### 3.143 $$\int \frac{(f x)^m (d+e x^n)}{(a+b x^n+c x^{2 n})^3} \, dx$$

Optimal. Leaf size=816 $-\frac{c \left (\left (-d (m-2 n+1) b^3+a e (m+1) b^2+2 a c d (2 m-7 n+2) b-4 a^2 c e (m-3 n+1)\right ) (m-n+1)+\frac{-d \left (m^2+(2-3 n) m+2 n^2-3 n+1\right ) b^4+a e (m+1) (m-n+1) b^3+6 a c d \left (m^2+(2-4 n) m+3 n^2-4 n+1\right ) b^2-4 a^2 c e \left (m^2+(2-n) m-3 n^2-n+1\right ) b-8 a^2 c^2 d \left (m^2+(2-6 n) m+8 n^2-6 n+1\right )}{\sqrt{b^2-4 a c}}\right ) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right ) (f x)^{m+1}}{2 a^2 \left (b^2-4 a c\right )^2 \left (b-\sqrt{b^2-4 a c}\right ) f (m+1) n^2}-\frac{c \left (\left (-d (m-2 n+1) b^3+a e (m+1) b^2+2 a c d (2 m-7 n+2) b-4 a^2 c e (m-3 n+1)\right ) (m-n+1)-\frac{-d \left (m^2+(2-3 n) m+2 n^2-3 n+1\right ) b^4+a e (m+1) (m-n+1) b^3+6 a c d \left (m^2+(2-4 n) m+3 n^2-4 n+1\right ) b^2-4 a^2 c e \left (m^2+(2-n) m-3 n^2-n+1\right ) b-8 a^2 c^2 d \left (m^2+(2-6 n) m+8 n^2-6 n+1\right )}{\sqrt{b^2-4 a c}}\right ) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right ) (f x)^{m+1}}{2 a^2 \left (b^2-4 a c\right )^2 \left (b+\sqrt{b^2-4 a c}\right ) f (m+1) n^2}+\frac{\left (c \left (-d (m-2 n+1) b^3+a e (m+1) b^2+2 a c d (2 m-7 n+2) b-4 a^2 c e (m-3 n+1)\right ) x^n+\left (b^2-2 a c\right ) \left (-d (m-2 n+1) b^2+a e (m+1) b+2 a c d (m-4 n+1)\right )+a b c (b d-2 a e) (m-3 n+1)\right ) (f x)^{m+1}}{2 a^2 \left (b^2-4 a c\right )^2 f n^2 \left (b x^n+c x^{2 n}+a\right )}+\frac{\left (c (b d-2 a e) x^n+b^2 d-2 a c d-a b e\right ) (f x)^{m+1}}{2 a \left (b^2-4 a c\right ) f n \left (b x^n+c x^{2 n}+a\right )^2}$

[Out]

