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

Optimal. Leaf size=138 \[ -\frac{2 a}{x (b-c)^2}+\frac{2 \sqrt{a+b x} \sqrt{a+c x}}{x (b-c)^2}+\frac{2 (b+c) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a+c x}}\right )}{(b-c)^2}-\frac{4 \sqrt{b} \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{b} \sqrt{a+c x}}\right )}{(b-c)^2}+\frac{(b+c) \log (x)}{(b-c)^2} \]

[Out]

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

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Rubi [A]  time = 0.113665, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {6690, 97, 157, 63, 217, 206, 93, 208} \[ -\frac{2 a}{x (b-c)^2}+\frac{2 \sqrt{a+b x} \sqrt{a+c x}}{x (b-c)^2}+\frac{2 (b+c) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a+c x}}\right )}{(b-c)^2}-\frac{4 \sqrt{b} \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{b} \sqrt{a+c x}}\right )}{(b-c)^2}+\frac{(b+c) \log (x)}{(b-c)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 6690

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

Rule 97

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p)/(b*(m + 1)), x] - Dist[1/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n
- 1)*(e + f*x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[m
, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[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}{\left (\sqrt{a+b x}+\sqrt{a+c x}\right )^2} \, dx &=\frac{\int \left (\frac{2 a}{x^2}+\frac{b \left (1+\frac{c}{b}\right )}{x}-\frac{2 \sqrt{a+b x} \sqrt{a+c x}}{x^2}\right ) \, dx}{(b-c)^2}\\ &=-\frac{2 a}{(b-c)^2 x}+\frac{(b+c) \log (x)}{(b-c)^2}-\frac{2 \int \frac{\sqrt{a+b x} \sqrt{a+c x}}{x^2} \, dx}{(b-c)^2}\\ &=-\frac{2 a}{(b-c)^2 x}+\frac{2 \sqrt{a+b x} \sqrt{a+c x}}{(b-c)^2 x}+\frac{(b+c) \log (x)}{(b-c)^2}-\frac{2 \int \frac{\frac{1}{2} a (b+c)+b c x}{x \sqrt{a+b x} \sqrt{a+c x}} \, dx}{(b-c)^2}\\ &=-\frac{2 a}{(b-c)^2 x}+\frac{2 \sqrt{a+b x} \sqrt{a+c x}}{(b-c)^2 x}+\frac{(b+c) \log (x)}{(b-c)^2}-\frac{(2 b c) \int \frac{1}{\sqrt{a+b x} \sqrt{a+c x}} \, dx}{(b-c)^2}-\frac{(a (b+c)) \int \frac{1}{x \sqrt{a+b x} \sqrt{a+c x}} \, dx}{(b-c)^2}\\ &=-\frac{2 a}{(b-c)^2 x}+\frac{2 \sqrt{a+b x} \sqrt{a+c x}}{(b-c)^2 x}+\frac{(b+c) \log (x)}{(b-c)^2}-\frac{(4 c) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-\frac{a c}{b}+\frac{c x^2}{b}}} \, dx,x,\sqrt{a+b x}\right )}{(b-c)^2}-\frac{(2 a (b+c)) \operatorname{Subst}\left (\int \frac{1}{-a+a x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{a+c x}}\right )}{(b-c)^2}\\ &=-\frac{2 a}{(b-c)^2 x}+\frac{2 \sqrt{a+b x} \sqrt{a+c x}}{(b-c)^2 x}+\frac{2 (b+c) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a+c x}}\right )}{(b-c)^2}+\frac{(b+c) \log (x)}{(b-c)^2}-\frac{(4 c) \operatorname{Subst}\left (\int \frac{1}{1-\frac{c x^2}{b}} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{a+c x}}\right )}{(b-c)^2}\\ &=-\frac{2 a}{(b-c)^2 x}+\frac{2 \sqrt{a+b x} \sqrt{a+c x}}{(b-c)^2 x}+\frac{2 (b+c) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a+c x}}\right )}{(b-c)^2}-\frac{4 \sqrt{b} \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{b} \sqrt{a+c x}}\right )}{(b-c)^2}+\frac{(b+c) \log (x)}{(b-c)^2}\\ \end{align*}

