\(\int \frac {(a+b x)^5}{(a c+b c x)^3} \, dx\) [1021]

Optimal result

Integrand size = 18, antiderivative size = 17 \[ \int \frac {(a+b x)^5}{(a c+b c x)^3} \, dx=\frac {(a+b x)^3}{3 b c^3} \]

[Out]

1/3*(b*x+a)^3/b/c^3

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {21, 32} \[ \int \frac {(a+b x)^5}{(a c+b c x)^3} \, dx=\frac {(a+b x)^3}{3 b c^3} \]

[In]

Int[(a + b*x)^5/(a*c + b*c*x)^3,x]

[Out]

(a + b*x)^3/(3*b*c^3)

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])

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\int (a+b x)^2 \, dx}{c^3} \\ & = \frac {(a+b x)^3}{3 b c^3} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^5}{(a c+b c x)^3} \, dx=\frac {(a+b x)^3}{3 b c^3} \]

[In]

Integrate[(a + b*x)^5/(a*c + b*c*x)^3,x]

[Out]

(a + b*x)^3/(3*b*c^3)

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94

method	result	size
default	\(\frac {\left (b x +a \right )^{3}}{3 b \,c^{3}}\)	\(16\)
gosper	\(\frac {x \left (b^{2} x^{2}+3 a b x +3 a^{2}\right )}{3 c^{3}}\)	\(25\)
parallelrisch	\(\frac {b^{2} x^{3}+3 a b \,x^{2}+3 a^{2} x}{3 c^{3}}\)	\(27\)
risch	\(\frac {b^{2} x^{3}}{3 c^{3}}+\frac {b a \,x^{2}}{c^{3}}+\frac {a^{2} x}{c^{3}}+\frac {a^{3}}{3 b \,c^{3}}\)	\(41\)
norman	\(\frac {\frac {b^{4} x^{5}}{3 c}+\frac {5 a \,b^{3} x^{4}}{3 c}+\frac {10 a^{2} b^{2} x^{3}}{3 c}-\frac {3 a^{5}}{b c}-\frac {5 a^{4} x}{c}}{c^{2} \left (b x +a \right )^{2}}\)	\(70\)

[In]

int((b*x+a)^5/(b*c*x+a*c)^3,x,method=_RETURNVERBOSE)

[Out]

1/3*(b*x+a)^3/b/c^3

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.53 \[ \int \frac {(a+b x)^5}{(a c+b c x)^3} \, dx=\frac {b^{2} x^{3} + 3 \, a b x^{2} + 3 \, a^{2} x}{3 \, c^{3}} \]

[In]

integrate((b*x+a)^5/(b*c*x+a*c)^3,x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 29 vs. \(2 (12) = 24\).

Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.71 \[ \int \frac {(a+b x)^5}{(a c+b c x)^3} \, dx=\frac {a^{2} x}{c^{3}} + \frac {a b x^{2}}{c^{3}} + \frac {b^{2} x^{3}}{3 c^{3}} \]

[In]

integrate((b*x+a)**5/(b*c*x+a*c)**3,x)

[Out]

a**2*x/c**3 + a*b*x**2/c**3 + b**2*x**3/(3*c**3)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.53 \[ \int \frac {(a+b x)^5}{(a c+b c x)^3} \, dx=\frac {b^{2} x^{3} + 3 \, a b x^{2} + 3 \, a^{2} x}{3 \, c^{3}} \]

[In]

integrate((b*x+a)^5/(b*c*x+a*c)^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.53 \[ \int \frac {(a+b x)^5}{(a c+b c x)^3} \, dx=\frac {b^{2} x^{3} + 3 \, a b x^{2} + 3 \, a^{2} x}{3 \, c^{3}} \]

[In]

integrate((b*x+a)^5/(b*c*x+a*c)^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.41 \[ \int \frac {(a+b x)^5}{(a c+b c x)^3} \, dx=\frac {x\,\left (3\,a^2+3\,a\,b\,x+b^2\,x^2\right )}{3\,c^3} \]

[In]

int((a + b*x)^5/(a*c + b*c*x)^3,x)

[Out]

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