\(\int \frac {x}{\sqrt {c x^2} (a+b x)^2} \, dx\) [913]

Optimal result

Integrand size = 18, antiderivative size = 22 \[ \int \frac {x}{\sqrt {c x^2} (a+b x)^2} \, dx=-\frac {x}{b \sqrt {c x^2} (a+b x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 22, 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 = {15, 32} \[ \int \frac {x}{\sqrt {c x^2} (a+b x)^2} \, dx=-\frac {x}{b \sqrt {c x^2} (a+b x)} \]

[In]

Int[x/(Sqrt[c*x^2]*(a + b*x)^2),x]

[Out]

-(x/(b*Sqrt[c*x^2]*(a + b*x)))

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[a^IntPart[m]*((a*x^n)^FracPart[m]/x^(n*FracPart[m])), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.00 \[ \int \frac {x}{\sqrt {c x^2} (a+b x)^2} \, dx=-\frac {x}{b \sqrt {c x^2} (a+b x)} \]

[In]

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

[Out]

-(x/(b*Sqrt[c*x^2]*(a + b*x)))

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95

method	result	size
gosper	\(-\frac {x}{b \left (b x +a \right ) \sqrt {c \,x^{2}}}\)	\(21\)
default	\(-\frac {x}{b \left (b x +a \right ) \sqrt {c \,x^{2}}}\)	\(21\)
risch	\(-\frac {x}{b \left (b x +a \right ) \sqrt {c \,x^{2}}}\)	\(21\)
trager	\(\frac {\left (-1+x \right ) \sqrt {c \,x^{2}}}{c \left (b x +a \right ) \left (a +b \right ) x}\)	\(30\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {x}{\sqrt {c x^2} (a+b x)^2} \, dx=-\frac {\sqrt {c x^{2}}}{b^{2} c x^{2} + a b c x} \]

[In]

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

[Out]

-sqrt(c*x^2)/(b^2*c*x^2 + a*b*c*x)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (17) = 34\).

Time = 0.43 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.91 \[ \int \frac {x}{\sqrt {c x^2} (a+b x)^2} \, dx=\begin {cases} - \frac {x}{a b \sqrt {c x^{2}} + b^{2} x \sqrt {c x^{2}}} & \text {for}\: b \neq 0 \\\frac {x^{2}}{a^{2} \sqrt {c x^{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-x/(a*b*sqrt(c*x**2) + b**2*x*sqrt(c*x**2)), Ne(b, 0)), (x**2/(a**2*sqrt(c*x**2)), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.95 \[ \int \frac {x}{\sqrt {c x^2} (a+b x)^2} \, dx=\frac {\sqrt {c x^{2}}}{a b c x + a^{2} c} \]

[In]

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

[Out]

sqrt(c*x^2)/(a*b*c*x + a^2*c)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45 \[ \int \frac {x}{\sqrt {c x^2} (a+b x)^2} \, dx=\frac {\mathrm {sgn}\left (x\right )}{a b \sqrt {c}} - \frac {1}{{\left (b x + a\right )} b \sqrt {c} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

sgn(x)/(a*b*sqrt(c)) - 1/((b*x + a)*b*sqrt(c)*sgn(x))

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.14 \[ \int \frac {x}{\sqrt {c x^2} (a+b x)^2} \, dx=-\frac {\sqrt {c\,x^2}}{b\,c\,x\,\left (a+b\,x\right )} \]

[In]

int(x/((c*x^2)^(1/2)*(a + b*x)^2),x)

[Out]

-(c*x^2)^(1/2)/(b*c*x*(a + b*x))