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

Optimal. Leaf size=135 \[ -\frac{2 \sqrt{a+b x} \sqrt{a+c x}}{(b-c)^2}+\frac{2 a \log (x)}{(b-c)^2}+\frac{4 a \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a+c x}}\right )}{(b-c)^2}-\frac{2 a (b+c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{b} \sqrt{a+c x}}\right )}{\sqrt{b} \sqrt{c} (b-c)^2}+\frac{x (b+c)}{(b-c)^2} \]

[Out]

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

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Rubi [A]  time = 0.181448, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {6690, 101, 157, 63, 217, 206, 93, 208} \[ -\frac{2 \sqrt{a+b x} \sqrt{a+c x}}{(b-c)^2}+\frac{2 a \log (x)}{(b-c)^2}+\frac{4 a \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a+c x}}\right )}{(b-c)^2}-\frac{2 a (b+c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{b} \sqrt{a+c x}}\right )}{\sqrt{b} \sqrt{c} (b-c)^2}+\frac{x (b+c)}{(b-c)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b + c)*x)/(b - c)^2 - (2*Sqrt[a + b*x]*Sqrt[a + c*x])/(b - c)^2 + (4*a*ArcTanh[Sqrt[a + b*x]/Sqrt[a + c*x]])
/(b - c)^2 - (2*a*(b + c)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[a + c*x])])/(Sqrt[b]*(b - c)^2*Sqrt[c]
) + (2*a*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 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{x}{\left (\sqrt{a+b x}+\sqrt{a+c x}\right )^2} \, dx &=\frac{\int \left (b \left (1+\frac{c}{b}\right )+\frac{2 a}{x}-\frac{2 \sqrt{a+b x} \sqrt{a+c x}}{x}\right ) \, dx}{(b-c)^2}\\ &=\frac{(b+c) x}{(b-c)^2}+\frac{2 a \log (x)}{(b-c)^2}-\frac{2 \int \frac{\sqrt{a+b x} \sqrt{a+c x}}{x} \, dx}{(b-c)^2}\\ &=\frac{(b+c) x}{(b-c)^2}-\frac{2 \sqrt{a+b x} \sqrt{a+c x}}{(b-c)^2}+\frac{2 a \log (x)}{(b-c)^2}+\frac{2 \int \frac{-a^2-\frac{1}{2} a (b+c) x}{x \sqrt{a+b x} \sqrt{a+c x}} \, dx}{(b-c)^2}\\ &=\frac{(b+c) x}{(b-c)^2}-\frac{2 \sqrt{a+b x} \sqrt{a+c x}}{(b-c)^2}+\frac{2 a \log (x)}{(b-c)^2}-\frac{\left (2 a^2\right ) \int \frac{1}{x \sqrt{a+b x} \sqrt{a+c x}} \, dx}{(b-c)^2}-\frac{(a (b+c)) \int \frac{1}{\sqrt{a+b x} \sqrt{a+c x}} \, dx}{(b-c)^2}\\ &=\frac{(b+c) x}{(b-c)^2}-\frac{2 \sqrt{a+b x} \sqrt{a+c x}}{(b-c)^2}+\frac{2 a \log (x)}{(b-c)^2}-\frac{\left (4 a^2\right ) \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)) \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 (b-c)^2}\\ &=\frac{(b+c) x}{(b-c)^2}-\frac{2 \sqrt{a+b x} \sqrt{a+c x}}{(b-c)^2}+\frac{4 a \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a+c x}}\right )}{(b-c)^2}+\frac{2 a \log (x)}{(b-c)^2}-\frac{(2 a (b+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 (b-c)^2}\\ &=\frac{(b+c) x}{(b-c)^2}-\frac{2 \sqrt{a+b x} \sqrt{a+c x}}{(b-c)^2}+\frac{4 a \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a+c x}}\right )}{(b-c)^2}-\frac{2 a (b+c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{b} \sqrt{a+c x}}\right )}{\sqrt{b} (b-c)^2 \sqrt{c}}+\frac{2 a \log (x)}{(b-c)^2}\\ \end{align*}

Mathematica [A]  time = 0.843061, size = 195, normalized size = 1.44 \[ \frac{\frac{2 (b+c) \sqrt{a (b-c)} (a+c x) \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x}}{\sqrt{a (b-c)}}\right )}{\sqrt{c} \sqrt{\frac{b (a+c x)}{a (b-c)}}}-(b-c) \left (-2 c x \sqrt{a+b x}+b x \sqrt{a+c x}+4 a \sqrt{a+c x} \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a+c x}}\right )-2 a \sqrt{a+b x}+c x \sqrt{a+c x}+2 a \log (x) \sqrt{a+c x}\right )}{(c-b)^3 \sqrt{a+c x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*Sqrt[a*(b - c)]*(b + c)*(a + c*x)*ArcSinh[(Sqrt[c]*Sqrt[a + b*x])/Sqrt[a*(b - c)]])/(Sqrt[c]*Sqrt[(b*(a +
c*x))/(a*(b - c))]) - (b - c)*(-2*a*Sqrt[a + b*x] - 2*c*x*Sqrt[a + b*x] + b*x*Sqrt[a + c*x] + c*x*Sqrt[a + c*x
] + 4*a*Sqrt[a + c*x]*ArcTanh[Sqrt[a + b*x]/Sqrt[a + c*x]] + 2*a*Sqrt[a + c*x]*Log[x]))/((-b + c)^3*Sqrt[a + c
*x])

