3.272 \(\int (g+h x)^m (a+b x+c x^2)^p (d+e x+f x^2) \, dx\)

Optimal. Leaf size=510 \[ \frac{(g+h x)^{m+1} \left (a+b x+c x^2\right )^p \left (1-\frac{2 c (g+h x)}{2 c g-h \left (b-\sqrt{b^2-4 a c}\right )}\right )^{-p} \left (1-\frac{2 c (g+h x)}{2 c g-h \left (\sqrt{b^2-4 a c}+b\right )}\right )^{-p} F_1\left (m+1;-p,-p;m+2;\frac{2 c (g+h x)}{2 c g-\left (b-\sqrt{b^2-4 a c}\right ) h},\frac{2 c (g+h x)}{2 c g-\left (b+\sqrt{b^2-4 a c}\right ) h}\right ) \left (f h (m+1) (b g-a h)+c \left (2 f g^2 (p+1)-h (m+2 p+3) (e g-d h)\right )\right )}{c h^3 (m+1) (m+2 p+3)}-\frac{(g+h x)^{m+2} \left (a+b x+c x^2\right )^p \left (1-\frac{2 c (g+h x)}{2 c g-h \left (b-\sqrt{b^2-4 a c}\right )}\right )^{-p} \left (1-\frac{2 c (g+h x)}{2 c g-h \left (\sqrt{b^2-4 a c}+b\right )}\right )^{-p} (b f h (m+p+2)+c (2 f g (p+1)-e h (m+2 p+3))) F_1\left (m+2;-p,-p;m+3;\frac{2 c (g+h x)}{2 c g-\left (b-\sqrt{b^2-4 a c}\right ) h},\frac{2 c (g+h x)}{2 c g-\left (b+\sqrt{b^2-4 a c}\right ) h}\right )}{c h^3 (m+2) (m+2 p+3)}+\frac{f (g+h x)^{m+1} \left (a+b x+c x^2\right )^{p+1}}{c h (m+2 p+3)} \]

[Out]

(f*(g + h*x)^(1 + m)*(a + b*x + c*x^2)^(1 + p))/(c*h*(3 + m + 2*p)) + ((f*h*(b*g - a*h)*(1 + m) + c*(2*f*g^2*(
1 + p) - h*(e*g - d*h)*(3 + m + 2*p)))*(g + h*x)^(1 + m)*(a + b*x + c*x^2)^p*AppellF1[1 + m, -p, -p, 2 + m, (2
*c*(g + h*x))/(2*c*g - (b - Sqrt[b^2 - 4*a*c])*h), (2*c*(g + h*x))/(2*c*g - (b + Sqrt[b^2 - 4*a*c])*h)])/(c*h^
3*(1 + m)*(3 + m + 2*p)*(1 - (2*c*(g + h*x))/(2*c*g - (b - Sqrt[b^2 - 4*a*c])*h))^p*(1 - (2*c*(g + h*x))/(2*c*
g - (b + Sqrt[b^2 - 4*a*c])*h))^p) - ((b*f*h*(2 + m + p) + c*(2*f*g*(1 + p) - e*h*(3 + m + 2*p)))*(g + h*x)^(2
+ m)*(a + b*x + c*x^2)^p*AppellF1[2 + m, -p, -p, 3 + m, (2*c*(g + h*x))/(2*c*g - (b - Sqrt[b^2 - 4*a*c])*h),
(2*c*(g + h*x))/(2*c*g - (b + Sqrt[b^2 - 4*a*c])*h)])/(c*h^3*(2 + m)*(3 + m + 2*p)*(1 - (2*c*(g + h*x))/(2*c*g
- (b - Sqrt[b^2 - 4*a*c])*h))^p*(1 - (2*c*(g + h*x))/(2*c*g - (b + Sqrt[b^2 - 4*a*c])*h))^p)

