\(\int \frac {(a+b x)^5}{(a c+b c x)^2} \, dx\) [1020]

Optimal result

Integrand size = 18, antiderivative size = 17 \[ \int \frac {(a+b x)^5}{(a c+b c x)^2} \, dx=\frac {(a+b x)^4}{4 b c^2} \]

[Out]

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

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

[In]

Int[(a + b*x)^5/(a*c + b*c*x)^2,x]

[Out]

(a + b*x)^4/(4*b*c^2)

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)^3 \, dx}{c^2} \\ & = \frac {(a+b x)^4}{4 b c^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {(a+b x)^5}{(a c+b c x)^2} \, dx=\frac {(a+b x)^4}{4 b c^2} \]

[In]

Integrate[(a + b*x)^5/(a*c + b*c*x)^2,x]

[Out]

(a + b*x)^4/(4*b*c^2)

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94

method	result	size
default	\(\frac {\left (b x +a \right )^{4}}{4 b \,c^{2}}\)	\(16\)
gosper	\(\frac {x \left (b^{3} x^{3}+4 a \,b^{2} x^{2}+6 a^{2} b x +4 a^{3}\right )}{4 c^{2}}\)	\(36\)
parallelrisch	\(\frac {b^{3} x^{4}+4 a \,b^{2} x^{3}+6 a^{2} b \,x^{2}+4 a^{3} x}{4 c^{2}}\)	\(38\)
risch	\(\frac {b^{3} x^{4}}{4 c^{2}}+\frac {b^{2} a \,x^{3}}{c^{2}}+\frac {3 b \,a^{2} x^{2}}{2 c^{2}}+\frac {a^{3} x}{c^{2}}+\frac {a^{4}}{4 b \,c^{2}}\)	\(55\)
norman	\(\frac {\frac {a^{4} x}{c}+\frac {b^{4} x^{5}}{4 c}+\frac {5 a \,b^{3} x^{4}}{4 c}+\frac {5 a^{2} b^{2} x^{3}}{2 c}+\frac {5 a^{3} b \,x^{2}}{2 c}}{c \left (b x +a \right )}\)	\(70\)

[In]

int((b*x+a)^5/(b*c*x+a*c)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (15) = 30\).

Time = 0.21 (sec) , antiderivative size = 37, normalized size of antiderivative = 2.18 \[ \int \frac {(a+b x)^5}{(a c+b c x)^2} \, dx=\frac {b^{3} x^{4} + 4 \, a b^{2} x^{3} + 6 \, a^{2} b x^{2} + 4 \, a^{3} x}{4 \, c^{2}} \]

[In]

integrate((b*x+a)^5/(b*c*x+a*c)^2,x, algorithm="fricas")

[Out]

1/4*(b^3*x^4 + 4*a*b^2*x^3 + 6*a^2*b*x^2 + 4*a^3*x)/c^2

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (12) = 24\).

Time = 0.06 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.71 \[ \int \frac {(a+b x)^5}{(a c+b c x)^2} \, dx=\frac {a^{3} x}{c^{2}} + \frac {3 a^{2} b x^{2}}{2 c^{2}} + \frac {a b^{2} x^{3}}{c^{2}} + \frac {b^{3} x^{4}}{4 c^{2}} \]

[In]

integrate((b*x+a)**5/(b*c*x+a*c)**2,x)

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (15) = 30\).

Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 2.18 \[ \int \frac {(a+b x)^5}{(a c+b c x)^2} \, dx=\frac {b^{3} x^{4} + 4 \, a b^{2} x^{3} + 6 \, a^{2} b x^{2} + 4 \, a^{3} x}{4 \, c^{2}} \]

[In]

integrate((b*x+a)^5/(b*c*x+a*c)^2,x, algorithm="maxima")

[Out]

1/4*(b^3*x^4 + 4*a*b^2*x^3 + 6*a^2*b*x^2 + 4*a^3*x)/c^2

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.06 \[ \int \frac {(a+b x)^5}{(a c+b c x)^2} \, dx=\frac {{\left (b c x + a c\right )}^{4}}{4 \, b c^{6}} \]

[In]

integrate((b*x+a)^5/(b*c*x+a*c)^2,x, algorithm="giac")

[Out]

1/4*(b*c*x + a*c)^4/(b*c^6)

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 43, normalized size of antiderivative = 2.53 \[ \int \frac {(a+b x)^5}{(a c+b c x)^2} \, dx=\frac {a^3\,x}{c^2}+\frac {b^3\,x^4}{4\,c^2}+\frac {3\,a^2\,b\,x^2}{2\,c^2}+\frac {a\,b^2\,x^3}{c^2} \]

[In]

int((a + b*x)^5/(a*c + b*c*x)^2,x)

[Out]

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