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

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Rubi [A]  time = 0.83, antiderivative size = 508, normalized size of antiderivative = 1.00, 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 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]

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]))

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*}

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Mathematica [F]  time = 2.28, size = 0, normalized size = 0.00 \[ \int (g+h x)^m \left (a+b x+c x^2\right )^p \left (d+e x+f x^2\right ) \, dx \]

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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fricas [F]  time = 0.93, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (f x^{2} + e x + d\right )} {\left (c x^{2} + b x + a\right )}^{p} {\left (h x + g\right )}^{m}, x\right ) \]

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

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)

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maple [F]  time = 0.16, size = 0, normalized size = 0.00 \[ \int \left (f \,x^{2}+e x +d \right ) \left (h x +g \right )^{m} \left (c \,x^{2}+b x +a \right )^{p}\, dx \]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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