Integrand size = 37, antiderivative size = 248 \[ \int \frac {A+B x+C x^2}{\sqrt {1-d x} \sqrt {1+d x} (e+f x)^3} \, dx=\frac {\left (C e^2-B e f+A f^2\right ) \sqrt {1-d^2 x^2}}{2 f \left (d^2 e^2-f^2\right ) (e+f x)^2}-\frac {\left (C d^2 e^3+B d^2 e^2 f-4 C e f^2-3 A d^2 e f^2+2 B f^3\right ) \sqrt {1-d^2 x^2}}{2 f \left (d^2 e^2-f^2\right )^2 (e+f x)}+\frac {\left (C \left (d^2 e^2+2 f^2\right )-d^2 \left (3 B e f-A \left (2 d^2 e^2+f^2\right )\right )\right ) \arctan \left (\frac {f+d^2 e x}{\sqrt {d^2 e^2-f^2} \sqrt {1-d^2 x^2}}\right )}{2 \left (d^2 e^2-f^2\right )^{5/2}} \] Output:
1/2*(A*f^2-B*e*f+C*e^2)*(-d^2*x^2+1)^(1/2)/f/(d^2*e^2-f^2)/(f*x+e)^2-1/2*( -3*A*d^2*e*f^2+B*d^2*e^2*f+C*d^2*e^3+2*B*f^3-4*C*e*f^2)*(-d^2*x^2+1)^(1/2) /f/(d^2*e^2-f^2)^2/(f*x+e)+1/2*(C*(d^2*e^2+2*f^2)-d^2*(3*B*e*f-A*(2*d^2*e^ 2+f^2)))*arctan((d^2*e*x+f)/(d^2*e^2-f^2)^(1/2)/(-d^2*x^2+1)^(1/2))/(d^2*e ^2-f^2)^(5/2)
Time = 1.23 (sec) , antiderivative size = 228, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x+C x^2}{\sqrt {1-d x} \sqrt {1+d x} (e+f x)^3} \, dx=-\frac {\frac {(d e-f) (d e+f) \sqrt {1-d^2 x^2} \left (A f^3+B d^2 e^2 (2 e+f x)+B f^2 (e+2 f x)-A d^2 e f (4 e+3 f x)+C e \left (-3 e f+d^2 e^2 x-4 f^2 x\right )\right )}{(e+f x)^2}+2 \sqrt {d^2 e^2-f^2} \left (C \left (d^2 e^2+2 f^2\right )+d^2 \left (-3 B e f+A \left (2 d^2 e^2+f^2\right )\right )\right ) \arctan \left (\frac {\sqrt {d^2 e^2-f^2} x}{e+f x-e \sqrt {1-d^2 x^2}}\right )}{2 (d e-f)^3 (d e+f)^3} \] Input:
Integrate[(A + B*x + C*x^2)/(Sqrt[1 - d*x]*Sqrt[1 + d*x]*(e + f*x)^3),x]
Output:
-1/2*(((d*e - f)*(d*e + f)*Sqrt[1 - d^2*x^2]*(A*f^3 + B*d^2*e^2*(2*e + f*x ) + B*f^2*(e + 2*f*x) - A*d^2*e*f*(4*e + 3*f*x) + C*e*(-3*e*f + d^2*e^2*x - 4*f^2*x)))/(e + f*x)^2 + 2*Sqrt[d^2*e^2 - f^2]*(C*(d^2*e^2 + 2*f^2) + d^ 2*(-3*B*e*f + A*(2*d^2*e^2 + f^2)))*ArcTan[(Sqrt[d^2*e^2 - f^2]*x)/(e + f* x - e*Sqrt[1 - d^2*x^2])])/((d*e - f)^3*(d*e + f)^3)
Time = 0.61 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {2112, 2182, 679, 488, 217}
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.
