3.8.88 \(\int \frac {(A+B x) (a+b x+c x^2)^2}{x^6} \, dx\)

Optimal. Leaf size=95 \[ -\frac {a^2 A}{5 x^5}-\frac {A \left (2 a c+b^2\right )+2 a b B}{3 x^3}-\frac {2 a B c+2 A b c+b^2 B}{2 x^2}-\frac {a (a B+2 A b)}{4 x^4}-\frac {c (A c+2 b B)}{x}+B c^2 \log (x) \]

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Rubi [A]  time = 0.05, antiderivative size = 95, 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} \begin {gather*} -\frac {a^2 A}{5 x^5}-\frac {2 a B c+2 A b c+b^2 B}{2 x^2}-\frac {A \left (2 a c+b^2\right )+2 a b B}{3 x^3}-\frac {a (a B+2 A b)}{4 x^4}-\frac {c (A c+2 b B)}{x}+B c^2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + b*x + c*x^2)^2)/x^6,x]

[Out]

-(a^2*A)/(5*x^5) - (a*(2*A*b + a*B))/(4*x^4) - (2*a*b*B + A*(b^2 + 2*a*c))/(3*x^3) - (b^2*B + 2*A*b*c + 2*a*B*
c)/(2*x^2) - (c*(2*b*B + A*c))/x + B*c^2*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^6} \, dx &=\int \left (\frac {a^2 A}{x^6}+\frac {a (2 A b+a B)}{x^5}+\frac {2 a b B+A \left (b^2+2 a c\right )}{x^4}+\frac {b^2 B+2 A b c+2 a B c}{x^3}+\frac {c (2 b B+A c)}{x^2}+\frac {B c^2}{x}\right ) \, dx\\ &=-\frac {a^2 A}{5 x^5}-\frac {a (2 A b+a B)}{4 x^4}-\frac {2 a b B+A \left (b^2+2 a c\right )}{3 x^3}-\frac {b^2 B+2 A b c+2 a B c}{2 x^2}-\frac {c (2 b B+A c)}{x}+B c^2 \log (x)\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 92, normalized size = 0.97 \begin {gather*} B c^2 \log (x)-\frac {3 a^2 (4 A+5 B x)+10 a x \left (3 A b+4 A c x+4 b B x+6 B c x^2\right )+10 x^2 \left (2 A \left (b^2+3 b c x+3 c^2 x^2\right )+3 b B x (b+4 c x)\right )}{60 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + b*x + c*x^2)^2)/x^6,x]

[Out]

-1/60*(3*a^2*(4*A + 5*B*x) + 10*a*x*(3*A*b + 4*b*B*x + 4*A*c*x + 6*B*c*x^2) + 10*x^2*(3*b*B*x*(b + 4*c*x) + 2*
A*(b^2 + 3*b*c*x + 3*c^2*x^2)))/x^5 + B*c^2*Log[x]

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(A+B x) \left (a+b x+c x^2\right )^2}{x^6} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((A + B*x)*(a + b*x + c*x^2)^2)/x^6,x]

[Out]

IntegrateAlgebraic[((A + B*x)*(a + b*x + c*x^2)^2)/x^6, x]

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fricas [A]  time = 0.40, size = 95, normalized size = 1.00 \begin {gather*} \frac {60 \, B c^{2} x^{5} \log \relax (x) - 60 \, {\left (2 \, B b c + A c^{2}\right )} x^{4} - 30 \, {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} x^{3} - 12 \, A a^{2} - 20 \, {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} - 15 \, {\left (B a^{2} + 2 \, A a b\right )} x}{60 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^2/x^6,x, algorithm="fricas")

[Out]

1/60*(60*B*c^2*x^5*log(x) - 60*(2*B*b*c + A*c^2)*x^4 - 30*(B*b^2 + 2*(B*a + A*b)*c)*x^3 - 12*A*a^2 - 20*(2*B*a
*b + A*b^2 + 2*A*a*c)*x^2 - 15*(B*a^2 + 2*A*a*b)*x)/x^5

