3.1.78 \(\int \frac {1}{(d+e x^n) (a+b x^n+c x^{2 n})^2} \, dx\) [78]

Optimal. Leaf size=726 \[ \frac {x \left (b^2 c d-2 a c^2 d-b^3 e+3 a b c e+c \left (b c d-b^2 e+2 a c e\right ) x^n\right )}{a \left (b^2-4 a c\right ) \left (c d^2-b d e+a e^2\right ) n \left (a+b x^n+c x^{2 n}\right )}-\frac {c e^2 \left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right )^2}-\frac {c \left (\frac {2 a b c e (2-3 n)-4 a c^2 d (1-2 n)+b^2 c d (1-n)-b^3 e (1-n)}{\sqrt {b^2-4 a c}}+\left (b c d-b^2 e+2 a c e\right ) (1-n)\right ) x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right ) \left (b-\sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right ) n}-\frac {c e^2 \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right )^2}+\frac {c \left (b c \left (2 a e (2-3 n)-\sqrt {b^2-4 a c} d (1-n)\right )-2 a c \left (2 c d (1-2 n)+\sqrt {b^2-4 a c} e (1-n)\right )-b^3 e (1-n)+b^2 \left (c d+\sqrt {b^2-4 a c} e\right ) (1-n)\right ) x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right ) \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) \left (c d^2-b d e+a e^2\right ) n}+\frac {e^4 x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d \left (c d^2-b d e+a e^2\right )^2} \]

[Out]

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

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Rubi [A]
time = 1.27, antiderivative size = 726, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1450, 251, 1444, 1436} \begin {gather*} -\frac {c e^2 x \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\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 )}{\left (-b \sqrt {b^2-4 a c}-4 a c+b^2\right ) \left (a e^2-b d e+c d^2\right )^2}-\frac {c e^2 x \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\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 )}{\left (b \sqrt {b^2-4 a c}-4 a c+b^2\right ) \left (a e^2-b d e+c d^2\right )^2}-\frac {c x \left ((1-n) \left (2 a c e+b^2 (-e)+b c d\right )+\frac {2 a b c e (2-3 n)-4 a c^2 d (1-2 n)+b^3 (-e) (1-n)+b^2 c d (1-n)}{\sqrt {b^2-4 a c}}\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 )}{a n \left (b^2-4 a c\right ) \left (b-\sqrt {b^2-4 a c}\right ) \left (a e^2-b d e+c d^2\right )}+\frac {x \left (c x^n \left (2 a c e+b^2 (-e)+b c d\right )+3 a b c e-2 a c^2 d-b^3 e+b^2 c d\right )}{a n \left (b^2-4 a c\right ) \left (a e^2-b d e+c d^2\right ) \left (a+b x^n+c x^{2 n}\right )}+\frac {c x \left (b^2 (1-n) \left (e \sqrt {b^2-4 a c}+c d\right )+b c \left (2 a e (2-3 n)-d (1-n) \sqrt {b^2-4 a c}\right )-2 a c \left (e (1-n) \sqrt {b^2-4 a c}+2 c d (1-2 n)\right )+b^3 (-e) (1-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 )}{a n \left (b^2-4 a c\right ) \left (b \sqrt {b^2-4 a c}-4 a c+b^2\right ) \left (a e^2-b d e+c d^2\right )}+\frac {e^4 x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {e x^n}{d}\right )}{d \left (a e^2-b d e+c d^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x^n)*(a + b*x^n + c*x^(2*n))^2),x]

[Out]

(x*(b^2*c*d - 2*a*c^2*d - b^3*e + 3*a*b*c*e + c*(b*c*d - b^2*e + 2*a*c*e)*x^n))/(a*(b^2 - 4*a*c)*(c*d^2 - b*d*
e + a*e^2)*n*(a + b*x^n + c*x^(2*n))) - (c*e^2*(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*x*Hypergeometric2F1[1, n^(-
1), 1 + n^(-1), (-2*c*x^n)/(b - Sqrt[b^2 - 4*a*c])])/((b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c])*(c*d^2 - b*d*e + a*e
^2)^2) - (c*((2*a*b*c*e*(2 - 3*n) - 4*a*c^2*d*(1 - 2*n) + b^2*c*d*(1 - n) - b^3*e*(1 - n))/Sqrt[b^2 - 4*a*c] +
(b*c*d - b^2*e + 2*a*c*e)*(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 - Sqrt[b^2 - 4*a*c])*(c*d^2 - b*d*e + a*e^2)*n) - (c*e^2*(2*c*d - (b - Sqrt[b^2 - 4*
a*c])*e)*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/((b^2 - 4*a*c + b*Sqr
t[b^2 - 4*a*c])*(c*d^2 - b*d*e + a*e^2)^2) + (c*(b*c*(2*a*e*(2 - 3*n) - Sqrt[b^2 - 4*a*c]*d*(1 - n)) - 2*a*c*(
2*c*d*(1 - 2*n) + Sqrt[b^2 - 4*a*c]*e*(1 - n)) - b^3*e*(1 - n) + b^2*(c*d + Sqrt[b^2 - 4*a*c]*e)*(1 - n))*x*Hy
pergeometric2F1[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])*(c*d^2 - b*d*e + a*e^2)*n) + (e^4*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), -((e*x^n)/d)])
/(d*(c*d^2 - b*d*e + a*e^2)^2)

