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]

1/4*b^2*arctanh((b*x+a)^(1/2)/a^(1/2))/a^(3/2)/(b-c)-1/4*c^2*arctanh((c*x+a)^(1/2)/a^(1/2))/a^(3/2)/(b-c)-1/2*
(b*x+a)^(1/2)/(b-c)/x^2-1/4*b*(b*x+a)^(1/2)/a/(b-c)/x+1/2*(c*x+a)^(1/2)/(b-c)/x^2+1/4*c*(c*x+a)^(1/2)/a/(b-c)/
x

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Rubi [A]  time = 0.11, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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]

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]

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*}

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Mathematica [C]  time = 0.09, 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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fricas [A]  time = 0.49, size = 243, normalized size = 1.42 \[ \left [-\frac {\sqrt {a} b^{2} x^{2} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + \sqrt {a} c^{2} x^{2} \log \left (\frac {c x + 2 \, \sqrt {c x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (a b x + 2 \, a^{2}\right )} \sqrt {b x + a} - 2 \, {\left (a c x + 2 \, a^{2}\right )} \sqrt {c x + a}}{8 \, {\left (a^{2} b - a^{2} c\right )} x^{2}}, -\frac {\sqrt {-a} b^{2} x^{2} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right ) - \sqrt {-a} c^{2} x^{2} \arctan \left (\frac {\sqrt {c x + a} \sqrt {-a}}{a}\right ) + {\left (a b x + 2 \, a^{2}\right )} \sqrt {b x + a} - {\left (a c x + 2 \, a^{2}\right )} \sqrt {c x + a}}{4 \, {\left (a^{2} b - a^{2} c\right )} x^{2}}\right ] \]

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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giac [B]  time = 37.26, size = 1895, normalized size = 11.08 \[ \text {result too large to display} \]

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]

