\(\int \frac {1}{x^2 (\sqrt {a+b x}+\sqrt {a+c x})} \, dx\) [31]

Optimal result

Integrand size = 25, antiderivative size = 171 \[ \int \frac {1}{x^2 \left (\sqrt {a+b x}+\sqrt {a+c x}\right )} \, dx=-\frac {\sqrt {a+b x}}{2 (b-c) x^2}-\frac {b \sqrt {a+b x}}{4 a (b-c) x}+\frac {\sqrt {a+c x}}{2 (b-c) x^2}+\frac {c \sqrt {a+c x}}{4 a (b-c) x}+\frac {b^2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{4 a^{3/2} (b-c)}-\frac {c^2 \text {arctanh}\left (\frac {\sqrt {a+c x}}{\sqrt {a}}\right )}{4 a^{3/2} (b-c)} \] Output:

-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+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)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 10.15 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.44 \[ \int \frac {1}{x^2 \left (\sqrt {a+b x}+\sqrt {a+c x}\right )} \, dx=\frac {-2 b^2 (a+b x)^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},3,\frac {5}{2},1+\frac {b x}{a}\right )+2 c^2 (a+c x)^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{2},3,\frac {5}{2},1+\frac {c x}{a}\right )}{3 a^3 (b-c)} \] Input:

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

Output:

(-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))
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.84, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2528, 51, 52, 73, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 2528

\(\displaystyle \frac {\int \frac {\sqrt {a+b x}}{x^3}dx}{b-c}-\frac {\int \frac {\sqrt {a+c x}}{x^3}dx}{b-c}\)

\(\Big \downarrow \) 51

\(\displaystyle \frac {\frac {1}{4} b \int \frac {1}{x^2 \sqrt {a+b x}}dx-\frac {\sqrt {a+b x}}{2 x^2}}{b-c}-\frac {\frac {1}{4} c \int \frac {1}{x^2 \sqrt {a+c x}}dx-\frac {\sqrt {a+c x}}{2 x^2}}{b-c}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {\frac {1}{4} b \left (-\frac {b \int \frac {1}{x \sqrt {a+b x}}dx}{2 a}-\frac {\sqrt {a+b x}}{a x}\right )-\frac {\sqrt {a+b x}}{2 x^2}}{b-c}-\frac {\frac {1}{4} c \left (-\frac {c \int \frac {1}{x \sqrt {a+c x}}dx}{2 a}-\frac {\sqrt {a+c x}}{a x}\right )-\frac {\sqrt {a+c x}}{2 x^2}}{b-c}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {1}{4} b \left (-\frac {\int \frac {1}{\frac {a+b x}{b}-\frac {a}{b}}d\sqrt {a+b x}}{a}-\frac {\sqrt {a+b x}}{a x}\right )-\frac {\sqrt {a+b x}}{2 x^2}}{b-c}-\frac {\frac {1}{4} c \left (-\frac {\int \frac {1}{\frac {a+c x}{c}-\frac {a}{c}}d\sqrt {a+c x}}{a}-\frac {\sqrt {a+c x}}{a x}\right )-\frac {\sqrt {a+c x}}{2 x^2}}{b-c}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {1}{4} b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\sqrt {a+b x}}{a x}\right )-\frac {\sqrt {a+b x}}{2 x^2}}{b-c}-\frac {\frac {1}{4} c \left (\frac {c \text {arctanh}\left (\frac {\sqrt {a+c x}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\sqrt {a+c x}}{a x}\right )-\frac {\sqrt {a+c x}}{2 x^2}}{b-c}\)

Input:

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

Output:

(-1/2*Sqrt[a + b*x]/x^2 + (b*(-(Sqrt[a + b*x]/(a*x)) + (b*ArcTanh[Sqrt[a + 
 b*x]/Sqrt[a]])/a^(3/2)))/4)/(b - c) - (-1/2*Sqrt[a + c*x]/x^2 + (c*(-(Sqr 
t[a + c*x]/(a*x)) + (c*ArcTanh[Sqrt[a + c*x]/Sqrt[a]])/a^(3/2)))/4)/(b - c 
)
 

