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

Optimal. Leaf size=141 \[ \frac{e x^{n+1} \left (3 c d^2-a e^2\right ) \, _2F_1\left (1,\frac{n+1}{2 n};\frac{1}{2} \left (3+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{a c (n+1)}+\frac{d x \left (c d^2-3 a e^2\right ) \, _2F_1\left (1,\frac{1}{2 n};\frac{1}{2} \left (2+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{a c}+\frac{3 d e^2 x}{c}+\frac{e^3 x^{n+1}}{c (n+1)} \]

[Out]

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

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Rubi [A]  time = 0.145232, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {1425, 1418, 245, 364} \[ \frac{e x^{n+1} \left (3 c d^2-a e^2\right ) \, _2F_1\left (1,\frac{n+1}{2 n};\frac{1}{2} \left (3+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{a c (n+1)}+\frac{d x \left (c d^2-3 a e^2\right ) \, _2F_1\left (1,\frac{1}{2 n};\frac{1}{2} \left (2+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{a c}+\frac{3 d e^2 x}{c}+\frac{e^3 x^{n+1}}{c (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x^n)^3/(a + c*x^(2*n)),x]

[Out]

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

Rule 1425

Int[((d_) + (e_.)*(x_)^(n_))^(q_)/((a_) + (c_.)*(x_)^(n2_)), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^n)^q/(a
+ c*x^(2*n)), x], x] /; FreeQ[{a, c, d, e, n}, x] && EqQ[n2, 2*n] && NeQ[c*d^2 + a*e^2, 0] && IntegerQ[q]

Rule 1418

Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (c_.)*(x_)^(n2_)), x_Symbol] :> Dist[d, Int[1/(a + c*x^(2*n)), x], x] + D
ist[e, Int[x^n/(a + c*x^(2*n)), x], x] /; FreeQ[{a, c, d, e, n}, x] && EqQ[n2, 2*n] && NeQ[c*d^2 + a*e^2, 0] &
& (PosQ[a*c] ||  !IntegerQ[n])

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])

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{\left (d+e x^n\right )^3}{a+c x^{2 n}} \, dx &=\int \left (\frac{3 d e^2}{c}+\frac{e^3 x^n}{c}+\frac{c d^3-3 a d e^2+\left (3 c d^2 e-a e^3\right ) x^n}{c \left (a+c x^{2 n}\right )}\right ) \, dx\\ &=\frac{3 d e^2 x}{c}+\frac{e^3 x^{1+n}}{c (1+n)}+\frac{\int \frac{c d^3-3 a d e^2+\left (3 c d^2 e-a e^3\right ) x^n}{a+c x^{2 n}} \, dx}{c}\\ &=\frac{3 d e^2 x}{c}+\frac{e^3 x^{1+n}}{c (1+n)}+\frac{\left (d \left (c d^2-3 a e^2\right )\right ) \int \frac{1}{a+c x^{2 n}} \, dx}{c}+\frac{\left (e \left (3 c d^2-a e^2\right )\right ) \int \frac{x^n}{a+c x^{2 n}} \, dx}{c}\\ &=\frac{3 d e^2 x}{c}+\frac{e^3 x^{1+n}}{c (1+n)}+\frac{d \left (c d^2-3 a e^2\right ) x \, _2F_1\left (1,\frac{1}{2 n};\frac{1}{2} \left (2+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{a c}+\frac{e \left (3 c d^2-a e^2\right ) x^{1+n} \, _2F_1\left (1,\frac{1+n}{2 n};\frac{1}{2} \left (3+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )}{a c (1+n)}\\ \end{align*}

Mathematica [A]  time = 0.247389, size = 127, normalized size = 0.9 \[ \frac{x \left (d (n+1) \left (c d^2-3 a e^2\right ) \, _2F_1\left (1,\frac{1}{2 n};\frac{1}{2} \left (2+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )+e \left (x^n \left (3 c d^2-a e^2\right ) \, _2F_1\left (1,\frac{n+1}{2 n};\frac{1}{2} \left (3+\frac{1}{n}\right );-\frac{c x^{2 n}}{a}\right )+a e \left (3 d (n+1)+e x^n\right )\right )\right )}{a c (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.078, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( d+e{x}^{n} \right ) ^{3}}{a+c{x}^{2\,n}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d+e*x^n)^3/(a+c*x^(2*n)),x)

