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

Optimal. Leaf size=83 \[ \frac {d x \, _2F_1\left (1,\frac {1}{2 n};\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a}+\frac {e x^{n+1} \, _2F_1\left (1,\frac {n+1}{2 n};\frac {1}{2} \left (3+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a (n+1)} \]

[Out]

d*x*hypergeom([1, 1/2/n],[1+1/2/n],-c*x^(2*n)/a)/a+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 = 0.03, antiderivative size = 83, normalized size of antiderivative = 1.00, 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 = {1418, 245, 364} \[ \frac {d x \, _2F_1\left (1,\frac {1}{2 n};\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a}+\frac {e x^{n+1} \, _2F_1\left (1,\frac {n+1}{2 n};\frac {1}{2} \left (3+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x^n)/(a + c*x^(2*n)),x]

[Out]

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

Rubi steps

\begin {align*} \int \frac {d+e x^n}{a+c x^{2 n}} \, dx &=d \int \frac {1}{a+c x^{2 n}} \, dx+e \int \frac {x^n}{a+c x^{2 n}} \, dx\\ &=\frac {d 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}+\frac {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 = 0.04, size = 83, normalized size = 1.00 \[ \frac {d x \, _2F_1\left (1,\frac {1}{2 n};\frac {1}{2} \left (2+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a}+\frac {e x^{n+1} \, _2F_1\left (1,\frac {n+1}{2 n};\frac {1}{2} \left (3+\frac {1}{n}\right );-\frac {c x^{2 n}}{a}\right )}{a (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d*x*Hypergeometric2F1[1, 1/(2*n), (2 + n^(-1))/2, -((c*x^(2*n))/a)])/a + (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 = 1.02, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {e x^{n} + d}{c x^{2 \, n} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {e \,x^{n}+d}{c \,x^{2 n}+a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x^n)/(a + c*x^(2*n)),x)

[Out]

int((d + e*x^n)/(a + c*x^(2*n)), x)

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sympy [C]  time = 5.54, size = 153, normalized size = 1.84 \[ \frac {d x \Phi \left (\frac {c x^{2 n} e^{i \pi }}{a}, 1, \frac {1}{2 n}\right ) \Gamma \left (\frac {1}{2 n}\right )}{4 a n^{2} \Gamma \left (1 + \frac {1}{2 n}\right )} + \frac {e x x^{n} \Phi \left (\frac {c x^{2 n} e^{i \pi }}{a}, 1, \frac {1}{2} + \frac {1}{2 n}\right ) \Gamma \left (\frac {1}{2} + \frac {1}{2 n}\right )}{4 a n \Gamma \left (\frac {3}{2} + \frac {1}{2 n}\right )} + \frac {e x x^{n} \Phi \left (\frac {c x^{2 n} e^{i \pi }}{a}, 1, \frac {1}{2} + \frac {1}{2 n}\right ) \Gamma \left (\frac {1}{2} + \frac {1}{2 n}\right )}{4 a n^{2} \Gamma \left (\frac {3}{2} + \frac {1}{2 n}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*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))) + 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))) +
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)))

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