((f*x)^(1 + m)*(b^2*d - 2*a*c*d - a*b*e + c*(b*d - 2*a*e)*x^n))/(2*a*(b^2 - 4*a*c)*f*n*(a + b*x^n + c*x^(2*n))
^2) + ((f*x)^(1 + m)*((b^2 - 2*a*c)*(a*b*e*(1 + m) + 2*a*c*d*(1 + m - 4*n) - b^2*d*(1 + m - 2*n)) + a*b*c*(b*d
- 2*a*e)*(1 + m - 3*n) + c*(a*b^2*e*(1 + m) + 2*a*b*c*d*(2 + 2*m - 7*n) - 4*a^2*c*e*(1 + m - 3*n) - b^3*d*(1
+ m - 2*n))*x^n))/(2*a^2*(b^2 - 4*a*c)^2*f*n^2*(a + b*x^n + c*x^(2*n))) - (c*((a*b^2*e*(1 + m) + 2*a*b*c*d*(2
+ 2*m - 7*n) - 4*a^2*c*e*(1 + m - 3*n) - b^3*d*(1 + m - 2*n))*(1 + m - n) + (a*b^3*e*(1 + m)*(1 + m - n) - 4*a
^2*b*c*e*(1 + m^2 + m*(2 - n) - n - 3*n^2) - b^4*d*(1 + m^2 + m*(2 - 3*n) - 3*n + 2*n^2) + 6*a*b^2*c*d*(1 + m^
2 + m*(2 - 4*n) - 4*n + 3*n^2) - 8*a^2*c^2*d*(1 + m^2 + m*(2 - 6*n) - 6*n + 8*n^2))/Sqrt[b^2 - 4*a*c])*(f*x)^(
1 + m)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(2*a^2*(b^2 - 4*a*c
)^2*(b - Sqrt[b^2 - 4*a*c])*f*(1 + m)*n^2) - (c*((a*b^2*e*(1 + m) + 2*a*b*c*d*(2 + 2*m - 7*n) - 4*a^2*c*e*(1 +
m - 3*n) - b^3*d*(1 + m - 2*n))*(1 + m - n) - (a*b^3*e*(1 + m)*(1 + m - n) - 4*a^2*b*c*e*(1 + m^2 + m*(2 - n)
- n - 3*n^2) - b^4*d*(1 + m^2 + m*(2 - 3*n) - 3*n + 2*n^2) + 6*a*b^2*c*d*(1 + m^2 + m*(2 - 4*n) - 4*n + 3*n^2
) - 8*a^2*c^2*d*(1 + m^2 + m*(2 - 6*n) - 6*n + 8*n^2))/Sqrt[b^2 - 4*a*c])*(f*x)^(1 + m)*Hypergeometric2F1[1, (
1 + m)/n, (1 + m + n)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(2*a^2*(b^2 - 4*a*c)^2*(b + Sqrt[b^2 - 4*a*c])*f
*(1 + m)*n^2)

________________________________________________________________________________________

Rubi [A]  time = 4.55205, antiderivative size = 816, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 29, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.103, Rules used = {1558, 1560, 364} $-\frac{c \left (\left (-d (m-2 n+1) b^3+a e (m+1) b^2+2 a c d (2 m-7 n+2) b-4 a^2 c e (m-3 n+1)\right ) (m-n+1)+\frac{-d \left (m^2+(2-3 n) m+2 n^2-3 n+1\right ) b^4+a e (m+1) (m-n+1) b^3+6 a c d \left (m^2+(2-4 n) m+3 n^2-4 n+1\right ) b^2-4 a^2 c e \left (m^2+(2-n) m-3 n^2-n+1\right ) b-8 a^2 c^2 d \left (m^2+(2-6 n) m+8 n^2-6 n+1\right )}{\sqrt{b^2-4 a c}}\right ) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right ) (f x)^{m+1}}{2 a^2 \left (b^2-4 a c\right )^2 \left (b-\sqrt{b^2-4 a c}\right ) f (m+1) n^2}-\frac{c \left (\left (-d (m-2 n+1) b^3+a e (m+1) b^2+2 a c d (2 m-7 n+2) b-4 a^2 c e (m-3 n+1)\right ) (m-n+1)-\frac{-d \left (m^2+(2-3 n) m+2 n^2-3 n+1\right ) b^4+a e (m+1) (m-n+1) b^3+6 a c d \left (m^2+(2-4 n) m+3 n^2-4 n+1\right ) b^2-4 a^2 c e \left (m^2+(2-n) m-3 n^2-n+1\right ) b-8 a^2 c^2 d \left (m^2+(2-6 n) m+8 n^2-6 n+1\right )}{\sqrt{b^2-4 a c}}\right ) \, _2F_1\left (1,\frac{m+1}{n};\frac{m+n+1}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right ) (f x)^{m+1}}{2 a^2 \left (b^2-4 a c\right )^2 \left (b+\sqrt{b^2-4 a c}\right ) f (m+1) n^2}+\frac{\left (c \left (-d (m-2 n+1) b^3+a e (m+1) b^2+2 a c d (2 m-7 n+2) b-4 a^2 c e (m-3 n+1)\right ) x^n+\left (b^2-2 a c\right ) \left (-d (m-2 n+1) b^2+a e (m+1) b+2 a c d (m-4 n+1)\right )+a b c (b d-2 a e) (m-3 n+1)\right ) (f x)^{m+1}}{2 a^2 \left (b^2-4 a c\right )^2 f n^2 \left (b x^n+c x^{2 n}+a\right )}+\frac{\left (c (b d-2 a e) x^n+b^2 d-2 a c d-a b e\right ) (f x)^{m+1}}{2 a \left (b^2-4 a c\right ) f n \left (b x^n+c x^{2 n}+a\right )^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[((f*x)^m*(d + e*x^n))/(a + b*x^n + c*x^(2*n))^3,x]

