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

Optimal. Leaf size=571 \[ -\frac {c \left (-c d^2 \left (-8 a^2 d^2-b \sqrt {b^2-4 a c}+5 b^2\right )-a b d^4 \left (\sqrt {b^2-4 a c}+b\right )+12 a c^2 d^2+4 c^3\right ) \tanh ^{-1}\left (\frac {d^2 x \left (b-\sqrt {b^2-4 a c}\right )+2 c}{\sqrt {2} \sqrt {1-d^2 x^2} \sqrt {-b d^2 \left (b-\sqrt {b^2-4 a c}\right )+2 a c d^2+2 c^2}}\right )}{\sqrt {2} \left (b^2-4 a c\right )^{3/2} \sqrt {-b d^2 \left (b-\sqrt {b^2-4 a c}\right )+2 a c d^2+2 c^2} \left (b^2 d^2-\left (a d^2+c\right )^2\right )}+\frac {c \left (-4 c d^2 \left (b^2-2 a^2 d^2\right )-b d^2 \left (\sqrt {b^2-4 a c}+b\right ) \left (c-a d^2\right )-2 a b^2 d^4+12 a c^2 d^2+4 c^3\right ) \tanh ^{-1}\left (\frac {d^2 x \left (\sqrt {b^2-4 a c}+b\right )+2 c}{\sqrt {2} \sqrt {1-d^2 x^2} \sqrt {-b d^2 \left (\sqrt {b^2-4 a c}+b\right )+2 a c d^2+2 c^2}}\right )}{\sqrt {2} \left (b^2-4 a c\right )^{3/2} \sqrt {-b d^2 \left (\sqrt {b^2-4 a c}+b\right )+2 a c d^2+2 c^2} \left (b^2 d^2-\left (a d^2+c\right )^2\right )}-\frac {\sqrt {1-d^2 x^2} \left (b \left (b^2 d^2-c \left (3 a d^2+c\right )\right )-c x \left (2 a c d^2-b^2 d^2+2 c^2\right )\right )}{\left (b^2-4 a c\right ) \left (b^2 d^2-\left (a d^2+c\right )^2\right ) \left (a+b x+c x^2\right )} \]

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Rubi [A]  time = 5.23, antiderivative size = 571, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {899, 975, 1034, 725, 206} \begin {gather*} -\frac {c \left (-c d^2 \left (-8 a^2 d^2-b \sqrt {b^2-4 a c}+5 b^2\right )-a b d^4 \left (\sqrt {b^2-4 a c}+b\right )+12 a c^2 d^2+4 c^3\right ) \tanh ^{-1}\left (\frac {d^2 x \left (b-\sqrt {b^2-4 a c}\right )+2 c}{\sqrt {2} \sqrt {1-d^2 x^2} \sqrt {-b d^2 \left (b-\sqrt {b^2-4 a c}\right )+2 a c d^2+2 c^2}}\right )}{\sqrt {2} \left (b^2-4 a c\right )^{3/2} \sqrt {-b d^2 \left (b-\sqrt {b^2-4 a c}\right )+2 a c d^2+2 c^2} \left (b^2 d^2-\left (a d^2+c\right )^2\right )}+\frac {c \left (-4 c d^2 \left (b^2-2 a^2 d^2\right )-b d^2 \left (\sqrt {b^2-4 a c}+b\right ) \left (c-a d^2\right )-2 a b^2 d^4+12 a c^2 d^2+4 c^3\right ) \tanh ^{-1}\left (\frac {d^2 x \left (\sqrt {b^2-4 a c}+b\right )+2 c}{\sqrt {2} \sqrt {1-d^2 x^2} \sqrt {-b d^2 \left (\sqrt {b^2-4 a c}+b\right )+2 a c d^2+2 c^2}}\right )}{\sqrt {2} \left (b^2-4 a c\right )^{3/2} \sqrt {-b d^2 \left (\sqrt {b^2-4 a c}+b\right )+2 a c d^2+2 c^2} \left (b^2 d^2-\left (a d^2+c\right )^2\right )}-\frac {\sqrt {1-d^2 x^2} \left (b \left (b^2 d^2-c \left (3 a d^2+c\right )\right )-c x \left (2 a c d^2-b^2 d^2+2 c^2\right )\right )}{\left (b^2-4 a c\right ) \left (b^2 d^2-\left (a d^2+c\right )^2\right ) \left (a+b x+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 206

