\(\int \frac {(d+e x^n)^3}{(a+b x^n+c x^{2 n})^2} \, dx\) [75]

Optimal result

Integrand size = 26, antiderivative size = 750 \[ \int \frac {\left (d+e x^n\right )^3}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\frac {x \left (b^2 c d^3-2 a c d \left (c d^2-3 a e^2\right )-a b e \left (3 c d^2+a e^2\right )-\left (a b^2 e^3+2 a c e \left (3 c d^2-a e^2\right )-b c d \left (c d^2+3 a e^2\right )\right ) x^n\right )}{a c \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {e^2 \left (e+\frac {6 c d-3 b e}{\sqrt {b^2-4 a c}}\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{c \left (b-\sqrt {b^2-4 a c}\right )}+\frac {\left (\left (a b^2 e^3+2 a c e \left (3 c d^2-a e^2\right )-b c d \left (c d^2+3 a e^2\right )\right ) (1-n)+\frac {b^2 c d \left (3 a e^2 (1-3 n)-c d^2 (1-n)\right )-a b^3 e^3 (1-3 n)+4 a c^2 d \left (c d^2-3 a e^2\right ) (1-2 n)+2 a b c e \left (a e^2 (2-5 n)+3 c d^2 n\right )}{\sqrt {b^2-4 a c}}\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a c \left (b^2-4 a c\right ) \left (b-\sqrt {b^2-4 a c}\right ) n}+\frac {e^2 \left (e-\frac {3 (2 c d-b e)}{\sqrt {b^2-4 a c}}\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{c \left (b+\sqrt {b^2-4 a c}\right )}+\frac {\left (\left (a b^2 e^3+2 a c e \left (3 c d^2-a e^2\right )-b c d \left (c d^2+3 a e^2\right )\right ) (1-n)-\frac {b^2 c d \left (3 a e^2 (1-3 n)-c d^2 (1-n)\right )-a b^3 e^3 (1-3 n)+4 a c^2 d \left (c d^2-3 a e^2\right ) (1-2 n)+2 a b c e \left (a e^2 (2-5 n)+3 c d^2 n\right )}{\sqrt {b^2-4 a c}}\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a c \left (b^2-4 a c\right ) \left (b+\sqrt {b^2-4 a c}\right ) n} \]

[Out]

x*(b^2*c*d^3-2*a*c*d*(-3*a*e^2+c*d^2)-a*b*e*(a*e^2+3*c*d^2)-(a*b^2*e^3+2*a*c*e*(-a*e^2+3*c*d^2)-b*c*d*(3*a*e^2
+c*d^2))*x^n)/a/c/(-4*a*c+b^2)/n/(a+b*x^n+c*x^(2*n))+e^2*x*hypergeom([1, 1/n],[1+1/n],-2*c*x^n/(b-(-4*a*c+b^2)
^(1/2)))*(e+(-3*b*e+6*c*d)/(-4*a*c+b^2)^(1/2))/c/(b-(-4*a*c+b^2)^(1/2))+x*hypergeom([1, 1/n],[1+1/n],-2*c*x^n/
(b-(-4*a*c+b^2)^(1/2)))*((a*b^2*e^3+2*a*c*e*(-a*e^2+3*c*d^2)-b*c*d*(3*a*e^2+c*d^2))*(1-n)+(b^2*c*d*(3*a*e^2*(1
-3*n)-c*d^2*(1-n))-a*b^3*e^3*(1-3*n)+4*a*c^2*d*(-3*a*e^2+c*d^2)*(1-2*n)+2*a*b*c*e*(a*e^2*(2-5*n)+3*c*d^2*n))/(
-4*a*c+b^2)^(1/2))/a/c/(-4*a*c+b^2)/n/(b-(-4*a*c+b^2)^(1/2))+e^2*x*hypergeom([1, 1/n],[1+1/n],-2*c*x^n/(b+(-4*
a*c+b^2)^(1/2)))*(e-3*(-b*e+2*c*d)/(-4*a*c+b^2)^(1/2))/c/(b+(-4*a*c+b^2)^(1/2))+x*hypergeom([1, 1/n],[1+1/n],-
2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))*((a*b^2*e^3+2*a*c*e*(-a*e^2+3*c*d^2)-b*c*d*(3*a*e^2+c*d^2))*(1-n)+(-b^2*c*d*(3
*a*e^2*(1-3*n)-c*d^2*(1-n))+a*b^3*e^3*(1-3*n)-4*a*c^2*d*(-3*a*e^2+c*d^2)*(1-2*n)-2*a*b*c*e*(a*e^2*(2-5*n)+3*c*
d^2*n))/(-4*a*c+b^2)^(1/2))/a/c/(-4*a*c+b^2)/n/(b+(-4*a*c+b^2)^(1/2))