[Out]

((f*x)^(1 + m)*(b^2*d - 2*a*c*d - a*b*e + c*(b*d - 2*a*e)*x^n))/(2*a*(b^2 - 4*a*c)*f*n*(a + b*x^n + c*x^(2*n))
^2) + ((f*x)^(1 + m)*((b^2 - 2*a*c)*(a*b*e*(1 + m) + 2*a*c*d*(1 + m - 4*n) - b^2*d*(1 + m - 2*n)) + a*b*c*(b*d
- 2*a*e)*(1 + m - 3*n) + c*(a*b^2*e*(1 + m) + 2*a*b*c*d*(2 + 2*m - 7*n) - 4*a^2*c*e*(1 + m - 3*n) - b^3*d*(1
+ m - 2*n))*x^n))/(2*a^2*(b^2 - 4*a*c)^2*f*n^2*(a + b*x^n + c*x^(2*n))) - (c*((a*b^2*e*(1 + m) + 2*a*b*c*d*(2
+ 2*m - 7*n) - 4*a^2*c*e*(1 + m - 3*n) - b^3*d*(1 + m - 2*n))*(1 + m - n) + (a*b^3*e*(1 + m)*(1 + m - n) - 4*a
^2*b*c*e*(1 + m^2 + m*(2 - n) - n - 3*n^2) - b^4*d*(1 + m^2 + m*(2 - 3*n) - 3*n + 2*n^2) + 6*a*b^2*c*d*(1 + m^
2 + m*(2 - 4*n) - 4*n + 3*n^2) - 8*a^2*c^2*d*(1 + m^2 + m*(2 - 6*n) - 6*n + 8*n^2))/Sqrt[b^2 - 4*a*c])*(f*x)^(
1 + m)*Hypergeometric2F1[1, (1 + m)/n, (1 + m + n)/n, (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/(2*a^2*(b^2 - 4*a*c
)^2*(b - Sqrt[b^2 - 4*a*c])*f*(1 + m)*n^2) - (c*((a*b^2*e*(1 + m) + 2*a*b*c*d*(2 + 2*m - 7*n) - 4*a^2*c*e*(1 +
m - 3*n) - b^3*d*(1 + m - 2*n))*(1 + m - n) - (a*b^3*e*(1 + m)*(1 + m - n) - 4*a^2*b*c*e*(1 + m^2 + m*(2 - n)
- n - 3*n^2) - b^4*d*(1 + m^2 + m*(2 - 3*n) - 3*n + 2*n^2) + 6*a*b^2*c*d*(1 + m^2 + m*(2 - 4*n) - 4*n + 3*n^2
) - 8*a^2*c^2*d*(1 + m^2 + m*(2 - 6*n) - 6*n + 8*n^2))/Sqrt[b^2 - 4*a*c])*(f*x)^(1 + m)*Hypergeometric2F1[1, (
1 + m)/n, (1 + m + n)/n, (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(2*a^2*(b^2 - 4*a*c)^2*(b + Sqrt[b^2 - 4*a*c])*f
*(1 + m)*n^2)

