3.1.71 \(\int \frac {d+e x^n}{a+b x^n+c x^{2 n}} \, dx\) [71]

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

[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 = 0.08, 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 = {1436, 251} \begin {gather*} \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} \end {gather*}

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 251

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 1436

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 = 0.37, size = 279, normalized size = 1.81 \begin {gather*} \frac {2^{-\frac {1+n}{n}} x \left (2^{1+\frac {1}{n}} \sqrt {b^2-4 a c} d-\left (b d+\sqrt {b^2-4 a c} d-2 a e\right ) \left (\frac {c x^n}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )^{-1/n} \, _2F_1\left (-\frac {1}{n},-\frac {1}{n};\frac {-1+n}{n};\frac {b-\sqrt {b^2-4 a c}}{b-\sqrt {b^2-4 a c}+2 c x^n}\right )+\left (b d-\sqrt {b^2-4 a c} d-2 a e\right ) \left (\frac {c x^n}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )^{-1/n} \, _2F_1\left (-\frac {1}{n},-\frac {1}{n};\frac {-1+n}{n};\frac {b+\sqrt {b^2-4 a c}}{b+\sqrt {b^2-4 a c}+2 c x^n}\right )\right )}{a \sqrt {b^2-4 a c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.01, 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]

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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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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((x^n*e + 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 \begin {gather*} \int \frac {d+e\,x^n}{a+b\,x^n+c\,x^{2\,n}} \,d x \end {gather*}

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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