\(\int \frac {(e+f x)^2 (1-d^2 x^2)^{3/2}}{(1+d x)^2} \, dx\) [40]

Optimal result

Integrand size = 29, antiderivative size = 145 \[ \int \frac {(e+f x)^2 \left (1-d^2 x^2\right )^{3/2}}{(1+d x)^2} \, dx=-\frac {f (8 d e+f) (1-d x)^2 \sqrt {1-d^2 x^2}}{12 d^3}+\frac {f^2 (1-d x)^3 \sqrt {1-d^2 x^2}}{4 d^3}+\frac {\left (12 d^2 e^2-16 d e f+7 f^2\right ) (4-d x) \sqrt {1-d^2 x^2}}{24 d^3}+\frac {\left (12 d^2 e^2-16 d e f+7 f^2\right ) \arcsin (d x)}{8 d^3} \] Output:

-1/12*f*(8*d*e+f)*(-d*x+1)^2*(-d^2*x^2+1)^(1/2)/d^3+1/4*f^2*(-d*x+1)^3*(-d 
^2*x^2+1)^(1/2)/d^3+1/24*(12*d^2*e^2-16*d*e*f+7*f^2)*(-d*x+4)*(-d^2*x^2+1) 
^(1/2)/d^3+1/8*(12*d^2*e^2-16*d*e*f+7*f^2)*arcsin(d*x)/d^3
 

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.91 \[ \int \frac {(e+f x)^2 \left (1-d^2 x^2\right )^{3/2}}{(1+d x)^2} \, dx=\frac {\sqrt {1-d^2 x^2} \left (32 f^2-d f (80 e+21 f x)+16 d^2 \left (3 e^2+3 e f x+f^2 x^2\right )-2 d^3 x \left (6 e^2+8 e f x+3 f^2 x^2\right )\right )+6 \left (12 d^2 e^2-16 d e f+7 f^2\right ) \arctan \left (\frac {d x}{-1+\sqrt {1-d^2 x^2}}\right )}{24 d^3} \] Input:

Integrate[((e + f*x)^2*(1 - d^2*x^2)^(3/2))/(1 + d*x)^2,x]
 

Output:

(Sqrt[1 - d^2*x^2]*(32*f^2 - d*f*(80*e + 21*f*x) + 16*d^2*(3*e^2 + 3*e*f*x 
 + f^2*x^2) - 2*d^3*x*(6*e^2 + 8*e*f*x + 3*f^2*x^2)) + 6*(12*d^2*e^2 - 16* 
d*e*f + 7*f^2)*ArcTan[(d*x)/(-1 + Sqrt[1 - d^2*x^2])])/(24*d^3)
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {711, 25, 27, 671, 466, 211, 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.

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

\(\Big \downarrow \) 711

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 671

\(\displaystyle \frac {\left (12 d^2 e^2-16 d e f+7 f^2\right ) \int \frac {\left (1-d^2 x^2\right )^{3/2}}{d x+1}dx+\frac {4 \left (1-d^2 x^2\right )^{5/2} (d e-f)^2}{d (d x+1)^2}}{4 d^2}-\frac {f^2 \left (1-d^2 x^2\right )^{5/2}}{4 d^3 (d x+1)}\)

\(\Big \downarrow \) 466

\(\displaystyle \frac {\left (12 d^2 e^2-16 d e f+7 f^2\right ) \left (\int \sqrt {1-d^2 x^2}dx+\frac {\left (1-d^2 x^2\right )^{3/2}}{3 d}\right )+\frac {4 \left (1-d^2 x^2\right )^{5/2} (d e-f)^2}{d (d x+1)^2}}{4 d^2}-\frac {f^2 \left (1-d^2 x^2\right )^{5/2}}{4 d^3 (d x+1)}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {\left (12 d^2 e^2-16 d e f+7 f^2\right ) \left (\frac {1}{2} \int \frac {1}{\sqrt {1-d^2 x^2}}dx+\frac {\left (1-d^2 x^2\right )^{3/2}}{3 d}+\frac {1}{2} x \sqrt {1-d^2 x^2}\right )+\frac {4 \left (1-d^2 x^2\right )^{5/2} (d e-f)^2}{d (d x+1)^2}}{4 d^2}-\frac {f^2 \left (1-d^2 x^2\right )^{5/2}}{4 d^3 (d x+1)}\)

