\(\int \frac {a+b x+c x^2}{x^5 \sqrt {1-d x} \sqrt {1+d x}} \, dx\) [8]

Optimal result

Integrand size = 33, antiderivative size = 133 \[ \int \frac {a+b x+c x^2}{x^5 \sqrt {1-d x} \sqrt {1+d x}} \, dx=-\frac {a \sqrt {1-d^2 x^2}}{4 x^4}-\frac {b \sqrt {1-d^2 x^2}}{3 x^3}-\frac {\left (4 c+3 a d^2\right ) \sqrt {1-d^2 x^2}}{8 x^2}-\frac {2 b d^2 \sqrt {1-d^2 x^2}}{3 x}-\frac {1}{8} d^2 \left (4 c+3 a d^2\right ) \text {arctanh}\left (\sqrt {1-d^2 x^2}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.80 \[ \int \frac {a+b x+c x^2}{x^5 \sqrt {1-d x} \sqrt {1+d x}} \, dx=\frac {\sqrt {1-d^2 x^2} \left (-6 a-8 b x-12 c x^2-9 a d^2 x^2-16 b d^2 x^3\right )}{24 x^4}-\frac {1}{8} d^2 \left (4 c+3 a d^2\right ) \log (x)+\frac {1}{8} d^2 \left (4 c+3 a d^2\right ) \log \left (-1+\sqrt {1-d^2 x^2}\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.06, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {2112, 2338, 25, 539, 25, 539, 25, 27, 534, 243, 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.

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

\(\Big \downarrow \) 2112

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

\(\Big \downarrow \) 2338

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 539

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 539

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 534

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

-1/4*(a*Sqrt[1 - d^2*x^2])/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*((-16*b*Sqrt[1 - d^2*x^2 
])/x - 3*(4*c + 3*a*d^2)*ArcTanh[Sqrt[1 - d^2*x^2]]))/2)/3)/4
 

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 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 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 2112

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

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 = 0.75 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.08

method	result	size
risch	\(\frac {\sqrt {x d +1}\, \left (x d -1\right ) \left (16 b \,d^{2} x^{3}+9 a \,d^{2} x^{2}+12 c \,x^{2}+8 b x +6 a \right ) \sqrt {\left (-x d +1\right ) \left (x d +1\right )}}{24 x^{4} \sqrt {-\left (x d +1\right ) \left (x d -1\right )}\, \sqrt {-x d +1}}-\frac {d^{2} \left (3 a \,d^{2}+4 c \right ) \operatorname {arctanh}\left (\frac {1}{\sqrt {-d^{2} x^{2}+1}}\right ) \sqrt {\left (-x d +1\right ) \left (x d +1\right )}}{8 \sqrt {-x d +1}\, \sqrt {x d +1}}\)	\(144\)
default	\(-\frac {\sqrt {-x d +1}\, \sqrt {x d +1}\, \operatorname {csgn}\left (d \right )^{2} \left (9 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-d^{2} x^{2}+1}}\right ) a \,d^{4} x^{4}+12 \,\operatorname {arctanh}\left (\frac {1}{\sqrt {-d^{2} x^{2}+1}}\right ) c \,d^{2} x^{4}+16 \sqrt {-d^{2} x^{2}+1}\, b \,d^{2} x^{3}+9 \sqrt {-d^{2} x^{2}+1}\, a \,d^{2} x^{2}+12 \sqrt {-d^{2} x^{2}+1}\, c \,x^{2}+8 \sqrt {-d^{2} x^{2}+1}\, b x +6 \sqrt {-d^{2} x^{2}+1}\, a \right )}{24 \sqrt {-d^{2} x^{2}+1}\, x^{4}}\)	\(173\)

