3.1573 \(\int \frac{(b+2 c x) (d+e x)^3}{\sqrt{a+b x+c x^2}} \, dx\)

Optimal. Leaf size=245 \[ \frac{3 e \left (b^2-4 a c\right ) \left (-4 c e (a e+4 b d)+5 b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{64 c^{7/2}}+\frac{\sqrt{a+b x+c x^2} \left (2 c e x \left (-4 c e (3 a e+2 b d)+5 b^2 e^2+8 c^2 d^2\right )-8 c^2 d e (16 a e+5 b d)+4 b c e^2 (13 a e+12 b d)-15 b^3 e^3+32 c^3 d^3\right )}{32 c^3}+\frac{1}{2} (d+e x)^3 \sqrt{a+b x+c x^2}+\frac{(d+e x)^2 \sqrt{a+b x+c x^2} (2 c d-b e)}{4 c} \]

[Out]

((2*c*d - b*e)*(d + e*x)^2*Sqrt[a + b*x + c*x^2])/(4*c) + ((d + e*x)^3*Sqrt[a +
b*x + c*x^2])/2 + ((32*c^3*d^3 - 15*b^3*e^3 + 4*b*c*e^2*(12*b*d + 13*a*e) - 8*c^
2*d*e*(5*b*d + 16*a*e) + 2*c*e*(8*c^2*d^2 + 5*b^2*e^2 - 4*c*e*(2*b*d + 3*a*e))*x
)*Sqrt[a + b*x + c*x^2])/(32*c^3) + (3*(b^2 - 4*a*c)*e*(16*c^2*d^2 + 5*b^2*e^2 -
4*c*e*(4*b*d + a*e))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(6
4*c^(7/2))

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Rubi [A]  time = 0.803074, antiderivative size = 245, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143 \[ \frac{3 e \left (b^2-4 a c\right ) \left (-4 c e (a e+4 b d)+5 b^2 e^2+16 c^2 d^2\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{64 c^{7/2}}+\frac{\sqrt{a+b x+c x^2} \left (2 c e x \left (-4 c e (3 a e+2 b d)+5 b^2 e^2+8 c^2 d^2\right )-8 c^2 d e (16 a e+5 b d)+4 b c e^2 (13 a e+12 b d)-15 b^3 e^3+32 c^3 d^3\right )}{32 c^3}+\frac{1}{2} (d+e x)^3 \sqrt{a+b x+c x^2}+\frac{(d+e x)^2 \sqrt{a+b x+c x^2} (2 c d-b e)}{4 c} \]

Antiderivative was successfully verified.

[In]  Int[((b + 2*c*x)*(d + e*x)^3)/Sqrt[a + b*x + c*x^2],x]

[Out]

((2*c*d - b*e)*(d + e*x)^2*Sqrt[a + b*x + c*x^2])/(4*c) + ((d + e*x)^3*Sqrt[a +
b*x + c*x^2])/2 + ((32*c^3*d^3 - 15*b^3*e^3 + 4*b*c*e^2*(12*b*d + 13*a*e) - 8*c^
2*d*e*(5*b*d + 16*a*e) + 2*c*e*(8*c^2*d^2 + 5*b^2*e^2 - 4*c*e*(2*b*d + 3*a*e))*x
)*Sqrt[a + b*x + c*x^2])/(32*c^3) + (3*(b^2 - 4*a*c)*e*(16*c^2*d^2 + 5*b^2*e^2 -
4*c*e*(4*b*d + a*e))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(6
4*c^(7/2))

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Rubi in Sympy [A]  time = 108.421, size = 260, normalized size = 1.06 \[ \frac{\left (d + e x\right )^{3} \sqrt{a + b x + c x^{2}}}{2} - \frac{\left (d + e x\right )^{2} \left (b e - 2 c d\right ) \sqrt{a + b x + c x^{2}}}{4 c} - \frac{\sqrt{a + b x + c x^{2}} \left (- 39 a b c e^{3} + 96 a c^{2} d e^{2} + \frac{45 b^{3} e^{3}}{4} - 36 b^{2} c d e^{2} + 30 b c^{2} d^{2} e - 24 c^{3} d^{3} - \frac{3 c e x \left (- 12 a c e^{2} + 5 b^{2} e^{2} - 8 b c d e + 8 c^{2} d^{2}\right )}{2}\right )}{24 c^{3}} + \frac{3 e \left (- 4 a c + b^{2}\right ) \left (- 4 a c e^{2} + 5 b^{2} e^{2} - 16 b c d e + 16 c^{2} d^{2}\right ) \operatorname{atanh}{\left (\frac{b + 2 c x}{2 \sqrt{c} \sqrt{a + b x + c x^{2}}} \right )}}{64 c^{\frac{7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((2*c*x+b)*(e*x+d)**3/(c*x**2+b*x+a)**(1/2),x)

