\(\int \frac {a+b x+c x^2}{x^5 \sqrt {-1+d x} \sqrt {1+d x}} \, dx\) [16]

Optimal result

Integrand size = 32, antiderivative size = 154 \[ \int \frac {a+b x+c x^2}{x^5 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {a \sqrt {-1+d x} \sqrt {1+d x}}{4 x^4}+\frac {b \sqrt {-1+d x} \sqrt {1+d x}}{3 x^3}+\frac {\left (4 c+3 a d^2\right ) \sqrt {-1+d x} \sqrt {1+d x}}{8 x^2}+\frac {2 b d^2 \sqrt {-1+d x} \sqrt {1+d x}}{3 x}+\frac {1}{8} d^2 \left (4 c+3 a d^2\right ) \arctan \left (\sqrt {-1+d x} \sqrt {1+d x}\right ) \] Output:

1/4*a*(d*x-1)^(1/2)*(d*x+1)^(1/2)/x^4+1/3*b*(d*x-1)^(1/2)*(d*x+1)^(1/2)/x^ 
3+1/8*(3*a*d^2+4*c)*(d*x-1)^(1/2)*(d*x+1)^(1/2)/x^2+2/3*b*d^2*(d*x-1)^(1/2 
)*(d*x+1)^(1/2)/x+1/8*d^2*(3*a*d^2+4*c)*arctan((d*x-1)^(1/2)*(d*x+1)^(1/2) 
)
 

Mathematica [A] (warning: unable to verify)

Time = 0.23 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.60 \[ \int \frac {a+b x+c x^2}{x^5 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {1}{24} \left (\frac {\sqrt {-1+d x} \sqrt {1+d x} \left (a \left (6+9 d^2 x^2\right )+4 x \left (3 c x+b \left (2+4 d^2 x^2\right )\right )\right )}{x^4}+6 d^2 \left (4 c+3 a d^2\right ) \arctan \left (\sqrt {\frac {-1+d x}{1+d x}}\right )\right ) \] Input:

Integrate[(a + b*x + c*x^2)/(x^5*Sqrt[-1 + d*x]*Sqrt[1 + d*x]),x]
 

Output:

((Sqrt[-1 + d*x]*Sqrt[1 + d*x]*(a*(6 + 9*d^2*x^2) + 4*x*(3*c*x + b*(2 + 4* 
d^2*x^2))))/x^4 + 6*d^2*(4*c + 3*a*d^2)*ArcTan[Sqrt[(-1 + d*x)/(1 + d*x)]] 
)/24
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.09, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.281, Rules used = {2113, 2338, 539, 539, 27, 534, 243, 73, 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.

\(\displaystyle \int \frac {a+b x+c x^2}{x^5 \sqrt {d x-1} \sqrt {d x+1}} \, dx\)

\(\Big \downarrow \) 2113

\(\displaystyle \frac {\sqrt {d^2 x^2-1} \int \frac {c x^2+b x+a}{x^5 \sqrt {d^2 x^2-1}}dx}{\sqrt {d x-1} \sqrt {d x+1}}\)

\(\Big \downarrow \) 2338

\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (\frac {1}{4} \int \frac {4 b+\left (3 a d^2+4 c\right ) x}{x^4 \sqrt {d^2 x^2-1}}dx+\frac {a \sqrt {d^2 x^2-1}}{4 x^4}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\)

\(\Big \downarrow \) 539

\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (\frac {1}{4} \left (\frac {1}{3} \int \frac {8 b x d^2+3 \left (3 a d^2+4 c\right )}{x^3 \sqrt {d^2 x^2-1}}dx+\frac {4 b \sqrt {d^2 x^2-1}}{3 x^3}\right )+\frac {a \sqrt {d^2 x^2-1}}{4 x^4}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\)

\(\Big \downarrow \) 539

\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int \frac {d^2 \left (16 b+3 \left (3 a d^2+4 c\right ) x\right )}{x^2 \sqrt {d^2 x^2-1}}dx+\frac {3 \sqrt {d^2 x^2-1} \left (3 a d^2+4 c\right )}{2 x^2}\right )+\frac {4 b \sqrt {d^2 x^2-1}}{3 x^3}\right )+\frac {a \sqrt {d^2 x^2-1}}{4 x^4}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} d^2 \int \frac {16 b+3 \left (3 a d^2+4 c\right ) x}{x^2 \sqrt {d^2 x^2-1}}dx+\frac {3 \sqrt {d^2 x^2-1} \left (3 a d^2+4 c\right )}{2 x^2}\right )+\frac {4 b \sqrt {d^2 x^2-1}}{3 x^3}\right )+\frac {a \sqrt {d^2 x^2-1}}{4 x^4}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\)