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giac [A]  time = 0.18, size = 93, normalized size = 0.98 \begin {gather*} B c^{2} \log \left ({\left | x \right |}\right ) - \frac {60 \, {\left (2 \, B b c + A c^{2}\right )} x^{4} + 30 \, {\left (B b^{2} + 2 \, B a c + 2 \, A b c\right )} x^{3} + 12 \, A a^{2} + 20 \, {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} + 15 \, {\left (B a^{2} + 2 \, A a b\right )} x}{60 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^2/x^6,x, algorithm="giac")

[Out]

B*c^2*log(abs(x)) - 1/60*(60*(2*B*b*c + A*c^2)*x^4 + 30*(B*b^2 + 2*B*a*c + 2*A*b*c)*x^3 + 12*A*a^2 + 20*(2*B*a
*b + A*b^2 + 2*A*a*c)*x^2 + 15*(B*a^2 + 2*A*a*b)*x)/x^5

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maple [A]  time = 0.05, size = 102, normalized size = 1.07 \begin {gather*} B \,c^{2} \ln \relax (x )-\frac {A \,c^{2}}{x}-\frac {2 B b c}{x}-\frac {A b c}{x^{2}}-\frac {B a c}{x^{2}}-\frac {B \,b^{2}}{2 x^{2}}-\frac {2 A a c}{3 x^{3}}-\frac {A \,b^{2}}{3 x^{3}}-\frac {2 B a b}{3 x^{3}}-\frac {A a b}{2 x^{4}}-\frac {B \,a^{2}}{4 x^{4}}-\frac {A \,a^{2}}{5 x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x+a)^2/x^6,x)

[Out]

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

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maxima [A]  time = 0.72, size = 92, normalized size = 0.97 \begin {gather*} B c^{2} \log \relax (x) - \frac {60 \, {\left (2 \, B b c + A c^{2}\right )} x^{4} + 30 \, {\left (B b^{2} + 2 \, {\left (B a + A b\right )} c\right )} x^{3} + 12 \, A a^{2} + 20 \, {\left (2 \, B a b + A b^{2} + 2 \, A a c\right )} x^{2} + 15 \, {\left (B a^{2} + 2 \, A a b\right )} x}{60 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^2/x^6,x, algorithm="maxima")

[Out]

B*c^2*log(x) - 1/60*(60*(2*B*b*c + A*c^2)*x^4 + 30*(B*b^2 + 2*(B*a + A*b)*c)*x^3 + 12*A*a^2 + 20*(2*B*a*b + A*
b^2 + 2*A*a*c)*x^2 + 15*(B*a^2 + 2*A*a*b)*x)/x^5

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mupad [B]  time = 0.07, size = 89, normalized size = 0.94 \begin {gather*} B\,c^2\,\ln \relax (x)-\frac {x^4\,\left (A\,c^2+2\,B\,b\,c\right )+\frac {A\,a^2}{5}+x^2\,\left (\frac {A\,b^2}{3}+\frac {2\,B\,a\,b}{3}+\frac {2\,A\,a\,c}{3}\right )+x^3\,\left (\frac {B\,b^2}{2}+A\,c\,b+B\,a\,c\right )+x\,\left (\frac {B\,a^2}{4}+\frac {A\,b\,a}{2}\right )}{x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x + c*x^2)^2)/x^6,x)

[Out]

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

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sympy [A]  time = 7.11, size = 105, normalized size = 1.11 \begin {gather*} B c^{2} \log {\relax (x )} + \frac {- 12 A a^{2} + x^{4} \left (- 60 A c^{2} - 120 B b c\right ) + x^{3} \left (- 60 A b c - 60 B a c - 30 B b^{2}\right ) + x^{2} \left (- 40 A a c - 20 A b^{2} - 40 B a b\right ) + x \left (- 30 A a b - 15 B a^{2}\right )}{60 x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x+a)**2/x**6,x)

[Out]

B*c**2*log(x) + (-12*A*a**2 + x**4*(-60*A*c**2 - 120*B*b*c) + x**3*(-60*A*b*c - 60*B*a*c - 30*B*b**2) + x**2*(
-40*A*a*c - 20*A*b**2 - 40*B*a*b) + x*(-30*A*a*b - 15*B*a**2))/(60*x**5)

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