\(\Big \downarrow \) 223

\(\displaystyle \frac {\left (\frac {\arcsin (d x)}{2 d}+\frac {\left (1-d^2 x^2\right )^{3/2}}{3 d}+\frac {1}{2} x \sqrt {1-d^2 x^2}\right ) \left (12 d^2 e^2-16 d e f+7 f^2\right )+\frac {4 \left (1-d^2 x^2\right )^{5/2} (d e-f)^2}{d (d x+1)^2}}{4 d^2}-\frac {f^2 \left (1-d^2 x^2\right )^{5/2}}{4 d^3 (d x+1)}\)

Input:

Int[((e + f*x)^2*(1 - d^2*x^2)^(3/2))/(1 + d*x)^2,x]
 

Output:

-1/4*(f^2*(1 - d^2*x^2)^(5/2))/(d^3*(1 + d*x)) + ((4*(d*e - f)^2*(1 - d^2* 
x^2)^(5/2))/(d*(1 + d*x)^2) + (12*d^2*e^2 - 16*d*e*f + 7*f^2)*((x*Sqrt[1 - 
 d^2*x^2])/2 + (1 - d^2*x^2)^(3/2)/(3*d) + ArcSin[d*x]/(2*d)))/(4*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 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

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 466

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(c + d*x)^(n + 1)*((a + b*x^2)^p/(d*(n + 2*p + 1))), x] - Simp[2*b*c*(p/(d^ 
2*(n + 2*p + 1)))   Int[(c + d*x)^(n + 1)*(a + b*x^2)^(p - 1), x], x] /; Fr 
eeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0] && (LeQ[-2, n, 0 
] || EqQ[n + p + 1, 0]) && NeQ[n + 2*p + 1, 0] && IntegerQ[2*p]
 

rule 671

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ 
), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + c*x^2)^(p + 1)/(2*c*d*(m 
 + p + 1))), x] + Simp[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d)*(m 
+ p + 1))   Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, 
e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p 
 + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p 
 + 1, 0]
 

rule 711

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_) + (g_.)*(x_))^(n_.)*((a_.) + (c_.)*(x_ 
)^2)^(p_), x_Symbol] :> Simp[g^n*(d + e*x)^(m + n - 1)*((a + c*x^2)^(p + 1) 
/(c*e^(n - 1)*(m + n + 2*p + 1))), x] + Simp[1/(c*e^n*(m + n + 2*p + 1)) 
Int[(d + e*x)^m*(a + c*x^2)^p*ExpandToSum[c*e^n*(m + n + 2*p + 1)*(f + g*x) 
^n - c*g^n*(m + n + 2*p + 1)*(d + e*x)^n - 2*e*g^n*(m + p + n)*(d + e*x)^(n 
 - 2)*(a*e - c*d*x), x], x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && Eq 
Q[c*d^2 + a*e^2, 0] && IGtQ[n, 0] && NeQ[m + n + 2*p + 1, 0]
 

