3.1180 \(\int (b d+2 c d x)^4 \sqrt{a+b x+c x^2} \, dx\)

Optimal. Leaf size=165 \[ -\frac{d^4 \left (b^2-4 a c\right )^3 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{64 c^{3/2}}-\frac{d^4 \left (b^2-4 a c\right ) (b+2 c x)^3 \sqrt{a+b x+c x^2}}{48 c}-\frac{d^4 \left (b^2-4 a c\right )^2 (b+2 c x) \sqrt{a+b x+c x^2}}{32 c}+\frac{d^4 (b+2 c x)^5 \sqrt{a+b x+c x^2}}{12 c} \]

[Out]

-((b^2 - 4*a*c)^2*d^4*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(32*c) - ((b^2 - 4*a*c)
*d^4*(b + 2*c*x)^3*Sqrt[a + b*x + c*x^2])/(48*c) + (d^4*(b + 2*c*x)^5*Sqrt[a + b
*x + c*x^2])/(12*c) - ((b^2 - 4*a*c)^3*d^4*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a
 + b*x + c*x^2])])/(64*c^(3/2))

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Rubi [A]  time = 0.275564, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154 \[ -\frac{d^4 \left (b^2-4 a c\right )^3 \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{64 c^{3/2}}-\frac{d^4 \left (b^2-4 a c\right ) (b+2 c x)^3 \sqrt{a+b x+c x^2}}{48 c}-\frac{d^4 \left (b^2-4 a c\right )^2 (b+2 c x) \sqrt{a+b x+c x^2}}{32 c}+\frac{d^4 (b+2 c x)^5 \sqrt{a+b x+c x^2}}{12 c} \]

Antiderivative was successfully verified.

[In]  Int[(b*d + 2*c*d*x)^4*Sqrt[a + b*x + c*x^2],x]

[Out]

-((b^2 - 4*a*c)^2*d^4*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(32*c) - ((b^2 - 4*a*c)
*d^4*(b + 2*c*x)^3*Sqrt[a + b*x + c*x^2])/(48*c) + (d^4*(b + 2*c*x)^5*Sqrt[a + b
*x + c*x^2])/(12*c) - ((b^2 - 4*a*c)^3*d^4*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a
 + b*x + c*x^2])])/(64*c^(3/2))

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Rubi in Sympy [A]  time = 49.9846, size = 151, normalized size = 0.92 \[ \frac{d^{4} \left (b + 2 c x\right )^{5} \sqrt{a + b x + c x^{2}}}{12 c} - \frac{d^{4} \left (b + 2 c x\right )^{3} \left (- 4 a c + b^{2}\right ) \sqrt{a + b x + c x^{2}}}{48 c} - \frac{d^{4} \left (b + 2 c x\right ) \left (- 4 a c + b^{2}\right )^{2} \sqrt{a + b x + c x^{2}}}{32 c} - \frac{d^{4} \left (- 4 a c + b^{2}\right )^{3} \operatorname{atanh}{\left (\frac{b + 2 c x}{2 \sqrt{c} \sqrt{a + b x + c x^{2}}} \right )}}{64 c^{\frac{3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((2*c*d*x+b*d)**4*(c*x**2+b*x+a)**(1/2),x)

[Out]

d**4*(b + 2*c*x)**5*sqrt(a + b*x + c*x**2)/(12*c) - d**4*(b + 2*c*x)**3*(-4*a*c
+ b**2)*sqrt(a + b*x + c*x**2)/(48*c) - d**4*(b + 2*c*x)*(-4*a*c + b**2)**2*sqrt
(a + b*x + c*x**2)/(32*c) - d**4*(-4*a*c + b**2)**3*atanh((b + 2*c*x)/(2*sqrt(c)
*sqrt(a + b*x + c*x**2)))/(64*c**(3/2))

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Mathematica [A]  time = 0.20335, size = 147, normalized size = 0.89 \[ d^4 \left (\frac{(b+2 c x) \sqrt{a+x (b+c x)} \left (16 c^2 \left (-3 a^2+2 a c x^2+8 c^2 x^4\right )+8 b^2 c \left (4 a+23 c x^2\right )+32 b c^2 x \left (a+8 c x^2\right )+3 b^4+56 b^3 c x\right )}{96 c}-\frac{\left (b^2-4 a c\right )^3 \log \left (2 \sqrt{c} \sqrt{a+x (b+c x)}+b+2 c x\right )}{64 c^{3/2}}\right ) \]

Antiderivative was successfully verified.

