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

Optimal. Leaf size=174 \[ \frac{2 (a+b x)^{3/2} (a+c x)^{3/2}}{3 a^2 x^3 (b-c)^2}-\frac{(b+c) \sqrt{a+b x} (a+c x)^{3/2}}{2 a^2 x^2 (b-c)^2}-\frac{(b+c) \sqrt{a+b x} \sqrt{a+c x}}{4 a^2 x (b-c)}+\frac{(b+c) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a+c x}}\right )}{4 a^2}-\frac{2 a}{3 x^3 (b-c)^2}-\frac{b+c}{2 x^2 (b-c)^2} \]

[Out]

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

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Rubi [A]  time = 0.223245, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {6690, 96, 94, 93, 208} \[ \frac{2 (a+b x)^{3/2} (a+c x)^{3/2}}{3 a^2 x^3 (b-c)^2}-\frac{(b+c) \sqrt{a+b x} (a+c x)^{3/2}}{2 a^2 x^2 (b-c)^2}-\frac{(b+c) \sqrt{a+b x} \sqrt{a+c x}}{4 a^2 x (b-c)}+\frac{(b+c) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a+c x}}\right )}{4 a^2}-\frac{2 a}{3 x^3 (b-c)^2}-\frac{b+c}{2 x^2 (b-c)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a)/(3*(b - c)^2*x^3) - (b + c)/(2*(b - c)^2*x^2) - ((b + c)*Sqrt[a + b*x]*Sqrt[a + c*x])/(4*a^2*(b - c)*x)
- ((b + c)*Sqrt[a + b*x]*(a + c*x)^(3/2))/(2*a^2*(b - c)^2*x^2) + (2*(a + b*x)^(3/2)*(a + c*x)^(3/2))/(3*a^2*
(b - c)^2*x^3) + ((b + c)*ArcTanh[Sqrt[a + b*x]/Sqrt[a + c*x]])/(4*a^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 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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^2 \left (\sqrt{a+b x}+\sqrt{a+c x}\right )^2} \, dx &=\frac{\int \left (\frac{2 a}{x^4}+\frac{b \left (1+\frac{c}{b}\right )}{x^3}-\frac{2 \sqrt{a+b x} \sqrt{a+c x}}{x^4}\right ) \, dx}{(b-c)^2}\\ &=-\frac{2 a}{3 (b-c)^2 x^3}-\frac{b+c}{2 (b-c)^2 x^2}-\frac{2 \int \frac{\sqrt{a+b x} \sqrt{a+c x}}{x^4} \, dx}{(b-c)^2}\\ &=-\frac{2 a}{3 (b-c)^2 x^3}-\frac{b+c}{2 (b-c)^2 x^2}+\frac{2 (a+b x)^{3/2} (a+c x)^{3/2}}{3 a^2 (b-c)^2 x^3}+\frac{(b+c) \int \frac{\sqrt{a+b x} \sqrt{a+c x}}{x^3} \, dx}{a (b-c)^2}\\ &=-\frac{2 a}{3 (b-c)^2 x^3}-\frac{b+c}{2 (b-c)^2 x^2}-\frac{(b+c) \sqrt{a+b x} (a+c x)^{3/2}}{2 a^2 (b-c)^2 x^2}+\frac{2 (a+b x)^{3/2} (a+c x)^{3/2}}{3 a^2 (b-c)^2 x^3}+\frac{(b+c) \int \frac{\sqrt{a+c x}}{x^2 \sqrt{a+b x}} \, dx}{4 a (b-c)}\\ &=-\frac{2 a}{3 (b-c)^2 x^3}-\frac{b+c}{2 (b-c)^2 x^2}-\frac{(b+c) \sqrt{a+b x} \sqrt{a+c x}}{4 a^2 (b-c) x}-\frac{(b+c) \sqrt{a+b x} (a+c x)^{3/2}}{2 a^2 (b-c)^2 x^2}+\frac{2 (a+b x)^{3/2} (a+c x)^{3/2}}{3 a^2 (b-c)^2 x^3}-\frac{(b+c) \int \frac{1}{x \sqrt{a+b x} \sqrt{a+c x}} \, dx}{8 a}\\ &=-\frac{2 a}{3 (b-c)^2 x^3}-\frac{b+c}{2 (b-c)^2 x^2}-\frac{(b+c) \sqrt{a+b x} \sqrt{a+c x}}{4 a^2 (b-c) x}-\frac{(b+c) \sqrt{a+b x} (a+c x)^{3/2}}{2 a^2 (b-c)^2 x^2}+\frac{2 (a+b x)^{3/2} (a+c x)^{3/2}}{3 a^2 (b-c)^2 x^3}-\frac{(b+c) \operatorname{Subst}\left (\int \frac{1}{-a+a x^2} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{a+c x}}\right )}{4 a}\\ &=-\frac{2 a}{3 (b-c)^2 x^3}-\frac{b+c}{2 (b-c)^2 x^2}-\frac{(b+c) \sqrt{a+b x} \sqrt{a+c x}}{4 a^2 (b-c) x}-\frac{(b+c) \sqrt{a+b x} (a+c x)^{3/2}}{2 a^2 (b-c)^2 x^2}+\frac{2 (a+b x)^{3/2} (a+c x)^{3/2}}{3 a^2 (b-c)^2 x^3}+\frac{(b+c) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a+c x}}\right )}{4 a^2}\\ \end{align*}

