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

Optimal. Leaf size=171 \[ \frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{4 a^{3/2} (b-c)}-\frac{c^2 \tanh ^{-1}\left (\frac{\sqrt{a+c x}}{\sqrt{a}}\right )}{4 a^{3/2} (b-c)}-\frac{\sqrt{a+b x}}{2 x^2 (b-c)}+\frac{\sqrt{a+c x}}{2 x^2 (b-c)}-\frac{b \sqrt{a+b x}}{4 a x (b-c)}+\frac{c \sqrt{a+c x}}{4 a x (b-c)} \]

[Out]

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

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Rubi [A]  time = 0.112481, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2103, 47, 51, 63, 208} \[ \frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{4 a^{3/2} (b-c)}-\frac{c^2 \tanh ^{-1}\left (\frac{\sqrt{a+c x}}{\sqrt{a}}\right )}{4 a^{3/2} (b-c)}-\frac{\sqrt{a+b x}}{2 x^2 (b-c)}+\frac{\sqrt{a+c x}}{2 x^2 (b-c)}-\frac{b \sqrt{a+b x}}{4 a x (b-c)}+\frac{c \sqrt{a+c x}}{4 a x (b-c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2103

Int[(u_)/((e_.)*Sqrt[(a_.) + (b_.)*(x_)] + (f_.)*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[c/(e*(b*c - a*d)
), Int[(u*Sqrt[a + b*x])/x, x], x] - Dist[a/(f*(b*c - a*d)), Int[(u*Sqrt[c + d*x])/x, x], x] /; FreeQ[{a, b, c
, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a*e^2 - c*f^2, 0]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, 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 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 )} \, dx &=\frac{\int \frac{\sqrt{a+b x}}{x^3} \, dx}{b-c}-\frac{\int \frac{\sqrt{a+c x}}{x^3} \, dx}{b-c}\\ &=-\frac{\sqrt{a+b x}}{2 (b-c) x^2}+\frac{\sqrt{a+c x}}{2 (b-c) x^2}+\frac{b \int \frac{1}{x^2 \sqrt{a+b x}} \, dx}{4 (b-c)}-\frac{c \int \frac{1}{x^2 \sqrt{a+c x}} \, dx}{4 (b-c)}\\ &=-\frac{\sqrt{a+b x}}{2 (b-c) x^2}-\frac{b \sqrt{a+b x}}{4 a (b-c) x}+\frac{\sqrt{a+c x}}{2 (b-c) x^2}+\frac{c \sqrt{a+c x}}{4 a (b-c) x}-\frac{b^2 \int \frac{1}{x \sqrt{a+b x}} \, dx}{8 a (b-c)}+\frac{c^2 \int \frac{1}{x \sqrt{a+c x}} \, dx}{8 a (b-c)}\\ &=-\frac{\sqrt{a+b x}}{2 (b-c) x^2}-\frac{b \sqrt{a+b x}}{4 a (b-c) x}+\frac{\sqrt{a+c x}}{2 (b-c) x^2}+\frac{c \sqrt{a+c x}}{4 a (b-c) x}-\frac{b \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x}\right )}{4 a (b-c)}+\frac{c \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{c}+\frac{x^2}{c}} \, dx,x,\sqrt{a+c x}\right )}{4 a (b-c)}\\ &=-\frac{\sqrt{a+b x}}{2 (b-c) x^2}-\frac{b \sqrt{a+b x}}{4 a (b-c) x}+\frac{\sqrt{a+c x}}{2 (b-c) x^2}+\frac{c \sqrt{a+c x}}{4 a (b-c) x}+\frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{a+b x}}{\sqrt{a}}\right )}{4 a^{3/2} (b-c)}-\frac{c^2 \tanh ^{-1}\left (\frac{\sqrt{a+c x}}{\sqrt{a}}\right )}{4 a^{3/2} (b-c)}\\ \end{align*}

Mathematica [C]  time = 0.0892831, size = 75, normalized size = 0.44 \[ \frac{2 c^2 (a+c x)^{3/2} \, _2F_1\left (\frac{3}{2},3;\frac{5}{2};\frac{c x}{a}+1\right )-2 b^2 (a+b x)^{3/2} \, _2F_1\left (\frac{3}{2},3;\frac{5}{2};\frac{b x}{a}+1\right )}{3 a^3 (b-c)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b^2*(a + b*x)^(3/2)*Hypergeometric2F1[3/2, 3, 5/2, 1 + (b*x)/a] + 2*c^2*(a + c*x)^(3/2)*Hypergeometric2F1[
3/2, 3, 5/2, 1 + (c*x)/a])/(3*a^3*(b - c))

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Maple [A]  time = 0.009, size = 120, normalized size = 0.7 \begin{align*} 2\,{\frac{{b}^{2}}{b-c} \left ({\frac{1}{{b}^{2}{x}^{2}} \left ( -1/8\,{\frac{ \left ( bx+a \right ) ^{3/2}}{a}}-1/8\,\sqrt{bx+a} \right ) }+1/8\,{\frac{1}{{a}^{3/2}}{\it Artanh} \left ({\frac{\sqrt{bx+a}}{\sqrt{a}}} \right ) } \right ) }-2\,{\frac{{c}^{2}}{b-c} \left ({\frac{1}{{c}^{2}{x}^{2}} \left ( -1/8\,{\frac{ \left ( cx+a \right ) ^{3/2}}{a}}-1/8\,\sqrt{cx+a} \right ) }+1/8\,{\frac{1}{{a}^{3/2}}{\it Artanh} \left ({\frac{\sqrt{cx+a}}{\sqrt{a}}} \right ) } \right ) } \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)),x)

[Out]

2/(b-c)*b^2*((-1/8/a*(b*x+a)^(3/2)-1/8*(b*x+a)^(1/2))/b^2/x^2+1/8/a^(3/2)*arctanh((b*x+a)^(1/2)/a^(1/2)))-2/(b
-c)*c^2*((-1/8/a*(c*x+a)^(3/2)-1/8*(c*x+a)^(1/2))/c^2/x^2+1/8/a^(3/2)*arctanh((c*x+a)^(1/2)/a^(1/2)))

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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 )}}\,{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)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.35671, size = 570, normalized size = 3.33 \begin{align*} \left [-\frac{\sqrt{a} b^{2} x^{2} \log \left (\frac{b x - 2 \, \sqrt{b x + a} \sqrt{a} + 2 \, a}{x}\right ) + \sqrt{a} c^{2} x^{2} \log \left (\frac{c x + 2 \, \sqrt{c x + a} \sqrt{a} + 2 \, a}{x}\right ) + 2 \,{\left (a b x + 2 \, a^{2}\right )} \sqrt{b x + a} - 2 \,{\left (a c x + 2 \, a^{2}\right )} \sqrt{c x + a}}{8 \,{\left (a^{2} b - a^{2} c\right )} x^{2}}, -\frac{\sqrt{-a} b^{2} x^{2} \arctan \left (\frac{\sqrt{b x + a} \sqrt{-a}}{a}\right ) - \sqrt{-a} c^{2} x^{2} \arctan \left (\frac{\sqrt{c x + a} \sqrt{-a}}{a}\right ) +{\left (a b x + 2 \, a^{2}\right )} \sqrt{b x + a} -{\left (a c x + 2 \, a^{2}\right )} \sqrt{c x + a}}{4 \,{\left (a^{2} b - a^{2} c\right )} x^{2}}\right ] \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)),x, algorithm="fricas")

[Out]

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

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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 )}\, 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)),x)

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \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)),x, algorithm="giac")

[Out]

Exception raised: TypeError