[Out]

(d + e*x)**3*sqrt(a + b*x + c*x**2)/2 - (d + e*x)**2*(b*e - 2*c*d)*sqrt(a + b*x
+ c*x**2)/(4*c) - sqrt(a + b*x + c*x**2)*(-39*a*b*c*e**3 + 96*a*c**2*d*e**2 + 45
*b**3*e**3/4 - 36*b**2*c*d*e**2 + 30*b*c**2*d**2*e - 24*c**3*d**3 - 3*c*e*x*(-12
*a*c*e**2 + 5*b**2*e**2 - 8*b*c*d*e + 8*c**2*d**2)/2)/(24*c**3) + 3*e*(-4*a*c +
b**2)*(-4*a*c*e**2 + 5*b**2*e**2 - 16*b*c*d*e + 16*c**2*d**2)*atanh((b + 2*c*x)/
(2*sqrt(c)*sqrt(a + b*x + c*x**2)))/(64*c**(7/2))

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Mathematica [A]  time = 0.333972, size = 199, normalized size = 0.81 \[ \frac{2 \sqrt{c} \sqrt{a+x (b+c x)} \left (-8 c^2 e \left (a e (16 d+3 e x)+b \left (6 d^2+4 d e x+e^2 x^2\right )\right )+2 b c e^2 (26 a e+24 b d+5 b e x)-15 b^3 e^3+16 c^3 \left (4 d^3+6 d^2 e x+4 d e^2 x^2+e^3 x^3\right )\right )+3 e \left (b^2-4 a c\right ) \left (-4 c e (a e+4 b d)+5 b^2 e^2+16 c^2 d^2\right ) \log \left (2 \sqrt{c} \sqrt{a+x (b+c x)}+b+2 c x\right )}{64 c^{7/2}} \]

Antiderivative was successfully verified.

[In]  Integrate[((b + 2*c*x)*(d + e*x)^3)/Sqrt[a + b*x + c*x^2],x]

[Out]

(2*Sqrt[c]*Sqrt[a + x*(b + c*x)]*(-15*b^3*e^3 + 2*b*c*e^2*(24*b*d + 26*a*e + 5*b
*e*x) + 16*c^3*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3) - 8*c^2*e*(a*e*(16*d
+ 3*e*x) + b*(6*d^2 + 4*d*e*x + e^2*x^2))) + 3*(b^2 - 4*a*c)*e*(16*c^2*d^2 + 5*b
^2*e^2 - 4*c*e*(4*b*d + a*e))*Log[b + 2*c*x + 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/
(64*c^(7/2))

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Maple [B]  time = 0.017, size = 539, normalized size = 2.2 \[ -{\frac{15\,{b}^{3}{e}^{3}}{32\,{c}^{3}}\sqrt{c{x}^{2}+bx+a}}+{\frac{15\,{b}^{4}{e}^{3}}{64}\ln \left ({1 \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{7}{2}}}}-{\frac{b{x}^{2}{e}^{3}}{4\,c}\sqrt{c{x}^{2}+bx+a}}+{\frac{5\,{b}^{2}x{e}^{3}}{16\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,{b}^{2}d{e}^{2}}{2\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}-{\frac{3\,d{b}^{3}{e}^{2}}{4}\ln \left ({1 \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{5}{2}}}}+2\,{d}^{3}\sqrt{c{x}^{2}+bx+a}-{\frac{bxd{e}^{2}}{c}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,{b}^{2}{d}^{2}e}{4}\ln \left ({1 \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}+2\,d{e}^{2}{x}^{2}\sqrt{c{x}^{2}+bx+a}+3\,{d}^{2}ex\sqrt{c{x}^{2}+bx+a}+{\frac{{e}^{3}{x}^{3}}{2}\sqrt{c{x}^{2}+bx+a}}-{\frac{9\,a{b}^{2}{e}^{3}}{8}\ln \left ({1 \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{5}{2}}}}+{\frac{13\,ab{e}^{3}}{8\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}-4\,{\frac{a\sqrt{c{x}^{2}+bx+a}d{e}^{2}}{c}}+{\frac{3\,{a}^{2}{e}^{3}}{4}\ln \left ({1 \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}+3\,{\frac{d{e}^{2}ab}{{c}^{3/2}}\ln \left ({\frac{b/2+cx}{\sqrt{c}}}+\sqrt{c{x}^{2}+bx+a} \right ) }-{\frac{3\,b{d}^{2}e}{2\,c}\sqrt{c{x}^{2}+bx+a}}-3\,{\frac{a{d}^{2}e}{\sqrt{c}}\ln \left ({\frac{b/2+cx}{\sqrt{c}}}+\sqrt{c{x}^{2}+bx+a} \right ) }-{\frac{3\,a{e}^{3}x}{4\,c}\sqrt{c{x}^{2}+bx+a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((2*c*x+b)*(e*x+d)^3/(c*x^2+b*x+a)^(1/2),x)

