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

Optimal. Leaf size=210 \[ -\frac{d x \left (a^2 d^2 \left (2 n^2-3 n+1\right )-2 a b c d \left (3 n^2-4 n+1\right )+b^2 c^2 \left (6 n^2-5 n+1\right )\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{d x^n}{c}\right )}{2 c^3 n^2 (b c-a d)^3}+\frac{b^3 x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{b x^n}{a}\right )}{a (b c-a d)^3}-\frac{d x (a d (1-2 n)-b (c-4 c n))}{2 c^2 n^2 (b c-a d)^2 \left (c+d x^n\right )}-\frac{d x}{2 c n (b c-a d) \left (c+d x^n\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.328671, antiderivative size = 210, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {414, 527, 522, 245} \[ -\frac{d x \left (a^2 d^2 \left (2 n^2-3 n+1\right )-2 a b c d \left (3 n^2-4 n+1\right )+b^2 c^2 \left (6 n^2-5 n+1\right )\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{d x^n}{c}\right )}{2 c^3 n^2 (b c-a d)^3}+\frac{b^3 x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{b x^n}{a}\right )}{a (b c-a d)^3}-\frac{d x (a d (1-2 n)-b (c-4 c n))}{2 c^2 n^2 (b c-a d)^2 \left (c+d x^n\right )}-\frac{d x}{2 c n (b c-a d) \left (c+d x^n\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*x^n)*(c + d*x^n)^3),x]

[Out]

-(d*x)/(2*c*(b*c - a*d)*n*(c + d*x^n)^2) - (d*(a*d*(1 - 2*n) - b*(c - 4*c*n))*x)/(2*c^2*(b*c - a*d)^2*n^2*(c +
d*x^n)) + (b^3*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), -((b*x^n)/a)])/(a*(b*c - a*d)^3) - (d*(a^2*d^2*(1 -
3*n + 2*n^2) - 2*a*b*c*d*(1 - 4*n + 3*n^2) + b^2*c^2*(1 - 5*n + 6*n^2))*x*Hypergeometric2F1[1, n^(-1), 1 + n^
(-1), -((d*x^n)/c)])/(2*c^3*(b*c - a*d)^3*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 ) \left (c+d x^n\right )^3} \, dx &=-\frac{d x}{2 c (b c-a d) n \left (c+d x^n\right )^2}+\frac{\int \frac{2 b c n+a (d-2 d n)+b d (1-2 n) x^n}{\left (a+b x^n\right ) \left (c+d x^n\right )^2} \, dx}{2 c (b c-a d) n}\\ &=-\frac{d x}{2 c (b c-a d) n \left (c+d x^n\right )^2}+\frac{d (b c (1-4 n)-a d (1-2 n)) x}{2 c^2 (b c-a d)^2 n^2 \left (c+d x^n\right )}+\frac{\int \frac{2 b^2 c^2 n^2+a^2 d^2 \left (1-3 n+2 n^2\right )-a b c d \left (1-5 n+4 n^2\right )-b d (b c (1-4 n)-a d (1-2 n)) (1-n) x^n}{\left (a+b x^n\right ) \left (c+d x^n\right )} \, dx}{2 c^2 (b c-a d)^2 n^2}\\ &=-\frac{d x}{2 c (b c-a d) n \left (c+d x^n\right )^2}+\frac{d (b c (1-4 n)-a d (1-2 n)) x}{2 c^2 (b c-a d)^2 n^2 \left (c+d x^n\right )}+\frac{b^3 \int \frac{1}{a+b x^n} \, dx}{(b c-a d)^3}-\frac{\left (d \left (a^2 d^2 \left (1-3 n+2 n^2\right )-2 a b c d \left (1-4 n+3 n^2\right )+b^2 c^2 \left (1-5 n+6 n^2\right )\right )\right ) \int \frac{1}{c+d x^n} \, dx}{2 c^2 (b c-a d)^3 n^2}\\ &=-\frac{d x}{2 c (b c-a d) n \left (c+d x^n\right )^2}+\frac{d (b c (1-4 n)-a d (1-2 n)) x}{2 c^2 (b c-a d)^2 n^2 \left (c+d x^n\right )}+\frac{b^3 x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{b x^n}{a}\right )}{a (b c-a d)^3}-\frac{d \left (a^2 d^2 \left (1-3 n+2 n^2\right )-2 a b c d \left (1-4 n+3 n^2\right )+b^2 c^2 \left (1-5 n+6 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)^3 n^2}\\ \end{align*}

Mathematica [A]  time = 0.197811, size = 210, normalized size = 1. \[ \frac{x \left (-a d \left (c+d x^n\right )^2 \left (a^2 d^2 \left (2 n^2-3 n+1\right )-2 a b c d \left (3 n^2-4 n+1\right )+b^2 c^2 \left (6 n^2-5 n+1\right )\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{d x^n}{c}\right )+2 b^3 c^3 n^2 \left (c+d x^n\right )^2 \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{b x^n}{a}\right )-a c^2 d n (b c-a d)^2+a c d (b c-a d) \left (c+d x^n\right ) (a d (2 n-1)+b (c-4 c n))\right )}{2 a c^3 n^2 (b c-a d)^3 \left (c+d x^n\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.757, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( a+b{x}^{n} \right ) \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)/(c+d*x^n)^3,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b^3*integrate(-1/(a*b^3*c^3 - 3*a^2*b^2*c^2*d + 3*a^3*b*c*d^2 - a^4*d^3 + (b^4*c^3 - 3*a*b^3*c^2*d + 3*a^2*b^
2*c*d^2 - a^3*b*d^3)*x^n), x) + ((6*n^2 - 5*n + 1)*b^2*c^2*d - 2*(3*n^2 - 4*n + 1)*a*b*c*d^2 + (2*n^2 - 3*n +
1)*a^2*d^3)*integrate(-1/2/(b^3*c^6*n^2 - 3*a*b^2*c^5*d*n^2 + 3*a^2*b*c^4*d^2*n^2 - a^3*c^3*d^3*n^2 + (b^3*c^5
*d*n^2 - 3*a*b^2*c^4*d^2*n^2 + 3*a^2*b*c^3*d^3*n^2 - a^3*c^2*d^4*n^2)*x^n), x) - 1/2*((b*c*d^2*(4*n - 1) - a*d
^3*(2*n - 1))*x*x^n + (b*c^2*d*(5*n - 1) - a*c*d^2*(3*n - 1))*x)/(b^2*c^6*n^2 - 2*a*b*c^5*d*n^2 + a^2*c^4*d^2*
n^2 + (b^2*c^4*d^2*n^2 - 2*a*b*c^3*d^3*n^2 + a^2*c^2*d^4*n^2)*x^(2*n) + 2*(b^2*c^5*d*n^2 - 2*a*b*c^4*d^2*n^2 +
a^2*c^3*d^3*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 d^{3} x^{4 \, n} + a c^{3} +{\left (3 \, b c d^{2} + a d^{3}\right )} x^{3 \, n} + 3 \,{\left (b c^{2} d + a c d^{2}\right )} x^{2 \, n} +{\left (b c^{3} + 3 \, a 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)/(c+d*x^n)^3,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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