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

Optimal. Leaf size=237 \[ \frac{\left (4 a b c d-5 (a d+b c)^2\right ) (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{64 b^{7/2} d^{7/2}}-\frac{\sqrt{a+b x} \sqrt{c+d x} \left (4 a b c d-5 (a d+b c)^2\right ) (b c-a d)}{64 b^3 d^3}-\frac{(a+b x)^{3/2} \sqrt{c+d x} \left (4 a b c d-5 (a d+b c)^2\right )}{32 b^3 d^2}-\frac{5 (a+b x)^{3/2} (c+d x)^{3/2} (a d+b c)}{24 b^2 d^2}+\frac{x (a+b x)^{3/2} (c+d x)^{3/2}}{4 b d} \]

[Out]

-((b*c - a*d)*(4*a*b*c*d - 5*(b*c + a*d)^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/(64*b^3
*d^3) - ((4*a*b*c*d - 5*(b*c + a*d)^2)*(a + b*x)^(3/2)*Sqrt[c + d*x])/(32*b^3*d^
2) - (5*(b*c + a*d)*(a + b*x)^(3/2)*(c + d*x)^(3/2))/(24*b^2*d^2) + (x*(a + b*x)
^(3/2)*(c + d*x)^(3/2))/(4*b*d) + ((b*c - a*d)^2*(4*a*b*c*d - 5*(b*c + a*d)^2)*A
rcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(64*b^(7/2)*d^(7/2))

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Rubi [A]  time = 0.472098, antiderivative size = 237, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227 \[ \frac{\left (4 a b c d-5 (a d+b c)^2\right ) (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x}}{\sqrt{b} \sqrt{c+d x}}\right )}{64 b^{7/2} d^{7/2}}-\frac{\sqrt{a+b x} \sqrt{c+d x} \left (4 a b c d-5 (a d+b c)^2\right ) (b c-a d)}{64 b^3 d^3}-\frac{(a+b x)^{3/2} \sqrt{c+d x} \left (4 a b c d-5 (a d+b c)^2\right )}{32 b^3 d^2}-\frac{5 (a+b x)^{3/2} (c+d x)^{3/2} (a d+b c)}{24 b^2 d^2}+\frac{x (a+b x)^{3/2} (c+d x)^{3/2}}{4 b d} \]

Antiderivative was successfully verified.

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

[Out]

-((b*c - a*d)*(4*a*b*c*d - 5*(b*c + a*d)^2)*Sqrt[a + b*x]*Sqrt[c + d*x])/(64*b^3
*d^3) - ((4*a*b*c*d - 5*(b*c + a*d)^2)*(a + b*x)^(3/2)*Sqrt[c + d*x])/(32*b^3*d^
2) - (5*(b*c + a*d)*(a + b*x)^(3/2)*(c + d*x)^(3/2))/(24*b^2*d^2) + (x*(a + b*x)
^(3/2)*(c + d*x)^(3/2))/(4*b*d) + ((b*c - a*d)^2*(4*a*b*c*d - 5*(b*c + a*d)^2)*A
rcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[c + d*x])])/(64*b^(7/2)*d^(7/2))

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Rubi in Sympy [A]  time = 39.7355, size = 219, normalized size = 0.92 \[ \frac{x \left (a + b x\right )^{\frac{3}{2}} \left (c + d x\right )^{\frac{3}{2}}}{4 b d} - \frac{5 \left (a + b x\right )^{\frac{3}{2}} \left (c + d x\right )^{\frac{3}{2}} \left (a d + b c\right )}{24 b^{2} d^{2}} - \frac{\left (a + b x\right )^{\frac{3}{2}} \sqrt{c + d x} \left (a b c d - \frac{5 \left (a d + b c\right )^{2}}{4}\right )}{8 b^{3} d^{2}} + \frac{\sqrt{a + b x} \sqrt{c + d x} \left (a d - b c\right ) \left (4 a b c d - 5 \left (a d + b c\right )^{2}\right )}{64 b^{3} d^{3}} + \frac{\left (a d - b c\right )^{2} \left (a b c d - \frac{5 \left (a d + b c\right )^{2}}{4}\right ) \operatorname{atanh}{\left (\frac{\sqrt{b} \sqrt{c + d x}}{\sqrt{d} \sqrt{a + b x}} \right )}}{16 b^{\frac{7}{2}} d^{\frac{7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

