Integrand size = 24, antiderivative size = 190 \[ \int (d+e x)^{-1-2 p} \left (a+b x+c x^2\right )^p \, dx=-\frac {(d+e x)^{-2 p} \left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 e p} \] Output:
-1/2*(c*x^2+b*x+a)^p*AppellF1(-2*p,-p,-p,1-2*p,2*c*(e*x+d)/(2*c*d-(b-(-4*a *c+b^2)^(1/2))*e),2*c*(e*x+d)/(2*c*d-(b+(-4*a*c+b^2)^(1/2))*e))/e/p/((e*x+ d)^(2*p))/((1-2*c*(e*x+d)/(2*c*d-(b-(-4*a*c+b^2)^(1/2))*e))^p)/((1-2*c*(e* x+d)/(2*c*d-(b+(-4*a*c+b^2)^(1/2))*e))^p)
Time = 0.35 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.09 \[ \int (d+e x)^{-1-2 p} \left (a+b x+c x^2\right )^p \, dx=-\frac {\left (\frac {e \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} \left (\frac {e \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^{-p} (d+e x)^{-2 p} (a+x (b+c x))^p \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 e p} \] Input:
Integrate[(d + e*x)^(-1 - 2*p)*(a + b*x + c*x^2)^p,x]
Output:
-1/2*((a + x*(b + c*x))^p*AppellF1[-2*p, -p, -p, 1 - 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*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*(d + e*x)^(2*p))
Time = 0.31 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1179, 150}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int (d+e x)^{-2 p-1} \left (a+b x+c x^2\right )^p \, dx\) |
\(\Big \downarrow \) 1179 |
\(\displaystyle \frac {\left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )^{-p} \int (d+e x)^{-2 p-1} \left (1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )^pd(d+e x)}{e}\) |
\(\Big \downarrow \) 150 |
\(\displaystyle -\frac {(d+e x)^{-2 p} \left (a+b x+c x^2\right )^p \left (1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}\right )^{-p} \left (1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}\right )^{-p} \operatorname {AppellF1}\left (-2 p,-p,-p,1-2 p,\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 e p}\) |
Input:
Int[(d + e*x)^(-1 - 2*p)*(a + b*x + c*x^2)^p,x]
Output:
-1/2*((a + b*x + c*x^2)^p*AppellF1[-2*p, -p, -p, 1 - 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*p*(d + e*x)^(2*p)*(1 - (2*c*(d + e*x))/(2*c*d - (b - S qrt[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)
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_ ] :> 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] && !In tegerQ[m] && !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(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]
\[\int \left (e x +d \right )^{-1-2 p} \left (c \,x^{2}+b x +a \right )^{p}d x\]
Input:
int((e*x+d)^(-1-2*p)*(c*x^2+b*x+a)^p,x)
Output:
int((e*x+d)^(-1-2*p)*(c*x^2+b*x+a)^p,x)
\[ \int (d+e x)^{-1-2 p} \left (a+b x+c x^2\right )^p \, dx=\int { {\left (c x^{2} + b x + a\right )}^{p} {\left (e x + d\right )}^{-2 \, p - 1} \,d x } \] Input:
integrate((e*x+d)^(-1-2*p)*(c*x^2+b*x+a)^p,x, algorithm="fricas")
Output:
integral((c*x^2 + b*x + a)^p*(e*x + d)^(-2*p - 1), x)
Timed out. \[ \int (d+e x)^{-1-2 p} \left (a+b x+c x^2\right )^p \, dx=\text {Timed out} \] Input:
integrate((e*x+d)**(-1-2*p)*(c*x**2+b*x+a)**p,x)
Output:
Timed out
\[ \int (d+e x)^{-1-2 p} \left (a+b x+c x^2\right )^p \, dx=\int { {\left (c x^{2} + b x + a\right )}^{p} {\left (e x + d\right )}^{-2 \, p - 1} \,d x } \] Input:
integrate((e*x+d)^(-1-2*p)*(c*x^2+b*x+a)^p,x, algorithm="maxima")
Output:
integrate((c*x^2 + b*x + a)^p*(e*x + d)^(-2*p - 1), x)
\[ \int (d+e x)^{-1-2 p} \left (a+b x+c x^2\right )^p \, dx=\int { {\left (c x^{2} + b x + a\right )}^{p} {\left (e x + d\right )}^{-2 \, p - 1} \,d x } \] Input:
integrate((e*x+d)^(-1-2*p)*(c*x^2+b*x+a)^p,x, algorithm="giac")
Output:
integrate((c*x^2 + b*x + a)^p*(e*x + d)^(-2*p - 1), x)
Timed out. \[ \int (d+e x)^{-1-2 p} \left (a+b x+c x^2\right )^p \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^p}{{\left (d+e\,x\right )}^{2\,p+1}} \,d x \] Input:
int((a + b*x + c*x^2)^p/(d + e*x)^(2*p + 1),x)
Output:
int((a + b*x + c*x^2)^p/(d + e*x)^(2*p + 1), x)
\[ \int (d+e x)^{-1-2 p} \left (a+b x+c x^2\right )^p \, dx=\int \frac {\left (c \,x^{2}+b x +a \right )^{p}}{\left (e x +d \right )^{2 p} d +\left (e x +d \right )^{2 p} e x}d x \] Input:
int((e*x+d)^(-1-2*p)*(c*x^2+b*x+a)^p,x)
Output:
int((a + b*x + c*x**2)**p/((d + e*x)**(2*p)*d + (d + e*x)**(2*p)*e*x),x)