\(\int \frac {1}{\sqrt {c (a+b x)^2}} \, dx\) [2805]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 13, antiderivative size = 28 \[ \int \frac {1}{\sqrt {c (a+b x)^2}} \, dx=\frac {(a+b x) \log (a+b x)}{b \sqrt {c (a+b x)^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {253, 15, 29} \[ \int \frac {1}{\sqrt {c (a+b x)^2}} \, dx=\frac {(a+b x) \log (a+b x)}{b \sqrt {c (a+b x)^2}} \]

[In]

Int[1/Sqrt[c*(a + b*x)^2],x]

[Out]

((a + b*x)*Log[a + b*x])/(b*Sqrt[c*(a + b*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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 253

Int[((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[1/Coefficient[v, x, 1], Subst[Int[(a + b*x^n)^p, x], x,
v], x] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && NeQ[v, x]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {c (a+b x)^2}} \, dx=\frac {(a+b x) \log (a+b x)}{b \sqrt {c (a+b x)^2}} \]

[In]

Integrate[1/Sqrt[c*(a + b*x)^2],x]

[Out]

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

Maple [A] (verified)

Time = 3.77 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96

method result size
default \(\frac {\left (b x +a \right ) \ln \left (b x +a \right )}{b \sqrt {c \left (b x +a \right )^{2}}}\) \(27\)
risch \(\frac {\left (b x +a \right ) \ln \left (b x +a \right )}{b \sqrt {c \left (b x +a \right )^{2}}}\) \(27\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.50 \[ \int \frac {1}{\sqrt {c (a+b x)^2}} \, dx=\frac {\sqrt {b^{2} c x^{2} + 2 \, a b c x + a^{2} c} \log \left (b x + a\right )}{b^{2} c x + a b c} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.93 \[ \int \frac {1}{\sqrt {c (a+b x)^2}} \, dx=\frac {\left (\frac {a}{b} + x\right ) \log {\left (\frac {a}{b} + x \right )}}{\sqrt {b^{2} c \left (\frac {a}{b} + x\right )^{2}}} \]

[In]

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

[Out]

(a/b + x)*log(a/b + x)/sqrt(b**2*c*(a/b + x)**2)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.54 \[ \int \frac {1}{\sqrt {c (a+b x)^2}} \, dx=\frac {\log \left (x + \frac {a}{b}\right )}{b \sqrt {c}} \]

[In]

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

[Out]

log(x + a/b)/(b*sqrt(c))

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.21 \[ \int \frac {1}{\sqrt {c (a+b x)^2}} \, dx=\frac {\log \left ({\left | b x + a \right |} \sqrt {{\left | c \right |}} {\left | \mathrm {sgn}\left (b x + a\right ) \right |}\right )}{b \sqrt {c} \mathrm {sgn}\left (b x + a\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 6.76 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.04 \[ \int \frac {1}{\sqrt {c (a+b x)^2}} \, dx=\frac {\ln \left (\sqrt {c}\,\left (a+b\,x\right )+\sqrt {c\,{\left (a+b\,x\right )}^2}\right )}{b\,\sqrt {c}} \]

[In]

int(1/(c*(a + b*x)^2)^(1/2),x)

[Out]

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