\(\int \frac {(a+b x+c x^2)^2}{\sqrt {1-d x} \sqrt {1+d x}} \, dx\) [793]

Optimal result

Integrand size = 32, antiderivative size = 166 \[ \int \frac {\left (a+b x+c x^2\right )^2}{\sqrt {1-d x} \sqrt {1+d x}} \, dx=-\frac {2 b \left (2 c+3 a d^2\right ) \sqrt {1-d^2 x^2}}{3 d^4}-\frac {\left (4 b^2+c \left (8 a+\frac {3 c}{d^2}\right )\right ) x \sqrt {1-d^2 x^2}}{8 d^2}-\frac {2 b c x^2 \sqrt {1-d^2 x^2}}{3 d^2}-\frac {c^2 x^3 \sqrt {1-d^2 x^2}}{4 d^2}+\frac {\left (3 c^2+4 b^2 d^2+8 a c d^2+8 a^2 d^4\right ) \arcsin (d x)}{8 d^5} \]

[Out]

1/8*(8*a^2*d^4+8*a*c*d^2+4*b^2*d^2+3*c^2)*arcsin(d*x)/d^5-2/3*b*(3*a*d^2+2*c)*(-d^2*x^2+1)^(1/2)/d^4-1/8*(4*b^
2+c*(8*a+3*c/d^2))*x*(-d^2*x^2+1)^(1/2)/d^2-2/3*b*c*x^2*(-d^2*x^2+1)^(1/2)/d^2-1/4*c^2*x^3*(-d^2*x^2+1)^(1/2)/
d^2

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {913, 1829, 655, 222} \[ \int \frac {\left (a+b x+c x^2\right )^2}{\sqrt {1-d x} \sqrt {1+d x}} \, dx=\frac {\arcsin (d x) \left (8 a^2 d^4+8 a c d^2+4 b^2 d^2+3 c^2\right )}{8 d^5}-\frac {x \sqrt {1-d^2 x^2} \left (c \left (8 a+\frac {3 c}{d^2}\right )+4 b^2\right )}{8 d^2}-\frac {2 b \sqrt {1-d^2 x^2} \left (3 a d^2+2 c\right )}{3 d^4}-\frac {2 b c x^2 \sqrt {1-d^2 x^2}}{3 d^2}-\frac {c^2 x^3 \sqrt {1-d^2 x^2}}{4 d^2} \]

[In]

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

[Out]

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

Rule 222

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

Rule 655

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

Rule 913

Int[((d_) + (e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :>
Int[(d*f + e*g*x^2)^m*(a + b*x + c*x^2)^p, x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[m - n, 0] &&
EqQ[e*f + d*g, 0] && (IntegerQ[m] || (GtQ[d, 0] && GtQ[f, 0]))