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)*(16*b*d^2*x^3+9*a*d^2*x^2+12*c*x^2+8*b*x+6*a)/x 
^4/(-(d*x+1)*(d*x-1))^(1/2)*((-d*x+1)*(d*x+1))^(1/2)/(-d*x+1)^(1/2)-1/8*d^ 
2*(3*a*d^2+4*c)*arctanh(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 = 96, normalized size of antiderivative = 0.72 \[ \int \frac {a+b x+c x^2}{x^5 \sqrt {1-d x} \sqrt {1+d x}} \, dx=\frac {3 \, {\left (3 \, a d^{4} + 4 \, c d^{2}\right )} x^{4} \log \left (\frac {\sqrt {d x + 1} \sqrt {-d x + 1} - 1}{x}\right ) - {\left (16 \, b d^{2} x^{3} + 3 \, {\left (3 \, a d^{2} + 4 \, c\right )} x^{2} + 8 \, b x + 6 \, a\right )} \sqrt {d x + 1} \sqrt {-d x + 1}}{24 \, x^{4}} \] Input:

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

Output:

1/24*(3*(3*a*d^4 + 4*c*d^2)*x^4*log((sqrt(d*x + 1)*sqrt(-d*x + 1) - 1)/x) 
- (16*b*d^2*x^3 + 3*(3*a*d^2 + 4*c)*x^2 + 8*b*x + 6*a)*sqrt(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 = 161, normalized size of antiderivative = 1.21 \[ \int \frac {a+b x+c x^2}{x^5 \sqrt {1-d x} \sqrt {1+d x}} \, dx=-\frac {3}{8} \, a d^{4} \log \left (\frac {2 \, \sqrt {-d^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) - \frac {1}{2} \, c d^{2} \log \left (\frac {2 \, \sqrt {-d^{2} x^{2} + 1}}{{\left | x \right |}} + \frac {2}{{\left | x \right |}}\right ) - \frac {2 \, \sqrt {-d^{2} x^{2} + 1} b d^{2}}{3 \, x} - \frac {3 \, \sqrt {-d^{2} x^{2} + 1} a d^{2}}{8 \, x^{2}} - \frac {\sqrt {-d^{2} x^{2} + 1} c}{2 \, x^{2}} - \frac {\sqrt {-d^{2} x^{2} + 1} b}{3 \, x^{3}} - \frac {\sqrt {-d^{2} x^{2} + 1} a}{4 \, x^{4}} \] Input:

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

Output:

-3/8*a*d^4*log(2*sqrt(-d^2*x^2 + 1)/abs(x) + 2/abs(x)) - 1/2*c*d^2*log(2*s 
qrt(-d^2*x^2 + 1)/abs(x) + 2/abs(x)) - 2/3*sqrt(-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. 861 vs. \(2 (113) = 226\).

Time = 0.28 (sec) , antiderivative size = 861, normalized size of antiderivative = 6.47 \[ \int \frac {a+b x+c x^2}{x^5 \sqrt {1-d x} \sqrt {1+d x}} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/24*(3*(3*a*d^5 + 4*c*d^3)*log(abs(-(sqrt(2) - sqrt(-d*x + 1))/sqrt(d*x 
+ 1) + sqrt(d*x + 1)/(sqrt(2) - sqrt(-d*x + 1)) + 2)) - 3*(3*a*d^5 + 4*c*d 
^3)*log(abs(-(sqrt(2) - sqrt(-d*x + 1))/sqrt(d*x + 1) + sqrt(d*x + 1)/(sqr 
t(2) - sqrt(-d*x + 1)) - 2)) - 4*(15*a*d^5*((sqrt(2) - sqrt(-d*x + 1))/sqr 
t(d*x + 1) - sqrt(d*x + 1)/(sqrt(2) - sqrt(-d*x + 1)))^7 - 24*b*d^4*((sqrt 
(2) - sqrt(-d*x + 1))/sqrt(d*x + 1) - sqrt(d*x + 1)/(sqrt(2) - sqrt(-d*x + 
 1)))^7 + 36*a*d^5*((sqrt(2) - sqrt(-d*x + 1))/sqrt(d*x + 1) - sqrt(d*x + 
1)/(sqrt(2) - sqrt(-d*x + 1)))^5 + 12*c*d^3*((sqrt(2) - sqrt(-d*x + 1))/sq 
rt(d*x + 1) - sqrt(d*x + 1)/(sqrt(2) - sqrt(-d*x + 1)))^7 + 160*b*d^4*((sq 
rt(2) - sqrt(-d*x + 1))/sqrt(d*x + 1) - sqrt(d*x + 1)/(sqrt(2) - sqrt(-d*x 
 + 1)))^5 + 144*a*d^5*((sqrt(2) - sqrt(-d*x + 1))/sqrt(d*x + 1) - sqrt(d*x 
 + 1)/(sqrt(2) - sqrt(-d*x + 1)))^3 - 48*c*d^3*((sqrt(2) - sqrt(-d*x + 1)) 
/sqrt(d*x + 1) - sqrt(d*x + 1)/(sqrt(2) - sqrt(-d*x + 1)))^5 - 640*b*d^4*( 
(sqrt(2) - sqrt(-d*x + 1))/sqrt(d*x + 1) - sqrt(d*x + 1)/(sqrt(2) - sqrt(- 
d*x + 1)))^3 + 960*a*d^5*((sqrt(2) - sqrt(-d*x + 1))/sqrt(d*x + 1) - sqrt( 
d*x + 1)/(sqrt(2) - sqrt(-d*x + 1))) - 192*c*d^3*((sqrt(2) - sqrt(-d*x + 1 
))/sqrt(d*x + 1) - sqrt(d*x + 1)/(sqrt(2) - sqrt(-d*x + 1)))^3 + 1536*b*d^ 
4*((sqrt(2) - sqrt(-d*x + 1))/sqrt(d*x + 1) - sqrt(d*x + 1)/(sqrt(2) - sqr 
t(-d*x + 1))) + 768*c*d^3*((sqrt(2) - sqrt(-d*x + 1))/sqrt(d*x + 1) - sqrt 
(d*x + 1)/(sqrt(2) - sqrt(-d*x + 1))))/(((sqrt(2) - sqrt(-d*x + 1))/sqr...
 