\(\displaystyle \int \frac {A+B x+C x^2}{\sqrt {1-d x} \sqrt {d x+1} (e+f x)^3} \, dx\) |
\(\Big \downarrow \) 2112 |
\(\displaystyle \int \frac {A+B x+C x^2}{\sqrt {1-d^2 x^2} (e+f x)^3}dx\) |
\(\Big \downarrow \) 2182 |
\(\displaystyle \frac {\int \frac {2 \left (A e d^2+C e-B f\right )+\left (B e d^2-A f d^2+\frac {C e^2 d^2}{f}-2 C f\right ) x}{(e+f x)^2 \sqrt {1-d^2 x^2}}dx}{2 \left (d^2 e^2-f^2\right )}+\frac {\sqrt {1-d^2 x^2} \left (A f^2-B e f+C e^2\right )}{2 f \left (d^2 e^2-f^2\right ) (e+f x)^2}\) |
\(\Big \downarrow \) 679 |
\(\displaystyle \frac {\frac {\left (C \left (d^2 e^2+2 f^2\right )-d^2 \left (3 B e f-A \left (2 d^2 e^2+f^2\right )\right )\right ) \int \frac {1}{(e+f x) \sqrt {1-d^2 x^2}}dx}{d^2 e^2-f^2}-\frac {\sqrt {1-d^2 x^2} \left (-3 A d^2 e f^2+B d^2 e^2 f+2 B f^3+C d^2 e^3-4 C e f^2\right )}{f \left (d^2 e^2-f^2\right ) (e+f x)}}{2 \left (d^2 e^2-f^2\right )}+\frac {\sqrt {1-d^2 x^2} \left (A f^2-B e f+C e^2\right )}{2 f \left (d^2 e^2-f^2\right ) (e+f x)^2}\) |
\(\Big \downarrow \) 488 |
\(\displaystyle \frac {-\frac {\left (C \left (d^2 e^2+2 f^2\right )-d^2 \left (3 B e f-A \left (2 d^2 e^2+f^2\right )\right )\right ) \int \frac {1}{-d^2 e^2+f^2-\frac {\left (e x d^2+f\right )^2}{1-d^2 x^2}}d\frac {e x d^2+f}{\sqrt {1-d^2 x^2}}}{d^2 e^2-f^2}-\frac {\sqrt {1-d^2 x^2} \left (-3 A d^2 e f^2+B d^2 e^2 f+2 B f^3+C d^2 e^3-4 C e f^2\right )}{f \left (d^2 e^2-f^2\right ) (e+f x)}}{2 \left (d^2 e^2-f^2\right )}+\frac {\sqrt {1-d^2 x^2} \left (A f^2-B e f+C e^2\right )}{2 f \left (d^2 e^2-f^2\right ) (e+f x)^2}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {\frac {\arctan \left (\frac {d^2 e x+f}{\sqrt {1-d^2 x^2} \sqrt {d^2 e^2-f^2}}\right ) \left (C \left (d^2 e^2+2 f^2\right )-d^2 \left (3 B e f-A \left (2 d^2 e^2+f^2\right )\right )\right )}{\left (d^2 e^2-f^2\right )^{3/2}}-\frac {\sqrt {1-d^2 x^2} \left (-3 A d^2 e f^2+B d^2 e^2 f+2 B f^3+C d^2 e^3-4 C e f^2\right )}{f \left (d^2 e^2-f^2\right ) (e+f x)}}{2 \left (d^2 e^2-f^2\right )}+\frac {\sqrt {1-d^2 x^2} \left (A f^2-B e f+C e^2\right )}{2 f \left (d^2 e^2-f^2\right ) (e+f x)^2}\) |
Input:
Int[(A + B*x + C*x^2)/(Sqrt[1 - d*x]*Sqrt[1 + d*x]*(e + f*x)^3),x]
Output:
((C*e^2 - B*e*f + A*f^2)*Sqrt[1 - d^2*x^2])/(2*f*(d^2*e^2 - f^2)*(e + f*x) ^2) + (-(((C*d^2*e^3 + B*d^2*e^2*f - 4*C*e*f^2 - 3*A*d^2*e*f^2 + 2*B*f^3)* Sqrt[1 - d^2*x^2])/(f*(d^2*e^2 - f^2)*(e + f*x))) + ((C*(d^2*e^2 + 2*f^2) - d^2*(3*B*e*f - A*(2*d^2*e^2 + f^2)))*ArcTan[(f + d^2*e*x)/(Sqrt[d^2*e^2 - f^2]*Sqrt[1 - d^2*x^2])])/(d^2*e^2 - f^2)^(3/2))/(2*(d^2*e^2 - f^2))
