\(\int (d+e x)^{-2 p} (a+b x+c x^2)^p \, dx\) [2574]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [F]
   Fricas [F]
   Sympy [F(-1)]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 22, antiderivative size = 195 \[ \int (d+e x)^{-2 p} \left (a+b x+c x^2\right )^p \, dx=\frac {(d+e x)^{1-2 p} \left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e (1-2 p)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {773, 138} \[ \int (d+e x)^{-2 p} \left (a+b x+c x^2\right )^p \, dx=\frac {(d+e x)^{1-2 p} \left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )^{-p} \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{e (1-2 p)} \]

[In]

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

[Out]

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

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 773

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

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

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.10 \[ \int (d+e x)^{-2 p} \left (a+b x+c x^2\right )^p \, dx=-\frac {\left (\frac {e \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} (d+e x)^{1-2 p} (a+x (b+c x))^p \operatorname {AppellF1}\left (1-2 p,-p,-p,2-2 p,\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )}{e (-1+2 p)} \]

[In]

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

[Out]

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

Maple [F]

\[\int \left (c \,x^{2}+b x +a \right )^{p} \left (e x +d \right )^{-2 p}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int (d+e x)^{-2 p} \left (a+b x+c x^2\right )^p \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{p}}{{\left (e x + d\right )}^{2 \, p}} \,d x } \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int (d+e x)^{-2 p} \left (a+b x+c x^2\right )^p \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int (d+e x)^{-2 p} \left (a+b x+c x^2\right )^p \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{p}}{{\left (e x + d\right )}^{2 \, p}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int (d+e x)^{-2 p} \left (a+b x+c x^2\right )^p \, dx=\int { \frac {{\left (c x^{2} + b x + a\right )}^{p}}{{\left (e x + d\right )}^{2 \, p}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int (d+e x)^{-2 p} \left (a+b x+c x^2\right )^p \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^p}{{\left (d+e\,x\right )}^{2\,p}} \,d x \]

[In]

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

[Out]

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