3.311 \(\int \frac{1}{(a+b x^n)^2 (c+d x^n)^3} \, dx\)

Optimal. Leaf size=299 \[ \frac{d^2 x \left (a^2 d^2 \left (2 n^2-3 n+1\right )-2 a b c d \left (4 n^2-5 n+1\right )+b^2 c^2 \left (12 n^2-7 n+1\right )\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{d x^n}{c}\right )}{2 c^3 n^2 (b c-a d)^4}-\frac{d x \left (-a^2 d^2 (1-2 n)+a b c d (1-6 n)-2 b^2 c^2 n\right )}{2 a c^2 n^2 (b c-a d)^3 \left (c+d x^n\right )}+\frac{b^3 x (a d (1-4 n)-b c (1-n)) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{b x^n}{a}\right )}{a^2 n (b c-a d)^4}+\frac{b x}{a n (b c-a d) \left (a+b x^n\right ) \left (c+d x^n\right )^2}+\frac{d x (a d+2 b c)}{2 a c n (b c-a d)^2 \left (c+d x^n\right )^2} \]

[Out]

(d*(2*b*c + a*d)*x)/(2*a*c*(b*c - a*d)^2*n*(c + d*x^n)^2) + (b*x)/(a*(b*c - a*d)*n*(a + b*x^n)*(c + d*x^n)^2)
- (d*(a*b*c*d*(1 - 6*n) - a^2*d^2*(1 - 2*n) - 2*b^2*c^2*n)*x)/(2*a*c^2*(b*c - a*d)^3*n^2*(c + d*x^n)) + (b^3*(
a*d*(1 - 4*n) - b*c*(1 - n))*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), -((b*x^n)/a)])/(a^2*(b*c - a*d)^4*n) +
(d^2*(a^2*d^2*(1 - 3*n + 2*n^2) - 2*a*b*c*d*(1 - 5*n + 4*n^2) + b^2*c^2*(1 - 7*n + 12*n^2))*x*Hypergeometric2
F1[1, n^(-1), 1 + n^(-1), -((d*x^n)/c)])/(2*c^3*(b*c - a*d)^4*n^2)

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Rubi [A]  time = 0.546893, antiderivative size = 299, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {414, 527, 522, 245} \[ \frac{d^2 x \left (a^2 d^2 \left (2 n^2-3 n+1\right )-2 a b c d \left (4 n^2-5 n+1\right )+b^2 c^2 \left (12 n^2-7 n+1\right )\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{d x^n}{c}\right )}{2 c^3 n^2 (b c-a d)^4}-\frac{d x \left (-a^2 d^2 (1-2 n)+a b c d (1-6 n)-2 b^2 c^2 n\right )}{2 a c^2 n^2 (b c-a d)^3 \left (c+d x^n\right )}+\frac{b^3 x (a d (1-4 n)-b c (1-n)) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{b x^n}{a}\right )}{a^2 n (b c-a d)^4}+\frac{b x}{a n (b c-a d) \left (a+b x^n\right ) \left (c+d x^n\right )^2}+\frac{d x (a d+2 b c)}{2 a c n (b c-a d)^2 \left (c+d x^n\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(2*b*c + a*d)*x)/(2*a*c*(b*c - a*d)^2*n*(c + d*x^n)^2) + (b*x)/(a*(b*c - a*d)*n*(a + b*x^n)*(c + d*x^n)^2)
- (d*(a*b*c*d*(1 - 6*n) - a^2*d^2*(1 - 2*n) - 2*b^2*c^2*n)*x)/(2*a*c^2*(b*c - a*d)^3*n^2*(c + d*x^n)) + (b^3*(
a*d*(1 - 4*n) - b*c*(1 - n))*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), -((b*x^n)/a)])/(a^2*(b*c - a*d)^4*n) +
(d^2*(a^2*d^2*(1 - 3*n + 2*n^2) - 2*a*b*c*d*(1 - 5*n + 4*n^2) + b^2*c^2*(1 - 7*n + 12*n^2))*x*Hypergeometric2
F1[1, n^(-1), 1 + n^(-1), -((d*x^n)/c)])/(2*c^3*(b*c - a*d)^4*n^2)

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 527

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]

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]

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])