Rule 1558

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

Rule 1560

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 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{(f x)^m \left (d+e x^n\right )}{\left (a+b x^n+c x^{2 n}\right )^3} \, dx &=\frac{(f x)^{1+m} \left (b^2 d-2 a c d-a b e+c (b d-2 a e) x^n\right )}{2 a \left (b^2-4 a c\right ) f n \left (a+b x^n+c x^{2 n}\right )^2}-\frac{\int \frac{(f x)^m \left (-a b e (1+m)-2 a c d (1+m-4 n)+b^2 d (1+m-2 n)+c (b d-2 a e) (1+m-3 n) x^n\right )}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx}{2 a \left (b^2-4 a c\right ) n}\\ &=\frac{(f x)^{1+m} \left (b^2 d-2 a c d-a b e+c (b d-2 a e) x^n\right )}{2 a \left (b^2-4 a c\right ) f n \left (a+b x^n+c x^{2 n}\right )^2}+\frac{(f x)^{1+m} \left (\left (b^2-2 a c\right ) \left (a b e (1+m)+2 a c d (1+m-4 n)-b^2 d (1+m-2 n)\right )+a b c (b d-2 a e) (1+m-3 n)+c \left (a b^2 e (1+m)+2 a b c d (2+2 m-7 n)-4 a^2 c e (1+m-3 n)-b^3 d (1+m-2 n)\right ) x^n\right )}{2 a^2 \left (b^2-4 a c\right )^2 f n^2 \left (a+b x^n+c x^{2 n}\right )}+\frac{\int \frac{(f x)^m \left (\left (a b e (1+m)+2 a c d (1+m-4 n)-b^2 d (1+m-2 n)\right ) \left (2 a c (1+m-2 n)-b^2 (1+m-n)\right )-a b c (b d-2 a e) (1+m) (1+m-3 n)-c \left (a b^2 e (1+m)+2 a b c d (2+2 m-7 n)-4 a^2 c e (1+m-3 n)-b^3 d (1+m-2 n)\right ) (1+m-n) x^n\right )}{a+b x^n+c x^{2 n}} \, dx}{2 a^2 \left (b^2-4 a c\right )^2 n^2}\\ &=\frac{(f x)^{1+m} \left (b^2 d-2 a c d-a b e+c (b d-2 a e) x^n\right )}{2 a \left (b^2-4 a c\right ) f n \left (a+b x^n+c x^{2 n}\right )^2}+\frac{(f x)^{1+m} \left (\left (b^2-2 a c\right ) \left (a b e (1+m)+2 a c d (1+m-4 n)-b^2 d (1+m-2 n)\right )+a b c (b d-2 a e) (1+m-3 n)+c \left (a b^2 e (1+m)+2 a b c d (2+2 m-7 n)-4 a^2 c e (1+m-3 n)-b^3 d (1+m-2 n)\right ) x^n\right )}{2 a^2 \left (b^2-4 a c\right )^2 f n^2 \left (a+b x^n+c x^{2 n}\right )}+\frac{\int \left (\frac{\left (-c \left (a b^2 e (1+m)+2 a b c d (2+2 m-7 n)-4 a^2 c e (1+m-3 n)-b^3 d (1+m-2 n)\right ) (1+m-n)+\frac{c \left (b^4 d-6 a b^2 c d+8 