac {3+n}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{3 a \left (b^2-4 a c\right ) \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) n}-\frac {2 b c^2 e (2-n) x^{2+n} \, _2F_1\left (1,\frac {2+n}{n};2 \left (1+\frac {1}{n}\right );-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) n (2+n)}+\frac {2 b c^2 e (2-n) x^{2+n} \, _2F_1\left (1,\frac {2+n}{n};2 \left (1+\frac {1}{n}\right );-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) n (2+n)}-\frac {2 b c^2 f (3-n) x^{3+n} \, _2F_1\left (1,\frac {3+n}{n};2+\frac {3}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) n (3+n)}+\frac {2 b c^2 f (3-n) x^{3+n} \, _2F_1\left (1,\frac {3+n}{n};2+\frac {3}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) n (3+n)} \]

[Out]

d*x*(b^2-2*a*c+b*c*x^n)/a/(-4*a*c+b^2)/n/(a+b*x^n+c*x^(2*n))+e*x^2*(b^2-2*a*c+b*c*x^n)/a/(-4*a*c+b^2)/n/(a+b*x
^n+c*x^(2*n))+f*x^3*(b^2-2*a*c+b*c*x^n)/a/(-4*a*c+b^2)/n/(a+b*x^n+c*x^(2*n))-2*b*c^2*e*(2-n)*x^(2+n)*hypergeom
([1, (2+n)/n],[2+2/n],-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))/a/(-4*a*c+b^2)^(3/2)/n/(2+n)/(b-(-4*a*c+b^2)^(1/2))-2*b
*c^2*f*(3-n)*x^(3+n)*hypergeom([1, (3+n)/n],[2+3/n],-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))/a/(-4*a*c+b^2)^(3/2)/n/(3
+n)/(b-(-4*a*c+b^2)^(1/2))+2*b*c^2*e*(2-n)*x^(2+n)*hypergeom([1, (2+n)/n],[2+2/n],-2*c*x^n/(b+(-4*a*c+b^2)^(1/
2)))/a/(-4*a*c+b^2)^(3/2)/n/(2+n)/(b+(-4*a*c+b^2)^(1/2))+2*b*c^2*f*(3-n)*x^(3+n)*hypergeom([1, (3+n)/n],[2+3/n
],-2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))/a/(-4*a*c+b^2)^(3/2)/n/(3+n)/(b+(-4*a*c+b^2)^(1/2))-c*e*(4*a*c*(1-n)-b^2*(2
-n))*x^2*hypergeom([1, 2/n],[(2+n)/n],-2*c*x^n/(b-(-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))-2/3*c*f*(2*a*c*(3-2*n)-b^2*(3-n))*x^3*hypergeom([1, 3/n],[(3+n)/n],-2*c*x^n/(b-(-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*e*(4*a*c*(1-n)-b^2*(2-n))*x^2*hypergeom([1, 2/n],[(2+n)/n
],-2*c*x^n/(b+(-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))-2/3*c*f*(2*a*c*(3-2*n)-b^
2*(3-n))*x^3*hypergeom([1, 3/n],[(3+n)/n],-2*c*x^n/(b+(-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*d*x*hypergeom([1, 1/n],[1+1/n],-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))*(4*a*c*(1-2*n)-b^2*(1-n)-b*(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*d*x*hypergeom([1, 1/n],[1+1/n],-2*
c*x^n/(b+(-4*a*c+b^2)^(1/2)))*(4*a*c*(1-2*n)-b^2*(1-n)+b*(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))

