3.39 \(\int \frac{(A+B x) (b x+c x^2)^3}{x^8} \, dx\)

Optimal. Leaf size=69 \[ -\frac{b^2 (3 A c+b B)}{3 x^3}-\frac{A b^3}{4 x^4}-\frac{c^2 (A c+3 b B)}{x}-\frac{3 b c (A c+b B)}{2 x^2}+B c^3 \log (x) \]

[Out]

-(A*b^3)/(4*x^4) - (b^2*(b*B + 3*A*c))/(3*x^3) - (3*b*c*(b*B + A*c))/(2*x^2) - (c^2*(3*b*B + A*c))/x + B*c^3*L
og[x]

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Rubi [A]  time = 0.0376182, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {765} \[ -\frac{b^2 (3 A c+b B)}{3 x^3}-\frac{A b^3}{4 x^4}-\frac{c^2 (A c+3 b B)}{x}-\frac{3 b c (A c+b B)}{2 x^2}+B c^3 \log (x) \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^3)/x^8,x]

[Out]

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

Mathematica [A]  time = 0.0329326, size = 71, normalized size = 1.03 \[ B c^3 \log (x)-\frac{3 A \left (4 b^2 c x+b^3+6 b c^2 x^2+4 c^3 x^3\right )+2 b B x \left (2 b^2+9 b c x+18 c^2 x^2\right )}{12 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^3)/x^8,x]

[Out]

-(2*b*B*x*(2*b^2 + 9*b*c*x + 18*c^2*x^2) + 3*A*(b^3 + 4*b^2*c*x + 6*b*c^2*x^2 + 4*c^3*x^3))/(12*x^4) + B*c^3*L
og[x]

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Maple [A]  time = 0.006, size = 76, normalized size = 1.1 \begin{align*} B{c}^{3}\ln \left ( x \right ) -{\frac{A{b}^{3}}{4\,{x}^{4}}}-{\frac{A{b}^{2}c}{{x}^{3}}}-{\frac{{b}^{3}B}{3\,{x}^{3}}}-{\frac{3\,Ab{c}^{2}}{2\,{x}^{2}}}-{\frac{3\,B{b}^{2}c}{2\,{x}^{2}}}-{\frac{A{c}^{3}}{x}}-3\,{\frac{Bb{c}^{2}}{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^3/x^8,x)

[Out]

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

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Maxima [A]  time = 1.11024, size = 97, normalized size = 1.41 \begin{align*} B c^{3} \log \left (x\right ) - \frac{3 \, A b^{3} + 12 \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{3} + 18 \,{\left (B b^{2} c + A b c^{2}\right )} x^{2} + 4 \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x}{12 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^3/x^8,x, algorithm="maxima")

[Out]

B*c^3*log(x) - 1/12*(3*A*b^3 + 12*(3*B*b*c^2 + A*c^3)*x^3 + 18*(B*b^2*c + A*b*c^2)*x^2 + 4*(B*b^3 + 3*A*b^2*c)
*x)/x^4

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Fricas [A]  time = 1.8299, size = 170, normalized size = 2.46 \begin{align*} \frac{12 \, B c^{3} x^{4} \log \left (x\right ) - 3 \, A b^{3} - 12 \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{3} - 18 \,{\left (B b^{2} c + A b c^{2}\right )} x^{2} - 4 \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x}{12 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^3/x^8,x, algorithm="fricas")

[Out]

1/12*(12*B*c^3*x^4*log(x) - 3*A*b^3 - 12*(3*B*b*c^2 + A*c^3)*x^3 - 18*(B*b^2*c + A*b*c^2)*x^2 - 4*(B*b^3 + 3*A
*b^2*c)*x)/x^4

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Sympy [A]  time = 1.52076, size = 75, normalized size = 1.09 \begin{align*} B c^{3} \log{\left (x \right )} - \frac{3 A b^{3} + x^{3} \left (12 A c^{3} + 36 B b c^{2}\right ) + x^{2} \left (18 A b c^{2} + 18 B b^{2} c\right ) + x \left (12 A b^{2} c + 4 B b^{3}\right )}{12 x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**3/x**8,x)

[Out]

B*c**3*log(x) - (3*A*b**3 + x**3*(12*A*c**3 + 36*B*b*c**2) + x**2*(18*A*b*c**2 + 18*B*b**2*c) + x*(12*A*b**2*c
+ 4*B*b**3))/(12*x**4)

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Giac [A]  time = 1.1156, size = 99, normalized size = 1.43 \begin{align*} B c^{3} \log \left ({\left | x \right |}\right ) - \frac{3 \, A b^{3} + 12 \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{3} + 18 \,{\left (B b^{2} c + A b c^{2}\right )} x^{2} + 4 \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x}{12 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^3/x^8,x, algorithm="giac")

[Out]

B*c^3*log(abs(x)) - 1/12*(3*A*b^3 + 12*(3*B*b*c^2 + A*c^3)*x^3 + 18*(B*b^2*c + A*b*c^2)*x^2 + 4*(B*b^3 + 3*A*b
^2*c)*x)/x^4