### 3.2561 $$\int (d x)^m (a+b x+c x^2)^p \, dx$$

Optimal. Leaf size=137 $\frac{(d x)^{m+1} \left (\frac{2 c x}{b-\sqrt{b^2-4 a c}}+1\right )^{-p} \left (\frac{2 c x}{\sqrt{b^2-4 a c}+b}+1\right )^{-p} \left (a+b x+c x^2\right )^p F_1\left (m+1;-p,-p;m+2;-\frac{2 c x}{b-\sqrt{b^2-4 a c}},-\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )}{d (m+1)}$

[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)

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Rubi [A]  time = 0.277654, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.111, Rules used = {759, 133} $\frac{(d x)^{m+1} \left (\frac{2 c x}{b-\sqrt{b^2-4 a c}}+1\right )^{-p} \left (\frac{2 c x}{\sqrt{b^2-4 a c}+b}+1\right )^{-p} \left (a+b x+c x^2\right )^p F_1\left (m+1;-p,-p;m+2;-\frac{2 c x}{b-\sqrt{b^2-4 a c}},-\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )}{d (m+1)}$

Antiderivative was successfully veriﬁed.

[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 759

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]

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rubi steps

\begin{align*} \int (d x)^m \left (a+b x+c x^2\right )^p \, dx &=\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 ) \operatorname{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]  time = 0.271972, size = 160, normalized size = 1.17 $\frac{x (d x)^m \left (\frac{-\sqrt{b^2-4 a c}+b+2 c x}{b-\sqrt{b^2-4 a c}}\right )^{-p} \left (\frac{\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}+b}\right )^{-p} (a+x (b+c x))^p F_1\left (m+1;-p,-p;m+2;-\frac{2 c x}{b+\sqrt{b^2-4 a c}},\frac{2 c x}{\sqrt{b^2-4 a c}-b}\right )}{m+1}$

Warning: Unable to verify antiderivative.

[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)

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Maple [F]  time = 1.112, size = 0, normalized size = 0. \begin{align*} \int \left ( dx \right ) ^{m} \left ( c{x}^{2}+bx+a \right ) ^{p}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x^{2} + b x + a\right )}^{p} \left (d x\right )^{m}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (c x^{2} + b x + a\right )}^{p} \left (d x\right )^{m}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (c x^{2} + b x + a\right )}^{p} \left (d x\right )^{m}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[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)