a^2 c^2 d-a b^3 e+4 a^2 b c e+2 b^4 d m-12 a b^2 c d m+16 a^2 c^2 d m-2 a b^3 e m+8 a^2 b c e m+b^4 d m^2-6 a b^2 c d m^2+8 a^2 c^2 d m^2-a b^3 e m^2+4 a^2 b c e m^2-3 b^4 d n+24 a b^2 c d n-48 a^2 c^2 d n+a b^3 e n-4 a^2 b c e n-3 b^4 d m n+24 a b^2 c d m n-48 a^2 c^2 d m n+a b^3 e m n-4 a^2 b c e m n+2 b^4 d n^2-18 a b^2 c d n^2+64 a^2 c^2 d n^2-12 a^2 b c e n^2\right )}{\sqrt{b^2-4 a c}}\right ) (f x)^m}{b-\sqrt{b^2-4 a c}+2 c x^n}+\frac{\left (-c \left (a b^2 e (1+m)+2 a b c d (2+2 m-7 n)-4 a^2 c e (1+m-3 n)-b^3 d (1+m-2 n)\right ) (1+m-n)-\frac{c \left (b^4 d-6 a b^2 c d+8 a^2 c^2 d-a b^3 e+4 a^2 b c e+2 b^4 d m-12 a b^2 c d m+16 a^2 c^2 d m-2 a b^3 e m+8 a^2 b c e m+b^4 d m^2-6 a b^2 c d m^2+8 a^2 c^2 d m^2-a b^3 e m^2+4 a^2 b c e m^2-3 b^4 d n+24 a b^2 c d n-48 a^2 c^2 d n+a b^3 e n-4 a^2 b c e n-3 b^4 d m n+24 a b^2 c d m n-48 a^2 c^2 d m n+a b^3 e m n-4 a^2 b c e m n+2 b^4 d n^2-18 a b^2 c d n^2+64 a^2 c^2 d n^2-12 a^2 b c e n^2\right )}{\sqrt{b^2-4 a c}}\right ) (f x)^m}{b+\sqrt{b^2-4 a c}+2 c x^n}\right ) \, dx}{2 a^2 \left (b^2-4 a c\right )^2 n^2}\\ &=\frac{(f x)^{1+m} \left (b^2 d-2 a c d-a b e+c (b d-2 a e) x^n\right )}{2 a \left (b^2-4 a c\right ) f n \left (a+b x^n+c x^{2 n}\right )^2}+\frac{(f x)^{1+m} \left (\left (b^2-2 a c\right ) \left (a b e (1+m)+2 a c d (1+m-4 n)-b^2 d (1+m-2 n)\right )+a b c (b d-2 a e) (1+m-3 n)+c \left (a b^2 e (1+m)+2 a b c d (2+2 m-7 n)-4 a^2 c e (1+m-3 n)-b^3 d (1+m-2 n)\right ) x^n\right )}{2 a^2 \left (b^2-4 a c\right )^2 f n^2 \left (a+b x^n+c x^{2 n}\right )}-\frac{\left (c \left (\left (a b^2 e (1+m)+2 a b c d (2+2 m-7 n)-4 a^2 c e (1+m-3 n)-b^3 d (1+m-2 n)\right ) (1+m-n)-\frac{a b^3 e (1+m) (1+m-n)-4 a^2 b c e \left (1+m^2+m (2-n)-n-3 n^2\right )-b^4 d \left (1+m^2+m (2-3 n)-3 n+2 n^2\right )+6 a b^2 c d \left (1+m^2+m (2-4 n)-4 n+3 n^2\right )-8 a^2 c^2 d \left (1+m^2+m (2-6 n)-6 n+8 n^2\right )}{\sqrt{b^2-4 a c}}\right )\right ) \int \frac{(f x)^m}{b+\sqrt{b^2-4 a c}+2 c x^n} \, dx}{2 a^2 \left (b^2-4 a c\right )^2 n^2}-\frac{\left (c \left (\left (a b^2 e (1+m)+2 a b c d (2+2 m-7 n)-4 a^2 c e (1+m-3 n)-b^3 d (1+m-2 n)\right ) (1+m-n)+\frac{a b^3 e (1+m) (1+m-n)-4 a^2 b c e \left (1+m^2+m (2-n)-n-3 n^2\right )-b^4 d \left (1+m^2+m (2-3 n)-3 n+2 n^2\right )+6 a b^2 c d \left (1+m^2+m (2-4 n)-4 n+3 n^2\right )-8 a^2 c^2 d \left (1+m^2+m (2-6 n)-6 n+8 n^2\right )}{\sqrt{b^2-4 a c}}\right )\right ) \int \frac{(f x)^m}{b-\sqrt{b^2-4 a c}+2 c x^n} \, dx}{2 a^2 \left (b^2-4 a c\right )^2 n^2}\\ &=\frac{(f x)^{1+m} \left (b^2 d-2 a c d-a b e+c (b d-2 a e) x^n\right )}{2 a \left (b^2-4 a c\right ) f n \left (a+b x^n+c x^{2 n}\right )^2}+\frac{(f x)^{1+m} \left (\left (b^2-2 a c\right ) \left (a b e (1+m)+2 a c d (1+m-4 n)-b^2 d (1+m-2 n)\right )+a b c (b d-2 a e) (1+m-3 n)+c \left (a b^2 e (1+m)+2 a b c d (2+2 m-7 n)-4 a^2 c e (1+m-3 n)-b^3 d (1+m-2 n)\right ) x^n\right )}{2 a^2 \left (b^2-4 a c\right )^2 f n^2 \left (a+b x^n+c x^{2 n}\right )}-\frac{c \left (\left (a b^2 e (1+m)+2 a b c d (2+2 m-7 n)-4 a^2 c e (1+m-3 n)-b^3 d (1+m-2 n)\right ) (1+m-n)+\frac{a b^3 e (1+m) (1+m-n)-4 a^2 b c e \left (1+m^2+m (2-n)-n-3 n^2\right )-b^4 d \left (1+m^2+m (2-3 n)-3 n+2 n^2\right )+6 a b^2 c d \left (1+m^2+m (2-4 n)-4 n+3 n^2\right )-8 a^2 c^2 d \left (1+m^2+m (2-6 n)-6 n+8 n^2\right )}{\sqrt{b^2-4 a c}}\right ) (f x)^{1+m} \, _2F_1\left (1,\frac{1+m}{n};\frac{1+m+n}{n};-\frac{2 c x^n}{b-\sqrt{b^2-4 a c}}\right )}{2 a^2 \left (b^2-4 a c\right )^2 \left (b-\sqrt{b^2-4 a c}\right ) f (1+m) n^2}-\frac{c \left (\left (a b^2 e (1+m)+2 a b c d (2+2 m-7 n)-4 a^2 c e (1+m-3 n)-b^3 d (1+m-2 n)\right ) (1+m-n)-\frac{a b^3 e (1+m) (1+m-n)-4 a^2 b c e \left (1+m^2+m (2-n)-n-3 n^2\right )-b^4 d \left (1+m^2+m (2-3 n)-3 n+2 n^2\right )+6 a b^2 c d \left (1+m^2+m (2-4 n)-4 n+3 n^2\right )-8 a^2 c^2 d \left (1+m^2+m (2-6 n)-6 n+8 n^2\right )}{\sqrt{b^2-4 a c}}\right ) (f x)^{1+m} \, _2F_1\left (1,\frac{1+m}{n};\frac{1+m+n}{n};-\frac{2 c x^n}{b+\sqrt{b^2-4 a c}}\right )}{2 a^2 \left (b^2-4 a c\right )^2 \left (b+\sqrt{b^2-4 a c}\right ) f (1+m) n^2}\\ \end{align*}

