3.800 \(\int \frac {1}{(1-d x)^{3/2} (1+d x)^{3/2} (a+b x+c x^2)} \, dx\)

Optimal. Leaf size=443 \[ \frac {c \left (-b d^2 \left (\sqrt {b^2-4 a c}+b\right )+2 a c d^2+2 c^2\right ) \tanh ^{-1}\left (\frac {d^2 x \left (b-\sqrt {b^2-4 a c}\right )+2 c}{\sqrt {2} \sqrt {1-d^2 x^2} \sqrt {-b d^2 \left (b-\sqrt {b^2-4 a c}\right )+2 a c d^2+2 c^2}}\right )}{\sqrt {2} \sqrt {b^2-4 a c} \sqrt {-b d^2 \left (b-\sqrt {b^2-4 a c}\right )+2 a c d^2+2 c^2} \left (b^2 d^2-\left (a d^2+c\right )^2\right )}-\frac {c \left (-b d^2 \left (b-\sqrt {b^2-4 a c}\right )+2 a c d^2+2 c^2\right ) \tanh ^{-1}\left (\frac {d^2 x \left (\sqrt {b^2-4 a c}+b\right )+2 c}{\sqrt {2} \sqrt {1-d^2 x^2} \sqrt {-b d^2 \left (\sqrt {b^2-4 a c}+b\right )+2 a c d^2+2 c^2}}\right )}{\sqrt {2} \sqrt {b^2-4 a c} \sqrt {-b d^2 \left (\sqrt {b^2-4 a c}+b\right )+2 a c d^2+2 c^2} \left (b^2 d^2-\left (a d^2+c\right )^2\right )}+\frac {d^2 \left (b-x \left (a d^2+c\right )\right )}{\sqrt {1-d^2 x^2} \left (b^2 d^2-\left (a d^2+c\right )^2\right )} \]

[Out]

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

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Rubi [A]  time = 1.44, antiderivative size = 443, 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, 976, 1034, 725, 206} \[ \frac {c \left (-b d^2 \left (\sqrt {b^2-4 a c}+b\right )+2 a c d^2+2 c^2\right ) \tanh ^{-1}\left (\frac {d^2 x \left (b-\sqrt {b^2-4 a c}\right )+2 c}{\sqrt {2} \sqrt {1-d^2 x^2} \sqrt {-b d^2 \left (b-\sqrt {b^2-4 a c}\right )+2 a c d^2+2 c^2}}\right )}{\sqrt {2} \sqrt {b^2-4 a c} \sqrt {-b d^2 \left (b-\sqrt {b^2-4 a c}\right )+2 a c d^2+2 c^2} \left (b^2 d^2-\left (a d^2+c\right )^2\right )}-\frac {c \left (-b d^2 \left (b-\sqrt {b^2-4 a c}\right )+2 a c d^2+2 c^2\right ) \tanh ^{-1}\left (\frac {d^2 x \left (\sqrt {b^2-4 a c}+b\right )+2 c}{\sqrt {2} \sqrt {1-d^2 x^2} \sqrt {-b d^2 \left (\sqrt {b^2-4 a c}+b\right )+2 a c d^2+2 c^2}}\right )}{\sqrt {2} \sqrt {b^2-4 a c} \sqrt {-b d^2 \left (\sqrt {b^2-4 a c}+b\right )+2 a c d^2+2 c^2} \left (b^2 d^2-\left (a d^2+c\right )^2\right )}+\frac {d^2 \left (b-x \left (a d^2+c\right )\right )}{\sqrt {1-d^2 x^2} \left (b^2 d^2-\left (a d^2+c\right )^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/((1 - d*x)^(3/2)*(1 + d*x)^(3/2)*(a + b*x + c*x^2)),x]

[Out]

(d^2*(b - (c + a*d^2)*x))/((b^2*d^2 - (c + a*d^2)^2)*Sqrt[1 - d^2*x^2]) + (c*(2*c^2 + 2*a*c*d^2 - b*(b + Sqrt[
b^2 - 4*a*c])*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]*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*d^2 - (c + a*d^2)^2)) - (c*(2*c^2 + 2*a*c*d^2 - b*(b - Sqrt[b^2 - 4*a*c])*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]*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*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 976

