### 3.189 $$\int \sqrt{a+b x+c x^2} (d+e x+f x^2) \, dx$$

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

[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))

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Rubi [A]  time = 0.166976, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.2, 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 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]

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 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 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])

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*}

Mathematica [A]  time = 0.286353, size = 173, normalized size = 0.99 $\frac{2 \sqrt{c} \sqrt{a+x (b+c x)} \left (4 b c \left (2 c \left (6 d+2 e x+f x^2\right )-13 a f\right )+8 c^2 \left (a (8 e+3 f x)+2 c x \left (6 d+4 e x+3 f x^2\right )\right )-2 b^2 c (12 e+5 f x)+15 b^3 f\right )-3 \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right ) \left (-4 c (a f+2 b e)+5 b^2 f+16 c^2 d\right )}{384 c^{7/2}}$

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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Maple [B]  time = 0.052, size = 453, normalized size = 2.6 \begin{align*}{\frac{fx}{4\,c} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}-{\frac{5\,bf}{24\,{c}^{2}} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{b}^{2}fx}{32\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{5\,f{b}^{3}}{64\,{c}^{3}}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,{b}^{2}fa}{16}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{5}{2}}}}-{\frac{5\,f{b}^{4}}{128}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{7}{2}}}}-{\frac{afx}{8\,c}\sqrt{c{x}^{2}+bx+a}}-{\frac{abf}{16\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}-{\frac{{a}^{2}f}{8}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}+{\frac{e}{3\,c} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}-{\frac{bxe}{4\,c}\sqrt{c{x}^{2}+bx+a}}-{\frac{{b}^{2}e}{8\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}-{\frac{aeb}{4}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}+{\frac{e{b}^{3}}{16}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{5}{2}}}}+{\frac{dx}{2}\sqrt{c{x}^{2}+bx+a}}+{\frac{bd}{4\,c}\sqrt{c{x}^{2}+bx+a}}+{\frac{ad}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){\frac{1}{\sqrt{c}}}}-{\frac{{b}^{2}d}{8}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [A]  time = 1.73372, size = 1061, normalized size = 6.06 \begin{align*} \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 ] \end{align*}

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

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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Giac [A]  time = 1.24786, size = 286, normalized size = 1.63 \begin{align*} \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}}} \end{align*}

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)