-1/4*b^2*arctan(sqrt(b*x + a)/sqrt(-a))/((a*b - a*c)*sqrt(-a)) - 1/2*((sqrt(b*c)*sqrt(b*x + a) - sqrt(a*b^2 +
(b*x + a)*b*c - a*b*c))*a^3*b^6*c^2*abs(b) - 3*(sqrt(b*c)*sqrt(b*x + a) - sqrt(a*b^2 + (b*x + a)*b*c - a*b*c))
*a^3*b^5*c^3*abs(b) + 3*(sqrt(b*c)*sqrt(b*x + a) - sqrt(a*b^2 + (b*x + a)*b*c - a*b*c))*a^3*b^4*c^4*abs(b) - (
sqrt(b*c)*sqrt(b*x + a) - sqrt(a*b^2 + (b*x + a)*b*c - a*b*c))*a^3*b^3*c^5*abs(b) + 7*(sqrt(b*c)*sqrt(b*x + a)
- sqrt(a*b^2 + (b*x + a)*b*c - a*b*c))^3*a^2*b^4*c^2*abs(b) - 10*(sqrt(b*c)*sqrt(b*x + a) - sqrt(a*b^2 + (b*x
+ a)*b*c - a*b*c))^3*a^2*b^3*c^3*abs(b) + 3*(sqrt(b*c)*sqrt(b*x + a) - sqrt(a*b^2 + (b*x + a)*b*c - a*b*c))^3
*a^2*b^2*c^4*abs(b) + 7*(sqrt(b*c)*sqrt(b*x + a) - sqrt(a*b^2 + (b*x + a)*b*c - a*b*c))^5*a*b^2*c^2*abs(b) - 3
*(sqrt(b*c)*sqrt(b*x + a) - sqrt(a*b^2 + (b*x + a)*b*c - a*b*c))^5*a*b*c^3*abs(b) + (sqrt(b*c)*sqrt(b*x + a) -
sqrt(a*b^2 + (b*x + a)*b*c - a*b*c))^7*c^2*abs(b))/((a^2*b^4 - 2*a^2*b^3*c + a^2*b^2*c^2 - 2*(sqrt(b*c)*sqrt(
b*x + a) - sqrt(a*b^2 + (b*x + a)*b*c - a*b*c))^2*a*b^2 - 2*(sqrt(b*c)*sqrt(b*x + a) - sqrt(a*b^2 + (b*x + a)*
b*c - a*b*c))^2*a*b*c + (sqrt(b*c)*sqrt(b*x + a) - sqrt(a*b^2 + (b*x + a)*b*c - a*b*c))^4)^2*(a*b - a*c)) - 1/
4*((b*x + a)^(3/2)*b^2 + sqrt(b*x + a)*a*b^2)/((a*b - a*c)*b^2*x^2) - 1/4*(2*(a*b^3*c^3 - a*b^2*c^4)*(a^2*b^2
- a^2*b*c)^2*sqrt(-a)*abs(b)*sgn(8*a*b - 8*a*c) + 2*(a*b^3*c^2 - a*b^2*c^3)*(a^2*b^2 - a^2*b*c)^2*sqrt(-a*b*c)
*abs(b) + (a^3*b^5*c^2 - 3*a^3*b^4*c^3 + 3*a^3*b^3*c^4 - a^3*b^2*c^5)*sqrt(-a*b*c)*abs(a^2*b^2 - a^2*b*c)*abs(