Int[((a_.) + (c_.)*(x_)^2)^(p_)*((d_.) + (e_.)*(x_) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp[((2*a*c^2*e + c*(2
*c^2*d - c*(2*a*f))*x)*(a + c*x^2)^(p + 1)*(d + e*x + f*x^2)^(q + 1))/((-4*a*c)*(a*c*e^2 + (c*d - a*f)^2)*(p +
1)), x] - Dist[1/((-4*a*c)*(a*c*e^2 + (c*d - a*f)^2)*(p + 1)), Int[(a + c*x^2)^(p + 1)*(d + e*x + f*x^2)^q*Si
mp[2*c*((c*d - a*f)^2 - (-(a*e))*(c*e))*(p + 1) - (2*c^2*d - c*(2*a*f))*(a*f*(p + 1) - c*d*(p + 2)) - e*(-2*a*
c^2*e)*(p + q + 2) + (2*f*(2*a*c^2*e)*(p + q + 2) - (2*c^2*d - c*(2*a*f))*(-(c*e*(2*p + q + 4))))*x + c*f*(2*c
^2*d - c*(2*a*f))*(2*p + 2*q + 5)*x^2, x], x], x] /; FreeQ[{a, c, d, e, f, q}, x] && NeQ[e^2 - 4*d*f, 0] && Lt
Q[p, -1] && NeQ[a*c*e^2 + (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}{(1-d x)^{3/2} (1+d x)^{3/2} \left (a+b x+c x^2\right )} \, dx &=\int \frac {1}{\left (a+b x+c x^2\right ) \left (1-d^2 x^2\right )^{3/2}} \, dx\\ &=\frac {d^2 \left (b-\left (c+a d^2\right ) x\right )}{\left (b^2 d^2-\left (c+a d^2\right )^2\right ) \sqrt {1-d^2 x^2}}-\frac {\int \frac {2 d^2 \left (c^2-b^2 d^2+a c d^2\right )-2 b c d^4 x}{\left (a+b x+c x^2\right ) \sqrt {1-d^2 x^2}} \, dx}{2 d^2 \left (b^2 d^2-\left (c+a d^2\right )^2\right )}\\ &=\frac {d^2 \left (b-\left (c+a d^2\right ) x\right )}{\left (b^2 d^2-\left (c+a d^2\right )^2\right ) \sqrt {1-d^2 x^2}}+\frac {\left (c \left (2 c^2+2 a c d^2-b \left (b-\sqrt {b^2-4 a c}\right ) d^2\right )\right ) \int \frac {1}{\left (b+\sqrt {b^2-4 a c}+2 c x\right ) \sqrt {1-d^2 x^2}} \, dx}{\sqrt {b^2-4 a c} \left (b^2 d^2-\left (c+a d^2\right )^2\right )}-\frac {\left (c \left (2 c^2+2 a c d^2-b \left (b+\sqrt {b^2-4 a c}\right ) d^2\right )\right ) \int \frac {1}{\left (b-\sqrt {b^2-4 a c}+2 c x\right ) \sqrt {1-d^2 x^2}} \, dx}{\sqrt {b^2-4 a c} \left (b^2 d^2-\left (c+a d^2\right )^2\right )}\\ &=\frac {d^2 \left (b-\left (c+a d^2\right ) x\right )}{\left (b^2 d^2-\left (c+a d^2\right )^2\right ) \sqrt {1-d^2 x^2}}-\frac {\left (c \left (2 c^2+2 a c d^2-b \left (b-\sqrt {b^2-4 a c}\right ) d^2\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 )}{\sqrt {b^2-4 a c} \left (b^2 d^2-\left (c+a d^2\right )^2\right )}+\frac {\left (c \left (2 c^2+2 a c d^2-b \left (b+\sqrt {b^2-4 a c}\right ) d^2\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 )}{\sqrt {b^2-4 a c} \left (b^2 d^2-\left (c+a d^2\right )^2\right )}\\ &=\frac {d^2 \left (b-\left (c+a d^2\right ) x\right )}{\left (b^2 d^2-\left (c+a d^2\right )^2\right ) \sqrt {1-d^2 x^2}}+\frac {c \left (2 c^2+2 a c d^2-b \left (b+\sqrt {b^2-4 a c}\right ) d^2\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} \sqrt {b^2-4 a c} \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 (2 c^2+2 a c d^2-b \left (b-\sqrt {b^2-4 a c}\right ) d^2\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} \sqrt {b^2-4 a c} \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 = 2.39, size = 335, normalized size = 0.76 \[ \frac {d^2 \left (x \left (a d^2+c\right )-b\right )}{\sqrt {1-d^2 x^2} \left (a^2 d^4+2 a c d^2-b^2 d^2+c^2\right )}-\frac {2 \sqrt {2} c^3 \tanh ^{-1}\left (\frac {d^2 x \left (b-\sqrt {b^2-4 a c}\right )+2 c}{\sqrt {1-d^2 x^2} \sqrt {2 b d^2 \left (\sqrt {b^2-4 a c}-b\right )+4 a c d^2+4 c^2}}\right )}{\sqrt {b^2-4 a c} \left (b d^2 \left (\sqrt {b^2-4 a c}-b\right )+2 a c d^2+2 c^2\right )^{3/2}}+\frac {2 \sqrt {2} c^3 \tanh ^{-1}\left (\frac {d^2 x \left (\sqrt {b^2-4 a c}+b\right )+2 c}{\sqrt {1-d^2 x^2} \sqrt {-2 b d^2 \left (\sqrt {b^2-4 a c}+b\right )+4 a c d^2+4 c^2}}\right )}{\sqrt {b^2-4 a c} \left (-b d^2 \left (\sqrt {b^2-4 a c}+b\right )+2 a c d^2+2 c^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((1 - d*x)^(3/2)*(1 + d*x)^(3/2)*(a + b*x + c*x^2)),x]

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-d*x+1)^(3/2)/(d*x+1)^(3/2)/(c*x^2+b*x+a),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-d*x+1)^(3/2)/(d*x+1)^(3/2)/(c*x^2+b*x+a),x, algorithm="giac")

[Out]

Timed out

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maple [C]  time = 0.12, size = 11141, normalized size = 25.15 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-d*x+1)^(3/2)/(d*x+1)^(3/2)/(c*x^2+b*x+a),x)

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (c x^{2} + b x + a\right )} {\left (d x + 1\right )}^{\frac {3}{2}} {\left (-d x + 1\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-d*x+1)^(3/2)/(d*x+1)^(3/2)/(c*x^2+b*x+a),x, algorithm="maxima")

[Out]

integrate(1/((c*x^2 + b*x + a)*(d*x + 1)^(3/2)*(-d*x + 1)^(3/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (1-d\,x\right )}^{3/2}\,{\left (d\,x+1\right )}^{3/2}\,\left (c\,x^2+b\,x+a\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((1 - d*x)^(3/2)*(d*x + 1)^(3/2)*(a + b*x + c*x^2)),x)

[Out]

int(1/((1 - d*x)^(3/2)*(d*x + 1)^(3/2)*(a + b*x + c*x^2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-d*x+1)**(3/2)/(d*x+1)**(3/2)/(c*x**2+b*x+a),x)

[Out]

Timed out

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