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)*(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
)))/e/(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)+B*(e*x)^(2+m)*(c*x^2+b*x
+a)^p*AppellF1(2+m,-p,-p,3+m,-2*c*x/(b-(-4*a*c+b^2)^(1/2)),-2*c*x/(b+(-4*a*c+b^2)^(1/2)))/e^2/(2+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)

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Rubi [A]  time = 0.40, antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, 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 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])

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 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]

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*}

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Mathematica [A]  time = 0.35, 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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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B x + A\right )} {\left (c x^{2} + b x + a\right )}^{p} \left (e x\right )^{m}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B x + A\right )} {\left (c x^{2} + b x + a\right )}^{p} \left (e x\right )^{m}\,{d x} \]

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)

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maple [F]  time = 1.25, size = 0, normalized size = 0.00 \[ \int \left (B x +A \right ) \left (e x \right )^{m} \left (c \,x^{2}+b x +a \right )^{p}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B x + A\right )} {\left (c x^{2} + b x + a\right )}^{p} \left (e x\right )^{m}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (e\,x\right )}^m\,\left (A+B\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^p \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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