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[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_))*((a_) + (c_.)*(x_)^2)^(p _.), x_Symbol] :> Simp[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1 )/(2*(p + 1)*(c*d^2 + a*e^2))), x] + Simp[(c*d*f + a*e*g)/(c*d^2 + a*e^2) Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && EqQ[Simplify[m + 2*p + 3], 0]
Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f _.)*(x_))^(p_.), x_Symbol] :> Int[Px*(a*c + b*d*x^2)^m*(e + f*x)^p, x] /; F reeQ[{a, b, c, d, e, f, m, n, p}, x] && PolyQ[Px, x] && EqQ[b*c + a*d, 0] & & EqQ[m, n] && (IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Qx = PolynomialQuotient[Pq, d + e*x, x], R = PolynomialRemainder[Pq, d + e*x, x]}, Simp[e*R*(d + e*x)^(m + 1)*((a + b*x^2)^(p + 1)/((m + 1)*(b* d^2 + a*e^2))), x] + Simp[1/((m + 1)*(b*d^2 + a*e^2)) Int[(d + e*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[(m + 1)*(b*d^2 + a*e^2)*Qx + b*d*R*(m + 1) - b *e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, d, e, p}, x] && PolyQ[Pq, x] && NeQ[b*d^2 + a*e^2, 0] && LtQ[m, -1]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.11 (sec) , antiderivative size = 1449, normalized size of antiderivative = 5.84
Input:
int((C*x^2+B*x+A)/(-d*x+1)^(1/2)/(d*x+1)^(1/2)/(f*x+e)^3,x,method=_RETURNV ERBOSE)
Output:
-1/2*(B*d^2*e^2*f^2*x*(-d^2*x^2+1)^(1/2)*(-(d^2*e^2-f^2)/f^2)^(1/2)+2*B*f^ 4*x*(-d^2*x^2+1)^(1/2)*(-(d^2*e^2-f^2)/f^2)^(1/2)+B*e*f^3*(-d^2*x^2+1)^(1/ 2)*(-(d^2*e^2-f^2)/f^2)^(1/2)-3*C*e^2*f^2*(-d^2*x^2+1)^(1/2)*(-(d^2*e^2-f^ 2)/f^2)^(1/2)+A*ln(2*(d^2*x*e+(-d^2*x^2+1)^(1/2)*(-(d^2*e^2-f^2)/f^2)^(1/2 )*f+f)/(f*x+e))*d^2*f^4*x^2+A*ln(2*(d^2*x*e+(-d^2*x^2+1)^(1/2)*(-(d^2*e^2- f^2)/f^2)^(1/2)*f+f)/(f*x+e))*d^2*e^2*f^2-3*B*ln(2*(d^2*x*e+(-d^2*x^2+1)^( 1/2)*(-(d^2*e^2-f^2)/f^2)^(1/2)*f+f)/(f*x+e))*d^2*e^3*f+4*C*ln(2*(d^2*x*e+ (-d^2*x^2+1)^(1/2)*(-(d^2*e^2-f^2)/f^2)^(1/2)*f+f)/(f*x+e))*e*f^3*x-3*B*ln (2*(d^2*x*e+(-d^2*x^2+1)^(1/2)*(-(d^2*e^2-f^2)/f^2)^(1/2)*f+f)/(f*x+e))*d^ 2*e*f^3*x^2+C*ln(2*(d^2*x*e+(-d^2*x^2+1)^(1/2)*(-(d^2*e^2-f^2)/f^2)^(1/2)* f+f)/(f*x+e))*d^2*e^2*f^2*x^2+2*A*ln(2*(d^2*x*e+(-d^2*x^2+1)^(1/2)*(-(d^2* e^2-f^2)/f^2)^(1/2)*f+f)/(f*x+e))*d^2*e*f^3*x-6*B*ln(2*(d^2*x*e+(-d^2*x^2+ 1)^(1/2)*(-(d^2*e^2-f^2)/f^2)^(1/2)*f+f)/(f*x+e))*d^2*e^2*f^2*x-4*A*d^2*e^ 2*f^2*(-d^2*x^2+1)^(1/2)*(-(d^2*e^2-f^2)/f^2)^(1/2)+2*B*d^2*e^3*f*(-d^2*x^ 2+1)^(1/2)*(-(d^2*e^2-f^2)/f^2)^(1/2)-4*C*e*f^3*x*(-d^2*x^2+1)^(1/2)*(-(d^ 2*e^2-f^2)/f^2)^(1/2)+2*C*ln(2*(d^2*x*e+(-d^2*x^2+1)^(1/2)*(-(d^2*e^2-f^2) /f^2)^(1/2)*f+f)/(f*x+e))*d^2*e^3*f*x+2*A*ln(2*(d^2*x*e+(-d^2*x^2+1)^(1/2) *(-(d^2*e^2-f^2)/f^2)^(1/2)*f+f)/(f*x+e))*d^4*e^2*f^2*x^2+4*A*ln(2*(d^2*x* e+(-d^2*x^2+1)^(1/2)*(-(d^2*e^2-f^2)/f^2)^(1/2)*f+f)/(f*x+e))*d^4*e^3*f*x+ 2*A*ln(2*(d^2*x*e+(-d^2*x^2+1)^(1/2)*(-(d^2*e^2-f^2)/f^2)^(1/2)*f+f)/(f...
