\(\int \frac {(e+f x)^2 \sqrt {1-d^2 x^2}}{(1-d x)^4} \, dx\) [57]

Optimal result

Integrand size = 30, antiderivative size = 120 \[ \int \frac {(e+f x)^2 \sqrt {1-d^2 x^2}}{(1-d x)^4} \, dx=\frac {2 f^2 \sqrt {1-d^2 x^2}}{d^3 (1-d x)}+\frac {(d e+f)^2 \left (1-d^2 x^2\right )^{3/2}}{5 d^3 (1-d x)^4}+\frac {(d e-9 f) (d e+f) \left (1-d^2 x^2\right )^{3/2}}{15 d^3 (1-d x)^3}-\frac {f^2 \arcsin (d x)}{d^3} \] Output:

2*f^2*(-d^2*x^2+1)^(1/2)/d^3/(-d*x+1)+1/5*(d*e+f)^2*(-d^2*x^2+1)^(3/2)/d^3 
/(-d*x+1)^4+1/15*(d*e-9*f)*(d*e+f)*(-d^2*x^2+1)^(3/2)/d^3/(-d*x+1)^3-f^2*a 
rcsin(d*x)/d^3
 

Mathematica [A] (verified)

Time = 0.48 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.02 \[ \int \frac {(e+f x)^2 \sqrt {1-d^2 x^2}}{(1-d x)^4} \, dx=\frac {\frac {\sqrt {1-d^2 x^2} \left (-24 f^2+d^4 e^2 x^2-d^3 e x (3 e+8 f x)+d f (2 e+57 f x)-d^2 \left (4 e^2+6 e f x+39 f^2 x^2\right )\right )}{(-1+d x)^3}-30 f^2 \arctan \left (\frac {d x}{-1+\sqrt {1-d^2 x^2}}\right )}{15 d^3} \] Input:

Integrate[((e + f*x)^2*Sqrt[1 - d^2*x^2])/(1 - d*x)^4,x]
 

Output:

((Sqrt[1 - d^2*x^2]*(-24*f^2 + d^4*e^2*x^2 - d^3*e*x*(3*e + 8*f*x) + d*f*( 
2*e + 57*f*x) - d^2*(4*e^2 + 6*e*f*x + 39*f^2*x^2)))/(-1 + d*x)^3 - 30*f^2 
*ArcTan[(d*x)/(-1 + Sqrt[1 - d^2*x^2])])/(15*d^3)
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.99, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {717, 100, 25, 27, 87, 57, 39, 223}

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.

\(\displaystyle \int \frac {\sqrt {1-d^2 x^2} (e+f x)^2}{(1-d x)^4} \, dx\)

\(\Big \downarrow \) 717

\(\displaystyle \int \frac {\sqrt {d x+1} (e+f x)^2}{(1-d x)^{7/2}}dx\)

\(\Big \downarrow \) 100

\(\displaystyle \frac {(d x+1)^{3/2} (d e+f)^2}{5 d^3 (1-d x)^{5/2}}-\frac {\int -\frac {d \sqrt {d x+1} \left (d^2 e^2-8 d f e-4 f^2-5 d f^2 x\right )}{(1-d x)^{5/2}}dx}{5 d^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {d \sqrt {d x+1} \left (d^2 e^2-8 d f e-4 f^2-5 d f^2 x\right )}{(1-d x)^{5/2}}dx}{5 d^3}+\frac {(d x+1)^{3/2} (d e+f)^2}{5 d^3 (1-d x)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sqrt {d x+1} \left (d^2 e^2-8 d f e-4 f^2-5 d f^2 x\right )}{(1-d x)^{5/2}}dx}{5 d^2}+\frac {(d x+1)^{3/2} (d e+f)^2}{5 d^3 (1-d x)^{5/2}}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {5 f^2 \int \frac {\sqrt {d x+1}}{(1-d x)^{3/2}}dx+\frac {(d x+1)^{3/2} (d e-9 f) (d e+f)}{3 d (1-d x)^{3/2}}}{5 d^2}+\frac {(d x+1)^{3/2} (d e+f)^2}{5 d^3 (1-d x)^{5/2}}\)

\(\Big \downarrow \) 57

\(\displaystyle \frac {5 f^2 \left (\frac {2 \sqrt {d x+1}}{d \sqrt {1-d x}}-\int \frac {1}{\sqrt {1-d x} \sqrt {d x+1}}dx\right )+\frac {(d x+1)^{3/2} (d e-9 f) (d e+f)}{3 d (1-d x)^{3/2}}}{5 d^2}+\frac {(d x+1)^{3/2} (d e+f)^2}{5 d^3 (1-d x)^{5/2}}\)

