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

Optimal. Leaf size=133 \[ \frac{2 b x}{(a-c)^2}-\frac{2 \sqrt{a+b x} \sqrt{b x+c}}{(a-c)^2}-\frac{2 (a+c) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{b x+c}}\right )}{(a-c)^2}+\frac{4 \sqrt{a} \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{b x+c}}\right )}{(a-c)^2}+\frac{(a+c) \log (x)}{(a-c)^2} \]

[Out]

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

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Rubi [A]  time = 0.229698, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {6689, 101, 157, 63, 217, 206, 93, 208} \[ \frac{2 b x}{(a-c)^2}-\frac{2 \sqrt{a+b x} \sqrt{b x+c}}{(a-c)^2}-\frac{2 (a+c) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{b x+c}}\right )}{(a-c)^2}+\frac{4 \sqrt{a} \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{b x+c}}\right )}{(a-c)^2}+\frac{(a+c) \log (x)}{(a-c)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6689

Int[(u_.)*((e_.)*Sqrt[(a_.) + (b_.)*(x_)^(n_.)] + (f_.)*Sqrt[(c_.) + (d_.)*(x_)^(n_.)])^(m_), x_Symbol] :> Dis
t[(a*e^2 - c*f^2)^m, Int[ExpandIntegrand[u/(e*Sqrt[a + b*x^n] - f*Sqrt[c + d*x^n])^m, x], x], x] /; FreeQ[{a,
b, c, d, e, f, n}, x] && ILtQ[m, 0] && EqQ[b*e^2 - d*f^2, 0]

Rule 101

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a +
b*x)^m*(c + d*x)^n*(e + f*x)^(p + 1))/(f*(m + n + p + 1)), x] - Dist[1/(f*(m + n + p + 1)), Int[(a + b*x)^(m -
1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[c*m*(b*e - a*f) + a*n*(d*e - c*f) + (d*m*(b*e - a*f) + b*n*(d*e - c*f))
*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && GtQ[m, 0] && GtQ[n, 0] && NeQ[m + n + p + 1, 0] && (Integ
ersQ[2*m, 2*n, 2*p] || (IntegersQ[m, n + p] || IntegersQ[p, m + n]))

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 208

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

Rubi steps

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

Mathematica [A]  time = 1.0643, size = 195, normalized size = 1.47 \[ \frac{\sqrt{b} \left (-2 \left (c \sqrt{a+b x}+b x \left (\sqrt{a+b x}-\sqrt{b x+c}\right )\right )+(a+c) \log (x) \sqrt{b x+c}+4 \sqrt{a} \sqrt{c} \sqrt{b x+c} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a} \sqrt{b x+c}}\right )\right )-2 (a+c) \sqrt{b (c-a)} \sqrt{-\frac{b x+c}{a-c}} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{a+b x}}{\sqrt{b (c-a)}}\right )}{\sqrt{b} (a-c)^2 \sqrt{b x+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[b*(-a + c)]*(a + c)*Sqrt[-((c + b*x)/(a - c))]*ArcSinh[(Sqrt[b]*Sqrt[a + b*x])/Sqrt[b*(-a + c)]] + Sq
rt[b]*(-2*(c*Sqrt[a + b*x] + b*x*(Sqrt[a + b*x] - Sqrt[c + b*x])) + 4*Sqrt[a]*Sqrt[c]*Sqrt[c + b*x]*ArcTanh[(S
qrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + b*x])] + (a + c)*Sqrt[c + b*x]*Log[x]))/(Sqrt[b]*(a - c)^2*Sqrt[c + b*
x])

