3.1549 \(\int (b+2 c x) (d+e x) \sqrt{a+b x+c x^2} \, dx\)

Optimal. Leaf size=122 \[ -\frac{e \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{64 c^{5/2}}+\frac{e \left (b^2-4 a c\right ) (b+2 c x) \sqrt{a+b x+c x^2}}{32 c^2}+\frac{\left (a+b x+c x^2\right )^{3/2} (-b e+8 c d+6 c e x)}{12 c} \]

[Out]

((b^2 - 4*a*c)*e*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(32*c^2) + ((8*c*d - b*e + 6
*c*e*x)*(a + b*x + c*x^2)^(3/2))/(12*c) - ((b^2 - 4*a*c)^2*e*ArcTanh[(b + 2*c*x)
/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(64*c^(5/2))

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Rubi [A]  time = 0.217473, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154 \[ -\frac{e \left (b^2-4 a c\right )^2 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{64 c^{5/2}}+\frac{e \left (b^2-4 a c\right ) (b+2 c x) \sqrt{a+b x+c x^2}}{32 c^2}+\frac{\left (a+b x+c x^2\right )^{3/2} (-b e+8 c d+6 c e x)}{12 c} \]

Antiderivative was successfully verified.

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

[Out]

((b^2 - 4*a*c)*e*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(32*c^2) + ((8*c*d - b*e + 6
*c*e*x)*(a + b*x + c*x^2)^(3/2))/(12*c) - ((b^2 - 4*a*c)^2*e*ArcTanh[(b + 2*c*x)
/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(64*c^(5/2))

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Rubi in Sympy [A]  time = 20.4766, size = 114, normalized size = 0.93 \[ - \frac{\left (a + b x + c x^{2}\right )^{\frac{3}{2}} \left (b e - 8 c d - 6 c e x\right )}{12 c} + \frac{e \left (b + 2 c x\right ) \left (- 4 a c + b^{2}\right ) \sqrt{a + b x + c x^{2}}}{32 c^{2}} - \frac{e \left (- 4 a c + b^{2}\right )^{2} \operatorname{atanh}{\left (\frac{b + 2 c x}{2 \sqrt{c} \sqrt{a + b x + c x^{2}}} \right )}}{64 c^{\frac{5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((2*c*x+b)*(e*x+d)*(c*x**2+b*x+a)**(1/2),x)

[Out]

-(a + b*x + c*x**2)**(3/2)*(b*e - 8*c*d - 6*c*e*x)/(12*c) + e*(b + 2*c*x)*(-4*a*
c + b**2)*sqrt(a + b*x + c*x**2)/(32*c**2) - e*(-4*a*c + b**2)**2*atanh((b + 2*c
*x)/(2*sqrt(c)*sqrt(a + b*x + c*x**2)))/(64*c**(5/2))

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Mathematica [A]  time = 0.269724, size = 132, normalized size = 1.08 \[ \frac{\sqrt{a+x (b+c x)} \left (4 a c (-5 b e+16 c d+6 c e x)+3 b^3 e-2 b^2 c e x+8 b c^2 x (8 d+5 e x)+16 c^3 x^2 (4 d+3 e x)\right )}{96 c^2}-\frac{e \left (b^2-4 a c\right )^2 \log \left (2 \sqrt{c} \sqrt{a+x (b+c x)}+b+2 c x\right )}{64 c^{5/2}} \]

Antiderivative was successfully verified.

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

[Out]

(Sqrt[a + x*(b + c*x)]*(3*b^3*e - 2*b^2*c*e*x + 16*c^3*x^2*(4*d + 3*e*x) + 8*b*c
^2*x*(8*d + 5*e*x) + 4*a*c*(16*c*d - 5*b*e + 6*c*e*x)))/(96*c^2) - ((b^2 - 4*a*c
)^2*e*Log[b + 2*c*x + 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/(64*c^(5/2))