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

Rule 725

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

Rule 899

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 975

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_.) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp[((b^3*f + b*c*(c*d
- 3*a*f) + c*(2*c^2*d + b^2*f - c*(2*a*f))*x)*(a + b*x + c*x^2)^(p + 1)*(d + f*x^2)^(q + 1))/((b^2 - 4*a*c)*(
b^2*d*f + (c*d - a*f)^2)*(p + 1)), x] - Dist[1/((b^2 - 4*a*c)*(b^2*d*f + (c*d - a*f)^2)*(p + 1)), Int[(a + b*x
+ c*x^2)^(p + 1)*(d + f*x^2)^q*Simp[2*c*(b^2*d*f + (c*d - a*f)^2)*(p + 1) - (2*c^2*d + b^2*f - c*(2*a*f))*(a*
f*(p + 1) - c*d*(p + 2)) + (2*f*(b^3*f + b*c*(c*d - 3*a*f))*(p + q + 2) - (2*c^2*d + b^2*f - c*(2*a*f))*(b*f*(
p + 1)))*x + c*f*(2*c^2*d + b^2*f - c*(2*a*f))*(2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, b, c, d, f, q}, x]
&& NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[b^2*d*f + (c*d - a*f)^2, 0] &&  !( !IntegerQ[p] && ILtQ[q, -1]) &
&  !IGtQ[q, 0]

