3.636 \(\int \frac{1}{x (a+b \sqrt{c+d x})} \, dx\)

Optimal. Leaf size=82 \[ -\frac{2 a \log \left (a+b \sqrt{c+d x}\right )}{a^2-b^2 c}+\frac{2 b \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{a^2-b^2 c}+\frac{a \log (x)}{a^2-b^2 c} \]

[Out]

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

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Rubi [A]  time = 0.0791696, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {371, 1398, 801, 635, 206, 260} \[ -\frac{2 a \log \left (a+b \sqrt{c+d x}\right )}{a^2-b^2 c}+\frac{2 b \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{a^2-b^2 c}+\frac{a \log (x)}{a^2-b^2 c} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*Sqrt[c + d*x])),x]

[Out]

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

Rule 371

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coefficient[v, x, 0], d = Coefficient[v,
x, 1]}, Dist[1/d^(m + 1), Subst[Int[SimplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]
] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]

Rule 1398

Int[((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{g = Denominator[n]}, D
ist[g, Subst[Int[x^(g - 1)*(d + e*x^(g*n))^q*(a + c*x^(2*g*n))^p, x], x, x^(1/g)], x]] /; FreeQ[{a, c, d, e, p
, q}, x] && EqQ[n2, 2*n] && FractionQ[n]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 260

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

Rubi steps

\begin{align*} \int \frac{1}{x \left (a+b \sqrt{c+d x}\right )} \, dx &=\operatorname{Subst}\left (\int \frac{1}{\left (a+b \sqrt{x}\right ) (-c+x)} \, dx,x,c+d x\right )\\ &=2 \operatorname{Subst}\left (\int \frac{x}{(a+b x) \left (-c+x^2\right )} \, dx,x,\sqrt{c+d x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (-\frac{a b}{\left (a^2-b^2 c\right ) (a+b x)}+\frac{b c-a x}{\left (a^2-b^2 c\right ) \left (c-x^2\right )}\right ) \, dx,x,\sqrt{c+d x}\right )\\ &=-\frac{2 a \log \left (a+b \sqrt{c+d x}\right )}{a^2-b^2 c}+\frac{2 \operatorname{Subst}\left (\int \frac{b c-a x}{c-x^2} \, dx,x,\sqrt{c+d x}\right )}{a^2-b^2 c}\\ &=-\frac{2 a \log \left (a+b \sqrt{c+d x}\right )}{a^2-b^2 c}-\frac{(2 a) \operatorname{Subst}\left (\int \frac{x}{c-x^2} \, dx,x,\sqrt{c+d x}\right )}{a^2-b^2 c}+\frac{(2 b c) \operatorname{Subst}\left (\int \frac{1}{c-x^2} \, dx,x,\sqrt{c+d x}\right )}{a^2-b^2 c}\\ &=\frac{2 b \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{a^2-b^2 c}+\frac{a \log (x)}{a^2-b^2 c}-\frac{2 a \log \left (a+b \sqrt{c+d x}\right )}{a^2-b^2 c}\\ \end{align*}

Mathematica [A]  time = 0.081121, size = 61, normalized size = 0.74 \[ \frac{-2 a \log \left (a+b \sqrt{c+d x}\right )+a \log (d x)+2 b \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{a^2-b^2 c} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*Sqrt[c + d*x])),x]

[Out]

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

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Maple [A]  time = 0.007, size = 77, normalized size = 0.9 \begin{align*} -2\,{\frac{a\ln \left ( a+b\sqrt{dx+c} \right ) }{-{b}^{2}c+{a}^{2}}}+{\frac{a\ln \left ( dx \right ) }{-{b}^{2}c+{a}^{2}}}+2\,{\frac{b\sqrt{c}}{-{b}^{2}c+{a}^{2}}{\it Artanh} \left ({\frac{\sqrt{dx+c}}{\sqrt{c}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(a+b*(d*x+c)^(1/2)),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.89238, size = 301, normalized size = 3.67 \begin{align*} \left [\frac{b \sqrt{c} \log \left (\frac{d x - 2 \, \sqrt{d x + c} \sqrt{c} + 2 \, c}{x}\right ) + 2 \, a \log \left (\sqrt{d x + c} b + a\right ) - a \log \left (x\right )}{b^{2} c - a^{2}}, \frac{2 \, b \sqrt{-c} \arctan \left (\frac{\sqrt{d x + c} \sqrt{-c}}{c}\right ) + 2 \, a \log \left (\sqrt{d x + c} b + a\right ) - a \log \left (x\right )}{b^{2} c - a^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(b*sqrt(c)*log((d*x - 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) + 2*a*log(sqrt(d*x + c)*b + a) - a*log(x))/(b^2*c - a
^2), (2*b*sqrt(-c)*arctan(sqrt(d*x + c)*sqrt(-c)/c) + 2*a*log(sqrt(d*x + c)*b + a) - a*log(x))/(b^2*c - a^2)]

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Sympy [A]  time = 8.24437, size = 85, normalized size = 1.04 \begin{align*} - \frac{2 a b \left (\begin{cases} \frac{\sqrt{c + d x}}{a} & \text{for}\: b = 0 \\\frac{\log{\left (a + b \sqrt{c + d x} \right )}}{b} & \text{otherwise} \end{cases}\right )}{a^{2} - b^{2} c} - \frac{2 \left (- \frac{a \log{\left (- d x \right )}}{2} + \frac{b c \operatorname{atan}{\left (\frac{\sqrt{c + d x}}{\sqrt{- c}} \right )}}{\sqrt{- c}}\right )}{a^{2} - b^{2} c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a*b*Piecewise((sqrt(c + d*x)/a, Eq(b, 0)), (log(a + b*sqrt(c + d*x))/b, True))/(a**2 - b**2*c) - 2*(-a*log(
-d*x)/2 + b*c*atan(sqrt(c + d*x)/sqrt(-c))/sqrt(-c))/(a**2 - b**2*c)

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Giac [A]  time = 1.20587, size = 155, normalized size = 1.89 \begin{align*} \frac{2 \, a b \log \left ({\left | \sqrt{d x + c} b + a \right |}\right )}{b^{3} c - a^{2} b} + \frac{2 \, b c \arctan \left (\frac{\sqrt{d x + c}}{\sqrt{-c}}\right )}{{\left (b^{2} c - a^{2}\right )} \sqrt{-c}} - \frac{a \log \left (d x\right )}{b^{2} c - a^{2}} + \frac{a \log \left (-c\right ) - 2 \, a \log \left ({\left | a \right |}\right )}{b^{2} c - a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*a*b*log(abs(sqrt(d*x + c)*b + a))/(b^3*c - a^2*b) + 2*b*c*arctan(sqrt(d*x + c)/sqrt(-c))/((b^2*c - a^2)*sqrt
(-c)) - a*log(d*x)/(b^2*c - a^2) + (a*log(-c) - 2*a*log(abs(a)))/(b^2*c - a^2)