\(\int \frac {\frac {b c}{d}+b x}{(c+d x)^3} \, dx\) [1011]

Optimal result

Integrand size = 18, antiderivative size = 13 \[ \int \frac {\frac {b c}{d}+b x}{(c+d x)^3} \, dx=-\frac {b}{d^2 (c+d x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, 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} \[ \int \frac {\frac {b c}{d}+b x}{(c+d x)^3} \, dx=-\frac {b}{d^2 (c+d x)} \]

[In]

Int[((b*c)/d + b*x)/(c + d*x)^3,x]

[Out]

-(b/(d^2*(c + d*x)))

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {\frac {b c}{d}+b x}{(c+d x)^3} \, dx=-\frac {b}{d^2 (c+d x)} \]

[In]

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

[Out]

-(b/(d^2*(c + d*x)))

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.08

method	result	size
gosper	\(-\frac {b}{d^{2} \left (d x +c \right )}\)	\(14\)
default	\(-\frac {b}{d^{2} \left (d x +c \right )}\)	\(14\)
risch	\(-\frac {b}{d^{2} \left (d x +c \right )}\)	\(14\)
parallelrisch	\(\frac {b x}{c \left (d x +c \right ) d}\)	\(17\)
norman	\(\frac {-\frac {c b}{d^{2}}-\frac {b x}{d}}{\left (d x +c \right )^{2}}\)	\(24\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.23 \[ \int \frac {\frac {b c}{d}+b x}{(c+d x)^3} \, dx=-\frac {b}{d^{3} x + c d^{2}} \]

[In]

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

[Out]

-b/(d^3*x + c*d^2)

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92 \[ \int \frac {\frac {b c}{d}+b x}{(c+d x)^3} \, dx=- \frac {b}{c d^{2} + d^{3} x} \]

[In]

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

[Out]

-b/(c*d**2 + d**3*x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.23 \[ \int \frac {\frac {b c}{d}+b x}{(c+d x)^3} \, dx=-\frac {b}{d^{3} x + c d^{2}} \]

[In]

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

[Out]

-b/(d^3*x + c*d^2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {\frac {b c}{d}+b x}{(c+d x)^3} \, dx=-\frac {b}{{\left (d x + c\right )} d^{2}} \]

[In]

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

[Out]

-b/((d*x + c)*d^2)

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {\frac {b c}{d}+b x}{(c+d x)^3} \, dx=-\frac {b}{d^2\,\left (c+d\,x\right )} \]

[In]

int((b*x + (b*c)/d)/(c + d*x)^3,x)

[Out]

-b/(d^2*(c + d*x))