Integrand size = 32, antiderivative size = 443 \[ \int \frac {1}{(1-d x)^{3/2} (1+d x)^{3/2} \left (a+b x+c x^2\right )} \, dx=\frac {d^2 \left (b-\left (c+a d^2\right ) x\right )}{\left (b^2 d^2-\left (c+a d^2\right )^2\right ) \sqrt {1-d^2 x^2}}+\frac {c \left (2 c^2+2 a c d^2-b \left (b+\sqrt {b^2-4 a c}\right ) d^2\right ) \text {arctanh}\left (\frac {2 c+\left (b-\sqrt {b^2-4 a c}\right ) d^2 x}{\sqrt {2} \sqrt {2 c^2+2 a c d^2-b \left (b-\sqrt {b^2-4 a c}\right ) d^2} \sqrt {1-d^2 x^2}}\right )}{\sqrt {2} \sqrt {b^2-4 a c} \sqrt {2 c^2+2 a c d^2-b \left (b-\sqrt {b^2-4 a c}\right ) d^2} \left (b^2 d^2-\left (c+a d^2\right )^2\right )}-\frac {c \left (2 c^2+2 a c d^2-b \left (b-\sqrt {b^2-4 a c}\right ) d^2\right ) \text {arctanh}\left (\frac {2 c+\left (b+\sqrt {b^2-4 a c}\right ) d^2 x}{\sqrt {2} \sqrt {2 c^2+2 a c d^2-b \left (b+\sqrt {b^2-4 a c}\right ) d^2} \sqrt {1-d^2 x^2}}\right )}{\sqrt {2} \sqrt {b^2-4 a c} \sqrt {2 c^2+2 a c d^2-b \left (b+\sqrt {b^2-4 a c}\right ) d^2} \left (b^2 d^2-\left (c+a d^2\right )^2\right )} \]
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)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.56 (sec) , antiderivative size = 456, normalized size of antiderivative = 1.03 \[ \int \frac {1}{(1-d x)^{3/2} (1+d x)^{3/2} \left (a+b x+c x^2\right )} \, dx=\frac {d^2 \left (b-\left (c+a d^2\right ) x\right ) \sqrt {1-d^2 x^2}+\left (1-d^2 x^2\right ) \text {RootSum}\left [a d^4-2 b d^2 \text {$\#$1}+4 c \text {$\#$1}^2+2 a d^2 \text {$\#$1}^2-2 b \text {$\#$1}^3+a \text {$\#$1}^4\&,\frac {-c^2 d^2 \log (x)+b^2 d^4 \log (x)-a c d^4 \log (x)+c^2 d^2 \log \left (-1+\sqrt {1-d^2 x^2}-x \text {$\#$1}\right )-b^2 d^4 \log \left (-1+\sqrt {1-d^2 x^2}-x \text {$\#$1}\right )+a c d^4 \log \left (-1+\sqrt {1-d^2 x^2}-x \text {$\#$1}\right )-2 b c d^2 \log (x) \text {$\#$1}+2 b c d^2 \log \left (-1+\sqrt {1-d^2 x^2}-x \text {$\#$1}\right ) \text {$\#$1}-c^2 \log (x) \text {$\#$1}^2+b^2 d^2 \log (x) \text {$\#$1}^2-a c d^2 \log (x) \text {$\#$1}^2+c^2 \log \left (-1+\sqrt {1-d^2 x^2}-x \text {$\#$1}\right ) \text {$\#$1}^2-b^2 d^2 \log \left (-1+\sqrt {1-d^2 x^2}-x \text {$\#$1}\right ) \text {$\#$1}^2+a c d^2 \log \left (-1+\sqrt {1-d^2 x^2}-x \text {$\#$1}\right ) \text {$\#$1}^2}{-b d^2+4 c \text {$\#$1}+2 a d^2 \text {$\#$1}-3 b \text {$\#$1}^2+2 a \text {$\#$1}^3}\&\right ]}{(c+d (-b+a d)) (c+d (b+a d)) (-1+d x) (1+d x)} \]
(d^2*(b - (c + a*d^2)*x)*Sqrt[1 - d^2*x^2] + (1 - d^2*x^2)*RootSum[a*d^4 - 2*b*d^2*#1 + 4*c*#1^2 + 2*a*d^2*#1^2 - 2*b*#1^3 + a*#1^4 & , (-(c^2*d^2*L og[x]) + b^2*d^4*Log[x] - a*c*d^4*Log[x] + c^2*d^2*Log[-1 + Sqrt[1 - d^2*x ^2] - x*#1] - b^2*d^4*Log[-1 + Sqrt[1 - d^2*x^2] - x*#1] + a*c*d^4*Log[-1 + Sqrt[1 - d^2*x^2] - x*#1] - 2*b*c*d^2*Log[x]*#1 + 2*b*c*d^2*Log[-1 + Sqr t[1 - d^2*x^2] - x*#1]*#1 - c^2*Log[x]*#1^2 + b^2*d^2*Log[x]*#1^2 - a*c*d^ 2*Log[x]*#1^2 + c^2*Log[-1 + Sqrt[1 - d^2*x^2] - x*#1]*#1^2 - b^2*d^2*Log[ -1 + Sqrt[1 - d^2*x^2] - x*#1]*#1^2 + a*c*d^2*Log[-1 + Sqrt[1 - d^2*x^2] - x*#1]*#1^2)/(-(b*d^2) + 4*c*#1 + 2*a*d^2*#1 - 3*b*#1^2 + 2*a*#1^3) & ])/( (c + d*(-b + a*d))*(c + d*(b + a*d))*(-1 + d*x)*(1 + d*x))
Time = 0.72 (sec) , antiderivative size = 425, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {1188, 1307, 27, 1367, 488, 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.
