\(\int \frac {x}{\sqrt {c x^2} (a+b x)} \, dx\) [880]

Optimal result

Integrand size = 18, antiderivative size = 20 \[ \int \frac {x}{\sqrt {c x^2} (a+b x)} \, dx=\frac {x \log (a+b x)}{b \sqrt {c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 20, 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, 31} \[ \int \frac {x}{\sqrt {c x^2} (a+b x)} \, dx=\frac {x \log (a+b x)}{b \sqrt {c x^2}} \]

[In]

Int[x/(Sqrt[c*x^2]*(a + b*x)),x]

[Out]

(x*Log[a + b*x])/(b*Sqrt[c*x^2])

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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {x \int \frac {1}{a+b x} \, dx}{\sqrt {c x^2}} \\ & = \frac {x \log (a+b x)}{b \sqrt {c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int \frac {x}{\sqrt {c x^2} (a+b x)} \, dx=\frac {x \log (a+b x)}{b \sqrt {c x^2}} \]

[In]

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

[Out]

(x*Log[a + b*x])/(b*Sqrt[c*x^2])

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95

method	result	size
default	\(\frac {x \ln \left (b x +a \right )}{b \sqrt {c \,x^{2}}}\)	\(19\)
risch	\(\frac {x \ln \left (b x +a \right )}{b \sqrt {c \,x^{2}}}\)	\(19\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.15 \[ \int \frac {x}{\sqrt {c x^2} (a+b x)} \, dx=\frac {\sqrt {c x^{2}} \log \left (b x + a\right )}{b c x} \]

[In]

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

[Out]

sqrt(c*x^2)*log(b*x + a)/(b*c*x)

Sympy [F]

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

[In]

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

[Out]

Integral(x/(sqrt(c*x**2)*(a + b*x)), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 46 vs. \(2 (18) = 36\).

Time = 0.23 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.30 \[ \int \frac {x}{\sqrt {c x^2} (a+b x)} \, dx=\frac {\left (-1\right )^{\frac {2 \, a c x}{b}} \log \left (-\frac {2 \, a c x}{b {\left | b x + a \right |}}\right )}{b \sqrt {c}} + \frac {\log \left (b x\right )}{b \sqrt {c}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.60 \[ \int \frac {x}{\sqrt {c x^2} (a+b x)} \, dx=-\frac {\log \left ({\left | a \right |}\right ) \mathrm {sgn}\left (x\right )}{b \sqrt {c}} + \frac {\log \left ({\left | b x + a \right |}\right )}{b \sqrt {c} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{\sqrt {c x^2} (a+b x)} \, dx=\int \frac {x}{\sqrt {c\,x^2}\,\left (a+b\,x\right )} \,d x \]

[In]

int(x/((c*x^2)^(1/2)*(a + b*x)),x)

[Out]

int(x/((c*x^2)^(1/2)*(a + b*x)), x)