### 3.273 $$\int (g+h x)^m \sqrt{a+b x+c x^2} (d+e x+f x^2) \, dx$$

Optimal. Leaf size=496 $\frac{\sqrt{a+b x+c x^2} (g+h x)^{m+1} F_1\left (m+1;-\frac{1}{2},-\frac{1}{2};m+2;\frac{2 c (g+h x)}{2 c g-\left (b-\sqrt{b^2-4 a c}\right ) h},\frac{2 c (g+h x)}{2 c g-\left (b+\sqrt{b^2-4 a c}\right ) h}\right ) \left (f h (m+1) (b g-a h)+c \left (3 f g^2-h (m+4) (e g-d h)\right )\right )}{c h^3 (m+1) (m+4) \sqrt{1-\frac{2 c (g+h x)}{2 c g-h \left (b-\sqrt{b^2-4 a c}\right )}} \sqrt{1-\frac{2 c (g+h x)}{2 c g-h \left (\sqrt{b^2-4 a c}+b\right )}}}-\frac{\sqrt{a+b x+c x^2} (g+h x)^{m+2} (b f h (2 m+5)+c (6 f g-2 e h (m+4))) F_1\left (m+2;-\frac{1}{2},-\frac{1}{2};m+3;\frac{2 c (g+h x)}{2 c g-\left (b-\sqrt{b^2-4 a c}\right ) h},\frac{2 c (g+h x)}{2 c g-\left (b+\sqrt{b^2-4 a c}\right ) h}\right )}{2 c h^3 (m+2) (m+4) \sqrt{1-\frac{2 c (g+h x)}{2 c g-h \left (b-\sqrt{b^2-4 a c}\right )}} \sqrt{1-\frac{2 c (g+h x)}{2 c g-h \left (\sqrt{b^2-4 a c}+b\right )}}}+\frac{f \left (a+b x+c x^2\right )^{3/2} (g+h x)^{m+1}}{c h (m+4)}$

[Out]

(f*(g + h*x)^(1 + m)*(a + b*x + c*x^2)^(3/2))/(c*h*(4 + m)) + ((f*h*(b*g - a*h)*(1 + m) + c*(3*f*g^2 - h*(e*g
- d*h)*(4 + m)))*(g + h*x)^(1 + m)*Sqrt[a + b*x + c*x^2]*AppellF1[1 + m, -1/2, -1/2, 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)*(4 + m
)*Sqrt[1 - (2*c*(g + h*x))/(2*c*g - (b - Sqrt[b^2 - 4*a*c])*h)]*Sqrt[1 - (2*c*(g + h*x))/(2*c*g - (b + Sqrt[b^
2 - 4*a*c])*h)]) - ((b*f*h*(5 + 2*m) + c*(6*f*g - 2*e*h*(4 + m)))*(g + h*x)^(2 + m)*Sqrt[a + b*x + c*x^2]*Appe
llF1[2 + m, -1/2, -1/2, 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)])/(2*c*h^3*(2 + m)*(4 + m)*Sqrt[1 - (2*c*(g + h*x))/(2*c*g - (b - Sqrt[b^2 - 4*a*c])
*h)]*Sqrt[1 - (2*c*(g + h*x))/(2*c*g - (b + Sqrt[b^2 - 4*a*c])*h)])

