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

Optimal. Leaf size=115 \[ \frac{x (b c-a d) (a d (1-n)-b c (n+1)) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{d x^n}{c}\right )}{c^2 d^2 n}-\frac{b x (a d-b c (n+1))}{c d^2 n}-\frac{x (b c-a d) \left (a+b x^n\right )}{c d n \left (c+d x^n\right )} \]

[Out]

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

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Rubi [A]  time = 0.092344, antiderivative size = 115, 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, 388, 245} \[ \frac{x (b c-a d) (a d (1-n)-b c (n+1)) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{d x^n}{c}\right )}{c^2 d^2 n}-\frac{b x (a d-b c (n+1))}{c d^2 n}-\frac{x (b c-a d) \left (a+b x^n\right )}{c d n \left (c+d x^n\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^n)^2/(c + d*x^n)^2,x]

[Out]

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

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 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 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 )^2} \, dx &=-\frac{(b c-a d) x \left (a+b x^n\right )}{c d n \left (c+d x^n\right )}+\frac{\int \frac{a (b c-a d (1-n))-b (a d-b c (1+n)) x^n}{c+d x^n} \, dx}{c d n}\\ &=-\frac{b (a d-b c (1+n)) x}{c d^2 n}-\frac{(b c-a d) x \left (a+b x^n\right )}{c d n \left (c+d x^n\right )}+\frac{((b c-a d) (a d (1-n)-b c (1+n))) \int \frac{1}{c+d x^n} \, dx}{c d^2 n}\\ &=-\frac{b (a d-b c (1+n)) x}{c d^2 n}-\frac{(b c-a d) x \left (a+b x^n\right )}{c d n \left (c+d x^n\right )}+\frac{(b c-a d) (a d (1-n)-b c (1+n)) x \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{d x^n}{c}\right )}{c^2 d^2 n}\\ \end{align*}

Mathematica [A]  time = 0.113189, size = 95, normalized size = 0.83 \[ \frac{x \left (\frac{c \left (a^2 d^2-2 a b c d+b^2 c \left (c n+c+d n x^n\right )\right )}{c+d x^n}-(b c-a d) (a d (n-1)+b c (n+1)) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{d x^n}{c}\right )\right )}{c^2 d^2 n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.385, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b{x}^{n} \right ) ^{2}}{ \left ( c+d{x}^{n} \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^{2} x^{2 \, n} + 2 \, c d x^{n} + c^{2}}, 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)^2,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )}^{2}}\,{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)^2,x, algorithm="giac")

[Out]

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