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.05

method	result	size
risch	\(\frac {\left (6 f^{2} d^{3} x^{3}+16 d^{3} e f \,x^{2}+12 d^{3} e^{2} x -16 d^{2} f^{2} x^{2}-48 d^{2} e f x -48 d^{2} e^{2}+21 d \,f^{2} x +80 d e f -32 f^{2}\right ) \left (d^{2} x^{2}-1\right )}{24 d^{3} \sqrt {-d^{2} x^{2}+1}}+\frac {\left (12 d^{2} e^{2}-16 d e f +7 f^{2}\right ) \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+1}}\right )}{8 d^{2} \sqrt {d^{2}}}\)	\(152\)
default	\(\frac {f^{2} \left (\frac {x \left (-d^{2} x^{2}+1\right )^{\frac {3}{2}}}{4}+\frac {3 x \sqrt {-d^{2} x^{2}+1}}{8}+\frac {3 \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+1}}\right )}{8 \sqrt {d^{2}}}\right )}{d^{2}}+\frac {\left (d^{2} e^{2}-2 d e f +f^{2}\right ) \left (\frac {\left (-d^{2} \left (x +\frac {1}{d}\right )^{2}+2 d \left (x +\frac {1}{d}\right )\right )^{\frac {5}{2}}}{d \left (x +\frac {1}{d}\right )^{2}}+3 d \left (\frac {\left (-d^{2} \left (x +\frac {1}{d}\right )^{2}+2 d \left (x +\frac {1}{d}\right )\right )^{\frac {3}{2}}}{3}+d \left (-\frac {\left (-2 d^{2} \left (x +\frac {1}{d}\right )+2 d \right ) \sqrt {-d^{2} \left (x +\frac {1}{d}\right )^{2}+2 d \left (x +\frac {1}{d}\right )}}{4 d^{2}}+\frac {\arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {1}{d}\right )^{2}+2 d \left (x +\frac {1}{d}\right )}}\right )}{2 \sqrt {d^{2}}}\right )\right )\right )}{d^{4}}+\frac {2 f \left (d e -f \right ) \left (\frac {\left (-d^{2} \left (x +\frac {1}{d}\right )^{2}+2 d \left (x +\frac {1}{d}\right )\right )^{\frac {3}{2}}}{3}+d \left (-\frac {\left (-2 d^{2} \left (x +\frac {1}{d}\right )+2 d \right ) \sqrt {-d^{2} \left (x +\frac {1}{d}\right )^{2}+2 d \left (x +\frac {1}{d}\right )}}{4 d^{2}}+\frac {\arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {1}{d}\right )^{2}+2 d \left (x +\frac {1}{d}\right )}}\right )}{2 \sqrt {d^{2}}}\right )\right )}{d^{3}}\)	\(356\)

Input:

int((f*x+e)^2*(-d^2*x^2+1)^(3/2)/(d*x+1)^2,x,method=_RETURNVERBOSE)
 

Output:

1/24/d^3*(6*d^3*f^2*x^3+16*d^3*e*f*x^2+12*d^3*e^2*x-16*d^2*f^2*x^2-48*d^2* 
e*f*x-48*d^2*e^2+21*d*f^2*x+80*d*e*f-32*f^2)*(d^2*x^2-1)/(-d^2*x^2+1)^(1/2 
)+1/8/d^2*(12*d^2*e^2-16*d*e*f+7*f^2)/(d^2)^(1/2)*arctan((d^2)^(1/2)*x/(-d 
^2*x^2+1)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.94 \[ \int \frac {(e+f x)^2 \left (1-d^2 x^2\right )^{3/2}}{(1+d x)^2} \, dx=-\frac {6 \, {\left (12 \, d^{2} e^{2} - 16 \, d e f + 7 \, f^{2}\right )} \arctan \left (\frac {\sqrt {-d^{2} x^{2} + 1} - 1}{d x}\right ) + {\left (6 \, d^{3} f^{2} x^{3} - 48 \, d^{2} e^{2} + 80 \, d e f + 16 \, {\left (d^{3} e f - d^{2} f^{2}\right )} x^{2} - 32 \, f^{2} + 3 \, {\left (4 \, d^{3} e^{2} - 16 \, d^{2} e f + 7 \, d f^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + 1}}{24 \, d^{3}} \] Input:

integrate((f*x+e)^2*(-d^2*x^2+1)^(3/2)/(d*x+1)^2,x, algorithm="fricas")
 

Output:

-1/24*(6*(12*d^2*e^2 - 16*d*e*f + 7*f^2)*arctan((sqrt(-d^2*x^2 + 1) - 1)/( 
d*x)) + (6*d^3*f^2*x^3 - 48*d^2*e^2 + 80*d*e*f + 16*(d^3*e*f - d^2*f^2)*x^ 
2 - 32*f^2 + 3*(4*d^3*e^2 - 16*d^2*e*f + 7*d*f^2)*x)*sqrt(-d^2*x^2 + 1))/d 
^3
 