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Rubi [A]  time = 0.833057, antiderivative size = 508, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {1653, 843, 759, 133} \[ \frac{(g+h x)^{m+1} \left (a+b x+c x^2\right )^p \left (1-\frac{2 c (g+h x)}{2 c g-h \left (b-\sqrt{b^2-4 a c}\right )}\right )^{-p} \left (1-\frac{2 c (g+h x)}{2 c g-h \left (\sqrt{b^2-4 a c}+b\right )}\right )^{-p} F_1\left (m+1;-p,-p;m+2;\frac{2 c (g+h x)}{2 c g-\left (b-\sqrt{b^2-4 a c}\right ) h},\frac{2 c (g+h x)}{2 c g-\left (b+\sqrt{b^2-4 a c}\right ) h}\right ) \left (f h (m+1) (b g-a h)-c h (m+2 p+3) (e g-d h)+2 c f g^2 (p+1)\right )}{c h^3 (m+1) (m+2 p+3)}-\frac{(g+h x)^{m+2} \left (a+b x+c x^2\right )^p \left (1-\frac{2 c (g+h x)}{2 c g-h \left (b-\sqrt{b^2-4 a c}\right )}\right )^{-p} \left (1-\frac{2 c (g+h x)}{2 c g-h \left (\sqrt{b^2-4 a c}+b\right )}\right )^{-p} (b f h (m+p+2)-c e h (m+2 p+3)+2 c f g (p+1)) F_1\left (m+2;-p,-p;m+3;\frac{2 c (g+h x)}{2 c g-\left (b-\sqrt{b^2-4 a c}\right ) h},\frac{2 c (g+h x)}{2 c g-\left (b+\sqrt{b^2-4 a c}\right ) h}\right )}{c h^3 (m+2) (m+2 p+3)}+\frac{f (g+h x)^{m+1} \left (a+b x+c x^2\right )^{p+1}}{c h (m+2 p+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f*(g + h*x)^(1 + m)*(a + b*x + c*x^2)^(1 + p))/(c*h*(3 + m + 2*p)) + ((f*h*(b*g - a*h)*(1 + m) + 2*c*f*g^2*(1
+ p) - c*h*(e*g - d*h)*(3 + m + 2*p))*(g + h*x)^(1 + m)*(a + b*x + c*x^2)^p*AppellF1[1 + m, -p, -p, 2 + m, (2
*c*(g + h*x))/(2*c*g - (b - Sqrt[b^2 - 4*a*c])*h), (2*c*(g + h*x))/(2*c*g - (b + Sqrt[b^2 - 4*a*c])*h)])/(c*h^
3*(1 + m)*(3 + m + 2*p)*(1 - (2*c*(g + h*x))/(2*c*g - (b - Sqrt[b^2 - 4*a*c])*h))^p*(1 - (2*c*(g + h*x))/(2*c*
g - (b + Sqrt[b^2 - 4*a*c])*h))^p) - ((2*c*f*g*(1 + p) + b*f*h*(2 + m + p) - c*e*h*(3 + m + 2*p))*(g + h*x)^(2
+ m)*(a + b*x + c*x^2)^p*AppellF1[2 + m, -p, -p, 3 + m, (2*c*(g + h*x))/(2*c*g - (b - Sqrt[b^2 - 4*a*c])*h),
(2*c*(g + h*x))/(2*c*g - (b + Sqrt[b^2 - 4*a*c])*h)])/(c*h^3*(2 + m)*(3 + m + 2*p)*(1 - (2*c*(g + h*x))/(2*c*g
- (b - Sqrt[b^2 - 4*a*c])*h))^p*(1 - (2*c*(g + h*x))/(2*c*g - (b + Sqrt[b^2 - 4*a*c])*h))^p)

Rule 1653

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq
, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*e^(q - 1)*(
m + q + 2*p + 1)), x] + Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^
q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(b*d*e*(p + 1) + a*e^2*(m + q
- 1) - c*d^2*(m + q + 2*p + 1) - e*(2*c*d - b*e)*(m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p +
1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] &&  !(IGtQ[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