Rule 1829

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Si
mp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*(q + 2*p + 1))), x] + Dist[1/(b*(q + 2*p + 1)), Int[(a + b*x^2)^p*Expan
dToSum[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x]
&& PolyQ[Pq, x] &&  !LeQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\left (a+b x+c x^2\right )^2}{\sqrt {1-d^2 x^2}} \, dx \\ & = -\frac {c^2 x^3 \sqrt {1-d^2 x^2}}{4 d^2}-\frac {\int \frac {-4 a^2 d^2-8 a b d^2 x-\left (3 c^2+4 b^2 d^2+8 a c d^2\right ) x^2-8 b c d^2 x^3}{\sqrt {1-d^2 x^2}} \, dx}{4 d^2} \\ & = -\frac {2 b c x^2 \sqrt {1-d^2 x^2}}{3 d^2}-\frac {c^2 x^3 \sqrt {1-d^2 x^2}}{4 d^2}+\frac {\int \frac {12 a^2 d^4+8 b d^2 \left (2 c+3 a d^2\right ) x+3 d^2 \left (3 c^2+4 b^2 d^2+8 a c d^2\right ) x^2}{\sqrt {1-d^2 x^2}} \, dx}{12 d^4} \\ & = -\frac {\left (3 c^2+4 b^2 d^2+8 a c d^2\right ) x \sqrt {1-d^2 x^2}}{8 d^4}-\frac {2 b c x^2 \sqrt {1-d^2 x^2}}{3 d^2}-\frac {c^2 x^3 \sqrt {1-d^2 x^2}}{4 d^2}-\frac {\int \frac {-3 d^2 \left (3 c^2+4 b^2 d^2+8 a c d^2+8 a^2 d^4\right )-16 b d^4 \left (2 c+3 a d^2\right ) x}{\sqrt {1-d^2 x^2}} \, dx}{24 d^6} \\ & = -\frac {2 b \left (2 c+3 a d^2\right ) \sqrt {1-d^2 x^2}}{3 d^4}-\frac {\left (3 c^2+4 b^2 d^2+8 a c d^2\right ) x \sqrt {1-d^2 x^2}}{8 d^4}-\frac {2 b c x^2 \sqrt {1-d^2 x^2}}{3 d^2}-\frac {c^2 x^3 \sqrt {1-d^2 x^2}}{4 d^2}+\frac {\left (3 c^2+4 b^2 d^2+8 a c d^2+8 a^2 d^4\right ) \int \frac {1}{\sqrt {1-d^2 x^2}} \, dx}{8 d^4} \\ & = -\frac {2 b \left (2 c+3 a d^2\right ) \sqrt {1-d^2 x^2}}{3 d^4}-\frac {\left (3 c^2+4 b^2 d^2+8 a c d^2\right ) x \sqrt {1-d^2 x^2}}{8 d^4}-\frac {2 b c x^2 \sqrt {1-d^2 x^2}}{3 d^2}-\frac {c^2 x^3 \sqrt {1-d^2 x^2}}{4 d^2}+\frac {\left (3 c^2+4 b^2 d^2+8 a c d^2+8 a^2 d^4\right ) \sin ^{-1}(d x)}{8 d^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.80 \[ \int \frac {\left (a+b x+c x^2\right )^2}{\sqrt {1-d x} \sqrt {1+d x}} \, dx=\frac {-d \sqrt {1-d^2 x^2} \left (12 b^2 d^2 x+16 b \left (3 a d^2+c \left (2+d^2 x^2\right )\right )+3 c x \left (8 a d^2+c \left (3+2 d^2 x^2\right )\right )\right )+6 \left (3 c^2+4 b^2 d^2+8 a c d^2+8 a^2 d^4\right ) \arctan \left (\frac {d x}{-1+\sqrt {1-d^2 x^2}}\right )}{24 d^5} \]

[In]

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

[Out]

(-(d*Sqrt[1 - d^2*x^2]*(12*b^2*d^2*x + 16*b*(3*a*d^2 + c*(2 + d^2*x^2)) + 3*c*x*(8*a*d^2 + c*(3 + 2*d^2*x^2)))
) + 6*(3*c^2 + 4*b^2*d^2 + 8*a*c*d^2 + 8*a^2*d^4)*ArcTan[(d*x)/(-1 + Sqrt[1 - d^2*x^2])])/(24*d^5)

