3.37 \(\int \frac{(A+B x) (b x+c x^2)^3}{x^6} \, dx\)

Optimal. Leaf size=65 \[ -\frac{b^2 (3 A c+b B)}{x}-\frac{A b^3}{2 x^2}+c^2 x (A c+3 b B)+3 b c \log (x) (A c+b B)+\frac{1}{2} B c^3 x^2 \]

[Out]

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

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Rubi [A]  time = 0.0448049, antiderivative size = 65, 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} \[ -\frac{b^2 (3 A c+b B)}{x}-\frac{A b^3}{2 x^2}+c^2 x (A c+3 b B)+3 b c \log (x) (A c+b B)+\frac{1}{2} B c^3 x^2 \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^3)/x^6,x]

[Out]

-(A*b^3)/(2*x^2) - (b^2*(b*B + 3*A*c))/x + c^2*(3*b*B + A*c)*x + (B*c^3*x^2)/2 + 3*b*c*(b*B + A*c)*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 (b x+c x^2\right )^3}{x^6} \, dx &=\int \left (c^2 (3 b B+A c)+\frac{A b^3}{x^3}+\frac{b^2 (b B+3 A c)}{x^2}+\frac{3 b c (b B+A c)}{x}+B c^3 x\right ) \, dx\\ &=-\frac{A b^3}{2 x^2}-\frac{b^2 (b B+3 A c)}{x}+c^2 (3 b B+A c) x+\frac{1}{2} B c^3 x^2+3 b c (b B+A c) \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0210096, size = 71, normalized size = 1.09 \[ 3 \log (x) \left (A b c^2+b^2 B c\right )+\frac{b^3 (-B)-3 A b^2 c}{x}-\frac{A b^3}{2 x^2}+c^2 x (A c+3 b B)+\frac{1}{2} B c^3 x^2 \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^3)/x^6,x]

[Out]

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

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Maple [A]  time = 0.007, size = 71, normalized size = 1.1 \begin{align*}{\frac{B{c}^{3}{x}^{2}}{2}}+A{c}^{3}x+3\,Bb{c}^{2}x+3\,A\ln \left ( x \right ) b{c}^{2}+3\,B\ln \left ( x \right ){b}^{2}c-{\frac{A{b}^{3}}{2\,{x}^{2}}}-3\,{\frac{A{b}^{2}c}{x}}-{\frac{{b}^{3}B}{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^6,x)

[Out]

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

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Maxima [A]  time = 1.08848, size = 93, normalized size = 1.43 \begin{align*} \frac{1}{2} \, B c^{3} x^{2} +{\left (3 \, B b c^{2} + A c^{3}\right )} x + 3 \,{\left (B b^{2} c + A b c^{2}\right )} \log \left (x\right ) - \frac{A b^{3} + 2 \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^3/x^6,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.79337, size = 159, normalized size = 2.45 \begin{align*} \frac{B c^{3} x^{4} - A b^{3} + 2 \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{3} + 6 \,{\left (B b^{2} c + A b c^{2}\right )} x^{2} \log \left (x\right ) - 2 \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^3/x^6,x, algorithm="fricas")

[Out]

1/2*(B*c^3*x^4 - A*b^3 + 2*(3*B*b*c^2 + A*c^3)*x^3 + 6*(B*b^2*c + A*b*c^2)*x^2*log(x) - 2*(B*b^3 + 3*A*b^2*c)*
x)/x^2

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Sympy [A]  time = 0.607529, size = 66, normalized size = 1.02 \begin{align*} \frac{B c^{3} x^{2}}{2} + 3 b c \left (A c + B b\right ) \log{\left (x \right )} + x \left (A c^{3} + 3 B b c^{2}\right ) - \frac{A b^{3} + x \left (6 A b^{2} c + 2 B b^{3}\right )}{2 x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**3/x**6,x)

[Out]

B*c**3*x**2/2 + 3*b*c*(A*c + B*b)*log(x) + x*(A*c**3 + 3*B*b*c**2) - (A*b**3 + x*(6*A*b**2*c + 2*B*b**3))/(2*x
**2)

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Giac [A]  time = 1.14856, size = 93, normalized size = 1.43 \begin{align*} \frac{1}{2} \, B c^{3} x^{2} + 3 \, B b c^{2} x + A c^{3} x + 3 \,{\left (B b^{2} c + A b c^{2}\right )} \log \left ({\left | x \right |}\right ) - \frac{A b^{3} + 2 \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^3/x^6,x, algorithm="giac")

[Out]

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