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Rubi [A]  time = 0.672379, antiderivative size = 494, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 32, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.125, Rules used = {1653, 843, 759, 133} $\frac{\sqrt{a+b x+c x^2} (g+h x)^{m+1} F_1\left (m+1;-\frac{1}{2},-\frac{1}{2};m+2;\frac{2 c (g+h x)}{2 c g-\left (b-\sqrt{b^2-4 a c}\right ) h},\frac{2 c (g+h x)}{2 c g-\left (b+\sqrt{b^2-4 a c}\right ) h}\right ) \left (f h (m+1) (b g-a h)-c h (m+4) (e g-d h)+3 c f g^2\right )}{c h^3 (m+1) (m+4) \sqrt{1-\frac{2 c (g+h x)}{2 c g-h \left (b-\sqrt{b^2-4 a c}\right )}} \sqrt{1-\frac{2 c (g+h x)}{2 c g-h \left (\sqrt{b^2-4 a c}+b\right )}}}-\frac{\sqrt{a+b x+c x^2} (g+h x)^{m+2} (b f h (2 m+5)-2 c e h (m+4)+6 c f g) F_1\left (m+2;-\frac{1}{2},-\frac{1}{2};m+3;\frac{2 c (g+h x)}{2 c g-\left (b-\sqrt{b^2-4 a c}\right ) h},\frac{2 c (g+h x)}{2 c g-\left (b+\sqrt{b^2-4 a c}\right ) h}\right )}{2 c h^3 (m+2) (m+4) \sqrt{1-\frac{2 c (g+h x)}{2 c g-h \left (b-\sqrt{b^2-4 a c}\right )}} \sqrt{1-\frac{2 c (g+h x)}{2 c g-h \left (\sqrt{b^2-4 a c}+b\right )}}}+\frac{f \left (a+b x+c x^2\right )^{3/2} (g+h x)^{m+1}}{c h (m+4)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(f*(g + h*x)^(1 + m)*(a + b*x + c*x^2)^(3/2))/(c*h*(4 + m)) + ((3*c*f*g^2 + f*h*(b*g - a*h)*(1 + m) - c*h*(e*g
- d*h)*(4 + m))*(g + h*x)^(1 + m)*Sqrt[a + b*x + c*x^2]*AppellF1[1 + m, -1/2, -1/2, 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)*(4 + m
)*Sqrt[1 - (2*c*(g + h*x))/(2*c*g - (b - Sqrt[b^2 - 4*a*c])*h)]*Sqrt[1 - (2*c*(g + h*x))/(2*c*g - (b + Sqrt[b^
2 - 4*a*c])*h)]) - ((6*c*f*g - 2*c*e*h*(4 + m) + b*f*h*(5 + 2*m))*(g + h*x)^(2 + m)*Sqrt[a + b*x + c*x^2]*Appe
llF1[2 + m, -1/2, -1/2, 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)])/(2*c*h^3*(2 + m)*(4 + m)*Sqrt[1 - (2*c*(g + h*x))/(2*c*g - (b - Sqrt[b^2 - 4*a*c])
*h)]*Sqrt[1 - (2*c*(g + h*x))/(2*c*g - (b + Sqrt[b^2 - 4*a*c])*h)])