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.20

method	result	size
risch	\(\frac {\left (6 c^{2} x^{3} d^{2}+16 b c \,x^{2} d^{2}+24 a c \,d^{2} x +12 b^{2} d^{2} x +48 b a \,d^{2}+9 c^{2} x +32 b c \right ) \left (d x -1\right ) \sqrt {d x +1}\, \sqrt {\left (-d x +1\right ) \left (d x +1\right )}}{24 d^{4} \sqrt {-\left (d x -1\right ) \left (d x +1\right )}\, \sqrt {-d x +1}}+\frac {\left (8 a^{2} d^{4}+8 c \,d^{2} a +4 b^{2} d^{2}+3 c^{2}\right ) \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+1}}\right ) \sqrt {\left (-d x +1\right ) \left (d x +1\right )}}{8 d^{4} \sqrt {d^{2}}\, \sqrt {-d x +1}\, \sqrt {d x +1}}\)	\(199\)
default	\(-\frac {\sqrt {-d x +1}\, \sqrt {d x +1}\, \left (6 \,\operatorname {csgn}\left (d \right ) c^{2} d^{3} x^{3} \sqrt {-d^{2} x^{2}+1}+16 \,\operatorname {csgn}\left (d \right ) b c \,d^{3} x^{2} \sqrt {-d^{2} x^{2}+1}+24 \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{3} a c x +12 \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d^{3} b^{2} x +48 \,\operatorname {csgn}\left (d \right ) d^{3} \sqrt {-d^{2} x^{2}+1}\, a b -24 \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) a^{2} d^{4}+9 \sqrt {-d^{2} x^{2}+1}\, \operatorname {csgn}\left (d \right ) d \,c^{2} x +32 \,\operatorname {csgn}\left (d \right ) d \sqrt {-d^{2} x^{2}+1}\, b c -24 \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) a c \,d^{2}-12 \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) b^{2} d^{2}-9 \arctan \left (\frac {\operatorname {csgn}\left (d \right ) d x}{\sqrt {-d^{2} x^{2}+1}}\right ) c^{2}\right ) \operatorname {csgn}\left (d \right )}{24 d^{5} \sqrt {-d^{2} x^{2}+1}}\)	\(291\)

[In]

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

[Out]

1/24*(6*c^2*d^2*x^3+16*b*c*d^2*x^2+24*a*c*d^2*x+12*b^2*d^2*x+48*a*b*d^2+9*c^2*x+32*b*c)*(d*x-1)*(d*x+1)^(1/2)/
d^4/(-(d*x-1)*(d*x+1))^(1/2)*((-d*x+1)*(d*x+1))^(1/2)/(-d*x+1)^(1/2)+1/8*(8*a^2*d^4+8*a*c*d^2+4*b^2*d^2+3*c^2)
/d^4/(d^2)^(1/2)*arctan((d^2)^(1/2)*x/(-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)

none

Time = 0.33 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.81 \[ \int \frac {\left (a+b x+c x^2\right )^2}{\sqrt {1-d x} \sqrt {1+d x}} \, dx=-\frac {{\left (6 \, c^{2} d^{3} x^{3} + 16 \, b c d^{3} x^{2} + 48 \, a b d^{3} + 32 \, b c d + 3 \, {\left (4 \, {\left (b^{2} + 2 \, a c\right )} d^{3} + 3 \, c^{2} d\right )} x\right )} \sqrt {d x + 1} \sqrt {-d x + 1} + 6 \, {\left (8 \, a^{2} d^{4} + 4 \, {\left (b^{2} + 2 \, a c\right )} d^{2} + 3 \, c^{2}\right )} \arctan \left (\frac {\sqrt {d x + 1} \sqrt {-d x + 1} - 1}{d x}\right )}{24 \, d^{5}} \]

[In]

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

[Out]

-1/24*((6*c^2*d^3*x^3 + 16*b*c*d^3*x^2 + 48*a*b*d^3 + 32*b*c*d + 3*(4*(b^2 + 2*a*c)*d^3 + 3*c^2*d)*x)*sqrt(d*x
+ 1)*sqrt(-d*x + 1) + 6*(8*a^2*d^4 + 4*(b^2 + 2*a*c)*d^2 + 3*c^2)*arctan((sqrt(d*x + 1)*sqrt(-d*x + 1) - 1)/(
d*x)))/d^5

