\(\int \frac {1}{(a+b x)^3 (\frac {a d}{b}+d x)^3} \, dx\) [1006]

Optimal result

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

[Out]

-1/5*b^2/d^3/(b*x+a)^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} \[ \int \frac {1}{(a+b x)^3 \left (\frac {a d}{b}+d x\right )^3} \, dx=-\frac {b^2}{5 d^3 (a+b x)^5} \]

[In]

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

[Out]

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

Mathematica [A] (verified)

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

[In]

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

[Out]

-1/5*b^2/(d^3*(a + b*x)^5)

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94

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

[In]

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

[Out]

-1/5*b^2/d^3/(b*x+a)^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 \[ \int \frac {1}{(a+b x)^3 \left (\frac {a d}{b}+d x\right )^3} \, dx=-\frac {b^{2}}{5 \, {\left (b^{5} d^{3} x^{5} + 5 \, a b^{4} d^{3} x^{4} + 10 \, a^{2} b^{3} d^{3} x^{3} + 10 \, a^{3} b^{2} d^{3} x^{2} + 5 \, a^{4} b d^{3} x + a^{5} d^{3}\right )}} \]

[In]

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

[Out]

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

Sympy [B] (verification not implemented)

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

Time = 0.34 (sec) , antiderivative size = 83, normalized size of antiderivative = 4.88 \[ \int \frac {1}{(a+b x)^3 \left (\frac {a d}{b}+d x\right )^3} \, dx=- \frac {b^{3}}{5 a^{5} b d^{3} + 25 a^{4} b^{2} d^{3} x + 50 a^{3} b^{3} d^{3} x^{2} + 50 a^{2} b^{4} d^{3} x^{3} + 25 a b^{5} d^{3} x^{4} + 5 b^{6} d^{3} x^{5}} \]

[In]

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

[Out]

-b**3/(5*a**5*b*d**3 + 25*a**4*b**2*d**3*x + 50*a**3*b**3*d**3*x**2 + 50*a**2*b**4*d**3*x**3 + 25*a*b**5*d**3*
x**4 + 5*b**6*d**3*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 \[ \int \frac {1}{(a+b x)^3 \left (\frac {a d}{b}+d x\right )^3} \, dx=-\frac {b^{2}}{5 \, {\left (b^{5} d^{3} x^{5} + 5 \, a b^{4} d^{3} x^{4} + 10 \, a^{2} b^{3} d^{3} x^{3} + 10 \, a^{3} b^{2} d^{3} x^{2} + 5 \, a^{4} b d^{3} x + a^{5} d^{3}\right )}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

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

[In]

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

[Out]

-1/5*b^2/((b*x + a)^5*d^3)

Mupad [B] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 77, normalized size of antiderivative = 4.53 \[ \int \frac {1}{(a+b x)^3 \left (\frac {a d}{b}+d x\right )^3} \, dx=-\frac {b^2}{5\,\left (a^5\,d^3+5\,a^4\,b\,d^3\,x+10\,a^3\,b^2\,d^3\,x^2+10\,a^2\,b^3\,d^3\,x^3+5\,a\,b^4\,d^3\,x^4+b^5\,d^3\,x^5\right )} \]

[In]

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

[Out]

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