Integrand size = 19, antiderivative size = 159 \[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^n \, dx=\frac {\text {c1} \left (a+2 b x+c x^2\right )^{1+n}}{2 c (1+n)}-\frac {2^n (\text {b1} c-b \text {c1}) \left (-\frac {b-\sqrt {b^2-a c}+c x}{\sqrt {b^2-a c}}\right )^{-1-n} \left (a+2 b x+c x^2\right )^{1+n} \operatorname {Hypergeometric2F1}\left (-n,1+n,2+n,\frac {b+\sqrt {b^2-a c}+c x}{2 \sqrt {b^2-a c}}\right )}{c \sqrt {b^2-a c} (1+n)} \]
1/2*c1*(c*x^2+2*b*x+a)^(1+n)/c/(1+n)-2^n*(-b*c1+b1*c)*(c*x^2+2*b*x+a)^(1+n )*hypergeom([-n, 1+n],[2+n],1/2*(b+c*x+(-a*c+b^2)^(1/2))/(-a*c+b^2)^(1/2)) *((-b-c*x+(-a*c+b^2)^(1/2))/(-a*c+b^2)^(1/2))^(-1-n)/c/(1+n)/(-a*c+b^2)^(1 /2)
Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
Time = 0.67 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.68 \[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^n \, dx=\frac {1}{2} (a+x (2 b+c x))^n \left (\text {c1} x^2 \left (\frac {b-\sqrt {b^2-a c}+c x}{b-\sqrt {b^2-a c}}\right )^{-n} \left (\frac {b+\sqrt {b^2-a c}+c x}{b+\sqrt {b^2-a c}}\right )^{-n} \operatorname {AppellF1}\left (2,-n,-n,3,-\frac {c x}{b+\sqrt {b^2-a c}},\frac {c x}{-b+\sqrt {b^2-a c}}\right )+\frac {2^{1+n} \text {b1} \left (b-\sqrt {b^2-a c}+c x\right ) \left (\frac {b+\sqrt {b^2-a c}+c x}{\sqrt {b^2-a c}}\right )^{-n} \operatorname {Hypergeometric2F1}\left (-n,1+n,2+n,\frac {-b+\sqrt {b^2-a c}-c x}{2 \sqrt {b^2-a c}}\right )}{c (1+n)}\right ) \]
((a + x*(2*b + c*x))^n*((c1*x^2*AppellF1[2, -n, -n, 3, -((c*x)/(b + Sqrt[b ^2 - a*c])), (c*x)/(-b + Sqrt[b^2 - a*c])])/(((b - Sqrt[b^2 - a*c] + c*x)/ (b - Sqrt[b^2 - a*c]))^n*((b + Sqrt[b^2 - a*c] + c*x)/(b + Sqrt[b^2 - a*c] ))^n) + (2^(1 + n)*b1*(b - Sqrt[b^2 - a*c] + c*x)*Hypergeometric2F1[-n, 1 + n, 2 + n, (-b + Sqrt[b^2 - a*c] - c*x)/(2*Sqrt[b^2 - a*c])])/(c*(1 + n)* ((b + Sqrt[b^2 - a*c] + c*x)/Sqrt[b^2 - a*c])^n)))/2
Time = 0.25 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {1160, 1096}
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.
\(\displaystyle \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^n \, dx\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {(\text {b1} c-b \text {c1}) \int \left (c x^2+2 b x+a\right )^ndx}{c}+\frac {\text {c1} \left (a+2 b x+c x^2\right )^{n+1}}{2 c (n+1)}\) |
\(\Big \downarrow \) 1096 |
\(\displaystyle \frac {\text {c1} \left (a+2 b x+c x^2\right )^{n+1}}{2 c (n+1)}-\frac {2^n (\text {b1} c-b \text {c1}) \left (-\frac {-\sqrt {b^2-a c}+b+c x}{\sqrt {b^2-a c}}\right )^{-n-1} \left (a+2 b x+c x^2\right )^{n+1} \operatorname {Hypergeometric2F1}\left (-n,n+1,n+2,\frac {b+c x+\sqrt {b^2-a c}}{2 \sqrt {b^2-a c}}\right )}{c (n+1) \sqrt {b^2-a c}}\) |
(c1*(a + 2*b*x + c*x^2)^(1 + n))/(2*c*(1 + n)) - (2^n*(b1*c - b*c1)*(-((b - Sqrt[b^2 - a*c] + c*x)/Sqrt[b^2 - a*c]))^(-1 - n)*(a + 2*b*x + c*x^2)^(1 + n)*Hypergeometric2F1[-n, 1 + n, 2 + n, (b + Sqrt[b^2 - a*c] + c*x)/(2*S qrt[b^2 - a*c])])/(c*Sqrt[b^2 - a*c]*(1 + n))