________________________________________________________________________________________

Rubi [A]
time = 1.42, antiderivative size = 1194, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {1810, 1359, 1436, 251, 1398, 1574, 1397, 371} \begin {gather*} -\frac {2 b c^2 e (2-n) \, _2F_1\left (1,\frac {n+2}{n};2 \left (1+\frac {1}{n}\right );-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right ) x^{n+2}}{a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) n (n+2)}+\frac {2 b c^2 e (2-n) \, _2F_1\left (1,\frac {n+2}{n};2 \left (1+\frac {1}{n}\right );-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right ) x^{n+2}}{a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) n (n+2)}-\frac {2 b c^2 f (3-n) \, _2F_1\left (1,\frac {n+3}{n};2+\frac {3}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right ) x^{n+3}}{a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) n (n+3)}+\frac {2 b c^2 f (3-n) \, _2F_1\left (1,\frac {n+3}{n};2+\frac {3}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right ) x^{n+3}}{a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) n (n+3)}-\frac {2 c f \left (2 a c (3-2 n)-b^2 (3-n)\right ) \, _2F_1\left (1,\frac {3}{n};\frac {n+3}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right ) x^3}{3 a \left (b^2-4 a c\right ) \left (b^2-\sqrt {b^2-4 a c} b-4 a c\right ) n}-\frac {2 c f \left (2 a c (3-2 n)-b^2 (3-n)\right ) \, _2F_1\left (1,\frac {3}{n};\frac {n+3}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right ) x^3}{3 a \left (b^2-4 a c\right ) \left (b^2+\sqrt {b^2-4 a c} b-4 a c\right ) n}+\frac {f \left (b c x^n+b^2-2 a c\right ) x^3}{a \left (b^2-4 a c\right ) n \left (b x^n+c x^{2 n}+a\right )}-\frac {c e \left (4 a c (1-n)-b^2 (2-n)\right ) \, _2F_1\left (1,\frac {2}{n};\frac {n+2}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right ) x^2}{a \left (b^2-4 a c\right ) \left (b^2-\sqrt {b^2-4 a c} b-4 a c\right ) n}-\frac {c e \left (4 a c (1-n)-b^2 (2-n)\right ) \, _2F_1\left (1,\frac {2}{n};\frac {n+2}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right ) x^2}{a \left (b^2-4 a c\right ) \left (b^2+\sqrt {b^2-4 a c} b-4 a c\right ) n}+\frac {e \left (b c x^n+b^2-2 a c\right ) x^2}{a \left (b^2-4 a c\right ) n \left (b x^n+c x^{2 n}+a\right )}-\frac {c d \left (-\left ((1-n) b^2\right )-\sqrt {b^2-4 a c} (1-n) b+4 a c (1-2 n)\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right ) x}{a \left (b^2-4 a c\right ) \left (b^2-\sqrt {b^2-4 a c} b-4 a c\right ) n}-\frac {c d \left (-\left ((1-n) b^2\right )+\sqrt {b^2-4 a c} (1-n) b+4 a c (1-2 n)\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right ) x}{a \left (b^2-4 a c\right ) \left (b^2+\sqrt {b^2-4 a c} b-4 a c\right ) n}+\frac {d \left (b c x^n+b^2-2 a c\right ) x}{a \left (b^2-4 a c\right ) n \left (b x^n+c x^{2 n}+a\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x + f*x^2)/(a + b*x^n + c*x^(2*n))^2,x]

[Out]

(d*x*(b^2 - 2*a*c + b*c*x^n))/(a*(b^2 - 4*a*c)*n*(a + b*x^n + c*x^(2*n))) + (e*x^2*(b^2 - 2*a*c + b*c*x^n))/(a
*(b^2 - 4*a*c)*n*(a + b*x^n + c*x^(2*n))) + (f*x^3*(b^2 - 2*a*c + b*c*x^n))/(a*(b^2 - 4*a*c)*n*(a + b*x^n + c*
x^(2*n))) - (c*d*(4*a*c*(1 - 2*n) - b^2*(1 - n) - b*Sqrt[b^2 - 4*a*c]*(1 - n))*x*Hypergeometric2F1[1, n^(-1),
1 + n^(-1), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)*(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c])*n) - (c*
d*(4*a*c*(1 - 2*n) - b^2*(1 - n) + b*Sqrt[b^2 - 4*a*c]*(1 - n))*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2
*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)*(b^2 - 4*a*c + b*Sqrt[b^2 - 4*a*c])*n) - (c*e*(4*a*c*(1 - n
) - b^2*(2 - n))*x^2*Hypergeometric2F1[1, 2/n, (2 + n)/n, (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c
)*(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c])*n) - (c*e*(4*a*c*(1 - n) - b^2*(2 - n))*x^2*Hypergeometric2F1[1, 2/n, (2
+ n)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)*(b^2 - 4*a*c + b*Sqrt[b^2 - 4*a*c])*n) - (2*c*f
*(2*a*c*(3 - 2*n) - b^2*(3 - n))*x^3*Hypergeometric2F1[1, 3/n, (3 + n)/n, (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])
/(3*a*(b^2 - 4*a*c)*(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c])*n) - (2*c*f*(2*a*c*(3 - 2*n) - b^2*(3 - n))*x^3*Hyperg
eometric2F1[1, 3/n, (3 + n)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(3*a*(b^2 - 4*a*c)*(b^2 - 4*a*c + b*Sqrt[b
^2 - 4*a*c])*n) - (2*b*c^2*e*(2 - n)*x^(2 + n)*Hypergeometric2F1[1, (2 + n)/n, 2*(1 + n^(-1)), (-2*c*x^n)/(b -
Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)^(3/2)*(b - Sqrt[b^2 - 4*a*c])*n*(2 + n)) + (2*b*c^2*e*(2 - n)*x^(2 + n)
*Hypergeometric2F1[1, (2 + n)/n, 2*(1 + n^(-1)), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)^(3/2)*(
b + Sqrt[b^2 - 4*a*c])*n*(2 + n)) - (2*b*c^2*f*(3 - n)*x^(3 + n)*Hypergeometric2F1[1, (3 + n)/n, 2 + 3/n, (-2*
c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)^(3/2)*(b - Sqrt[b^2 - 4*a*c])*n*(3 + n)) + (2*b*c^2*f*(3 - n
)*x^(3 + n)*Hypergeometric2F1[1, (3 + n)/n, 2 + 3/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(a*(b^2 - 4*a*c)^(3/
2)*(b + Sqrt[b^2 - 4*a*c])*n*(3 + n))

