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

Optimal. Leaf size=84 \[ \frac{x (b c-a d)^2 \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{b x^n}{a}\right )}{a b^2}-\frac{d x (a d (n+1)-b (2 c n+c))}{b^2 (n+1)}+\frac{d x \left (c+d x^n\right )}{b (n+1)} \]

[Out]

-((d*(a*d*(1 + n) - b*(c + 2*c*n))*x)/(b^2*(1 + n))) + (d*x*(c + d*x^n))/(b*(1 + n)) + ((b*c - a*d)^2*x*Hyperg
eometric2F1[1, n^(-1), 1 + n^(-1), -((b*x^n)/a)])/(a*b^2)

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Rubi [A]  time = 0.0974269, antiderivative size = 84, 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 = {416, 388, 245} \[ \frac{x (b c-a d)^2 \, _2F_1\left (1,\frac{1}{n};1+\frac{1}{n};-\frac{b x^n}{a}\right )}{a b^2}-\frac{d x (a d (n+1)-b (2 c n+c))}{b^2 (n+1)}+\frac{d x \left (c+d x^n\right )}{b (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^n)^2/(a + b*x^n),x]

[Out]

-((d*(a*d*(1 + n) - b*(c + 2*c*n))*x)/(b^2*(1 + n))) + (d*x*(c + d*x^n))/(b*(1 + n)) + ((b*c - a*d)^2*x*Hyperg
eometric2F1[1, n^(-1), 1 + n^(-1), -((b*x^n)/a)])/(a*b^2)

Rule 416

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

Mathematica [C]  time = 0.295013, size = 75, normalized size = 0.89 \[ \frac{x \left (c^2 \Phi \left (-\frac{b x^n}{a},1,\frac{1}{n}\right )+2 c d x^n \Phi \left (-\frac{b x^n}{a},1,1+\frac{1}{n}\right )+d^2 x^{2 n} \Phi \left (-\frac{b x^n}{a},1,2+\frac{1}{n}\right )\right )}{a n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.429, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c+d{x}^{n} \right ) ^{2}}{a+b{x}^{n}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 x^{n} + a}, 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),x, algorithm="fricas")

[Out]

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

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Sympy [C]  time = 3.18218, size = 170, normalized size = 2.02 \begin{align*} - \frac{2 c d x \Phi \left (\frac{a x^{- n} e^{i \pi }}{b}, 1, \frac{e^{i \pi }}{n}\right ) \Gamma \left (\frac{1}{n}\right )}{b n^{2} \Gamma \left (1 + \frac{1}{n}\right )} + \frac{c^{2} x \Phi \left (\frac{b x^{n} e^{i \pi }}{a}, 1, \frac{1}{n}\right ) \Gamma \left (\frac{1}{n}\right )}{a n^{2} \Gamma \left (1 + \frac{1}{n}\right )} + \frac{2 d^{2} x x^{2 n} \Phi \left (\frac{b x^{n} e^{i \pi }}{a}, 1, 2 + \frac{1}{n}\right ) \Gamma \left (2 + \frac{1}{n}\right )}{a n \Gamma \left (3 + \frac{1}{n}\right )} + \frac{d^{2} x x^{2 n} \Phi \left (\frac{b x^{n} e^{i \pi }}{a}, 1, 2 + \frac{1}{n}\right ) \Gamma \left (2 + \frac{1}{n}\right )}{a n^{2} \Gamma \left (3 + \frac{1}{n}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*c*d*x*lerchphi(a*x**(-n)*exp_polar(I*pi)/b, 1, exp_polar(I*pi)/n)*gamma(1/n)/(b*n**2*gamma(1 + 1/n)) + c**2
*x*lerchphi(b*x**n*exp_polar(I*pi)/a, 1, 1/n)*gamma(1/n)/(a*n**2*gamma(1 + 1/n)) + 2*d**2*x*x**(2*n)*lerchphi(
b*x**n*exp_polar(I*pi)/a, 1, 2 + 1/n)*gamma(2 + 1/n)/(a*n*gamma(3 + 1/n)) + d**2*x*x**(2*n)*lerchphi(b*x**n*ex
p_polar(I*pi)/a, 1, 2 + 1/n)*gamma(2 + 1/n)/(a*n**2*gamma(3 + 1/n))

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

[Out]

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