3.1.38 \(\int \frac {(A+B x) (b x+c x^2)^3}{x^7} \, dx\)

Optimal. Leaf size=64 \[ -\frac {A b^3}{3 x^3}-\frac {b^2 (3 A c+b B)}{2 x^2}+c^2 \log (x) (A c+3 b B)-\frac {3 b c (A c+b B)}{x}+B c^3 x \]

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Rubi [A]  time = 0.04, antiderivative size = 64, 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)}{2 x^2}-\frac {A b^3}{3 x^3}+c^2 \log (x) (A c+3 b B)-\frac {3 b c (A c+b B)}{x}+B c^3 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2)^3)/x^7,x]

[Out]

-(A*b^3)/(3*x^3) - (b^2*(b*B + 3*A*c))/(2*x^2) - (3*b*c*(b*B + A*c))/x + B*c^3*x + c^2*(3*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^7} \, dx &=\int \left (B c^3+\frac {A b^3}{x^4}+\frac {b^2 (b B+3 A c)}{x^3}+\frac {3 b c (b B+A c)}{x^2}+\frac {c^2 (3 b B+A c)}{x}\right ) \, dx\\ &=-\frac {A b^3}{3 x^3}-\frac {b^2 (b B+3 A c)}{2 x^2}-\frac {3 b c (b B+A c)}{x}+B c^3 x+c^2 (3 b B+A c) \log (x)\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 73, normalized size = 1.14 \begin {gather*} -\frac {A b^3}{3 x^3}-\frac {3 \left (A b c^2+b^2 B c\right )}{x}+\frac {b^3 (-B)-3 A b^2 c}{2 x^2}+\log (x) \left (A c^3+3 b B c^2\right )+B c^3 x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2)^3)/x^7,x]

[Out]

-1/3*(A*b^3)/x^3 + (-(b^3*B) - 3*A*b^2*c)/(2*x^2) - (3*(b^2*B*c + A*b*c^2))/x + B*c^3*x + (3*b*B*c^2 + A*c^3)*
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^7} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^3)/x^7,x]

[Out]

IntegrateAlgebraic[((A + B*x)*(b*x + c*x^2)^3)/x^7, x]

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fricas [A]  time = 0.40, size = 75, normalized size = 1.17 \begin {gather*} \frac {6 \, B c^{3} x^{4} + 6 \, {\left (3 \, B b c^{2} + A c^{3}\right )} x^{3} \log \relax (x) - 2 \, A b^{3} - 18 \, {\left (B b^{2} c + A b c^{2}\right )} x^{2} - 3 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x}{6 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^3/x^7,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 70, normalized size = 1.09 \begin {gather*} B c^{3} x + {\left (3 \, B b c^{2} + A c^{3}\right )} \log \left ({\left | x \right |}\right ) - \frac {2 \, A b^{3} + 18 \, {\left (B b^{2} c + A b c^{2}\right )} x^{2} + 3 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x}{6 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^3/x^7,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 72, normalized size = 1.12 \begin {gather*} A \,c^{3} \ln \relax (x )+3 B b \,c^{2} \ln \relax (x )+B \,c^{3} x -\frac {3 A b \,c^{2}}{x}-\frac {3 B \,b^{2} c}{x}-\frac {3 A \,b^{2} c}{2 x^{2}}-\frac {B \,b^{3}}{2 x^{2}}-\frac {A \,b^{3}}{3 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)^3/x^7,x)

[Out]

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

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maxima [A]  time = 0.82, size = 69, normalized size = 1.08 \begin {gather*} B c^{3} x + {\left (3 \, B b c^{2} + A c^{3}\right )} \log \relax (x) - \frac {2 \, A b^{3} + 18 \, {\left (B b^{2} c + A b c^{2}\right )} x^{2} + 3 \, {\left (B b^{3} + 3 \, A b^{2} c\right )} x}{6 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)^3/x^7,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.07, size = 70, normalized size = 1.09 \begin {gather*} \ln \relax (x)\,\left (A\,c^3+3\,B\,b\,c^2\right )-\frac {x^2\,\left (3\,B\,b^2\,c+3\,A\,b\,c^2\right )+x\,\left (\frac {B\,b^3}{2}+\frac {3\,A\,c\,b^2}{2}\right )+\frac {A\,b^3}{3}}{x^3}+B\,c^3\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x + c*x^2)^3*(A + B*x))/x^7,x)

[Out]

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

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sympy [A]  time = 0.66, size = 73, normalized size = 1.14 \begin {gather*} B c^{3} x + c^{2} \left (A c + 3 B b\right ) \log {\relax (x )} + \frac {- 2 A b^{3} + x^{2} \left (- 18 A b c^{2} - 18 B b^{2} c\right ) + x \left (- 9 A b^{2} c - 3 B b^{3}\right )}{6 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)**3/x**7,x)

[Out]

B*c**3*x + c**2*(A*c + 3*B*b)*log(x) + (-2*A*b**3 + x**2*(-18*A*b*c**2 - 18*B*b**2*c) + x*(-9*A*b**2*c - 3*B*b
**3))/(6*x**3)

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