[In]  Integrate[(b*d + 2*c*d*x)^4*Sqrt[a + b*x + c*x^2],x]

[Out]

d^4*(((b + 2*c*x)*Sqrt[a + x*(b + c*x)]*(3*b^4 + 56*b^3*c*x + 32*b*c^2*x*(a + 8*
c*x^2) + 8*b^2*c*(4*a + 23*c*x^2) + 16*c^2*(-3*a^2 + 2*a*c*x^2 + 8*c^2*x^4)))/(9
6*c) - ((b^2 - 4*a*c)^3*Log[b + 2*c*x + 2*Sqrt[c]*Sqrt[a + x*(b + c*x)]])/(64*c^
(3/2)))

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Maple [B]  time = 0.024, size = 413, normalized size = 2.5 \[ 4\,{d}^{4}{c}^{2}b{x}^{2} \left ( c{x}^{2}+bx+a \right ) ^{3/2}+{\frac{5\,{d}^{4}x{b}^{2}c}{2} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}-{d}^{4}cba \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}-2\,{d}^{4}{c}^{2}ax \left ( c{x}^{2}+bx+a \right ) ^{3/2}+{d}^{4}{c}^{2}{a}^{2}\sqrt{c{x}^{2}+bx+a}x+{\frac{c{d}^{4}{a}^{2}b}{2}\sqrt{c{x}^{2}+bx+a}}+{d}^{4}{c}^{{\frac{3}{2}}}{a}^{3}\ln \left ({1 \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ) +{\frac{3\,{d}^{4}{b}^{4}a}{16}\ln \left ({1 \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){\frac{1}{\sqrt{c}}}}+{\frac{8\,{d}^{4}{c}^{3}{x}^{3}}{3} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}-{\frac{{d}^{4}{b}^{3}a}{4}\sqrt{c{x}^{2}+bx+a}}+{\frac{7\,{d}^{4}{b}^{3}}{12} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{{d}^{4}{b}^{4}x}{16}\sqrt{c{x}^{2}+bx+a}}+{\frac{{d}^{4}{b}^{5}}{32\,c}\sqrt{c{x}^{2}+bx+a}}-{\frac{{d}^{4}{b}^{6}}{64}\ln \left ({1 \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}-{\frac{c{d}^{4}{b}^{2}ax}{2}\sqrt{c{x}^{2}+bx+a}}-{\frac{3\,{d}^{4}{b}^{2}{a}^{2}}{4}\sqrt{c}\ln \left ({1 \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ) } \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((2*c*d*x+b*d)^4*(c*x^2+b*x+a)^(1/2),x)

[Out]