Rubi steps

\begin{align*} \int \frac{1}{\left (a+b x^n\right )^2 \left (c+d x^n\right )^3} \, dx &=\frac{b x}{a (b c-a d) n \left (a+b x^n\right ) \left (c+d x^n\right )^2}-\frac{\int \frac{a d n+b (c-c n)+b d (1-3 n) x^n}{\left (a+b x^n\right ) \left (c+d x^n\right )^3} \, dx}{a (b c-a d) n}\\ &=\frac{d (2 b c+a d) x}{2 a c (b c-a d)^2 n \left (c+d x^n\right )^2}+\frac{b x}{a (b c-a d) n \left (a+b x^n\right ) \left (c+d x^n\right )^2}-\frac{\int \frac{n \left (a^2 d^2 (1-2 n)+2 b^2 c^2 (1-n)+4 a b c d n\right )+b d (2 b c+a d) (1-2 n) n x^n}{\left (a+b x^n\right ) \left (c+d x^n\right )^2} \, dx}{2 a c (b c-a d)^2 n^2}\\ &=\frac{d (2 b c+a d) x}{2 a c (b c-a d)^2 n \left (c+d x^n\right )^2}+\frac{b x}{a (b c-a d) n \left (a+b x^n\right ) \left (c+d x^n\right )^2}-\frac{d \left (a b c d (1-6 n)-a^2 d^2 (1-2 n)-2 b^2 c^2 n\right ) x}{2 a c^2 (b c-a d)^3 n^2 \left (c+d x^n\right )}-\frac{\int \frac{n \left (2 b^3 c^3 (1-n) n+6 a b^2 c^2 d n^2+a^3 d^3 \left (1-3 n+2 n^2\right )-a^2 b c d^2 \left (1-7 n+6 n^2\right )\right )-b d (1-n) n \left (a b c d (1-6 n)-a^2 d^2 (1-2 n)-2 b^2 c^2 n\right ) x^n}{\left (a+b x^n\right ) \left (c+d x^n\right )} \, dx}{2 a c^2 (b c-a d)^3 n^3}\\ &=\frac{d (2 b c+a d) x}{2 a c (b c-a d)^2 n \left (c+d x^n\right )^2}+\frac{b x}{a (b c-a d) n \left (a+b x^n\right ) \left (c+d x^n\right )^2}-\frac{d \left (a b c d (1-6 n)-a^2 d^2 (1-2 n)-2 b^2 c^2 n\right ) x}{2 a c^2 (b c-a d)^3 n^2 \left (c+d x^n\right )}+\frac{\left (b^3 (a d (1-4 n)-b c (1-n))\right ) \int \frac{1}{a+b x^n} \, dx}{a (b c-a d)^4 n}+\frac{\left (d^2 \left (a^2 d^2 \left (1-3 n+2 n^2\right )-2 a b c d \left (1-5 n+4 n^2\right )+b^2 c^2 \left (1-7 n+12 n^2\right )\right )\right ) \int \frac{1}{c+d x^n} \, dx}{2 c^2 (b c-a d)^4 n^2}\\ &=\frac{d (2 b c+a d) x}{2 a c (b c-a d)^2 n \left (c+d x^n\right )^2}+\frac{b x}{a (b c-a d) n \left (a+b x^n\right ) \left (c+d x^n\right )^2}-\frac{d \left (a b c d (1-6 n)-a^2 d^2 (1-2 n)-2 b^2 c^2 n\right ) x}{2 a c^2 (b c-a d)^3 n^2 \left (c+d x^n\right )}+\frac{b^3 (a d (1-4 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)^4 n}+\frac{d^2 \left (a^2 d^2 \left (1-3 n+2 n^2\right )-2 a b c d \left (1-5 n+4 n^2\right )+b^2 c^2 \left (1-7 n+12 n^2\right )\right ) x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{d x^n}{c}\right )}{2 c^3 (b c-a d)^4 n^2}\\ \end{align*}

Mathematica [A]  time = 0.324012, size = 233, normalized size = 0.78 \[ \frac{x \left (\frac{d^2 \left (a^2 d^2 \left (2 n^2-3 n+1\right )-2 a b c d \left (4 n^2-5 n+1\right )+b^2 c^2 \left (12 n^2-7 n+1\right )\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{d x^n}{c}\right )}{c^3}+\frac{2 b^3 n (a d (1-4 n)+b c (n-1)) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{b x^n}{a}\right )}{a^2}+\frac{2 b^3 n (b c-a d)}{a \left (a+b x^n\right )}+\frac{d^2 (a d-b c) (a d (2 n-1)+b (c-6 c n))}{c^2 \left (c+d x^n\right )}+\frac{d^2 n (b c-a d)^2}{c \left (c+d x^n\right )^2}\right )}{2 n^2 (b c-a d)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*((2*b^3*(b*c - a*d)*n)/(a*(a + b*x^n)) + (d^2*(b*c - a*d)^2*n)/(c*(c + d*x^n)^2) + (d^2*(-(b*c) + a*d)*(a*d
*(-1 + 2*n) + b*(c - 6*c*n)))/(c^2*(c + d*x^n)) + (2*b^3*(a*d*(1 - 4*n) + b*c*(-1 + n))*n*Hypergeometric2F1[1,
n^(-1), 1 + n^(-1), -((b*x^n)/a)])/a^2 + (d^2*(a^2*d^2*(1 - 3*n + 2*n^2) - 2*a*b*c*d*(1 - 5*n + 4*n^2) + b^2*
c^2*(1 - 7*n + 12*n^2))*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), -((d*x^n)/c)])/c^3))/(2*(b*c - a*d)^4*n^2)