b)*sgn(8*a*b - 8*a*c) + (a^3*b^6*c^2 - 3*a^3*b^5*c^3 + 3*a^3*b^4*c^4 - a^3*b^3*c^5)*sqrt(-a)*abs(a^2*b^2 - a^2
*b*c)*abs(b) + (a^5*b^7*c^3 - 2*a^5*b^6*c^4 + 2*a^5*b^4*c^6 - a^5*b^3*c^7)*sqrt(-a)*abs(b)*sgn(8*a*b - 8*a*c)
+ (a^5*b^7*c^2 - 2*a^5*b^6*c^3 + 2*a^5*b^4*c^5 - a^5*b^3*c^6)*sqrt(-a*b*c)*abs(b))*arctan(-(sqrt(b*c)*sqrt(b*x
+ a) - sqrt(a*b^2 + (b*x + a)*b*c - a*b*c))/sqrt(-(a^2*b^3 - a^2*b*c^2 + sqrt((a^2*b^3 - a^2*b*c^2)^2 - (a^3*
b^5 - 3*a^3*b^4*c + 3*a^3*b^3*c^2 - a^3*b^2*c^3)*(a*b - a*c)))/(a*b - a*c)))/((b^8 - 5*b^7*c + 10*b^6*c^2 - 10
*b^5*c^3 + 5*b^4*c^4 - b^3*c^5)*a^5*abs(a^2*b^2 - a^2*b*c)) + 1/4*(2*(a*b^3*c^3 - a*b^2*c^4)*(a^2*b^2 - a^2*b*
c)^2*sqrt(-a)*abs(b)*sgn(8*a*b - 8*a*c) + 2*(a*b^3*c^2 - a*b^2*c^3)*(a^2*b^2 - a^2*b*c)^2*sqrt(-a*b*c)*abs(b)
+ (a^3*b^5*c^2 - 3*a^3*b^4*c^3 + 3*a^3*b^3*c^4 - a^3*b^2*c^5)*sqrt(-a*b*c)*abs(a^2*b^2 - a^2*b*c)*abs(b)*sgn(8
*a*b - 8*a*c) + (a^3*b^6*c^2 - 3*a^3*b^5*c^3 + 3*a^3*b^4*c^4 - a^3*b^3*c^5)*sqrt(-a)*abs(a^2*b^2 - a^2*b*c)*ab
s(b) + (a^5*b^7*c^3 - 2*a^5*b^6*c^4 + 2*a^5*b^4*c^6 - a^5*b^3*c^7)*sqrt(-a)*abs(b)*sgn(8*a*b - 8*a*c) + (a^5*b
^7*c^2 - 2*a^5*b^6*c^3 + 2*a^5*b^4*c^5 - a^5*b^3*c^6)*sqrt(-a*b*c)*abs(b))*arctan(-(sqrt(b*c)*sqrt(b*x + a) -
sqrt(a*b^2 + (b*x + a)*b*c - a*b*c))/sqrt(-(a^2*b^3 - a^2*b*c^2 - sqrt((a^2*b^3 - a^2*b*c^2)^2 - (a^3*b^5 - 3*
a^3*b^4*c + 3*a^3*b^3*c^2 - a^3*b^2*c^3)*(a*b - a*c)))/(a*b - a*c)))/((b^8 - 5*b^7*c + 10*b^6*c^2 - 10*b^5*c^3
+ 5*b^4*c^4 - b^3*c^5)*a^5*abs(a^2*b^2 - a^2*b*c))

