\(\int \frac {(a+b x) (a c-b c x)^4}{x^2} \, dx\) [19]

Optimal result

Integrand size = 20, antiderivative size = 73 \[ \int \frac {(a+b x) (a c-b c x)^4}{x^2} \, dx=-\frac {a^5 c^4}{x}+2 a^3 b^2 c^4 x+a^2 b^3 c^4 x^2-a b^4 c^4 x^3+\frac {1}{4} b^5 c^4 x^4-3 a^4 b c^4 \log (x) \]

[Out]

-a^5*c^4/x+2*a^3*b^2*c^4*x+a^2*b^3*c^4*x^2-a*b^4*c^4*x^3+1/4*b^5*c^4*x^4-3*a^4*b*c^4*ln(x)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {76} \[ \int \frac {(a+b x) (a c-b c x)^4}{x^2} \, dx=-\frac {a^5 c^4}{x}-3 a^4 b c^4 \log (x)+2 a^3 b^2 c^4 x+a^2 b^3 c^4 x^2-a b^4 c^4 x^3+\frac {1}{4} b^5 c^4 x^4 \]

[In]

Int[((a + b*x)*(a*c - b*c*x)^4)/x^2,x]

[Out]

-((a^5*c^4)/x) + 2*a^3*b^2*c^4*x + a^2*b^3*c^4*x^2 - a*b^4*c^4*x^3 + (b^5*c^4*x^4)/4 - 3*a^4*b*c^4*Log[x]

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && EqQ[b*e + a*f, 0] &&  !(ILtQ[n
+ p + 2, 0] && GtQ[n + 2*p, 0])

Rubi steps \begin{align*} \text {integral}& = \int \left (2 a^3 b^2 c^4+\frac {a^5 c^4}{x^2}-\frac {3 a^4 b c^4}{x}+2 a^2 b^3 c^4 x-3 a b^4 c^4 x^2+b^5 c^4 x^3\right ) \, dx \\ & = -\frac {a^5 c^4}{x}+2 a^3 b^2 c^4 x+a^2 b^3 c^4 x^2-a b^4 c^4 x^3+\frac {1}{4} b^5 c^4 x^4-3 a^4 b c^4 \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x) (a c-b c x)^4}{x^2} \, dx=-\frac {a^5 c^4}{x}+2 a^3 b^2 c^4 x+a^2 b^3 c^4 x^2-a b^4 c^4 x^3+\frac {1}{4} b^5 c^4 x^4-3 a^4 b c^4 \log (x) \]

[In]

Integrate[((a + b*x)*(a*c - b*c*x)^4)/x^2,x]

[Out]

-((a^5*c^4)/x) + 2*a^3*b^2*c^4*x + a^2*b^3*c^4*x^2 - a*b^4*c^4*x^3 + (b^5*c^4*x^4)/4 - 3*a^4*b*c^4*Log[x]

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.79

method	result	size
default	\(c^{4} \left (\frac {b^{5} x^{4}}{4}-a \,b^{4} x^{3}+a^{2} b^{3} x^{2}+2 a^{3} b^{2} x -3 a^{4} b \ln \left (x \right )-\frac {a^{5}}{x}\right )\)	\(58\)
risch	\(-\frac {a^{5} c^{4}}{x}+2 a^{3} b^{2} c^{4} x +a^{2} b^{3} c^{4} x^{2}-a \,b^{4} c^{4} x^{3}+\frac {b^{5} c^{4} x^{4}}{4}-3 a^{4} b \,c^{4} \ln \left (x \right )\)	\(72\)
norman	\(\frac {a^{2} b^{3} c^{4} x^{3}-a^{5} c^{4}+\frac {1}{4} b^{5} c^{4} x^{5}-a \,b^{4} c^{4} x^{4}+2 a^{3} b^{2} c^{4} x^{2}}{x}-3 a^{4} b \,c^{4} \ln \left (x \right )\)	\(76\)
parallelrisch	\(-\frac {-b^{5} c^{4} x^{5}+4 a \,b^{4} c^{4} x^{4}-4 a^{2} b^{3} c^{4} x^{3}+12 a^{4} c^{4} b \ln \left (x \right ) x -8 a^{3} b^{2} c^{4} x^{2}+4 a^{5} c^{4}}{4 x}\)	\(78\)