x*(a + b*x)**(3/2)*(c + d*x)**(3/2)/(4*b*d) - 5*(a + b*x)**(3/2)*(c + d*x)**(3/2
)*(a*d + b*c)/(24*b**2*d**2) - (a + b*x)**(3/2)*sqrt(c + d*x)*(a*b*c*d - 5*(a*d
+ b*c)**2/4)/(8*b**3*d**2) + sqrt(a + b*x)*sqrt(c + d*x)*(a*d - b*c)*(4*a*b*c*d
- 5*(a*d + b*c)**2)/(64*b**3*d**3) + (a*d - b*c)**2*(a*b*c*d - 5*(a*d + b*c)**2/
4)*atanh(sqrt(b)*sqrt(c + d*x)/(sqrt(d)*sqrt(a + b*x)))/(16*b**(7/2)*d**(7/2))

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Mathematica [A]  time = 0.19506, size = 208, normalized size = 0.88 \[ \frac{2 \sqrt{b} \sqrt{d} \sqrt{a+b x} \sqrt{c+d x} \left (15 a^3 d^3-a^2 b d^2 (7 c+10 d x)+a b^2 d \left (-7 c^2+4 c d x+8 d^2 x^2\right )+b^3 \left (15 c^3-10 c^2 d x+8 c d^2 x^2+48 d^3 x^3\right )\right )-3 (b c-a d)^2 \left (5 a^2 d^2+6 a b c d+5 b^2 c^2\right ) \log \left (2 \sqrt{b} \sqrt{d} \sqrt{a+b x} \sqrt{c+d x}+a d+b c+2 b d x\right )}{384 b^{7/2} d^{7/2}} \]

Antiderivative was successfully verified.

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

[Out]

(2*Sqrt[b]*Sqrt[d]*Sqrt[a + b*x]*Sqrt[c + d*x]*(15*a^3*d^3 - a^2*b*d^2*(7*c + 10
*d*x) + a*b^2*d*(-7*c^2 + 4*c*d*x + 8*d^2*x^2) + b^3*(15*c^3 - 10*c^2*d*x + 8*c*
d^2*x^2 + 48*d^3*x^3)) - 3*(b*c - a*d)^2*(5*b^2*c^2 + 6*a*b*c*d + 5*a^2*d^2)*Log
[b*c + a*d + 2*b*d*x + 2*Sqrt[b]*Sqrt[d]*Sqrt[a + b*x]*Sqrt[c + d*x]])/(384*b^(7
/2)*d^(7/2))

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Maple [B]  time = 0.022, size = 686, normalized size = 2.9 \[ -{\frac{1}{384\,{b}^{3}{d}^{3}}\sqrt{bx+a}\sqrt{dx+c} \left ( -96\,{x}^{3}{b}^{3}{d}^{3}\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}-16\,{x}^{2}a{b}^{2}{d}^{3}\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}-16\,{x}^{2}{b}^{3}c{d}^{2}\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+15\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{4}{d}^{4}-12\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{3}bc{d}^{3}-6\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{2}{b}^{2}{c}^{2}{d}^{2}-12\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) a{b}^{3}{c}^{3}d+15\,\ln \left ( 1/2\,{\frac{2\,bdx+2\,\sqrt{d{x}^{2}b+adx+bcx+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){b}^{4}{c}^{4}+20\,\sqrt{bd}\sqrt{d{x}^{2}b+adx+bcx+ac}x{a}^{2}b{d}^{3}-8\,\sqrt{bd}\sqrt{d{x}^{2}b+adx+bcx+ac}xa{b}^{2}c{d}^{2}+20\,\sqrt{bd}\sqrt{d{x}^{2}b+adx+bcx+ac}x{b}^{3}{c}^{2}d-30\,\sqrt{bd}\sqrt{d{x}^{2}b+adx+bcx+ac}{a}^{3}{d}^{3}+14\,\sqrt{bd}\sqrt{d{x}^{2}b+adx+bcx+ac}{a}^{2}bc{d}^{2}+14\,\sqrt{bd}\sqrt{d{x}^{2}b+adx+bcx+ac}a{b}^{2}{c}^{2}d-30\,\sqrt{bd}\sqrt{d{x}^{2}b+adx+bcx+ac}{b}^{3}{c}^{3} \right ){\frac{1}{\sqrt{d{x}^{2}b+adx+bcx+ac}}}{\frac{1}{\sqrt{bd}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

