\(\int \frac {1}{(a+b x^n)^2 (c+d x^n)} \, dx\) [309]

Optimal result

Integrand size = 19, antiderivative size = 122 \[ \int \frac {1}{\left (a+b x^n\right )^2 \left (c+d x^n\right )} \, dx=\frac {b x}{a (b c-a d) n \left (a+b x^n\right )}+\frac {b (a d (1-2 n)-b c (1-n)) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{a^2 (b c-a d)^2 n}+\frac {d^2 x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {d x^n}{c}\right )}{c (b c-a d)^2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {425, 536, 251} \[ \int \frac {1}{\left (a+b x^n\right )^2 \left (c+d x^n\right )} \, dx=\frac {b x (a d (1-2 n)-b c (1-n)) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{a^2 n (b c-a d)^2}+\frac {d^2 x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {d x^n}{c}\right )}{c (b c-a d)^2}+\frac {b x}{a n (b c-a d) \left (a+b x^n\right )} \]

[In]

Int[1/((a + b*x^n)^2*(c + d*x^n)),x]

[Out]

(b*x)/(a*(b*c - a*d)*n*(a + b*x^n)) + (b*(a*d*(1 - 2*n) - b*c*(1 - n))*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-
1), -((b*x^n)/a)])/(a^2*(b*c - a*d)^2*n) + (d^2*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), -((d*x^n)/c)])/(c*(
b*c - a*d)^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 425

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*
((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1
)*(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c,
d, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomi
alQ[a, b, c, d, n, p, q, x]

Rule 536

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {b x}{a (b c-a d) n \left (a+b x^n\right )}-\frac {\int \frac {a d n+b (c-c n)+b d (1-n) x^n}{\left (a+b x^n\right ) \left (c+d x^n\right )} \, dx}{a (b c-a d) n} \\ & = \frac {b x}{a (b c-a d) n \left (a+b x^n\right )}+\frac {d^2 \int \frac {1}{c+d x^n} \, dx}{(b c-a d)^2}+\frac {(b (a d (1-2 n)-b c (1-n))) \int \frac {1}{a+b x^n} \, dx}{a (b c-a d)^2 n} \\ & = \frac {b x}{a (b c-a d) n \left (a+b x^n\right )}+\frac {b (a d (1-2 n)-b c (1-n)) x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {b x^n}{a}\right )}{a^2 (b c-a d)^2 n}+\frac {d^2 x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {d x^n}{c}\right )}{c (b c-a d)^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.89 \[ \int \frac {1}{\left (a+b x^n\right )^2 \left (c+d x^n\right )} \, dx=\frac {x \left (\frac {b^2 c-a b d}{a^2 n+a b n x^n}+\frac {b (a d (1-2 n)+b c (-1+n)) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{a^2 n}+\frac {d^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {d x^n}{c}\right )}{c}\right )}{(b c-a d)^2} \]

[In]

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

[Out]

(x*((b^2*c - a*b*d)/(a^2*n + a*b*n*x^n) + (b*(a*d*(1 - 2*n) + b*c*(-1 + n))*Hypergeometric2F1[1, n^(-1), 1 + n
^(-1), -((b*x^n)/a)])/(a^2*n) + (d^2*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), -((d*x^n)/c)])/c))/(b*c - a*d)^2

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-2)]

Exception generated. \[ \int \frac {1}{\left (a+b x^n\right )^2 \left (c+d x^n\right )} \, dx=\text {Exception raised: HeuristicGCDFailed} \]

[In]

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

[Out]

Exception raised: HeuristicGCDFailed >> no luck

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/((a + b*x^n)^2*(c + d*x^n)),x)

[Out]

int(1/((a + b*x^n)^2*(c + d*x^n)), x)