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Maple [B]  time = 0.01, size = 235, normalized size = 1.9 \[ -{\frac{be}{12\,c} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{2\,d}{3} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{{b}^{2}ex}{16\,c}\sqrt{c{x}^{2}+bx+a}}+{\frac{{b}^{3}e}{32\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{ae{b}^{2}}{8}\ln \left ({1 \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}-{\frac{{b}^{4}e}{64}\ln \left ({1 \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{5}{2}}}}+{\frac{ex}{2} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}-{\frac{aex}{4}\sqrt{c{x}^{2}+bx+a}}-{\frac{bea}{8\,c}\sqrt{c{x}^{2}+bx+a}}-{\frac{e{a}^{2}}{4}\ln \left ({1 \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){\frac{1}{\sqrt{c}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-1/12*(c*x^2+b*x+a)^(3/2)/c*b*e+2/3*d*(c*x^2+b*x+a)^(3/2)+1/16*b^2/c*(c*x^2+b*x+
a)^(1/2)*x*e+1/32*b^3/c^2*(c*x^2+b*x+a)^(1/2)*e+1/8*b^2/c^(3/2)*ln((1/2*b+c*x)/c
^(1/2)+(c*x^2+b*x+a)^(1/2))*a*e-1/64*b^4/c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b
*x+a)^(1/2))*e+1/2*e*x*(c*x^2+b*x+a)^(3/2)-1/4*e*a*(c*x^2+b*x+a)^(1/2)*x-1/8/c*e
*a*(c*x^2+b*x+a)^(1/2)*b-1/4/c^(1/2)*e*a^2*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^
(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \text{Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(sqrt(c*x^2 + b*x + a)*(2*c*x + b)*(e*x + d),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.313735, size = 1, normalized size = 0.01 \[ \left [\frac{3 \,{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} e \log \left (4 \,{\left (2 \, c^{2} x + b c\right )} \sqrt{c x^{2} + b x + a} -{\left (8 \, c^{2} x^{2} + 8 \, b c x + b^{2} + 4 \, a c\right )} \sqrt{c}\right ) + 4 \,{\left (48 \, c^{3} e x^{3} + 64 \, a c^{2} d + 8 \,{\left (8 \, c^{3} d + 5 \, b c^{2} e\right )} x^{2} +{\left (3 \, b^{3} - 20 \, a b c\right )} e + 2 \,{\left (32 \, b c^{2} d -{\left (b^{2} c - 12 \, a c^{2}\right )} e\right )} x\right )} \sqrt{c x^{2} + b x + a} \sqrt{c}}{384 \, c^{\frac{5}{2}}}, -\frac{3 \,{\left (b^{4} - 8 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} e \arctan \left (\frac{{\left (2 \, c x + b\right )} \sqrt{-c}}{2 \, \sqrt{c x^{2} + b x + a} c}\right ) - 2 \,{\left (48 \, c^{3} e x^{3} + 64 \, a c^{2} d + 8 \,{\left (8 \, c^{3} d + 5 \, b c^{2} e\right )} x^{2} +{\left (3 \, b^{3} - 20 \, a b c\right )} e + 2 \,{\left (32 \, b c^{2} d -{\left (b^{2} c - 12 \, a c^{2}\right )} e\right )} x\right )} \sqrt{c x^{2} + b x + a} \sqrt{-c}}{192 \, \sqrt{-c} c^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(sqrt(c*x^2 + b*x + a)*(2*c*x + b)*(e*x + d),x, algorithm="fricas")

[Out]

[1/384*(3*(b^4 - 8*a*b^2*c + 16*a^2*c^2)*e*log(4*(2*c^2*x + b*c)*sqrt(c*x^2 + b*
x + a) - (8*c^2*x^2 + 8*b*c*x + b^2 + 4*a*c)*sqrt(c)) + 4*(48*c^3*e*x^3 + 64*a*c
^2*d + 8*(8*c^3*d + 5*b*c^2*e)*x^2 + (3*b^3 - 20*a*b*c)*e + 2*(32*b*c^2*d - (b^2
*c - 12*a*c^2)*e)*x)*sqrt(c*x^2 + b*x + a)*sqrt(c))/c^(5/2), -1/192*(3*(b^4 - 8*
a*b^2*c + 16*a^2*c^2)*e*arctan(1/2*(2*c*x + b)*sqrt(-c)/(sqrt(c*x^2 + b*x + a)*c
)) - 2*(48*c^3*e*x^3 + 64*a*c^2*d + 8*(8*c^3*d + 5*b*c^2*e)*x^2 + (3*b^3 - 20*a*
b*c)*e + 2*(32*b*c^2*d - (b^2*c - 12*a*c^2)*e)*x)*sqrt(c*x^2 + b*x + a)*sqrt(-c)
)/(sqrt(-c)*c^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((2*c*x+b)*(e*x+d)*(c*x**2+b*x+a)**(1/2),x)

[Out]

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

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GIAC/XCAS [A]  time = 0.282369, size = 230, normalized size = 1.89 \[ \frac{1}{96} \, \sqrt{c x^{2} + b x + a}{\left (2 \,{\left (4 \,{\left (6 \, c x e + \frac{8 \, c^{4} d + 5 \, b c^{3} e}{c^{3}}\right )} x + \frac{32 \, b c^{3} d - b^{2} c^{2} e + 12 \, a c^{3} e}{c^{3}}\right )} x + \frac{64 \, a c^{3} d + 3 \, b^{3} c e - 20 \, a b c^{2} e}{c^{3}}\right )} + \frac{{\left (b^{4} e - 8 \, a b^{2} c e + 16 \, a^{2} c^{2} e\right )}{\rm ln}\left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{64 \, c^{\frac{5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(sqrt(c*x^2 + b*x + a)*(2*c*x + b)*(e*x + d),x, algorithm="giac")

[Out]

1/96*sqrt(c*x^2 + b*x + a)*(2*(4*(6*c*x*e + (8*c^4*d + 5*b*c^3*e)/c^3)*x + (32*b
*c^3*d - b^2*c^2*e + 12*a*c^3*e)/c^3)*x + (64*a*c^3*d + 3*b^3*c*e - 20*a*b*c^2*e
)/c^3) + 1/64*(b^4*e - 8*a*b^2*c*e + 16*a^2*c^2*e)*ln(abs(-2*(sqrt(c)*x - sqrt(c
*x^2 + b*x + a))*sqrt(c) - b))/c^(5/2)