Mupad [B] (verification not implemented)

Time = 7.39 (sec) , antiderivative size = 684, normalized size of antiderivative = 5.14 \[ \int \frac {a+b x+c x^2}{x^5 \sqrt {1-d x} \sqrt {1+d x}} \, dx =\text {Too large to display} \] Input:

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

Output:

((a*d^4)/4 + (6*a*d^4*((1 - d*x)^(1/2) - 1)^2)/((d*x + 1)^(1/2) - 1)^2 - ( 
53*a*d^4*((1 - d*x)^(1/2) - 1)^4)/(2*((d*x + 1)^(1/2) - 1)^4) - (87*a*d^4* 
((1 - d*x)^(1/2) - 1)^6)/((d*x + 1)^(1/2) - 1)^6 + (657*a*d^4*((1 - d*x)^( 
1/2) - 1)^8)/(4*((d*x + 1)^(1/2) - 1)^8) - (121*a*d^4*((1 - d*x)^(1/2) - 1 
)^10)/((d*x + 1)^(1/2) - 1)^10)/((256*((1 - d*x)^(1/2) - 1)^4)/((d*x + 1)^ 
(1/2) - 1)^4 - (1024*((1 - d*x)^(1/2) - 1)^6)/((d*x + 1)^(1/2) - 1)^6 + (1 
536*((1 - d*x)^(1/2) - 1)^8)/((d*x + 1)^(1/2) - 1)^8 - (1024*((1 - d*x)^(1 
/2) - 1)^10)/((d*x + 1)^(1/2) - 1)^10 + (256*((1 - d*x)^(1/2) - 1)^12)/((d 
*x + 1)^(1/2) - 1)^12) - ((c*d^2*((1 - d*x)^(1/2) - 1)^2)/((d*x + 1)^(1/2) 
 - 1)^2 - (c*d^2)/2 + (15*c*d^2*((1 - d*x)^(1/2) - 1)^4)/(2*((d*x + 1)^(1/ 
2) - 1)^4))/((16*((1 - d*x)^(1/2) - 1)^2)/((d*x + 1)^(1/2) - 1)^2 - (32*(( 
1 - d*x)^(1/2) - 1)^4)/((d*x + 1)^(1/2) - 1)^4 + (16*((1 - d*x)^(1/2) - 1) 
^6)/((d*x + 1)^(1/2) - 1)^6) + (3*a*d^4*log(((1 - d*x)^(1/2) - 1)^2/((d*x 
+ 1)^(1/2) - 1)^2 - 1))/8 + (c*d^2*log(((1 - d*x)^(1/2) - 1)^2/((d*x + 1)^ 
(1/2) - 1)^2 - 1))/2 - (3*a*d^4*log(((1 - d*x)^(1/2) - 1)/((d*x + 1)^(1/2) 
 - 1)))/8 - (c*d^2*log(((1 - d*x)^(1/2) - 1)/((d*x + 1)^(1/2) - 1)))/2 - ( 
(1 - d*x)^(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)) + (7*a*d^4*((1 - d*x)^(1/2) - 1)^2)/(256*((d*x + 1)^(1/ 
2) - 1)^2) + (a*d^4*((1 - d*x)^(1/2) - 1)^4)/(1024*((d*x + 1)^(1/2) - 1)^4 
) + (c*d^2*((1 - d*x)^(1/2) - 1)^2)/(32*((d*x + 1)^(1/2) - 1)^2)
