\(\int \frac {(a+b x)^5}{(a c+b c x)^4} \, dx\) [1022]

Optimal result

Integrand size = 18, antiderivative size = 18 \[ \int \frac {(a+b x)^5}{(a c+b c x)^4} \, dx=\frac {a x}{c^4}+\frac {b x^2}{2 c^4} \]

[Out]

a*x/c^4+1/2*b*x^2/c^4

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {21} \[ \int \frac {(a+b x)^5}{(a c+b c x)^4} \, dx=\frac {a x}{c^4}+\frac {b x^2}{2 c^4} \]

[In]

Int[(a + b*x)^5/(a*c + b*c*x)^4,x]

[Out]

(a*x)/c^4 + (b*x^2)/(2*c^4)

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rubi steps \begin{align*} \text {integral}& = \frac {\int (a+b x) \, dx}{c^4} \\ & = \frac {a x}{c^4}+\frac {b x^2}{2 c^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int \frac {(a+b x)^5}{(a c+b c x)^4} \, dx=\frac {a x+\frac {b x^2}{2}}{c^4} \]

[In]

Integrate[(a + b*x)^5/(a*c + b*c*x)^4,x]

[Out]

(a*x + (b*x^2)/2)/c^4

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78

method	result	size
gosper	\(\frac {x \left (b x +2 a \right )}{2 c^{4}}\)	\(14\)
default	\(\frac {a x +\frac {1}{2} b \,x^{2}}{c^{4}}\)	\(15\)
parallelrisch	\(\frac {b \,x^{2}+2 a x}{2 c^{4}}\)	\(16\)
risch	\(\frac {a x}{c^{4}}+\frac {b \,x^{2}}{2 c^{4}}\)	\(17\)
norman	\(\frac {\frac {b^{4} x^{5}}{2 c}+\frac {5 a \,b^{3} x^{4}}{2 c}-\frac {9 a^{5}}{2 b c}-\frac {10 a^{3} b \,x^{2}}{c}-\frac {25 a^{4} x}{2 c}}{c^{3} \left (b x +a \right )^{3}}\)	\(68\)

[In]

int((b*x+a)^5/(b*c*x+a*c)^4,x,method=_RETURNVERBOSE)

[Out]

1/2*x*(b*x+2*a)/c^4

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {(a+b x)^5}{(a c+b c x)^4} \, dx=\frac {b x^{2} + 2 \, a x}{2 \, c^{4}} \]

[In]

integrate((b*x+a)^5/(b*c*x+a*c)^4,x, algorithm="fricas")

[Out]

1/2*(b*x^2 + 2*a*x)/c^4

Sympy [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {(a+b x)^5}{(a c+b c x)^4} \, dx=\frac {a x}{c^{4}} + \frac {b x^{2}}{2 c^{4}} \]

[In]

integrate((b*x+a)**5/(b*c*x+a*c)**4,x)

[Out]

a*x/c**4 + b*x**2/(2*c**4)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {(a+b x)^5}{(a c+b c x)^4} \, dx=\frac {b x^{2} + 2 \, a x}{2 \, c^{4}} \]

[In]

integrate((b*x+a)^5/(b*c*x+a*c)^4,x, algorithm="maxima")

[Out]

1/2*(b*x^2 + 2*a*x)/c^4

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {(a+b x)^5}{(a c+b c x)^4} \, dx=\frac {b x^{2} + 2 \, a x}{2 \, c^{4}} \]

[In]

integrate((b*x+a)^5/(b*c*x+a*c)^4,x, algorithm="giac")

[Out]

1/2*(b*x^2 + 2*a*x)/c^4

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.72 \[ \int \frac {(a+b x)^5}{(a c+b c x)^4} \, dx=\frac {x\,\left (2\,a+b\,x\right )}{2\,c^4} \]

[In]

int((a + b*x)^5/(a*c + b*c*x)^4,x)

[Out]

(x*(2*a + b*x))/(2*c^4)