Leaf count of result is larger than twice the leaf count of optimal. 778 vs. \(2 (232) = 464\).
Time = 0.14 (sec) , antiderivative size = 1580, normalized size of antiderivative = 6.37 \[ \int \frac {A+B x+C x^2}{\sqrt {1-d x} \sqrt {1+d x} (e+f x)^3} \, dx=\text {Too large to display} \] Input:
integrate((C*x^2+B*x+A)/(-d*x+1)^(1/2)/(d*x+1)^(1/2)/(f*x+e)^3,x, algorith m="fricas")
Output:
[-1/2*(2*B*d^4*e^7 - B*d^2*e^5*f^2 - (4*A*d^4 + 3*C*d^2)*e^6*f + (5*A*d^2 + 3*C)*e^4*f^3 - B*e^3*f^4 - A*e^2*f^5 + (2*B*d^4*e^5*f^2 - B*d^2*e^3*f^4 - (4*A*d^4 + 3*C*d^2)*e^4*f^3 + (5*A*d^2 + 3*C)*e^2*f^5 - B*e*f^6 - A*f^7) *x^2 - (3*B*d^2*e^5*f - (2*A*d^4 + C*d^2)*e^6 - (A*d^2 + 2*C)*e^4*f^2 + (3 *B*d^2*e^3*f^3 - (2*A*d^4 + C*d^2)*e^4*f^2 - (A*d^2 + 2*C)*e^2*f^4)*x^2 + 2*(3*B*d^2*e^4*f^2 - (2*A*d^4 + C*d^2)*e^5*f - (A*d^2 + 2*C)*e^3*f^3)*x)*s qrt(-d^2*e^2 + f^2)*log((d^2*e*f*x + f^2 - sqrt(-d^2*e^2 + f^2)*(d^2*e*x + f) - (sqrt(-d^2*e^2 + f^2)*sqrt(-d*x + 1)*f + (d^2*e^2 - f^2)*sqrt(-d*x + 1))*sqrt(d*x + 1))/(f*x + e)) + (2*B*d^4*e^7 - B*d^2*e^5*f^2 - (4*A*d^4 + 3*C*d^2)*e^6*f + (5*A*d^2 + 3*C)*e^4*f^3 - B*e^3*f^4 - A*e^2*f^5 + (C*d^4 *e^7 + B*d^4*e^6*f + B*d^2*e^4*f^3 - (3*A*d^4 + 5*C*d^2)*e^5*f^2 + (3*A*d^ 2 + 4*C)*e^3*f^4 - 2*B*e^2*f^5)*x)*sqrt(d*x + 1)*sqrt(-d*x + 1) + 2*(2*B*d ^4*e^6*f - B*d^2*e^4*f^3 - (4*A*d^4 + 3*C*d^2)*e^5*f^2 + (5*A*d^2 + 3*C)*e ^3*f^4 - B*e^2*f^5 - A*e*f^6)*x)/(d^6*e^10 - 3*d^4*e^8*f^2 + 3*d^2*e^6*f^4 - e^4*f^6 + (d^6*e^8*f^2 - 3*d^4*e^6*f^4 + 3*d^2*e^4*f^6 - e^2*f^8)*x^2 + 2*(d^6*e^9*f - 3*d^4*e^7*f^3 + 3*d^2*e^5*f^5 - e^3*f^7)*x), -1/2*(2*B*d^4 *e^7 - B*d^2*e^5*f^2 - (4*A*d^4 + 3*C*d^2)*e^6*f + (5*A*d^2 + 3*C)*e^4*f^3 - B*e^3*f^4 - A*e^2*f^5 + (2*B*d^4*e^5*f^2 - B*d^2*e^3*f^4 - (4*A*d^4 + 3 *C*d^2)*e^4*f^3 + (5*A*d^2 + 3*C)*e^2*f^5 - B*e*f^6 - A*f^7)*x^2 + 2*(3*B* d^2*e^5*f - (2*A*d^4 + C*d^2)*e^6 - (A*d^2 + 2*C)*e^4*f^2 + (3*B*d^2*e^...
