3.26.68 \(\int \frac {(a+b x+c x^2)^p}{(d+e x)^2} \, dx\) [2568]

Optimal. Leaf size=196 \[ -\frac {4^p \left (\frac {e \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{-p} \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{-p} \left (a+b x+c x^2\right )^p F_1\left (1-2 p;-p,-p;2 (1-p);\frac {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}{2 c (d+e x)},\frac {2 d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{c}}{2 (d+e x)}\right )}{e (1-2 p) (d+e x)} \]

[Out]

-4^p*(c*x^2+b*x+a)^p*AppellF1(1-2*p,-p,-p,2-2*p,1/2*(2*d-e*(b+(-4*a*c+b^2)^(1/2))/c)/(e*x+d),1/2*(2*c*d-e*(b-(
-4*a*c+b^2)^(1/2)))/c/(e*x+d))/e/(1-2*p)/(e*x+d)/((e*(b+2*c*x-(-4*a*c+b^2)^(1/2))/c/(e*x+d))^p)/((e*(b+2*c*x+(
-4*a*c+b^2)^(1/2))/c/(e*x+d))^p)

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Rubi [A]
time = 0.07, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {772, 138} \begin {gather*} -\frac {4^p \left (a+b x+c x^2\right )^p \left (\frac {e \left (-\sqrt {b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{-p} \left (\frac {e \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{c (d+e x)}\right )^{-p} F_1\left (1-2 p;-p,-p;2 (1-p);\frac {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}{2 c (d+e x)},\frac {2 d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{c}}{2 (d+e x)}\right )}{e (1-2 p) (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((4^p*(a + b*x + c*x^2)^p*AppellF1[1 - 2*p, -p, -p, 2*(1 - p), (2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)/(2*c*(d +
e*x)), (2*d - ((b + Sqrt[b^2 - 4*a*c])*e)/c)/(2*(d + e*x))])/(e*(1 - 2*p)*((e*(b - Sqrt[b^2 - 4*a*c] + 2*c*x))
/(c*(d + e*x)))^p*((e*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(c*(d + e*x)))^p*(d + e*x)))

Rule 138

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[c^n*e^p*((b*x)^(m +
1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 772

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c,
2]}, Dist[(-(1/(d + e*x))^(2*p))*((a + b*x + c*x^2)^p/(e*(e*((b - q + 2*c*x)/(2*c*(d + e*x))))^p*(e*((b + q +
2*c*x)/(2*c*(d + e*x))))^p)), Subst[Int[x^(-m - 2*(p + 1))*Simp[1 - (d - e*((b - q)/(2*c)))*x, x]^p*Simp[1 -
(d - e*((b + q)/(2*c)))*x, x]^p, x], x, 1/(d + e*x)], x]] /; FreeQ[{a, b, c, d, e, p}, 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[p] && ILtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {\left (a+b x+c x^2\right )^p}{(d+e x)^2} \, dx &=-\frac {\left (2^{2 p} \left (\frac {1}{d+e x}\right )^{2 p} \left (\frac {e \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{-p} \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{-p} \left (a+b x+c x^2\right )^p\right ) \text {Subst}\left (\int x^{2-2 (1+p)} \left (1-\frac {1}{2} \left (2 d-\frac {\left (b-\sqrt {b^2-4 a c}\right ) e}{c}\right ) x\right )^p \left (1-\frac {1}{2} \left (2 d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{c}\right ) x\right )^p \, dx,x,\frac {1}{d+e x}\right )}{e}\\ &=-\frac {4^p \left (\frac {e \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{-p} \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{-p} \left (a+b x+c x^2\right )^p F_1\left (1-2 p;-p,-p;2 (1-p);\frac {2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}{2 c (d+e x)},\frac {2 d-\frac {\left (b+\sqrt {b^2-4 a c}\right ) e}{c}}{2 (d+e x)}\right )}{e (1-2 p) (d+e x)}\\ \end {align*}

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Mathematica [A]
time = 0.38, size = 191, normalized size = 0.97 \begin {gather*} \frac {4^p \left (\frac {e \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{-p} \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{c (d+e x)}\right )^{-p} (a+x (b+c x))^p F_1\left (1-2 p;-p,-p;2-2 p;\frac {2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}{2 c (d+e x)},\frac {2 c d-b e+\sqrt {b^2-4 a c} e}{2 c d+2 c e x}\right )}{e (-1+2 p) (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4^p*(a + x*(b + c*x))^p*AppellF1[1 - 2*p, -p, -p, 2 - 2*p, (2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)/(2*c*(d + e*x)
), (2*c*d - b*e + Sqrt[b^2 - 4*a*c]*e)/(2*c*d + 2*c*e*x)])/(e*(-1 + 2*p)*((e*(b - Sqrt[b^2 - 4*a*c] + 2*c*x))/
(c*(d + e*x)))^p*((e*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(c*(d + e*x)))^p*(d + e*x))

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Maple [F]
time = 0.42, size = 0, normalized size = 0.00 \[\int \frac {\left (c \,x^{2}+b x +a \right )^{p}}{\left (e x +d \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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