Rule 1034

Int[((g_.) + (h_.)*(x_))/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*Sqrt[(d_) + (f_.)*(x_)^2]), x_Symbol] :> With[{q
= Rt[b^2 - 4*a*c, 2]}, Dist[(2*c*g - h*(b - q))/q, Int[1/((b - q + 2*c*x)*Sqrt[d + f*x^2]), x], x] - Dist[(2*c
*g - h*(b + q))/q, Int[1/((b + q + 2*c*x)*Sqrt[d + f*x^2]), x], x]] /; FreeQ[{a, b, c, d, f, g, h}, x] && NeQ[
b^2 - 4*a*c, 0] && PosQ[b^2 - 4*a*c]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {1-d x} \sqrt {1+d x} \left (a+b x+c x^2\right )^2} \, dx &=\int \frac {1}{\left (a+b x+c x^2\right )^2 \sqrt {1-d^2 x^2}} \, dx\\ &=-\frac {\left (b \left (b^2 d^2-c \left (c+3 a d^2\right )\right )-c \left (2 c^2-b^2 d^2+2 a c d^2\right ) x\right ) \sqrt {1-d^2 x^2}}{\left (b^2-4 a c\right ) \left (b^2 d^2-\left (c+a d^2\right )^2\right ) \left (a+b x+c x^2\right )}-\frac {\int \frac {-2 c^3-6 a c^2 d^2+a b^2 d^4+2 c d^2 \left (b^2-2 a^2 d^2\right )-b c d^2 \left (c-a d^2\right ) x}{\left (a+b x+c x^2\right ) \sqrt {1-d^2 x^2}} \, dx}{\left (b^2-4 a c\right ) \left (b^2 d^2-\left (c+a d^2\right )^2\right )}\\ &=-\frac {\left (b \left (b^2 d^2-c \left (c+3 a d^2\right )\right )-c \left (2 c^2-b^2 d^2+2 a c d^2\right ) x\right ) \sqrt {1-d^2 x^2}}{\left (b^2-4 a c\right ) \left (b^2 d^2-\left (c+a d^2\right )^2\right ) \left (a+b x+c x^2\right )}+\frac {\left (c \left (4 c^3+12 a c^2 d^2-a b \left (b+\sqrt {b^2-4 a c}\right ) d^4-c d^2 \left (5 b^2-b \sqrt {b^2-4 a c}-8 a^2 d^2\right )\right )\right ) \int \frac {1}{\left (b-\sqrt {b^2-4 a c}+2 c x\right ) \sqrt {1-d^2 x^2}} \, dx}{\left (b^2-4 a c\right )^{3/2} \left (b^2 d^2-\left (c+a d^2\right )^2\right )}+\frac {\left (b c \left (b+\sqrt {b^2-4 a c}\right ) d^2 \left (c-a d^2\right )+2 c \left (-2 c^3-6 a c^2 d^2+a b^2 d^4+2 c d^2 \left (b^2-2 a^2 d^2\right )\right )\right ) \int \frac {1}{\left (b+\sqrt {b^2-4 a c}+2 c x\right ) \sqrt {1-d^2 x^2}} \, dx}{\left (b^2-4 a c\right )^{3/2} \left (b^2 d^2-\left (c+a d^2\right )^2\right )}\\ &=-\frac {\left (b \left (b^2 d^2-c \left (c+3 a d^2\right )\right )-c \left (2 c^2-b^2 d^2+2 a c d^2\right ) x\right ) \sqrt {1-d^2 x^2}}{\left (b^2-4 a c\right ) \left (b^2 d^2-\left (c+a d^2\right )^2\right ) \left (a+b x+c x^2\right )}-\frac {\left (c \left (4 c^3+12 a c^2 d^2-a b \left (b+\sqrt {b^2-4 a c}\right ) d^4-c d^2 \left (5 b^2-b \sqrt {b^2-4 a c}-8 a^2 d^2\right )\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 c^2-\left (b-\sqrt {b^2-4 a c}\right )^2 d^2-x^2} \, dx,x,\frac {2 c+\left (b-\sqrt {b^2-4 a c}\right ) d^2 x}{\sqrt {1-d^2 x^2}}\right )}{\left (b^2-4 a c\right )^{3/2} \left (b^2 d^2-\left (c+a d^2\right )^2\right )}-\frac {\left (b c \left (b+\sqrt {b^2-4 a c}\right ) d^2 \left (c-a d^2\right )+2 c \left (-2 c^3-6 a c^2 d^2+a b^2 d^4+2 c d^2 \left (b^2-2 a^2 d^2\right )\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 c^2-\left (b+\sqrt {b^2-4 a c}\right )^2 d^2-x^2} \, dx,x,\frac {2 c+\left (b+\sqrt {b^2-4 a c}\right ) d^2 x}{\sqrt {1-d^2 x^2}}\right )}{\left (b^2-4 a c\right )^{3/2} \left (b^2 d^2-\left (c+a d^2\right )^2\right )}\\ &=-\frac {\left (b \left (b^2 d^2-c \left (c+3 a d^2\right )\right )-c \left (2 c^2-b^2 d^2+2 a c d^2\right ) x\right ) \sqrt {1-d^2 x^2}}{\left (b^2-4 a c\right ) \left (b^2 d^2-\left (c+a d^2\right )^2\right ) \left (a+b x+c x^2\right )}-\frac {c \left (4 c^3+12 a c^2 d^2-a b \left (b+\sqrt {b^2-4 a c}\right ) d^4-c d^2 \left (5 b^2-b \sqrt {b^2-4 a c}-8 a^2 d^2\right )\right ) \tanh ^{-1}\left (\frac {2 c+\left (b-\sqrt {b^2-4 a c}\right ) d^2 x}{\sqrt {2} \sqrt {2 c^2+2 a c d^2-b \left (b-\sqrt {b^2-4 a c}\right ) d^2} \sqrt {1-d^2 x^2}}\right )}{\sqrt {2} \left (b^2-4 a c\right )^{3/2} \sqrt {2 c^2+2 a c d^2-b \left (b-\sqrt {b^2-4 a c}\right ) d^2} \left (b^2 d^2-\left (c+a d^2\right )^2\right )}-\frac {c \left (b \left (b+\sqrt {b^2-4 a c}\right ) d^2 \left (c-a d^2\right )-2 \left (2 c^3+6 a c^2 d^2-a b^2 d^4-2 c d^2 \left (b^2-2 a^2 d^2\right )\right )\right ) \tanh ^{-1}\left (\frac {2 c+\left (b+\sqrt {b^2-4 a c}\right ) d^2 x}{\sqrt {2} \sqrt {2 c^2+2 a c d^2-b \left (b+\sqrt {b^2-4 a c}\right ) d^2} \sqrt {1-d^2 x^2}}\right )}{\sqrt {2} \left (b^2-4 a c\right )^{3/2} \sqrt {2 c^2+2 a c d^2-b \left (b+\sqrt {b^2-4 a c}\right ) d^2} \left (b^2 d^2-\left (c+a d^2\right )^2\right )}\\ \end {align*}

