\(\int \frac {1}{(\frac {b c}{d}+b x)^2 (c+d x)^3} \, dx\) [1013]

Optimal result

Integrand size = 20, antiderivative size = 15 \[ \int \frac {1}{\left (\frac {b c}{d}+b x\right )^2 (c+d x)^3} \, dx=-\frac {d}{4 b^2 (c+d x)^4} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, 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} \[ \int \frac {1}{\left (\frac {b c}{d}+b x\right )^2 (c+d x)^3} \, dx=-\frac {d}{4 b^2 (c+d x)^4} \]

[In]

Int[1/(((b*c)/d + b*x)^2*(c + d*x)^3),x]

[Out]

-1/4*d/(b^2*(c + d*x)^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])

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^2 \int \frac {1}{(c+d x)^5} \, dx}{b^2} \\ & = -\frac {d}{4 b^2 (c+d x)^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (\frac {b c}{d}+b x\right )^2 (c+d x)^3} \, dx=-\frac {d}{4 b^2 (c+d x)^4} \]

[In]

Integrate[1/(((b*c)/d + b*x)^2*(c + d*x)^3),x]

[Out]

-1/4*d/(b^2*(c + d*x)^4)

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.93

method	result	size
gosper	\(-\frac {d}{4 b^{2} \left (d x +c \right )^{4}}\)	\(14\)
default	\(-\frac {d}{4 b^{2} \left (d x +c \right )^{4}}\)	\(14\)
norman	\(-\frac {d}{4 b^{2} \left (d x +c \right )^{4}}\)	\(14\)
risch	\(-\frac {d}{4 b^{2} \left (d x +c \right )^{4}}\)	\(14\)
parallelrisch	\(-\frac {d}{4 b^{2} \left (d x +c \right )^{4}}\)	\(14\)

[In]

int(1/(b*c/d+b*x)^2/(d*x+c)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (13) = 26\).

Time = 0.22 (sec) , antiderivative size = 59, normalized size of antiderivative = 3.93 \[ \int \frac {1}{\left (\frac {b c}{d}+b x\right )^2 (c+d x)^3} \, dx=-\frac {d}{4 \, {\left (b^{2} d^{4} x^{4} + 4 \, b^{2} c d^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} x^{2} + 4 \, b^{2} c^{3} d x + b^{2} c^{4}\right )}} \]

[In]

integrate(1/(b*c/d+b*x)^2/(d*x+c)^3,x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (14) = 28\).

Time = 0.23 (sec) , antiderivative size = 68, normalized size of antiderivative = 4.53 \[ \int \frac {1}{\left (\frac {b c}{d}+b x\right )^2 (c+d x)^3} \, dx=- \frac {d^{2}}{4 b^{2} c^{4} d + 16 b^{2} c^{3} d^{2} x + 24 b^{2} c^{2} d^{3} x^{2} + 16 b^{2} c d^{4} x^{3} + 4 b^{2} d^{5} x^{4}} \]

[In]

integrate(1/(b*c/d+b*x)**2/(d*x+c)**3,x)

[Out]

-d**2/(4*b**2*c**4*d + 16*b**2*c**3*d**2*x + 24*b**2*c**2*d**3*x**2 + 16*b**2*c*d**4*x**3 + 4*b**2*d**5*x**4)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (13) = 26\).

Time = 0.21 (sec) , antiderivative size = 59, normalized size of antiderivative = 3.93 \[ \int \frac {1}{\left (\frac {b c}{d}+b x\right )^2 (c+d x)^3} \, dx=-\frac {d}{4 \, {\left (b^{2} d^{4} x^{4} + 4 \, b^{2} c d^{3} x^{3} + 6 \, b^{2} c^{2} d^{2} x^{2} + 4 \, b^{2} c^{3} d x + b^{2} c^{4}\right )}} \]

[In]

integrate(1/(b*c/d+b*x)^2/(d*x+c)^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.33 \[ \int \frac {1}{\left (\frac {b c}{d}+b x\right )^2 (c+d x)^3} \, dx=-\frac {b^{2}}{4 \, {\left (b x + \frac {b c}{d}\right )}^{4} d^{3}} \]

[In]

integrate(1/(b*c/d+b*x)^2/(d*x+c)^3,x, algorithm="giac")

[Out]

-1/4*b^2/((b*x + b*c/d)^4*d^3)

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 61, normalized size of antiderivative = 4.07 \[ \int \frac {1}{\left (\frac {b c}{d}+b x\right )^2 (c+d x)^3} \, dx=-\frac {d}{4\,\left (b^2\,c^4+4\,b^2\,c^3\,d\,x+6\,b^2\,c^2\,d^2\,x^2+4\,b^2\,c\,d^3\,x^3+b^2\,d^4\,x^4\right )} \]

[In]

int(1/((b*x + (b*c)/d)^2*(c + d*x)^3),x)

[Out]

-d/(4*(b^2*c^4 + b^2*d^4*x^4 + 4*b^2*c*d^3*x^3 + 6*b^2*c^2*d^2*x^2 + 4*b^2*c^3*d*x))