\(\int \frac {1}{(a+b x^n)^2 (c+d x^n)^2} \, dx\) [310]

Optimal result

Integrand size = 19, antiderivative size = 193 \[ \int \frac {1}{\left (a+b x^n\right )^2 \left (c+d x^n\right )^2} \, dx=\frac {d (b c+a d) x}{a c (b c-a d)^2 n \left (c+d x^n\right )}+\frac {b x}{a (b c-a d) n \left (a+b x^n\right ) \left (c+d x^n\right )}+\frac {b^2 (a d (1-3 n)-b (c-c n)) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{a^2 (b c-a d)^3 n}-\frac {d^2 (b c (1-3 n)-a d (1-n)) x \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {d x^n}{c}\right )}{c^2 (b c-a d)^3 n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {425, 541, 536, 251} \[ \int \frac {1}{\left (a+b x^n\right )^2 \left (c+d x^n\right )^2} \, dx=\frac {b^2 x (a d (1-3 n)-b (c-c n)) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{a^2 n (b c-a d)^3}-\frac {d^2 x (b c (1-3 n)-a d (1-n)) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {d x^n}{c}\right )}{c^2 n (b c-a d)^3}+\frac {b x}{a n (b c-a d) \left (a+b x^n\right ) \left (c+d x^n\right )}+\frac {d x (a d+b c)}{a c n (b c-a d)^2 \left (c+d x^n\right )} \]

[In]

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

[Out]

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

Rule 541

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

Rubi steps \begin{align*} \text {integral}& = \frac {b x}{a (b c-a d) n \left (a+b x^n\right ) \left (c+d x^n\right )}-\frac {\int \frac {a d n+b (c-c n)+b d (1-2 n) x^n}{\left (a+b x^n\right ) \left (c+d x^n\right )^2} \, dx}{a (b c-a d) n} \\ & = \frac {d (b c+a d) x}{a c (b c-a d)^2 n \left (c+d x^n\right )}+\frac {b x}{a (b c-a d) n \left (a+b x^n\right ) \left (c+d x^n\right )}-\frac {\int \frac {n \left (b^2 c^2 (1-n)+a^2 d^2 (1-n)+2 a b c d n\right )+b d (b c+a d) (1-n) n x^n}{\left (a+b x^n\right ) \left (c+d x^n\right )} \, dx}{a c (b c-a d)^2 n^2} \\ & = \frac {d (b c+a d) x}{a c (b c-a d)^2 n \left (c+d x^n\right )}+\frac {b x}{a (b c-a d) n \left (a+b x^n\right ) \left (c+d x^n\right )}+\frac {\left (d^2 (a d (1-n)-b (c-3 c n))\right ) \int \frac {1}{c+d x^n} \, dx}{c (b c-a d)^3 n}+\frac {\left (b^2 (a d (1-3 n)-b (c-c n))\right ) \int \frac {1}{a+b x^n} \, dx}{a (b c-a d)^3 n} \\ & = \frac {d (b c+a d) x}{a c (b c-a d)^2 n \left (c+d x^n\right )}+\frac {b x}{a (b c-a d) n \left (a+b x^n\right ) \left (c+d x^n\right )}+\frac {b^2 (a d (1-3 n)-b (c-c 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)^3 n}-\frac {d^2 (b c (1-3 n)-a d (1-n)) x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {d x^n}{c}\right )}{c^2 (b c-a d)^3 n} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.76 \[ \int \frac {1}{\left (a+b x^n\right )^2 \left (c+d x^n\right )^2} \, dx=\frac {x \left (\frac {b^2 (b c-a d)}{a \left (a+b x^n\right )}+\frac {d^2 (b c-a d)}{c \left (c+d x^n\right )}+\frac {b^2 (a d (1-3 n)+b c (-1+n)) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {b x^n}{a}\right )}{a^2}+\frac {d^2 (-a d (-1+n)+b c (-1+3 n)) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{n},1+\frac {1}{n},-\frac {d x^n}{c}\right )}{c^2}\right )}{(b c-a d)^3 n} \]

[In]

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

[Out]

(x*((b^2*(b*c - a*d))/(a*(a + b*x^n)) + (d^2*(b*c - a*d))/(c*(c + d*x^n)) + (b^2*(a*d*(1 - 3*n) + b*c*(-1 + n)
)*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), -((b*x^n)/a)])/a^2 + (d^2*(-(a*d*(-1 + n)) + b*c*(-1 + 3*n))*Hyperg
eometric2F1[1, n^(-1), 1 + n^(-1), -((d*x^n)/c)])/c^2))/((b*c - a*d)^3*n)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-2)]

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

[In]

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

[Out]

Exception raised: HeuristicGCDFailed >> no luck

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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