Rubi [A] (verified)

Time = 1.77 (sec) , antiderivative size = 750, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1450, 1444, 1436, 251} \[ \int \frac {\left (d+e x^n\right )^3}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\frac {x \left (-\left (x^n \left (a b^2 e^3-b c d \left (3 a e^2+c d^2\right )+2 a c e \left (3 c d^2-a e^2\right )\right )\right )-a b e \left (a e^2+3 c d^2\right )-2 a c d \left (c d^2-3 a e^2\right )+b^2 c d^3\right )}{a c n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )}+\frac {e^2 x \left (\frac {6 c d-3 b e}{\sqrt {b^2-4 a c}}+e\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{c \left (b-\sqrt {b^2-4 a c}\right )}+\frac {e^2 x \left (e-\frac {3 (2 c d-b e)}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{c \left (\sqrt {b^2-4 a c}+b\right )}+\frac {x \left ((1-n) \left (a b^2 e^3-b c d \left (3 a e^2+c d^2\right )+2 a c e \left (3 c d^2-a e^2\right )\right )+\frac {-a b^3 e^3 (1-3 n)+b^2 c d \left (3 a e^2 (1-3 n)-c d^2 (1-n)\right )+2 a b c e \left (a e^2 (2-5 n)+3 c d^2 n\right )+4 a c^2 d (1-2 n) \left (c d^2-3 a e^2\right )}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a c n \left (b^2-4 a c\right ) \left (b-\sqrt {b^2-4 a c}\right )}+\frac {x \left ((1-n) \left (a b^2 e^3-b c d \left (3 a e^2+c d^2\right )+2 a c e \left (3 c d^2-a e^2\right )\right )-\frac {-a b^3 e^3 (1-3 n)+b^2 c d \left (3 a e^2 (1-3 n)-c d^2 (1-n)\right )+2 a b c e \left (a e^2 (2-5 n)+3 c d^2 n\right )+4 a c^2 d (1-2 n) \left (c d^2-3 a e^2\right )}{\sqrt {b^2-4 a c}}\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a c n \left (b^2-4 a c\right ) \left (\sqrt {b^2-4 a c}+b\right )} \]

[In]

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

[Out]

(x*(b^2*c*d^3 - 2*a*c*d*(c*d^2 - 3*a*e^2) - a*b*e*(3*c*d^2 + a*e^2) - (a*b^2*e^3 + 2*a*c*e*(3*c*d^2 - a*e^2) -
b*c*d*(c*d^2 + 3*a*e^2))*x^n))/(a*c*(b^2 - 4*a*c)*n*(a + b*x^n + c*x^(2*n))) + (e^2*(e + (6*c*d - 3*b*e)/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])])/(c*(b - Sqrt[b^
2 - 4*a*c])) + (((a*b^2*e^3 + 2*a*c*e*(3*c*d^2 - a*e^2) - b*c*d*(c*d^2 + 3*a*e^2))*(1 - n) + (b^2*c*d*(3*a*e^2
*(1 - 3*n) - c*d^2*(1 - n)) - a*b^3*e^3*(1 - 3*n) + 4*a*c^2*d*(c*d^2 - 3*a*e^2)*(1 - 2*n) + 2*a*b*c*e*(a*e^2*(
2 - 5*n) + 3*c*d^2*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])])/(a*c*(b^2 - 4*a*c)*(b - Sqrt[b^2 - 4*a*c])*n) + (e^2*(e - (3*(2*c*d - b*e))/Sqrt[b^2 - 4*a*c])*x*H
ypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(c*(b + Sqrt[b^2 - 4*a*c])) + (((
a*b^2*e^3 + 2*a*c*e*(3*c*d^2 - a*e^2) - b*c*d*(c*d^2 + 3*a*e^2))*(1 - n) - (b^2*c*d*(3*a*e^2*(1 - 3*n) - c*d^2
*(1 - n)) - a*b^3*e^3*(1 - 3*n) + 4*a*c^2*d*(c*d^2 - 3*a*e^2)*(1 - 2*n) + 2*a*b*c*e*(a*e^2*(2 - 5*n) + 3*c*d^2
*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])])/(a*c*(
b^2 - 4*a*c)*(b + Sqrt[b^2 - 4*a*c])*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 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 1444

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] + Di
st[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] && Eq
Q[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && ILtQ[p, -1]

Rule 1450

Int[((d_) + (e_.)*(x_)^(n_))^(q_)*((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Int[ExpandInt
egrand[(d + e*x^n)^q*(a + b*x^n + c*x^(2*n))^p, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] &
& NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && ((IntegersQ[p, q] &&  !IntegerQ[n]) || IGtQ[p, 0] ||
(IGtQ[q, 0] &&  !IntegerQ[n]))