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Mathematica [A]  time = 1.28, size = 508, normalized size = 0.89 \begin {gather*} \frac {\frac {c \left (c d^2 \left (8 a^2 d^2+b \sqrt {b^2-4 a c}-5 b^2\right )-a b d^4 \left (\sqrt {b^2-4 a c}+b\right )+12 a c^2 d^2+4 c^3\right ) \tanh ^{-1}\left (\frac {d^2 x \left (b-\sqrt {b^2-4 a c}\right )+2 c}{\sqrt {1-d^2 x^2} \sqrt {2 b d^2 \left (\sqrt {b^2-4 a c}-b\right )+4 a c d^2+4 c^2}}\right )}{\sqrt {2} \sqrt {b^2-4 a c} \sqrt {b d^2 \left (\sqrt {b^2-4 a c}-b\right )+2 a c d^2+2 c^2}}+\frac {c \left (c d^2 \left (-8 a^2 d^2+b \sqrt {b^2-4 a c}+5 b^2\right )+a b d^4 \left (b-\sqrt {b^2-4 a c}\right )-12 a c^2 d^2-4 c^3\right ) \tanh ^{-1}\left (\frac {d^2 x \left (\sqrt {b^2-4 a c}+b\right )+2 c}{\sqrt {1-d^2 x^2} \sqrt {-2 b d^2 \left (\sqrt {b^2-4 a c}+b\right )+4 a c d^2+4 c^2}}\right )}{\sqrt {2} \sqrt {b^2-4 a c} \sqrt {-b d^2 \left (\sqrt {b^2-4 a c}+b\right )+2 a c d^2+2 c^2}}+\frac {\sqrt {1-d^2 x^2} \left (-b c \left (3 a d^2+c\right )-2 c^2 x \left (a d^2+c\right )+b^3 d^2+b^2 c d^2 x\right )}{a+x (b+c x)}}{\left (b^2-4 a c\right ) \left (\left (a d^2+c\right )^2-b^2 d^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((b^3*d^2 - b*c*(c + 3*a*d^2) + b^2*c*d^2*x - 2*c^2*(c + a*d^2)*x)*Sqrt[1 - d^2*x^2])/(a + x*(b + c*x)) + (c*
(4*c^3 + 12*a*c^2*d^2 - a*b*(b + Sqrt[b^2 - 4*a*c])*d^4 + c*d^2*(-5*b^2 + b*Sqrt[b^2 - 4*a*c] + 8*a^2*d^2))*Ar
cTanh[(2*c + (b - Sqrt[b^2 - 4*a*c])*d^2*x)/(Sqrt[4*c^2 + 4*a*c*d^2 + 2*b*(-b + Sqrt[b^2 - 4*a*c])*d^2]*Sqrt[1
- d^2*x^2])])/(Sqrt[2]*Sqrt[b^2 - 4*a*c]*Sqrt[2*c^2 + 2*a*c*d^2 + b*(-b + Sqrt[b^2 - 4*a*c])*d^2]) + (c*(-4*c
^3 - 12*a*c^2*d^2 + a*b*(b - Sqrt[b^2 - 4*a*c])*d^4 + c*d^2*(5*b^2 + b*Sqrt[b^2 - 4*a*c] - 8*a^2*d^2))*ArcTanh
[(2*c + (b + Sqrt[b^2 - 4*a*c])*d^2*x)/(Sqrt[4*c^2 + 4*a*c*d^2 - 2*b*(b + Sqrt[b^2 - 4*a*c])*d^2]*Sqrt[1 - d^2
*x^2])])/(Sqrt[2]*Sqrt[b^2 - 4*a*c]*Sqrt[2*c^2 + 2*a*c*d^2 - b*(b + Sqrt[b^2 - 4*a*c])*d^2]))/((b^2 - 4*a*c)*(
-(b^2*d^2) + (c + a*d^2)^2))