\(\Big \downarrow \) 534

\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} d^2 \left (3 \left (3 a d^2+4 c\right ) \int \frac {1}{x \sqrt {d^2 x^2-1}}dx+\frac {16 b \sqrt {d^2 x^2-1}}{x}\right )+\frac {3 \sqrt {d^2 x^2-1} \left (3 a d^2+4 c\right )}{2 x^2}\right )+\frac {4 b \sqrt {d^2 x^2-1}}{3 x^3}\right )+\frac {a \sqrt {d^2 x^2-1}}{4 x^4}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} d^2 \left (\frac {3}{2} \left (3 a d^2+4 c\right ) \int \frac {1}{x^2 \sqrt {d^2 x^2-1}}dx^2+\frac {16 b \sqrt {d^2 x^2-1}}{x}\right )+\frac {3 \sqrt {d^2 x^2-1} \left (3 a d^2+4 c\right )}{2 x^2}\right )+\frac {4 b \sqrt {d^2 x^2-1}}{3 x^3}\right )+\frac {a \sqrt {d^2 x^2-1}}{4 x^4}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} d^2 \left (\frac {3 \left (3 a d^2+4 c\right ) \int \frac {1}{\frac {x^4}{d^2}+\frac {1}{d^2}}d\sqrt {d^2 x^2-1}}{d^2}+\frac {16 b \sqrt {d^2 x^2-1}}{x}\right )+\frac {3 \sqrt {d^2 x^2-1} \left (3 a d^2+4 c\right )}{2 x^2}\right )+\frac {4 b \sqrt {d^2 x^2-1}}{3 x^3}\right )+\frac {a \sqrt {d^2 x^2-1}}{4 x^4}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\sqrt {d^2 x^2-1} \left (\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} d^2 \left (3 \left (3 a d^2+4 c\right ) \arctan \left (\sqrt {d^2 x^2-1}\right )+\frac {16 b \sqrt {d^2 x^2-1}}{x}\right )+\frac {3 \sqrt {d^2 x^2-1} \left (3 a d^2+4 c\right )}{2 x^2}\right )+\frac {4 b \sqrt {d^2 x^2-1}}{3 x^3}\right )+\frac {a \sqrt {d^2 x^2-1}}{4 x^4}\right )}{\sqrt {d x-1} \sqrt {d x+1}}\)

Input:

Int[(a + b*x + c*x^2)/(x^5*Sqrt[-1 + d*x]*Sqrt[1 + d*x]),x]
 

Output:

(Sqrt[-1 + d^2*x^2]*((a*Sqrt[-1 + d^2*x^2])/(4*x^4) + ((4*b*Sqrt[-1 + d^2* 
x^2])/(3*x^3) + ((3*(4*c + 3*a*d^2)*Sqrt[-1 + d^2*x^2])/(2*x^2) + (d^2*((1 
6*b*Sqrt[-1 + d^2*x^2])/x + 3*(4*c + 3*a*d^2)*ArcTan[Sqrt[-1 + d^2*x^2]])) 
/2)/3)/4))/(Sqrt[-1 + d*x]*Sqrt[1 + d*x])
 

Defintions of rubi rules used

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 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 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 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 534

Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[(-c)*x^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[d   Int[ 
x^(m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[m, 
0] && GtQ[p, -1] && EqQ[m + 2*p + 3, 0]
 

rule 539

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

rule 2113

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_. 
)*(x_))^(p_.), x_Symbol] :> Simp[(a + b*x)^FracPart[m]*((c + d*x)^FracPart[ 
m]/(a*c + b*d*x^2)^FracPart[m])   Int[Px*(a*c + b*d*x^2)^m*(e + f*x)^p, x], 
 x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && PolyQ[Px, x] && EqQ[b*c + a 
*d, 0] && EqQ[m, n] &&  !IntegerQ[m]
 

rule 2338

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{ 
Q = PolynomialQuotient[Pq, c*x, x], R = PolynomialRemainder[Pq, c*x, x]}, S 
imp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Simp[1/(a*c*( 
m + 1))   Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*( 
m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && Lt 
Q[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])
 