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {c^2 d^3-3 a c d e^2+a b e^3+\left (3 c^2 d^2 e-3 b c d e^2+b^2 e^3-a c e^3\right ) x^n}{c^2 \left (a+b x^n+c x^{2 n}\right )^2}+\frac {e^2 \left (3 c d-b e+c e x^n\right )}{c^2 \left (a+b x^n+c x^{2 n}\right )}\right ) \, dx \\ & = \frac {\int \frac {c^2 d^3-3 a c d e^2+a b e^3+\left (3 c^2 d^2 e-3 b c d e^2+b^2 e^3-a c e^3\right ) x^n}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx}{c^2}+\frac {e^2 \int \frac {3 c d-b e+c e x^n}{a+b x^n+c x^{2 n}} \, dx}{c^2} \\ & = \frac {x \left (b^2 c d^3-2 a c d \left (c d^2-3 a e^2\right )-a b e \left (3 c d^2+a e^2\right )-\left (a b^2 e^3+2 a c e \left (3 c d^2-a e^2\right )-b c d \left (c d^2+3 a e^2\right )\right ) x^n\right )}{a c \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {\left (e^2 \left (e+\frac {6 c d-3 b e}{\sqrt {b^2-4 a c}}\right )\right ) \int \frac {1}{\frac {b}{2}-\frac {1}{2} \sqrt {b^2-4 a c}+c x^n} \, dx}{2 c}+\frac {\left (e^2 \left (e-\frac {3 (2 c d-b e)}{\sqrt {b^2-4 a c}}\right )\right ) \int \frac {1}{\frac {b}{2}+\frac {1}{2} \sqrt {b^2-4 a c}+c x^n} \, dx}{2 c}-\frac {\int \frac {-a b c e \left (3 c d^2+a e^2 (1-4 n)\right )-2 a c^2 d \left (c d^2-3 a e^2\right ) (1-2 n)-a b^3 e^3 n+b^2 c d \left (c d^2 (1-n)+3 a e^2 n\right )-c \left (a b^2 e^3+2 a c e \left (3 c d^2-a e^2\right )-b c d \left (c d^2+3 a e^2\right )\right ) (1-n) x^n}{a+b x^n+c x^{2 n}} \, dx}{a c^2 \left (b^2-4 a c\right ) n} \\ & = \frac {x \left (b^2 c d^3-2 a c d \left (c d^2-3 a e^2\right )-a b e \left (3 c d^2+a e^2\right )-\left (a b^2 e^3+2 a c e \left (3 c d^2-a e^2\right )-b c d \left (c d^2+3 a e^2\right )\right ) x^n\right )}{a c \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {e^2 \left (e+\frac {6 c d-3 b e}{\sqrt {b^2-4 a c}}\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 )}{c \left (b-\sqrt {b^2-4 a c}\right )}+\frac {e^2 \left (e-\frac {3 (2 c d-b e)}{\sqrt {b^2-4 a c}}\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 )}{c \left (b+\sqrt {b^2-4 a c}\right )}+\frac {\left (\left (a b^2 e^3+2 a c e \left (3 c d^2-a e^2\right )-b c d \left (c d^2+3 a e^2\right )\right ) (1-n)-\frac {b^2 c d \left (3 a e^2 (1-3 n)-c d^2 (1-n)\right )-a b^3 e^3 (1-3 n)+4 a c^2 d \left (c d^2-3 a e^2\right ) (1-2 n)+2 a b c e \left (a e^2 (2-5 n)+3 c d^2 n\right )}{\sqrt {b^2-4 a c}}\right ) \int \frac {1}{\frac {b}{2}+\frac {1}{2} \sqrt {b^2-4 a c}+c x^n} \, dx}{2 a c \left (b^2-4 a c\right ) n}+\frac {\left (\left (a b^2 e^3+2 a c e \left (3 c d^2-a e^2\right )-b c d \left (c d^2+3 a e^2\right )\right ) (1-n)+\frac {b^2 c d \left (3 a e^2 (1-3 n)-c d^2 (1-n)\right )-a b^3 e^3 (1-3 n)+4 a c^2 d \left (c d^2-3 a e^2\right ) (1-2 n)+2 a b c e \left (a e^2 (2-5 n)+3 c d^2 n\right )}{\sqrt {b^2-4 a c}}\right ) \int \frac {1}{\frac {b}{2}-\frac {1}{2} \sqrt {b^2-4 a c}+c x^n} \, dx}{2 a c \left (b^2-4 a c\right ) n} \\ & = \frac {x \left (b^2 c d^3-2 a c d \left (c d^2-3 a e^2\right )-a b e \left (3 c d^2+a e^2\right )-\left (a b^2 e^3+2 a c e \left (3 c d^2-a e^2\right )-b c d \left (c d^2+3 a e^2\right )\right ) x^n\right )}{a c \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}+\frac {e^2 \left (e+\frac {6 c d-3 b e}{\sqrt {b^2-4 a c}}\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 )}{c \left (b-\sqrt {b^2-4 a c}\right )}+\frac {\left (\left (a b^2 e^3+2 a c e \left (3 c d^2-a e^2\right )-b c d \left (c d^2+3 a e^2\right )\right ) (1-n)+\frac {b^2 c d \left (3 a e^2 (1-3 n)-c d^2 (1-n)\right )-a b^3 e^3 (1-3 n)+4 a c^2 d \left (c d^2-3 a e^2\right ) (1-2 n)+2 a b c e \left (a e^2 (2-5 n)+3 c d^2 n\right )}{\sqrt {b^2-4 a c}}\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 c \left (b^2-4 a c\right ) \left (b-\sqrt {b^2-4 a c}\right ) n}+\frac {e^2 \left (e-\frac {3 (2 c d-b e)}{\sqrt {b^2-4 a c}}\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 )}{c \left (b+\sqrt {b^2-4 a c}\right )}+\frac {\left (\left (a b^2 e^3+2 a c e \left (3 c d^2-a e^2\right )-b c d \left (c d^2+3 a e^2\right )\right ) (1-n)-\frac {b^2 c d \left (3 a e^2 (1-3 n)-c d^2 (1-n)\right )-a b^3 e^3 (1-3 n)+4 a c^2 d \left (c d^2-3 a e^2\right ) (1-2 n)+2 a b c e \left (a e^2 (2-5 n)+3 c d^2 n\right )}{\sqrt {b^2-4 a c}}\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 c \left (b^2-4 a c\right ) \left (b+\sqrt {b^2-4 a c}\right ) n} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(5537\) vs. \(2(750)=1500\).

