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-1/2*arctanh((b*x+a)^(1/2)/(c*x+a)^(1/2))/a+(c*x+a)^(3/2)*(b*x+a)^(1/2)/a/(b-c)
^2/x^2+1/2*(b*x+a)^(1/2)*(c*x+a)^(1/2)/a/(b-c)/x

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Rubi [A]  time = 0.20, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, 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 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 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 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]

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]

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*}

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Mathematica [A]  time = 0.15, 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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fricas [A]  time = 0.52, size = 126, normalized size = 1.02 \[ \frac {4 \, {\left (b^{2} - 2 \, b c + c^{2}\right )} x^{2} \log \left (-\frac {{\left (b + c\right )} x - 2 \, \sqrt {b x + a} \sqrt {c x + a} + 2 \, a}{x}\right ) + {\left (b^{2} + 2 \, b c + c^{2}\right )} x^{2} + 8 \, {\left ({\left (b + c\right )} x + 2 \, a\right )} \sqrt {b x + a} \sqrt {c x + a} - 16 \, a^{2} - 16 \, {\left (a b + a c\right )} x}{16 \, {\left (a b^{2} - 2 \, a b c + a c^{2}\right )} x^{2}} \]

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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giac [B]  time = 12.25, size = 532, normalized size = 4.33 \[ \frac {\sqrt {b c} {\left | b \right |} \arctan \left (\frac {a b^{2} + a b c - {\left (\sqrt {b c} \sqrt {b x + a} - \sqrt {a b^{2} + {\left (b x + a\right )} b c - a b c}\right )}^{2}}{2 \, \sqrt {-b c} a b}\right )}{2 \, \sqrt {-b c} a b} - \frac {{\left (b^{2} + 6 \, b c + c^{2}\right )} \sqrt {b c} {\left (\sqrt {b c} \sqrt {b x + a} - \sqrt {a b^{2} + {\left (b x + a\right )} b c - a b c}\right )}^{6} {\left | b \right |} - {\left (3 \, b^{4} + 5 \, b^{3} c + 5 \, b^{2} c^{2} + 3 \, b c^{3}\right )} \sqrt {b c} {\left (\sqrt {b c} \sqrt {b x + a} - \sqrt {a b^{2} + {\left (b x + a\right )} b c - a b c}\right )}^{4} a {\left | b \right |} + {\left (3 \, b^{6} - 4 \, b^{5} c + 2 \, b^{4} c^{2} - 4 \, b^{3} c^{3} + 3 \, b^{2} c^{4}\right )} \sqrt {b c} {\left (\sqrt {b c} \sqrt {b x + a} - \sqrt {a b^{2} + {\left (b x + a\right )} b c - a b c}\right )}^{2} a^{2} {\left | b \right |} - {\left (b^{8} - 3 \, b^{7} c + 2 \, b^{6} c^{2} + 2 \, b^{5} c^{3} - 3 \, b^{4} c^{4} + b^{3} c^{5}\right )} \sqrt {b c} a^{3} {\left | b \right |}}{{\left ({\left (\sqrt {b c} \sqrt {b x + a} - \sqrt {a b^{2} + {\left (b x + a\right )} b c - a b c}\right )}^{4} - 2 \, {\left (b^{2} + b c\right )} {\left (\sqrt {b c} \sqrt {b x + a} - \sqrt {a b^{2} + {\left (b x + a\right )} b c - a b c}\right )}^{2} a + {\left (b^{4} - 2 \, b^{3} c + b^{2} c^{2}\right )} a^{2}\right )}^{2} {\left (b^{2} - 2 \, b c + c^{2}\right )}} - \frac {{\left (b x + a\right )} b^{2} + {\left (b x + a\right )} b c - a b c}{{\left (b^{2} - 2 \, b c + c^{2}\right )} b^{2} x^{2}} \]

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]

