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

Optimal. Leaf size=160 \[ -\frac{x \left (-a^2 d^2 \left (2 n^2-3 n+1\right )+2 a b c d (1-n)-b^2 c^2 (n+1)\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{d x^n}{c}\right )}{2 c^3 d^2 n^2}+\frac{x (b c-a d) (a d (1-2 n)-b c (n+1))}{2 c^2 d^2 n^2 \left (c+d x^n\right )}-\frac{x (b c-a d) \left (a+b x^n\right )}{2 c d n \left (c+d x^n\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.159603, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {413, 385, 245} \[ -\frac{x \left (-a^2 d^2 \left (2 n^2-3 n+1\right )+2 a b c d (1-n)-b^2 c^2 (n+1)\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{d x^n}{c}\right )}{2 c^3 d^2 n^2}+\frac{x (b c-a d) (a d (1-2 n)-b c (n+1))}{2 c^2 d^2 n^2 \left (c+d x^n\right )}-\frac{x (b c-a d) \left (a+b x^n\right )}{2 c d n \left (c+d x^n\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^n)^2/(c + d*x^n)^3,x]

[Out]

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

Rule 413

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

Rule 385

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

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

Mathematica [A]  time = 0.106566, size = 133, normalized size = 0.83 \[ \frac{x \left (\left (a^2 d^2 \left (2 n^2-3 n+1\right )+2 a b c d (n-1)+b^2 c^2 (n+1)\right ) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{d x^n}{c}\right )+\frac{c^2 n (b c-a d)^2}{\left (c+d x^n\right )^2}-\frac{c (b c-a d) (a d (2 n-1)+b (2 c n+c))}{c+d x^n}\right )}{2 c^3 d^2 n^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.353, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \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((a+b*x^n)^2/(c+d*x^n)^3,x)

[Out]

int((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*}{\left ({\left (2 \, n^{2} - 3 \, n + 1\right )} a^{2} d^{2} + b^{2} c^{2}{\left (n + 1\right )} + 2 \, a b c d{\left (n - 1\right )}\right )} \int \frac{1}{2 \,{\left (c^{2} d^{3} n^{2} x^{n} + c^{3} d^{2} n^{2}\right )}}\,{d x} - \frac{{\left (b^{2} c^{2} d{\left (2 \, n + 1\right )} - a^{2} d^{3}{\left (2 \, n - 1\right )} - 2 \, a b c d^{2}\right )} x x^{n} -{\left (a^{2} c d^{2}{\left (3 \, n - 1\right )} - b^{2} c^{3}{\left (n + 1\right )} - 2 \, a b c^{2} d{\left (n - 1\right )}\right )} x}{2 \,{\left (c^{2} d^{4} n^{2} x^{2 \, n} + 2 \, c^{3} d^{3} n^{2} x^{n} + c^{4} d^{2} n^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*x^(2*n) + 2*a*b*x^n + a^2)/(d^3*x^(3*n) + 3*c*d^2*x^(2*n) + 3*c^2*d*x^n + c^3), 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((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{{\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((a+b*x^n)^2/(c+d*x^n)^3,x, algorithm="giac")

[Out]

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