Time = 7.56 (sec) , antiderivative size = 5537, normalized size of antiderivative = 7.38 \[ \int \frac {\left (d+e x^n\right )^3}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\text {Result too large to show} \]

[In]

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

[Out]

Result too large to show

Maple [F]

\[\int \frac {\left (d +e \,x^{n}\right )^{3}}{\left (a +b \,x^{n}+c \,x^{2 n}\right )^{2}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {\left (d+e x^n\right )^3}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int { \frac {{\left (e x^{n} + d\right )}^{3}}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}} \,d x } \]

[In]

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

[Out]

integral((e^3*x^(3*n) + 3*d*e^2*x^(2*n) + 3*d^2*e*x^n + d^3)/(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)

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^n\right )^3}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {\left (d+e x^n\right )^3}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int { \frac {{\left (e x^{n} + d\right )}^{3}}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}} \,d x } \]

[In]

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

[Out]

((b*c^2*d^3 + 2*a^2*c*e^3 - (6*c^2*d^2*e - 3*b*c*d*e^2 + b^2*e^3)*a)*x*x^n + (b^2*c*d^3 + (6*c*d*e^2 - b*e^3)*
a^2 - (2*c^2*d^3 + 3*b*c*d^2*e)*a)*x)/(a^2*b^2*c*n - 4*a^3*c^2*n + (a*b^2*c^2*n - 4*a^2*c^3*n)*x^(2*n) + (a*b^
3*c*n - 4*a^2*b*c^2*n)*x^n) + integrate((b^2*c*d^3*(n - 1) - (6*c*d*e^2 - b*e^3)*a^2 - (2*c^2*d^3*(2*n - 1) -
3*b*c*d^2*e)*a - (2*a^2*c*e^3*(n + 1) - b*c^2*d^3*(n - 1) + (6*c^2*d^2*e*(n - 1) - 3*b*c*d*e^2*(n - 1) - b^2*e
^3)*a)*x^n)/(a^2*b^2*c*n - 4*a^3*c^2*n + (a*b^2*c^2*n - 4*a^2*c^3*n)*x^(2*n) + (a*b^3*c*n - 4*a^2*b*c^2*n)*x^n
), x)

Giac [F]

\[ \int \frac {\left (d+e x^n\right )^3}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int { \frac {{\left (e x^{n} + d\right )}^{3}}{{\left (c x^{2 \, n} + b x^{n} + a\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (d+e x^n\right )^3}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\int \frac {{\left (d+e\,x^n\right )}^3}{{\left (a+b\,x^n+c\,x^{2\,n}\right )}^2} \,d x \]

[In]

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

[Out]

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