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Maple [F]  time = 0.734, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( a+b{x}^{n} \right ) ^{2} \left ( c+d{x}^{n} \right ) ^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*x^n)^2/(c+d*x^n)^3,x, algorithm="maxima")

[Out]

((12*n^2 - 7*n + 1)*b^2*c^2*d^2 - 2*(4*n^2 - 5*n + 1)*a*b*c*d^3 + (2*n^2 - 3*n + 1)*a^2*d^4)*integrate(1/2/(b^
4*c^7*n^2 - 4*a*b^3*c^6*d*n^2 + 6*a^2*b^2*c^5*d^2*n^2 - 4*a^3*b*c^4*d^3*n^2 + a^4*c^3*d^4*n^2 + (b^4*c^6*d*n^2
- 4*a*b^3*c^5*d^2*n^2 + 6*a^2*b^2*c^4*d^3*n^2 - 4*a^3*b*c^3*d^4*n^2 + a^4*c^2*d^5*n^2)*x^n), x) - (a*b^3*d*(4
*n - 1) - b^4*c*(n - 1))*integrate(1/(a^2*b^4*c^4*n - 4*a^3*b^3*c^3*d*n + 6*a^4*b^2*c^2*d^2*n - 4*a^5*b*c*d^3*
n + a^6*d^4*n + (a*b^5*c^4*n - 4*a^2*b^4*c^3*d*n + 6*a^3*b^3*c^2*d^2*n - 4*a^4*b^2*c*d^3*n + a^5*b*d^4*n)*x^n)
, x) + 1/2*((a*b^2*c*d^3*(6*n - 1) - a^2*b*d^4*(2*n - 1) + 2*b^3*c^2*d^2*n)*x*x^(2*n) + (a*b^2*c^2*d^2*(7*n -
1) - a^3*d^4*(2*n - 1) + 4*b^3*c^3*d*n + 3*a^2*b*c*d^3*n)*x*x^n + (a^2*b*c^2*d^2*(7*n - 1) - a^3*c*d^3*(3*n -
1) + 2*b^3*c^4*n)*x)/(a^2*b^3*c^7*n^2 - 3*a^3*b^2*c^6*d*n^2 + 3*a^4*b*c^5*d^2*n^2 - a^5*c^4*d^3*n^2 + (a*b^4*c
^5*d^2*n^2 - 3*a^2*b^3*c^4*d^3*n^2 + 3*a^3*b^2*c^3*d^4*n^2 - a^4*b*c^2*d^5*n^2)*x^(3*n) + (2*a*b^4*c^6*d*n^2 -
5*a^2*b^3*c^5*d^2*n^2 + 3*a^3*b^2*c^4*d^3*n^2 + a^4*b*c^3*d^4*n^2 - a^5*c^2*d^5*n^2)*x^(2*n) + (a*b^4*c^7*n^2
- a^2*b^3*c^6*d*n^2 - 3*a^3*b^2*c^5*d^2*n^2 + 5*a^4*b*c^4*d^3*n^2 - 2*a^5*c^3*d^4*n^2)*x^n)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{b^{2} d^{3} x^{5 \, n} + a^{2} c^{3} +{\left (3 \, b^{2} c d^{2} + 2 \, a b d^{3}\right )} x^{4 \, n} +{\left (3 \, b^{2} c^{2} d + 6 \, a b c d^{2} + a^{2} d^{3}\right )} x^{3 \, n} +{\left (b^{2} c^{3} + 6 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x^{2 \, n} +{\left (2 \, a b c^{3} + 3 \, a^{2} c^{2} d\right )} x^{n}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*x^n)^2/(c+d*x^n)^3,x, algorithm="fricas")

[Out]

integral(1/(b^2*d^3*x^(5*n) + a^2*c^3 + (3*b^2*c*d^2 + 2*a*b*d^3)*x^(4*n) + (3*b^2*c^2*d + 6*a*b*c*d^2 + a^2*d
^3)*x^(3*n) + (b^2*c^3 + 6*a*b*c^2*d + 3*a^2*c*d^2)*x^(2*n) + (2*a*b*c^3 + 3*a^2*c^2*d)*x^n), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{n} + a\right )}^{2}{\left (d x^{n} + c\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*x^n)^2/(c+d*x^n)^3,x, algorithm="giac")

[Out]

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