\[ \int \frac {A+B x+C x^2}{\sqrt {1-d x} \sqrt {1+d x} (e+f x)^3} \, dx=\int \frac {A + B x + C x^{2}}{\left (e + f x\right )^{3} \sqrt {- d x + 1} \sqrt {d x + 1}}\, dx \] Input:
integrate((C*x**2+B*x+A)/(-d*x+1)**(1/2)/(d*x+1)**(1/2)/(f*x+e)**3,x)
Output:
Integral((A + B*x + C*x**2)/((e + f*x)**3*sqrt(-d*x + 1)*sqrt(d*x + 1)), x )
Exception generated. \[ \int \frac {A+B x+C x^2}{\sqrt {1-d x} \sqrt {1+d x} (e+f x)^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((C*x^2+B*x+A)/(-d*x+1)^(1/2)/(d*x+1)^(1/2)/(f*x+e)^3,x, algorith m="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume((f-d*e)*(f+d*e)>0)', see `assume ?` for mor
Exception generated. \[ \int \frac {A+B x+C x^2}{\sqrt {1-d x} \sqrt {1+d x} (e+f x)^3} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((C*x^2+B*x+A)/(-d*x+1)^(1/2)/(d*x+1)^(1/2)/(f*x+e)^3,x, algorith m="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Limit: Max order reached or unable to make series expansion Error: Bad Argument Value
Time = 42.01 (sec) , antiderivative size = 9097, normalized size of antiderivative = 36.68 \[ \int \frac {A+B x+C x^2}{\sqrt {1-d x} \sqrt {1+d x} (e+f x)^3} \, dx=\text {Too large to display} \] Input:
int((A + B*x + C*x^2)/((e + f*x)^3*(1 - d*x)^(1/2)*(d*x + 1)^(1/2)),x)
Output:
((12*(2*C*f^3 + C*d^2*e^2*f)*((1 - d*x)^(1/2) - 1)^2)/(((d*x + 1)^(1/2) - 1)^2*(f^4 + d^4*e^4 - 2*d^2*e^2*f^2)) - (24*(2*C*f^3 - C*d^2*e^2*f)*((1 - d*x)^(1/2) - 1)^4)/(((d*x + 1)^(1/2) - 1)^4*(f^4 + d^4*e^4 - 2*d^2*e^2*f^2 )) + (12*(2*C*f^3 + C*d^2*e^2*f)*((1 - d*x)^(1/2) - 1)^6)/(((d*x + 1)^(1/2 ) - 1)^6*(f^4 + d^4*e^4 - 2*d^2*e^2*f^2)) - (2*((1 - d*x)^(1/2) - 1)^7*(C* d^3*e^3 + 2*C*d*e*f^2))/(((d*x + 1)^(1/2) - 1)^7*(f^4 + d^4*e^4 - 2*d^2*e^ 2*f^2)) - (2*((1 - d*x)^(1/2) - 1)^3*(7*C*d^3*e^3 - 34*C*d*e*f^2))/(((d*x + 1)^(1/2) - 1)^3*(f^4 + d^4*e^4 - 2*d^2*e^2*f^2)) + (2*((1 - d*x)^(1/2) - 1)^5*(7*C*d^3*e^3 - 34*C*d*e*f^2))/(((d*x + 1)^(1/2) - 1)^5*(f^4 + d^4*e^ 4 - 2*d^2*e^2*f^2)) + (2*d*e*((1 - d*x)^(1/2) - 1)*(2*C*f^2 + C*d^2*e^2))/ (((d*x + 1)^(1/2) - 1)*(f^4 + d^4*e^4 - 2*d^2*e^2*f^2)))/(d^2*e^2 + (((1 - d*x)^(1/2) - 1)^2*(16*f^2 + 4*d^2*e^2))/((d*x + 1)^(1/2) - 1)^2 + (((1 - d*x)^(1/2) - 1)^6*(16*f^2 + 4*d^2*e^2))/((d*x + 1)^(1/2) - 1)^6 - (((1 - d *x)^(1/2) - 1)^4*(32*f^2 - 6*d^2*e^2))/((d*x + 1)^(1/2) - 1)^4 + (d^2*e^2* ((1 - d*x)^(1/2) - 1)^8)/((d*x + 1)^(1/2) - 1)^8 + (8*d*e*f*((1 - d*x)^(1/ 2) - 1)^3)/((d*x + 1)^(1/2) - 1)^3 - (8*d*e*f*((1 - d*x)^(1/2) - 1)^5)/((d *x + 1)^(1/2) - 1)^5 - (8*d*e*f*((1 - d*x)^(1/2) - 1)^7)/((d*x + 1)^(1/2) - 1)^7 + (8*d*e*f*((1 - d*x)^(1/2) - 1))/((d*x + 1)^(1/2) - 1)) + ((4*((1 - d*x)^(1/2) - 1)^2*(4*A*d^4*e^4*f - 2*A*f^5 + 7*A*d^2*e^2*f^3))/(e^2*((d* x + 1)^(1/2) - 1)^2*(f^4 + d^4*e^4 - 2*d^2*e^2*f^2)) + (8*((1 - d*x)^(1...
