\(\int \frac {\sqrt {c x^2}}{x (a+b x)^2} \, dx\) [897]

Optimal result

Integrand size = 20, antiderivative size = 24 \[ \int \frac {\sqrt {c x^2}}{x (a+b x)^2} \, dx=-\frac {\sqrt {c x^2}}{b x (a+b x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 24, 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 = {15, 32} \[ \int \frac {\sqrt {c x^2}}{x (a+b x)^2} \, dx=-\frac {\sqrt {c x^2}}{b x (a+b x)} \]

[In]

Int[Sqrt[c*x^2]/(x*(a + b*x)^2),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {\sqrt {c x^2}}{x (a+b x)^2} \, dx=-\frac {c x}{b \sqrt {c x^2} (a+b x)} \]

[In]

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

[Out]

-((c*x)/(b*Sqrt[c*x^2]*(a + b*x)))

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96

method	result	size
gosper	\(-\frac {\sqrt {c \,x^{2}}}{b x \left (b x +a \right )}\)	\(23\)
default	\(-\frac {\sqrt {c \,x^{2}}}{b x \left (b x +a \right )}\)	\(23\)
risch	\(-\frac {\sqrt {c \,x^{2}}}{b x \left (b x +a \right )}\)	\(23\)
trager	\(\frac {\left (-1+x \right ) \sqrt {c \,x^{2}}}{\left (b x +a \right ) \left (a +b \right ) x}\)	\(27\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {\sqrt {c x^2}}{x (a+b x)^2} \, dx=-\frac {\sqrt {c x^{2}}}{b^{2} x^{2} + a b x} \]

[In]

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

[Out]

-sqrt(c*x^2)/(b^2*x^2 + a*b*x)

Sympy [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.33 \[ \int \frac {\sqrt {c x^2}}{x (a+b x)^2} \, dx=\begin {cases} - \frac {\sqrt {c x^{2}}}{a b x + b^{2} x^{2}} & \text {for}\: b \neq 0 \\\frac {\sqrt {c x^{2}}}{a^{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.67 \[ \int \frac {\sqrt {c x^2}}{x (a+b x)^2} \, dx=-\frac {\sqrt {c}}{b^{2} x + a b} \]

[In]

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

[Out]

-sqrt(c)/(b^2*x + a*b)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {\sqrt {c x^2}}{x (a+b x)^2} \, dx=-\sqrt {c} {\left (\frac {\mathrm {sgn}\left (x\right )}{{\left (b x + a\right )} b} - \frac {\mathrm {sgn}\left (x\right )}{a b}\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {\sqrt {c x^2}}{x (a+b x)^2} \, dx=-\frac {\sqrt {c\,x^2}}{b\,x\,\left (a+b\,x\right )} \]

[In]

int((c*x^2)^(1/2)/(x*(a + b*x)^2),x)

[Out]

-(c*x^2)^(1/2)/(b*x*(a + b*x))