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Maple [C]  time = 0.013, size = 266, normalized size = 2. \begin{align*}{\frac{bx}{ \left ( b-c \right ) ^{2}}}+{\frac{cx}{ \left ( b-c \right ) ^{2}}}+2\,{\frac{a\ln \left ( x \right ) }{ \left ( b-c \right ) ^{2}}}-{\frac{{\it csgn} \left ( a \right ) }{ \left ( b-c \right ) ^{2}}\sqrt{bx+a}\sqrt{cx+a} \left ({\it csgn} \left ( a \right ) \ln \left ({\frac{1}{2} \left ( 2\,bcx+2\,\sqrt{bc{x}^{2}+abx+acx+{a}^{2}}\sqrt{bc}+ab+ac \right ){\frac{1}{\sqrt{bc}}}} \right ) ab+{\it csgn} \left ( a \right ) \ln \left ({\frac{1}{2} \left ( 2\,bcx+2\,\sqrt{bc{x}^{2}+abx+acx+{a}^{2}}\sqrt{bc}+ab+ac \right ){\frac{1}{\sqrt{bc}}}} \right ) ac+2\,{\it csgn} \left ( a \right ) \sqrt{bc}\sqrt{bc{x}^{2}+abx+acx+{a}^{2}}-2\,\ln \left ({\frac{a \left ( 2\,{\it csgn} \left ( a \right ) \sqrt{bc{x}^{2}+abx+acx+{a}^{2}}+bx+cx+2\,a \right ) }{x}} \right ) \sqrt{bc}a \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(x/((b*x+a)^(1/2)+(c*x+a)^(1/2))^2,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{{\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(x/((b*x+a)^(1/2)+(c*x+a)^(1/2))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.45708, size = 829, normalized size = 6.14 \begin{align*} \left [\frac{2 \, a b c \log \left (x\right ) - 2 \, a b c \log \left (-\frac{{\left (b + c\right )} x - 2 \, \sqrt{b x + a} \sqrt{c x + a} + 2 \, a}{x}\right ) - 2 \, \sqrt{b x + a} \sqrt{c x + a} b c +{\left (a b + a c\right )} \sqrt{b c} \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^{2} c + b c^{2}\right )} x}{b^{3} c - 2 \, b^{2} c^{2} + b c^{3}}, \frac{2 \, a b c \log \left (x\right ) - 2 \, a b c \log \left (-\frac{{\left (b + c\right )} x - 2 \, \sqrt{b x + a} \sqrt{c x + a} + 2 \, a}{x}\right ) - 2 \, \sqrt{b x + a} \sqrt{c x + a} b c + 2 \,{\left (a b + a c\right )} \sqrt{-b c} \arctan \left (\frac{\sqrt{-b c} \sqrt{b x + a} \sqrt{c x + a} - \sqrt{-b c} a}{b c x}\right ) +{\left (b^{2} c + b c^{2}\right )} x}{b^{3} c - 2 \, b^{2} c^{2} + b c^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[(2*a*b*c*log(x) - 2*a*b*c*log(-((b + c)*x - 2*sqrt(b*x + a)*sqrt(c*x + a) + 2*a)/x) - 2*sqrt(b*x + a)*sqrt(c*
x + a)*b*c + (a*b + a*c)*sqrt(b*c)*log(a*b^2 + 2*a*b*c + a*c^2 + 2*(2*b*c - sqrt(b*c)*(b + c))*sqrt(b*x + a)*s
qrt(c*x + a) + 2*(b^2*c + b*c^2)*x - 2*(2*b*c*x + a*b + a*c)*sqrt(b*c)) + (b^2*c + b*c^2)*x)/(b^3*c - 2*b^2*c^
2 + b*c^3), (2*a*b*c*log(x) - 2*a*b*c*log(-((b + c)*x - 2*sqrt(b*x + a)*sqrt(c*x + a) + 2*a)/x) - 2*sqrt(b*x +
a)*sqrt(c*x + a)*b*c + 2*(a*b + a*c)*sqrt(-b*c)*arctan((sqrt(-b*c)*sqrt(b*x + a)*sqrt(c*x + a) - sqrt(-b*c)*a
)/(b*c*x)) + (b^2*c + b*c^2)*x)/(b^3*c - 2*b^2*c^2 + b*c^3)]

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

[Out]

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

[Out]

Timed out