\(\int \frac {1}{(d+e x^n) (a+c x^{2 n})^2} \, dx\) [51]

Optimal result

Integrand size = 21, antiderivative size = 333 \[ \int \frac {1}{\left (d+e x^n\right ) \left (a+c x^{2 n}\right )^2} \, dx=\frac {c x \left (d-e x^n\right )}{2 a \left (c d^2+a e^2\right ) n \left (a+c x^{2 n}\right )}+\frac {c d e^2 x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a \left (c d^2+a e^2\right )^2}-\frac {c d (1-2 n) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{2 a^2 \left (c d^2+a e^2\right ) n}+\frac {e^4 x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{d \left (c d^2+a e^2\right )^2}-\frac {c e^3 x^{1+n} \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a \left (c d^2+a e^2\right )^2 (1+n)}+\frac {c e (1-n) x^{1+n} \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{2 a^2 \left (c d^2+a e^2\right ) n (1+n)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 333, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {1451, 251, 1445, 1432, 371} \[ \int \frac {1}{\left (d+e x^n\right ) \left (a+c x^{2 n}\right )^2} \, dx=\frac {c e (1-n) x^{n+1} \operatorname {Hypergeometric2F1}\left (1,\frac {n+1}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{2 a^2 n (n+1) \left (a e^2+c d^2\right )}-\frac {c d (1-2 n) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{2 a^2 n \left (a e^2+c d^2\right )}+\frac {c d e^2 x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a \left (a e^2+c d^2\right )^2}+\frac {c x \left (d-e x^n\right )}{2 a n \left (a e^2+c d^2\right ) \left (a+c x^{2 n}\right )}+\frac {e^4 x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )}{d \left (a e^2+c d^2\right )^2}-\frac {c e^3 x^{n+1} \operatorname {Hypergeometric2F1}\left (1,\frac {n+1}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )}{a (n+1) \left (a e^2+c d^2\right )^2} \]

[In]

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

[Out]

(c*x*(d - e*x^n))/(2*a*(c*d^2 + a*e^2)*n*(a + c*x^(2*n))) + (c*d*e^2*x*Hypergeometric2F1[1, 1/(2*n), (2 + n^(-
1))/2, -((c*x^(2*n))/a)])/(a*(c*d^2 + a*e^2)^2) - (c*d*(1 - 2*n)*x*Hypergeometric2F1[1, 1/(2*n), (2 + n^(-1))/
2, -((c*x^(2*n))/a)])/(2*a^2*(c*d^2 + a*e^2)*n) + (e^4*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), -((e*x^n)/d)
])/(d*(c*d^2 + a*e^2)^2) - (c*e^3*x^(1 + n)*Hypergeometric2F1[1, (1 + n)/(2*n), (3 + n^(-1))/2, -((c*x^(2*n))/
a)])/(a*(c*d^2 + a*e^2)^2*(1 + n)) + (c*e*(1 - n)*x^(1 + n)*Hypergeometric2F1[1, (1 + n)/(2*n), (3 + n^(-1))/2
, -((c*x^(2*n))/a)])/(2*a^2*(c*d^2 + a*e^2)*n*(1 + 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 1432

Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (c_.)*(x_)^(n2_)), x_Symbol] :> Dist[d, Int[1/(a + c*x^(2*n)), x], x] + D
ist[e, Int[x^n/(a + c*x^(2*n)), x], x] /; FreeQ[{a, c, d, e, n}, x] && EqQ[n2, 2*n] && NeQ[c*d^2 + a*e^2, 0] &
& (PosQ[a*c] ||  !IntegerQ[n])

