3.198 \(\int (\text{b1}+\text{c1} x) (a+2 b x+c x^2)^{-n} \, dx\)

Optimal. Leaf size=169 \[ \frac{\text{c1} \left (a+2 b x+c x^2\right )^{1-n}}{2 c (1-n)}-\frac{2^{-n} (\text{b1} c-b \text{c1}) \left (-\frac{-\sqrt{b^2-a c}+b+c x}{\sqrt{b^2-a c}}\right )^{n-1} \left (a+2 b x+c x^2\right )^{1-n} \, _2F_1\left (1-n,n;2-n;\frac{b+c x+\sqrt{b^2-a c}}{2 \sqrt{b^2-a c}}\right )}{c (1-n) \sqrt{b^2-a c}} \]

[Out]

(c1*(a + 2*b*x + c*x^2)^(1 - n))/(2*c*(1 - n)) - ((b1*c - b*c1)*(-((b - Sqrt[b^2 - a*c] + c*x)/Sqrt[b^2 - a*c]
))^(-1 + n)*(a + 2*b*x + c*x^2)^(1 - n)*Hypergeometric2F1[1 - n, n, 2 - n, (b + Sqrt[b^2 - a*c] + c*x)/(2*Sqrt
[b^2 - a*c])])/(2^n*c*Sqrt[b^2 - a*c]*(1 - n))

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Rubi [A]  time = 0.0793389, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {640, 624} \[ \frac{\text{c1} \left (a+2 b x+c x^2\right )^{1-n}}{2 c (1-n)}-\frac{2^{-n} (\text{b1} c-b \text{c1}) \left (-\frac{-\sqrt{b^2-a c}+b+c x}{\sqrt{b^2-a c}}\right )^{n-1} \left (a+2 b x+c x^2\right )^{1-n} \text{Hypergeometric2F1}\left (1-n,n,2-n,\frac{\sqrt{b^2-a c}+b+c x}{2 \sqrt{b^2-a c}}\right )}{c (1-n) \sqrt{b^2-a c}} \]

Antiderivative was successfully verified.

[In]

Int[(b1 + c1*x)/(a + 2*b*x + c*x^2)^n,x]

[Out]

(c1*(a + 2*b*x + c*x^2)^(1 - n))/(2*c*(1 - n)) - ((b1*c - b*c1)*(-((b - Sqrt[b^2 - a*c] + c*x)/Sqrt[b^2 - a*c]
))^(-1 + n)*(a + 2*b*x + c*x^2)^(1 - n)*Hypergeometric2F1[1 - n, n, 2 - n, (b + Sqrt[b^2 - a*c] + c*x)/(2*Sqrt
[b^2 - a*c])])/(2^n*c*Sqrt[b^2 - a*c]*(1 - n))

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 624

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, -Simp[((a + b*x + c*
x^2)^(p + 1)*Hypergeometric2F1[-p, p + 1, p + 2, (b + q + 2*c*x)/(2*q)])/(q*(p + 1)*((q - b - 2*c*x)/(2*q))^(p
+ 1)), x]] /; FreeQ[{a, b, c, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !IntegerQ[4*p]

Rubi steps

\begin{align*} \int (\text{b1}+\text{c1} x) \left (a+2 b x+c x^2\right )^{-n} \, dx &=\frac{\text{c1} \left (a+2 b x+c x^2\right )^{1-n}}{2 c (1-n)}+\frac{(2 \text{b1} c-2 b \text{c1}) \int \left (a+2 b x+c x^2\right )^{-n} \, dx}{2 c}\\ &=\frac{\text{c1} \left (a+2 b x+c x^2\right )^{1-n}}{2 c (1-n)}-\frac{2^{-n} (\text{b1} c-b \text{c1}) \left (-\frac{b-\sqrt{b^2-a c}+c x}{\sqrt{b^2-a c}}\right )^{-1+n} \left (a+2 b x+c x^2\right )^{1-n} \, _2F_1\left (1-n,n;2-n;\frac{b+\sqrt{b^2-a c}+c x}{2 \sqrt{b^2-a c}}\right )}{c \sqrt{b^2-a c} (1-n)}\\ \end{align*}

Mathematica [C]  time = 0.344348, size = 264, normalized size = 1.56 \[ \frac{1}{2} (a+x (2 b+c x))^{-n} \left (\text{c1} x^2 \left (\frac{-\sqrt{b^2-a c}+b+c x}{b-\sqrt{b^2-a c}}\right )^n \left (\frac{\sqrt{b^2-a c}+b+c x}{\sqrt{b^2-a c}+b}\right )^n F_1\left (2;n,n;3;-\frac{c x}{b+\sqrt{b^2-a c}},\frac{c x}{\sqrt{b^2-a c}-b}\right )-\frac{\text{b1} 2^{1-n} \left (-\sqrt{b^2-a c}+b+c x\right ) \left (\frac{\sqrt{b^2-a c}+b+c x}{\sqrt{b^2-a c}}\right )^n \, _2F_1\left (1-n,n;2-n;\frac{-b-c x+\sqrt{b^2-a c}}{2 \sqrt{b^2-a c}}\right )}{c (n-1)}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(b1 + c1*x)/(a + 2*b*x + c*x^2)^n,x]

[Out]

(c1*x^2*((b - Sqrt[b^2 - a*c] + c*x)/(b - Sqrt[b^2 - a*c]))^n*((b + Sqrt[b^2 - a*c] + c*x)/(b + Sqrt[b^2 - a*c
]))^n*AppellF1[2, n, n, 3, -((c*x)/(b + Sqrt[b^2 - a*c])), (c*x)/(-b + Sqrt[b^2 - a*c])] - (2^(1 - n)*b1*(b -
Sqrt[b^2 - a*c] + c*x)*((b + Sqrt[b^2 - a*c] + c*x)/Sqrt[b^2 - a*c])^n*Hypergeometric2F1[1 - n, n, 2 - n, (-b
+ Sqrt[b^2 - a*c] - c*x)/(2*Sqrt[b^2 - a*c])])/(c*(-1 + n)))/(2*(a + x*(2*b + c*x))^n)

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Maple [F]  time = 0.081, size = 0, normalized size = 0. \begin{align*} \int{\frac{{\it c1}\,x+{\it b1}}{ \left ( c{x}^{2}+2\,bx+a \right ) ^{n}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c1*x+b1)/((c*x^2+2*b*x+a)^n),x)

[Out]

int((c1*x+b1)/((c*x^2+2*b*x+a)^n),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{c_{1} x + b_{1}}{{\left (c x^{2} + 2 \, b x + a\right )}^{n}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c1*x+b1)/((c*x^2+2*b*x+a)^n),x, algorithm="maxima")

[Out]

integrate((c1*x + b1)/(c*x^2 + 2*b*x + a)^n, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{c_{1} x + b_{1}}{{\left (c x^{2} + 2 \, b x + a\right )}^{n}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c1*x+b1)/((c*x^2+2*b*x+a)^n),x, algorithm="fricas")

[Out]

integral((c1*x + b1)/(c*x^2 + 2*b*x + a)^n, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c1*x+b1)/((c*x**2+2*b*x+a)**n),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{c_{1} x + b_{1}}{{\left (c x^{2} + 2 \, b x + a\right )}^{n}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c1*x+b1)/((c*x^2+2*b*x+a)^n),x, algorithm="giac")

[Out]

integrate((c1*x + b1)/(c*x^2 + 2*b*x + a)^n, x)