\(\Big \downarrow \) 39

\(\displaystyle \frac {5 f^2 \left (\frac {2 \sqrt {d x+1}}{d \sqrt {1-d x}}-\int \frac {1}{\sqrt {1-d^2 x^2}}dx\right )+\frac {(d x+1)^{3/2} (d e-9 f) (d e+f)}{3 d (1-d x)^{3/2}}}{5 d^2}+\frac {(d x+1)^{3/2} (d e+f)^2}{5 d^3 (1-d x)^{5/2}}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {5 f^2 \left (\frac {2 \sqrt {d x+1}}{d \sqrt {1-d x}}-\frac {\arcsin (d x)}{d}\right )+\frac {(d x+1)^{3/2} (d e-9 f) (d e+f)}{3 d (1-d x)^{3/2}}}{5 d^2}+\frac {(d x+1)^{3/2} (d e+f)^2}{5 d^3 (1-d x)^{5/2}}\)

Input:

Int[((e + f*x)^2*Sqrt[1 - d^2*x^2])/(1 - d*x)^4,x]
 

Output:

((d*e + f)^2*(1 + d*x)^(3/2))/(5*d^3*(1 - d*x)^(5/2)) + (((d*e - 9*f)*(d*e 
 + f)*(1 + d*x)^(3/2))/(3*d*(1 - d*x)^(3/2)) + 5*f^2*((2*Sqrt[1 + d*x])/(d 
*Sqrt[1 - d*x]) - ArcSin[d*x]/d))/(5*d^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 39

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(m_.), x_Symbol] :> Int[( 
a*c + b*d*x^2)^m, x] /; FreeQ[{a, b, c, d, m}, x] && EqQ[b*c + a*d, 0] && ( 
IntegerQ[m] || (GtQ[a, 0] && GtQ[c, 0]))
 

rule 57

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & 
& GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m 
+ n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c 
, d, m, n, x]
 

rule 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 100

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d 
*e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1))   Int[(c + d*x)^ 
(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( 
p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n 
 + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))
 

rule 223

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 717

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(269\) vs. \(2(110)=220\).

Time = 0.75 (sec) , antiderivative size = 270, normalized size of antiderivative = 2.25

method	result	size
default	\(\frac {f^{2} \left (\frac {\left (-d^{2} \left (x -\frac {1}{d}\right )^{2}-2 d \left (x -\frac {1}{d}\right )\right )^{\frac {3}{2}}}{d \left (x -\frac {1}{d}\right )^{2}}+d \left (\sqrt {-d^{2} \left (x -\frac {1}{d}\right )^{2}-2 d \left (x -\frac {1}{d}\right )}-\frac {d \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x -\frac {1}{d}\right )^{2}-2 d \left (x -\frac {1}{d}\right )}}\right )}{\sqrt {d^{2}}}\right )\right )}{d^{4}}+\frac {\left (d^{2} e^{2}+2 d e f +f^{2}\right ) \left (\frac {\left (-d^{2} \left (x -\frac {1}{d}\right )^{2}-2 d \left (x -\frac {1}{d}\right )\right )^{\frac {3}{2}}}{5 d \left (x -\frac {1}{d}\right )^{4}}-\frac {\left (-d^{2} \left (x -\frac {1}{d}\right )^{2}-2 d \left (x -\frac {1}{d}\right )\right )^{\frac {3}{2}}}{15 \left (x -\frac {1}{d}\right )^{3}}\right )}{d^{6}}+\frac {2 f \left (d e +f \right ) \left (-d^{2} \left (x -\frac {1}{d}\right )^{2}-2 d \left (x -\frac {1}{d}\right )\right )^{\frac {3}{2}}}{3 d^{6} \left (x -\frac {1}{d}\right )^{3}}\)	\(270\)

Input:

int((f*x+e)^2*(-d^2*x^2+1)^(1/2)/(-d*x+1)^4,x,method=_RETURNVERBOSE)
 

Output:

1/d^4*f^2*(1/d/(x-1/d)^2*(-d^2*(x-1/d)^2-2*d*(x-1/d))^(3/2)+d*((-d^2*(x-1/ 
d)^2-2*d*(x-1/d))^(1/2)-d/(d^2)^(1/2)*arctan((d^2)^(1/2)*x/(-d^2*(x-1/d)^2 
-2*d*(x-1/d))^(1/2))))+(d^2*e^2+2*d*e*f+f^2)/d^6*(1/5/d/(x-1/d)^4*(-d^2*(x 
-1/d)^2-2*d*(x-1/d))^(3/2)-1/15/(x-1/d)^3*(-d^2*(x-1/d)^2-2*d*(x-1/d))^(3/ 
2))+2/3*f*(d*e+f)/d^6/(x-1/d)^3*(-d^2*(x-1/d)^2-2*d*(x-1/d))^(3/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 275 vs. \(2 (107) = 214\).

Time = 0.09 (sec) , antiderivative size = 275, normalized size of antiderivative = 2.29 \[ \int \frac {(e+f x)^2 \sqrt {1-d^2 x^2}}{(1-d x)^4} \, dx=-\frac {4 \, d^{2} e^{2} - 2 \, {\left (2 \, d^{5} e^{2} - d^{4} e f + 12 \, d^{3} f^{2}\right )} x^{3} - 2 \, d e f + 6 \, {\left (2 \, d^{4} e^{2} - d^{3} e f + 12 \, d^{2} f^{2}\right )} x^{2} + 24 \, f^{2} - 6 \, {\left (2 \, d^{3} e^{2} - d^{2} e f + 12 \, d f^{2}\right )} x - 30 \, {\left (d^{3} f^{2} x^{3} - 3 \, d^{2} f^{2} x^{2} + 3 \, d f^{2} x - f^{2}\right )} \arctan \left (\frac {\sqrt {-d^{2} x^{2} + 1} - 1}{d x}\right ) + {\left (4 \, d^{2} e^{2} - 2 \, d e f - {\left (d^{4} e^{2} - 8 \, d^{3} e f - 39 \, d^{2} f^{2}\right )} x^{2} + 24 \, f^{2} + 3 \, {\left (d^{3} e^{2} + 2 \, d^{2} e f - 19 \, d f^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + 1}}{15 \, {\left (d^{6} x^{3} - 3 \, d^{5} x^{2} + 3 \, d^{4} x - d^{3}\right )}} \] Input:

integrate((f*x+e)^2*(-d^2*x^2+1)^(1/2)/(-d*x+1)^4,x, algorithm="fricas")
 

Output:

-1/15*(4*d^2*e^2 - 2*(2*d^5*e^2 - d^4*e*f + 12*d^3*f^2)*x^3 - 2*d*e*f + 6* 
(2*d^4*e^2 - d^3*e*f + 12*d^2*f^2)*x^2 + 24*f^2 - 6*(2*d^3*e^2 - d^2*e*f + 
 12*d*f^2)*x - 30*(d^3*f^2*x^3 - 3*d^2*f^2*x^2 + 3*d*f^2*x - f^2)*arctan(( 
sqrt(-d^2*x^2 + 1) - 1)/(d*x)) + (4*d^2*e^2 - 2*d*e*f - (d^4*e^2 - 8*d^3*e 
*f - 39*d^2*f^2)*x^2 + 24*f^2 + 3*(d^3*e^2 + 2*d^2*e*f - 19*d*f^2)*x)*sqrt 
(-d^2*x^2 + 1))/(d^6*x^3 - 3*d^5*x^2 + 3*d^4*x - d^3)
 

Sympy [F]

\[ \int \frac {(e+f x)^2 \sqrt {1-d^2 x^2}}{(1-d x)^4} \, dx=\int \frac {\sqrt {- \left (d x - 1\right ) \left (d x + 1\right )} \left (e + f x\right )^{2}}{\left (d x - 1\right )^{4}}\, dx \] Input:

integrate((f*x+e)**2*(-d**2*x**2+1)**(1/2)/(-d*x+1)**4,x)
 

Output:

Integral(sqrt(-(d*x - 1)*(d*x + 1))*(e + f*x)**2/(d*x - 1)**4, x)
 

Maxima [F]

\[ \int \frac {(e+f x)^2 \sqrt {1-d^2 x^2}}{(1-d x)^4} \, dx=\int { \frac {\sqrt {-d^{2} x^{2} + 1} {\left (f x + e\right )}^{2}}{{\left (d x - 1\right )}^{4}} \,d x } \] Input:

integrate((f*x+e)^2*(-d^2*x^2+1)^(1/2)/(-d*x+1)^4,x, algorithm="maxima")
 

Output:

integrate(sqrt(-d^2*x^2 + 1)*(f*x + e)^2/(d*x - 1)^4, x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 393 vs. \(2 (107) = 214\).

