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

Optimal. Leaf size=154 \[ \frac {x \left (\frac {2 c d-b e}{\sqrt {b^2-4 a c}}+e\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{b-\sqrt {b^2-4 a c}}+\frac {x \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}+b} \]

[Out]

x*hypergeom([1, 1/n],[1+1/n],-2*c*x^n/(b-(-4*a*c+b^2)^(1/2)))*(e+(-b*e+2*c*d)/(-4*a*c+b^2)^(1/2))/(b-(-4*a*c+b
^2)^(1/2))+x*hypergeom([1, 1/n],[1+1/n],-2*c*x^n/(b+(-4*a*c+b^2)^(1/2)))*(e+(b*e-2*c*d)/(-4*a*c+b^2)^(1/2))/(b
+(-4*a*c+b^2)^(1/2))

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Rubi [A]  time = 0.12, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1422, 245} \[ \frac {x \left (\frac {2 c d-b e}{\sqrt {b^2-4 a c}}+e\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{b-\sqrt {b^2-4 a c}}+\frac {x \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{\sqrt {b^2-4 a c}+b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((e + (2*c*d - b*e)/Sqrt[b^2 - 4*a*c])*x*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b - Sqrt[b^2 - 4
*a*c])])/(b - Sqrt[b^2 - 4*a*c]) + ((e - (2*c*d - b*e)/Sqrt[b^2 - 4*a*c])*x*Hypergeometric2F1[1, n^(-1), 1 + n
^(-1), (-2*c*x^n)/(b + Sqrt[b^2 - 4*a*c])])/(b + Sqrt[b^2 - 4*a*c])

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 1422

Int[((d_) + (e_.)*(x_)^(n_))/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> With[{q = Rt[b^2 - 4*a*
c, 2]}, Dist[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^n), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), In
t[1/(b/2 + q/2 + c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ
[c*d^2 - b*d*e + a*e^2, 0] && (PosQ[b^2 - 4*a*c] ||  !IGtQ[n/2, 0])

Rubi steps

\begin {align*} \int \frac {d+e x^n}{a+b x^n+c x^{2 n}} \, dx &=\frac {1}{2} \left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \int \frac {1}{\frac {b}{2}+\frac {1}{2} \sqrt {b^2-4 a c}+c x^n} \, dx+\frac {1}{2} \left (e+\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) \int \frac {1}{\frac {b}{2}-\frac {1}{2} \sqrt {b^2-4 a c}+c x^n} \, dx\\ &=\frac {\left (e+\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b-\sqrt {b^2-4 a c}}\right )}{b-\sqrt {b^2-4 a c}}+\frac {\left (e-\frac {2 c d-b e}{\sqrt {b^2-4 a c}}\right ) x \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )}{b+\sqrt {b^2-4 a c}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 134, normalized size = 0.87 \[ \frac {x \left (\left (d \sqrt {b^2-4 a c}-2 a e+b d\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};\frac {2 c x^n}{\sqrt {b^2-4 a c}-b}\right )+\left (d \sqrt {b^2-4 a c}+2 a e-b d\right ) \, _2F_1\left (1,\frac {1}{n};1+\frac {1}{n};-\frac {2 c x^n}{b+\sqrt {b^2-4 a c}}\right )\right )}{2 a \sqrt {b^2-4 a c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*((b*d + Sqrt[b^2 - 4*a*c]*d - 2*a*e)*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (2*c*x^n)/(-b + Sqrt[b^2 - 4*
a*c])] + (-(b*d) + Sqrt[b^2 - 4*a*c]*d + 2*a*e)*Hypergeometric2F1[1, n^(-1), 1 + n^(-1), (-2*c*x^n)/(b + Sqrt[
b^2 - 4*a*c])]))/(2*a*Sqrt[b^2 - 4*a*c])

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fricas [F]  time = 0.80, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {e x^{n} + d}{c x^{2 \, n} + b x^{n} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \[ \int \frac {e \,x^{n}+d}{b \,x^{n}+c \,x^{2 n}+a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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