3.1.72 \(\int \frac {(a+b x)^3}{x^4} \, dx\) [72]

Optimal. Leaf size=37 \[ -\frac {a^3}{3 x^3}-\frac {3 a^2 b}{2 x^2}-\frac {3 a b^2}{x}+b^3 \log (x) \]

[Out]

-1/3*a^3/x^3-3/2*a^2*b/x^2-3*a*b^2/x+b^3*ln(x)

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Rubi [A]
time = 0.01, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {45} \begin {gather*} -\frac {a^3}{3 x^3}-\frac {3 a^2 b}{2 x^2}-\frac {3 a b^2}{x}+b^3 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^3/x^4,x]

[Out]

-1/3*a^3/x^3 - (3*a^2*b)/(2*x^2) - (3*a*b^2)/x + b^3*Log[x]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin {align*} \int \frac {(a+b x)^3}{x^4} \, dx &=\int \left (\frac {a^3}{x^4}+\frac {3 a^2 b}{x^3}+\frac {3 a b^2}{x^2}+\frac {b^3}{x}\right ) \, dx\\ &=-\frac {a^3}{3 x^3}-\frac {3 a^2 b}{2 x^2}-\frac {3 a b^2}{x}+b^3 \log (x)\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 37, normalized size = 1.00 \begin {gather*} -\frac {a^3}{3 x^3}-\frac {3 a^2 b}{2 x^2}-\frac {3 a b^2}{x}+b^3 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^3/x^4,x]

[Out]

-1/3*a^3/x^3 - (3*a^2*b)/(2*x^2) - (3*a*b^2)/x + b^3*Log[x]

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Maple [A]
time = 0.09, size = 34, normalized size = 0.92

method	result	size
default	\(-\frac {a^{3}}{3 x^{3}}-\frac {3 a^{2} b}{2 x^{2}}-\frac {3 a \,b^{2}}{x}+b^{3} \ln \left (x \right )\)	\(34\)
norman	\(\frac {-\frac {1}{3} a^{3}-3 a \,b^{2} x^{2}-\frac {3}{2} a^{2} b x}{x^{3}}+b^{3} \ln \left (x \right )\)	\(34\)
risch	\(\frac {-\frac {1}{3} a^{3}-3 a \,b^{2} x^{2}-\frac {3}{2} a^{2} b x}{x^{3}}+b^{3} \ln \left (x \right )\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^3/x^4,x,method=_RETURNVERBOSE)

[Out]

-1/3*a^3/x^3-3/2*a^2*b/x^2-3*a*b^2/x+b^3*ln(x)

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Maxima [A]
time = 0.29, size = 34, normalized size = 0.92 \begin {gather*} b^{3} \log \left (x\right ) - \frac {18 \, a b^{2} x^{2} + 9 \, a^{2} b x + 2 \, a^{3}}{6 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3/x^4,x, algorithm="maxima")

[Out]

b^3*log(x) - 1/6*(18*a*b^2*x^2 + 9*a^2*b*x + 2*a^3)/x^3

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Fricas [A]
time = 1.24, size = 37, normalized size = 1.00 \begin {gather*} \frac {6 \, b^{3} x^{3} \log \left (x\right ) - 18 \, a b^{2} x^{2} - 9 \, a^{2} b x - 2 \, a^{3}}{6 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3/x^4,x, algorithm="fricas")

[Out]

1/6*(6*b^3*x^3*log(x) - 18*a*b^2*x^2 - 9*a^2*b*x - 2*a^3)/x^3

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Sympy [A]
time = 0.07, size = 36, normalized size = 0.97 \begin {gather*} b^{3} \log {\left (x \right )} + \frac {- 2 a^{3} - 9 a^{2} b x - 18 a b^{2} x^{2}}{6 x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**3/x**4,x)

[Out]

b**3*log(x) + (-2*a**3 - 9*a**2*b*x - 18*a*b**2*x**2)/(6*x**3)

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Giac [A]
time = 1.38, size = 35, normalized size = 0.95 \begin {gather*} b^{3} \log \left ({\left | x \right |}\right ) - \frac {18 \, a b^{2} x^{2} + 9 \, a^{2} b x + 2 \, a^{3}}{6 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^3/x^4,x, algorithm="giac")

[Out]

b^3*log(abs(x)) - 1/6*(18*a*b^2*x^2 + 9*a^2*b*x + 2*a^3)/x^3

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Mupad [B]
time = 0.07, size = 34, normalized size = 0.92 \begin {gather*} b^3\,\ln \left (x\right )-\frac {\frac {a^3}{3}+\frac {3\,a^2\,b\,x}{2}+3\,a\,b^2\,x^2}{x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^3/x^4,x)

[Out]

b^3*log(x) - (a^3/3 + 3*a*b^2*x^2 + (3*a^2*b*x)/2)/x^3

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