Time = 0.20 (sec) , antiderivative size = 393, normalized size of antiderivative = 3.28 \[ \int \frac {(e+f x)^2 \sqrt {1-d^2 x^2}}{(1-d x)^4} \, dx=-\frac {f^{2} \arcsin \left (d x\right ) \mathrm {sgn}\left (d\right )}{d^{2} {\left | d \right |}} + \frac {2 \, {\left (4 \, d^{2} e^{2} - 2 \, d e f + 24 \, f^{2} - \frac {5 \, {\left (\sqrt {-d^{2} x^{2} + 1} {\left | d \right |} + d\right )} e^{2}}{x} + \frac {10 \, {\left (\sqrt {-d^{2} x^{2} + 1} {\left | d \right |} + d\right )} e f}{d x} + \frac {25 \, {\left (\sqrt {-d^{2} x^{2} + 1} {\left | d \right |} + d\right )}^{2} e^{2}}{d^{2} x^{2}} - \frac {105 \, {\left (\sqrt {-d^{2} x^{2} + 1} {\left | d \right |} + d\right )} f^{2}}{d^{2} x} + \frac {10 \, {\left (\sqrt {-d^{2} x^{2} + 1} {\left | d \right |} + d\right )}^{2} e f}{d^{3} x^{2}} - \frac {15 \, {\left (\sqrt {-d^{2} x^{2} + 1} {\left | d \right |} + d\right )}^{3} e^{2}}{d^{4} x^{3}} + \frac {165 \, {\left (\sqrt {-d^{2} x^{2} + 1} {\left | d \right |} + d\right )}^{2} f^{2}}{d^{4} x^{2}} + \frac {30 \, {\left (\sqrt {-d^{2} x^{2} + 1} {\left | d \right |} + d\right )}^{3} e f}{d^{5} x^{3}} + \frac {15 \, {\left (\sqrt {-d^{2} x^{2} + 1} {\left | d \right |} + d\right )}^{4} e^{2}}{d^{6} x^{4}} - \frac {75 \, {\left (\sqrt {-d^{2} x^{2} + 1} {\left | d \right |} + d\right )}^{3} f^{2}}{d^{6} x^{3}} + \frac {15 \, {\left (\sqrt {-d^{2} x^{2} + 1} {\left | d \right |} + d\right )}^{4} f^{2}}{d^{8} x^{4}}\right )}}{15 \, d^{2} {\left (\frac {\sqrt {-d^{2} x^{2} + 1} {\left | d \right |} + d}{d^{2} x} - 1\right )}^{5} {\left | d \right |}} \] Input:

integrate((f*x+e)^2*(-d^2*x^2+1)^(1/2)/(-d*x+1)^4,x, algorithm="giac")
 

Output:

-f^2*arcsin(d*x)*sgn(d)/(d^2*abs(d)) + 2/15*(4*d^2*e^2 - 2*d*e*f + 24*f^2 
- 5*(sqrt(-d^2*x^2 + 1)*abs(d) + d)*e^2/x + 10*(sqrt(-d^2*x^2 + 1)*abs(d) 
+ d)*e*f/(d*x) + 25*(sqrt(-d^2*x^2 + 1)*abs(d) + d)^2*e^2/(d^2*x^2) - 105* 
(sqrt(-d^2*x^2 + 1)*abs(d) + d)*f^2/(d^2*x) + 10*(sqrt(-d^2*x^2 + 1)*abs(d 
) + d)^2*e*f/(d^3*x^2) - 15*(sqrt(-d^2*x^2 + 1)*abs(d) + d)^3*e^2/(d^4*x^3 
) + 165*(sqrt(-d^2*x^2 + 1)*abs(d) + d)^2*f^2/(d^4*x^2) + 30*(sqrt(-d^2*x^ 
2 + 1)*abs(d) + d)^3*e*f/(d^5*x^3) + 15*(sqrt(-d^2*x^2 + 1)*abs(d) + d)^4* 
e^2/(d^6*x^4) - 75*(sqrt(-d^2*x^2 + 1)*abs(d) + d)^3*f^2/(d^6*x^3) + 15*(s 
qrt(-d^2*x^2 + 1)*abs(d) + d)^4*f^2/(d^8*x^4))/(d^2*((sqrt(-d^2*x^2 + 1)*a 
bs(d) + d)/(d^2*x) - 1)^5*abs(d))
 

