\(\int (\text {b1}+\text {c1} x) (a+2 b x+c x^2)^n \, dx\) [193]

Optimal result

Integrand size = 19, antiderivative size = 159 \[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^n \, dx=\frac {\text {c1} \left (a+2 b x+c x^2\right )^{1+n}}{2 c (1+n)}-\frac {2^n (\text {b1} c-b \text {c1}) \left (-\frac {b-\sqrt {b^2-a c}+c x}{\sqrt {b^2-a c}}\right )^{-1-n} \left (a+2 b x+c x^2\right )^{1+n} \operatorname {Hypergeometric2F1}\left (-n,1+n,2+n,\frac {b+\sqrt {b^2-a c}+c x}{2 \sqrt {b^2-a c}}\right )}{c \sqrt {b^2-a c} (1+n)} \]

[Out]

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)

Rubi [A] (verified)

Time = 0.08 (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 = {654, 638} \[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^n \, dx=\frac {\text {c1} \left (a+2 b x+c x^2\right )^{n+1}}{2 c (n+1)}-\frac {2^n (\text {b1} c-b \text {c1}) \left (-\frac {-\sqrt {b^2-a c}+b+c x}{\sqrt {b^2-a c}}\right )^{-n-1} \left (a+2 b x+c x^2\right )^{n+1} \operatorname {Hypergeometric2F1}\left (-n,n+1,n+2,\frac {b+c x+\sqrt {b^2-a c}}{2 \sqrt {b^2-a c}}\right )}{c (n+1) \sqrt {b^2-a c}} \]

[In]

Int[(b1 + c1*x)*(a + 2*b*x + c*x^2)^n,x]

[Out]

(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
*Sqrt[b^2 - a*c])])/(c*Sqrt[b^2 - a*c]*(1 + n))

Rule 638

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] && NeQ[b^2 - 4*a*c, 0] &&  !IntegerQ[4*p]

Rule 654

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] + Dist[(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[2*c*d - b*e, 0] && NeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {c1} \left (a+2 b x+c x^2\right )^{1+n}}{2 c (1+n)}+\frac {(2 \text {b1} c-2 b \text {c1}) \int \left (a+2 b x+c x^2\right )^n \, dx}{2 c} \\ & = \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)} \\ \end{align*}

Mathematica [C] (verified)

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 \[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^n \, dx=\frac {1}{2} (a+x (2 b+c x))^n \left (\text {c1} x^2 \left (\frac {b-\sqrt {b^2-a c}+c x}{b-\sqrt {b^2-a c}}\right )^{-n} \left (\frac {b+\sqrt {b^2-a c}+c x}{b+\sqrt {b^2-a c}}\right )^{-n} \operatorname {AppellF1}\left (2,-n,-n,3,-\frac {c x}{b+\sqrt {b^2-a c}},\frac {c x}{-b+\sqrt {b^2-a c}}\right )+\frac {2^{1+n} \text {b1} \left (b-\sqrt {b^2-a c}+c x\right ) \left (\frac {b+\sqrt {b^2-a c}+c x}{\sqrt {b^2-a c}}\right )^{-n} \operatorname {Hypergeometric2F1}\left (-n,1+n,2+n,\frac {-b+\sqrt {b^2-a c}-c x}{2 \sqrt {b^2-a c}}\right )}{c (1+n)}\right ) \]

[In]

Integrate[(b1 + c1*x)*(a + 2*b*x + c*x^2)^n,x]

[Out]

((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

Maple [F]

\[\int \left (\operatorname {c1} x +\operatorname {b1} \right ) \left (c \,x^{2}+2 b x +a \right )^{n}d x\]

[In]

int((c1*x+b1)*(c*x^2+2*b*x+a)^n,x)

[Out]

int((c1*x+b1)*(c*x^2+2*b*x+a)^n,x)

Fricas [F]

\[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^n \, dx=\int { {\left (c_{1} x + b_{1}\right )} {\left (c x^{2} + 2 \, b x + a\right )}^{n} \,d x } \]

[In]

integrate((c1*x+b1)*(c*x^2+2*b*x+a)^n,x, algorithm="fricas")

[Out]

integral((c1*x + b1)*(c*x^2 + 2*b*x + a)^n, x)

Sympy [F]

\[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^n \, dx=\int \left (b_{1} + c_{1} x\right ) \left (a + 2 b x + c x^{2}\right )^{n}\, dx \]

[In]

integrate((c1*x+b1)*(c*x**2+2*b*x+a)**n,x)

[Out]

Integral((b1 + c1*x)*(a + 2*b*x + c*x**2)**n, x)

Maxima [F]

\[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^n \, dx=\int { {\left (c_{1} x + b_{1}\right )} {\left (c x^{2} + 2 \, b x + a\right )}^{n} \,d x } \]

[In]

integrate((c1*x+b1)*(c*x^2+2*b*x+a)^n,x, algorithm="maxima")

[Out]

integrate((c1*x + b1)*(c*x^2 + 2*b*x + a)^n, x)

Giac [F]

\[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^n \, dx=\int { {\left (c_{1} x + b_{1}\right )} {\left (c x^{2} + 2 \, b x + a\right )}^{n} \,d x } \]

[In]

integrate((c1*x+b1)*(c*x^2+2*b*x+a)^n,x, algorithm="giac")

[Out]

integrate((c1*x + b1)*(c*x^2 + 2*b*x + a)^n, x)

Mupad [F(-1)]

Timed out. \[ \int (\text {b1}+\text {c1} x) \left (a+2 b x+c x^2\right )^n \, dx=\int \left (b_{1}+c_{1}\,x\right )\,{\left (c\,x^2+2\,b\,x+a\right )}^n \,d x \]

[In]

int((b1 + c1*x)*(a + 2*b*x + c*x^2)^n,x)

[Out]

int((b1 + c1*x)*(a + 2*b*x + c*x^2)^n, x)