Defintions of rubi rules used

rule 51

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] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
 

rule 52

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] - Simp[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] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[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] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 2528

Int[(u_)/((e_.)*Sqrt[(a_.) + (b_.)*(x_)] + (f_.)*Sqrt[(c_.) + (d_.)*(x_)]), 
 x_Symbol] :> Simp[c/(e*(b*c - a*d))   Int[(u*Sqrt[a + b*x])/x, x], x] - Si 
mp[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]
 

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.70

method	result	size
default	\(\frac {2 b^{2} \left (\frac {-\frac {\left (b x +a \right )^{\frac {3}{2}}}{8 a}-\frac {\sqrt {b x +a}}{8}}{x^{2} b^{2}}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{8 a^{\frac {3}{2}}}\right )}{b -c}-\frac {2 c^{2} \left (\frac {-\frac {\left (c x +a \right )^{\frac {3}{2}}}{8 a}-\frac {\sqrt {c x +a}}{8}}{x^{2} c^{2}}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {c x +a}}{\sqrt {a}}\right )}{8 a^{\frac {3}{2}}}\right )}{b -c}\)	\(120\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.39 \[ \int \frac {1}{x^2 \left (\sqrt {a+b x}+\sqrt {a+c x}\right )} \, dx=\left [-\frac {\sqrt {a} b^{2} x^{2} \log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right ) + \sqrt {a} c^{2} x^{2} \log \left (\frac {c x + 2 \, \sqrt {c x + a} \sqrt {a} + 2 \, a}{x}\right ) + 2 \, {\left (a b x + 2 \, a^{2}\right )} \sqrt {b x + a} - 2 \, {\left (a c x + 2 \, a^{2}\right )} \sqrt {c x + a}}{8 \, {\left (a^{2} b - a^{2} c\right )} x^{2}}, -\frac {\sqrt {-a} b^{2} x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x + a}}\right ) - \sqrt {-a} c^{2} x^{2} \arctan \left (\frac {\sqrt {-a}}{\sqrt {c x + a}}\right ) + {\left (a b x + 2 \, a^{2}\right )} \sqrt {b x + a} - {\left (a c x + 2 \, a^{2}\right )} \sqrt {c x + a}}{4 \, {\left (a^{2} b - a^{2} c\right )} x^{2}}\right ] \] Input:

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

Output:

[-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(-a)/sqrt(b*x + a)) - sqrt(-a)*c^2*x^2* 
arctan(sqrt(-a)/sqrt(c*x + 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)]
 

Sympy [F]

\[ \int \frac {1}{x^2 \left (\sqrt {a+b x}+\sqrt {a+c x}\right )} \, dx=\int \frac {1}{x^{2} \left (\sqrt {a + b x} + \sqrt {a + c x}\right )}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {1}{x^2 \left (\sqrt {a+b x}+\sqrt {a+c x}\right )} \, dx=\int { \frac {1}{x^{2} {\left (\sqrt {b x + a} + \sqrt {c x + a}\right )}} \,d x } \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1895 vs. \(2 (139) = 278\).

Time = 5.86 (sec) , antiderivative size = 1895, normalized size of antiderivative = 11.08 \[ \int \frac {1}{x^2 \left (\sqrt {a+b x}+\sqrt {a+c x}\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/4*b^2*arctan(sqrt(b*x + a)/sqrt(-a))/((a*b - a*c)*sqrt(-a)) - 1/2*((sqr 
t(b*c)*sqrt(b*x + a) - sqrt(a*b^2 + (b*x + a)*b*c - a*b*c))*a^3*b^6*c^2*ab 
s(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) - s 
qrt(a*b^2 + (b*x + a)*b*c - a*b*c))^3*a^2*b^4*c^2*abs(b) - 10*(sqrt(b*c)*s 
qrt(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*...
 

