\(\int \frac {1}{(\frac {b c}{d}+b x)^3 (c+d x)^3} \, dx\) [1014]

Optimal result

Integrand size = 20, antiderivative size = 17 \[ \int \frac {1}{\left (\frac {b c}{d}+b x\right )^3 (c+d x)^3} \, dx=-\frac {d^2}{5 b^3 (c+d x)^5} \]

[Out]

-1/5*d^2/b^3/(d*x+c)^5

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, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {21, 32} \[ \int \frac {1}{\left (\frac {b c}{d}+b x\right )^3 (c+d x)^3} \, dx=-\frac {d^2}{5 b^3 (c+d x)^5} \]

[In]

Int[1/(((b*c)/d + b*x)^3*(c + d*x)^3),x]

[Out]

-1/5*d^2/(b^3*(c + d*x)^5)

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 {d^3 \int \frac {1}{(c+d x)^6} \, dx}{b^3} \\ & = -\frac {d^2}{5 b^3 (c+d x)^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (\frac {b c}{d}+b x\right )^3 (c+d x)^3} \, dx=-\frac {d^2}{5 b^3 (c+d x)^5} \]

[In]

Integrate[1/(((b*c)/d + b*x)^3*(c + d*x)^3),x]

[Out]

-1/5*d^2/(b^3*(c + d*x)^5)

Maple [A] (verified)

Time = 0.17 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94

method	result	size
gosper	\(-\frac {d^{2}}{5 b^{3} \left (d x +c \right )^{5}}\)	\(16\)
default	\(-\frac {d^{2}}{5 b^{3} \left (d x +c \right )^{5}}\)	\(16\)
norman	\(-\frac {d^{2}}{5 b^{3} \left (d x +c \right )^{5}}\)	\(16\)
risch	\(-\frac {d^{2}}{5 b^{3} \left (d x +c \right )^{5}}\)	\(16\)
parallelrisch	\(-\frac {d^{2}}{5 b^{3} \left (d x +c \right )^{5}}\)	\(16\)

[In]

int(1/(b*c/d+b*x)^3/(d*x+c)^3,x,method=_RETURNVERBOSE)

[Out]

-1/5*d^2/b^3/(d*x+c)^5

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (15) = 30\).

Time = 0.23 (sec) , antiderivative size = 75, normalized size of antiderivative = 4.41 \[ \int \frac {1}{\left (\frac {b c}{d}+b x\right )^3 (c+d x)^3} \, dx=-\frac {d^{2}}{5 \, {\left (b^{3} d^{5} x^{5} + 5 \, b^{3} c d^{4} x^{4} + 10 \, b^{3} c^{2} d^{3} x^{3} + 10 \, b^{3} c^{3} d^{2} x^{2} + 5 \, b^{3} c^{4} d x + b^{3} c^{5}\right )}} \]

[In]

integrate(1/(b*c/d+b*x)^3/(d*x+c)^3,x, algorithm="fricas")

[Out]

-1/5*d^2/(b^3*d^5*x^5 + 5*b^3*c*d^4*x^4 + 10*b^3*c^2*d^3*x^3 + 10*b^3*c^3*d^2*x^2 + 5*b^3*c^4*d*x + b^3*c^5)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 83 vs. \(2 (15) = 30\).

Time = 0.32 (sec) , antiderivative size = 83, normalized size of antiderivative = 4.88 \[ \int \frac {1}{\left (\frac {b c}{d}+b x\right )^3 (c+d x)^3} \, dx=- \frac {d^{3}}{5 b^{3} c^{5} d + 25 b^{3} c^{4} d^{2} x + 50 b^{3} c^{3} d^{3} x^{2} + 50 b^{3} c^{2} d^{4} x^{3} + 25 b^{3} c d^{5} x^{4} + 5 b^{3} d^{6} x^{5}} \]

[In]

integrate(1/(b*c/d+b*x)**3/(d*x+c)**3,x)

[Out]

-d**3/(5*b**3*c**5*d + 25*b**3*c**4*d**2*x + 50*b**3*c**3*d**3*x**2 + 50*b**3*c**2*d**4*x**3 + 25*b**3*c*d**5*
x**4 + 5*b**3*d**6*x**5)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (15) = 30\).

Time = 0.20 (sec) , antiderivative size = 75, normalized size of antiderivative = 4.41 \[ \int \frac {1}{\left (\frac {b c}{d}+b x\right )^3 (c+d x)^3} \, dx=-\frac {d^{2}}{5 \, {\left (b^{3} d^{5} x^{5} + 5 \, b^{3} c d^{4} x^{4} + 10 \, b^{3} c^{2} d^{3} x^{3} + 10 \, b^{3} c^{3} d^{2} x^{2} + 5 \, b^{3} c^{4} d x + b^{3} c^{5}\right )}} \]

[In]

integrate(1/(b*c/d+b*x)^3/(d*x+c)^3,x, algorithm="maxima")

[Out]

-1/5*d^2/(b^3*d^5*x^5 + 5*b^3*c*d^4*x^4 + 10*b^3*c^2*d^3*x^3 + 10*b^3*c^3*d^2*x^2 + 5*b^3*c^4*d*x + b^3*c^5)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\left (\frac {b c}{d}+b x\right )^3 (c+d x)^3} \, dx=-\frac {d^{2}}{5 \, {\left (d x + c\right )}^{5} b^{3}} \]

[In]

integrate(1/(b*c/d+b*x)^3/(d*x+c)^3,x, algorithm="giac")

[Out]

-1/5*d^2/((d*x + c)^5*b^3)

Mupad [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 77, normalized size of antiderivative = 4.53 \[ \int \frac {1}{\left (\frac {b c}{d}+b x\right )^3 (c+d x)^3} \, dx=-\frac {d^2}{5\,\left (b^3\,c^5+5\,b^3\,c^4\,d\,x+10\,b^3\,c^3\,d^2\,x^2+10\,b^3\,c^2\,d^3\,x^3+5\,b^3\,c\,d^4\,x^4+b^3\,d^5\,x^5\right )} \]

[In]

int(1/((b*x + (b*c)/d)^3*(c + d*x)^3),x)

[Out]

-d^2/(5*(b^3*c^5 + b^3*d^5*x^5 + 5*b^3*c*d^4*x^4 + 10*b^3*c^3*d^2*x^2 + 10*b^3*c^2*d^3*x^3 + 5*b^3*c^4*d*x))