Integrand size = 23, antiderivative size = 277 \[ \int (e x)^m (A+B x) \left (a+b x+c x^2\right )^p \, dx=\frac {A (e x)^{1+m} \left (1+\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x+c x^2\right )^p \operatorname {AppellF1}\left (1+m,-p,-p,2+m,-\frac {2 c x}{b-\sqrt {b^2-4 a c}},-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{e (1+m)}+\frac {B (e x)^{2+m} \left (1+\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x+c x^2\right )^p \operatorname {AppellF1}\left (2+m,-p,-p,3+m,-\frac {2 c x}{b-\sqrt {b^2-4 a c}},-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{e^2 (2+m)} \]
A*(e*x)^(1+m)*(c*x^2+b*x+a)^p*AppellF1(1+m,-p,-p,2+m,-2*c*x/(b-(-4*a*c+b^2 )^(1/2)),-2*c*x/(b+(-4*a*c+b^2)^(1/2)))/e/(1+m)/((1+2*c*x/(b-(-4*a*c+b^2)^ (1/2)))^p)/((1+2*c*x/(b+(-4*a*c+b^2)^(1/2)))^p)+B*(e*x)^(2+m)*(c*x^2+b*x+a )^p*AppellF1(2+m,-p,-p,3+m,-2*c*x/(b-(-4*a*c+b^2)^(1/2)),-2*c*x/(b+(-4*a*c +b^2)^(1/2)))/e^2/(2+m)/((1+2*c*x/(b-(-4*a*c+b^2)^(1/2)))^p)/((1+2*c*x/(b+ (-4*a*c+b^2)^(1/2)))^p)
Time = 0.47 (sec) , antiderivative size = 232, normalized size of antiderivative = 0.84 \[ \int (e x)^m (A+B x) \left (a+b x+c x^2\right )^p \, dx=\frac {x (e x)^m \left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{b+\sqrt {b^2-4 a c}}\right )^{-p} (a+x (b+c x))^p \left (A (2+m) \operatorname {AppellF1}\left (1+m,-p,-p,2+m,-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )+B (1+m) x \operatorname {AppellF1}\left (2+m,-p,-p,3+m,-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )\right )}{(1+m) (2+m)} \]
(x*(e*x)^m*(a + x*(b + c*x))^p*(A*(2 + m)*AppellF1[1 + m, -p, -p, 2 + m, ( -2*c*x)/(b + Sqrt[b^2 - 4*a*c]), (2*c*x)/(-b + Sqrt[b^2 - 4*a*c])] + B*(1 + m)*x*AppellF1[2 + m, -p, -p, 3 + m, (-2*c*x)/(b + Sqrt[b^2 - 4*a*c]), (2 *c*x)/(-b + Sqrt[b^2 - 4*a*c])]))/((1 + m)*(2 + m)*((b - Sqrt[b^2 - 4*a*c] + 2*c*x)/(b - Sqrt[b^2 - 4*a*c]))^p*((b + Sqrt[b^2 - 4*a*c] + 2*c*x)/(b + Sqrt[b^2 - 4*a*c]))^p)
Time = 0.41 (sec) , antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {1269, 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.
\(\displaystyle \int (A+B x) (e x)^m \left (a+b x+c x^2\right )^p \, dx\) |
\(\Big \downarrow \) 1269 |
\(\displaystyle A \int (e x)^m \left (c x^2+b x+a\right )^pdx+\frac {B \int (e x)^{m+1} \left (c x^2+b x+a\right )^pdx}{e}\) |
\(\Big \downarrow \) 1179 |
\(\displaystyle \frac {A \left (\frac {2 c x}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (a+b x+c x^2\right )^p \left (\frac {2 c x}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \int (e x)^m \left (\frac {2 c x}{b-\sqrt {b^2-4 a c}}+1\right )^p \left (\frac {2 c x}{b+\sqrt {b^2-4 a c}}+1\right )^pd(e x)}{e}+\frac {B \left (\frac {2 c x}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (a+b x+c x^2\right )^p \left (\frac {2 c x}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \int (e x)^{m+1} \left (\frac {2 c x}{b-\sqrt {b^2-4 a c}}+1\right )^p \left (\frac {2 c x}{b+\sqrt {b^2-4 a c}}+1\right )^pd(e x)}{e^2}\) |
\(\Big \downarrow \) 150 |
\(\displaystyle \frac {A (e x)^{m+1} \left (\frac {2 c x}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x+c x^2\right )^p \operatorname {AppellF1}\left (m+1,-p,-p,m+2,-\frac {2 c x}{b-\sqrt {b^2-4 a c}},-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{e (m+1)}+\frac {B (e x)^{m+2} \left (\frac {2 c x}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x+c x^2\right )^p \operatorname {AppellF1}\left (m+2,-p,-p,m+3,-\frac {2 c x}{b-\sqrt {b^2-4 a c}},-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{e^2 (m+2)}\) |
(A*(e*x)^(1 + m)*(a + b*x + c*x^2)^p*AppellF1[1 + m, -p, -p, 2 + m, (-2*c* x)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x)/(b + Sqrt[b^2 - 4*a*c])])/(e*(1 + m)* (1 + (2*c*x)/(b - Sqrt[b^2 - 4*a*c]))^p*(1 + (2*c*x)/(b + Sqrt[b^2 - 4*a*c ]))^p) + (B*(e*x)^(2 + m)*(a + b*x + c*x^2)^p*AppellF1[2 + m, -p, -p, 3 + m, (-2*c*x)/(b - Sqrt[b^2 - 4*a*c]), (-2*c*x)/(b + Sqrt[b^2 - 4*a*c])])/(e ^2*(2 + m)*(1 + (2*c*x)/(b - Sqrt[b^2 - 4*a*c]))^p*(1 + (2*c*x)/(b + Sqrt[ b^2 - 4*a*c]))^p)
3.11.94.3.1 Defintions of rubi rules used
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[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g/e Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Simp[(e*f - d*g)/e Int[(d + e*x)^m*(a + b*x + c*x^2)^ p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && !IGtQ[m, 0]
\[\int \left (e x \right )^{m} \left (B x +A \right ) \left (c \,x^{2}+b x +a \right )^{p}d x\]
\[ \int (e x)^m (A+B x) \left (a+b x+c x^2\right )^p \, dx=\int { {\left (B x + A\right )} {\left (c x^{2} + b x + a\right )}^{p} \left (e x\right )^{m} \,d x } \]
Timed out. \[ \int (e x)^m (A+B x) \left (a+b x+c x^2\right )^p \, dx=\text {Timed out} \]
\[ \int (e x)^m (A+B x) \left (a+b x+c x^2\right )^p \, dx=\int { {\left (B x + A\right )} {\left (c x^{2} + b x + a\right )}^{p} \left (e x\right )^{m} \,d x } \]
\[ \int (e x)^m (A+B x) \left (a+b x+c x^2\right )^p \, dx=\int { {\left (B x + A\right )} {\left (c x^{2} + b x + a\right )}^{p} \left (e x\right )^{m} \,d x } \]
Timed out. \[ \int (e x)^m (A+B x) \left (a+b x+c x^2\right )^p \, dx=\int {\left (e\,x\right )}^m\,\left (A+B\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^p \,d x \]