\(\int \frac {x^2 (a+b x+c x^2)}{\sqrt {-1+d x} \sqrt {1+d x}} \, dx\) [9]

Optimal result

Integrand size = 32, antiderivative size = 159 \[ \int \frac {x^2 \left (a+b x+c x^2\right )}{\sqrt {-1+d x} \sqrt {1+d x}} \, dx=-\frac {(3 c-4 d (4 b+3 a d)) \sqrt {-1+d x} \sqrt {1+d x}}{24 d^5}+\frac {(9 c+4 b d) x^2 \sqrt {-1+d x} \sqrt {1+d x}}{12 d^3}-\frac {\left (3 c-4 a d^2\right ) (-1+d x)^{3/2} \sqrt {1+d x}}{8 d^5}+\frac {c (-1+d x)^{7/2} \sqrt {1+d x}}{4 d^5}+\frac {\left (3 c+4 a d^2\right ) \text {arccosh}(d x)}{8 d^5} \] Output:

-1/24*(3*c-4*d*(3*a*d+4*b))*(d*x-1)^(1/2)*(d*x+1)^(1/2)/d^5+1/12*(4*b*d+9* 
c)*x^2*(d*x-1)^(1/2)*(d*x+1)^(1/2)/d^3-1/8*(-4*a*d^2+3*c)*(d*x-1)^(3/2)*(d 
*x+1)^(1/2)/d^5+1/4*c*(d*x-1)^(7/2)*(d*x+1)^(1/2)/d^5+1/8*(4*a*d^2+3*c)*ar 
ccosh(d*x)/d^5
 

Mathematica [A] (warning: unable to verify)

Time = 0.20 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.58 \[ \int \frac {x^2 \left (a+b x+c x^2\right )}{\sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {d \sqrt {-1+d x} \sqrt {1+d x} \left (8 b \left (2+d^2 x^2\right )+3 x \left (4 a d^2+c \left (3+2 d^2 x^2\right )\right )\right )+6 \left (3 c+4 a d^2\right ) \text {arctanh}\left (\sqrt {\frac {-1+d x}{1+d x}}\right )}{24 d^5} \] Input:

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

Output:

(d*Sqrt[-1 + d*x]*Sqrt[1 + d*x]*(8*b*(2 + d^2*x^2) + 3*x*(4*a*d^2 + c*(3 + 
 2*d^2*x^2))) + 6*(3*c + 4*a*d^2)*ArcTanh[Sqrt[(-1 + d*x)/(1 + d*x)]])/(24 
*d^5)
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.10, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {2113, 2340, 533, 25, 27, 533, 25, 455, 224, 219}

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.

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

\(\Big \downarrow \) 2113

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

\(\Big \downarrow \) 2340

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

\(\Big \downarrow \) 533

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 533

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 224

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

rule 455

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

rule 533

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
 Simp[d*x^m*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 2))), x] - Simp[1/(b*(m + 2* 
p + 2))   Int[x^(m - 1)*(a + b*x^2)^p*Simp[a*d*m - b*c*(m + 2*p + 2)*x, x], 
 x], x] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[m, 0] && GtQ[p, -1] && Integer 
Q[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 2340

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(c*x)^(m + q - 1 
)*((a + b*x^2)^(p + 1)/(b*c^(q - 1)*(m + q + 2*p + 1))), x] + Simp[1/(b*(m 
+ q + 2*p + 1))   Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1) 
*Pq - b*f*(m + q + 2*p + 1)*x^q - a*f*(m + q - 1)*x^(q - 2), x], x], x] /; 
GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ 
[Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])
 

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.78

method	result	size
risch	\(\frac {\left (6 c \,x^{3} d^{2}+8 b \,d^{2} x^{2}+12 x a \,d^{2}+9 c x +16 b \right ) \sqrt {x d +1}\, \sqrt {x d -1}}{24 d^{4}}+\frac {\left (4 a \,d^{2}+3 c \right ) \ln \left (\frac {d^{2} x}{\sqrt {d^{2}}}+\sqrt {d^{2} x^{2}-1}\right ) \sqrt {\left (x d +1\right ) \left (x d -1\right )}}{8 d^{4} \sqrt {d^{2}}\, \sqrt {x d -1}\, \sqrt {x d +1}}\)	\(124\)
default	\(\frac {\sqrt {x d -1}\, \sqrt {x d +1}\, \left (6 \,\operatorname {csgn}\left (d \right ) c \,d^{3} x^{3} \sqrt {d^{2} x^{2}-1}+8 \,\operatorname {csgn}\left (d \right ) b \,d^{3} x^{2} \sqrt {d^{2} x^{2}-1}+12 \,\operatorname {csgn}\left (d \right ) d^{3} \sqrt {d^{2} x^{2}-1}\, a x +9 \,\operatorname {csgn}\left (d \right ) d \sqrt {d^{2} x^{2}-1}\, c x +16 \,\operatorname {csgn}\left (d \right ) d \sqrt {d^{2} x^{2}-1}\, b +12 \ln \left (\left (\sqrt {d^{2} x^{2}-1}\, \operatorname {csgn}\left (d \right )+x d \right ) \operatorname {csgn}\left (d \right )\right ) a \,d^{2}+9 \ln \left (\left (\sqrt {d^{2} x^{2}-1}\, \operatorname {csgn}\left (d \right )+x d \right ) \operatorname {csgn}\left (d \right )\right ) c \right ) \operatorname {csgn}\left (d \right )}{24 d^{5} \sqrt {d^{2} x^{2}-1}}\)	\(186\)