| \(\displaystyle \int \frac {1}{(1-d x)^{3/2} (d x+1)^{3/2} \left (a+b x+c x^2\right )} \, dx\) |
| \(\Big \downarrow \) 1188 |
| \(\displaystyle \int \frac {1}{\left (1-d^2 x^2\right )^{3/2} \left (a+b x+c x^2\right )}dx\) |
| \(\Big \downarrow \) 1307 |
| \(\displaystyle \frac {d^2 \left (b-x \left (a d^2+c\right )\right )}{\sqrt {1-d^2 x^2} \left (b^2 d^2-\left (a d^2+c\right )^2\right )}-\frac {\int \frac {2 d^2 \left (c^2+a d^2 c-b d^2 x c-b^2 d^2\right )}{\left (c x^2+b x+a\right ) \sqrt {1-d^2 x^2}}dx}{2 d^2 \left (b^2 d^2-\left (a d^2+c\right )^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^2 \left (b-x \left (a d^2+c\right )\right )}{\sqrt {1-d^2 x^2} \left (b^2 d^2-\left (a d^2+c\right )^2\right )}-\frac {\int \frac {c^2+a d^2 c-b d^2 x c-b^2 d^2}{\left (c x^2+b x+a\right ) \sqrt {1-d^2 x^2}}dx}{b^2 d^2-\left (a d^2+c\right )^2}\) |
| \(\Big \downarrow \) 1367 |
| \(\displaystyle \frac {d^2 \left (b-x \left (a d^2+c\right )\right )}{\sqrt {1-d^2 x^2} \left (b^2 d^2-\left (a d^2+c\right )^2\right )}-\frac {\frac {c \left (-b d^2 \left (\sqrt {b^2-4 a c}+b\right )+2 a c d^2+2 c^2\right ) \int \frac {1}{\left (b+2 c x-\sqrt {b^2-4 a c}\right ) \sqrt {1-d^2 x^2}}dx}{\sqrt {b^2-4 a c}}-\frac {c \left (-b d^2 \left (b-\sqrt {b^2-4 a c}\right )+2 a c d^2+2 c^2\right ) \int \frac {1}{\left (b+2 c x+\sqrt {b^2-4 a c}\right ) \sqrt {1-d^2 x^2}}dx}{\sqrt {b^2-4 a c}}}{b^2 d^2-\left (a d^2+c\right )^2}\) |
| \(\Big \downarrow \) 488 |
| \(\displaystyle \frac {d^2 \left (b-x \left (a d^2+c\right )\right )}{\sqrt {1-d^2 x^2} \left (b^2 d^2-\left (a d^2+c\right )^2\right )}-\frac {\frac {c \left (-b d^2 \left (b-\sqrt {b^2-4 a c}\right )+2 a c d^2+2 c^2\right ) \int \frac {1}{4 c^2-\left (b+\sqrt {b^2-4 a c}\right )^2 d^2-\frac {\left (\left (b+\sqrt {b^2-4 a c}\right ) x d^2+2 c\right )^2}{1-d^2 x^2}}d\frac {\left (b+\sqrt {b^2-4 a c}\right ) x d^2+2 c}{\sqrt {1-d^2 x^2}}}{\sqrt {b^2-4 a c}}-\frac {c \left (-b d^2 \left (\sqrt {b^2-4 a c}+b\right )+2 a c d^2+2 c^2\right ) \int \frac {1}{4 c^2-\left (b-\sqrt {b^2-4 a c}\right )^2 d^2-\frac {\left (\left (b-\sqrt {b^2-4 a c}\right ) x d^2+2 c\right )^2}{1-d^2 x^2}}d\frac {\left (b-\sqrt {b^2-4 a c}\right ) x d^2+2 c}{\sqrt {1-d^2 x^2}}}{\sqrt {b^2-4 a c}}}{b^2 d^2-\left (a d^2+c\right )^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {d^2 \left (b-x \left (a d^2+c\right )\right )}{\sqrt {1-d^2 x^2} \left (b^2 d^2-\left (a d^2+c\right )^2\right )}-\frac {\frac {c \left (-b d^2 \left (b-\sqrt {b^2-4 a c}\right )+2 a c d^2+2 c^2\right ) \text {arctanh}\left (\frac {d^2 x \left (\sqrt {b^2-4 a c}+b\right )+2 c}{\sqrt {2} \sqrt {1-d^2 x^2} \sqrt {-b d^2 \left (\sqrt {b^2-4 a c}+b\right )+2 a c d^2+2 c^2}}\right )}{\sqrt {2} \sqrt {b^2-4 a c} \sqrt {-b d^2 \left (\sqrt {b^2-4 a c}+b\right )+2 a c d^2+2 c^2}}-\frac {c \left (-b d^2 \left (\sqrt {b^2-4 a c}+b\right )+2 a c d^2+2 c^2\right ) \text {arctanh}\left (\frac {d^2 x \left (b-\sqrt {b^2-4 a c}\right )+2 c}{\sqrt {2} \sqrt {1-d^2 x^2} \sqrt {-b d^2 \left (b-\sqrt {b^2-4 a c}\right )+2 a c d^2+2 c^2}}\right )}{\sqrt {2} \sqrt {b^2-4 a c} \sqrt {-b d^2 \left (b-\sqrt {b^2-4 a c}\right )+2 a c d^2+2 c^2}}}{b^2 d^2-\left (a d^2+c\right )^2}\) |
(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 - Sqr t[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*(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]*S qrt[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)
3.8.100.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
Int[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> -Subst[ Int[1/(b*c^2 + a*d^2 - x^2), x], x, (a*d - b*c*x)/Sqrt[a + b*x^2]] /; FreeQ [{a, b, c, d}, x]
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] && EqQ[e*f + d*g, 0] && (IntegerQ[m] || (GtQ[d, 0] && GtQ[f, 0]))
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] - Simp[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*Simp[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] && LtQ[p, -1] && Ne Q[a*c*e^2 + (c*d - a*f)^2, 0] && !( !IntegerQ[p] && ILtQ[q, -1]) && !IGtQ [q, 0]
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]}, Simp[(2*c*g - h*( b - q))/q Int[1/((b - q + 2*c*x)*Sqrt[d + f*x^2]), x], x] - Simp[(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]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.61 (sec) , antiderivative size = 11142, normalized size of antiderivative = 25.15
Leaf count of result is larger than twice the leaf count of optimal. 21628 vs. \(2 (404) = 808\).
Time = 6.57 (sec) , antiderivative size = 21628, normalized size of antiderivative = 48.82 \[ \int \frac {1}{(1-d x)^{3/2} (1+d x)^{3/2} \left (a+b x+c x^2\right )} \, dx=\text {Too large to display} \]
\[ \int \frac {1}{(1-d x)^{3/2} (1+d x)^{3/2} \left (a+b x+c x^2\right )} \, dx=\int \frac {1}{\left (- d x + 1\right )^{\frac {3}{2}} \left (d x + 1\right )^{\frac {3}{2}} \left (a + b x + c x^{2}\right )}\, dx \]
\[ \int \frac {1}{(1-d x)^{3/2} (1+d x)^{3/2} \left (a+b x+c x^2\right )} \, dx=\int { \frac {1}{{\left (c x^{2} + b x + a\right )} {\left (d x + 1\right )}^{\frac {3}{2}} {\left (-d x + 1\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{(1-d x)^{3/2} (1+d x)^{3/2} \left (a+b x+c x^2\right )} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{(1-d x)^{3/2} (1+d x)^{3/2} \left (a+b x+c x^2\right )} \, dx=\int \frac {1}{{\left (1-d\,x\right )}^{3/2}\,{\left (d\,x+1\right )}^{3/2}\,\left (c\,x^2+b\,x+a\right )} \,d x \]