3.101 \(\int \sqrt {a+b x+c x^2} (d+e x+f x^2) \, dx\)

Optimal. Leaf size=175 \[ -\frac {\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (-4 c (a f+2 b e)+5 b^2 f+16 c^2 d\right )}{128 c^{7/2}}+\frac {(b+2 c x) \sqrt {a+b x+c x^2} \left (-4 a c f+5 b^2 f-8 b c e+16 c^2 d\right )}{64 c^3}+\frac {\left (a+b x+c x^2\right )^{3/2} (8 c e-5 b f)}{24 c^2}+\frac {f x \left (a+b x+c x^2\right )^{3/2}}{4 c} \]

[Out]

1/24*(-5*b*f+8*c*e)*(c*x^2+b*x+a)^(3/2)/c^2+1/4*f*x*(c*x^2+b*x+a)^(3/2)/c-1/128*(-4*a*c+b^2)*(16*c^2*d+5*b^2*f
-4*c*(a*f+2*b*e))*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(7/2)+1/64*(-4*a*c*f+5*b^2*f-8*b*c*e+16
*c^2*d)*(2*c*x+b)*(c*x^2+b*x+a)^(1/2)/c^3

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Rubi [A]  time = 0.16, antiderivative size = 175, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1661, 640, 612, 621, 206} \[ \frac {(b+2 c x) \sqrt {a+b x+c x^2} \left (-4 a c f+5 b^2 f-8 b c e+16 c^2 d\right )}{64 c^3}-\frac {\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (-4 c (a f+2 b e)+5 b^2 f+16 c^2 d\right )}{128 c^{7/2}}+\frac {\left (a+b x+c x^2\right )^{3/2} (8 c e-5 b f)}{24 c^2}+\frac {f x \left (a+b x+c x^2\right )^{3/2}}{4 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((16*c^2*d - 8*b*c*e + 5*b^2*f - 4*a*c*f)*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(64*c^3) + ((8*c*e - 5*b*f)*(a +
b*x + c*x^2)^(3/2))/(24*c^2) + (f*x*(a + b*x + c*x^2)^(3/2))/(4*c) - ((b^2 - 4*a*c)*(16*c^2*d + 5*b^2*f - 4*c*
(2*b*e + a*f))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(128*c^(7/2))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rule 1661

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expo
n[Pq, x]]}, Simp[(e*x^(q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*(q + 2*p + 1)), x] + Dist[1/(c*(q + 2*p + 1)), Int
[(a + b*x + c*x^2)^p*ExpandToSum[c*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + p)*x^(q - 1) - c*e*(q +
2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rubi steps

\begin {align*} \int \sqrt {a+b x+c x^2} \left (d+e x+f x^2\right ) \, dx &=\frac {f x \left (a+b x+c x^2\right )^{3/2}}{4 c}+\frac {\int \left (4 c d-a f+\frac {1}{2} (8 c e-5 b f) x\right ) \sqrt {a+b x+c x^2} \, dx}{4 c}\\ &=\frac {(8 c e-5 b f) \left (a+b x+c x^2\right )^{3/2}}{24 c^2}+\frac {f x \left (a+b x+c x^2\right )^{3/2}}{4 c}+\frac {\left (16 c^2 d-8 b c e+5 b^2 f-4 a c f\right ) \int \sqrt {a+b x+c x^2} \, dx}{16 c^2}\\ &=\frac {\left (16 c^2 d-8 b c e+5 b^2 f-4 a c f\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{64 c^3}+\frac {(8 c e-5 b f) \left (a+b x+c x^2\right )^{3/2}}{24 c^2}+\frac {f x \left (a+b x+c x^2\right )^{3/2}}{4 c}-\frac {\left (\left (b^2-4 a c\right ) \left (16 c^2 d+5 b^2 f-4 c (2 b e+a f)\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{128 c^3}\\ &=\frac {\left (16 c^2 d-8 b c e+5 b^2 f-4 a c f\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{64 c^3}+\frac {(8 c e-5 b f) \left (a+b x+c x^2\right )^{3/2}}{24 c^2}+\frac {f x \left (a+b x+c x^2\right )^{3/2}}{4 c}-\frac {\left (\left (b^2-4 a c\right ) \left (16 c^2 d+5 b^2 f-4 c (2 b e+a f)\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{64 c^3}\\ &=\frac {\left (16 c^2 d-8 b c e+5 b^2 f-4 a c f\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{64 c^3}+\frac {(8 c e-5 b f) \left (a+b x+c x^2\right )^{3/2}}{24 c^2}+\frac {f x \left (a+b x+c x^2\right )^{3/2}}{4 c}-\frac {\left (b^2-4 a c\right ) \left (16 c^2 d+5 b^2 f-4 c (2 b e+a f)\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 173, normalized size = 0.99 \[ \frac {2 \sqrt {c} \sqrt {a+x (b+c x)} \left (4 b c \left (2 c \left (6 d+2 e x+f x^2\right )-13 a f\right )+8 c^2 \left (a (8 e+3 f x)+2 c x \left (6 d+4 e x+3 f x^2\right )\right )+15 b^3 f-2 b^2 c (12 e+5 f x)\right )-3 \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right ) \left (-4 c (a f+2 b e)+5 b^2 f+16 c^2 d\right )}{384 c^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c]*Sqrt[a + x*(b + c*x)]*(15*b^3*f - 2*b^2*c*(12*e + 5*f*x) + 4*b*c*(-13*a*f + 2*c*(6*d + 2*e*x + f*x^
2)) + 8*c^2*(a*(8*e + 3*f*x) + 2*c*x*(6*d + 4*e*x + 3*f*x^2))) - 3*(b^2 - 4*a*c)*(16*c^2*d + 5*b^2*f - 4*c*(2*
b*e + a*f))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(384*c^(7/2))