Mathematica [A]  time = 0.179233, size = 153, normalized size = 0.88 \[ \frac{a^2 \left (8 \sqrt{a+b x} \sqrt{a+c x}-6 b x-6 c x\right )-8 a^3+x^2 \left (-3 b^2+2 b c-3 c^2\right ) \sqrt{a+b x} \sqrt{a+c x}+3 x^3 (b-c)^2 (b+c) \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a+c x}}\right )+2 a x (b+c) \sqrt{a+b x} \sqrt{a+c x}}{12 a^2 x^3 (b-c)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*a^3 + 2*a*(b + c)*x*Sqrt[a + b*x]*Sqrt[a + c*x] + (-3*b^2 + 2*b*c - 3*c^2)*x^2*Sqrt[a + b*x]*Sqrt[a + c*x]
+ a^2*(-6*b*x - 6*c*x + 8*Sqrt[a + b*x]*Sqrt[a + c*x]) + 3*(b - c)^2*(b + c)*x^3*ArcTanh[Sqrt[a + b*x]/Sqrt[a
+ c*x]])/(12*a^2*(b - c)^2*x^3)

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Maple [C]  time = 0.011, size = 457, normalized size = 2.6 \begin{align*} -{\frac{b}{2\,{x}^{2} \left ( b-c \right ) ^{2}}}-{\frac{c}{2\,{x}^{2} \left ( b-c \right ) ^{2}}}-{\frac{2\,a}{3\, \left ( b-c \right ) ^{2}{x}^{3}}}-{\frac{{\it csgn} \left ( a \right ) }{24\, \left ( b-c \right ) ^{2}{a}^{2}{x}^{3}}\sqrt{bx+a}\sqrt{cx+a} \left ( -3\,\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}^{3}{b}^{3}+3\,\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}^{3}{b}^{2}c+3\,\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}^{3}b{c}^{2}-3\,\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}^{3}{c}^{3}+6\,{\it csgn} \left ( a \right ) \sqrt{bc{x}^{2}+abx+acx+{a}^{2}}{x}^{2}{b}^{2}-4\,{\it csgn} \left ( a \right ) \sqrt{bc{x}^{2}+abx+acx+{a}^{2}}{x}^{2}bc+6\,{\it csgn} \left ( a \right ) \sqrt{bc{x}^{2}+abx+acx+{a}^{2}}{x}^{2}{c}^{2}-4\,{\it csgn} \left ( a \right ) a\sqrt{bc{x}^{2}+abx+acx+{a}^{2}}xb-4\,{\it csgn} \left ( a \right ) a\sqrt{bc{x}^{2}+abx+acx+{a}^{2}}xc-16\,\sqrt{bc{x}^{2}+abx+acx+{a}^{2}}{a}^{2}{\it csgn} \left ( a \right ) \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^2/((b*x+a)^(1/2)+(c*x+a)^(1/2))^2,x)

[Out]

-1/2/x^2/(b-c)^2*b-1/2/x^2/(b-c)^2*c-2/3*a/(b-c)^2/x^3-1/24/(b-c)^2*(b*x+a)^(1/2)*(c*x+a)^(1/2)/a^2*(-3*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^3*b^3+3*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^3*b^2*c+3*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^3*b*c^2-3*
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^3*c^3+6*csgn(a)*(b*c*x^2+a*b*x+a*c*x+a^2)^(1
/2)*x^2*b^2-4*csgn(a)*(b*c*x^2+a*b*x+a*c*x+a^2)^(1/2)*x^2*b*c+6*csgn(a)*(b*c*x^2+a*b*x+a*c*x+a^2)^(1/2)*x^2*c^
2-4*csgn(a)*a*(b*c*x^2+a*b*x+a*c*x+a^2)^(1/2)*x*b-4*csgn(a)*a*(b*c*x^2+a*b*x+a*c*x+a^2)^(1/2)*x*c-16*(b*c*x^2+
a*b*x+a*c*x+a^2)^(1/2)*a^2*csgn(a))*csgn(a)/(b*c*x^2+a*b*x+a*c*x+a^2)^(1/2)/x^3

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

[Out]

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

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Fricas [A]  time = 1.3039, size = 408, normalized size = 2.34 \begin{align*} -\frac{12 \,{\left (b^{3} - b^{2} c - b c^{2} + c^{3}\right )} x^{3} \log \left (-\frac{{\left (b + c\right )} x - 2 \, \sqrt{b x + a} \sqrt{c x + a} + 2 \, a}{x}\right ) +{\left (5 \, b^{3} + 3 \, b^{2} c + 3 \, b c^{2} + 5 \, c^{3}\right )} x^{3} + 64 \, a^{3} + 8 \,{\left ({\left (3 \, b^{2} - 2 \, b c + 3 \, c^{2}\right )} x^{2} - 8 \, a^{2} - 2 \,{\left (a b + a c\right )} x\right )} \sqrt{b x + a} \sqrt{c x + a} + 48 \,{\left (a^{2} b + a^{2} c\right )} x}{96 \,{\left (a^{2} b^{2} - 2 \, a^{2} b c + a^{2} c^{2}\right )} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(12*(b^3 - b^2*c - b*c^2 + c^3)*x^3*log(-((b + c)*x - 2*sqrt(b*x + a)*sqrt(c*x + a) + 2*a)/x) + (5*b^3 +
3*b^2*c + 3*b*c^2 + 5*c^3)*x^3 + 64*a^3 + 8*((3*b^2 - 2*b*c + 3*c^2)*x^2 - 8*a^2 - 2*(a*b + a*c)*x)*sqrt(b*x
+ a)*sqrt(c*x + a) + 48*(a^2*b + a^2*c)*x)/((a^2*b^2 - 2*a^2*b*c + a^2*c^2)*x^3)

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

[Out]

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

[Out]

Timed out