1/2*sqrt(b*c)*abs(b)*arctan(1/2*(a*b^2 + a*b*c - (sqrt(b*c)*sqrt(b*x + a) - sqrt(a*b^2 + (b*x + a)*b*c - a*b*c
))^2)/(sqrt(-b*c)*a*b))/(sqrt(-b*c)*a*b) - ((b^2 + 6*b*c + c^2)*sqrt(b*c)*(sqrt(b*c)*sqrt(b*x + a) - sqrt(a*b^
2 + (b*x + a)*b*c - a*b*c))^6*abs(b) - (3*b^4 + 5*b^3*c + 5*b^2*c^2 + 3*b*c^3)*sqrt(b*c)*(sqrt(b*c)*sqrt(b*x +
a) - sqrt(a*b^2 + (b*x + a)*b*c - a*b*c))^4*a*abs(b) + (3*b^6 - 4*b^5*c + 2*b^4*c^2 - 4*b^3*c^3 + 3*b^2*c^4)*
sqrt(b*c)*(sqrt(b*c)*sqrt(b*x + a) - sqrt(a*b^2 + (b*x + a)*b*c - a*b*c))^2*a^2*abs(b) - (b^8 - 3*b^7*c + 2*b^
6*c^2 + 2*b^5*c^3 - 3*b^4*c^4 + b^3*c^5)*sqrt(b*c)*a^3*abs(b))/(((sqrt(b*c)*sqrt(b*x + a) - sqrt(a*b^2 + (b*x
+ a)*b*c - a*b*c))^4 - 2*(b^2 + b*c)*(sqrt(b*c)*sqrt(b*x + a) - sqrt(a*b^2 + (b*x + a)*b*c - a*b*c))^2*a + (b^
4 - 2*b^3*c + b^2*c^2)*a^2)^2*(b^2 - 2*b*c + c^2)) - ((b*x + a)*b^2 + (b*x + a)*b*c - a*b*c)/((b^2 - 2*b*c + c
^2)*b^2*x^2)

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maple [C]  time = 0.01, size = 313, normalized size = 2.54 \[ -\frac {b}{\left (b -c \right )^{2} x}-\frac {c}{\left (b -c \right )^{2} x}-\frac {a}{\left (b -c \right )^{2} x^{2}}+\frac {\sqrt {b x +a}\, \sqrt {c x +a}\, \left (-b^{2} x^{2} \ln \left (\frac {\left (b x +c x +2 a +2 \sqrt {b c \,x^{2}+a b x +a c x +a^{2}}\, \mathrm {csgn}\relax (a )\right ) a}{x}\right )+2 b c \,x^{2} \ln \left (\frac {\left (b x +c x +2 a +2 \sqrt {b c \,x^{2}+a b x +a c x +a^{2}}\, \mathrm {csgn}\relax (a )\right ) a}{x}\right )-c^{2} x^{2} \ln \left (\frac {\left (b x +c x +2 a +2 \sqrt {b c \,x^{2}+a b x +a c x +a^{2}}\, \mathrm {csgn}\relax (a )\right ) a}{x}\right )+2 \sqrt {b c \,x^{2}+a b x +a c x +a^{2}}\, b x \,\mathrm {csgn}\relax (a )+2 \sqrt {b c \,x^{2}+a b x +a c x +a^{2}}\, c x \,\mathrm {csgn}\relax (a )+4 \sqrt {b c \,x^{2}+a b x +a c x +a^{2}}\, a \,\mathrm {csgn}\relax (a )\right ) \mathrm {csgn}\relax (a )}{4 \left (b -c \right )^{2} \sqrt {b c \,x^{2}+a b x +a c x +a^{2}}\, a \,x^{2}} \]

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((b*x+c*x+2*a+2*(b*c*
x^2+a*b*x+a*c*x+a^2)^(1/2)*csgn(a))*a/x)*x^2*b^2+2*ln((b*x+c*x+2*a+2*(b*c*x^2+a*b*x+a*c*x+a^2)^(1/2)*csgn(a))*
a/x)*x^2*b*c-ln((b*x+c*x+2*a+2*(b*c*x^2+a*b*x+a*c*x+a^2)^(1/2)*csgn(a))*a/x)*x^2*c^2+2*(b*c*x^2+a*b*x+a*c*x+a^
2)^(1/2)*csgn(a)*x*b+2*(b*c*x^2+a*b*x+a*c*x+a^2)^(1/2)*csgn(a)*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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x {\left (\sqrt {b x + a} + \sqrt {c x + a}\right )}^{2}}\,{d x} \]