Mathematica [A]  time = 0.721369, size = 178, normalized size = 1.29 \[ \frac{2 c \sqrt{a+b x}+\frac{2 a \left (\sqrt{a+b x}-\sqrt{a+c x}\right )}{x}+(b+c) \log (x) \sqrt{a+c x}-\frac{4 b \sqrt{c} (a+c x) \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a (b-c)}}\right )}{\sqrt{a (b-c)} \sqrt{\frac{b (a+c x)}{a (b-c)}}}+2 (b+c) \sqrt{a+c x} \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a+c x}}\right )}{(b-c)^2 \sqrt{a+c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.012, size = 272, normalized size = 2. \begin{align*}{\frac{b\ln \left ( x \right ) }{ \left ( b-c \right ) ^{2}}}+{\frac{c\ln \left ( x \right ) }{ \left ( b-c \right ) ^{2}}}-2\,{\frac{a}{ \left ( b-c \right ) ^{2}x}}-{\frac{{\it csgn} \left ( a \right ) }{ \left ( b-c \right ) ^{2}x}\sqrt{bx+a}\sqrt{cx+a} \left ( 2\,{\it csgn} \left ( a \right ) \ln \left ( 1/2\,{\frac{2\,bcx+2\,\sqrt{bc{x}^{2}+abx+acx+{a}^{2}}\sqrt{bc}+ab+ac}{\sqrt{bc}}} \right ) xbc-\ln \left ({\frac{a}{x} \left ( 2\,{\it csgn} \left ( a \right ) \sqrt{bc{x}^{2}+abx+acx+{a}^{2}}+bx+cx+2\,a \right ) } \right ) xb\sqrt{bc}-\ln \left ({\frac{a}{x} \left ( 2\,{\it csgn} \left ( a \right ) \sqrt{bc{x}^{2}+abx+acx+{a}^{2}}+bx+cx+2\,a \right ) } \right ) xc\sqrt{bc}-2\,{\it csgn} \left ( a \right ) \sqrt{bc}\sqrt{bc{x}^{2}+abx+acx+{a}^{2}} \right ){\frac{1}{\sqrt{bc{x}^{2}+abx+acx+{a}^{2}}}}{\frac{1}{\sqrt{bc}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.38796, size = 810, normalized size = 5.87 \begin{align*} \left [\frac{2 \,{\left (b + c\right )} x \log \left (x\right ) - 2 \,{\left (b + c\right )} x \log \left (-\frac{{\left (b + c\right )} x - 2 \, \sqrt{b x + a} \sqrt{c x + a} + 2 \, a}{x}\right ) + 4 \, \sqrt{b c} x \log \left (a b^{2} + 2 \, a b c + a c^{2} + 2 \,{\left (2 \, b c - \sqrt{b c}{\left (b + c\right )}\right )} \sqrt{b x + a} \sqrt{c x + a} + 2 \,{\left (b^{2} c + b c^{2}\right )} x - 2 \,{\left (2 \, b c x + a b + a c\right )} \sqrt{b c}\right ) +{\left (b + c\right )} x + 4 \, \sqrt{b x + a} \sqrt{c x + a} - 4 \, a}{2 \,{\left (b^{2} - 2 \, b c + c^{2}\right )} x}, \frac{2 \,{\left (b + c\right )} x \log \left (x\right ) - 2 \,{\left (b + c\right )} x \log \left (-\frac{{\left (b + c\right )} x - 2 \, \sqrt{b x + a} \sqrt{c x + a} + 2 \, a}{x}\right ) + 8 \, \sqrt{-b c} x \arctan \left (\frac{\sqrt{-b c} \sqrt{b x + a} \sqrt{c x + a} - \sqrt{-b c} a}{b c x}\right ) +{\left (b + c\right )} x + 4 \, \sqrt{b x + a} \sqrt{c x + a} - 4 \, a}{2 \,{\left (b^{2} - 2 \, b c + c^{2}\right )} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*(b + c)*x*log(x) - 2*(b + c)*x*log(-((b + c)*x - 2*sqrt(b*x + a)*sqrt(c*x + a) + 2*a)/x) + 4*sqrt(b*c)
*x*log(a*b^2 + 2*a*b*c + a*c^2 + 2*(2*b*c - sqrt(b*c)*(b + c))*sqrt(b*x + a)*sqrt(c*x + a) + 2*(b^2*c + b*c^2)
*x - 2*(2*b*c*x + a*b + a*c)*sqrt(b*c)) + (b + c)*x + 4*sqrt(b*x + a)*sqrt(c*x + a) - 4*a)/((b^2 - 2*b*c + c^2
)*x), 1/2*(2*(b + c)*x*log(x) - 2*(b + c)*x*log(-((b + c)*x - 2*sqrt(b*x + a)*sqrt(c*x + a) + 2*a)/x) + 8*sqrt
(-b*c)*x*arctan((sqrt(-b*c)*sqrt(b*x + a)*sqrt(c*x + a) - sqrt(-b*c)*a)/(b*c*x)) + (b + c)*x + 4*sqrt(b*x + a)
*sqrt(c*x + a) - 4*a)/((b^2 - 2*b*c + c^2)*x)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out