3.643 \(\int \frac{1}{x (a+b \sqrt{c+d x})^2} \, dx\)

Optimal. Leaf size=129 \[ \frac{2 a}{\left (a^2-b^2 c\right ) \left (a+b \sqrt{c+d x}\right )}-\frac{2 \left (a^2+b^2 c\right ) \log \left (a+b \sqrt{c+d x}\right )}{\left (a^2-b^2 c\right )^2}+\frac{4 a b \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{\left (a^2-b^2 c\right )^2}+\frac{\log (x) \left (a^2+b^2 c\right )}{\left (a^2-b^2 c\right )^2} \]

[Out]

(2*a)/((a^2 - b^2*c)*(a + b*Sqrt[c + d*x])) + (4*a*b*Sqrt[c]*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/(a^2 - b^2*c)^2 +
((a^2 + b^2*c)*Log[x])/(a^2 - b^2*c)^2 - (2*(a^2 + b^2*c)*Log[a + b*Sqrt[c + d*x]])/(a^2 - b^2*c)^2

________________________________________________________________________________________

Rubi [A]  time = 0.119484, antiderivative size = 129, 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} \[ \frac{2 a}{\left (a^2-b^2 c\right ) \left (a+b \sqrt{c+d x}\right )}-\frac{2 \left (a^2+b^2 c\right ) \log \left (a+b \sqrt{c+d x}\right )}{\left (a^2-b^2 c\right )^2}+\frac{4 a b \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{\left (a^2-b^2 c\right )^2}+\frac{\log (x) \left (a^2+b^2 c\right )}{\left (a^2-b^2 c\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a + b*Sqrt[c + d*x])^2),x]

[Out]

(2*a)/((a^2 - b^2*c)*(a + b*Sqrt[c + d*x])) + (4*a*b*Sqrt[c]*ArcTanh[Sqrt[c + d*x]/Sqrt[c]])/(a^2 - b^2*c)^2 +
((a^2 + b^2*c)*Log[x])/(a^2 - b^2*c)^2 - (2*(a^2 + b^2*c)*Log[a + b*Sqrt[c + d*x]])/(a^2 - b^2*c)^2

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 )^2} \, dx &=\operatorname{Subst}\left (\int \frac{1}{\left (a+b \sqrt{x}\right )^2 (-c+x)} \, dx,x,c+d x\right )\\ &=2 \operatorname{Subst}\left (\int \frac{x}{(a+b x)^2 \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)^2}-\frac{b \left (a^2+b^2 c\right )}{\left (a^2-b^2 c\right )^2 (a+b x)}+\frac{2 a b c-\left (a^2+b^2 c\right ) x}{\left (a^2-b^2 c\right )^2 \left (c-x^2\right )}\right ) \, dx,x,\sqrt{c+d x}\right )\\ &=\frac{2 a}{\left (a^2-b^2 c\right ) \left (a+b \sqrt{c+d x}\right )}-\frac{2 \left (a^2+b^2 c\right ) \log \left (a+b \sqrt{c+d x}\right )}{\left (a^2-b^2 c\right )^2}+\frac{2 \operatorname{Subst}\left (\int \frac{2 a b c-\left (a^2+b^2 c\right ) x}{c-x^2} \, dx,x,\sqrt{c+d x}\right )}{\left (a^2-b^2 c\right )^2}\\ &=\frac{2 a}{\left (a^2-b^2 c\right ) \left (a+b \sqrt{c+d x}\right )}-\frac{2 \left (a^2+b^2 c\right ) \log \left (a+b \sqrt{c+d x}\right )}{\left (a^2-b^2 c\right )^2}+\frac{(4 a b c) \operatorname{Subst}\left (\int \frac{1}{c-x^2} \, dx,x,\sqrt{c+d x}\right )}{\left (a^2-b^2 c\right )^2}-\frac{\left (2 \left (a^2+b^2 c\right )\right ) \operatorname{Subst}\left (\int \frac{x}{c-x^2} \, dx,x,\sqrt{c+d x}\right )}{\left (a^2-b^2 c\right )^2}\\ &=\frac{2 a}{\left (a^2-b^2 c\right ) \left (a+b \sqrt{c+d x}\right )}+\frac{4 a b \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c+d x}}{\sqrt{c}}\right )}{\left (a^2-b^2 c\right )^2}+\frac{\left (a^2+b^2 c\right ) \log (x)}{\left (a^2-b^2 c\right )^2}-\frac{2 \left (a^2+b^2 c\right ) \log \left (a+b \sqrt{c+d x}\right )}{\left (a^2-b^2 c\right )^2}\\ \end{align*}