Mupad [B] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 464, normalized size of antiderivative = 3.87 \[ \int \frac {(e+f x)^2 \sqrt {1-d^2 x^2}}{(1-d x)^4} \, dx=\frac {\sqrt {1-d^2\,x^2}\,\left (\frac {4\,\left (f^2\,{\left (-d^2\right )}^{3/2}+d^2\,e^2\,{\left (-d^2\right )}^{3/2}+2\,d\,e\,f\,{\left (-d^2\right )}^{3/2}\right )}{15\,d^2\,{\left (x\,\sqrt {-d^2}-\frac {\sqrt {-d^2}}{d}\right )}^2}+\frac {4\,d\,\left (f^2\,{\left (-d^2\right )}^{3/2}+d^2\,e^2\,{\left (-d^2\right )}^{3/2}+2\,d\,e\,f\,{\left (-d^2\right )}^{3/2}\right )}{15\,\left (x\,\sqrt {-d^2}-\frac {\sqrt {-d^2}}{d}\right )\,{\left (-d^2\right )}^{3/2}}+\frac {2\,\left (f^2\,{\left (-d^2\right )}^{3/2}+d^2\,e^2\,{\left (-d^2\right )}^{3/2}+2\,d\,e\,f\,{\left (-d^2\right )}^{3/2}\right )}{5\,d\,{\left (x\,\sqrt {-d^2}-\frac {\sqrt {-d^2}}{d}\right )}^3\,\sqrt {-d^2}}\right )}{d^3\,\sqrt {-d^2}}-\frac {\sqrt {1-d^2\,x^2}\,\left (\frac {d^4\,e^2+6\,d^3\,e\,f+5\,d^2\,f^2}{3\,d^2\,\left (x\,\sqrt {-d^2}-\frac {\sqrt {-d^2}}{d}\right )}+\frac {d^4\,e^2+6\,d^3\,e\,f+5\,d^2\,f^2}{3\,d\,{\left (x\,\sqrt {-d^2}-\frac {\sqrt {-d^2}}{d}\right )}^2\,\sqrt {-d^2}}\right )}{d^2\,\sqrt {-d^2}}-\frac {f^2\,\mathrm {asinh}\left (x\,\sqrt {-d^2}\right )}{d^2\,\sqrt {-d^2}}-\frac {2\,\left (2\,f^2\,\sqrt {-d^2}+d\,e\,f\,\sqrt {-d^2}\right )\,\sqrt {1-d^2\,x^2}}{d^4\,\left (x\,\sqrt {-d^2}-\frac {\sqrt {-d^2}}{d}\right )} \] Input:

int(((e + f*x)^2*(1 - d^2*x^2)^(1/2))/(d*x - 1)^4,x)
 

Output:

((1 - d^2*x^2)^(1/2)*((4*(f^2*(-d^2)^(3/2) + d^2*e^2*(-d^2)^(3/2) + 2*d*e* 
f*(-d^2)^(3/2)))/(15*d^2*(x*(-d^2)^(1/2) - (-d^2)^(1/2)/d)^2) + (4*d*(f^2* 
(-d^2)^(3/2) + d^2*e^2*(-d^2)^(3/2) + 2*d*e*f*(-d^2)^(3/2)))/(15*(x*(-d^2) 
^(1/2) - (-d^2)^(1/2)/d)*(-d^2)^(3/2)) + (2*(f^2*(-d^2)^(3/2) + d^2*e^2*(- 
d^2)^(3/2) + 2*d*e*f*(-d^2)^(3/2)))/(5*d*(x*(-d^2)^(1/2) - (-d^2)^(1/2)/d) 
^3*(-d^2)^(1/2))))/(d^3*(-d^2)^(1/2)) - ((1 - d^2*x^2)^(1/2)*((d^4*e^2 + 5 
*d^2*f^2 + 6*d^3*e*f)/(3*d^2*(x*(-d^2)^(1/2) - (-d^2)^(1/2)/d)) + (d^4*e^2 
 + 5*d^2*f^2 + 6*d^3*e*f)/(3*d*(x*(-d^2)^(1/2) - (-d^2)^(1/2)/d)^2*(-d^2)^ 
(1/2))))/(d^2*(-d^2)^(1/2)) - (f^2*asinh(x*(-d^2)^(1/2)))/(d^2*(-d^2)^(1/2 
)) - (2*(2*f^2*(-d^2)^(1/2) + d*e*f*(-d^2)^(1/2))*(1 - d^2*x^2)^(1/2))/(d^ 
4*(x*(-d^2)^(1/2) - (-d^2)^(1/2)/d))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 436, normalized size of antiderivative = 3.63 \[ \int \frac {(e+f x)^2 \sqrt {1-d^2 x^2}}{(1-d x)^4} \, dx=\frac {15 \mathit {atan} \left (\frac {2 \sqrt {-d^{2} x^{2}+1}\, d^{2} x^{2}-\sqrt {-d^{2} x^{2}+1}}{2 d^{3} x^{3}-2 d x}\right ) d^{3} f^{2} x^{3}-45 \mathit {atan} \left (\frac {2 \sqrt {-d^{2} x^{2}+1}\, d^{2} x^{2}-\sqrt {-d^{2} x^{2}+1}}{2 d^{3} x^{3}-2 d x}\right ) d^{2} f^{2} x^{2}+45 \mathit {atan} \left (\frac {2 \sqrt {-d^{2} x^{2}+1}\, d^{2} x^{2}-\sqrt {-d^{2} x^{2}+1}}{2 d^{3} x^{3}-2 d x}\right ) d \,f^{2} x -15 \mathit {atan} \left (\frac {2 \sqrt {-d^{2} x^{2}+1}\, d^{2} x^{2}-\sqrt {-d^{2} x^{2}+1}}{2 d^{3} x^{3}-2 d x}\right ) f^{2}+2 \sqrt {-d^{2} x^{2}+1}\, d^{4} e^{2} x^{2}-6 \sqrt {-d^{2} x^{2}+1}\, d^{3} e^{2} x -16 \sqrt {-d^{2} x^{2}+1}\, d^{3} e f \,x^{2}-8 \sqrt {-d^{2} x^{2}+1}\, d^{2} e^{2}-12 \sqrt {-d^{2} x^{2}+1}\, d^{2} e f x -78 \sqrt {-d^{2} x^{2}+1}\, d^{2} f^{2} x^{2}+4 \sqrt {-d^{2} x^{2}+1}\, d e f +114 \sqrt {-d^{2} x^{2}+1}\, d \,f^{2} x -48 \sqrt {-d^{2} x^{2}+1}\, f^{2}}{30 d^{3} \left (d^{3} x^{3}-3 d^{2} x^{2}+3 d x -1\right )} \] Input:

int((f*x+e)^2*(-d^2*x^2+1)^(1/2)/(-d*x+1)^4,x)
 

Output:

(15*atan((2*sqrt( - d**2*x**2 + 1)*d**2*x**2 - sqrt( - d**2*x**2 + 1))/(2* 
d**3*x**3 - 2*d*x))*d**3*f**2*x**3 - 45*atan((2*sqrt( - d**2*x**2 + 1)*d** 
2*x**2 - sqrt( - d**2*x**2 + 1))/(2*d**3*x**3 - 2*d*x))*d**2*f**2*x**2 + 4 
5*atan((2*sqrt( - d**2*x**2 + 1)*d**2*x**2 - sqrt( - d**2*x**2 + 1))/(2*d* 
*3*x**3 - 2*d*x))*d*f**2*x - 15*atan((2*sqrt( - d**2*x**2 + 1)*d**2*x**2 - 
 sqrt( - d**2*x**2 + 1))/(2*d**3*x**3 - 2*d*x))*f**2 + 2*sqrt( - d**2*x**2 
 + 1)*d**4*e**2*x**2 - 6*sqrt( - d**2*x**2 + 1)*d**3*e**2*x - 16*sqrt( - d 
**2*x**2 + 1)*d**3*e*f*x**2 - 8*sqrt( - d**2*x**2 + 1)*d**2*e**2 - 12*sqrt 
( - d**2*x**2 + 1)*d**2*e*f*x - 78*sqrt( - d**2*x**2 + 1)*d**2*f**2*x**2 + 
 4*sqrt( - d**2*x**2 + 1)*d*e*f + 114*sqrt( - d**2*x**2 + 1)*d*f**2*x - 48 
*sqrt( - d**2*x**2 + 1)*f**2)/(30*d**3*(d**3*x**3 - 3*d**2*x**2 + 3*d*x - 
1))