Sympy [F]

\[ \int \frac {(e+f x)^2 \left (1-d^2 x^2\right )^{3/2}}{(1+d x)^2} \, dx=\int \frac {\left (- \left (d x - 1\right ) \left (d x + 1\right )\right )^{\frac {3}{2}} \left (e + f x\right )^{2}}{\left (d x + 1\right )^{2}}\, dx \] Input:

integrate((f*x+e)**2*(-d**2*x**2+1)**(3/2)/(d*x+1)**2,x)
 

Output:

Integral((-(d*x - 1)*(d*x + 1))**(3/2)*(e + f*x)**2/(d*x + 1)**2, x)
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 375, normalized size of antiderivative = 2.59 \[ \int \frac {(e+f x)^2 \left (1-d^2 x^2\right )^{3/2}}{(1+d x)^2} \, dx=\frac {{\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}} e^{2}}{2 \, {\left (d^{2} x + d\right )}} - \frac {{\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}} e f}{d^{3} x + d^{2}} + \frac {{\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}} f^{2}}{2 \, {\left (d^{4} x + d^{3}\right )}} + \frac {\sqrt {d^{2} x^{2} + 4 \, d x + 3} e f x}{d} + \frac {{\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}} f^{2} x}{4 \, d^{2}} + \frac {3 \, e^{2} \arcsin \left (d x\right )}{2 \, d} + \frac {3 \, \sqrt {-d^{2} x^{2} + 1} e^{2}}{2 \, d} + \frac {2 \, {\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}} e f}{3 \, d^{2}} - \frac {\sqrt {d^{2} x^{2} + 4 \, d x + 3} f^{2} x}{d^{2}} + \frac {3 \, \sqrt {-d^{2} x^{2} + 1} f^{2} x}{8 \, d^{2}} - \frac {i \, e f \arcsin \left (d x + 2\right )}{d^{2}} - \frac {3 \, e f \arcsin \left (d x\right )}{d^{2}} + \frac {2 \, \sqrt {d^{2} x^{2} + 4 \, d x + 3} e f}{d^{2}} - \frac {3 \, \sqrt {-d^{2} x^{2} + 1} e f}{d^{2}} - \frac {2 \, {\left (-d^{2} x^{2} + 1\right )}^{\frac {3}{2}} f^{2}}{3 \, d^{3}} + \frac {i \, f^{2} \arcsin \left (d x + 2\right )}{d^{3}} + \frac {15 \, f^{2} \arcsin \left (d x\right )}{8 \, d^{3}} - \frac {2 \, \sqrt {d^{2} x^{2} + 4 \, d x + 3} f^{2}}{d^{3}} + \frac {3 \, \sqrt {-d^{2} x^{2} + 1} f^{2}}{2 \, d^{3}} \] Input:

integrate((f*x+e)^2*(-d^2*x^2+1)^(3/2)/(d*x+1)^2,x, algorithm="maxima")
 

Output:

1/2*(-d^2*x^2 + 1)^(3/2)*e^2/(d^2*x + d) - (-d^2*x^2 + 1)^(3/2)*e*f/(d^3*x 
 + d^2) + 1/2*(-d^2*x^2 + 1)^(3/2)*f^2/(d^4*x + d^3) + sqrt(d^2*x^2 + 4*d* 
x + 3)*e*f*x/d + 1/4*(-d^2*x^2 + 1)^(3/2)*f^2*x/d^2 + 3/2*e^2*arcsin(d*x)/ 
d + 3/2*sqrt(-d^2*x^2 + 1)*e^2/d + 2/3*(-d^2*x^2 + 1)^(3/2)*e*f/d^2 - sqrt 
(d^2*x^2 + 4*d*x + 3)*f^2*x/d^2 + 3/8*sqrt(-d^2*x^2 + 1)*f^2*x/d^2 - I*e*f 
*arcsin(d*x + 2)/d^2 - 3*e*f*arcsin(d*x)/d^2 + 2*sqrt(d^2*x^2 + 4*d*x + 3) 
*e*f/d^2 - 3*sqrt(-d^2*x^2 + 1)*e*f/d^2 - 2/3*(-d^2*x^2 + 1)^(3/2)*f^2/d^3 
 + I*f^2*arcsin(d*x + 2)/d^3 + 15/8*f^2*arcsin(d*x)/d^3 - 2*sqrt(d^2*x^2 + 
 4*d*x + 3)*f^2/d^3 + 3/2*sqrt(-d^2*x^2 + 1)*f^2/d^3
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 455 vs. \(2 (128) = 256\).