Mathematica [A]  time = 0.282711, size = 164, normalized size = 1.27 \[ \frac{-2 \left (a^2+b^2 c\right ) \left (a+b \sqrt{c+d x}\right ) \log \left (a+b \sqrt{c+d x}\right )+2 a^3-2 a b^2 c+\left (a-b \sqrt{c}\right )^2 \log \left (\sqrt{c}-\sqrt{c+d x}\right ) \left (a+b \sqrt{c+d x}\right )+\left (a+b \sqrt{c}\right )^2 \log \left (\sqrt{c+d x}+\sqrt{c}\right ) \left (a+b \sqrt{c+d x}\right )}{\left (a^2-b^2 c\right )^2 \left (a+b \sqrt{c+d x}\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a + b*Sqrt[c + d*x])^2),x]

[Out]

(2*a^3 - 2*a*b^2*c + (a - b*Sqrt[c])^2*(a + b*Sqrt[c + d*x])*Log[Sqrt[c] - Sqrt[c + d*x]] + (a + b*Sqrt[c])^2*
(a + b*Sqrt[c + d*x])*Log[Sqrt[c] + Sqrt[c + d*x]] - 2*(a^2 + b^2*c)*(a + b*Sqrt[c + d*x])*Log[a + b*Sqrt[c +
d*x]])/((a^2 - b^2*c)^2*(a + b*Sqrt[c + d*x]))

________________________________________________________________________________________

Maple [A]  time = 0.008, size = 161, normalized size = 1.3 \begin{align*} 2\,{\frac{a}{ \left ( -{b}^{2}c+{a}^{2} \right ) \left ( a+b\sqrt{dx+c} \right ) }}-2\,{\frac{\ln \left ( a+b\sqrt{dx+c} \right ){b}^{2}c}{ \left ( -{b}^{2}c+{a}^{2} \right ) ^{2}}}-2\,{\frac{\ln \left ( a+b\sqrt{dx+c} \right ){a}^{2}}{ \left ( -{b}^{2}c+{a}^{2} \right ) ^{2}}}+{\frac{\ln \left ( dx \right ){b}^{2}c}{ \left ( -{b}^{2}c+{a}^{2} \right ) ^{2}}}+{\frac{\ln \left ( dx \right ){a}^{2}}{ \left ( -{b}^{2}c+{a}^{2} \right ) ^{2}}}+4\,{\frac{\sqrt{c}ab}{ \left ( -{b}^{2}c+{a}^{2} \right ) ^{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))^2,x)

[Out]

2*a/(-b^2*c+a^2)/(a+b*(d*x+c)^(1/2))-2/(-b^2*c+a^2)^2*ln(a+b*(d*x+c)^(1/2))*b^2*c-2/(-b^2*c+a^2)^2*ln(a+b*(d*x
+c)^(1/2))*a^2+1/(-b^2*c+a^2)^2*ln(d*x)*b^2*c+1/(-b^2*c+a^2)^2*ln(d*x)*a^2+4*a*b*arctanh((d*x+c)^(1/2)/c^(1/2)
)*c^(1/2)/(-b^2*c+a^2)^2

