3.858 \(\int \frac{(A+B x) \left (a+b x+c x^2\right )^2}{x} \, dx\)

Optimal. Leaf size=92 \[ a^2 A \log (x)+\frac{1}{3} x^3 \left (2 a B c+2 A b c+b^2 B\right )+\frac{1}{2} x^2 \left (A \left (2 a c+b^2\right )+2 a b B\right )+a x (a B+2 A b)+\frac{1}{4} c x^4 (A c+2 b B)+\frac{1}{5} B c^2 x^5 \]

[Out]

a*(2*A*b + a*B)*x + ((2*a*b*B + A*(b^2 + 2*a*c))*x^2)/2 + ((b^2*B + 2*A*b*c + 2*
a*B*c)*x^3)/3 + (c*(2*b*B + A*c)*x^4)/4 + (B*c^2*x^5)/5 + a^2*A*Log[x]

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Rubi [A]  time = 0.122062, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.048 \[ a^2 A \log (x)+\frac{1}{3} x^3 \left (2 a B c+2 A b c+b^2 B\right )+\frac{1}{2} x^2 \left (A \left (2 a c+b^2\right )+2 a b B\right )+a x (a B+2 A b)+\frac{1}{4} c x^4 (A c+2 b B)+\frac{1}{5} B c^2 x^5 \]

Antiderivative was successfully verified.

[In]  Int[((A + B*x)*(a + b*x + c*x^2)^2)/x,x]

[Out]

a*(2*A*b + a*B)*x + ((2*a*b*B + A*(b^2 + 2*a*c))*x^2)/2 + ((b^2*B + 2*A*b*c + 2*
a*B*c)*x^3)/3 + (c*(2*b*B + A*c)*x^4)/4 + (B*c^2*x^5)/5 + a^2*A*Log[x]

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Rubi in Sympy [F]  time = 0., size = 0, normalized size = 0. \[ A a^{2} \log{\left (x \right )} + \frac{B c^{2} x^{5}}{5} + \frac{c x^{4} \left (A c + 2 B b\right )}{4} + x^{3} \left (\frac{2 A b c}{3} + \frac{2 B a c}{3} + \frac{B b^{2}}{3}\right ) + \left (2 A b + B a\right ) \int a\, dx + \left (2 A a c + A b^{2} + 2 B a b\right ) \int x\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((B*x+A)*(c*x**2+b*x+a)**2/x,x)

[Out]

A*a**2*log(x) + B*c**2*x**5/5 + c*x**4*(A*c + 2*B*b)/4 + x**3*(2*A*b*c/3 + 2*B*a
*c/3 + B*b**2/3) + (2*A*b + B*a)*Integral(a, x) + (2*A*a*c + A*b**2 + 2*B*a*b)*I
ntegral(x, x)

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Mathematica [A]  time = 0.0660736, size = 92, normalized size = 1. \[ a^2 A \log (x)+\frac{1}{3} x^3 \left (2 a B c+2 A b c+b^2 B\right )+\frac{1}{2} x^2 \left (2 a A c+2 a b B+A b^2\right )+a x (a B+2 A b)+\frac{1}{4} c x^4 (A c+2 b B)+\frac{1}{5} B c^2 x^5 \]

Antiderivative was successfully verified.

[In]  Integrate[((A + B*x)*(a + b*x + c*x^2)^2)/x,x]

[Out]

a*(2*A*b + a*B)*x + ((A*b^2 + 2*a*b*B + 2*a*A*c)*x^2)/2 + ((b^2*B + 2*A*b*c + 2*
a*B*c)*x^3)/3 + (c*(2*b*B + A*c)*x^4)/4 + (B*c^2*x^5)/5 + a^2*A*Log[x]

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Maple [A]  time = 0.004, size = 95, normalized size = 1. \[{\frac{B{c}^{2}{x}^{5}}{5}}+{\frac{A{c}^{2}{x}^{4}}{4}}+{\frac{B{x}^{4}bc}{2}}+{\frac{2\,A{x}^{3}bc}{3}}+{\frac{2\,aBc{x}^{3}}{3}}+{\frac{B{b}^{2}{x}^{3}}{3}}+aAc{x}^{2}+{\frac{A{b}^{2}{x}^{2}}{2}}+B{x}^{2}ab+2\,aAbx+{a}^{2}Bx+{a}^{2}A\ln \left ( x \right ) \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