4*d^4*c^2*b*x^2*(c*x^2+b*x+a)^(3/2)+5/2*d^4*c*b^2*x*(c*x^2+b*x+a)^(3/2)-d^4*c*b*
a*(c*x^2+b*x+a)^(3/2)-2*d^4*c^2*a*x*(c*x^2+b*x+a)^(3/2)+d^4*c^2*a^2*(c*x^2+b*x+a
)^(1/2)*x+1/2*d^4*c*a^2*(c*x^2+b*x+a)^(1/2)*b+d^4*c^(3/2)*a^3*ln((1/2*b+c*x)/c^(
1/2)+(c*x^2+b*x+a)^(1/2))+3/16*d^4*b^4/c^(1/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x
+a)^(1/2))*a+8/3*d^4*c^3*x^3*(c*x^2+b*x+a)^(3/2)-1/4*d^4*b^3*a*(c*x^2+b*x+a)^(1/
2)+7/12*d^4*b^3*(c*x^2+b*x+a)^(3/2)+1/16*d^4*b^4*(c*x^2+b*x+a)^(1/2)*x+1/32*d^4*
b^5/c*(c*x^2+b*x+a)^(1/2)-1/64*d^4*b^6/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x
+a)^(1/2))-1/2*d^4*c*b^2*a*(c*x^2+b*x+a)^(1/2)*x-3/4*d^4*c^(1/2)*b^2*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((2*c*d*x + b*d)^4*sqrt(c*x^2 + b*x + a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.258152, size = 1, normalized size = 0.01 \[ \left [-\frac{3 \,{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{4} \log \left (-4 \,{\left (2 \, c^{2} x + b c\right )} \sqrt{c x^{2} + b x + a} -{\left (8 \, c^{2} x^{2} + 8 \, b c x + b^{2} + 4 \, a c\right )} \sqrt{c}\right ) - 4 \,{\left (256 \, c^{5} d^{4} x^{5} + 640 \, b c^{4} d^{4} x^{4} + 16 \,{\left (39 \, b^{2} c^{3} + 4 \, a c^{4}\right )} d^{4} x^{3} + 8 \,{\left (37 \, b^{3} c^{2} + 12 \, a b c^{3}\right )} d^{4} x^{2} + 2 \,{\left (31 \, b^{4} c + 48 \, a b^{2} c^{2} - 48 \, a^{2} c^{3}\right )} d^{4} x +{\left (3 \, b^{5} + 32 \, a b^{3} c - 48 \, a^{2} b c^{2}\right )} d^{4}\right )} \sqrt{c x^{2} + b x + a} \sqrt{c}}{384 \, c^{\frac{3}{2}}}, -\frac{3 \,{\left (b^{6} - 12 \, a b^{4} c + 48 \, a^{2} b^{2} c^{2} - 64 \, a^{3} c^{3}\right )} d^{4} \arctan \left (\frac{{\left (2 \, c x + b\right )} \sqrt{-c}}{2 \, \sqrt{c x^{2} + b x + a} c}\right ) - 2 \,{\left (256 \, c^{5} d^{4} x^{5} + 640 \, b c^{4} d^{4} x^{4} + 16 \,{\left (39 \, b^{2} c^{3} + 4 \, a c^{4}\right )} d^{4} x^{3} + 8 \,{\left (37 \, b^{3} c^{2} + 12 \, a b c^{3}\right )} d^{4} x^{2} + 2 \,{\left (31 \, b^{4} c + 48 \, a b^{2} c^{2} - 48 \, a^{2} c^{3}\right )} d^{4} x +{\left (3 \, b^{5} + 32 \, a b^{3} c - 48 \, a^{2} b c^{2}\right )} d^{4}\right )} \sqrt{c x^{2} + b x + a} \sqrt{-c}}{192 \, \sqrt{-c} c}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((2*c*d*x + b*d)^4*sqrt(c*x^2 + b*x + a),x, algorithm="fricas")

[Out]