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.69

method	result	size
risch	\(\frac {\sqrt {x d +1}\, \sqrt {x d -1}\, \left (16 b \,d^{2} x^{3}+9 a \,d^{2} x^{2}+12 c \,x^{2}+8 b x +6 a \right )}{24 x^{4}}-\frac {d^{2} \left (3 a \,d^{2}+4 c \right ) \arctan \left (\frac {1}{\sqrt {d^{2} x^{2}-1}}\right ) \sqrt {\left (x d +1\right ) \left (x d -1\right )}}{8 \sqrt {x d -1}\, \sqrt {x d +1}}\)	\(107\)
default	\(-\frac {\sqrt {x d -1}\, \sqrt {x d +1}\, \operatorname {csgn}\left (d \right )^{2} \left (9 \arctan \left (\frac {1}{\sqrt {d^{2} x^{2}-1}}\right ) a \,d^{4} x^{4}+12 \arctan \left (\frac {1}{\sqrt {d^{2} x^{2}-1}}\right ) c \,d^{2} x^{4}-16 \sqrt {d^{2} x^{2}-1}\, b \,d^{2} x^{3}-9 \sqrt {d^{2} x^{2}-1}\, a \,d^{2} x^{2}-12 \sqrt {d^{2} x^{2}-1}\, c \,x^{2}-8 \sqrt {d^{2} x^{2}-1}\, b x -6 \sqrt {d^{2} x^{2}-1}\, a \right )}{24 \sqrt {d^{2} x^{2}-1}\, x^{4}}\)	\(164\)

Input:

int((c*x^2+b*x+a)/x^5/(d*x-1)^(1/2)/(d*x+1)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/24*(d*x+1)^(1/2)*(d*x-1)^(1/2)*(16*b*d^2*x^3+9*a*d^2*x^2+12*c*x^2+8*b*x+ 
6*a)/x^4-1/8*d^2*(3*a*d^2+4*c)*arctan(1/(d^2*x^2-1)^(1/2))*((d*x+1)*(d*x-1 
))^(1/2)/(d*x-1)^(1/2)/(d*x+1)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.66 \[ \int \frac {a+b x+c x^2}{x^5 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {16 \, b d^{3} x^{4} + 6 \, {\left (3 \, a d^{4} + 4 \, c d^{2}\right )} x^{4} \arctan \left (-d x + \sqrt {d x + 1} \sqrt {d x - 1}\right ) + {\left (16 \, b d^{2} x^{3} + 3 \, {\left (3 \, a d^{2} + 4 \, c\right )} x^{2} + 8 \, b x + 6 \, a\right )} \sqrt {d x + 1} \sqrt {d x - 1}}{24 \, x^{4}} \] Input:

integrate((c*x^2+b*x+a)/x^5/(d*x-1)^(1/2)/(d*x+1)^(1/2),x, algorithm="fric 
as")
 

Output:

1/24*(16*b*d^3*x^4 + 6*(3*a*d^4 + 4*c*d^2)*x^4*arctan(-d*x + sqrt(d*x + 1) 
*sqrt(d*x - 1)) + (16*b*d^2*x^3 + 3*(3*a*d^2 + 4*c)*x^2 + 8*b*x + 6*a)*sqr 
t(d*x + 1)*sqrt(d*x - 1))/x^4
 

Sympy [F(-1)]

Timed out. \[ \int \frac {a+b x+c x^2}{x^5 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\text {Timed out} \] Input:

integrate((c*x**2+b*x+a)/x**5/(d*x-1)**(1/2)/(d*x+1)**(1/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.79 \[ \int \frac {a+b x+c x^2}{x^5 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=-\frac {3}{8} \, a d^{4} \arcsin \left (\frac {1}{d {\left | x \right |}}\right ) - \frac {1}{2} \, c d^{2} \arcsin \left (\frac {1}{d {\left | x \right |}}\right ) + \frac {2 \, \sqrt {d^{2} x^{2} - 1} b d^{2}}{3 \, x} + \frac {3 \, \sqrt {d^{2} x^{2} - 1} a d^{2}}{8 \, x^{2}} + \frac {\sqrt {d^{2} x^{2} - 1} c}{2 \, x^{2}} + \frac {\sqrt {d^{2} x^{2} - 1} b}{3 \, x^{3}} + \frac {\sqrt {d^{2} x^{2} - 1} a}{4 \, x^{4}} \] Input:

integrate((c*x^2+b*x+a)/x^5/(d*x-1)^(1/2)/(d*x+1)^(1/2),x, algorithm="maxi 
ma")
 

Output:

-3/8*a*d^4*arcsin(1/(d*abs(x))) - 1/2*c*d^2*arcsin(1/(d*abs(x))) + 2/3*sqr 
t(d^2*x^2 - 1)*b*d^2/x + 3/8*sqrt(d^2*x^2 - 1)*a*d^2/x^2 + 1/2*sqrt(d^2*x^ 
2 - 1)*c/x^2 + 1/3*sqrt(d^2*x^2 - 1)*b/x^3 + 1/4*sqrt(d^2*x^2 - 1)*a/x^4
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 325 vs. \(2 (124) = 248\).