________________________________________________________________________________________

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))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.89787, size = 936, normalized size = 7.26 \begin{align*} \left [\frac{2 \, a^{2} b^{2} c - 2 \, a^{4} + 2 \,{\left (a b^{3} d x + a b^{3} c - a^{3} b\right )} \sqrt{c} \log \left (\frac{d x + 2 \, \sqrt{d x + c} \sqrt{c} + 2 \, c}{x}\right ) - 2 \,{\left (b^{4} c^{2} - a^{4} +{\left (b^{4} c + a^{2} b^{2}\right )} d x\right )} \log \left (\sqrt{d x + c} b + a\right ) +{\left (b^{4} c^{2} - a^{4} +{\left (b^{4} c + a^{2} b^{2}\right )} d x\right )} \log \left (x\right ) - 2 \,{\left (a b^{3} c - a^{3} b\right )} \sqrt{d x + c}}{b^{6} c^{3} - 3 \, a^{2} b^{4} c^{2} + 3 \, a^{4} b^{2} c - a^{6} +{\left (b^{6} c^{2} - 2 \, a^{2} b^{4} c + a^{4} b^{2}\right )} d x}, \frac{2 \, a^{2} b^{2} c - 2 \, a^{4} - 4 \,{\left (a b^{3} d x + a b^{3} c - a^{3} b\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{d x + c} \sqrt{-c}}{c}\right ) - 2 \,{\left (b^{4} c^{2} - a^{4} +{\left (b^{4} c + a^{2} b^{2}\right )} d x\right )} \log \left (\sqrt{d x + c} b + a\right ) +{\left (b^{4} c^{2} - a^{4} +{\left (b^{4} c + a^{2} b^{2}\right )} d x\right )} \log \left (x\right ) - 2 \,{\left (a b^{3} c - a^{3} b\right )} \sqrt{d x + c}}{b^{6} c^{3} - 3 \, a^{2} b^{4} c^{2} + 3 \, a^{4} b^{2} c - a^{6} +{\left (b^{6} c^{2} - 2 \, a^{2} b^{4} c + a^{4} b^{2}\right )} d x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(2*a^2*b^2*c - 2*a^4 + 2*(a*b^3*d*x + a*b^3*c - a^3*b)*sqrt(c)*log((d*x + 2*sqrt(d*x + c)*sqrt(c) + 2*c)/x) -
2*(b^4*c^2 - a^4 + (b^4*c + a^2*b^2)*d*x)*log(sqrt(d*x + c)*b + a) + (b^4*c^2 - a^4 + (b^4*c + a^2*b^2)*d*x)*
log(x) - 2*(a*b^3*c - a^3*b)*sqrt(d*x + c))/(b^6*c^3 - 3*a^2*b^4*c^2 + 3*a^4*b^2*c - a^6 + (b^6*c^2 - 2*a^2*b^
4*c + a^4*b^2)*d*x), (2*a^2*b^2*c - 2*a^4 - 4*(a*b^3*d*x + a*b^3*c - a^3*b)*sqrt(-c)*arctan(sqrt(d*x + c)*sqrt
(-c)/c) - 2*(b^4*c^2 - a^4 + (b^4*c + a^2*b^2)*d*x)*log(sqrt(d*x + c)*b + a) + (b^4*c^2 - a^4 + (b^4*c + a^2*b
^2)*d*x)*log(x) - 2*(a*b^3*c - a^3*b)*sqrt(d*x + c))/(b^6*c^3 - 3*a^2*b^4*c^2 + 3*a^4*b^2*c - a^6 + (b^6*c^2 -
2*a^2*b^4*c + a^4*b^2)*d*x)]

________________________________________________________________________________________

Sympy [A]  time = 26.9701, size = 153, normalized size = 1.19 \begin{align*} - \frac{2 a b \left (\begin{cases} \frac{\sqrt{c + d x}}{a^{2}} & \text{for}\: b = 0 \\- \frac{1}{b \left (a + b \sqrt{c + d x}\right )} & \text{otherwise} \end{cases}\right )}{a^{2} - b^{2} c} - \frac{2 b \left (a^{2} + b^{2} c\right ) \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 )}{\left (a^{2} - b^{2} c\right )^{2}} - \frac{2 \left (\frac{2 a b c \operatorname{atan}{\left (\frac{\sqrt{c + d x}}{\sqrt{- c}} \right )}}{\sqrt{- c}} + \left (- \frac{a^{2}}{2} - \frac{b^{2} c}{2}\right ) \log{\left (- d x \right )}\right )}{\left (a^{2} - b^{2} c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a*b*Piecewise((sqrt(c + d*x)/a**2, Eq(b, 0)), (-1/(b*(a + b*sqrt(c + d*x))), True))/(a**2 - b**2*c) - 2*b*(
a**2 + b**2*c)*Piecewise((sqrt(c + d*x)/a, Eq(b, 0)), (log(a + b*sqrt(c + d*x))/b, True))/(a**2 - b**2*c)**2 -
2*(2*a*b*c*atan(sqrt(c + d*x)/sqrt(-c))/sqrt(-c) + (-a**2/2 - b**2*c/2)*log(-d*x))/(a**2 - b**2*c)**2

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{undef} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

undef