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

Optimal. Leaf size=107 \[ \frac {x \left (c d^2-a e^2\right ) \, _2F_1\left (1,\frac {1}{2 n};\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a c}+\frac {2 d e x^{n+1} \, _2F_1\left (1,\frac {n+1}{2 n};\frac {1}{2} \left (3+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a (n+1)}+\frac {e^2 x}{c} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {1425, 1418, 245, 364} \[ \frac {x \left (c d^2-a e^2\right ) \, _2F_1\left (1,\frac {1}{2 n};\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a c}+\frac {2 d e x^{n+1} \, _2F_1\left (1,\frac {n+1}{2 n};\frac {1}{2} \left (3+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a (n+1)}+\frac {e^2 x}{c} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x^n)^2/(a + c*x^(2*n)),x]

[Out]

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

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 107, normalized size = 1.00 \[ \frac {x \left (c d^2-a e^2\right ) \, _2F_1\left (1,\frac {1}{2 n};\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a c}+\frac {2 d e x^{n+1} \, _2F_1\left (1,\frac {n+1}{2 n};\frac {1}{2} \left (3+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a (n+1)}+\frac {e^2 x}{c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.97, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {e^{2} x^{2 \, n} + 2 \, d e x^{n} + d^{2}}{c x^{2 \, n} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x^{n} + d\right )}^{2}}{c x^{2 \, n} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.09, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \,x^{n}+d \right )^{2}}{c \,x^{2 n}+a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {e^{2} x}{c} + \int \frac {2 \, c d e x^{n} + c d^{2} - a e^{2}}{c^{2} x^{2 \, n} + a c}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d+e\,x^n\right )}^2}{a+c\,x^{2\,n}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x^n)^2/(a + c*x^(2*n)),x)

[Out]

int((d + e*x^n)^2/(a + c*x^(2*n)), x)

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sympy [C]  time = 7.65, size = 207, normalized size = 1.93 \[ - \frac {e^{2} x \Phi \left (\frac {a x^{- 2 n} e^{i \pi }}{c}, 1, \frac {e^{i \pi }}{2 n}\right ) \Gamma \left (\frac {1}{2 n}\right )}{4 c n^{2} \Gamma \left (1 + \frac {1}{2 n}\right )} + \frac {d^{2} x \Phi \left (\frac {c x^{2 n} e^{i \pi }}{a}, 1, \frac {1}{2 n}\right ) \Gamma \left (\frac {1}{2 n}\right )}{4 a n^{2} \Gamma \left (1 + \frac {1}{2 n}\right )} + \frac {d e x x^{n} \Phi \left (\frac {c x^{2 n} e^{i \pi }}{a}, 1, \frac {1}{2} + \frac {1}{2 n}\right ) \Gamma \left (\frac {1}{2} + \frac {1}{2 n}\right )}{2 a n \Gamma \left (\frac {3}{2} + \frac {1}{2 n}\right )} + \frac {d e x x^{n} \Phi \left (\frac {c x^{2 n} e^{i \pi }}{a}, 1, \frac {1}{2} + \frac {1}{2 n}\right ) \Gamma \left (\frac {1}{2} + \frac {1}{2 n}\right )}{2 a n^{2} \Gamma \left (\frac {3}{2} + \frac {1}{2 n}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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**2*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
))) + d*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))/(2*a*n*gamma(3/
2 + 1/(2*n))) + d*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))/(2*a*
n**2*gamma(3/2 + 1/(2*n)))

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