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.03 \[ \int \frac {\left (a+b x+c x^2\right )^2}{\sqrt {1-d x} \sqrt {1+d x}} \, dx=-\frac {\sqrt {-d^{2} x^{2} + 1} c^{2} x^{3}}{4 \, d^{2}} - \frac {2 \, \sqrt {-d^{2} x^{2} + 1} b c x^{2}}{3 \, d^{2}} + \frac {a^{2} \arcsin \left (d x\right )}{d} - \frac {2 \, \sqrt {-d^{2} x^{2} + 1} a b}{d^{2}} - \frac {\sqrt {-d^{2} x^{2} + 1} {\left (b^{2} + 2 \, a c\right )} x}{2 \, d^{2}} - \frac {3 \, \sqrt {-d^{2} x^{2} + 1} c^{2} x}{8 \, d^{4}} + \frac {{\left (b^{2} + 2 \, a c\right )} \arcsin \left (d x\right )}{2 \, d^{3}} - \frac {4 \, \sqrt {-d^{2} x^{2} + 1} b c}{3 \, d^{4}} + \frac {3 \, c^{2} \arcsin \left (d x\right )}{8 \, d^{5}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a+b x+c x^2\right )^2}{\sqrt {1-d x} \sqrt {1+d x}} \, dx=-\frac {{\left (48 \, a b d^{3} - 12 \, b^{2} d^{2} - 24 \, a c d^{2} + 48 \, b c d + {\left (12 \, b^{2} d^{2} + 24 \, a c d^{2} - 32 \, b c d + 2 \, {\left (3 \, {\left (d x + 1\right )} c^{2} + 8 \, b c d - 9 \, c^{2}\right )} {\left (d x + 1\right )} + 27 \, c^{2}\right )} {\left (d x + 1\right )} - 15 \, c^{2}\right )} \sqrt {d x + 1} \sqrt {-d x + 1} - 6 \, {\left (8 \, a^{2} d^{4} + 4 \, b^{2} d^{2} + 8 \, a c d^{2} + 3 \, c^{2}\right )} \arcsin \left (\frac {1}{2} \, \sqrt {2} \sqrt {d x + 1}\right )}{24 \, d^{5}} \]

[In]

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

[Out]

-1/24*((48*a*b*d^3 - 12*b^2*d^2 - 24*a*c*d^2 + 48*b*c*d + (12*b^2*d^2 + 24*a*c*d^2 - 32*b*c*d + 2*(3*(d*x + 1)
*c^2 + 8*b*c*d - 9*c^2)*(d*x + 1) + 27*c^2)*(d*x + 1) - 15*c^2)*sqrt(d*x + 1)*sqrt(-d*x + 1) - 6*(8*a^2*d^4 +
4*b^2*d^2 + 8*a*c*d^2 + 3*c^2)*arcsin(1/2*sqrt(2)*sqrt(d*x + 1)))/d^5

Mupad [B] (verification not implemented)

