3.6.45 \(\int \frac {(e+f x)^n}{a+b x+c x^2} \, dx\) [545]

Optimal. Leaf size=191 \[ -\frac {2 c (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {2 c (e+f x)}{2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f}\right )}{\sqrt {b^2-4 a c} \left (2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f\right ) (1+n)}+\frac {2 c (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{\sqrt {b^2-4 a c} \left (2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f\right ) (1+n)} \]

[Out]

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

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Rubi [A]
time = 0.18, antiderivative size = 191, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {725, 70} \begin {gather*} \frac {2 c (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {2 c (e+f x)}{2 c e-\left (b+\sqrt {b^2-4 a c}\right ) f}\right )}{(n+1) \sqrt {b^2-4 a c} \left (2 c e-f \left (\sqrt {b^2-4 a c}+b\right )\right )}-\frac {2 c (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {2 c (e+f x)}{2 c e-\left (b-\sqrt {b^2-4 a c}\right ) f}\right )}{(n+1) \sqrt {b^2-4 a c} \left (2 c e-f \left (b-\sqrt {b^2-4 a c}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e + f*x)^n/(a + b*x + c*x^2),x]

[Out]

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

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 725

Int[((d_.) + (e_.)*(x_))^(m_)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^
m, 1/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a
*e^2, 0] && NeQ[2*c*d - b*e, 0] &&  !IntegerQ[m]

Rubi steps

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

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Mathematica [A]
time = 0.24, size = 245, normalized size = 1.28 \begin {gather*} \frac {2^{-n} f (e+f x)^n \left (\left (\frac {c (e+f x)}{b f-\sqrt {\left (b^2-4 a c\right ) f^2}+2 c f x}\right )^{-n} \, _2F_1\left (-n,-n;1-n;\frac {2 c e-b f+\sqrt {\left (b^2-4 a c\right ) f^2}}{-b f+\sqrt {\left (b^2-4 a c\right ) f^2}-2 c f x}\right )-\left (\frac {c (e+f x)}{b f+\sqrt {\left (b^2-4 a c\right ) f^2}+2 c f x}\right )^{-n} \, _2F_1\left (-n,-n;1-n;\frac {-2 c e+b f+\sqrt {\left (b^2-4 a c\right ) f^2}}{b f+\sqrt {\left (b^2-4 a c\right ) f^2}+2 c f x}\right )\right )}{\sqrt {\left (b^2-4 a c\right ) f^2} n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\left (f x +e \right )^{n}}{c \,x^{2}+b x +a}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (e+f\,x\right )}^n}{c\,x^2+b\,x+a} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e + f*x)^n/(a + b*x + c*x^2),x)

[Out]

int((e + f*x)^n/(a + b*x + c*x^2), x)

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