3.10.45 \(\int (d+e x)^m (f+g x)^2 \sqrt {a+b x+c x^2} \, dx\) [945]

Optimal. Leaf size=509 \[ \frac {g^2 (d+e x)^{1+m} \left (a+b x+c x^2\right )^{3/2}}{c e (4+m)}+\frac {\left (e (b d-a e) g^2 (1+m)+c \left (3 d^2 g^2+e^2 f^2 (4+m)-2 d e f g (4+m)\right )\right ) (d+e x)^{1+m} \sqrt {a+b x+c x^2} F_1\left (1+m;-\frac {1}{2},-\frac {1}{2};2+m;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{c e^3 (1+m) (4+m) \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}}}-\frac {g (b e g (5+2 m)+2 c (3 d g-2 e f (4+m))) (d+e x)^{2+m} \sqrt {a+b x+c x^2} F_1\left (2+m;-\frac {1}{2},-\frac {1}{2};3+m;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 c e^3 (2+m) (4+m) \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}}} \]

[Out]

g^2*(e*x+d)^(1+m)*(c*x^2+b*x+a)^(3/2)/c/e/(4+m)+(e*(-a*e+b*d)*g^2*(1+m)+c*(3*d^2*g^2+e^2*f^2*(4+m)-2*d*e*f*g*(
4+m)))*(e*x+d)^(1+m)*AppellF1(1+m,-1/2,-1/2,2+m,2*c*(e*x+d)/(2*c*d-e*(b-(-4*a*c+b^2)^(1/2))),2*c*(e*x+d)/(2*c*
d-e*(b+(-4*a*c+b^2)^(1/2))))*(c*x^2+b*x+a)^(1/2)/c/e^3/(1+m)/(4+m)/(1-2*c*(e*x+d)/(2*c*d-e*(b-(-4*a*c+b^2)^(1/
2))))^(1/2)/(1-2*c*(e*x+d)/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2))))^(1/2)-1/2*g*(b*e*g*(5+2*m)+2*c*(3*d*g-2*e*f*(4+m)
))*(e*x+d)^(2+m)*AppellF1(2+m,-1/2,-1/2,3+m,2*c*(e*x+d)/(2*c*d-e*(b-(-4*a*c+b^2)^(1/2))),2*c*(e*x+d)/(2*c*d-e*
(b+(-4*a*c+b^2)^(1/2))))*(c*x^2+b*x+a)^(1/2)/c/e^3/(2+m)/(4+m)/(1-2*c*(e*x+d)/(2*c*d-e*(b-(-4*a*c+b^2)^(1/2)))
)^(1/2)/(1-2*c*(e*x+d)/(2*c*d-e*(b+(-4*a*c+b^2)^(1/2))))^(1/2)

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Rubi [A]
time = 0.57, antiderivative size = 506, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {1667, 857, 773, 138} \begin {gather*} \frac {\sqrt {a+b x+c x^2} (d+e x)^{m+1} F_1\left (m+1;-\frac {1}{2},-\frac {1}{2};m+2;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right ) \left (g^2 (b d-a e)+\frac {c \left (3 d^2 g^2-2 d e f g (m+4)+e^2 f^2 (m+4)\right )}{e (m+1)}\right )}{c e^2 (m+4) \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}-\frac {g \sqrt {a+b x+c x^2} (d+e x)^{m+2} (b e g (2 m+5)+6 c d g-4 c e f (m+4)) F_1\left (m+2;-\frac {1}{2},-\frac {1}{2};m+3;\frac {2 c (d+e x)}{2 c d-\left (b-\sqrt {b^2-4 a c}\right ) e},\frac {2 c (d+e x)}{2 c d-\left (b+\sqrt {b^2-4 a c}\right ) e}\right )}{2 c e^3 (m+2) (m+4) \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (b-\sqrt {b^2-4 a c}\right )}} \sqrt {1-\frac {2 c (d+e x)}{2 c d-e \left (\sqrt {b^2-4 a c}+b\right )}}}+\frac {g^2 \left (a+b x+c x^2\right )^{3/2} (d+e x)^{m+1}}{c e (m+4)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 138

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[c^n*e^p*((b*x)^(m +
1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 773

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 857

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 1667

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

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Mathematica [F]
time = 1.43, size = 0, normalized size = 0.00 \begin {gather*} \int (d+e x)^m (f+g x)^2 \sqrt {a+b x+c x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \left (e x +d \right )^{m} \left (g x +f \right )^{2} \sqrt {c \,x^{2}+b x +a}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m*(g*x+f)^2*(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m*(g*x+f)^2*(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (d + e x\right )^{m} \left (f + g x\right )^{2} \sqrt {a + b x + c x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m*(g*x+f)^2*(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (f+g\,x\right )}^2\,{\left (d+e\,x\right )}^m\,\sqrt {c\,x^2+b\,x+a} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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