Time = 27.55 (sec) , antiderivative size = 897, normalized size of antiderivative = 5.40 \[ \int \frac {\left (a+b x+c x^2\right )^2}{\sqrt {1-d x} \sqrt {1+d x}} \, dx=-\frac {\frac {{\left (\sqrt {1-d\,x}-1\right )}^{15}\,\left (2\,b^2\,d^2+\frac {3\,c^2}{2}+4\,a\,c\,d^2\right )}{{\left (\sqrt {d\,x+1}-1\right )}^{15}}+\frac {{\left (\sqrt {1-d\,x}-1\right )}^3\,\left (6\,b^2\,d^2-\frac {23\,c^2}{2}+12\,a\,c\,d^2\right )}{{\left (\sqrt {d\,x+1}-1\right )}^3}-\frac {{\left (\sqrt {1-d\,x}-1\right )}^{13}\,\left (6\,b^2\,d^2-\frac {23\,c^2}{2}+12\,a\,c\,d^2\right )}{{\left (\sqrt {d\,x+1}-1\right )}^{13}}+\frac {{\left (\sqrt {1-d\,x}-1\right )}^5\,\left (30\,b^2\,d^2+\frac {333\,c^2}{2}+60\,a\,c\,d^2\right )}{{\left (\sqrt {d\,x+1}-1\right )}^5}-\frac {{\left (\sqrt {1-d\,x}-1\right )}^{11}\,\left (30\,b^2\,d^2+\frac {333\,c^2}{2}+60\,a\,c\,d^2\right )}{{\left (\sqrt {d\,x+1}-1\right )}^{11}}+\frac {{\left (\sqrt {1-d\,x}-1\right )}^7\,\left (22\,b^2\,d^2-\frac {671\,c^2}{2}+44\,a\,c\,d^2\right )}{{\left (\sqrt {d\,x+1}-1\right )}^7}-\frac {{\left (\sqrt {1-d\,x}-1\right )}^9\,\left (22\,b^2\,d^2-\frac {671\,c^2}{2}+44\,a\,c\,d^2\right )}{{\left (\sqrt {d\,x+1}-1\right )}^9}+\frac {{\left (\sqrt {1-d\,x}-1\right )}^4\,\left (96\,a\,b\,d^3+128\,b\,c\,d\right )}{{\left (\sqrt {d\,x+1}-1\right )}^4}+\frac {{\left (\sqrt {1-d\,x}-1\right )}^{12}\,\left (96\,a\,b\,d^3+128\,b\,c\,d\right )}{{\left (\sqrt {d\,x+1}-1\right )}^{12}}+\frac {{\left (\sqrt {1-d\,x}-1\right )}^8\,\left (320\,a\,b\,d^3+\frac {256\,b\,c\,d}{3}\right )}{{\left (\sqrt {d\,x+1}-1\right )}^8}+\frac {{\left (\sqrt {1-d\,x}-1\right )}^6\,\left (240\,a\,b\,d^3+\frac {512\,b\,c\,d}{3}\right )}{{\left (\sqrt {d\,x+1}-1\right )}^6}+\frac {{\left (\sqrt {1-d\,x}-1\right )}^{10}\,\left (240\,a\,b\,d^3+\frac {512\,b\,c\,d}{3}\right )}{{\left (\sqrt {d\,x+1}-1\right )}^{10}}-\frac {\left (\sqrt {1-d\,x}-1\right )\,\left (2\,b^2\,d^2+\frac {3\,c^2}{2}+4\,a\,c\,d^2\right )}{\sqrt {d\,x+1}-1}+\frac {16\,a\,b\,d^3\,{\left (\sqrt {1-d\,x}-1\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}+\frac {16\,a\,b\,d^3\,{\left (\sqrt {1-d\,x}-1\right )}^{14}}{{\left (\sqrt {d\,x+1}-1\right )}^{14}}}{d^5+\frac {8\,d^5\,{\left (\sqrt {1-d\,x}-1\right )}^2}{{\left (\sqrt {d\,x+1}-1\right )}^2}+\frac {28\,d^5\,{\left (\sqrt {1-d\,x}-1\right )}^4}{{\left (\sqrt {d\,x+1}-1\right )}^4}+\frac {56\,d^5\,{\left (\sqrt {1-d\,x}-1\right )}^6}{{\left (\sqrt {d\,x+1}-1\right )}^6}+\frac {70\,d^5\,{\left (\sqrt {1-d\,x}-1\right )}^8}{{\left (\sqrt {d\,x+1}-1\right )}^8}+\frac {56\,d^5\,{\left (\sqrt {1-d\,x}-1\right )}^{10}}{{\left (\sqrt {d\,x+1}-1\right )}^{10}}+\frac {28\,d^5\,{\left (\sqrt {1-d\,x}-1\right )}^{12}}{{\left (\sqrt {d\,x+1}-1\right )}^{12}}+\frac {8\,d^5\,{\left (\sqrt {1-d\,x}-1\right )}^{14}}{{\left (\sqrt {d\,x+1}-1\right )}^{14}}+\frac {d^5\,{\left (\sqrt {1-d\,x}-1\right )}^{16}}{{\left (\sqrt {d\,x+1}-1\right )}^{16}}}-\frac {\mathrm {atan}\left (\frac {\sqrt {1-d\,x}-1}{\sqrt {d\,x+1}-1}\right )\,\left (8\,a^2\,d^4+8\,a\,c\,d^2+4\,b^2\,d^2+3\,c^2\right )}{2\,d^5} \]

[In]

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

[Out]