1/5*B*c^2*x^5+1/4*A*c^2*x^4+1/2*B*x^4*b*c+2/3*A*x^3*b*c+2/3*a*B*c*x^3+1/3*B*b^2*
x^3+a*A*c*x^2+1/2*A*b^2*x^2+B*x^2*a*b+2*a*A*b*x+a^2*B*x+a^2*A*ln(x)

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Maxima [A]  time = 0.689928, size = 119, normalized size = 1.29 \[ \frac{1}{5} \, B c^{2} x^{5} + \frac{1}{4} \,{\left (2 \, B b c + A c^{2}\right )} x^{4} + \frac{1}{3} \,{\left (B b^{2} + 2 \,{\left (B a + A b\right )} c\right )} x^{3} + A a^{2} \log \left (x\right ) + \frac{1}{2} \,{\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} +{\left (B a^{2} + 2 \, A a b\right )} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((c*x^2 + b*x + a)^2*(B*x + A)/x,x, algorithm="maxima")

[Out]

1/5*B*c^2*x^5 + 1/4*(2*B*b*c + A*c^2)*x^4 + 1/3*(B*b^2 + 2*(B*a + A*b)*c)*x^3 +
A*a^2*log(x) + 1/2*(2*B*a*b + A*b^2 + 2*A*a*c)*x^2 + (B*a^2 + 2*A*a*b)*x

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Fricas [A]  time = 0.287325, size = 119, normalized size = 1.29 \[ \frac{1}{5} \, B c^{2} x^{5} + \frac{1}{4} \,{\left (2 \, B b c + A c^{2}\right )} x^{4} + \frac{1}{3} \,{\left (B b^{2} + 2 \,{\left (B a + A b\right )} c\right )} x^{3} + A a^{2} \log \left (x\right ) + \frac{1}{2} \,{\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} +{\left (B a^{2} + 2 \, A a b\right )} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((c*x^2 + b*x + a)^2*(B*x + A)/x,x, algorithm="fricas")

[Out]

1/5*B*c^2*x^5 + 1/4*(2*B*b*c + A*c^2)*x^4 + 1/3*(B*b^2 + 2*(B*a + A*b)*c)*x^3 +
A*a^2*log(x) + 1/2*(2*B*a*b + A*b^2 + 2*A*a*c)*x^2 + (B*a^2 + 2*A*a*b)*x

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Sympy [A]  time = 1.39288, size = 95, normalized size = 1.03 \[ A a^{2} \log{\left (x \right )} + \frac{B c^{2} x^{5}}{5} + x^{4} \left (\frac{A c^{2}}{4} + \frac{B b c}{2}\right ) + x^{3} \left (\frac{2 A b c}{3} + \frac{2 B a c}{3} + \frac{B b^{2}}{3}\right ) + x^{2} \left (A a c + \frac{A b^{2}}{2} + B a b\right ) + x \left (2 A a b + B a^{2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((B*x+A)*(c*x**2+b*x+a)**2/x,x)

[Out]

A*a**2*log(x) + B*c**2*x**5/5 + x**4*(A*c**2/4 + B*b*c/2) + x**3*(2*A*b*c/3 + 2*
B*a*c/3 + B*b**2/3) + x**2*(A*a*c + A*b**2/2 + B*a*b) + x*(2*A*a*b + B*a**2)

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GIAC/XCAS [A]  time = 0.272252, size = 128, normalized size = 1.39 \[ \frac{1}{5} \, B c^{2} x^{5} + \frac{1}{2} \, B b c x^{4} + \frac{1}{4} \, A c^{2} x^{4} + \frac{1}{3} \, B b^{2} x^{3} + \frac{2}{3} \, B a c x^{3} + \frac{2}{3} \, A b c x^{3} + B a b x^{2} + \frac{1}{2} \, A b^{2} x^{2} + A a c x^{2} + B a^{2} x + 2 \, A a b x + A a^{2}{\rm ln}\left ({\left | x \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((c*x^2 + b*x + a)^2*(B*x + A)/x,x, algorithm="giac")

[Out]

1/5*B*c^2*x^5 + 1/2*B*b*c*x^4 + 1/4*A*c^2*x^4 + 1/3*B*b^2*x^3 + 2/3*B*a*c*x^3 +
2/3*A*b*c*x^3 + B*a*b*x^2 + 1/2*A*b^2*x^2 + A*a*c*x^2 + B*a^2*x + 2*A*a*b*x + A*
a^2*ln(abs(x))