Rule 251

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-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])

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 1359

Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*(b^2 - 2*a*c + b*c*x^n)*((a + b*
x^n + c*x^(2*n))^(p + 1)/(a*n*(p + 1)*(b^2 - 4*a*c))), x] + Dist[1/(a*n*(p + 1)*(b^2 - 4*a*c)), Int[(b^2 - 2*a
*c + n*(p + 1)*(b^2 - 4*a*c) + b*c*(n*(2*p + 3) + 1)*x^n)*(a + b*x^n + c*x^(2*n))^(p + 1), x], x] /; FreeQ[{a,
b, c, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && ILtQ[p, -1]

Rule 1397

Int[((d_.)*(x_))^(m_.)/((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]
}, Dist[2*(c/q), Int[(d*x)^m/(b - q + 2*c*x^n), x], x] - Dist[2*(c/q), Int[(d*x)^m/(b + q + 2*c*x^n), x], x]]
/; FreeQ[{a, b, c, d, m, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0]

Rule 1398

Int[((d_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-(d*x)^(m + 1))*(
b^2 - 2*a*c + b*c*x^n)*((a + b*x^n + c*x^(2*n))^(p + 1)/(a*d*n*(p + 1)*(b^2 - 4*a*c))), x] + Dist[1/(a*n*(p +
1)*(b^2 - 4*a*c)), Int[(d*x)^m*(a + b*x^n + c*x^(2*n))^(p + 1)*Simp[b^2*(n*(p + 1) + m + 1) - 2*a*c*(m + 2*n*(
p + 1) + 1) + b*c*(2*n*p + 3*n + m + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, m, n}, x] && EqQ[n2, 2*n] && NeQ
[b^2 - 4*a*c, 0] && ILtQ[p + 1, 0]

Rule 1436

Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*
c, 2]}, Dist[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^n), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), In
t[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])

Rule 1574

Int[((f_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Sy
mbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^n)^q*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, d, e,
f, m, n, p, q}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && (IGtQ[p, 0] || IGtQ[q, 0])

Rule 1810

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^n +
c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && PolyQ[Pq, x] && ILtQ[p, -1]