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 350, normalized size of antiderivative = 2.63 \[ \int \frac {a+b x+c x^2}{x^5 \sqrt {1-d x} \sqrt {1+d x}} \, dx=\frac {-9 \sqrt {d x +1}\, \sqrt {-d x +1}\, a \,d^{2} x^{2}-6 \sqrt {d x +1}\, \sqrt {-d x +1}\, a -16 \sqrt {d x +1}\, \sqrt {-d x +1}\, b \,d^{2} x^{3}-8 \sqrt {d x +1}\, \sqrt {-d x +1}\, b x -12 \sqrt {d x +1}\, \sqrt {-d x +1}\, c \,x^{2}-9 \,\mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-d x +1}}{\sqrt {2}}\right )}{2}\right )-1\right ) a \,d^{4} x^{4}-12 \,\mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-d x +1}}{\sqrt {2}}\right )}{2}\right )-1\right ) c \,d^{2} x^{4}+9 \,\mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-d x +1}}{\sqrt {2}}\right )}{2}\right )+1\right ) a \,d^{4} x^{4}+12 \,\mathrm {log}\left (-\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-d x +1}}{\sqrt {2}}\right )}{2}\right )+1\right ) c \,d^{2} x^{4}-9 \,\mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-d x +1}}{\sqrt {2}}\right )}{2}\right )-1\right ) a \,d^{4} x^{4}-12 \,\mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-d x +1}}{\sqrt {2}}\right )}{2}\right )-1\right ) c \,d^{2} x^{4}+9 \,\mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-d x +1}}{\sqrt {2}}\right )}{2}\right )+1\right ) a \,d^{4} x^{4}+12 \,\mathrm {log}\left (\sqrt {2}+\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {-d x +1}}{\sqrt {2}}\right )}{2}\right )+1\right ) c \,d^{2} x^{4}}{24 x^{4}} \] Input:

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

Output:

( - 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 - 9*lo 
g( - sqrt(2) + tan(asin(sqrt( - d*x + 1)/sqrt(2))/2) - 1)*a*d**4*x**4 - 12 
*log( - sqrt(2) + tan(asin(sqrt( - d*x + 1)/sqrt(2))/2) - 1)*c*d**2*x**4 + 
 9*log( - sqrt(2) + tan(asin(sqrt( - d*x + 1)/sqrt(2))/2) + 1)*a*d**4*x**4 
 + 12*log( - sqrt(2) + tan(asin(sqrt( - d*x + 1)/sqrt(2))/2) + 1)*c*d**2*x 
**4 - 9*log(sqrt(2) + tan(asin(sqrt( - d*x + 1)/sqrt(2))/2) - 1)*a*d**4*x* 
*4 - 12*log(sqrt(2) + tan(asin(sqrt( - d*x + 1)/sqrt(2))/2) - 1)*c*d**2*x* 
*4 + 9*log(sqrt(2) + tan(asin(sqrt( - d*x + 1)/sqrt(2))/2) + 1)*a*d**4*x** 
4 + 12*log(sqrt(2) + tan(asin(sqrt( - d*x + 1)/sqrt(2))/2) + 1)*c*d**2*x** 
4)/(24*x**4)