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

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 = 0.052385, 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, Rules used = {765} \[ a^2 A \log (x)+\frac{1}{3} x^3 \left (2 a B c+2 A b c+b^2 B\right )+\frac{1}{2} x^2 \left (A \left (2 a c+b^2\right )+2 a b B\right )+a x (a B+2 A b)+\frac{1}{4} c x^4 (A c+2 b B)+\frac{1}{5} B c^2 x^5 \]

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]

Rule 765

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand
Integrand[(e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, e, f, g, m}, x] && IntegerQ[p] && (
GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin{align*} \int \frac{(A+B x) \left (a+b x+c x^2\right )^2}{x} \, dx &=\int \left (a (2 A b+a B)+\frac{a^2 A}{x}+\left (2 a b B+A \left (b^2+2 a c\right )\right ) x+\left (b^2 B+2 A b c+2 a B c\right ) x^2+c (2 b B+A c) x^3+B c^2 x^4\right ) \, dx\\ &=a (2 A b+a B) x+\frac{1}{2} \left (2 a b B+A \left (b^2+2 a c\right )\right ) x^2+\frac{1}{3} \left (b^2 B+2 A b c+2 a B c\right ) x^3+\frac{1}{4} c (2 b B+A c) x^4+\frac{1}{5} B c^2 x^5+a^2 A \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0329251, 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 = 0.001, size = 95, normalized size = 1. \begin{align*}{\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}^{2}B{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 ) \end{align*}

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^2*B*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 = 1.02091, size = 119, normalized size = 1.29 \begin{align*} \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 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^2/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 = 1.21463, size = 208, normalized size = 2.26 \begin{align*} \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 \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^2/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 = 0.648749, size = 95, normalized size = 1.03 \begin{align*} 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 ) \end{align*}

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 [A]  time = 1.2817, size = 128, normalized size = 1.39 \begin{align*} \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} \log \left ({\left | x \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^2/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*log(abs(x))