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fricas [A]  time = 0.77, size = 465, normalized size = 2.66 \[ \left [\frac {3 \, {\left (16 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d - 8 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} e + {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} f\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} + 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (48 \, c^{4} f x^{3} + 48 \, b c^{3} d + 8 \, {\left (8 \, c^{4} e + b c^{3} f\right )} x^{2} - 8 \, {\left (3 \, b^{2} c^{2} - 8 \, a c^{3}\right )} e + {\left (15 \, b^{3} c - 52 \, a b c^{2}\right )} f + 2 \, {\left (48 \, c^{4} d + 8 \, b c^{3} e - {\left (5 \, b^{2} c^{2} - 12 \, a c^{3}\right )} f\right )} x\right )} \sqrt {c x^{2} + b x + a}}{768 \, c^{4}}, \frac {3 \, {\left (16 \, {\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d - 8 \, {\left (b^{3} c - 4 \, a b c^{2}\right )} e + {\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} f\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, {\left (48 \, c^{4} f x^{3} + 48 \, b c^{3} d + 8 \, {\left (8 \, c^{4} e + b c^{3} f\right )} x^{2} - 8 \, {\left (3 \, b^{2} c^{2} - 8 \, a c^{3}\right )} e + {\left (15 \, b^{3} c - 52 \, a b c^{2}\right )} f + 2 \, {\left (48 \, c^{4} d + 8 \, b c^{3} e - {\left (5 \, b^{2} c^{2} - 12 \, a c^{3}\right )} f\right )} x\right )} \sqrt {c x^{2} + b x + a}}{384 \, c^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/768*(3*(16*(b^2*c^2 - 4*a*c^3)*d - 8*(b^3*c - 4*a*b*c^2)*e + (5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*f)*sqrt(c)*l
og(-8*c^2*x^2 - 8*b*c*x - b^2 + 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 4*(48*c^4*f*x^3 + 48*b*
c^3*d + 8*(8*c^4*e + b*c^3*f)*x^2 - 8*(3*b^2*c^2 - 8*a*c^3)*e + (15*b^3*c - 52*a*b*c^2)*f + 2*(48*c^4*d + 8*b*
c^3*e - (5*b^2*c^2 - 12*a*c^3)*f)*x)*sqrt(c*x^2 + b*x + a))/c^4, 1/384*(3*(16*(b^2*c^2 - 4*a*c^3)*d - 8*(b^3*c
- 4*a*b*c^2)*e + (5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*f)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*s
qrt(-c)/(c^2*x^2 + b*c*x + a*c)) + 2*(48*c^4*f*x^3 + 48*b*c^3*d + 8*(8*c^4*e + b*c^3*f)*x^2 - 8*(3*b^2*c^2 - 8
*a*c^3)*e + (15*b^3*c - 52*a*b*c^2)*f + 2*(48*c^4*d + 8*b*c^3*e - (5*b^2*c^2 - 12*a*c^3)*f)*x)*sqrt(c*x^2 + b*
x + a))/c^4]

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giac [A]  time = 0.30, size = 212, normalized size = 1.21 \[ \frac {1}{192} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (6 \, f x + \frac {b c^{2} f + 8 \, c^{3} e}{c^{3}}\right )} x + \frac {48 \, c^{3} d - 5 \, b^{2} c f + 12 \, a c^{2} f + 8 \, b c^{2} e}{c^{3}}\right )} x + \frac {48 \, b c^{2} d + 15 \, b^{3} f - 52 \, a b c f - 24 \, b^{2} c e + 64 \, a c^{2} e}{c^{3}}\right )} + \frac {{\left (16 \, b^{2} c^{2} d - 64 \, a c^{3} d + 5 \, b^{4} f - 24 \, a b^{2} c f + 16 \, a^{2} c^{2} f - 8 \, b^{3} c e + 32 \, a b c^{2} e\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{128 \, c^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*sqrt(c*x^2 + b*x + a)*(2*(4*(6*f*x + (b*c^2*f + 8*c^3*e)/c^3)*x + (48*c^3*d - 5*b^2*c*f + 12*a*c^2*f + 8
*b*c^2*e)/c^3)*x + (48*b*c^2*d + 15*b^3*f - 52*a*b*c*f - 24*b^2*c*e + 64*a*c^2*e)/c^3) + 1/128*(16*b^2*c^2*d -
64*a*c^3*d + 5*b^4*f - 24*a*b^2*c*f + 16*a^2*c^2*f - 8*b^3*c*e + 32*a*b*c^2*e)*log(abs(-2*(sqrt(c)*x - sqrt(c
*x^2 + b*x + a))*sqrt(c) - b))/c^(7/2)