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maple [A]  time = 0.01, size = 120, normalized size = 0.70 \[ \frac {2 \left (\frac {\arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 a^{\frac {3}{2}}}+\frac {-\frac {\left (b x +a \right )^{\frac {3}{2}}}{8 a}-\frac {\sqrt {b x +a}}{8}}{b^{2} x^{2}}\right ) b^{2}}{b -c}-\frac {2 \left (\frac {\arctanh \left (\frac {\sqrt {c x +a}}{\sqrt {a}}\right )}{8 a^{\frac {3}{2}}}+\frac {-\frac {\left (c x +a \right )^{\frac {3}{2}}}{8 a}-\frac {\sqrt {c x +a}}{8}}{c^{2} x^{2}}\right ) c^{2}}{b -c} \]

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

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

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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mupad [B]  time = 11.85, size = 1610, normalized size = 9.42 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a^(3/2)*b^3)/(16*(a^3*c^2 - a^3*b*c)) + (a^(3/2)*((a + b*x)^(1/2) - a^(1/2))^2*((b*c^2)/4 - (7*b^2*c)/16 + b
^3/4))/((a^3*c^2 - a^3*b*c)*((a + c*x)^(1/2) - a^(1/2))^2) - (a^(3/2)*((b^2*c)/16 + b^3/16)*((a + b*x)^(1/2) -
a^(1/2)))/((a^3*c^2 - a^3*b*c)*((a + c*x)^(1/2) - a^(1/2))) + ((b^2/8 - c^2/8)*((a + b*x)^(1/2) - a^(1/2))^3)
/(a^(3/2)*c*((a + c*x)^(1/2) - a^(1/2))^3))/(((a + b*x)^(1/2) - a^(1/2))^4/((a + c*x)^(1/2) - a^(1/2))^4 - ((b
+ c)*((a + b*x)^(1/2) - a^(1/2))^3)/(c*((a + c*x)^(1/2) - a^(1/2))^3) + (b*((a + b*x)^(1/2) - a^(1/2))^2)/(c*
((a + c*x)^(1/2) - a^(1/2))^2)) - (((c*(b + c))/(4*a^(3/2)*(b - c)) - (c*(b^2 - c^2))/(4*a^(3/2)*(b - c)^2))*(
(a + b*x)^(1/2) - a^(1/2)))/((a + c*x)^(1/2) - a^(1/2)) - (log(((a + b*x)^(1/2) - a^(1/2))/((a + c*x)^(1/2) -
a^(1/2)))*(a^(3/2)*b^2 + a^(3/2)*c^2))/(8*a^3*b - 8*a^3*c) + (atan((((b + c)*(((b + c)*((64*a^6*b^3 - 64*a^6*b
*c^2)/(64*(a^6*c^3 - a^6*b*c^2)) - (((a + b*x)^(1/2) - a^(1/2))*(64*a^6*b^3 - 64*a^6*c^3 + 128*a^6*b*c^2 - 128
*a^6*b^2*c))/(32*(a^6*c^3 - a^6*b*c^2)*((a + c*x)^(1/2) - a^(1/2)))))/(8*a^3) - (16*a^3*b^4 + 16*a^3*b*c^3)/(6
4*(a^6*c^3 - a^6*b*c^2)) + ((8*a^3*b^4 + 8*a^3*c^4)*((a + b*x)^(1/2) - a^(1/2)))/(32*(a^6*c^3 - a^6*b*c^2)*((a
+ c*x)^(1/2) - a^(1/2))))*1i)/(8*a^3) - ((b + c)*((16*a^3*b^4 + 16*a^3*b*c^3)/(64*(a^6*c^3 - a^6*b*c^2)) + ((
b + c)*((64*a^6*b^3 - 64*a^6*b*c^2)/(64*(a^6*c^3 - a^6*b*c^2)) - (((a + b*x)^(1/2) - a^(1/2))*(64*a^6*b^3 - 64
*a^6*c^3 + 128*a^6*b*c^2 - 128*a^6*b^2*c))/(32*(a^6*c^3 - a^6*b*c^2)*((a + c*x)^(1/2) - a^(1/2)))))/(8*a^3) -
((8*a^3*b^4 + 8*a^3*c^4)*((a + b*x)^(1/2) - a^(1/2)))/(32*(a^6*c^3 - a^6*b*c^2)*((a + c*x)^(1/2) - a^(1/2))))*
1i)/(8*a^3))/(((b + c)*(((b + c)*((64*a^6*b^3 - 64*a^6*b*c^2)/(64*(a^6*c^3 - a^6*b*c^2)) - (((a + b*x)^(1/2) -
a^(1/2))*(64*a^6*b^3 - 64*a^6*c^3 + 128*a^6*b*c^2 - 128*a^6*b^2*c))/(32*(a^6*c^3 - a^6*b*c^2)*((a + c*x)^(1/2
) - a^(1/2)))))/(8*a^3) - (16*a^3*b^4 + 16*a^3*b*c^3)/(64*(a^6*c^3 - a^6*b*c^2)) + ((8*a^3*b^4 + 8*a^3*c^4)*((
a + b*x)^(1/2) - a^(1/2)))/(32*(a^6*c^3 - a^6*b*c^2)*((a + c*x)^(1/2) - a^(1/2)))))/(8*a^3) - (b*c^4 - b^5)/(3
2*(a^6*c^3 - a^6*b*c^2)) + ((b + c)*((16*a^3*b^4 + 16*a^3*b*c^3)/(64*(a^6*c^3 - a^6*b*c^2)) + ((b + c)*((64*a^
6*b^3 - 64*a^6*b*c^2)/(64*(a^6*c^3 - a^6*b*c^2)) - (((a + b*x)^(1/2) - a^(1/2))*(64*a^6*b^3 - 64*a^6*c^3 + 128
*a^6*b*c^2 - 128*a^6*b^2*c))/(32*(a^6*c^3 - a^6*b*c^2)*((a + c*x)^(1/2) - a^(1/2)))))/(8*a^3) - ((8*a^3*b^4 +
8*a^3*c^4)*((a + b*x)^(1/2) - a^(1/2)))/(32*(a^6*c^3 - a^6*b*c^2)*((a + c*x)^(1/2) - a^(1/2)))))/(8*a^3) + (((
a + b*x)^(1/2) - a^(1/2))*(b*c^4 - b^4*c + b^2*c^3 - b^3*c^2))/(16*(a^6*c^3 - a^6*b*c^2)*((a + c*x)^(1/2) - a^
(1/2)))))*(a^(3/2)*b + a^(3/2)*c)*1i)/(4*a^3) + (c^2*((a + b*x)^(1/2) - a^(1/2))^2)/(16*a^(3/2)*(b - c)*((a +
c*x)^(1/2) - a^(1/2))^2)

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

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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