Rubi steps

\begin {align*} \int \frac {d+e x+f x^2}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx &=\int \left (\frac {d}{\left (a+b x^n+c x^{2 n}\right )^2}+\frac {e x}{\left (a+b x^n+c x^{2 n}\right )^2}+\frac {f x^2}{\left (a+b x^n+c x^{2 n}\right )^2}\right ) \, dx\\ &=d \int \frac {1}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx+e \int \frac {x}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx+f \int \frac {x^2}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx\\ &=\frac {d x \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {e x^2 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {f x^3 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}-\frac {d \int \frac {b^2-2 a c-\left (b^2-4 a c\right ) n+b c (1-n) x^n}{a+b x^n+c x^{2 n}} \, dx}{a \left (b^2-4 a c\right ) n}-\frac {e \int \frac {x \left (-4 a c (1-n)+b^2 (2-n)+b c (2-n) x^n\right )}{a+b x^n+c x^{2 n}} \, dx}{a \left (b^2-4 a c\right ) n}-\frac {f \int \frac {x^2 \left (-2 a c (3-2 n)+b^2 (3-n)+b c (3-n) x^n\right )}{a+b x^n+c x^{2 n}} \, dx}{a \left (b^2-4 a c\right ) n}\\ &=\frac {d x \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {e x^2 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {f x^3 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}-\frac {e \int \left (-\frac {b^2 \left (1-\frac {4 a c (-1+n)}{b^2 (-2+n)}\right ) (-2+n) x}{a+b x^n+c x^{2 n}}-\frac {b c (-2+n) x^{1+n}}{a+b x^n+c x^{2 n}}\right ) \, dx}{a \left (b^2-4 a c\right ) n}-\frac {f \int \left (-\frac {b^2 (-3+n) \left (1-\frac {2 a c (-3+2 n)}{b^2 (-3+n)}\right ) x^2}{a+b x^n+c x^{2 n}}-\frac {b c (-3+n) x^{2+n}}{a+b x^n+c x^{2 n}}\right ) \, dx}{a \left (b^2-4 a c\right ) n}+\frac {\left (c d \left (4 a c (1-2 n)-b^2 (1-n)-b \sqrt {b^2-4 a c} (1-n)\right )\right ) \int \frac {1}{\frac {b}{2}-\frac {1}{2} \sqrt {b^2-4 a c}+c x^n} \, dx}{2 a \left (b^2-4 a c\right )^{3/2} n}-\frac {\left (c d \left (4 a c (1-2 n)-b^2 (1-n)+b \sqrt {b^2-4 a c} (1-n)\right )\right ) \int \frac {1}{\frac {b}{2}+\frac {1}{2} \sqrt {b^2-4 a c}+c x^n} \, dx}{2 a \left (b^2-4 a c\right )^{3/2} n}\\ &=\frac {d x \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {e x^2 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {f x^3 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {c d \left (4 a c (1-2 n)-b^2 (1-n)-b \sqrt {b^2-4 a c} (1-n)\right ) x \, _2F_1\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 )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) n}-\frac {c d \left (4 a c (1-2 n)-b^2 (1-n)+b \sqrt {b^2-4 a c} (1-n)\right ) x \, _2F_1\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 )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) n}+\frac {\left (e \left (4 a c (1-n)-b^2 (2-n)\right )\right ) \int \frac {x}{a+b x^n+c x^{2 n}} \, dx}{a \left (b^2-4 a c\right ) n}+\frac {\left (f \left (2 a c (3-2 n)-b^2 (3-n)\right )\right ) \int \frac {x^2}{a+b x^n+c x^{2 n}} \, dx}{a \left (b^2-4 a c\right ) n}-\frac {(b c e (2-n)) \int \frac {x^{1+n}}{a+b x^n+c x^{2 n}} \, dx}{a \left (b^2-4 a c\right ) n}-\frac {(b c f (3-n)) \int \frac {x^{2+n}}{a+b x^n+c x^{2 n}} \, dx}{a \left (b^2-4 a c\right ) n}\\ &=\frac {d x \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {e x^2 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {f x^3 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {c d \left (4 a c (1-2 n)-b^2 (1-n)-b \sqrt {b^2-4 a c} (1-n)\right ) x \, _2F_1\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 )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) n}-\frac {c d \left (4 a c (1-2 n)-b^2 (1-n)+b \sqrt {b^2-4 a c} (1-n)\right ) x \, _2F_1\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 )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) n}+\frac {\left (2 c e \left (4 a c (1-n)-b^2 (2-n)\right )\right ) \int \frac {x}{b-\sqrt {b^2-4 a c}+2 c x^n} \, dx}{a \left (b^2-4 a c\right )^{3/2} n}-\frac {\left (2 c e \left (4 a c (1-n)-b^2 (2-n)\right )\right ) \int \frac {x}{b+\sqrt {b^2-4 a c}+2 c x^n} \, dx}{a \left (b^2-4 a c\right )^{3/2} n}+\frac {\left (2 c f \left (2 a c (3-2 n)-b^2 (3-n)\right )\right ) \int \frac {x^2}{b-\sqrt {b^2-4 a c}+2 c x^n} \, dx}{a \left (b^2-4 a c\right )^{3/2} n}-\frac {\left (2 c f \left (2 a c (3-2 n)-b^2 (3-n)\right )\right ) \int \frac {x^2}{b+\sqrt {b^2-4 a c}+2 c x^n} \, dx}{a \left (b^2-4 a c\right )^{3/2} n}-\frac {\left (2 b c^2 e (2-n)\right ) \int \frac {x^{1+n}}{b-\sqrt {b^2-4 a c}+2 c x^n} \, dx}{a \left (b^2-4 a c\right )^{3/2} n}+\frac {\left (2 b c^2 e (2-n)\right ) \int \frac {x^{1+n}}{b+\sqrt {b^2-4 a c}+2 c x^n} \, dx}{a \left (b^2-4 a c\right )^{3/2} n}-\frac {\left (2 b c^2 f (3-n)\right ) \int \frac {x^{2+n}}{b-\sqrt {b^2-4 a c}+2 c x^n} \, dx}{a \left (b^2-4 a c\right )^{3/2} n}+\frac {\left (2 b c^2 f (3-n)\right ) \int \frac {x^{2+n}}{b+\sqrt {b^2-4 a c}+2 c x^n} \, dx}{a \left (b^2-4 a c\right )^{3/2} n}\\ &=\frac {d x \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {e x^2 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {f x^3 \left (b^2-2 a c+b c x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {c d \left (4 a c (1-2 n)-b^2 (1-n)-b \sqrt {b^2-4 a c} (1-n)\right ) x \, _2F_1\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 )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) n}-\frac {c d \left (4 a c (1-2 n)-b^2 (1-n)+b \sqrt {b^2-4 a c} (1-n)\right ) x \, _2F_1\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 )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) n}+\frac {c e \left (4 a c (1-n)-b^2 (2-n)\right ) x^2 \, _2F_1\left (1,\frac {2}{n};\frac {2+n}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) n}-\frac {c e \left (4 a c (1-n)-b^2 (2-n)\right ) x^2 \, _2F_1\left (1,\frac {2}{n};\frac {2+n}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) n}+\frac {2 c f \left (2 a c (3-2 n)-b^2 (3-n)\right ) x^3 \, _2F_1\left (1,\frac {3}{n};\frac {3+n}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{3 a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) n}-\frac {2 c f \left (2 a c (3-2 n)-b^2 (3-n)\right ) x^3 \, _2F_1\left (1,\frac {3}{n};\frac {3+n}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{3 a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) n}-\frac {2 b c^2 e (2-n) x^{2+n} \, _2F_1\left (1,\frac {2+n}{n};2 \left (1+\frac {1}{n}\right );-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) n (2+n)}+\frac {2 b c^2 e (2-n) x^{2+n} \, _2F_1\left (1,\frac {2+n}{n};2 \left (1+\frac {1}{n}\right );-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) n (2+n)}-\frac {2 b c^2 f (3-n) x^{3+n} \, _2F_1\left (1,\frac {3+n}{n};2+\frac {3}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right )^{3/2} \left (b-\sqrt {b^2-4 a c}\right ) n (3+n)}+\frac {2 b c^2 f (3-n) x^{3+n} \, _2F_1\left (1,\frac {3+n}{n};2+\frac {3}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right )^{3/2} \left (b+\sqrt {b^2-4 a c}\right ) n (3+n)}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(6525\) vs. \(2(1194)=2388\).
time = 6.47, size = 6525, normalized size = 5.46 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(d + e*x + f*x^2)/(a + b*x^n + c*x^(2*n))^2,x]