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maple [B]  time = 0.01, size = 453, normalized size = 2.59 \[ -\frac {a^{2} f \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}+\frac {3 a \,b^{2} f \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {5}{2}}}-\frac {a b e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{4 c^{\frac {3}{2}}}+\frac {a d \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 \sqrt {c}}-\frac {5 b^{4} f \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{128 c^{\frac {7}{2}}}+\frac {b^{3} e \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {5}{2}}}-\frac {b^{2} d \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}-\frac {\sqrt {c \,x^{2}+b x +a}\, a f x}{8 c}+\frac {5 \sqrt {c \,x^{2}+b x +a}\, b^{2} f x}{32 c^{2}}-\frac {\sqrt {c \,x^{2}+b x +a}\, b e x}{4 c}+\frac {\sqrt {c \,x^{2}+b x +a}\, d x}{2}-\frac {\sqrt {c \,x^{2}+b x +a}\, a b f}{16 c^{2}}+\frac {5 \sqrt {c \,x^{2}+b x +a}\, b^{3} f}{64 c^{3}}-\frac {\sqrt {c \,x^{2}+b x +a}\, b^{2} e}{8 c^{2}}+\frac {\sqrt {c \,x^{2}+b x +a}\, b d}{4 c}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} f x}{4 c}-\frac {5 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} b f}{24 c^{2}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} e}{3 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*f*x*(c*x^2+b*x+a)^(3/2)/c-5/24*f*b/c^2*(c*x^2+b*x+a)^(3/2)+5/32*f*b^2/c^2*(c*x^2+b*x+a)^(1/2)*x+5/64*f*b^3
/c^3*(c*x^2+b*x+a)^(1/2)+3/16*f*b^2/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a-5/128*f*b^4/c^(7/2)*
ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-1/8*f*a/c*(c*x^2+b*x+a)^(1/2)*x-1/16*f*a/c^2*(c*x^2+b*x+a)^(1/2)*b
-1/8*f*a^2/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+1/3*e*(c*x^2+b*x+a)^(3/2)/c-1/4*e*b/c*(c*x^2+b*
x+a)^(1/2)*x-1/8*e*b^2/c^2*(c*x^2+b*x+a)^(1/2)-1/4*e*b/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a+1
/16*e*b^3/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+1/2*d*(c*x^2+b*x+a)^(1/2)*x+1/4*d/c*(c*x^2+b*x+a
)^(1/2)*b+1/2*d/c^(1/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a-1/8*d/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*
x^2+b*x+a)^(1/2))*b^2

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive, negative or zero?

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mupad [B]  time = 3.91, size = 320, normalized size = 1.83 \[ d\,\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}-\frac {a\,f\,\left (\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )}{4\,c}+\frac {d\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}+\frac {e\,\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}-\frac {5\,b\,f\,\left (\frac {\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}+\frac {\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2}\right )}{8\,c}+\frac {e\,\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2}+\frac {f\,x\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{4\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

d*(x/2 + b/(4*c))*(a + b*x + c*x^2)^(1/2) - (a*f*((x/2 + b/(4*c))*(a + b*x + c*x^2)^(1/2) + (log((b/2 + c*x)/c
^(1/2) + (a + b*x + c*x^2)^(1/2))*(a*c - b^2/4))/(2*c^(3/2))))/(4*c) + (d*log((b/2 + c*x)/c^(1/2) + (a + b*x +
c*x^2)^(1/2))*(a*c - b^2/4))/(2*c^(3/2)) + (e*log((b + 2*c*x)/c^(1/2) + 2*(a + b*x + c*x^2)^(1/2))*(b^3 - 4*a
*b*c))/(16*c^(5/2)) - (5*b*f*((log((b + 2*c*x)/c^(1/2) + 2*(a + b*x + c*x^2)^(1/2))*(b^3 - 4*a*b*c))/(16*c^(5/
2)) + ((8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*(a + b*x + c*x^2)^(1/2))/(24*c^2)))/(8*c) + (e*(8*c*(a + c*x^2) - 3
*b^2 + 2*b*c*x)*(a + b*x + c*x^2)^(1/2))/(24*c^2) + (f*x*(a + b*x + c*x^2)^(3/2))/(4*c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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