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IntegrateAlgebraic [A]  time = 50.06, size = 1064, normalized size = 1.86 \begin {gather*} \frac {2 d \sqrt {1-d x} \left (\frac {2 (1-d x) c^3}{d x+1}-2 c^3-2 a d^2 c^2-b d c^2+\frac {2 a d^2 (1-d x) c^2}{d x+1}-\frac {b d (1-d x) c^2}{d x+1}-3 a b d^3 c+b^2 d^2 c-\frac {3 a b d^3 (1-d x) c}{d x+1}-\frac {b^2 d^2 (1-d x) c}{d x+1}+b^3 d^3+\frac {b^3 d^3 (1-d x)}{d x+1}\right )}{\left (b^2-4 a c\right ) \left (-a d^2+b d-c\right ) \left (a d^2+b d+c\right ) \sqrt {d x+1} \left (-a d^2-\frac {2 a (1-d x) d^2}{d x+1}-\frac {a (1-d x)^2 d^2}{(d x+1)^2}-b d+\frac {b (1-d x)^2 d}{(d x+1)^2}-c+\frac {2 c (1-d x)}{d x+1}-\frac {c (1-d x)^2}{(d x+1)^2}\right )}+\frac {\left (a b^3 d^5-6 a^2 b c d^5-a b^2 \sqrt {b^2-4 a c} d^5+4 a^2 c \sqrt {b^2-4 a c} d^5+8 a^2 c^2 d^4-a b^2 c d^4+a b c \sqrt {b^2-4 a c} d^4-4 a b c^2 d^3+2 b^3 c d^3+6 a c^2 \sqrt {b^2-4 a c} d^3-2 b^2 c \sqrt {b^2-4 a c} d^3+12 a c^3 d^2-5 b^2 c^2 d^2-b c^2 \sqrt {b^2-4 a c} d^2-2 b c^3 d+2 c^3 \sqrt {b^2-4 a c} d+4 c^4\right ) \tan ^{-1}\left (\frac {\sqrt {a d^2-b d+c} \sqrt {1-d x}}{\sqrt {a d^2-\sqrt {b^2-4 a c} d-c} \sqrt {d x+1}}\right )}{\left (b^2-4 a c\right )^{3/2} \left (a d^2-b d+c\right )^{3/2} \left (a d^2+b d+c\right ) \sqrt {a d^2-\sqrt {b^2-4 a c} d-c}}+\frac {\left (-a b^3 d^5+6 a^2 b c d^5-a b^2 \sqrt {b^2-4 a c} d^5+4 a^2 c \sqrt {b^2-4 a c} d^5-8 a^2 c^2 d^4+a b^2 c d^4+a b c \sqrt {b^2-4 a c} d^4+4 a b c^2 d^3-2 b^3 c d^3+6 a c^2 \sqrt {b^2-4 a c} d^3-2 b^2 c \sqrt {b^2-4 a c} d^3-12 a c^3 d^2+5 b^2 c^2 d^2-b c^2 \sqrt {b^2-4 a c} d^2+2 b c^3 d+2 c^3 \sqrt {b^2-4 a c} d-4 c^4\right ) \tan ^{-1}\left (\frac {\sqrt {a d^2-b d+c} \sqrt {1-d x}}{\sqrt {a d^2+\sqrt {b^2-4 a c} d-c} \sqrt {d x+1}}\right )}{\left (b^2-4 a c\right )^{3/2} \left (a d^2-b d+c\right )^{3/2} \left (a d^2+b d+c\right ) \sqrt {a d^2+\sqrt {b^2-4 a c} d-c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [C]  time = 0.81, size = 41837, normalized size = 73.27 \begin {gather*} \text {output too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (c x^{2} + b x + a\right )}^{2} \sqrt {d x + 1} \sqrt {-d x + 1}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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