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

Optimal. Leaf size=123 \[ \frac{\sqrt{a+b x} (a+c x)^{3/2}}{a x^2 (b-c)^2}-\frac{a}{x^2 (b-c)^2}+\frac{\sqrt{a+b x} \sqrt{a+c x}}{2 a x (b-c)}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a+c x}}\right )}{2 a}-\frac{b+c}{x (b-c)^2} \]

[Out]

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

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Rubi [A]  time = 0.201447, antiderivative size = 123, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {6690, 94, 93, 208} \[ \frac{\sqrt{a+b x} (a+c x)^{3/2}}{a x^2 (b-c)^2}-\frac{a}{x^2 (b-c)^2}+\frac{\sqrt{a+b x} \sqrt{a+c x}}{2 a x (b-c)}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a+c x}}\right )}{2 a}-\frac{b+c}{x (b-c)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 94

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 + 1))/((m + 1)*(b*e - a*f)), x] - Dist[(n*(d*e - c*f))/((m + 1)*(b*e - a*
f)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[
m + n + p + 2, 0] && GtQ[n, 0] &&  !(SumSimplerQ[p, 1] &&  !SumSimplerQ[m, 1])

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{a+c x}\right )^2} \, dx &=\frac{\int \left (\frac{2 a}{x^3}+\frac{b \left (1+\frac{c}{b}\right )}{x^2}-\frac{2 \sqrt{a+b x} \sqrt{a+c x}}{x^3}\right ) \, dx}{(b-c)^2}\\ &=-\frac{a}{(b-c)^2 x^2}-\frac{b+c}{(b-c)^2 x}-\frac{2 \int \frac{\sqrt{a+b x} \sqrt{a+c x}}{x^3} \, dx}{(b-c)^2}\\ &=-\frac{a}{(b-c)^2 x^2}-\frac{b+c}{(b-c)^2 x}+\frac{\sqrt{a+b x} (a+c x)^{3/2}}{a (b-c)^2 x^2}-\frac{\int \frac{\sqrt{a+c x}}{x^2 \sqrt{a+b x}} \, dx}{2 (b-c)}\\ &=-\frac{a}{(b-c)^2 x^2}-\frac{b+c}{(b-c)^2 x}+\frac{\sqrt{a+b x} \sqrt{a+c x}}{2 a (b-c) x}+\frac{\sqrt{a+b x} (a+c x)^{3/2}}{a (b-c)^2 x^2}+\frac{1}{4} \int \frac{1}{x \sqrt{a+b x} \sqrt{a+c x}} \, dx\\ &=-\frac{a}{(b-c)^2 x^2}-\frac{b+c}{(b-c)^2 x}+\frac{\sqrt{a+b x} \sqrt{a+c x}}{2 a (b-c) x}+\frac{\sqrt{a+b x} (a+c x)^{3/2}}{a (b-c)^2 x^2}+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{-a+a x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{a+c x}}\right )\\ &=-\frac{a}{(b-c)^2 x^2}-\frac{b+c}{(b-c)^2 x}+\frac{\sqrt{a+b x} \sqrt{a+c x}}{2 a (b-c) x}+\frac{\sqrt{a+b x} (a+c x)^{3/2}}{a (b-c)^2 x^2}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a+c x}}\right )}{2 a}\\ \end{align*}

Mathematica [A]  time = 0.147289, size = 109, normalized size = 0.89 \[ \frac{-2 a^2-x^2 (b-c)^2 \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a+c x}}\right )+2 a \left (\sqrt{a+b x} \sqrt{a+c x}-b x-c x\right )+x (b+c) \sqrt{a+b x} \sqrt{a+c x}}{2 a x^2 (b-c)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.015, size = 313, normalized size = 2.5 \begin{align*} -{\frac{b}{x \left ( b-c \right ) ^{2}}}-{\frac{c}{x \left ( b-c \right ) ^{2}}}-{\frac{a}{ \left ( b-c \right ) ^{2}{x}^{2}}}+{\frac{{\it csgn} \left ( a \right ) }{4\, \left ( b-c \right ) ^{2}a{x}^{2}}\sqrt{bx+a}\sqrt{cx+a} \left ( -\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 ){x}^{2}{b}^{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 ){x}^{2}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 ){x}^{2}{c}^{2}+2\,{\it csgn} \left ( a \right ) \sqrt{bc{x}^{2}+abx+acx+{a}^{2}}xb+2\,{\it csgn} \left ( a \right ) \sqrt{bc{x}^{2}+abx+acx+{a}^{2}}xc+4\,{\it csgn} \left ( a \right ) a\sqrt{bc{x}^{2}+abx+acx+{a}^{2}} \right ){\frac{1}{\sqrt{bc{x}^{2}+abx+acx+{a}^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/x/(b-c)^2*b-1/x/(b-c)^2*c-a/(b-c)^2/x^2+1/4/(b-c)^2*(b*x+a)^(1/2)*(c*x+a)^(1/2)/a*(-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^2*b^2+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^2*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^2*c^2+2*csgn(a)*(b*c*x^2+a*b*x+
a*c*x+a^2)^(1/2)*x*b+2*csgn(a)*(b*c*x^2+a*b*x+a*c*x+a^2)^(1/2)*x*c+4*csgn(a)*a*(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^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{c x + a}\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)+(c*x+a)^(1/2))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.30788, size = 308, normalized size = 2.5 \begin{align*} \frac{4 \,{\left (b^{2} - 2 \, b c + c^{2}\right )} x^{2} \log \left (-\frac{{\left (b + c\right )} x - 2 \, \sqrt{b x + a} \sqrt{c x + a} + 2 \, a}{x}\right ) +{\left (b^{2} + 2 \, b c + c^{2}\right )} x^{2} + 8 \,{\left ({\left (b + c\right )} x + 2 \, a\right )} \sqrt{b x + a} \sqrt{c x + a} - 16 \, a^{2} - 16 \,{\left (a b + a c\right )} x}{16 \,{\left (a b^{2} - 2 \, a b c + a c^{2}\right )} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(4*(b^2 - 2*b*c + c^2)*x^2*log(-((b + c)*x - 2*sqrt(b*x + a)*sqrt(c*x + a) + 2*a)/x) + (b^2 + 2*b*c + c^2
)*x^2 + 8*((b + c)*x + 2*a)*sqrt(b*x + a)*sqrt(c*x + a) - 16*a^2 - 16*(a*b + a*c)*x)/((a*b^2 - 2*a*b*c + a*c^2
)*x^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{a + c x}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out