Time = 0.15 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.11 \[ \int \frac {a+b x+c x^2}{x^5 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=-\frac {3 \, {\left (3 \, a d^{5} + 4 \, c d^{3}\right )} \arctan \left (\frac {1}{2} \, {\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{2}\right ) + \frac {2 \, {\left (9 \, a d^{5} {\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{14} + 12 \, c d^{3} {\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{14} + 132 \, a d^{5} {\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{10} + 48 \, c d^{3} {\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{10} - 384 \, b d^{4} {\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{8} - 528 \, a d^{5} {\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{6} - 192 \, c d^{3} {\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{6} - 2048 \, b d^{4} {\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{4} - 576 \, a d^{5} {\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{2} - 768 \, c d^{3} {\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{2} - 2048 \, b d^{4}\right )}}{{\left ({\left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}^{4} + 4\right )}^{4}}}{12 \, d} \] Input:

integrate((c*x^2+b*x+a)/x^5/(d*x-1)^(1/2)/(d*x+1)^(1/2),x, algorithm="giac 
")
 

Output:

-1/12*(3*(3*a*d^5 + 4*c*d^3)*arctan(1/2*(sqrt(d*x + 1) - sqrt(d*x - 1))^2) 
 + 2*(9*a*d^5*(sqrt(d*x + 1) - sqrt(d*x - 1))^14 + 12*c*d^3*(sqrt(d*x + 1) 
 - sqrt(d*x - 1))^14 + 132*a*d^5*(sqrt(d*x + 1) - sqrt(d*x - 1))^10 + 48*c 
*d^3*(sqrt(d*x + 1) - sqrt(d*x - 1))^10 - 384*b*d^4*(sqrt(d*x + 1) - sqrt( 
d*x - 1))^8 - 528*a*d^5*(sqrt(d*x + 1) - sqrt(d*x - 1))^6 - 192*c*d^3*(sqr 
t(d*x + 1) - sqrt(d*x - 1))^6 - 2048*b*d^4*(sqrt(d*x + 1) - sqrt(d*x - 1)) 
^4 - 576*a*d^5*(sqrt(d*x + 1) - sqrt(d*x - 1))^2 - 768*c*d^3*(sqrt(d*x + 1 
) - sqrt(d*x - 1))^2 - 2048*b*d^4)/((sqrt(d*x + 1) - sqrt(d*x - 1))^4 + 4) 
^4)/d
 

Mupad [B] (verification not implemented)

Time = 19.24 (sec) , antiderivative size = 695, normalized size of antiderivative = 4.51 \[ \int \frac {a+b x+c x^2}{x^5 \sqrt {-1+d x} \sqrt {1+d x}} \, dx =\text {Too large to display} \] Input:

int((a + b*x + c*x^2)/(x^5*(d*x - 1)^(1/2)*(d*x + 1)^(1/2)),x)
 

Output:

((c*d^2*1i)/32 + (c*d^2*((d*x - 1)^(1/2) - 1i)^2*1i)/(16*((d*x + 1)^(1/2) 
- 1)^2) - (c*d^2*((d*x - 1)^(1/2) - 1i)^4*15i)/(32*((d*x + 1)^(1/2) - 1)^4 
))/(((d*x - 1)^(1/2) - 1i)^2/((d*x + 1)^(1/2) - 1)^2 + (2*((d*x - 1)^(1/2) 
 - 1i)^4)/((d*x + 1)^(1/2) - 1)^4 + ((d*x - 1)^(1/2) - 1i)^6/((d*x + 1)^(1 
/2) - 1)^6) - ((a*d^4*1i)/1024 - (a*d^4*((d*x - 1)^(1/2) - 1i)^2*3i)/(128* 
((d*x + 1)^(1/2) - 1)^2) - (a*d^4*((d*x - 1)^(1/2) - 1i)^4*53i)/(512*((d*x 
 + 1)^(1/2) - 1)^4) + (a*d^4*((d*x - 1)^(1/2) - 1i)^6*87i)/(256*((d*x + 1) 
^(1/2) - 1)^6) + (a*d^4*((d*x - 1)^(1/2) - 1i)^8*657i)/(1024*((d*x + 1)^(1 
/2) - 1)^8) + (a*d^4*((d*x - 1)^(1/2) - 1i)^10*121i)/(256*((d*x + 1)^(1/2) 
 - 1)^10))/(((d*x - 1)^(1/2) - 1i)^4/((d*x + 1)^(1/2) - 1)^4 + (4*((d*x - 
1)^(1/2) - 1i)^6)/((d*x + 1)^(1/2) - 1)^6 + (6*((d*x - 1)^(1/2) - 1i)^8)/( 
(d*x + 1)^(1/2) - 1)^8 + (4*((d*x - 1)^(1/2) - 1i)^10)/((d*x + 1)^(1/2) - 
1)^10 + ((d*x - 1)^(1/2) - 1i)^12/((d*x + 1)^(1/2) - 1)^12) - (a*d^4*log(( 
(d*x - 1)^(1/2) - 1i)^2/((d*x + 1)^(1/2) - 1)^2 + 1)*3i)/8 - (c*d^2*log((( 
d*x - 1)^(1/2) - 1i)^2/((d*x + 1)^(1/2) - 1)^2 + 1)*1i)/2 + (a*d^4*log(((d 
*x - 1)^(1/2) - 1i)/((d*x + 1)^(1/2) - 1))*3i)/8 + (c*d^2*log(((d*x - 1)^( 
1/2) - 1i)/((d*x + 1)^(1/2) - 1))*1i)/2 + ((d*x - 1)^(1/2)*(b/3 + (2*b*d^2 
*x^2)/3 + (2*b*d^3*x^3)/3 + (b*d*x)/3))/(x^3*(d*x + 1)^(1/2)) + (a*d^4*((d 
*x - 1)^(1/2) - 1i)^2*7i)/(256*((d*x + 1)^(1/2) - 1)^2) - (a*d^4*((d*x - 1 
)^(1/2) - 1i)^4*1i)/(1024*((d*x + 1)^(1/2) - 1)^4) + (c*d^2*((d*x - 1)^...
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.31 \[ \int \frac {a+b x+c x^2}{x^5 \sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {18 \mathit {atan} \left (\sqrt {d x -1}+\sqrt {d x +1}-1\right ) a \,d^{4} x^{4}+24 \mathit {atan} \left (\sqrt {d x -1}+\sqrt {d x +1}-1\right ) c \,d^{2} x^{4}-18 \mathit {atan} \left (\sqrt {d x -1}+\sqrt {d x +1}+1\right ) a \,d^{4} x^{4}-24 \mathit {atan} \left (\sqrt {d x -1}+\sqrt {d x +1}+1\right ) c \,d^{2} x^{4}+9 \sqrt {d x +1}\, \sqrt {d x -1}\, a \,d^{2} x^{2}+6 \sqrt {d x +1}\, \sqrt {d x -1}\, a +16 \sqrt {d x +1}\, \sqrt {d x -1}\, b \,d^{2} x^{3}+8 \sqrt {d x +1}\, \sqrt {d x -1}\, b x +12 \sqrt {d x +1}\, \sqrt {d x -1}\, c \,x^{2}-16 b \,d^{3} x^{4}}{24 x^{4}} \] Input:

int((c*x^2+b*x+a)/x^5/(d*x-1)^(1/2)/(d*x+1)^(1/2),x)
 

Output:

(18*atan(sqrt(d*x - 1) + sqrt(d*x + 1) - 1)*a*d**4*x**4 + 24*atan(sqrt(d*x 
 - 1) + sqrt(d*x + 1) - 1)*c*d**2*x**4 - 18*atan(sqrt(d*x - 1) + sqrt(d*x 
+ 1) + 1)*a*d**4*x**4 - 24*atan(sqrt(d*x - 1) + sqrt(d*x + 1) + 1)*c*d**2* 
x**4 + 9*sqrt(d*x + 1)*sqrt(d*x - 1)*a*d**2*x**2 + 6*sqrt(d*x + 1)*sqrt(d* 
x - 1)*a + 16*sqrt(d*x + 1)*sqrt(d*x - 1)*b*d**2*x**3 + 8*sqrt(d*x + 1)*sq 
rt(d*x - 1)*b*x + 12*sqrt(d*x + 1)*sqrt(d*x - 1)*c*x**2 - 16*b*d**3*x**4)/ 
(24*x**4)