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

Optimal. Leaf size=28 \[ \frac {(a+b x) \log (a+b x)}{b \sqrt {c (a+b x)^2}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {253, 15, 29} \begin {gather*} \frac {(a+b x) \log (a+b x)}{b \sqrt {c (a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

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

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Mathematica [A]
time = 0.01, size = 28, normalized size = 1.00 \begin {gather*} \frac {(a+b x) \log (a+b x)}{b \sqrt {c (a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[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])

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Maple [A]
time = 0.27, size = 27, normalized size = 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	\(\frac {\left (b x +a \right ) \ln \left (b x +a \right )}{b \sqrt {c \left (b x +a \right )^{2}}}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [A]
time = 0.28, size = 15, normalized size = 0.54 \begin {gather*} \frac {\log \left (x + \frac {a}{b}\right )}{b \sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(x + a/b)/(b*sqrt(c))

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Fricas [A]
time = 0.37, size = 42, normalized size = 1.50 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {c \left (a + b x\right )^{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(c*(a + b*x)**2), x)

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Giac [A]
time = 1.19, size = 34, normalized size = 1.21 \begin {gather*} \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 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Mupad [B]
time = 1.33, size = 29, normalized size = 1.04 \begin {gather*} \frac {\ln \left (\sqrt {c}\,\left (a+b\,x\right )+\sqrt {c\,{\left (a+b\,x\right )}^2}\right )}{b\,\sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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