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

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Rubi [A]  time = 0.05, antiderivative size = 92, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.03, size = 92, normalized size = 1.00 \[ a^2 A \log (x)+\frac {1}{3} x^3 \left (2 a B c+2 A b c+b^2 B\right )+\frac {1}{2} x^2 \left (2 a A c+2 a b B+A b^2\right )+a x (a B+2 A b)+\frac {1}{4} c x^4 (A c+2 b B)+\frac {1}{5} B c^2 x^5 \]

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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fricas [A]  time = 0.86, size = 88, normalized size = 0.96 \[ \frac {1}{5} \, B c^{2} x^{5} + \frac {1}{4} \, {\left (2 \, B b c + A c^{2}\right )} x^{4} + \frac {1}{3} \, {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} x^{3} + A a^{2} \log \relax (x) + \frac {1}{2} \, {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} + {\left (B a^{2} + 2 \, A a b\right )} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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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giac [A]  time = 0.16, size = 95, normalized size = 1.03 \[ \frac {1}{5} \, B c^{2} x^{5} + \frac {1}{2} \, B b c x^{4} + \frac {1}{4} \, A c^{2} x^{4} + \frac {1}{3} \, B b^{2} x^{3} + \frac {2}{3} \, B a c x^{3} + \frac {2}{3} \, A b c x^{3} + B a b x^{2} + \frac {1}{2} \, A b^{2} x^{2} + A a c x^{2} + B a^{2} x + 2 \, A a b x + A a^{2} \log \left ({\left | x \right |}\right ) \]

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))

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maple [A]  time = 0.05, size = 95, normalized size = 1.03 \[ \frac {B \,c^{2} x^{5}}{5}+\frac {A \,c^{2} x^{4}}{4}+\frac {B b c \,x^{4}}{2}+\frac {2 A b c \,x^{3}}{3}+\frac {2 B a c \,x^{3}}{3}+\frac {B \,b^{2} x^{3}}{3}+A a c \,x^{2}+\frac {A \,b^{2} x^{2}}{2}+B a b \,x^{2}+A \,a^{2} \ln \relax (x )+2 A a b x +B \,a^{2} x \]

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

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maxima [A]  time = 0.45, size = 88, normalized size = 0.96 \[ \frac {1}{5} \, B c^{2} x^{5} + \frac {1}{4} \, {\left (2 \, B b c + A c^{2}\right )} x^{4} + \frac {1}{3} \, {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} x^{3} + A a^{2} \log \relax (x) + \frac {1}{2} \, {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} + {\left (B a^{2} + 2 \, A a b\right )} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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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mupad [B]  time = 0.04, size = 86, normalized size = 0.93 \[ x^4\,\left (\frac {A\,c^2}{4}+\frac {B\,b\,c}{2}\right )+x^2\,\left (\frac {A\,b^2}{2}+B\,a\,b+A\,a\,c\right )+x^3\,\left (\frac {B\,b^2}{3}+\frac {2\,A\,c\,b}{3}+\frac {2\,B\,a\,c}{3}\right )+x\,\left (B\,a^2+2\,A\,b\,a\right )+\frac {B\,c^2\,x^5}{5}+A\,a^2\,\ln \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.21, size = 95, normalized size = 1.03 \[ A a^{2} \log {\relax (x )} + \frac {B c^{2} x^{5}}{5} + x^{4} \left (\frac {A c^{2}}{4} + \frac {B b c}{2}\right ) + x^{3} \left (\frac {2 A b c}{3} + \frac {2 B a c}{3} + \frac {B b^{2}}{3}\right ) + x^{2} \left (A a c + \frac {A b^{2}}{2} + B a b\right ) + x \left (2 A a b + B a^{2}\right ) \]

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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