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(11767\) vs. \(2(726)=1452\).
time = 6.80, size = 11767, normalized size = 16.21 \begin {gather*} \text {Result too large to show} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x^n)*(a + b*x^n + c*x^(2*n))^2),x]

[Out]

Result too large to show

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {1}{\left (d +e \,x^{n}\right ) \left (a +b \,x^{n}+c \,x^{2 n}\right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d+e*x^n)/(a+b*x^n+c*x^(2*n))^2,x)

[Out]

int(1/(d+e*x^n)/(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(1/(d+e*x^n)/(a+b*x^n+c*x^(2*n))^2,x, algorithm="maxima")

[Out]

e^4*integrate(1/(c^2*d^5 - 2*b*c*d^4*e - 2*a*b*d^2*e^3 + a^2*d*e^4 + (c^2*d^4*e - 2*b*c*d^3*e^2 - 2*a*b*d*e^4
+ a^2*e^5 + (b^2*d^2 + 2*a*c*d^2)*e^3)*x^n + (b^2*d^3 + 2*a*c*d^3)*e^2), x) + ((b*c^2*d - (b^2*c - 2*a*c^2)*e)
*x*x^n + (b^2*c*d - 2*a*c^2*d - (b^3 - 3*a*b*c)*e)*x)/(a^2*b^2*c*d^2*n - 4*a^3*c^2*d^2*n + (a*b^2*c^2*d^2*n -
4*a^2*c^3*d^2*n + (a^2*b^2*c*n - 4*a^3*c^2*n)*e^2 - (a*b^3*c*d*n - 4*a^2*b*c^2*d*n)*e)*x^(2*n) + (a*b^3*c*d^2*
n - 4*a^2*b*c^2*d^2*n + (a^2*b^3*n - 4*a^3*b*c*n)*e^2 - (a*b^4*d*n - 4*a^2*b^2*c*d*n)*e)*x^n + (a^3*b^2*n - 4*
a^4*c*n)*e^2 - (a^2*b^3*d*n - 4*a^3*b*c*d*n)*e) - integrate((2*a*c^3*d^3*(2*n - 1) - b^2*c^2*d^3*(n - 1) - (b*
c^3*d^3*(n - 1) + (2*a^2*c^2*(3*n - 1) - a*b^2*c*(2*n - 1))*e^3 + (b^3*c*d*(n - 1) - a*b*c^2*d*(n - 1))*e^2 -
2*(b^2*c^2*d^2*(n - 1) - a*c^3*d^2*(n - 1))*e)*x^n - (a^2*b*c*(8*n - 3) - a*b^3*(2*n - 1))*e^3 + (2*a^2*c^2*d*
(4*n - 1) - b^4*d*(n - 1) + 2*a*b^2*c*d*(n - 1))*e^2 - (a*b*c^2*d^2*(8*n - 5) - 2*b^3*c*d^2*(n - 1))*e)/(a^2*b
^2*c^2*d^4*n - 4*a^3*c^3*d^4*n + (a*b^2*c^3*d^4*n - 4*a^2*c^4*d^4*n + (a^3*b^2*c*n - 4*a^4*c^2*n)*e^4 - 2*(a^2
*b^3*c*d*n - 4*a^3*b*c^2*d*n)*e^3 + (a*b^4*c*d^2*n - 2*a^2*b^2*c^2*d^2*n - 8*a^3*c^3*d^2*n)*e^2 - 2*(a*b^3*c^2
*d^3*n - 4*a^2*b*c^3*d^3*n)*e)*x^(2*n) + (a*b^3*c^2*d^4*n - 4*a^2*b*c^3*d^4*n + (a^3*b^3*n - 4*a^4*b*c*n)*e^4
- 2*(a^2*b^4*d*n - 4*a^3*b^2*c*d*n)*e^3 + (a*b^5*d^2*n - 2*a^2*b^3*c*d^2*n - 8*a^3*b*c^2*d^2*n)*e^2 - 2*(a*b^4
*c*d^3*n - 4*a^2*b^2*c^2*d^3*n)*e)*x^n + (a^4*b^2*n - 4*a^5*c*n)*e^4 - 2*(a^3*b^3*d*n - 4*a^4*b*c*d*n)*e^3 + (
a^2*b^4*d^2*n - 2*a^3*b^2*c*d^2*n - 8*a^4*c^2*d^2*n)*e^2 - 2*(a^2*b^3*c*d^3*n - 4*a^3*b*c^2*d^3*n)*e), 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(1/(d+e*x^n)/(a+b*x^n+c*x^(2*n))^2,x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: HeuristicGCDFailed} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d+e*x**n)/(a+b*x**n+c*x**(2*n))**2,x)

[Out]

Exception raised: HeuristicGCDFailed >> no luck

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

[Out]

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

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

[Out]

int(1/((d + e*x^n)*(a + b*x^n + c*x^(2*n))^2), x)

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