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Maple [C]  time = 0.013, size = 258, normalized size = 1.9 \begin{align*}{\frac{a\ln \left ( x \right ) }{ \left ( a-c \right ) ^{2}}}+{\frac{c\ln \left ( x \right ) }{ \left ( a-c \right ) ^{2}}}+2\,{\frac{bx}{ \left ( a-c \right ) ^{2}}}+{\frac{{\it csgn} \left ( b \right ) }{ \left ( a-c \right ) ^{2}}\sqrt{bx+a}\sqrt{bx+c} \left ( 2\,{\it csgn} \left ( b \right ) \ln \left ({\frac{abx+bcx+2\,\sqrt{ac}\sqrt{{b}^{2}{x}^{2}+abx+bcx+ac}+2\,ac}{x}} \right ) ac-2\,{\it csgn} \left ( b \right ) \sqrt{ac}\sqrt{{b}^{2}{x}^{2}+abx+bcx+ac}-\ln \left ({\frac{{\it csgn} \left ( b \right ) }{2} \left ( 2\,{\it csgn} \left ( b \right ) \sqrt{{b}^{2}{x}^{2}+abx+bcx+ac}+2\,bx+a+c \right ) } \right ) \sqrt{ac}a-\ln \left ({\frac{{\it csgn} \left ( b \right ) }{2} \left ( 2\,{\it csgn} \left ( b \right ) \sqrt{{b}^{2}{x}^{2}+abx+bcx+ac}+2\,bx+a+c \right ) } \right ) \sqrt{ac}c \right ){\frac{1}{\sqrt{ac}}}{\frac{1}{\sqrt{{b}^{2}{x}^{2}+abx+bcx+ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x{\left (\sqrt{b x + a} + \sqrt{b x + c}\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.05974, size = 730, normalized size = 5.49 \begin{align*} \left [\frac{2 \, b x +{\left (a + c\right )} \log \left (-2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b x + c} - a - c\right ) +{\left (a + c\right )} \log \left (x\right ) + 2 \, \sqrt{a c} \log \left (\frac{2 \, a^{2} c + 2 \, a c^{2} + 2 \,{\left (2 \, a c + \sqrt{a c}{\left (a + c\right )}\right )} \sqrt{b x + a} \sqrt{b x + c} +{\left (a^{2} b + 2 \, a b c + b c^{2}\right )} x + 2 \,{\left (2 \, a c +{\left (a b + b c\right )} x\right )} \sqrt{a c}}{x}\right ) - 2 \, \sqrt{b x + a} \sqrt{b x + c}}{a^{2} - 2 \, a c + c^{2}}, \frac{2 \, b x +{\left (a + c\right )} \log \left (-2 \, b x + 2 \, \sqrt{b x + a} \sqrt{b x + c} - a - c\right ) +{\left (a + c\right )} \log \left (x\right ) - 4 \, \sqrt{-a c} \arctan \left (-\frac{\sqrt{-a c} b x - \sqrt{-a c} \sqrt{b x + a} \sqrt{b x + c}}{a c}\right ) - 2 \, \sqrt{b x + a} \sqrt{b x + c}}{a^{2} - 2 \, a c + c^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(2*b*x + (a + c)*log(-2*b*x + 2*sqrt(b*x + a)*sqrt(b*x + c) - a - c) + (a + c)*log(x) + 2*sqrt(a*c)*log((2*a^
2*c + 2*a*c^2 + 2*(2*a*c + sqrt(a*c)*(a + c))*sqrt(b*x + a)*sqrt(b*x + c) + (a^2*b + 2*a*b*c + b*c^2)*x + 2*(2
*a*c + (a*b + b*c)*x)*sqrt(a*c))/x) - 2*sqrt(b*x + a)*sqrt(b*x + c))/(a^2 - 2*a*c + c^2), (2*b*x + (a + c)*log
(-2*b*x + 2*sqrt(b*x + a)*sqrt(b*x + c) - a - c) + (a + c)*log(x) - 4*sqrt(-a*c)*arctan(-(sqrt(-a*c)*b*x - sqr
t(-a*c)*sqrt(b*x + a)*sqrt(b*x + c))/(a*c)) - 2*sqrt(b*x + a)*sqrt(b*x + c))/(a^2 - 2*a*c + c^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out