Mathematica [B]  time = 7.35683, size = 13117, normalized size = 16.07 $\text{Result too large to show}$

Warning: Unable to verify antiderivative.

[In]

Integrate[((f*x)^m*(d + e*x^n))/(a + b*x^n + c*x^(2*n))^3,x]

[Out]

Result too large to show

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Maple [F]  time = 0.055, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx \right ) ^{m} \left ( d+e{x}^{n} \right ) }{ \left ( a+b{x}^{n}+c{x}^{2\,n} \right ) ^{3}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((f*x)^m*(d+e*x^n)/(a+b*x^n+c*x^(2*n))^3,x)

[Out]

int((f*x)^m*(d+e*x^n)/(a+b*x^n+c*x^(2*n))^3,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*(d+e*x^n)/(a+b*x^n+c*x^(2*n))^3,x, algorithm="maxima")

[Out]

-1/2*((a*b^4*d*f^m*(m - 3*n + 1) + 2*(b*c*e*f^m*(2*m - 5*n + 2) + 2*c^2*d*f^m*(m - 6*n + 1))*a^3 - (b^2*c*d*f^
m*(5*m - 21*n + 5) + b^3*e*f^m*(m - n + 1))*a^2)*x*x^m + (b^3*c^2*d*f^m*(m - 2*n + 1) + 4*a^2*c^3*e*f^m*(m - 3
*n + 1) - (2*b*c^3*d*f^m*(2*m - 7*n + 2) + b^2*c^2*e*f^m*(m + 1))*a)*x*e^(m*log(x) + 3*n*log(x)) + (2*b^4*c*d*
f^m*(m - 2*n + 1) + 2*(b*c^2*e*f^m*(4*m - 9*n + 4) + 2*c^3*d*f^m*(m - 4*n + 1))*a^2 - (b^2*c^2*d*f^m*(9*m - 29
*n + 9) + 2*b^3*c*e*f^m*(m + 1))*a)*x*e^(m*log(x) + 2*n*log(x)) + (b^5*d*f^m*(m - 2*n + 1) + 4*a^3*c^2*e*f^m*(
m - 5*n + 1) + (b^2*c*e*f^m*(3*m - 4*n + 3) + 2*b*c^2*d*f^m*n)*a^2 - (4*b^3*c*d*f^m*(m - 3*n + 1) + b^4*e*f^m*
(m + 1))*a)*x*e^(m*log(x) + n*log(x)))/(a^4*b^4*n^2 - 8*a^5*b^2*c*n^2 + 16*a^6*c^2*n^2 + (a^2*b^4*c^2*n^2 - 8*
a^3*b^2*c^3*n^2 + 16*a^4*c^4*n^2)*x^(4*n) + 2*(a^2*b^5*c*n^2 - 8*a^3*b^3*c^2*n^2 + 16*a^4*b*c^3*n^2)*x^(3*n) +
(a^2*b^6*n^2 - 6*a^3*b^4*c*n^2 + 32*a^5*c^3*n^2)*x^(2*n) + 2*(a^3*b^5*n^2 - 8*a^4*b^3*c*n^2 + 16*a^5*b*c^2*n^
2)*x^n) + integrate(1/2*(((m^2 - m*(3*n - 2) + 2*n^2 - 3*n + 1)*b^4*d*f^m + 2*(2*(m^2 - 2*m*(3*n - 1) + 8*n^2
- 6*n + 1)*c^2*d*f^m + (2*m^2 - m*(5*n - 4) - 5*n + 2)*b*c*e*f^m)*a^2 - ((5*m^2 - m*(21*n - 10) + 16*n^2 - 21*
n + 5)*b^2*c*d*f^m + (m^2 - m*(n - 2) - n + 1)*b^3*e*f^m)*a)*x^m + ((m^2 - m*(3*n - 2) + 2*n^2 - 3*n + 1)*b^3*
c*d*f^m + 4*(m^2 - 2*m*(2*n - 1) + 3*n^2 - 4*n + 1)*a^2*c^2*e*f^m - (2*(2*m^2 - m*(9*n - 4) + 7*n^2 - 9*n + 2)
*b*c^2*d*f^m + (m^2 - m*(n - 2) - n + 1)*b^2*c*e*f^m)*a)*e^(m*log(x) + n*log(x)))/(a^3*b^4*n^2 - 8*a^4*b^2*c*n
^2 + 16*a^5*c^2*n^2 + (a^2*b^4*c*n^2 - 8*a^3*b^2*c^2*n^2 + 16*a^4*c^3*n^2)*x^(2*n) + (a^2*b^5*n^2 - 8*a^3*b^3*
c*n^2 + 16*a^4*b*c^2*n^2)*x^n), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (e x^{n} + d\right )} \left (f x\right )^{m}}{c^{3} x^{6 \, n} + b^{3} x^{3 \, n} + 3 \, a b^{2} x^{2 \, n} + 3 \, a^{2} b x^{n} + a^{3} + 3 \,{\left (b c^{2} x^{n} + a c^{2}\right )} x^{4 \, n} + 3 \,{\left (b^{2} c x^{2 \, n} + 2 \, a b c x^{n} + a^{2} c\right )} x^{2 \, n}}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*(d+e*x^n)/(a+b*x^n+c*x^(2*n))^3,x, algorithm="fricas")

[Out]

integral((e*x^n + d)*(f*x)^m/(c^3*x^(6*n) + b^3*x^(3*n) + 3*a*b^2*x^(2*n) + 3*a^2*b*x^n + a^3 + 3*(b*c^2*x^n +
a*c^2)*x^(4*n) + 3*(b^2*c*x^(2*n) + 2*a*b*c*x^n + a^2*c)*x^(2*n)), x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)**m*(d+e*x**n)/(a+b*x**n+c*x**(2*n))**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x^{n} + d\right )} \left (f x\right )^{m}}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{3}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x)^m*(d+e*x^n)/(a+b*x^n+c*x^(2*n))^3,x, algorithm="giac")

[Out]

integrate((e*x^n + d)*(f*x)^m/(c*x^(2*n) + b*x^n + a)^3, x)