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 (g+h x)^m \left (a+b x+c x^2\right )^p \left (d+e x+f x^2\right ) \, dx &=\frac{f (g+h x)^{1+m} \left (a+b x+c x^2\right )^{1+p}}{c h (3+m+2 p)}+\frac{\int (g+h x)^m (-h (a f h (1+m)+b f g (1+p)-c d h (3+m+2 p))-h (2 c f g (1+p)+b f h (2+m+p)-c e h (3+m+2 p)) x) \left (a+b x+c x^2\right )^p \, dx}{c h^2 (3+m+2 p)}\\ &=\frac{f (g+h x)^{1+m} \left (a+b x+c x^2\right )^{1+p}}{c h (3+m+2 p)}-\frac{(2 c f g (1+p)+b f h (2+m+p)-c e h (3+m+2 p)) \int (g+h x)^{1+m} \left (a+b x+c x^2\right )^p \, dx}{c h^2 (3+m+2 p)}+\frac{\left (f h (b g-a h) (1+m)+2 c f g^2 (1+p)-c h (e g-d h) (3+m+2 p)\right ) \int (g+h x)^m \left (a+b x+c x^2\right )^p \, dx}{c h^2 (3+m+2 p)}\\ &=\frac{f (g+h x)^{1+m} \left (a+b x+c x^2\right )^{1+p}}{c h (3+m+2 p)}-\frac{\left ((2 c f g (1+p)+b f h (2+m+p)-c e h (3+m+2 p)) \left (a+b x+c x^2\right )^p \left (1-\frac{g+h x}{g-\frac{\left (b-\sqrt{b^2-4 a c}\right ) h}{2 c}}\right )^{-p} \left (1-\frac{g+h x}{g-\frac{\left (b+\sqrt{b^2-4 a c}\right ) h}{2 c}}\right )^{-p}\right ) \operatorname{Subst}\left (\int x^{1+m} \left (1-\frac{2 c x}{2 c g-\left (b-\sqrt{b^2-4 a c}\right ) h}\right )^p \left (1-\frac{2 c x}{2 c g-\left (b+\sqrt{b^2-4 a c}\right ) h}\right )^p \, dx,x,g+h x\right )}{c h^3 (3+m+2 p)}+\frac{\left (\left (f h (b g-a h) (1+m)+2 c f g^2 (1+p)-c h (e g-d h) (3+m+2 p)\right ) \left (a+b x+c x^2\right )^p \left (1-\frac{g+h x}{g-\frac{\left (b-\sqrt{b^2-4 a c}\right ) h}{2 c}}\right )^{-p} \left (1-\frac{g+h x}{g-\frac{\left (b+\sqrt{b^2-4 a c}\right ) h}{2 c}}\right )^{-p}\right ) \operatorname{Subst}\left (\int x^m \left (1-\frac{2 c x}{2 c g-\left (b-\sqrt{b^2-4 a c}\right ) h}\right )^p \left (1-\frac{2 c x}{2 c g-\left (b+\sqrt{b^2-4 a c}\right ) h}\right )^p \, dx,x,g+h x\right )}{c h^3 (3+m+2 p)}\\ &=\frac{f (g+h x)^{1+m} \left (a+b x+c x^2\right )^{1+p}}{c h (3+m+2 p)}+\frac{\left (f h (b g-a h) (1+m)+2 c f g^2 (1+p)-c h (e g-d h) (3+m+2 p)\right ) (g+h x)^{1+m} \left (a+b x+c x^2\right )^p \left (1-\frac{2 c (g+h x)}{2 c g-\left (b-\sqrt{b^2-4 a c}\right ) h}\right )^{-p} \left (1-\frac{2 c (g+h x)}{2 c g-\left (b+\sqrt{b^2-4 a c}\right ) h}\right )^{-p} F_1\left (1+m;-p,-p;2+m;\frac{2 c (g+h x)}{2 c g-\left (b-\sqrt{b^2-4 a c}\right ) h},\frac{2 c (g+h x)}{2 c g-\left (b+\sqrt{b^2-4 a c}\right ) h}\right )}{c h^3 (1+m) (3+m+2 p)}-\frac{(2 c f g (1+p)+b f h (2+m+p)-c e h (3+m+2 p)) (g+h x)^{2+m} \left (a+b x+c x^2\right )^p \left (1-\frac{2 c (g+h x)}{2 c g-\left (b-\sqrt{b^2-4 a c}\right ) h}\right )^{-p} \left (1-\frac{2 c (g+h x)}{2 c g-\left (b+\sqrt{b^2-4 a c}\right ) h}\right )^{-p} F_1\left (2+m;-p,-p;3+m;\frac{2 c (g+h x)}{2 c g-\left (b-\sqrt{b^2-4 a c}\right ) h},\frac{2 c (g+h x)}{2 c g-\left (b+\sqrt{b^2-4 a c}\right ) h}\right )}{c h^3 (2+m) (3+m+2 p)}\\ \end{align*}

Mathematica [F]  time = 2.31765, size = 0, normalized size = 0. \[ \int (g+h x)^m \left (a+b x+c x^2\right )^p \left (d+e x+f x^2\right ) \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 1.428, size = 0, normalized size = 0. \begin{align*} \int \left ( hx+g \right ) ^{m} \left ( c{x}^{2}+bx+a \right ) ^{p} \left ( f{x}^{2}+ex+d \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((h*x+g)^m*(c*x^2+b*x+a)^p*(f*x^2+e*x+d),x)

[Out]

int((h*x+g)^m*(c*x^2+b*x+a)^p*(f*x^2+e*x+d),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((f*x^2 + e*x + d)*(c*x^2 + b*x + a)^p*(h*x + g)^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((h*x+g)**m*(c*x**2+b*x+a)**p*(f*x**2+e*x+d),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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