3.1.26 \(\int \frac {(A+B x) (b x+c x^2)^2}{x^8} \, dx\)

Optimal. Leaf size=55 \[ -\frac {A b^2}{5 x^5}-\frac {b (2 A c+b B)}{4 x^4}-\frac {c (A c+2 b B)}{3 x^3}-\frac {B c^2}{2 x^2} \]

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Rubi [A]  time = 0.03, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {765} \begin {gather*} -\frac {A b^2}{5 x^5}-\frac {b (2 A c+b B)}{4 x^4}-\frac {c (A c+2 b B)}{3 x^3}-\frac {B c^2}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^2)/x^8,x]

[Out]

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

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 (b x+c x^2\right )^2}{x^8} \, dx &=\int \left (\frac {A b^2}{x^6}+\frac {b (b B+2 A c)}{x^5}+\frac {c (2 b B+A c)}{x^4}+\frac {B c^2}{x^3}\right ) \, dx\\ &=-\frac {A b^2}{5 x^5}-\frac {b (b B+2 A c)}{4 x^4}-\frac {c (2 b B+A c)}{3 x^3}-\frac {B c^2}{2 x^2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 53, normalized size = 0.96 \begin {gather*} -\frac {2 A \left (6 b^2+15 b c x+10 c^2 x^2\right )+5 B x \left (3 b^2+8 b c x+6 c^2 x^2\right )}{60 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^2)/x^8,x]

[Out]

-1/60*(5*B*x*(3*b^2 + 8*b*c*x + 6*c^2*x^2) + 2*A*(6*b^2 + 15*b*c*x + 10*c^2*x^2))/x^5

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(A+B x) \left (b x+c x^2\right )^2}{x^8} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^2)/x^8,x]

[Out]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^2)/x^8, x]

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fricas [A]  time = 0.39, size = 51, normalized size = 0.93 \begin {gather*} -\frac {30 \, B c^{2} x^{3} + 12 \, A b^{2} + 20 \, {\left (2 \, B b c + A c^{2}\right )} x^{2} + 15 \, {\left (B b^{2} + 2 \, A b c\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)^2/x^8,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 51, normalized size = 0.93 \begin {gather*} -\frac {30 \, B c^{2} x^{3} + 40 \, B b c x^{2} + 20 \, A c^{2} x^{2} + 15 \, B b^{2} x + 30 \, A b c x + 12 \, A b^{2}}{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)^2/x^8,x, algorithm="giac")

[Out]

-1/60*(30*B*c^2*x^3 + 40*B*b*c*x^2 + 20*A*c^2*x^2 + 15*B*b^2*x + 30*A*b*c*x + 12*A*b^2)/x^5

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maple [A]  time = 0.05, size = 48, normalized size = 0.87 \begin {gather*} -\frac {B \,c^{2}}{2 x^{2}}-\frac {A \,b^{2}}{5 x^{5}}-\frac {\left (A c +2 b B \right ) c}{3 x^{3}}-\frac {\left (2 A c +b B \right ) b}{4 x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^2/x^8,x)

[Out]

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

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maxima [A]  time = 0.84, size = 51, normalized size = 0.93 \begin {gather*} -\frac {30 \, B c^{2} x^{3} + 12 \, A b^{2} + 20 \, {\left (2 \, B b c + A c^{2}\right )} x^{2} + 15 \, {\left (B b^{2} + 2 \, A b c\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)^2/x^8,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.04, size = 51, normalized size = 0.93 \begin {gather*} -\frac {x^2\,\left (\frac {A\,c^2}{3}+\frac {2\,B\,b\,c}{3}\right )+\frac {A\,b^2}{5}+x\,\left (\frac {B\,b^2}{4}+\frac {A\,c\,b}{2}\right )+\frac {B\,c^2\,x^3}{2}}{x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x + c*x^2)^2*(A + B*x))/x^8,x)

[Out]

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

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sympy [A]  time = 0.86, size = 56, normalized size = 1.02 \begin {gather*} \frac {- 12 A b^{2} - 30 B c^{2} x^{3} + x^{2} \left (- 20 A c^{2} - 40 B b c\right ) + x \left (- 30 A b c - 15 B b^{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)**2/x**8,x)

[Out]

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

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