3.11.56 \(\int \frac {(a c-b c x)^3}{(a+b x)^2} \, dx\) [1056]

Optimal. Leaf size=54 \[ 5 a c^3 x-\frac {1}{2} b c^3 x^2-\frac {8 a^3 c^3}{b (a+b x)}-\frac {12 a^2 c^3 \log (a+b x)}{b} \]

[Out]

5*a*c^3*x-1/2*b*c^3*x^2-8*a^3*c^3/b/(b*x+a)-12*a^2*c^3*ln(b*x+a)/b

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Rubi [A]
time = 0.02, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {45} \begin {gather*} -\frac {8 a^3 c^3}{b (a+b x)}-\frac {12 a^2 c^3 \log (a+b x)}{b}+5 a c^3 x-\frac {1}{2} b c^3 x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*c - b*c*x)^3/(a + b*x)^2,x]

[Out]

5*a*c^3*x - (b*c^3*x^2)/2 - (8*a^3*c^3)/(b*(a + b*x)) - (12*a^2*c^3*Log[a + b*x])/b

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 c-b c x)^3}{(a+b x)^2} \, dx &=\int \left (5 a c^3-b c^3 x+\frac {8 a^3 c^3}{(a+b x)^2}-\frac {12 a^2 c^3}{a+b x}\right ) \, dx\\ &=5 a c^3 x-\frac {1}{2} b c^3 x^2-\frac {8 a^3 c^3}{b (a+b x)}-\frac {12 a^2 c^3 \log (a+b x)}{b}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 46, normalized size = 0.85 \begin {gather*} c^3 \left (5 a x-\frac {b x^2}{2}-\frac {8 a^3}{b (a+b x)}-\frac {12 a^2 \log (a+b x)}{b}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*c - b*c*x)^3/(a + b*x)^2,x]

[Out]

c^3*(5*a*x - (b*x^2)/2 - (8*a^3)/(b*(a + b*x)) - (12*a^2*Log[a + b*x])/b)

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Mathics [A]
time = 1.89, size = 53, normalized size = 0.98 \begin {gather*} \frac {c^3 \left (-24 a^2 \text {Log}\left [a+b x\right ] \left (a+b x\right )-16 a^3+b x \left (a+b x\right ) \left (10 a-b x\right )\right )}{2 b \left (a+b x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[1/(a + b*x)^2*(a*c - b*c*x)^3,x]')

[Out]

c ^ 3 (-24 a ^ 2 Log[a + b x] (a + b x) - 16 a ^ 3 + b x (a + b x) (10 a - b x)) / (2 b (a + b x))

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Maple [A]
time = 0.15, size = 45, normalized size = 0.83

method	result	size
default	\(c^{3} \left (-\frac {x^{2} b}{2}+5 a x -\frac {8 a^{3}}{b \left (b x +a \right )}-\frac {12 a^{2} \ln \left (b x +a \right )}{b}\right )\)	\(45\)
risch	\(5 a \,c^{3} x -\frac {b \,c^{3} x^{2}}{2}-\frac {8 a^{3} c^{3}}{b \left (b x +a \right )}-\frac {12 a^{2} c^{3} \ln \left (b x +a \right )}{b}\)	\(53\)
norman	\(\frac {13 a^{2} c^{3} x -\frac {1}{2} b^{2} c^{3} x^{3}+\frac {9}{2} a \,c^{3} b \,x^{2}}{b x +a}-\frac {12 a^{2} c^{3} \ln \left (b x +a \right )}{b}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-b*c*x+a*c)^3/(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

c^3*(-1/2*x^2*b+5*a*x-8/b*a^3/(b*x+a)-12*a^2/b*ln(b*x+a))

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Maxima [A]
time = 0.26, size = 53, normalized size = 0.98 \begin {gather*} -\frac {1}{2} \, b c^{3} x^{2} - \frac {8 \, a^{3} c^{3}}{b^{2} x + a b} + 5 \, a c^{3} x - \frac {12 \, a^{2} c^{3} \log \left (b x + a\right )}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*c*x+a*c)^3/(b*x+a)^2,x, algorithm="maxima")

[Out]

-1/2*b*c^3*x^2 - 8*a^3*c^3/(b^2*x + a*b) + 5*a*c^3*x - 12*a^2*c^3*log(b*x + a)/b

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Fricas [A]
time = 0.29, size = 79, normalized size = 1.46 \begin {gather*} -\frac {b^{3} c^{3} x^{3} - 9 \, a b^{2} c^{3} x^{2} - 10 \, a^{2} b c^{3} x + 16 \, a^{3} c^{3} + 24 \, {\left (a^{2} b c^{3} x + a^{3} c^{3}\right )} \log \left (b x + a\right )}{2 \, {\left (b^{2} x + a b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*c*x+a*c)^3/(b*x+a)^2,x, algorithm="fricas")

[Out]

-1/2*(b^3*c^3*x^3 - 9*a*b^2*c^3*x^2 - 10*a^2*b*c^3*x + 16*a^3*c^3 + 24*(a^2*b*c^3*x + a^3*c^3)*log(b*x + a))/(
b^2*x + a*b)

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Sympy [A]
time = 0.12, size = 51, normalized size = 0.94 \begin {gather*} - \frac {8 a^{3} c^{3}}{a b + b^{2} x} - \frac {12 a^{2} c^{3} \log {\left (a + b x \right )}}{b} + 5 a c^{3} x - \frac {b c^{3} x^{2}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*c*x+a*c)**3/(b*x+a)**2,x)

[Out]

-8*a**3*c**3/(a*b + b**2*x) - 12*a**2*c**3*log(a + b*x)/b + 5*a*c**3*x - b*c**3*x**2/2

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Giac [A]
time = 0.00, size = 66, normalized size = 1.22 \begin {gather*} \frac {-\frac {1}{2} x^{2} b^{5} c^{3}+5 x b^{4} c^{3} a}{b^{4}}-\frac {8 c^{3} a^{3}}{b \left (x b+a\right )}-\frac {12 c^{3} a^{2} \ln \left |x b+a\right |}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*c*x+a*c)^3/(b*x+a)^2,x)

[Out]

-12*a^2*c^3*log(abs(b*x + a))/b - 8*a^3*c^3/((b*x + a)*b) - 1/2*(b^5*c^3*x^2 - 10*a*b^4*c^3*x)/b^4

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Mupad [B]
time = 0.05, size = 52, normalized size = 0.96 \begin {gather*} 5\,a\,c^3\,x-\frac {b\,c^3\,x^2}{2}-\frac {12\,a^2\,c^3\,\ln \left (a+b\,x\right )}{b}-\frac {8\,a^3\,c^3}{b\,\left (a+b\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*c - b*c*x)^3/(a + b*x)^2,x)

[Out]

5*a*c^3*x - (b*c^3*x^2)/2 - (12*a^2*c^3*log(a + b*x))/b - (8*a^3*c^3)/(b*(a + b*x))

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