3.1.36 \(\int \frac {(A+B x) (b x+c x^2)^3}{x^5} \, dx\)

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

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Rubi [A]  time = 0.05, 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*} b^2 \log (x) (3 A c+b B)-\frac {A b^3}{x}+\frac {1}{2} c^2 x^2 (A c+3 b B)+3 b c x (A c+b B)+\frac {1}{3} B c^3 x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^3)/x^5,x]

[Out]

-((A*b^3)/x) + 3*b*c*(b*B + A*c)*x + (c^2*(3*b*B + A*c)*x^2)/2 + (B*c^3*x^3)/3 + b^2*(b*B + 3*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^5} \, dx &=\int \left (3 b c (b B+A c)+\frac {A b^3}{x^2}+\frac {b^2 (b B+3 A c)}{x}+c^2 (3 b B+A c) x+B c^3 x^2\right ) \, dx\\ &=-\frac {A b^3}{x}+3 b c (b B+A c) x+\frac {1}{2} c^2 (3 b B+A c) x^2+\frac {1}{3} B c^3 x^3+b^2 (b B+3 A c) \log (x)\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 67, normalized size = 1.03 \begin {gather*} -\frac {A b^3}{x}+\log (x) \left (3 A b^2 c+b^3 B\right )+\frac {1}{2} c^2 x^2 (A c+3 b B)+3 b c x (A c+b B)+\frac {1}{3} B c^3 x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^3)/x^5,x]

[Out]

-((A*b^3)/x) + 3*b*c*(b*B + A*c)*x + (c^2*(3*b*B + A*c)*x^2)/2 + (B*c^3*x^3)/3 + (b^3*B + 3*A*b^2*c)*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^5} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^3)/x^5,x]

[Out]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^3)/x^5, x]

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fricas [A]  time = 0.38, size = 75, normalized size = 1.15 \begin {gather*} \frac {2 \, B c^{3} x^{4} - 6 \, A b^{3} + 3 \, {\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} + 6 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x \log \relax (x)}{6 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^3/x^5,x, algorithm="fricas")

[Out]

1/6*(2*B*c^3*x^4 - 6*A*b^3 + 3*(3*B*b*c^2 + A*c^3)*x^3 + 18*(B*b^2*c + A*b*c^2)*x^2 + 6*(B*b^3 + 3*A*b^2*c)*x*
log(x))/x

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giac [A]  time = 0.18, size = 71, normalized size = 1.09 \begin {gather*} \frac {1}{3} \, B c^{3} x^{3} + \frac {3}{2} \, B b c^{2} x^{2} + \frac {1}{2} \, A c^{3} x^{2} + 3 \, B b^{2} c x + 3 \, A b c^{2} x - \frac {A b^{3}}{x} + {\left (B b^{3} + 3 \, A b^{2} c\right )} \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^3/x^5,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 71, normalized size = 1.09 \begin {gather*} \frac {B \,c^{3} x^{3}}{3}+\frac {A \,c^{3} x^{2}}{2}+\frac {3 B b \,c^{2} x^{2}}{2}+3 A \,b^{2} c \ln \relax (x )+3 A b \,c^{2} x +B \,b^{3} \ln \relax (x )+3 B \,b^{2} c x -\frac {A \,b^{3}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^3/x^5,x)

[Out]

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

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maxima [A]  time = 0.91, size = 69, normalized size = 1.06 \begin {gather*} \frac {1}{3} \, B c^{3} x^{3} - \frac {A b^{3}}{x} + \frac {1}{2} \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{2} + 3 \, {\left (B b^{2} c + A b c^{2}\right )} x + {\left (B b^{3} + 3 \, A b^{2} c\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^3/x^5,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.04, size = 65, normalized size = 1.00 \begin {gather*} x^2\,\left (\frac {A\,c^3}{2}+\frac {3\,B\,b\,c^2}{2}\right )+\ln \relax (x)\,\left (B\,b^3+3\,A\,c\,b^2\right )-\frac {A\,b^3}{x}+\frac {B\,c^3\,x^3}{3}+3\,b\,c\,x\,\left (A\,c+B\,b\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x + c*x^2)^3*(A + B*x))/x^5,x)

[Out]

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

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sympy [A]  time = 0.22, size = 70, normalized size = 1.08 \begin {gather*} - \frac {A b^{3}}{x} + \frac {B c^{3} x^{3}}{3} + b^{2} \left (3 A c + B b\right ) \log {\relax (x )} + x^{2} \left (\frac {A c^{3}}{2} + \frac {3 B b c^{2}}{2}\right ) + x \left (3 A b c^{2} + 3 B b^{2} c\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**3/x**5,x)

[Out]

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

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