[-1/384*(3*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^4*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*
(256*c^5*d^4*x^5 + 640*b*c^4*d^4*x^4 + 16*(39*b^2*c^3 + 4*a*c^4)*d^4*x^3 + 8*(37
*b^3*c^2 + 12*a*b*c^3)*d^4*x^2 + 2*(31*b^4*c + 48*a*b^2*c^2 - 48*a^2*c^3)*d^4*x
+ (3*b^5 + 32*a*b^3*c - 48*a^2*b*c^2)*d^4)*sqrt(c*x^2 + b*x + a)*sqrt(c))/c^(3/2
), -1/192*(3*(b^6 - 12*a*b^4*c + 48*a^2*b^2*c^2 - 64*a^3*c^3)*d^4*arctan(1/2*(2*
c*x + b)*sqrt(-c)/(sqrt(c*x^2 + b*x + a)*c)) - 2*(256*c^5*d^4*x^5 + 640*b*c^4*d^
4*x^4 + 16*(39*b^2*c^3 + 4*a*c^4)*d^4*x^3 + 8*(37*b^3*c^2 + 12*a*b*c^3)*d^4*x^2
+ 2*(31*b^4*c + 48*a*b^2*c^2 - 48*a^2*c^3)*d^4*x + (3*b^5 + 32*a*b^3*c - 48*a^2*
b*c^2)*d^4)*sqrt(c*x^2 + b*x + a)*sqrt(-c))/(sqrt(-c)*c)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \[ d^{4} \left (\int b^{4} \sqrt{a + b x + c x^{2}}\, dx + \int 16 c^{4} x^{4} \sqrt{a + b x + c x^{2}}\, dx + \int 32 b c^{3} x^{3} \sqrt{a + b x + c x^{2}}\, dx + \int 24 b^{2} c^{2} x^{2} \sqrt{a + b x + c x^{2}}\, dx + \int 8 b^{3} c x \sqrt{a + b x + c x^{2}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((2*c*d*x+b*d)**4*(c*x**2+b*x+a)**(1/2),x)

[Out]

d**4*(Integral(b**4*sqrt(a + b*x + c*x**2), x) + Integral(16*c**4*x**4*sqrt(a +
b*x + c*x**2), x) + Integral(32*b*c**3*x**3*sqrt(a + b*x + c*x**2), x) + Integra
l(24*b**2*c**2*x**2*sqrt(a + b*x + c*x**2), x) + Integral(8*b**3*c*x*sqrt(a + b*
x + c*x**2), x))

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GIAC/XCAS [A]  time = 0.230716, size = 350, normalized size = 2.12 \[ \frac{1}{96} \, \sqrt{c x^{2} + b x + a}{\left (2 \,{\left (4 \,{\left (2 \,{\left (8 \,{\left (2 \, c^{4} d^{4} x + 5 \, b c^{3} d^{4}\right )} x + \frac{39 \, b^{2} c^{7} d^{4} + 4 \, a c^{8} d^{4}}{c^{5}}\right )} x + \frac{37 \, b^{3} c^{6} d^{4} + 12 \, a b c^{7} d^{4}}{c^{5}}\right )} x + \frac{31 \, b^{4} c^{5} d^{4} + 48 \, a b^{2} c^{6} d^{4} - 48 \, a^{2} c^{7} d^{4}}{c^{5}}\right )} x + \frac{3 \, b^{5} c^{4} d^{4} + 32 \, a b^{3} c^{5} d^{4} - 48 \, a^{2} b c^{6} d^{4}}{c^{5}}\right )} + \frac{{\left (b^{6} d^{4} - 12 \, a b^{4} c d^{4} + 48 \, a^{2} b^{2} c^{2} d^{4} - 64 \, a^{3} c^{3} d^{4}\right )}{\rm ln}\left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{64 \, c^{\frac{3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((2*c*d*x + b*d)^4*sqrt(c*x^2 + b*x + a),x, algorithm="giac")

[Out]

1/96*sqrt(c*x^2 + b*x + a)*(2*(4*(2*(8*(2*c^4*d^4*x + 5*b*c^3*d^4)*x + (39*b^2*c
^7*d^4 + 4*a*c^8*d^4)/c^5)*x + (37*b^3*c^6*d^4 + 12*a*b*c^7*d^4)/c^5)*x + (31*b^
4*c^5*d^4 + 48*a*b^2*c^6*d^4 - 48*a^2*c^7*d^4)/c^5)*x + (3*b^5*c^4*d^4 + 32*a*b^
3*c^5*d^4 - 48*a^2*b*c^6*d^4)/c^5) + 1/64*(b^6*d^4 - 12*a*b^4*c*d^4 + 48*a^2*b^2
*c^2*d^4 - 64*a^3*c^3*d^4)*ln(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c)
 - b))/c^(3/2)