[Out]

Result too large to show

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Maple [F]
time = 0.00, size = 0, normalized size = 0.00 \[\int \frac {f \,x^{2}+e x +d}{\left (a +b \,x^{n}+c \,x^{2 n}\right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x^2+e*x+d)/(a+b*x^n+c*x^(2*n))^2,x)

[Out]

int((f*x^2+e*x+d)/(a+b*x^n+c*x^(2*n))^2,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^2+e*x+d)/(a+b*x^n+c*x^(2*n))^2,x, algorithm="maxima")

[Out]

((b^2*f - 2*a*c*f)*x^3 + (b^2 - 2*a*c)*x^2*e + (b*c*f*x^3 + b*c*x^2*e + b*c*d*x)*x^n + (b^2*d - 2*a*c*d)*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((2*a*c*d*(
2*n - 1) - b^2*d*(n - 1) + (2*a*c*f*(2*n - 3) - b^2*f*(n - 3))*x^2 + (4*a*c*(n - 1) - b^2*(n - 2))*x*e - (b*c*
f*(n - 3)*x^2 + b*c*(n - 2)*x*e + b*c*d*(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)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^2+e*x+d)/(a+b*x^n+c*x^(2*n))^2,x, algorithm="fricas")

[Out]

integral((f*x^2 + x*e + 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)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x**2+e*x+d)/(a+b*x**n+c*x**(2*n))**2,x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^2+e*x+d)/(a+b*x^n+c*x^(2*n))^2,x, algorithm="giac")

[Out]

integrate((f*x^2 + x*e + d)/(c*x^(2*n) + b*x^n + a)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {f\,x^2+e\,x+d}{{\left (a+b\,x^n+c\,x^{2\,n}\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x + f*x^2)/(a + b*x^n + c*x^(2*n))^2,x)

[Out]

int((d + e*x + f*x^2)/(a + b*x^n + c*x^(2*n))^2, x)

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