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

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

[Out]

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

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Rubi [A]  time = 0.0939848, 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} \[ -\frac{x (b c-a d) (b c (1-n)-a d (n+1)) \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{b x^n}{a}\right )}{a^2 b^2 n}-\frac{d x (b c-a d (n+1))}{a b^2 n}+\frac{x (b c-a d) \left (c+d x^n\right )}{a b n \left (a+b x^n\right )} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^n)^2/(a + b*x^n)^2,x]

[Out]

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

Mathematica [C]  time = 1.57463, size = 666, normalized size = 5.79 \[ \frac{x \left (-2 b c^2 n^6 x^n \text{HypergeometricPFQ}\left (\left \{2,2,2,\frac{1}{n}+1\right \},\left \{1,1,\frac{1}{n}+4\right \},-\frac{b x^n}{a}\right )-4 b c d n^6 x^{2 n} \text{HypergeometricPFQ}\left (\left \{2,2,2,\frac{1}{n}+1\right \},\left \{1,1,\frac{1}{n}+4\right \},-\frac{b x^n}{a}\right )-2 b d^2 n^6 x^{3 n} \text{HypergeometricPFQ}\left (\left \{2,2,2,\frac{1}{n}+1\right \},\left \{1,1,\frac{1}{n}+4\right \},-\frac{b x^n}{a}\right )-2 a \left (6 n^3+11 n^2+6 n+1\right ) \left (c^2 (n+1)^3+2 c d \left (n^3+4 n^2+3 n+1\right ) x^n+d^2 (n+1)^3 x^{2 n}\right ) \Phi \left (-\frac{b x^n}{a},1,1+\frac{1}{n}\right )+a \left (6 n^3+11 n^2+6 n+1\right ) \left (c^2 (2 n+1)^3+2 c d (2 n+1)^3 x^n+d^2 \left (6 n^3+10 n^2+6 n+1\right ) x^{2 n}\right ) \Phi \left (-\frac{b x^n}{a},1,2+\frac{1}{n}\right )+12 a c^2 n^6 \Phi \left (-\frac{b x^n}{a},1,\frac{1}{n}\right )+10 a c^2 n^5 \Phi \left (-\frac{b x^n}{a},1,\frac{1}{n}\right )-10 a c^2 n^4 \Phi \left (-\frac{b x^n}{a},1,\frac{1}{n}\right )-4 a c^2 n^3 \Phi \left (-\frac{b x^n}{a},1,\frac{1}{n}\right )+9 a c^2 n^2 \Phi \left (-\frac{b x^n}{a},1,\frac{1}{n}\right )+a c^2 \Phi \left (-\frac{b x^n}{a},1,\frac{1}{n}\right )+6 a c^2 n \Phi \left (-\frac{b x^n}{a},1,\frac{1}{n}\right )+12 a c d n^3 x^n \Phi \left (-\frac{b x^n}{a},1,\frac{1}{n}\right )+22 a c d n^2 x^n \Phi \left (-\frac{b x^n}{a},1,\frac{1}{n}\right )+2 a c d x^n \Phi \left (-\frac{b x^n}{a},1,\frac{1}{n}\right )+12 a c d n x^n \Phi \left (-\frac{b x^n}{a},1,\frac{1}{n}\right )+6 a d^2 n^3 x^{2 n} \Phi \left (-\frac{b x^n}{a},1,\frac{1}{n}\right )+11 a d^2 n^2 x^{2 n} \Phi \left (-\frac{b x^n}{a},1,\frac{1}{n}\right )+a d^2 x^{2 n} \Phi \left (-\frac{b x^n}{a},1,\frac{1}{n}\right )+6 a d^2 n x^{2 n} \Phi \left (-\frac{b x^n}{a},1,\frac{1}{n}\right )\right )}{2 a^3 n^4 \left (6 n^3+11 n^2+6 n+1\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(-2*a*(1 + 6*n + 11*n^2 + 6*n^3)*(c^2*(1 + n)^3 + 2*c*d*(1 + 3*n + 4*n^2 + n^3)*x^n + d^2*(1 + n)^3*x^(2*n)
)*HurwitzLerchPhi[-((b*x^n)/a), 1, 1 + n^(-1)] + a*(1 + 6*n + 11*n^2 + 6*n^3)*(c^2*(1 + 2*n)^3 + 2*c*d*(1 + 2*
n)^3*x^n + d^2*(1 + 6*n + 10*n^2 + 6*n^3)*x^(2*n))*HurwitzLerchPhi[-((b*x^n)/a), 1, 2 + n^(-1)] + a*c^2*Hurwit
zLerchPhi[-((b*x^n)/a), 1, n^(-1)] + 6*a*c^2*n*HurwitzLerchPhi[-((b*x^n)/a), 1, n^(-1)] + 9*a*c^2*n^2*HurwitzL
erchPhi[-((b*x^n)/a), 1, n^(-1)] - 4*a*c^2*n^3*HurwitzLerchPhi[-((b*x^n)/a), 1, n^(-1)] - 10*a*c^2*n^4*Hurwitz
LerchPhi[-((b*x^n)/a), 1, n^(-1)] + 10*a*c^2*n^5*HurwitzLerchPhi[-((b*x^n)/a), 1, n^(-1)] + 12*a*c^2*n^6*Hurwi
tzLerchPhi[-((b*x^n)/a), 1, n^(-1)] + 2*a*c*d*x^n*HurwitzLerchPhi[-((b*x^n)/a), 1, n^(-1)] + 12*a*c*d*n*x^n*Hu
rwitzLerchPhi[-((b*x^n)/a), 1, n^(-1)] + 22*a*c*d*n^2*x^n*HurwitzLerchPhi[-((b*x^n)/a), 1, n^(-1)] + 12*a*c*d*
n^3*x^n*HurwitzLerchPhi[-((b*x^n)/a), 1, n^(-1)] + a*d^2*x^(2*n)*HurwitzLerchPhi[-((b*x^n)/a), 1, n^(-1)] + 6*
a*d^2*n*x^(2*n)*HurwitzLerchPhi[-((b*x^n)/a), 1, n^(-1)] + 11*a*d^2*n^2*x^(2*n)*HurwitzLerchPhi[-((b*x^n)/a),
1, n^(-1)] + 6*a*d^2*n^3*x^(2*n)*HurwitzLerchPhi[-((b*x^n)/a), 1, n^(-1)] - 2*b*c^2*n^6*x^n*HypergeometricPFQ[
{2, 2, 2, 1 + n^(-1)}, {1, 1, 4 + n^(-1)}, -((b*x^n)/a)] - 4*b*c*d*n^6*x^(2*n)*HypergeometricPFQ[{2, 2, 2, 1 +
n^(-1)}, {1, 1, 4 + n^(-1)}, -((b*x^n)/a)] - 2*b*d^2*n^6*x^(3*n)*HypergeometricPFQ[{2, 2, 2, 1 + n^(-1)}, {1,
1, 4 + n^(-1)}, -((b*x^n)/a)]))/(2*a^3*n^4*(1 + 6*n + 11*n^2 + 6*n^3))

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Maple [F]  time = 0.369, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c+d{x}^{n} \right ) ^{2}}{ \left ( a+b{x}^{n} \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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