\(\int \frac {1}{x^2 (a c+b c x^2+d \sqrt {a+b x^2})} \, dx\) [56]

Optimal result

Integrand size = 29, antiderivative size = 119 \[ \int \frac {1}{x^2 \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=-\frac {c}{\left (a c^2-d^2\right ) x}+\frac {d \sqrt {a+b x^2}}{a \left (a c^2-d^2\right ) x}-\frac {\sqrt {b} c^2 \arctan \left (\frac {\sqrt {b} \sqrt {a c^2-d^2} x}{a c+d \sqrt {a+b x^2}}\right )}{\left (a c^2-d^2\right )^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.88 \[ \int \frac {1}{x^2 \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=-\frac {a c-d \sqrt {a+b x^2}}{a^2 c^2 x-a d^2 x}+\frac {2 \sqrt {b} c^2 \arctan \left (\frac {d+c \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{\sqrt {a c^2-d^2}}\right )}{\left (a c^2-d^2\right )^{3/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.41, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {2587, 27, 264, 218, 382, 25, 27, 291, 218}

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.

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

\(\Big \downarrow \) 2587

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 264

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

\(\Big \downarrow \) 218

\(\displaystyle c \left (-\frac {\sqrt {b} c \arctan \left (\frac {\sqrt {b} c x}{\sqrt {a c^2-d^2}}\right )}{\left (a c^2-d^2\right )^{3/2}}-\frac {1}{x \left (a c^2-d^2\right )}\right )-d \int \frac {1}{x^2 \sqrt {b x^2+a} \left (b x^2 c^2+a c^2-d^2\right )}dx\)

\(\Big \downarrow \) 382

\(\displaystyle c \left (-\frac {\sqrt {b} c \arctan \left (\frac {\sqrt {b} c x}{\sqrt {a c^2-d^2}}\right )}{\left (a c^2-d^2\right )^{3/2}}-\frac {1}{x \left (a c^2-d^2\right )}\right )-d \left (\frac {\int -\frac {a b c^2}{\sqrt {b x^2+a} \left (b x^2 c^2+a c^2-d^2\right )}dx}{a \left (a c^2-d^2\right )}-\frac {\sqrt {a+b x^2}}{a x \left (a c^2-d^2\right )}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle c \left (-\frac {\sqrt {b} c \arctan \left (\frac {\sqrt {b} c x}{\sqrt {a c^2-d^2}}\right )}{\left (a c^2-d^2\right )^{3/2}}-\frac {1}{x \left (a c^2-d^2\right )}\right )-d \left (-\frac {\int \frac {a b c^2}{\sqrt {b x^2+a} \left (b x^2 c^2+a c^2-d^2\right )}dx}{a \left (a c^2-d^2\right )}-\frac {\sqrt {a+b x^2}}{a x \left (a c^2-d^2\right )}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle c \left (-\frac {\sqrt {b} c \arctan \left (\frac {\sqrt {b} c x}{\sqrt {a c^2-d^2}}\right )}{\left (a c^2-d^2\right )^{3/2}}-\frac {1}{x \left (a c^2-d^2\right )}\right )-d \left (-\frac {b c^2 \int \frac {1}{\sqrt {b x^2+a} \left (b x^2 c^2+a c^2-d^2\right )}dx}{a c^2-d^2}-\frac {\sqrt {a+b x^2}}{a x \left (a c^2-d^2\right )}\right )\)

\(\Big \downarrow \) 291

\(\displaystyle c \left (-\frac {\sqrt {b} c \arctan \left (\frac {\sqrt {b} c x}{\sqrt {a c^2-d^2}}\right )}{\left (a c^2-d^2\right )^{3/2}}-\frac {1}{x \left (a c^2-d^2\right )}\right )-d \left (-\frac {b c^2 \int \frac {1}{a c^2-d^2-\frac {\left (b \left (a c^2-d^2\right )-a b c^2\right ) x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{a c^2-d^2}-\frac {\sqrt {a+b x^2}}{a x \left (a c^2-d^2\right )}\right )\)

\(\Big \downarrow \) 218

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

Input:

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

Output:

c*(-(1/((a*c^2 - d^2)*x)) - (Sqrt[b]*c*ArcTan[(Sqrt[b]*c*x)/Sqrt[a*c^2 - d 
^2]])/(a*c^2 - d^2)^(3/2)) - d*(-(Sqrt[a + b*x^2]/(a*(a*c^2 - d^2)*x)) - ( 
Sqrt[b]*c^2*ArcTan[(Sqrt[b]*d*x)/(Sqrt[a*c^2 - d^2]*Sqrt[a + b*x^2])])/(d* 
(a*c^2 - d^2)^(3/2)))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 218

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

rule 264

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( 
m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c 
^2*(m + 1)))   Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p 
}, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
 

rule 291

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst 
[Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, 
d}, x] && NeQ[b*c - a*d, 0]
 

rule 382

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_) 
, x_Symbol] :> Simp[(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/ 
(a*c*e*(m + 1))), x] - Simp[1/(a*c*e^2*(m + 1))   Int[(e*x)^(m + 2)*(a + b* 
x^2)^p*(c + d*x^2)^q*Simp[(b*c + a*d)*(m + 3) + 2*(b*c*p + a*d*q) + b*d*(m 
+ 2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[ 
b*c - a*d, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, d, e, m, 2, p, q, x]
 

rule 2587

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1810\) vs. \(2(107)=214\).

Time = 0.04 (sec) , antiderivative size = 1811, normalized size of antiderivative = 15.22

method	result	size
default	\(\text {Expression too large to display}\)	\(1811\)

Input:

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

Output:

b*c^2/d^2/(b*(a*c^2-d^2))^(1/2)*arctan(x*b*c/(b*(a*c^2-d^2))^(1/2))-a*c^4/ 
(a*c^2-d^2)*b/d^2/(b*(a*c^2-d^2))^(1/2)*arctan(x*b*c/(b*(a*c^2-d^2))^(1/2) 
)-c/(a*c^2-d^2)/x-d*(1/(a*c^2-d^2)/a*(-1/a/x*(b*x^2+a)^(3/2)+2*b/a*(1/2*x* 
(b*x^2+a)^(1/2)+1/2*a/b^(1/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))))-1/2*b^2*c^2/ 
a/(-a*b)^(1/2)/((-a*b)^(1/2)*c^2+(-(a*c^2-d^2)*b*c^2)^(1/2))/((-a*b)^(1/2) 
*c^2-(-(a*c^2-d^2)*b*c^2)^(1/2))*(((x-(-a*b)^(1/2)/b)^2*b+2*(-a*b)^(1/2)*( 
x-(-a*b)^(1/2)/b))^(1/2)+(-a*b)^(1/2)*ln(((x-(-a*b)^(1/2)/b)*b+(-a*b)^(1/2 
))/b^(1/2)+((x-(-a*b)^(1/2)/b)^2*b+2*(-a*b)^(1/2)*(x-(-a*b)^(1/2)/b))^(1/2 
))/b^(1/2))+1/2*b^2*c^2/a/(-a*b)^(1/2)/((-a*b)^(1/2)*c^2+(-(a*c^2-d^2)*b*c 
^2)^(1/2))/((-a*b)^(1/2)*c^2-(-(a*c^2-d^2)*b*c^2)^(1/2))*(((x+(-a*b)^(1/2) 
/b)^2*b-2*(-a*b)^(1/2)*(x+(-a*b)^(1/2)/b))^(1/2)-(-a*b)^(1/2)*ln(((x+(-a*b 
)^(1/2)/b)*b-(-a*b)^(1/2))/b^(1/2)+((x+(-a*b)^(1/2)/b)^2*b-2*(-a*b)^(1/2)* 
(x+(-a*b)^(1/2)/b))^(1/2))/b^(1/2))+1/2*b^2*c^6/(a*c^2-d^2)/((-a*b)^(1/2)* 
c^2+(-(a*c^2-d^2)*b*c^2)^(1/2))/((-a*b)^(1/2)*c^2-(-(a*c^2-d^2)*b*c^2)^(1/ 
2))/(-(a*c^2-d^2)*b*c^2)^(1/2)*(((x-(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)^2*b+ 
2/c^2*(-(a*c^2-d^2)*b*c^2)^(1/2)*(x-(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)+d^2/ 
c^2)^(1/2)+1/c^2*(-(a*c^2-d^2)*b*c^2)^(1/2)*ln((1/c^2*(-(a*c^2-d^2)*b*c^2) 
^(1/2)+(x-(-(a*c^2-d^2)*b*c^2)^(1/2)/b/c^2)*b)/b^(1/2)+((x-(-(a*c^2-d^2)*b 
*c^2)^(1/2)/b/c^2)^2*b+2/c^2*(-(a*c^2-d^2)*b*c^2)^(1/2)*(x-(-(a*c^2-d^2)*b 
*c^2)^(1/2)/b/c^2)+d^2/c^2)^(1/2))/b^(1/2)-d^2/c^2/(d^2/c^2)^(1/2)*ln((...
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 581, normalized size of antiderivative = 4.88 \[ \int \frac {1}{x^2 \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\left [-\frac {a c^{2} x \sqrt {-\frac {b}{a c^{2} - d^{2}}} \log \left (\frac {a^{4} c^{4} - 2 \, a^{3} c^{2} d^{2} + a^{2} d^{4} + {\left (a^{2} b^{2} c^{4} - 8 \, a b^{2} c^{2} d^{2} + 8 \, b^{2} d^{4}\right )} x^{4} + 2 \, {\left (a^{3} b c^{4} - 5 \, a^{2} b c^{2} d^{2} + 4 \, a b d^{4}\right )} x^{2} + 4 \, {\left ({\left (a^{2} b c^{4} d - 3 \, a b c^{2} d^{3} + 2 \, b d^{5}\right )} x^{3} + {\left (a^{3} c^{4} d - 2 \, a^{2} c^{2} d^{3} + a d^{5}\right )} x\right )} \sqrt {b x^{2} + a} \sqrt {-\frac {b}{a c^{2} - d^{2}}}}{b^{2} c^{4} x^{4} + a^{2} c^{4} - 2 \, a c^{2} d^{2} + d^{4} + 2 \, {\left (a b c^{4} - b c^{2} d^{2}\right )} x^{2}}\right ) + 2 \, a c^{2} x \sqrt {-\frac {b}{a c^{2} - d^{2}}} \log \left (\frac {b c^{2} x^{2} - a c^{2} + 2 \, {\left (a c^{3} - c d^{2}\right )} x \sqrt {-\frac {b}{a c^{2} - d^{2}}} + d^{2}}{b c^{2} x^{2} + a c^{2} - d^{2}}\right ) + 4 \, a c - 4 \, \sqrt {b x^{2} + a} d}{4 \, {\left (a^{2} c^{2} - a d^{2}\right )} x}, -\frac {2 \, a c^{2} x \sqrt {\frac {b}{a c^{2} - d^{2}}} \arctan \left (c x \sqrt {\frac {b}{a c^{2} - d^{2}}}\right ) - a c^{2} x \sqrt {\frac {b}{a c^{2} - d^{2}}} \arctan \left (-\frac {{\left (a^{2} c^{2} - a d^{2} + {\left (a b c^{2} - 2 \, b d^{2}\right )} x^{2}\right )} \sqrt {b x^{2} + a} \sqrt {\frac {b}{a c^{2} - d^{2}}}}{2 \, {\left (b^{2} d x^{3} + a b d x\right )}}\right ) + 2 \, a c - 2 \, \sqrt {b x^{2} + a} d}{2 \, {\left (a^{2} c^{2} - a d^{2}\right )} x}\right ] \] Input:

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

Output:

[-1/4*(a*c^2*x*sqrt(-b/(a*c^2 - d^2))*log((a^4*c^4 - 2*a^3*c^2*d^2 + a^2*d 
^4 + (a^2*b^2*c^4 - 8*a*b^2*c^2*d^2 + 8*b^2*d^4)*x^4 + 2*(a^3*b*c^4 - 5*a^ 
2*b*c^2*d^2 + 4*a*b*d^4)*x^2 + 4*((a^2*b*c^4*d - 3*a*b*c^2*d^3 + 2*b*d^5)* 
x^3 + (a^3*c^4*d - 2*a^2*c^2*d^3 + a*d^5)*x)*sqrt(b*x^2 + a)*sqrt(-b/(a*c^ 
2 - d^2)))/(b^2*c^4*x^4 + a^2*c^4 - 2*a*c^2*d^2 + d^4 + 2*(a*b*c^4 - b*c^2 
*d^2)*x^2)) + 2*a*c^2*x*sqrt(-b/(a*c^2 - d^2))*log((b*c^2*x^2 - a*c^2 + 2* 
(a*c^3 - c*d^2)*x*sqrt(-b/(a*c^2 - d^2)) + d^2)/(b*c^2*x^2 + a*c^2 - d^2)) 
 + 4*a*c - 4*sqrt(b*x^2 + a)*d)/((a^2*c^2 - a*d^2)*x), -1/2*(2*a*c^2*x*sqr 
t(b/(a*c^2 - d^2))*arctan(c*x*sqrt(b/(a*c^2 - d^2))) - a*c^2*x*sqrt(b/(a*c 
^2 - d^2))*arctan(-1/2*(a^2*c^2 - a*d^2 + (a*b*c^2 - 2*b*d^2)*x^2)*sqrt(b* 
x^2 + a)*sqrt(b/(a*c^2 - d^2))/(b^2*d*x^3 + a*b*d*x)) + 2*a*c - 2*sqrt(b*x 
^2 + a)*d)/((a^2*c^2 - a*d^2)*x)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.77 \[ \int \frac {1}{x^2 \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=-b^{\frac {3}{2}} d {\left (\frac {c^{2} \arctan \left (\frac {{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} c^{2} + a c^{2} - 2 \, d^{2}}{2 \, \sqrt {a c^{2} - d^{2}} d}\right )}{{\left (a b c^{2} - b d^{2}\right )} \sqrt {a c^{2} - d^{2}} d} + \frac {2}{{\left (a b c^{2} - b d^{2}\right )} {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}}\right )} - \frac {b c^{2} \arctan \left (\frac {b c x}{\sqrt {a b c^{2} - b d^{2}}}\right )}{\sqrt {a b c^{2} - b d^{2}} {\left (a c^{2} - d^{2}\right )}} - \frac {c}{{\left (a c^{2} - d^{2}\right )} x} \] Input:

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

Output:

-b^(3/2)*d*(c^2*arctan(1/2*((sqrt(b)*x - sqrt(b*x^2 + a))^2*c^2 + a*c^2 - 
2*d^2)/(sqrt(a*c^2 - d^2)*d))/((a*b*c^2 - b*d^2)*sqrt(a*c^2 - d^2)*d) + 2/ 
((a*b*c^2 - b*d^2)*((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a))) - b*c^2*arctan( 
b*c*x/sqrt(a*b*c^2 - b*d^2))/(sqrt(a*b*c^2 - b*d^2)*(a*c^2 - d^2)) - c/((a 
*c^2 - d^2)*x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.19 \[ \int \frac {1}{x^2 \left (a c+b c x^2+d \sqrt {a+b x^2}\right )} \, dx=\frac {-2 \sqrt {b}\, \sqrt {a \,c^{2}-d^{2}}\, \mathit {atan} \left (\frac {\sqrt {b \,x^{2}+a}\, c +\sqrt {b}\, c x +d}{\sqrt {a \,c^{2}-d^{2}}}\right ) a \,c^{2} x +\sqrt {b \,x^{2}+a}\, a \,c^{2} d -\sqrt {b \,x^{2}+a}\, d^{3}+\sqrt {b}\, a \,c^{2} d x -\sqrt {b}\, d^{3} x -a^{2} c^{3}+a c \,d^{2}}{a x \left (a^{2} c^{4}-2 a \,c^{2} d^{2}+d^{4}\right )} \] Input:

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

Output:

( - 2*sqrt(b)*sqrt(a*c**2 - d**2)*atan((sqrt(a + b*x**2)*c + sqrt(b)*c*x + 
 d)/sqrt(a*c**2 - d**2))*a*c**2*x + sqrt(a + b*x**2)*a*c**2*d - sqrt(a + b 
*x**2)*d**3 + sqrt(b)*a*c**2*d*x - sqrt(b)*d**3*x - a**2*c**3 + a*c*d**2)/ 
(a*x*(a**2*c**4 - 2*a*c**2*d**2 + d**4))