Mupad [B] (verification not implemented)

Time = 33.57 (sec) , antiderivative size = 1610, normalized size of antiderivative = 9.42 \[ \int \frac {1}{x^2 \left (\sqrt {a+b x}+\sqrt {a+c x}\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

((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)/(6 
4*(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))))*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*...
 

Reduce [B] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 868, normalized size of antiderivative = 5.08 \[ \int \frac {1}{x^2 \left (\sqrt {a+b x}+\sqrt {a+c x}\right )} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 2*sqrt(b)*sqrt(a)*sqrt(2*sqrt(c)*sqrt(b) - b - c)*atan((sqrt(c)*sqrt(a 
 + b*x) + sqrt(b)*sqrt(a + c*x))/(sqrt(a)*sqrt(2*sqrt(c)*sqrt(b) - b - c)) 
)*b**2*x**2 - 2*sqrt(b)*sqrt(a)*sqrt(2*sqrt(c)*sqrt(b) - b - c)*atan((sqrt 
(c)*sqrt(a + b*x) + sqrt(b)*sqrt(a + c*x))/(sqrt(a)*sqrt(2*sqrt(c)*sqrt(b) 
 - b - c)))*c**2*x**2 - 2*sqrt(c)*sqrt(a)*sqrt(2*sqrt(c)*sqrt(b) - b - c)* 
atan((sqrt(c)*sqrt(a + b*x) + sqrt(b)*sqrt(a + c*x))/(sqrt(a)*sqrt(2*sqrt( 
c)*sqrt(b) - b - c)))*b**2*x**2 - 2*sqrt(c)*sqrt(a)*sqrt(2*sqrt(c)*sqrt(b) 
 - b - c)*atan((sqrt(c)*sqrt(a + b*x) + sqrt(b)*sqrt(a + c*x))/(sqrt(a)*sq 
rt(2*sqrt(c)*sqrt(b) - b - c)))*c**2*x**2 - sqrt(b)*sqrt(a)*sqrt(2*sqrt(c) 
*sqrt(b) + b + c)*log(( - sqrt(a)*sqrt(2*sqrt(c)*sqrt(b) + b + c) + sqrt(c 
)*sqrt(a + b*x) + sqrt(b)*sqrt(a + c*x))/sqrt(a))*b**2*x**2 + sqrt(b)*sqrt 
(a)*sqrt(2*sqrt(c)*sqrt(b) + b + c)*log(( - sqrt(a)*sqrt(2*sqrt(c)*sqrt(b) 
 + b + c) + sqrt(c)*sqrt(a + b*x) + sqrt(b)*sqrt(a + c*x))/sqrt(a))*c**2*x 
**2 + sqrt(b)*sqrt(a)*sqrt(2*sqrt(c)*sqrt(b) + b + c)*log((sqrt(a)*sqrt(2* 
sqrt(c)*sqrt(b) + b + c) + sqrt(c)*sqrt(a + b*x) + sqrt(b)*sqrt(a + c*x))/ 
sqrt(a))*b**2*x**2 - sqrt(b)*sqrt(a)*sqrt(2*sqrt(c)*sqrt(b) + b + c)*log(( 
sqrt(a)*sqrt(2*sqrt(c)*sqrt(b) + b + c) + sqrt(c)*sqrt(a + b*x) + sqrt(b)* 
sqrt(a + c*x))/sqrt(a))*c**2*x**2 + sqrt(c)*sqrt(a)*sqrt(2*sqrt(c)*sqrt(b) 
 + b + c)*log(( - sqrt(a)*sqrt(2*sqrt(c)*sqrt(b) + b + c) + sqrt(c)*sqrt(a 
 + b*x) + sqrt(b)*sqrt(a + c*x))/sqrt(a))*b**2*x**2 - sqrt(c)*sqrt(a)*s...