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

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

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Rubi [A]  time = 0.04, antiderivative size = 65, 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 {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 \end {gather*}

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*}

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Mathematica [A]  time = 0.02, size = 71, normalized size = 1.09 \begin {gather*} -\frac {A b^3}{2 x^2}+3 \log (x) \left (A b c^2+b^2 B c\right )+\frac {b^3 (-B)-3 A b^2 c}{x}+c^2 x (A c+3 b B)+\frac {1}{2} B c^3 x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(A*b^3)/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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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 )^3}{x^6} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 74, normalized size = 1.14 \begin {gather*} \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 \relax (x) - 2 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x}{2 \, x^{2}} \end {gather*}

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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giac [A]  time = 0.15, size = 69, normalized size = 1.06 \begin {gather*} \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 {gather*}

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

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maple [A]  time = 0.06, size = 71, normalized size = 1.09 \begin {gather*} \frac {B \,c^{3} x^{2}}{2}+3 A b \,c^{2} \ln \relax (x )+A \,c^{3} x +3 B \,b^{2} c \ln \relax (x )+3 B b \,c^{2} x -\frac {3 A \,b^{2} c}{x}-\frac {B \,b^{3}}{x}-\frac {A \,b^{3}}{2 x^{2}} \end {gather*}

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

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maxima [A]  time = 0.90, size = 69, normalized size = 1.06 \begin {gather*} \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 \relax (x) - \frac {A b^{3} + 2 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x}{2 \, x^{2}} \end {gather*}

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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mupad [B]  time = 1.06, size = 70, normalized size = 1.08 \begin {gather*} \ln \relax (x)\,\left (3\,B\,b^2\,c+3\,A\,b\,c^2\right )-\frac {x\,\left (B\,b^3+3\,A\,c\,b^2\right )+\frac {A\,b^3}{2}}{x^2}+x\,\left (A\,c^3+3\,B\,b\,c^2\right )+\frac {B\,c^3\,x^2}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.36, size = 68, normalized size = 1.05 \begin {gather*} \frac {B c^{3} x^{2}}{2} + 3 b c \left (A c + B b\right ) \log {\relax (x )} + 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 {gather*}

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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