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

Optimal. Leaf size=123 \[ \frac {b^2 x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {b x^n}{a}\right )}{a (b c-a d)^2}+\frac {d x (b c (1-2 n)-a d (1-n)) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {d x^n}{c}\right )}{c^2 n (b c-a d)^2}-\frac {d x}{c n (b c-a d) \left (c+d x^n\right )} \]

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {414, 522, 245} \[ \frac {b^2 x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {b x^n}{a}\right )}{a (b c-a d)^2}+\frac {d x (b c (1-2 n)-a d (1-n)) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {d x^n}{c}\right )}{c^2 n (b c-a d)^2}-\frac {d x}{c n (b c-a d) \left (c+d x^n\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x^n)*(c + d*x^n)^2),x]

[Out]

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

Rule 245

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 414

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]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 522

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

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Mathematica [A]  time = 0.22, size = 121, normalized size = 0.98 \[ \frac {x \left (b^2 c^2 n \left (c+d x^n\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {b x^n}{a}\right )+a d \left (\left (c+d x^n\right ) (a d (n-1)+b (c-2 c n)) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {d x^n}{c}\right )+c (a d-b c)\right )\right )}{a c^2 n (b c-a d)^2 \left (c+d x^n\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.23, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{b d^{2} x^{3 \, n} + a c^{2} + {\left (2 \, b c d + a d^{2}\right )} x^{2 \, n} + {\left (b c^{2} + 2 \, a c d\right )} x^{n}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.93, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \,x^{n}+a \right ) \left (d \,x^{n}+c \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ b^{2} \int \frac {1}{a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2} + {\left (b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2}\right )} x^{n}}\,{d x} - {\left (b c d {\left (2 \, n - 1\right )} - a d^{2} {\left (n - 1\right )}\right )} \int \frac {1}{b^{2} c^{4} n - 2 \, a b c^{3} d n + a^{2} c^{2} d^{2} n + {\left (b^{2} c^{3} d n - 2 \, a b c^{2} d^{2} n + a^{2} c d^{3} n\right )} x^{n}}\,{d x} - \frac {d x}{b c^{3} n - a c^{2} d n + {\left (b c^{2} d n - a c d^{2} n\right )} x^{n}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\left (a+b\,x^n\right )\,{\left (c+d\,x^n\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + b*x^n)*(c + d*x^n)^2),x)

[Out]

int(1/((a + b*x^n)*(c + d*x^n)^2), x)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: HeuristicGCDFailed} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: HeuristicGCDFailed

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