\(\int \frac {d+e x^n}{(a+b x^n+c x^{2 n})^2} \, dx\) [77]

Optimal result

Integrand size = 24, antiderivative size = 362 \[ \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\frac {x \left (b^2 d-2 a c d-a b e+c (b d-2 a e) x^n\right )}{a \left (b^2-4 a c\right ) n \left (a+b x^n+c x^{2 n}\right )}-\frac {c \left (2 a \left (2 c d (1-2 n)+\sqrt {b^2-4 a c} e (1-n)\right )-b^2 (d-d n)-b \left (\sqrt {b^2-4 a c} d (1-n)-2 a e n\right )\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right ) \left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) n}-\frac {c \left (2 a \left (c d (2-4 n)-\sqrt {b^2-4 a c} e (1-n)\right )-b^2 d (1-n)+b \left (\sqrt {b^2-4 a c} d (1-n)+2 a e n\right )\right ) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a \left (b^2-4 a c\right ) \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) n} \]

[Out]

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))

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 328, normalized size of antiderivative = 0.91, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1444, 1436, 251} \[ \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=-\frac {c x \left (-(1-n) \sqrt {b^2-4 a c} (b d-2 a e)+2 a b e n+2 a c d (2-4 n)+b^2 (-d) (1-n)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{a n \left (b^2-4 a c\right ) \left (-b \sqrt {b^2-4 a c}-4 a c+b^2\right )}-\frac {c x \left ((1-n) \sqrt {b^2-4 a c} (b d-2 a e)+2 a b e n+4 a c d (1-2 n)+b^2 (-d) (1-n)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{a n \left (b^2-4 a c\right ) \left (b \sqrt {b^2-4 a c}-4 a c+b^2\right )}+\frac {x \left (c x^n (b d-2 a e)-a b e-2 a c d+b^2 d\right )}{a n \left (b^2-4 a c\right ) \left (a+b x^n+c x^{2 n}\right )} \]

[In]

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

[Out]

(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])])/(a*(b^2 - 4*a*c)*(b^2 - 4*a*c - b*Sqrt[b^2 - 4*a*c])*n) - (
c*(4*a*c*d*(1 - 2*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*Hypergeometric2F
1[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)

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]

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

Mathematica [A] (verified)

Time = 4.18 (sec) , antiderivative size = 603, normalized size of antiderivative = 1.67 \[ \int \frac {d+e x^n}{\left (a+b x^n+c x^{2 n}\right )^2} \, dx=\frac {c x \left (\frac {4 \left (b^2-4 a c\right ) \left (b^2 d (-1+n) x^n \left (b+c x^n\right )-2 a^2 c \left (2 d n+e x^n\right )+a \left (-2 c^2 d (-1+2 n) x^{2 n}+b c x^n \left (3 d-4 d n+e x^n\right )+b^2 \left (d n+e x^n\right )\right )\right )}{\left (b^2-4 a c-b \sqrt {b^2-4 a c}\right ) \left (b^2-4 a c+b \sqrt {b^2-4 a c}\right ) \left (a+x^n \left (b+c x^n\right )\right )}+\frac {2^{-1/n} \left (4 a c \left (\sqrt {b^2-4 a c} d (1-2 n)+2 a e (-1+n)\right )+b^3 d (-1+n)+b^2 \left (\sqrt {b^2-4 a c} d-2 a e\right ) (-1+n)+2 a b \left (-2 c d (-1+n)+\sqrt {b^2-4 a c} e n\right )\right ) \left (\frac {c x^n}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )^{-1/n} \operatorname {Hypergeometric2F1}\left (-\frac {1}{n},-\frac {1}{n},\frac {-1+n}{n},\frac {b-\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )}{\sqrt {b^2-4 a c} \left (-b^2+4 a c+b \sqrt {b^2-4 a c}\right )}+\frac {2^{-1/n} \left (b \sqrt {b^2-4 a c} d (-1+n)-2 a \sqrt {b^2-4 a c} e (-1+n)-2 a b e n+4 a c d (-1+2 n)+b^2 (d-d n)\right ) \left (\frac {c x^n}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )^{-1/n} \operatorname {Hypergeometric2F1}\left (-\frac {1}{n},-\frac {1}{n},\frac {-1+n}{n},\frac {b+\sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}\right )}\right )}{a \left (-b^2+4 a c\right ) n} \]

[In]

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

[Out]

(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))*Hyp
ergeometric2F1[-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*Sqrt[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)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

((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)

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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