[Out]

-15/32*b^3/c^3*(c*x^2+b*x+a)^(1/2)*e^3+15/64*b^4/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+
(c*x^2+b*x+a)^(1/2))*e^3-1/4*x^2/c*(c*x^2+b*x+a)^(1/2)*b*e^3+5/16*b^2/c^2*x*(c*x
^2+b*x+a)^(1/2)*e^3+3/2*b^2/c^2*(c*x^2+b*x+a)^(1/2)*d*e^2-3/4*b^3/c^(5/2)*ln((1/
2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*d*e^2+2*d^3*(c*x^2+b*x+a)^(1/2)-b/c*x*(c*x
^2+b*x+a)^(1/2)*d*e^2+3/4*b^2/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2)
)*d^2*e+2*d*e^2*x^2*(c*x^2+b*x+a)^(1/2)+3*d^2*e*x*(c*x^2+b*x+a)^(1/2)+1/2*e^3*x^
3*(c*x^2+b*x+a)^(1/2)-9/8*b^2/c^(5/2)*a*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/
2))*e^3+13/8*a/c^2*(c*x^2+b*x+a)^(1/2)*b*e^3-4*a/c*(c*x^2+b*x+a)^(1/2)*d*e^2+3/4
*e^3/c^(3/2)*a^2*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+3*b/c^(3/2)*a*ln((1
/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*d*e^2-3/2*b/c*(c*x^2+b*x+a)^(1/2)*d^2*e-3
*a/c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*d^2*e-3/4*e^3/c*a*x*(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((2*c*x + b)*(e*x + d)^3/sqrt(c*x^2 + b*x + a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.481019, size = 1, normalized size = 0. \[ \left [\frac{4 \,{\left (16 \, c^{3} e^{3} x^{3} + 64 \, c^{3} d^{3} - 48 \, b c^{2} d^{2} e + 16 \,{\left (3 \, b^{2} c - 8 \, a c^{2}\right )} d e^{2} -{\left (15 \, b^{3} - 52 \, a b c\right )} e^{3} + 8 \,{\left (8 \, c^{3} d e^{2} - b c^{2} e^{3}\right )} x^{2} + 2 \,{\left (48 \, c^{3} d^{2} e - 16 \, b c^{2} d e^{2} +{\left (5 \, b^{2} c - 12 \, a c^{2}\right )} e^{3}\right )} x\right )} \sqrt{c x^{2} + b x + a} \sqrt{c} + 3 \,{\left (16 \,{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d^{2} e - 16 \,{\left (b^{3} c - 4 \, a b c^{2}\right )} d e^{2} +{\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} e^{3}\right )} \log \left (-4 \,{\left (2 \, c^{2} x + b c\right )} \sqrt{c x^{2} + b x + a} -{\left (8 \, c^{2} x^{2} + 8 \, b c x + b^{2} + 4 \, a c\right )} \sqrt{c}\right )}{128 \, c^{\frac{7}{2}}}, \frac{2 \,{\left (16 \, c^{3} e^{3} x^{3} + 64 \, c^{3} d^{3} - 48 \, b c^{2} d^{2} e + 16 \,{\left (3 \, b^{2} c - 8 \, a c^{2}\right )} d e^{2} -{\left (15 \, b^{3} - 52 \, a b c\right )} e^{3} + 8 \,{\left (8 \, c^{3} d e^{2} - b c^{2} e^{3}\right )} x^{2} + 2 \,{\left (48 \, c^{3} d^{2} e - 16 \, b c^{2} d e^{2} +{\left (5 \, b^{2} c - 12 \, a c^{2}\right )} e^{3}\right )} x\right )} \sqrt{c x^{2} + b x + a} \sqrt{-c} + 3 \,{\left (16 \,{\left (b^{2} c^{2} - 4 \, a c^{3}\right )} d^{2} e - 16 \,{\left (b^{3} c - 4 \, a b c^{2}\right )} d e^{2} +{\left (5 \, b^{4} - 24 \, a b^{2} c + 16 \, a^{2} c^{2}\right )} e^{3}\right )} \arctan \left (\frac{{\left (2 \, c x + b\right )} \sqrt{-c}}{2 \, \sqrt{c x^{2} + b x + a} c}\right )}{64 \, \sqrt{-c} c^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((2*c*x + b)*(e*x + d)^3/sqrt(c*x^2 + b*x + a),x, algorithm="fricas")

