3.1.39 \(\int \frac {(A+B x) (b x+c x^2)^3}{x^8} \, dx\) [39]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 69, 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 = {779} \begin {gather*} -\frac {A b^3}{4 x^4}-\frac {b^2 (3 A c+b B)}{3 x^3}-\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) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(A*b^3)/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 779

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*}

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Mathematica [A]
time = 0.02, size = 71, normalized size = 1.03 \begin {gather*} -\frac {2 b B x \left (2 b^2+9 b c x+18 c^2 x^2\right )+3 A \left (b^3+4 b^2 c x+6 b c^2 x^2+4 c^3 x^3\right )}{12 x^4}+B c^3 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(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))/x^4 + B*c^3*L
og[x]

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Maple [A]
time = 0.48, size = 64, normalized size = 0.93

method	result	size
default	\(-\frac {A \,b^{3}}{4 x^{4}}-\frac {b^{2} \left (3 A c +B b \right )}{3 x^{3}}-\frac {3 b c \left (A c +B b \right )}{2 x^{2}}-\frac {c^{2} \left (A c +3 B b \right )}{x}+B \,c^{3} \ln \left (x \right )\)	\(64\)
risch	\(\frac {\left (-A \,c^{3}-3 B b \,c^{2}\right ) x^{3}+\left (-\frac {3}{2} A b \,c^{2}-\frac {3}{2} B \,b^{2} c \right ) x^{2}+\left (-A \,b^{2} c -\frac {1}{3} B \,b^{3}\right ) x -\frac {A \,b^{3}}{4}}{x^{4}}+B \,c^{3} \ln \left (x \right )\)	\(73\)
norman	\(\frac {\left (-\frac {3}{2} A b \,c^{2}-\frac {3}{2} B \,b^{2} c \right ) x^{5}+\left (-A \,b^{2} c -\frac {1}{3} B \,b^{3}\right ) x^{4}+\left (-A \,c^{3}-3 B b \,c^{2}\right ) x^{6}-\frac {A \,b^{3} x^{3}}{4}}{x^{7}}+B \,c^{3} \ln \left (x \right )\)	\(78\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 72, normalized size = 1.04 \begin {gather*} 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 {gather*}

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 = 0.65, size = 75, normalized size = 1.09 \begin {gather*} \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 {gather*}

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 = 0.58, size = 80, normalized size = 1.16 \begin {gather*} 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 {gather*}

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 = 1.56, size = 73, normalized size = 1.06 \begin {gather*} 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 {gather*}

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

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Mupad [B]
time = 0.08, size = 71, normalized size = 1.03 \begin {gather*} B\,c^3\,\ln \left (x\right )-\frac {x^2\,\left (\frac {3\,B\,b^2\,c}{2}+\frac {3\,A\,b\,c^2}{2}\right )+x\,\left (\frac {B\,b^3}{3}+A\,c\,b^2\right )+\frac {A\,b^3}{4}+x^3\,\left (A\,c^3+3\,B\,b\,c^2\right )}{x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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