3.2.93.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(-(a + b*x + c*x^2)^(p + 1)/(q*(p + 1)*((q - b - 2*c*x) /(2*q))^(p + 1)))*Hypergeometric2F1[-p, p + 1, p + 2, (b + q + 2*c*x)/(2*q) ], x]] /; FreeQ[{a, b, c, p}, x] && !IntegerQ[4*p] && !IntegerQ[3*p]
Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b *e)/(2*c) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]
\[\int \left (\operatorname {c1} x +\operatorname {b1} \right ) \left (c \,x^{2}+2 b x +a \right )^{n}d x\]
\[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^n \, dx=\int { {\left (c_{1} x + b_{1}\right )} {\left (c x^{2} + 2 \, b x + a\right )}^{n} \,d x } \]
\[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^n \, dx=\int \left (b_{1} + c_{1} x\right ) \left (a + 2 b x + c x^{2}\right )^{n}\, dx \]
\[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^n \, dx=\int { {\left (c_{1} x + b_{1}\right )} {\left (c x^{2} + 2 \, b x + a\right )}^{n} \,d x } \]
\[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^n \, dx=\int { {\left (c_{1} x + b_{1}\right )} {\left (c x^{2} + 2 \, b x + a\right )}^{n} \,d x } \]
Timed out. \[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^n \, dx=\int \left (b_{1}+c_{1}\,x\right )\,{\left (c\,x^2+2\,b\,x+a\right )}^n \,d x \]
\[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^n \, dx =\text {Too large to display} \]
( - (a + 2*b*x + c*x**2)**n*a*b*c1 + 2*(a + 2*b*x + c*x**2)**n*a*b1*c*n + 2*(a + 2*b*x + c*x**2)**n*a*b1*c + 2*(a + 2*b*x + c*x**2)**n*b**2*c1*n*x + 2*(a + 2*b*x + c*x**2)**n*b*b1*c*n*x + 2*(a + 2*b*x + c*x**2)**n*b*b1*c*x + 2*(a + 2*b*x + c*x**2)**n*b*c*c1*n*x**2 + (a + 2*b*x + c*x**2)**n*b*c*c 1*x**2 + 8*int(((a + 2*b*x + c*x**2)**n*x)/(2*a*n + a + 4*b*n*x + 2*b*x + 2*c*n*x**2 + c*x**2),x)*a*b*c*c1*n**3 + 12*int(((a + 2*b*x + c*x**2)**n*x) /(2*a*n + a + 4*b*n*x + 2*b*x + 2*c*n*x**2 + c*x**2),x)*a*b*c*c1*n**2 + 4* int(((a + 2*b*x + c*x**2)**n*x)/(2*a*n + a + 4*b*n*x + 2*b*x + 2*c*n*x**2 + c*x**2),x)*a*b*c*c1*n - 8*int(((a + 2*b*x + c*x**2)**n*x)/(2*a*n + a + 4 *b*n*x + 2*b*x + 2*c*n*x**2 + c*x**2),x)*a*b1*c**2*n**3 - 12*int(((a + 2*b *x + c*x**2)**n*x)/(2*a*n + a + 4*b*n*x + 2*b*x + 2*c*n*x**2 + c*x**2),x)* a*b1*c**2*n**2 - 4*int(((a + 2*b*x + c*x**2)**n*x)/(2*a*n + a + 4*b*n*x + 2*b*x + 2*c*n*x**2 + c*x**2),x)*a*b1*c**2*n - 8*int(((a + 2*b*x + c*x**2)* *n*x)/(2*a*n + a + 4*b*n*x + 2*b*x + 2*c*n*x**2 + c*x**2),x)*b**3*c1*n**3 - 12*int(((a + 2*b*x + c*x**2)**n*x)/(2*a*n + a + 4*b*n*x + 2*b*x + 2*c*n* x**2 + c*x**2),x)*b**3*c1*n**2 - 4*int(((a + 2*b*x + c*x**2)**n*x)/(2*a*n + a + 4*b*n*x + 2*b*x + 2*c*n*x**2 + c*x**2),x)*b**3*c1*n + 8*int(((a + 2* b*x + c*x**2)**n*x)/(2*a*n + a + 4*b*n*x + 2*b*x + 2*c*n*x**2 + c*x**2),x) *b**2*b1*c*n**3 + 12*int(((a + 2*b*x + c*x**2)**n*x)/(2*a*n + a + 4*b*n*x + 2*b*x + 2*c*n*x**2 + c*x**2),x)*b**2*b1*c*n**2 + 4*int(((a + 2*b*x + ...