[Out]

[1/128*(4*(16*c^3*e^3*x^3 + 64*c^3*d^3 - 48*b*c^2*d^2*e + 16*(3*b^2*c - 8*a*c^2)
*d*e^2 - (15*b^3 - 52*a*b*c)*e^3 + 8*(8*c^3*d*e^2 - b*c^2*e^3)*x^2 + 2*(48*c^3*d
^2*e - 16*b*c^2*d*e^2 + (5*b^2*c - 12*a*c^2)*e^3)*x)*sqrt(c*x^2 + b*x + a)*sqrt(
c) + 3*(16*(b^2*c^2 - 4*a*c^3)*d^2*e - 16*(b^3*c - 4*a*b*c^2)*d*e^2 + (5*b^4 - 2
4*a*b^2*c + 16*a^2*c^2)*e^3)*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)))/c^(7/2), 1/64*(2*(16*c^3*e^3*x^3 + 64*
c^3*d^3 - 48*b*c^2*d^2*e + 16*(3*b^2*c - 8*a*c^2)*d*e^2 - (15*b^3 - 52*a*b*c)*e^
3 + 8*(8*c^3*d*e^2 - b*c^2*e^3)*x^2 + 2*(48*c^3*d^2*e - 16*b*c^2*d*e^2 + (5*b^2*
c - 12*a*c^2)*e^3)*x)*sqrt(c*x^2 + b*x + a)*sqrt(-c) + 3*(16*(b^2*c^2 - 4*a*c^3)
*d^2*e - 16*(b^3*c - 4*a*b*c^2)*d*e^2 + (5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*e^3)*a
rctan(1/2*(2*c*x + b)*sqrt(-c)/(sqrt(c*x^2 + b*x + a)*c)))/(sqrt(-c)*c^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{\left (b + 2 c x\right ) \left (d + e x\right )^{3}}{\sqrt{a + b x + c x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((2*c*x+b)*(e*x+d)**3/(c*x**2+b*x+a)**(1/2),x)

[Out]

Integral((b + 2*c*x)*(d + e*x)**3/sqrt(a + b*x + c*x**2), x)

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GIAC/XCAS [A]  time = 0.292593, size = 340, normalized size = 1.39 \[ \frac{1}{32} \, \sqrt{c x^{2} + b x + a}{\left (2 \,{\left (4 \,{\left (2 \, x e^{3} + \frac{8 \, c^{3} d e^{2} - b c^{2} e^{3}}{c^{3}}\right )} x + \frac{48 \, c^{3} d^{2} e - 16 \, b c^{2} d e^{2} + 5 \, b^{2} c e^{3} - 12 \, a c^{2} e^{3}}{c^{3}}\right )} x + \frac{64 \, c^{3} d^{3} - 48 \, b c^{2} d^{2} e + 48 \, b^{2} c d e^{2} - 128 \, a c^{2} d e^{2} - 15 \, b^{3} e^{3} + 52 \, a b c e^{3}}{c^{3}}\right )} - \frac{3 \,{\left (16 \, b^{2} c^{2} d^{2} e - 64 \, a c^{3} d^{2} e - 16 \, b^{3} c d e^{2} + 64 \, a b c^{2} d e^{2} + 5 \, b^{4} e^{3} - 24 \, a b^{2} c e^{3} + 16 \, a^{2} c^{2} e^{3}\right )}{\rm ln}\left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{64 \, c^{\frac{7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((2*c*x + b)*(e*x + d)^3/sqrt(c*x^2 + b*x + a),x, algorithm="giac")

[Out]

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