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 \sqrt{a+b x+c x^2} \left (d+e x+f x^2\right ) \, dx &=\frac{f (g+h x)^{1+m} \left (a+b x+c x^2\right )^{3/2}}{c h (4+m)}+\frac{\int (g+h x)^m \left (-\frac{1}{2} h (3 b f g+2 a f h (1+m)-2 c d h (4+m))-\frac{1}{2} h (6 c f g-2 c e h (4+m)+b f h (5+2 m)) x\right ) \sqrt{a+b x+c x^2} \, dx}{c h^2 (4+m)}\\ &=\frac{f (g+h x)^{1+m} \left (a+b x+c x^2\right )^{3/2}}{c h (4+m)}+\frac{\left (3 c f g^2+f h (b g-a h) (1+m)-c h (e g-d h) (4+m)\right ) \int (g+h x)^m \sqrt{a+b x+c x^2} \, dx}{c h^2 (4+m)}-\frac{(6 c f g-2 c e h (4+m)+b f h (5+2 m)) \int (g+h x)^{1+m} \sqrt{a+b x+c x^2} \, dx}{2 c h^2 (4+m)}\\ &=\frac{f (g+h x)^{1+m} \left (a+b x+c x^2\right )^{3/2}}{c h (4+m)}+\frac{\left (\left (3 c f g^2+f h (b g-a h) (1+m)-c h (e g-d h) (4+m)\right ) \sqrt{a+b x+c x^2}\right ) \operatorname{Subst}\left (\int x^m \sqrt{1-\frac{2 c x}{2 c g-\left (b-\sqrt{b^2-4 a c}\right ) h}} \sqrt{1-\frac{2 c x}{2 c g-\left (b+\sqrt{b^2-4 a c}\right ) h}} \, dx,x,g+h x\right )}{c h^3 (4+m) \sqrt{1-\frac{g+h x}{g-\frac{\left (b-\sqrt{b^2-4 a c}\right ) h}{2 c}}} \sqrt{1-\frac{g+h x}{g-\frac{\left (b+\sqrt{b^2-4 a c}\right ) h}{2 c}}}}-\frac{\left ((6 c f g-2 c e h (4+m)+b f h (5+2 m)) \sqrt{a+b x+c x^2}\right ) \operatorname{Subst}\left (\int x^{1+m} \sqrt{1-\frac{2 c x}{2 c g-\left (b-\sqrt{b^2-4 a c}\right ) h}} \sqrt{1-\frac{2 c x}{2 c g-\left (b+\sqrt{b^2-4 a c}\right ) h}} \, dx,x,g+h x\right )}{2 c h^3 (4+m) \sqrt{1-\frac{g+h x}{g-\frac{\left (b-\sqrt{b^2-4 a c}\right ) h}{2 c}}} \sqrt{1-\frac{g+h x}{g-\frac{\left (b+\sqrt{b^2-4 a c}\right ) h}{2 c}}}}\\ &=\frac{f (g+h x)^{1+m} \left (a+b x+c x^2\right )^{3/2}}{c h (4+m)}+\frac{\left (3 c f g^2+f h (b g-a h) (1+m)-c h (e g-d h) (4+m)\right ) (g+h x)^{1+m} \sqrt{a+b x+c x^2} F_1\left (1+m;-\frac{1}{2},-\frac{1}{2};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) (4+m) \sqrt{1-\frac{2 c (g+h x)}{2 c g-\left (b-\sqrt{b^2-4 a c}\right ) h}} \sqrt{1-\frac{2 c (g+h x)}{2 c g-\left (b+\sqrt{b^2-4 a c}\right ) h}}}-\frac{(6 c f g-2 c e h (4+m)+b f h (5+2 m)) (g+h x)^{2+m} \sqrt{a+b x+c x^2} F_1\left (2+m;-\frac{1}{2},-\frac{1}{2};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 )}{2 c h^3 (2+m) (4+m) \sqrt{1-\frac{2 c (g+h x)}{2 c g-\left (b-\sqrt{b^2-4 a c}\right ) h}} \sqrt{1-\frac{2 c (g+h x)}{2 c g-\left (b+\sqrt{b^2-4 a c}\right ) h}}}\\ \end{align*}

Mathematica [F]  time = 1.46647, size = 0, normalized size = 0. $\int (g+h x)^m \sqrt{a+b x+c x^2} \left (d+e x+f x^2\right ) \, dx$

Veriﬁcation is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 1.338, size = 0, normalized size = 0. \begin{align*} \int \left ( hx+g \right ) ^{m} \left ( f{x}^{2}+ex+d \right ) \sqrt{c{x}^{2}+bx+a}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((h*x+g)^m*(f*x^2+e*x+d)*(c*x^2+b*x+a)^(1/2),x)

[Out]

int((h*x+g)^m*(f*x^2+e*x+d)*(c*x^2+b*x+a)^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{c x^{2} + b x + a}{\left (f x^{2} + e x + d\right )}{\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*(f*x^2+e*x+d)*(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{c x^{2} + b x + a}{\left (f x^{2} + e x + d\right )}{\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*(f*x^2+e*x+d)*(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(c*x^2 + b*x + a)*(f*x^2 + e*x + d)*(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*(f*x**2+e*x+d)*(c*x**2+b*x+a)**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{c x^{2} + b x + a}{\left (f x^{2} + e x + d\right )}{\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*(f*x^2+e*x+d)*(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

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