-1/384*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(-96*x^3*b^3*d^3*(b*d*x^2+a*d*x+b*c*x+a*c)^(1
/2)*(b*d)^(1/2)-16*x^2*a*b^2*d^3*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)-16*
x^2*b^3*c*d^2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+15*ln(1/2*(2*b*d*x+2*(
b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a^4*d^4-12*ln(1
/2*(2*b*d*x+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*
a^3*b*c*d^3-6*ln(1/2*(2*b*d*x+2*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+a*d+
b*c)/(b*d)^(1/2))*a^2*b^2*c^2*d^2-12*ln(1/2*(2*b*d*x+2*(b*d*x^2+a*d*x+b*c*x+a*c)
^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*a*b^3*c^3*d+15*ln(1/2*(2*b*d*x+2*(b*d*x
^2+a*d*x+b*c*x+a*c)^(1/2)*(b*d)^(1/2)+a*d+b*c)/(b*d)^(1/2))*b^4*c^4+20*(b*d)^(1/
2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x*a^2*b*d^3-8*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*
x+a*c)^(1/2)*x*a*b^2*c*d^2+20*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*x*b^3*
c^2*d-30*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^3*d^3+14*(b*d)^(1/2)*(b*d
*x^2+a*d*x+b*c*x+a*c)^(1/2)*a^2*b*c*d^2+14*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)
^(1/2)*a*b^2*c^2*d-30*(b*d)^(1/2)*(b*d*x^2+a*d*x+b*c*x+a*c)^(1/2)*b^3*c^3)/(b*d*
x^2+a*d*x+b*c*x+a*c)^(1/2)/b^3/d^3/(b*d)^(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(b*x + a)*sqrt(d*x + c)*x^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.272486, size = 1, normalized size = 0. \[ \left [\frac{4 \,{\left (48 \, b^{3} d^{3} x^{3} + 15 \, b^{3} c^{3} - 7 \, a b^{2} c^{2} d - 7 \, a^{2} b c d^{2} + 15 \, a^{3} d^{3} + 8 \,{\left (b^{3} c d^{2} + a b^{2} d^{3}\right )} x^{2} - 2 \,{\left (5 \, b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + 5 \, a^{2} b d^{3}\right )} x\right )} \sqrt{b d} \sqrt{b x + a} \sqrt{d x + c} + 3 \,{\left (5 \, b^{4} c^{4} - 4 \, a b^{3} c^{3} d - 2 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + 5 \, a^{4} d^{4}\right )} \log \left (-4 \,{\left (2 \, b^{2} d^{2} x + b^{2} c d + a b d^{2}\right )} \sqrt{b x + a} \sqrt{d x + c} +{\left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \,{\left (b^{2} c d + a b d^{2}\right )} x\right )} \sqrt{b d}\right )}{768 \, \sqrt{b d} b^{3} d^{3}}, \frac{2 \,{\left (48 \, b^{3} d^{3} x^{3} + 15 \, b^{3} c^{3} - 7 \, a b^{2} c^{2} d - 7 \, a^{2} b c d^{2} + 15 \, a^{3} d^{3} + 8 \,{\left (b^{3} c d^{2} + a b^{2} d^{3}\right )} x^{2} - 2 \,{\left (5 \, b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + 5 \, a^{2} b d^{3}\right )} x\right )} \sqrt{-b d} \sqrt{b x + a} \sqrt{d x + c} - 3 \,{\left (5 \, b^{4} c^{4} - 4 \, a b^{3} c^{3} d - 2 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + 5 \, a^{4} d^{4}\right )} \arctan \left (\frac{{\left (2 \, b d x + b c + a d\right )} \sqrt{-b d}}{2 \, \sqrt{b x + a} \sqrt{d x + c} b d}\right )}{384 \, \sqrt{-b d} b^{3} d^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