[Out]

int((d+e*x^n)^3/(a+c*x^(2*n)),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((e^3*x^(3*n) + 3*d*e^2*x^(2*n) + 3*d^2*e*x^n + d^3)/(c*x^(2*n) + a), x)

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Sympy [C]  time = 10.3546, size = 337, normalized size = 2.39 \begin{align*} - \frac{3 d e^{2} x \Phi \left (\frac{a x^{- 2 n} e^{i \pi }}{c}, 1, \frac{e^{i \pi }}{2 n}\right ) \Gamma \left (\frac{1}{2 n}\right )}{4 c n^{2} \Gamma \left (1 + \frac{1}{2 n}\right )} + \frac{d^{3} x \Phi \left (\frac{c x^{2 n} e^{i \pi }}{a}, 1, \frac{1}{2 n}\right ) \Gamma \left (\frac{1}{2 n}\right )}{4 a n^{2} \Gamma \left (1 + \frac{1}{2 n}\right )} + \frac{3 d^{2} e x x^{n} \Phi \left (\frac{c x^{2 n} e^{i \pi }}{a}, 1, \frac{1}{2} + \frac{1}{2 n}\right ) \Gamma \left (\frac{1}{2} + \frac{1}{2 n}\right )}{4 a n \Gamma \left (\frac{3}{2} + \frac{1}{2 n}\right )} + \frac{3 d^{2} e x x^{n} \Phi \left (\frac{c x^{2 n} e^{i \pi }}{a}, 1, \frac{1}{2} + \frac{1}{2 n}\right ) \Gamma \left (\frac{1}{2} + \frac{1}{2 n}\right )}{4 a n^{2} \Gamma \left (\frac{3}{2} + \frac{1}{2 n}\right )} + \frac{3 e^{3} x x^{3 n} \Phi \left (\frac{c x^{2 n} e^{i \pi }}{a}, 1, \frac{3}{2} + \frac{1}{2 n}\right ) \Gamma \left (\frac{3}{2} + \frac{1}{2 n}\right )}{4 a n \Gamma \left (\frac{5}{2} + \frac{1}{2 n}\right )} + \frac{e^{3} x x^{3 n} \Phi \left (\frac{c x^{2 n} e^{i \pi }}{a}, 1, \frac{3}{2} + \frac{1}{2 n}\right ) \Gamma \left (\frac{3}{2} + \frac{1}{2 n}\right )}{4 a n^{2} \Gamma \left (\frac{5}{2} + \frac{1}{2 n}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*d*e**2*x*lerchphi(a*x**(-2*n)*exp_polar(I*pi)/c, 1, exp_polar(I*pi)/(2*n))*gamma(1/(2*n))/(4*c*n**2*gamma(1
+ 1/(2*n))) + d**3*x*lerchphi(c*x**(2*n)*exp_polar(I*pi)/a, 1, 1/(2*n))*gamma(1/(2*n))/(4*a*n**2*gamma(1 + 1/
(2*n))) + 3*d**2*e*x*x**n*lerchphi(c*x**(2*n)*exp_polar(I*pi)/a, 1, 1/2 + 1/(2*n))*gamma(1/2 + 1/(2*n))/(4*a*n
*gamma(3/2 + 1/(2*n))) + 3*d**2*e*x*x**n*lerchphi(c*x**(2*n)*exp_polar(I*pi)/a, 1, 1/2 + 1/(2*n))*gamma(1/2 +
1/(2*n))/(4*a*n**2*gamma(3/2 + 1/(2*n))) + 3*e**3*x*x**(3*n)*lerchphi(c*x**(2*n)*exp_polar(I*pi)/a, 1, 3/2 + 1
/(2*n))*gamma(3/2 + 1/(2*n))/(4*a*n*gamma(5/2 + 1/(2*n))) + e**3*x*x**(3*n)*lerchphi(c*x**(2*n)*exp_polar(I*pi
)/a, 1, 3/2 + 1/(2*n))*gamma(3/2 + 1/(2*n))/(4*a*n**2*gamma(5/2 + 1/(2*n)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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