Time = 0.23 (sec) , antiderivative size = 455, normalized size of antiderivative = 3.14 \[ \int \frac {(e+f x)^2 \left (1-d^2 x^2\right )^{3/2}}{(1+d x)^2} \, dx =\text {Too large to display} \] Input:

integrate((f*x+e)^2*(-d^2*x^2+1)^(3/2)/(d*x+1)^2,x, algorithm="giac")
 

Output:

1/192*((60*d^7*e^2*(2/(d*x + 1) - 1)^(7/2)*sgn(1/(d*x + 1))*sgn(d) + 156*d 
^7*e^2*(2/(d*x + 1) - 1)^(5/2)*sgn(1/(d*x + 1))*sgn(d) - 144*d^6*e*f*(2/(d 
*x + 1) - 1)^(7/2)*sgn(1/(d*x + 1))*sgn(d) + 132*d^7*e^2*(2/(d*x + 1) - 1) 
^(3/2)*sgn(1/(d*x + 1))*sgn(d) - 272*d^6*e*f*(2/(d*x + 1) - 1)^(5/2)*sgn(1 
/(d*x + 1))*sgn(d) + 75*d^5*f^2*(2/(d*x + 1) - 1)^(7/2)*sgn(1/(d*x + 1))*s 
gn(d) + 36*d^7*e^2*sqrt(2/(d*x + 1) - 1)*sgn(1/(d*x + 1))*sgn(d) - 176*d^6 
*e*f*(2/(d*x + 1) - 1)^(3/2)*sgn(1/(d*x + 1))*sgn(d) + 83*d^5*f^2*(2/(d*x 
+ 1) - 1)^(5/2)*sgn(1/(d*x + 1))*sgn(d) - 48*d^6*e*f*sqrt(2/(d*x + 1) - 1) 
*sgn(1/(d*x + 1))*sgn(d) + 77*d^5*f^2*(2/(d*x + 1) - 1)^(3/2)*sgn(1/(d*x + 
 1))*sgn(d) + 21*d^5*f^2*sqrt(2/(d*x + 1) - 1)*sgn(1/(d*x + 1))*sgn(d))*(d 
*x + 1)^4 - 48*(12*d^7*e^2*sgn(1/(d*x + 1))*sgn(d) - 16*d^6*e*f*sgn(1/(d*x 
 + 1))*sgn(d) + 7*d^5*f^2*sgn(1/(d*x + 1))*sgn(d))*arctan(sqrt(2/(d*x + 1) 
 - 1)))*abs(d)/d^9
 

Mupad [B] (verification not implemented)

Time = 6.37 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.67 \[ \int \frac {(e+f x)^2 \left (1-d^2 x^2\right )^{3/2}}{(1+d x)^2} \, dx=\frac {\sqrt {1-d^2\,x^2}\,\left (\frac {2\,\left (2\,d\,f^2\,\sqrt {-d^2}-2\,d^2\,e\,f\,\sqrt {-d^2}\right )}{3\,d^4}-\frac {2\,d\,e^2\,{\left (-d^2\right )}^{3/2}-2\,e\,f\,{\left (-d^2\right )}^{3/2}}{d^4}+\frac {x^2\,\left (2\,d\,f^2\,\sqrt {-d^2}-2\,d^2\,e\,f\,\sqrt {-d^2}\right )}{3\,d^2}-x\,\left (\frac {d^4\,e^2-4\,d^3\,e\,f+d^2\,f^2}{2\,d^4}+\frac {3\,f^2}{8\,d^2}\right )\,\sqrt {-d^2}+\frac {f^2\,x^3\,{\left (-d^2\right )}^{3/2}}{4\,d^2}\right )}{\sqrt {-d^2}}+\frac {\mathrm {asinh}\left (x\,\sqrt {-d^2}\right )\,\left (\frac {d^4\,e^2-4\,d^3\,e\,f+d^2\,f^2}{2\,d^4}+e^2+\frac {3\,f^2}{8\,d^2}\right )}{\sqrt {-d^2}} \] Input:

int(((e + f*x)^2*(1 - d^2*x^2)^(3/2))/(d*x + 1)^2,x)
 

Output:

((1 - d^2*x^2)^(1/2)*((2*(2*d*f^2*(-d^2)^(1/2) - 2*d^2*e*f*(-d^2)^(1/2)))/ 
(3*d^4) - (2*d*e^2*(-d^2)^(3/2) - 2*e*f*(-d^2)^(3/2))/d^4 + (x^2*(2*d*f^2* 
(-d^2)^(1/2) - 2*d^2*e*f*(-d^2)^(1/2)))/(3*d^2) - x*((d^4*e^2 + d^2*f^2 - 
4*d^3*e*f)/(2*d^4) + (3*f^2)/(8*d^2))*(-d^2)^(1/2) + (f^2*x^3*(-d^2)^(3/2) 
)/(4*d^2)))/(-d^2)^(1/2) + (asinh(x*(-d^2)^(1/2))*((d^4*e^2 + d^2*f^2 - 4* 
d^3*e*f)/(2*d^4) + e^2 + (3*f^2)/(8*d^2)))/(-d^2)^(1/2)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.57 \[ \int \frac {(e+f x)^2 \left (1-d^2 x^2\right )^{3/2}}{(1+d x)^2} \, dx=\frac {36 \mathit {asin} \left (d x \right ) d^{2} e^{2}-48 \mathit {asin} \left (d x \right ) d e f +21 \mathit {asin} \left (d x \right ) f^{2}-12 \sqrt {-d^{2} x^{2}+1}\, d^{3} e^{2} x -16 \sqrt {-d^{2} x^{2}+1}\, d^{3} e f \,x^{2}-6 \sqrt {-d^{2} x^{2}+1}\, d^{3} f^{2} x^{3}+48 \sqrt {-d^{2} x^{2}+1}\, d^{2} e^{2}+48 \sqrt {-d^{2} x^{2}+1}\, d^{2} e f x +16 \sqrt {-d^{2} x^{2}+1}\, d^{2} f^{2} x^{2}-80 \sqrt {-d^{2} x^{2}+1}\, d e f -21 \sqrt {-d^{2} x^{2}+1}\, d \,f^{2} x +32 \sqrt {-d^{2} x^{2}+1}\, f^{2}-48 d^{2} e^{2}+80 d e f -32 f^{2}}{24 d^{3}} \] Input:

int((f*x+e)^2*(-d^2*x^2+1)^(3/2)/(d*x+1)^2,x)
 

Output:

(36*asin(d*x)*d**2*e**2 - 48*asin(d*x)*d*e*f + 21*asin(d*x)*f**2 - 12*sqrt 
( - d**2*x**2 + 1)*d**3*e**2*x - 16*sqrt( - d**2*x**2 + 1)*d**3*e*f*x**2 - 
 6*sqrt( - d**2*x**2 + 1)*d**3*f**2*x**3 + 48*sqrt( - d**2*x**2 + 1)*d**2* 
e**2 + 48*sqrt( - d**2*x**2 + 1)*d**2*e*f*x + 16*sqrt( - d**2*x**2 + 1)*d* 
*2*f**2*x**2 - 80*sqrt( - d**2*x**2 + 1)*d*e*f - 21*sqrt( - d**2*x**2 + 1) 
*d*f**2*x + 32*sqrt( - d**2*x**2 + 1)*f**2 - 48*d**2*e**2 + 80*d*e*f - 32* 
f**2)/(24*d**3)