3.1094 \(\int (e x)^m (A+B x) (a+b x+c x^2)^p \, dx\)

Optimal. Leaf size=277 \[ \frac{A (e x)^{m+1} \left (\frac{2 c x}{b-\sqrt{b^2-4 a c}}+1\right )^{-p} \left (\frac{2 c x}{\sqrt{b^2-4 a c}+b}+1\right )^{-p} \left (a+b x+c x^2\right )^p F_1\left (m+1;-p,-p;m+2;-\frac{2 c x}{b-\sqrt{b^2-4 a c}},-\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )}{e (m+1)}+\frac{B (e x)^{m+2} \left (\frac{2 c x}{b-\sqrt{b^2-4 a c}}+1\right )^{-p} \left (\frac{2 c x}{\sqrt{b^2-4 a c}+b}+1\right )^{-p} \left (a+b x+c x^2\right )^p F_1\left (m+2;-p,-p;m+3;-\frac{2 c x}{b-\sqrt{b^2-4 a c}},-\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )}{e^2 (m+2)} \]

[Out]

(A*(e*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])])/(e*(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) + (B*(e*x)^(2 + m)*(a + b*x + c*x^2)^p*AppellF1[2 + m, -p, -p, 3 + m, (-2*c*x)/(b - Sqrt[b^2 - 4*a*c]
), (-2*c*x)/(b + Sqrt[b^2 - 4*a*c])])/(e^2*(2 + m)*(1 + (2*c*x)/(b - Sqrt[b^2 - 4*a*c]))^p*(1 + (2*c*x)/(b + S
qrt[b^2 - 4*a*c]))^p)

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Rubi [A]  time = 0.404451, antiderivative size = 277, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {843, 759, 133} \[ \frac{A (e x)^{m+1} \left (\frac{2 c x}{b-\sqrt{b^2-4 a c}}+1\right )^{-p} \left (\frac{2 c x}{\sqrt{b^2-4 a c}+b}+1\right )^{-p} \left (a+b x+c x^2\right )^p F_1\left (m+1;-p,-p;m+2;-\frac{2 c x}{b-\sqrt{b^2-4 a c}},-\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )}{e (m+1)}+\frac{B (e x)^{m+2} \left (\frac{2 c x}{b-\sqrt{b^2-4 a c}}+1\right )^{-p} \left (\frac{2 c x}{\sqrt{b^2-4 a c}+b}+1\right )^{-p} \left (a+b x+c x^2\right )^p F_1\left (m+2;-p,-p;m+3;-\frac{2 c x}{b-\sqrt{b^2-4 a c}},-\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )}{e^2 (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*(e*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])])/(e*(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) + (B*(e*x)^(2 + m)*(a + b*x + c*x^2)^p*AppellF1[2 + m, -p, -p, 3 + m, (-2*c*x)/(b - Sqrt[b^2 - 4*a*c]
), (-2*c*x)/(b + Sqrt[b^2 - 4*a*c])])/(e^2*(2 + m)*(1 + (2*c*x)/(b - Sqrt[b^2 - 4*a*c]))^p*(1 + (2*c*x)/(b + S
qrt[b^2 - 4*a*c]))^p)

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

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 (e x)^m (A+B x) \left (a+b x+c x^2\right )^p \, dx &=A \int (e x)^m \left (a+b x+c x^2\right )^p \, dx+\frac{B \int (e x)^{1+m} \left (a+b x+c x^2\right )^p \, dx}{e}\\ &=\frac{\left (B \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^{1+m} \left (1+\frac{2 c x}{\left (b-\sqrt{b^2-4 a c}\right ) e}\right )^p \left (1+\frac{2 c x}{\left (b+\sqrt{b^2-4 a c}\right ) e}\right )^p \, dx,x,e x\right )}{e^2}+\frac{\left (A \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 ) e}\right )^p \left (1+\frac{2 c x}{\left (b+\sqrt{b^2-4 a c}\right ) e}\right )^p \, dx,x,e x\right )}{e}\\ &=\frac{A (e 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 )}{e (1+m)}+\frac{B (e x)^{2+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 (2+m;-p,-p;3+m;-\frac{2 c x}{b-\sqrt{b^2-4 a c}},-\frac{2 c x}{b+\sqrt{b^2-4 a c}}\right )}{e^2 (2+m)}\\ \end{align*}

Mathematica [A]  time = 0.354815, size = 232, normalized size = 0.84 \[ \frac{x (e x)^m \left (\frac{-\sqrt{b^2-4 a c}+b+2 c x}{b-\sqrt{b^2-4 a c}}\right )^{-p} \left (\frac{\sqrt{b^2-4 a c}+b+2 c x}{\sqrt{b^2-4 a c}+b}\right )^{-p} (a+x (b+c x))^p \left (A (m+2) F_1\left (m+1;-p,-p;m+2;-\frac{2 c x}{b+\sqrt{b^2-4 a c}},\frac{2 c x}{\sqrt{b^2-4 a c}-b}\right )+B (m+1) x F_1\left (m+2;-p,-p;m+3;-\frac{2 c x}{b+\sqrt{b^2-4 a c}},\frac{2 c x}{\sqrt{b^2-4 a c}-b}\right )\right )}{(m+1) (m+2)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(e*x)^m*(a + x*(b + c*x))^p*(A*(2 + m)*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])] + B*(1 + m)*x*AppellF1[2 + m, -p, -p, 3 + m, (-2*c*x)/(b + Sqrt[b^2 - 4*a*c]),
(2*c*x)/(-b + Sqrt[b^2 - 4*a*c])]))/((1 + m)*(2 + 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 = 0.096, size = 0, normalized size = 0. \begin{align*} \int \left ( ex \right ) ^{m} \left ( Bx+A \right ) \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)^m*(B*x+A)*(c*x^2+b*x+a)^p,x)

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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