Integrand size = 24, antiderivative size = 332 \[ \int (d+e x)^{-3-2 p} \left (a+b x+c x^2\right )^p \, dx=-\frac {e (d+e x)^{-2 (1+p)} \left (a+b x+c x^2\right )^{1+p}}{2 \left (c d^2-b d e+a e^2\right ) (1+p)}+\frac {(2 c d-b e) \left (b-\sqrt {b^2-4 a c}+2 c x\right ) \left (\frac {\left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b-\sqrt {b^2-4 a c}+2 c x\right )}\right )^{-p} (d+e x)^{-1-2 p} \left (a+b x+c x^2\right )^p \operatorname {Hypergeometric2F1}\left (-1-2 p,-p,-2 p,-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b-\sqrt {b^2-4 a c}+2 c x\right )}\right )}{2 \left (2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e\right ) \left (c d^2-b d e+a e^2\right ) (1+2 p)} \]
-1/2*e*(c*x^2+b*x+a)^(p+1)/(a*e^2-b*d*e+c*d^2)/(p+1)/((e*x+d)^(2+2*p))+1/2 *(-b*e+2*c*d)*(e*x+d)^(-1-2*p)*(c*x^2+b*x+a)^p*hypergeom([-p, -1-2*p],[-2* p],-4*c*(e*x+d)*(-4*a*c+b^2)^(1/2)/(b+2*c*x-(-4*a*c+b^2)^(1/2))/(2*c*d-e*( b+(-4*a*c+b^2)^(1/2))))*(b+2*c*x-(-4*a*c+b^2)^(1/2))/(a*e^2-b*d*e+c*d^2)/( 1+2*p)/(2*c*d-e*(b-(-4*a*c+b^2)^(1/2)))/(((2*c*d-e*(b-(-4*a*c+b^2)^(1/2))) *(b+2*c*x+(-4*a*c+b^2)^(1/2))/(b+2*c*x-(-4*a*c+b^2)^(1/2))/(2*c*d-e*(b+(-4 *a*c+b^2)^(1/2))))^p)
Time = 0.38 (sec) , antiderivative size = 294, normalized size of antiderivative = 0.89 \[ \int (d+e x)^{-3-2 p} \left (a+b x+c x^2\right )^p \, dx=\frac {(d+e x)^{-2 (1+p)} (a+x (b+c x))^p \left (-\frac {e (a+x (b+c x))}{1+p}+\frac {(-2 c d+b e) \left (b+\sqrt {b^2-4 a c}+2 c x\right ) \left (\frac {\left (2 c d+\left (-b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}\right )^{-1-p} (d+e x) \operatorname {Hypergeometric2F1}\left (-1-2 p,-p,-2 p,-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (-b+\sqrt {b^2-4 a c}-2 c x\right )}\right )}{\left (-2 c d+\left (b+\sqrt {b^2-4 a c}\right ) e\right ) (1+2 p)}\right )}{2 \left (c d^2+e (-b d+a e)\right )} \]
((a + x*(b + c*x))^p*(-((e*(a + x*(b + c*x)))/(1 + p)) + ((-2*c*d + b*e)*( b + Sqrt[b^2 - 4*a*c] + 2*c*x)*(((2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/((-2*c*d + (b + Sqrt[b^2 - 4*a*c])*e)*(-b + S qrt[b^2 - 4*a*c] - 2*c*x)))^(-1 - p)*(d + e*x)*Hypergeometric2F1[-1 - 2*p, -p, -2*p, (-4*c*Sqrt[b^2 - 4*a*c]*(d + e*x))/((-2*c*d + (b + Sqrt[b^2 - 4 *a*c])*e)*(-b + Sqrt[b^2 - 4*a*c] - 2*c*x))])/((-2*c*d + (b + Sqrt[b^2 - 4 *a*c])*e)*(1 + 2*p))))/(2*(c*d^2 + e*(-(b*d) + a*e))*(d + e*x)^(2*(1 + p)) )
Time = 0.38 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {1157, 1155}
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.