Time = 0.26 (sec) , antiderivative size = 2499, normalized size of antiderivative = 10.08 \[ \int \frac {A+B x+C x^2}{\sqrt {1-d x} \sqrt {1+d x} (e+f x)^3} \, dx =\text {Too large to display} \] Input:
int((C*x^2+B*x+A)/(-d*x+1)^(1/2)/(d*x+1)^(1/2)/(f*x+e)^3,x)
Output:
( - 4*sqrt(d*e + f)*sqrt(d*e - f)*atan((sqrt(d*e + f)*tan(asin(sqrt( - d*x + 1)/sqrt(2))/2) - sqrt(f)*sqrt(2))/sqrt(d*e - f))*a*d**4*e**4 - 8*sqrt(d *e + f)*sqrt(d*e - f)*atan((sqrt(d*e + f)*tan(asin(sqrt( - d*x + 1)/sqrt(2 ))/2) - sqrt(f)*sqrt(2))/sqrt(d*e - f))*a*d**4*e**3*f*x - 4*sqrt(d*e + f)* sqrt(d*e - f)*atan((sqrt(d*e + f)*tan(asin(sqrt( - d*x + 1)/sqrt(2))/2) - sqrt(f)*sqrt(2))/sqrt(d*e - f))*a*d**4*e**2*f**2*x**2 - 2*sqrt(d*e + f)*sq rt(d*e - f)*atan((sqrt(d*e + f)*tan(asin(sqrt( - d*x + 1)/sqrt(2))/2) - sq rt(f)*sqrt(2))/sqrt(d*e - f))*a*d**2*e**2*f**2 - 4*sqrt(d*e + f)*sqrt(d*e - f)*atan((sqrt(d*e + f)*tan(asin(sqrt( - d*x + 1)/sqrt(2))/2) - sqrt(f)*s qrt(2))/sqrt(d*e - f))*a*d**2*e*f**3*x - 2*sqrt(d*e + f)*sqrt(d*e - f)*ata n((sqrt(d*e + f)*tan(asin(sqrt( - d*x + 1)/sqrt(2))/2) - sqrt(f)*sqrt(2))/ sqrt(d*e - f))*a*d**2*f**4*x**2 + 6*sqrt(d*e + f)*sqrt(d*e - f)*atan((sqrt (d*e + f)*tan(asin(sqrt( - d*x + 1)/sqrt(2))/2) - sqrt(f)*sqrt(2))/sqrt(d* e - f))*b*d**2*e**3*f + 12*sqrt(d*e + f)*sqrt(d*e - f)*atan((sqrt(d*e + f) *tan(asin(sqrt( - d*x + 1)/sqrt(2))/2) - sqrt(f)*sqrt(2))/sqrt(d*e - f))*b *d**2*e**2*f**2*x + 6*sqrt(d*e + f)*sqrt(d*e - f)*atan((sqrt(d*e + f)*tan( asin(sqrt( - d*x + 1)/sqrt(2))/2) - sqrt(f)*sqrt(2))/sqrt(d*e - f))*b*d**2 *e*f**3*x**2 - 2*sqrt(d*e + f)*sqrt(d*e - f)*atan((sqrt(d*e + f)*tan(asin( sqrt( - d*x + 1)/sqrt(2))/2) - sqrt(f)*sqrt(2))/sqrt(d*e - f))*c*d**2*e**4 - 4*sqrt(d*e + f)*sqrt(d*e - f)*atan((sqrt(d*e + f)*tan(asin(sqrt( - d...