Input:

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

Output:

1/24*(6*c*d^2*x^3+8*b*d^2*x^2+12*a*d^2*x+9*c*x+16*b)*(d*x+1)^(1/2)*(d*x-1) 
^(1/2)/d^4+1/8*(4*a*d^2+3*c)/d^4*ln(d^2*x/(d^2)^(1/2)+(d^2*x^2-1)^(1/2))/( 
d^2)^(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 = 91, normalized size of antiderivative = 0.57 \[ \int \frac {x^2 \left (a+b x+c x^2\right )}{\sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {{\left (6 \, c d^{3} x^{3} + 8 \, b d^{3} x^{2} + 16 \, b d + 3 \, {\left (4 \, a d^{3} + 3 \, c d\right )} x\right )} \sqrt {d x + 1} \sqrt {d x - 1} - 3 \, {\left (4 \, a d^{2} + 3 \, c\right )} \log \left (-d x + \sqrt {d x + 1} \sqrt {d x - 1}\right )}{24 \, d^{5}} \] Input:

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

Output:

1/24*((6*c*d^3*x^3 + 8*b*d^3*x^2 + 16*b*d + 3*(4*a*d^3 + 3*c*d)*x)*sqrt(d* 
x + 1)*sqrt(d*x - 1) - 3*(4*a*d^2 + 3*c)*log(-d*x + sqrt(d*x + 1)*sqrt(d*x 
 - 1)))/d^5
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.94 \[ \int \frac {x^2 \left (a+b x+c x^2\right )}{\sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {\sqrt {d^{2} x^{2} - 1} c x^{3}}{4 \, d^{2}} + \frac {\sqrt {d^{2} x^{2} - 1} b x^{2}}{3 \, d^{2}} + \frac {\sqrt {d^{2} x^{2} - 1} a x}{2 \, d^{2}} + \frac {a \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - 1} d\right )}{2 \, d^{3}} + \frac {3 \, \sqrt {d^{2} x^{2} - 1} c x}{8 \, d^{4}} + \frac {2 \, \sqrt {d^{2} x^{2} - 1} b}{3 \, d^{4}} + \frac {3 \, c \log \left (2 \, d^{2} x + 2 \, \sqrt {d^{2} x^{2} - 1} d\right )}{8 \, d^{5}} \] Input:

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

Output:

1/4*sqrt(d^2*x^2 - 1)*c*x^3/d^2 + 1/3*sqrt(d^2*x^2 - 1)*b*x^2/d^2 + 1/2*sq 
rt(d^2*x^2 - 1)*a*x/d^2 + 1/2*a*log(2*d^2*x + 2*sqrt(d^2*x^2 - 1)*d)/d^3 + 
 3/8*sqrt(d^2*x^2 - 1)*c*x/d^4 + 2/3*sqrt(d^2*x^2 - 1)*b/d^4 + 3/8*c*log(2 
*d^2*x + 2*sqrt(d^2*x^2 - 1)*d)/d^5
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.91 \[ \int \frac {x^2 \left (a+b x+c x^2\right )}{\sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {{\left ({\left (d x + 1\right )} {\left (2 \, {\left (d x + 1\right )} {\left (\frac {3 \, {\left (d x + 1\right )} c}{d^{4}} + \frac {4 \, b d^{17} - 9 \, c d^{16}}{d^{20}}\right )} + \frac {12 \, a d^{18} - 16 \, b d^{17} + 27 \, c d^{16}}{d^{20}}\right )} - \frac {3 \, {\left (4 \, a d^{18} - 8 \, b d^{17} + 5 \, c d^{16}\right )}}{d^{20}}\right )} \sqrt {d x + 1} \sqrt {d x - 1} - \frac {6 \, {\left (4 \, a d^{2} + 3 \, c\right )} \log \left (\sqrt {d x + 1} - \sqrt {d x - 1}\right )}{d^{4}}}{24 \, d} \] Input:

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

Output:

1/24*(((d*x + 1)*(2*(d*x + 1)*(3*(d*x + 1)*c/d^4 + (4*b*d^17 - 9*c*d^16)/d 
^20) + (12*a*d^18 - 16*b*d^17 + 27*c*d^16)/d^20) - 3*(4*a*d^18 - 8*b*d^17 
+ 5*c*d^16)/d^20)*sqrt(d*x + 1)*sqrt(d*x - 1) - 6*(4*a*d^2 + 3*c)*log(sqrt 
(d*x + 1) - sqrt(d*x - 1))/d^4)/d
 

Mupad [B] (verification not implemented)

Time = 18.04 (sec) , antiderivative size = 767, normalized size of antiderivative = 4.82 \[ \int \frac {x^2 \left (a+b x+c x^2\right )}{\sqrt {-1+d x} \sqrt {1+d x}} \, dx =\text {Too large to display} \] Input:

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

Output:

((23*c*((d*x - 1)^(1/2) - 1i)^3)/(2*((d*x + 1)^(1/2) - 1)^3) + (333*c*((d* 
x - 1)^(1/2) - 1i)^5)/(2*((d*x + 1)^(1/2) - 1)^5) + (671*c*((d*x - 1)^(1/2 
) - 1i)^7)/(2*((d*x + 1)^(1/2) - 1)^7) + (671*c*((d*x - 1)^(1/2) - 1i)^9)/ 
(2*((d*x + 1)^(1/2) - 1)^9) + (333*c*((d*x - 1)^(1/2) - 1i)^11)/(2*((d*x + 
 1)^(1/2) - 1)^11) + (23*c*((d*x - 1)^(1/2) - 1i)^13)/(2*((d*x + 1)^(1/2) 
- 1)^13) - (3*c*((d*x - 1)^(1/2) - 1i)^15)/(2*((d*x + 1)^(1/2) - 1)^15) - 
(3*c*((d*x - 1)^(1/2) - 1i))/(2*((d*x + 1)^(1/2) - 1)))/(d^5 - (8*d^5*((d* 
x - 1)^(1/2) - 1i)^2)/((d*x + 1)^(1/2) - 1)^2 + (28*d^5*((d*x - 1)^(1/2) - 
 1i)^4)/((d*x + 1)^(1/2) - 1)^4 - (56*d^5*((d*x - 1)^(1/2) - 1i)^6)/((d*x 
+ 1)^(1/2) - 1)^6 + (70*d^5*((d*x - 1)^(1/2) - 1i)^8)/((d*x + 1)^(1/2) - 1 
)^8 - (56*d^5*((d*x - 1)^(1/2) - 1i)^10)/((d*x + 1)^(1/2) - 1)^10 + (28*d^ 
5*((d*x - 1)^(1/2) - 1i)^12)/((d*x + 1)^(1/2) - 1)^12 - (8*d^5*((d*x - 1)^ 
(1/2) - 1i)^14)/((d*x + 1)^(1/2) - 1)^14 + (d^5*((d*x - 1)^(1/2) - 1i)^16) 
/((d*x + 1)^(1/2) - 1)^16) - ((14*a*((d*x - 1)^(1/2) - 1i)^3)/((d*x + 1)^( 
1/2) - 1)^3 + (14*a*((d*x - 1)^(1/2) - 1i)^5)/((d*x + 1)^(1/2) - 1)^5 + (2 
*a*((d*x - 1)^(1/2) - 1i)^7)/((d*x + 1)^(1/2) - 1)^7 + (2*a*((d*x - 1)^(1/ 
2) - 1i))/((d*x + 1)^(1/2) - 1))/(d^3 - (4*d^3*((d*x - 1)^(1/2) - 1i)^2)/( 
(d*x + 1)^(1/2) - 1)^2 + (6*d^3*((d*x - 1)^(1/2) - 1i)^4)/((d*x + 1)^(1/2) 
 - 1)^4 - (4*d^3*((d*x - 1)^(1/2) - 1i)^6)/((d*x + 1)^(1/2) - 1)^6 + (d^3* 
((d*x - 1)^(1/2) - 1i)^8)/((d*x + 1)^(1/2) - 1)^8) + (2*a*atanh(((d*x -...
 

Reduce [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.92 \[ \int \frac {x^2 \left (a+b x+c x^2\right )}{\sqrt {-1+d x} \sqrt {1+d x}} \, dx=\frac {12 \sqrt {d x +1}\, \sqrt {d x -1}\, a \,d^{3} x +8 \sqrt {d x +1}\, \sqrt {d x -1}\, b \,d^{3} x^{2}+16 \sqrt {d x +1}\, \sqrt {d x -1}\, b d +6 \sqrt {d x +1}\, \sqrt {d x -1}\, c \,d^{3} x^{3}+9 \sqrt {d x +1}\, \sqrt {d x -1}\, c d x +24 \,\mathrm {log}\left (\frac {\sqrt {d x -1}+\sqrt {d x +1}}{\sqrt {2}}\right ) a \,d^{2}+18 \,\mathrm {log}\left (\frac {\sqrt {d x -1}+\sqrt {d x +1}}{\sqrt {2}}\right ) c}{24 d^{5}} \] Input:

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

Output:

(12*sqrt(d*x + 1)*sqrt(d*x - 1)*a*d**3*x + 8*sqrt(d*x + 1)*sqrt(d*x - 1)*b 
*d**3*x**2 + 16*sqrt(d*x + 1)*sqrt(d*x - 1)*b*d + 6*sqrt(d*x + 1)*sqrt(d*x 
 - 1)*c*d**3*x**3 + 9*sqrt(d*x + 1)*sqrt(d*x - 1)*c*d*x + 24*log((sqrt(d*x 
 - 1) + sqrt(d*x + 1))/sqrt(2))*a*d**2 + 18*log((sqrt(d*x - 1) + sqrt(d*x 
+ 1))/sqrt(2))*c)/(24*d**5)