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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mupad [B]  time = 12.32, size = 787, normalized size = 6.40 \[ \frac {\ln \left (\frac {\left (\sqrt {a+b\,x}-\sqrt {a+c\,x}\right )\,\left (b-\frac {c\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\sqrt {a+c\,x}-\sqrt {a}}\right )}{\sqrt {a+c\,x}-\sqrt {a}}\right )}{4\,a}-\frac {\frac {b^4}{2}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4\,\left (-\frac {b^4}{2}+4\,b^3\,c+\frac {3\,b^2\,c^2}{2}+4\,b\,c^3-\frac {c^4}{2}\right )}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^4}-\frac {\left (2\,b^4+2\,c\,b^3\right )\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{\sqrt {a+c\,x}-\sqrt {a}}-\frac {\left (b^2\,c^2+b\,c^3\right )\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^5}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^5}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2\,\left (\frac {5\,b^4}{2}+6\,b^3\,c+\frac {5\,b^2\,c^2}{2}\right )}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^2}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3\,\left (b^4+6\,b^3\,c+6\,b^2\,c^2+b\,c^3\right )}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^3}}{\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4\,\left (8\,a\,b^4+16\,a\,b^3\,c-48\,a\,b^2\,c^2+16\,a\,b\,c^3+8\,a\,c^4\right )}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^4}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3\,\left (16\,a\,b^4-16\,a\,b^3\,c-16\,a\,b^2\,c^2+16\,a\,b\,c^3\right )}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^3}-\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^5\,\left (16\,a\,b^3\,c-16\,a\,b^2\,c^2-16\,a\,b\,c^3+16\,a\,c^4\right )}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^5}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2\,\left (8\,a\,b^4-16\,a\,b^3\,c+8\,a\,b^2\,c^2\right )}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^2}+\frac {{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6\,\left (8\,a\,b^2\,c^2-16\,a\,b\,c^3+8\,a\,c^4\right )}{{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^6}}-\frac {\ln \left (\frac {\sqrt {a+b\,x}-\sqrt {a}}{\sqrt {a+c\,x}-\sqrt {a}}\right )}{4\,a}-\frac {a+x\,\left (b+c\right )}{x^2\,\left (b^2-2\,b\,c+c^2\right )}-\frac {c^2\,{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}{16\,a\,{\left (b-c\right )}^2\,{\left (\sqrt {a+c\,x}-\sqrt {a}\right )}^2}+\frac {c\,\left (b+c\right )\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}{8\,a\,{\left (b-c\right )}^2\,\left (\sqrt {a+c\,x}-\sqrt {a}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log((((a + b*x)^(1/2) - (a + c*x)^(1/2))*(b - (c*((a + b*x)^(1/2) - a^(1/2)))/((a + c*x)^(1/2) - a^(1/2))))/((
a + c*x)^(1/2) - a^(1/2)))/(4*a) - (b^4/2 + (((a + b*x)^(1/2) - a^(1/2))^4*(4*b*c^3 + 4*b^3*c - b^4/2 - c^4/2
+ (3*b^2*c^2)/2))/((a + c*x)^(1/2) - a^(1/2))^4 - ((2*b^3*c + 2*b^4)*((a + b*x)^(1/2) - a^(1/2)))/((a + c*x)^(
1/2) - a^(1/2)) - ((b*c^3 + b^2*c^2)*((a + b*x)^(1/2) - a^(1/2))^5)/((a + c*x)^(1/2) - a^(1/2))^5 + (((a + b*x
)^(1/2) - a^(1/2))^2*(6*b^3*c + (5*b^4)/2 + (5*b^2*c^2)/2))/((a + c*x)^(1/2) - a^(1/2))^2 - (((a + b*x)^(1/2)
- a^(1/2))^3*(b*c^3 + 6*b^3*c + b^4 + 6*b^2*c^2))/((a + c*x)^(1/2) - a^(1/2))^3)/((((a + b*x)^(1/2) - a^(1/2))
^4*(8*a*b^4 + 8*a*c^4 - 48*a*b^2*c^2 + 16*a*b*c^3 + 16*a*b^3*c))/((a + c*x)^(1/2) - a^(1/2))^4 - (((a + b*x)^(
1/2) - a^(1/2))^3*(16*a*b^4 - 16*a*b^2*c^2 + 16*a*b*c^3 - 16*a*b^3*c))/((a + c*x)^(1/2) - a^(1/2))^3 - (((a +
b*x)^(1/2) - a^(1/2))^5*(16*a*c^4 - 16*a*b^2*c^2 - 16*a*b*c^3 + 16*a*b^3*c))/((a + c*x)^(1/2) - a^(1/2))^5 + (
((a + b*x)^(1/2) - a^(1/2))^2*(8*a*b^4 + 8*a*b^2*c^2 - 16*a*b^3*c))/((a + c*x)^(1/2) - a^(1/2))^2 + (((a + b*x
)^(1/2) - a^(1/2))^6*(8*a*c^4 + 8*a*b^2*c^2 - 16*a*b*c^3))/((a + c*x)^(1/2) - a^(1/2))^6) - log(((a + b*x)^(1/
2) - a^(1/2))/((a + c*x)^(1/2) - a^(1/2)))/(4*a) - (a + x*(b + c))/(x^2*(b^2 - 2*b*c + c^2)) - (c^2*((a + b*x)
^(1/2) - a^(1/2))^2)/(16*a*(b - c)^2*((a + c*x)^(1/2) - a^(1/2))^2) + (c*(b + c)*((a + b*x)^(1/2) - a^(1/2)))/
(8*a*(b - c)^2*((a + c*x)^(1/2) - a^(1/2)))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x \left (\sqrt {a + b x} + \sqrt {a + c x}\right )^{2}}\, dx \]

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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