- ((((1 - d*x)^(1/2) - 1)^15*((3*c^2)/2 + 2*b^2*d^2 + 4*a*c*d^2))/((d*x + 1)^(1/2) - 1)^15 + (((1 - d*x)^(1/2)
- 1)^3*(6*b^2*d^2 - (23*c^2)/2 + 12*a*c*d^2))/((d*x + 1)^(1/2) - 1)^3 - (((1 - d*x)^(1/2) - 1)^13*(6*b^2*d^2
- (23*c^2)/2 + 12*a*c*d^2))/((d*x + 1)^(1/2) - 1)^13 + (((1 - d*x)^(1/2) - 1)^5*((333*c^2)/2 + 30*b^2*d^2 + 60
*a*c*d^2))/((d*x + 1)^(1/2) - 1)^5 - (((1 - d*x)^(1/2) - 1)^11*((333*c^2)/2 + 30*b^2*d^2 + 60*a*c*d^2))/((d*x
+ 1)^(1/2) - 1)^11 + (((1 - d*x)^(1/2) - 1)^7*(22*b^2*d^2 - (671*c^2)/2 + 44*a*c*d^2))/((d*x + 1)^(1/2) - 1)^7
- (((1 - d*x)^(1/2) - 1)^9*(22*b^2*d^2 - (671*c^2)/2 + 44*a*c*d^2))/((d*x + 1)^(1/2) - 1)^9 + (((1 - d*x)^(1/
2) - 1)^4*(128*b*c*d + 96*a*b*d^3))/((d*x + 1)^(1/2) - 1)^4 + (((1 - d*x)^(1/2) - 1)^12*(128*b*c*d + 96*a*b*d^
3))/((d*x + 1)^(1/2) - 1)^12 + (((1 - d*x)^(1/2) - 1)^8*((256*b*c*d)/3 + 320*a*b*d^3))/((d*x + 1)^(1/2) - 1)^8
+ (((1 - d*x)^(1/2) - 1)^6*((512*b*c*d)/3 + 240*a*b*d^3))/((d*x + 1)^(1/2) - 1)^6 + (((1 - d*x)^(1/2) - 1)^10
*((512*b*c*d)/3 + 240*a*b*d^3))/((d*x + 1)^(1/2) - 1)^10 - (((1 - d*x)^(1/2) - 1)*((3*c^2)/2 + 2*b^2*d^2 + 4*a
*c*d^2))/((d*x + 1)^(1/2) - 1) + (16*a*b*d^3*((1 - d*x)^(1/2) - 1)^2)/((d*x + 1)^(1/2) - 1)^2 + (16*a*b*d^3*((
1 - d*x)^(1/2) - 1)^14)/((d*x + 1)^(1/2) - 1)^14)/(d^5 + (8*d^5*((1 - d*x)^(1/2) - 1)^2)/((d*x + 1)^(1/2) - 1)
^2 + (28*d^5*((1 - d*x)^(1/2) - 1)^4)/((d*x + 1)^(1/2) - 1)^4 + (56*d^5*((1 - d*x)^(1/2) - 1)^6)/((d*x + 1)^(1
/2) - 1)^6 + (70*d^5*((1 - d*x)^(1/2) - 1)^8)/((d*x + 1)^(1/2) - 1)^8 + (56*d^5*((1 - d*x)^(1/2) - 1)^10)/((d*
x + 1)^(1/2) - 1)^10 + (28*d^5*((1 - d*x)^(1/2) - 1)^12)/((d*x + 1)^(1/2) - 1)^12 + (8*d^5*((1 - d*x)^(1/2) -
1)^14)/((d*x + 1)^(1/2) - 1)^14 + (d^5*((1 - d*x)^(1/2) - 1)^16)/((d*x + 1)^(1/2) - 1)^16) - (atan(((1 - d*x)^
(1/2) - 1)/((d*x + 1)^(1/2) - 1))*(3*c^2 + 8*a^2*d^4 + 4*b^2*d^2 + 8*a*c*d^2))/(2*d^5)