[1/768*(4*(48*b^3*d^3*x^3 + 15*b^3*c^3 - 7*a*b^2*c^2*d - 7*a^2*b*c*d^2 + 15*a^3*
d^3 + 8*(b^3*c*d^2 + a*b^2*d^3)*x^2 - 2*(5*b^3*c^2*d - 2*a*b^2*c*d^2 + 5*a^2*b*d
^3)*x)*sqrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 3*(5*b^4*c^4 - 4*a*b^3*c^3*d - 2*
a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + 5*a^4*d^4)*log(-4*(2*b^2*d^2*x + b^2*c*d + a*b
*d^2)*sqrt(b*x + a)*sqrt(d*x + c) + (8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d
^2 + 8*(b^2*c*d + a*b*d^2)*x)*sqrt(b*d)))/(sqrt(b*d)*b^3*d^3), 1/384*(2*(48*b^3*
d^3*x^3 + 15*b^3*c^3 - 7*a*b^2*c^2*d - 7*a^2*b*c*d^2 + 15*a^3*d^3 + 8*(b^3*c*d^2
+ a*b^2*d^3)*x^2 - 2*(5*b^3*c^2*d - 2*a*b^2*c*d^2 + 5*a^2*b*d^3)*x)*sqrt(-b*d)*
sqrt(b*x + a)*sqrt(d*x + c) - 3*(5*b^4*c^4 - 4*a*b^3*c^3*d - 2*a^2*b^2*c^2*d^2 -
4*a^3*b*c*d^3 + 5*a^4*d^4)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)/(sqrt(b*
x + a)*sqrt(d*x + c)*b*d)))/(sqrt(-b*d)*b^3*d^3)]

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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

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GIAC/XCAS [A]  time = 0.234062, size = 386, normalized size = 1.63 \[ \frac{{\left (\sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d}{\left (2 \,{\left (b x + a\right )}{\left (4 \,{\left (b x + a\right )}{\left (\frac{6 \,{\left (b x + a\right )}}{b^{2}} + \frac{b^{7} c d^{5} - 17 \, a b^{6} d^{6}}{b^{8} d^{6}}\right )} - \frac{5 \, b^{8} c^{2} d^{4} + 6 \, a b^{7} c d^{5} - 59 \, a^{2} b^{6} d^{6}}{b^{8} d^{6}}\right )} + \frac{3 \,{\left (5 \, b^{9} c^{3} d^{3} + a b^{8} c^{2} d^{4} - a^{2} b^{7} c d^{5} - 5 \, a^{3} b^{6} d^{6}\right )}}{b^{8} d^{6}}\right )} \sqrt{b x + a} + \frac{3 \,{\left (5 \, b^{4} c^{4} - 4 \, a b^{3} c^{3} d - 2 \, a^{2} b^{2} c^{2} d^{2} - 4 \, a^{3} b c d^{3} + 5 \, a^{4} d^{4}\right )}{\rm ln}\left ({\left | -\sqrt{b d} \sqrt{b x + a} + \sqrt{b^{2} c +{\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt{b d} b d^{3}}\right )}{\left | b \right |}}{192 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

1/192*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(b*x + a)*(4*(b*x + a)*(6*(b*x + a
)/b^2 + (b^7*c*d^5 - 17*a*b^6*d^6)/(b^8*d^6)) - (5*b^8*c^2*d^4 + 6*a*b^7*c*d^5 -
59*a^2*b^6*d^6)/(b^8*d^6)) + 3*(5*b^9*c^3*d^3 + a*b^8*c^2*d^4 - a^2*b^7*c*d^5 -
5*a^3*b^6*d^6)/(b^8*d^6))*sqrt(b*x + a) + 3*(5*b^4*c^4 - 4*a*b^3*c^3*d - 2*a^2*
b^2*c^2*d^2 - 4*a^3*b*c*d^3 + 5*a^4*d^4)*ln(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(
b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*b*d^3))*abs(b)/b^3