\(\int (d x)^m (a+b x+c x^2)^p \, dx\) [2561]

Optimal result

Integrand size = 18, antiderivative size = 137 \[ \int (d x)^m \left (a+b x+c x^2\right )^p \, dx=\frac {(d x)^{1+m} \left (1+\frac {2 c x}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (1+\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )^{-p} \left (a+b x+c x^2\right )^p \operatorname {AppellF1}\left (1+m,-p,-p,2+m,-\frac {2 c x}{b-\sqrt {b^2-4 a c}},-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{d (1+m)} \]

[Out]

(d*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))
)/d/(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)

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {773, 138} \[ \int (d x)^m \left (a+b x+c x^2\right )^p \, dx=\frac {(d x)^{m+1} \left (\frac {2 c x}{b-\sqrt {b^2-4 a c}}+1\right )^{-p} \left (\frac {2 c x}{\sqrt {b^2-4 a c}+b}+1\right )^{-p} \left (a+b x+c x^2\right )^p \operatorname {AppellF1}\left (m+1,-p,-p,m+2,-\frac {2 c x}{b-\sqrt {b^2-4 a c}},-\frac {2 c x}{b+\sqrt {b^2-4 a c}}\right )}{d (m+1)} \]

[In]

Int[(d*x)^m*(a + b*x + c*x^2)^p,x]

[Out]

((d*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])])/(d*(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)

Rule 138

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> 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] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 773

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c,
2]}, Dist[(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] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] &
& NeQ[2*c*d - b*e, 0] &&  !IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (\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\right ) \text {Subst}\left (\int x^m \left (1+\frac {2 c x}{\left (b-\sqrt {b^2-4 a c}\right ) d}\right )^p \left (1+\frac {2 c x}{\left (b+\sqrt {b^2-4 a c}\right ) d}\right )^p \, dx,x,d x\right )}{d} \\ & = \frac {(d 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 F_1\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 )}{d (1+m)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.17 \[ \int (d x)^m \left (a+b x+c x^2\right )^p \, dx=\frac {x (d x)^m \left (\frac {b-\sqrt {b^2-4 a c}+2 c x}{b-\sqrt {b^2-4 a c}}\right )^{-p} \left (\frac {b+\sqrt {b^2-4 a c}+2 c x}{b+\sqrt {b^2-4 a c}}\right )^{-p} (a+x (b+c x))^p \operatorname {AppellF1}\left (1+m,-p,-p,2+m,-\frac {2 c x}{b+\sqrt {b^2-4 a c}},\frac {2 c x}{-b+\sqrt {b^2-4 a c}}\right )}{1+m} \]

[In]

Integrate[(d*x)^m*(a + b*x + c*x^2)^p,x]

[Out]

(x*(d*x)^m*(a + x*(b + c*x))^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])])/((1 + 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)

Maple [F]

\[\int \left (d x \right )^{m} \left (c \,x^{2}+b x +a \right )^{p}d x\]

[In]

int((d*x)^m*(c*x^2+b*x+a)^p,x)

[Out]

int((d*x)^m*(c*x^2+b*x+a)^p,x)

Fricas [F]

\[ \int (d x)^m \left (a+b x+c x^2\right )^p \, dx=\int { {\left (c x^{2} + b x + a\right )}^{p} \left (d x\right )^{m} \,d x } \]

[In]

integrate((d*x)^m*(c*x^2+b*x+a)^p,x, algorithm="fricas")

[Out]

integral((c*x^2 + b*x + a)^p*(d*x)^m, x)

Sympy [F(-1)]

Timed out. \[ \int (d x)^m \left (a+b x+c x^2\right )^p \, dx=\text {Timed out} \]

[In]

integrate((d*x)**m*(c*x**2+b*x+a)**p,x)

[Out]

Timed out

Maxima [F]

\[ \int (d x)^m \left (a+b x+c x^2\right )^p \, dx=\int { {\left (c x^{2} + b x + a\right )}^{p} \left (d x\right )^{m} \,d x } \]

[In]

integrate((d*x)^m*(c*x^2+b*x+a)^p,x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^p*(d*x)^m, x)

Giac [F]

\[ \int (d x)^m \left (a+b x+c x^2\right )^p \, dx=\int { {\left (c x^{2} + b x + a\right )}^{p} \left (d x\right )^{m} \,d x } \]

[In]

integrate((d*x)^m*(c*x^2+b*x+a)^p,x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^p*(d*x)^m, x)

Mupad [F(-1)]

Timed out. \[ \int (d x)^m \left (a+b x+c x^2\right )^p \, dx=\int {\left (d\,x\right )}^m\,{\left (c\,x^2+b\,x+a\right )}^p \,d x \]

[In]

int((d*x)^m*(a + b*x + c*x^2)^p,x)

[Out]

int((d*x)^m*(a + b*x + c*x^2)^p, x)