### 3.2562 $$\int (d+e x)^m (a+b x+c x^2)^p \, dx$$

Optimal. Leaf size=187 $\frac{(d+e x)^{m+1} \left (a+b x+c x^2\right )^p \left (1-\frac{2 c (d+e x)}{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}\right )^{-p} \left (1-\frac{2 c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}\right )^{-p} F_1\left (m+1;-p,-p;m+2;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e},\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{e (m+1)}$

[Out]

((d + e*x)^(1 + m)*(a + b*x + c*x^2)^p*AppellF1[1 + m, -p, -p, 2 + m, (2*c*(d + e*x))/(2*c*d - (b - Sqrt[b^2 -
4*a*c])*e), (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/(e*(1 + m)*(1 - (2*c*(d + e*x))/(2*c*d - (b
- Sqrt[b^2 - 4*a*c])*e))^p*(1 - (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e))^p)

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Rubi [A]  time = 0.110415, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 20, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.1, Rules used = {759, 133} $\frac{(d+e x)^{m+1} \left (a+b x+c x^2\right )^p \left (1-\frac{2 c (d+e x)}{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}\right )^{-p} \left (1-\frac{2 c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}\right )^{-p} F_1\left (m+1;-p,-p;m+2;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e},\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{e (m+1)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((d + e*x)^(1 + m)*(a + b*x + c*x^2)^p*AppellF1[1 + m, -p, -p, 2 + m, (2*c*(d + e*x))/(2*c*d - (b - Sqrt[b^2 -
4*a*c])*e), (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/(e*(1 + m)*(1 - (2*c*(d + e*x))/(2*c*d - (b
- Sqrt[b^2 - 4*a*c])*e))^p*(1 - (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e))^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+e x)^m \left (a+b x+c x^2\right )^p \, dx &=\frac{\left (\left (a+b x+c x^2\right )^p \left (1-\frac{d+e x}{d-\frac{\left (b-\sqrt{b^2-4 a c}\right ) e}{2 c}}\right )^{-p} \left (1-\frac{d+e x}{d-\frac{\left (b+\sqrt{b^2-4 a c}\right ) e}{2 c}}\right )^{-p}\right ) \operatorname{Subst}\left (\int x^m \left (1-\frac{2 c x}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right )^p \left (1-\frac{2 c x}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )^p \, dx,x,d+e x\right )}{e}\\ &=\frac{(d+e x)^{1+m} \left (a+b x+c x^2\right )^p \left (1-\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right )^{-p} \left (1-\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )^{-p} F_1\left (1+m;-p,-p;2+m;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e},\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{e (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.4013, size = 205, normalized size = 1.1 $\frac{(d+e x)^{m+1} (a+x (b+c x))^p \left (\frac{e \left (\sqrt{b^2-4 a c}-b-2 c x\right )}{e \left (\sqrt{b^2-4 a c}-b\right )+2 c d}\right )^{-p} \left (\frac{e \left (\sqrt{b^2-4 a c}+b+2 c x\right )}{e \left (\sqrt{b^2-4 a c}+b\right )-2 c d}\right )^{-p} F_1\left (m+1;-p,-p;m+2;\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e},\frac{2 c (d+e x)}{2 c d+\left (\sqrt{b^2-4 a c}-b\right ) e}\right )}{e (m+1)}$

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((d + e*x)^(1 + m)*(a + x*(b + c*x))^p*AppellF1[1 + m, -p, -p, 2 + m, (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 -
4*a*c])*e), (2*c*(d + e*x))/(2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e)])/(e*(1 + m)*((e*(-b + Sqrt[b^2 - 4*a*c] - 2
*c*x))/(2*c*d + (-b + Sqrt[b^2 - 4*a*c])*e))^p*((e*(b + Sqrt[b^2 - 4*a*c] + 2*c*x))/(-2*c*d + (b + Sqrt[b^2 -
4*a*c])*e))^p)

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Maple [F]  time = 1.395, size = 0, normalized size = 0. \begin{align*} \int \left ( ex+d \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((e*x+d)^m*(c*x^2+b*x+a)^p,x)

[Out]

int((e*x+d)^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 (e x + d\right )}^{m}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^2 + b*x + a)^p*(e*x + d)^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 (e x + d\right )}^{m}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*x^2 + b*x + a)^p*(e*x + d)^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((e*x+d)**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 (e x + d\right )}^{m}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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