Rule 1445

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

Rule 1451

Int[((d_) + (e_.)*(x_)^(n_))^(q_)*((a_) + (c_.)*(x_)^(n2_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^n)
^q*(a + c*x^(2*n))^p, x], x] /; FreeQ[{a, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && NeQ[c*d^2 + 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 {e^4}{\left (c d^2+a e^2\right )^2 \left (d+e x^n\right )}-\frac {c \left (-d+e x^n\right )}{\left (c d^2+a e^2\right ) \left (a+c x^{2 n}\right )^2}-\frac {c e^2 \left (-d+e x^n\right )}{\left (c d^2+a e^2\right )^2 \left (a+c x^{2 n}\right )}\right ) \, dx \\ & = -\frac {\left (c e^2\right ) \int \frac {-d+e x^n}{a+c x^{2 n}} \, dx}{\left (c d^2+a e^2\right )^2}+\frac {e^4 \int \frac {1}{d+e x^n} \, dx}{\left (c d^2+a e^2\right )^2}-\frac {c \int \frac {-d+e x^n}{\left (a+c x^{2 n}\right )^2} \, dx}{c d^2+a e^2} \\ & = \frac {c x \left (d-e x^n\right )}{2 a \left (c d^2+a e^2\right ) n \left (a+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+a e^2\right )^2}+\frac {\left (c d e^2\right ) \int \frac {1}{a+c x^{2 n}} \, dx}{\left (c d^2+a e^2\right )^2}-\frac {\left (c e^3\right ) \int \frac {x^n}{a+c x^{2 n}} \, dx}{\left (c d^2+a e^2\right )^2}+\frac {c \int \frac {-d (1-2 n)+e (1-n) x^n}{a+c x^{2 n}} \, dx}{2 a \left (c d^2+a e^2\right ) n} \\ & = \frac {c x \left (d-e x^n\right )}{2 a \left (c d^2+a e^2\right ) n \left (a+c x^{2 n}\right )}+\frac {c d e^2 x \, _2F_1\left (1,\frac {1}{2 n};\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a \left (c d^2+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+a e^2\right )^2}-\frac {c e^3 x^{1+n} \, _2F_1\left (1,\frac {1+n}{2 n};\frac {1}{2} \left (3+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a \left (c d^2+a e^2\right )^2 (1+n)}-\frac {(c d (1-2 n)) \int \frac {1}{a+c x^{2 n}} \, dx}{2 a \left (c d^2+a e^2\right ) n}+\frac {(c e (1-n)) \int \frac {x^n}{a+c x^{2 n}} \, dx}{2 a \left (c d^2+a e^2\right ) n} \\ & = \frac {c x \left (d-e x^n\right )}{2 a \left (c d^2+a e^2\right ) n \left (a+c x^{2 n}\right )}+\frac {c d e^2 x \, _2F_1\left (1,\frac {1}{2 n};\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a \left (c d^2+a e^2\right )^2}-\frac {c d (1-2 n) x \, _2F_1\left (1,\frac {1}{2 n};\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{2 a^2 \left (c d^2+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+a e^2\right )^2}-\frac {c e^3 x^{1+n} \, _2F_1\left (1,\frac {1+n}{2 n};\frac {1}{2} \left (3+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a \left (c d^2+a e^2\right )^2 (1+n)}+\frac {c e (1-n) x^{1+n} \, _2F_1\left (1,\frac {1+n}{2 n};\frac {1}{2} \left (3+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{2 a^2 \left (c d^2+a e^2\right ) n (1+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.68 \[ \int \frac {1}{\left (d+e x^n\right ) \left (a+c x^{2 n}\right )^2} \, dx=\frac {x \left (a c d^2 e^2 (1+n) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )+a^2 e^4 (1+n) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {e x^n}{d}\right )+c d \left (-a e^3 x^n \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )+\left (c d^2+a e^2\right ) \left (d (1+n) \operatorname {Hypergeometric2F1}\left (2,\frac {1}{2 n},\frac {1}{2} \left (2+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )-e x^n \operatorname {Hypergeometric2F1}\left (2,\frac {1+n}{2 n},\frac {1}{2} \left (3+\frac {1}{n}\right ),-\frac {c x^{2 n}}{a}\right )\right )\right )\right )}{a^2 d \left (c d^2+a e^2\right )^2 (1+n)} \]

[In]

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

[Out]

(x*(a*c*d^2*e^2*(1 + n)*Hypergeometric2F1[1, 1/(2*n), (2 + n^(-1))/2, -((c*x^(2*n))/a)] + a^2*e^4*(1 + n)*Hype
rgeometric2F1[1, n^(-1), 1 + n^(-1), -((e*x^n)/d)] + c*d*(-(a*e^3*x^n*Hypergeometric2F1[1, (1 + n)/(2*n), (3 +
n^(-1))/2, -((c*x^(2*n))/a)]) + (c*d^2 + a*e^2)*(d*(1 + n)*Hypergeometric2F1[2, 1/(2*n), (2 + n^(-1))/2, -((c
*x^(2*n))/a)] - e*x^n*Hypergeometric2F1[2, (1 + n)/(2*n), (3 + n^(-1))/2, -((c*x^(2*n))/a)]))))/(a^2*d*(c*d^2
+ a*e^2)^2*(1 + n))

Maple [F]

\[\int \frac {1}{\left (d +e \,x^{n}\right ) \left (a +c \,x^{2 n}\right )^{2}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {1}{\left (d+e x^n\right ) \left (a+c x^{2 n}\right )^2} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + a\right )}^{2} {\left (e x^{n} + d\right )}} \,d x } \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{\left (d+e x^n\right ) \left (a+c x^{2 n}\right )^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {1}{\left (d+e x^n\right ) \left (a+c x^{2 n}\right )^2} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + a\right )}^{2} {\left (e x^{n} + d\right )}} \,d x } \]

[In]

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

[Out]

e^4*integrate(1/(c^2*d^5 + 2*a*c*d^3*e^2 + a^2*d*e^4 + (c^2*d^4*e + 2*a*c*d^2*e^3 + a^2*e^5)*x^n), x) - 1/2*(c
*e*x*x^n - c*d*x)/(a^2*c*d^2*n + a^3*e^2*n + (a*c^2*d^2*n + a^2*c*e^2*n)*x^(2*n)) - integrate(-1/2*(a*c*d*e^2*
(4*n - 1) + c^2*d^3*(2*n - 1) - (a*c*e^3*(3*n - 1) + c^2*d^2*e*(n - 1))*x^n)/(a^2*c^2*d^4*n + 2*a^3*c*d^2*e^2*
n + a^4*e^4*n + (a*c^3*d^4*n + 2*a^2*c^2*d^2*e^2*n + a^3*c*e^4*n)*x^(2*n)), x)

Giac [F]

\[ \int \frac {1}{\left (d+e x^n\right ) \left (a+c x^{2 n}\right )^2} \, dx=\int { \frac {1}{{\left (c x^{2 \, n} + a\right )}^{2} {\left (e x^{n} + d\right )}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\left (d+e x^n\right ) \left (a+c x^{2 n}\right )^2} \, dx=\int \frac {1}{{\left (a+c\,x^{2\,n}\right )}^2\,\left (d+e\,x^n\right )} \,d x \]

[In]

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

[Out]

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