\(\displaystyle \int (d+e x)^{-2 p-3} \left (a+b x+c x^2\right )^p \, dx\) |
\(\Big \downarrow \) 1157 |
\(\displaystyle \frac {(2 c d-b e) \int (d+e x)^{-2 (p+1)} \left (c x^2+b x+a\right )^pdx}{2 \left (a e^2-b d e+c d^2\right )}-\frac {e (d+e x)^{-2 (p+1)} \left (a+b x+c x^2\right )^{p+1}}{2 (p+1) \left (a e^2-b d e+c d^2\right )}\) |
\(\Big \downarrow \) 1155 |
\(\displaystyle \frac {\left (-\sqrt {b^2-4 a c}+b+2 c x\right ) (2 c d-b e) (d+e x)^{-2 p-1} \left (a+b x+c x^2\right )^p \left (\frac {\left (\sqrt {b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right )}{\left (-\sqrt {b^2-4 a c}+b+2 c x\right ) \left (2 c d-e \left (\sqrt {b^2-4 a c}+b\right )\right )}\right )^{-p} \operatorname {Hypergeometric2F1}\left (-2 p-1,-p,-2 p,-\frac {4 c \sqrt {b^2-4 a c} (d+e x)}{\left (2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e\right ) \left (b+2 c x-\sqrt {b^2-4 a c}\right )}\right )}{2 (2 p+1) \left (2 c d-e \left (b-\sqrt {b^2-4 a c}\right )\right ) \left (a e^2-b d e+c d^2\right )}-\frac {e (d+e x)^{-2 (p+1)} \left (a+b x+c x^2\right )^{p+1}}{2 (p+1) \left (a e^2-b d e+c d^2\right )}\) |
-1/2*(e*(a + b*x + c*x^2)^(1 + p))/((c*d^2 - b*d*e + a*e^2)*(1 + p)*(d + e *x)^(2*(1 + p))) + ((2*c*d - b*e)*(b - Sqrt[b^2 - 4*a*c] + 2*c*x)*(d + e*x )^(-1 - 2*p)*(a + b*x + c*x^2)^p*Hypergeometric2F1[-1 - 2*p, -p, -2*p, (-4 *c*Sqrt[b^2 - 4*a*c]*(d + e*x))/((2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*(b - Sqrt[b^2 - 4*a*c] + 2*c*x))])/(2*(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)*(c*d^ 2 - b*d*e + a*e^2)*(1 + 2*p)*(((2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)*(b + Sq rt[b^2 - 4*a*c] + 2*c*x))/((2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)*(b - Sqrt[b ^2 - 4*a*c] + 2*c*x)))^p)
3.26.77.3.1 Defintions of rubi rules used
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[(-(b - q + 2*c*x))*(d + e*x)^ (m + 1)*((a + b*x + c*x^2)^p/((m + 1)*(2*c*d - b*e + e*q)*((2*c*d - b*e + e *q)*((b + q + 2*c*x)/((2*c*d - b*e - e*q)*(b - q + 2*c*x))))^p))*Hypergeome tric2F1[m + 1, -p, m + 2, -4*c*q*((d + e*x)/((2*c*d - b*e - e*q)*(b - q + 2 *c*x)))], x]] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[m + 2*p + 2, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/((m + 1)*(c*d ^2 - b*d*e + a*e^2))), x] + Simp[(2*c*d - b*e)/(2*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e , m, p}, x] && EqQ[m + 2*p + 3, 0]
\[\int \left (e x +d \right )^{-3-2 p} \left (c \,x^{2}+b x +a \right )^{p}d x\]
\[ \int (d+e x)^{-3-2 p} \left (a+b x+c x^2\right )^p \, dx=\int { {\left (c x^{2} + b x + a\right )}^{p} {\left (e x + d\right )}^{-2 \, p - 3} \,d x } \]
Timed out. \[ \int (d+e x)^{-3-2 p} \left (a+b x+c x^2\right )^p \, dx=\text {Timed out} \]
\[ \int (d+e x)^{-3-2 p} \left (a+b x+c x^2\right )^p \, dx=\int { {\left (c x^{2} + b x + a\right )}^{p} {\left (e x + d\right )}^{-2 \, p - 3} \,d x } \]
\[ \int (d+e x)^{-3-2 p} \left (a+b x+c x^2\right )^p \, dx=\int { {\left (c x^{2} + b x + a\right )}^{p} {\left (e x + d\right )}^{-2 \, p - 3} \,d x } \]
Timed out. \[ \int (d+e x)^{-3-2 p} \left (a+b x+c x^2\right )^p \, dx=\int \frac {{\left (c\,x^2+b\,x+a\right )}^p}{{\left (d+e\,x\right )}^{2\,p+3}} \,d x \]