[In]

int((b*x+a)*(-b*c*x+a*c)^4/x^2,x,method=_RETURNVERBOSE)

[Out]

c^4*(1/4*b^5*x^4-a*b^4*x^3+a^2*b^3*x^2+2*a^3*b^2*x-3*a^4*b*ln(x)-a^5/x)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.04 \[ \int \frac {(a+b x) (a c-b c x)^4}{x^2} \, dx=\frac {b^{5} c^{4} x^{5} - 4 \, a b^{4} c^{4} x^{4} + 4 \, a^{2} b^{3} c^{4} x^{3} + 8 \, a^{3} b^{2} c^{4} x^{2} - 12 \, a^{4} b c^{4} x \log \left (x\right ) - 4 \, a^{5} c^{4}}{4 \, x} \]

[In]

integrate((b*x+a)*(-b*c*x+a*c)^4/x^2,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.97 \[ \int \frac {(a+b x) (a c-b c x)^4}{x^2} \, dx=- \frac {a^{5} c^{4}}{x} - 3 a^{4} b c^{4} \log {\left (x \right )} + 2 a^{3} b^{2} c^{4} x + a^{2} b^{3} c^{4} x^{2} - a b^{4} c^{4} x^{3} + \frac {b^{5} c^{4} x^{4}}{4} \]

[In]

integrate((b*x+a)*(-b*c*x+a*c)**4/x**2,x)

[Out]

-a**5*c**4/x - 3*a**4*b*c**4*log(x) + 2*a**3*b**2*c**4*x + a**2*b**3*c**4*x**2 - a*b**4*c**4*x**3 + b**5*c**4*
x**4/4

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.97 \[ \int \frac {(a+b x) (a c-b c x)^4}{x^2} \, dx=\frac {1}{4} \, b^{5} c^{4} x^{4} - a b^{4} c^{4} x^{3} + a^{2} b^{3} c^{4} x^{2} + 2 \, a^{3} b^{2} c^{4} x - 3 \, a^{4} b c^{4} \log \left (x\right ) - \frac {a^{5} c^{4}}{x} \]

[In]

integrate((b*x+a)*(-b*c*x+a*c)^4/x^2,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.99 \[ \int \frac {(a+b x) (a c-b c x)^4}{x^2} \, dx=\frac {1}{4} \, b^{5} c^{4} x^{4} - a b^{4} c^{4} x^{3} + a^{2} b^{3} c^{4} x^{2} + 2 \, a^{3} b^{2} c^{4} x - 3 \, a^{4} b c^{4} \log \left ({\left | x \right |}\right ) - \frac {a^{5} c^{4}}{x} \]

[In]

integrate((b*x+a)*(-b*c*x+a*c)^4/x^2,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.97 \[ \int \frac {(a+b x) (a c-b c x)^4}{x^2} \, dx=\frac {b^5\,c^4\,x^4}{4}-\frac {a^5\,c^4}{x}+2\,a^3\,b^2\,c^4\,x-a\,b^4\,c^4\,x^3-3\,a^4\,b\,c^4\,\ln \left (x\right )+a^2\,b^3\,c^4\,x^2 \]

[In]

int(((a*c - b*c*x)^4*(a + b*x))/x^2,x)

[Out]

(b^5*c^4*x^4)/4 - (a^5*c^4)/x + 2*a^3*b^2*c^4*x - a*b^4*c^4*x^3 - 3*a^4*b*c^4*log(x) + a^2*b^3*c^4*x^2