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

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

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Rubi [A]  time = 0.08, antiderivative size = 169, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [C]  time = 0.41, 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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^{-n} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

Could not integrate

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fricas [F]  time = 1.32, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {c_{1} x + b_{1}}{{\left (c x^{2} + 2 \, b x + a\right )}^{n}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {c_{1} x + b_{1}}{{\left (c x^{2} + 2 \, b x + a\right )}^{n}}\,{d x} \]

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)

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[\int \left (\mathit {c1} x +\mathit {b1} \right ) \left (c \,x^{2}+2 b x +a \right )^{-n}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {c_{1} x + b_{1}}{{\left (c x^{2} + 2 \, b x + a\right )}^{n}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {b_{1}+c_{1}\,x}{{\left (c\,x^2+2\,b\,x+a\right )}^n} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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