\(\int \frac {(c+d x)^2 (A+B x+C x^2+D x^3)}{(c^2-d^2 x^2)^{5/2}} \, dx\) [187]

Optimal result

Integrand size = 39, antiderivative size = 169 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {\left (c^2 C d+B c d^2+A d^3+c^3 D\right ) (c+d x)^2}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\left (5 c^2 C d+2 B c d^2-A d^3+8 c^3 D\right ) (c+d x)}{3 c^2 d^4 \sqrt {c^2-d^2 x^2}}-\frac {D \sqrt {c^2-d^2 x^2}}{d^4}+\frac {(C d+2 c D) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{d^4} \] Output:

1/3*(A*d^3+B*c*d^2+C*c^2*d+D*c^3)*(d*x+c)^2/c/d^4/(-d^2*x^2+c^2)^(3/2)-1/3 
*(-A*d^3+2*B*c*d^2+5*C*c^2*d+8*D*c^3)*(d*x+c)/c^2/d^4/(-d^2*x^2+c^2)^(1/2) 
-D*(-d^2*x^2+c^2)^(1/2)/d^4+(C*d+2*D*c)*arctan(d*x/(-d^2*x^2+c^2)^(1/2))/d 
^4
 

Mathematica [A] (verified)

Time = 1.07 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.80 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {\frac {\sqrt {c^2-d^2 x^2} \left (10 c^4 D+A d^4 x-2 c d^3 (A+B x)+2 c^3 d (2 C-7 D x)+c^2 d^2 (B+x (-5 C+3 D x))\right )}{c^2 (c-d x)^2}+6 (C d+2 c D) \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{3 d^4} \] Input:

Integrate[((c + d*x)^2*(A + B*x + C*x^2 + D*x^3))/(c^2 - d^2*x^2)^(5/2),x]
 

Output:

-1/3*((Sqrt[c^2 - d^2*x^2]*(10*c^4*D + A*d^4*x - 2*c*d^3*(A + B*x) + 2*c^3 
*d*(2*C - 7*D*x) + c^2*d^2*(B + x*(-5*C + 3*D*x))))/(c^2*(c - d*x)^2) + 6* 
(C*d + 2*c*D)*ArcTan[(d*x)/(Sqrt[c^2] - Sqrt[c^2 - d^2*x^2])])/d^4
 

Rubi [A] (verified)

Time = 0.95 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2166, 2166, 27, 455, 224, 216}

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.

\(\displaystyle \int \frac {(c+d x)^2 \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 2166

\(\displaystyle \frac {(c+d x)^2 \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\int \frac {(c+d x) \left (\frac {3 c D x^2}{d}+\frac {3 c (C d+c D) x}{d^2}+\frac {2 D c^3+2 C d c^2+2 B d^2 c-A d^3}{d^3}\right )}{\left (c^2-d^2 x^2\right )^{3/2}}dx}{3 c}\)

\(\Big \downarrow \) 2166

\(\displaystyle \frac {(c+d x)^2 \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {(c+d x) \left (-A d^3+2 B c d^2+8 c^3 D+5 c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\int \frac {3 c^2 (C d+D x d+2 c D)}{d^3 \sqrt {c^2-d^2 x^2}}dx}{c}}{3 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(c+d x)^2 \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {(c+d x) \left (-A d^3+2 B c d^2+8 c^3 D+5 c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {3 c \int \frac {C d+D x d+2 c D}{\sqrt {c^2-d^2 x^2}}dx}{d^3}}{3 c}\)

\(\Big \downarrow \) 455

\(\displaystyle \frac {(c+d x)^2 \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {(c+d x) \left (-A d^3+2 B c d^2+8 c^3 D+5 c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {3 c \left ((2 c D+C d) \int \frac {1}{\sqrt {c^2-d^2 x^2}}dx-\frac {D \sqrt {c^2-d^2 x^2}}{d}\right )}{d^3}}{3 c}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {(c+d x)^2 \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {(c+d x) \left (-A d^3+2 B c d^2+8 c^3 D+5 c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {3 c \left ((2 c D+C d) \int \frac {1}{\frac {d^2 x^2}{c^2-d^2 x^2}+1}d\frac {x}{\sqrt {c^2-d^2 x^2}}-\frac {D \sqrt {c^2-d^2 x^2}}{d}\right )}{d^3}}{3 c}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {(c+d x)^2 \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {(c+d x) \left (-A d^3+2 B c d^2+8 c^3 D+5 c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {3 c \left (\frac {\arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right ) (2 c D+C d)}{d}-\frac {D \sqrt {c^2-d^2 x^2}}{d}\right )}{d^3}}{3 c}\)

Input:

Int[((c + d*x)^2*(A + B*x + C*x^2 + D*x^3))/(c^2 - d^2*x^2)^(5/2),x]
 

Output:

((c^2*C*d + B*c*d^2 + A*d^3 + c^3*D)*(c + d*x)^2)/(3*c*d^4*(c^2 - d^2*x^2) 
^(3/2)) - (((5*c^2*C*d + 2*B*c*d^2 - A*d^3 + 8*c^3*D)*(c + d*x))/(c*d^4*Sq 
rt[c^2 - d^2*x^2]) - (3*c*(-((D*Sqrt[c^2 - d^2*x^2])/d) + ((C*d + 2*c*D)*A 
rcTan[(d*x)/Sqrt[c^2 - d^2*x^2]])/d))/d^3)/(3*c)
 

Defintions of rubi rules used

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 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 455

Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( 
a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c   Int[(a + b*x^2)^p, x], x] 
/; FreeQ[{a, b, c, d, p}, x] &&  !LeQ[p, -1]
 

rule 2166

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{Qx = PolynomialQuotient[Pq, a*e + b*d*x, x], R = PolynomialRemainde 
r[Pq, a*e + b*d*x, x]}, Simp[(-d)*R*(d + e*x)^m*((a + b*x^2)^(p + 1)/(2*a*e 
*(p + 1))), x] + Simp[d/(2*a*(p + 1))   Int[(d + e*x)^(m - 1)*(a + b*x^2)^( 
p + 1)*ExpandToSum[2*a*e*(p + 1)*Qx + R*(m + 2*p + 2), x], x], x]] /; FreeQ 
[{a, b, d, e}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 0] && ILtQ[p + 1/2, 
 0] && GtQ[m, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(393\) vs. \(2(157)=314\).

Time = 0.50 (sec) , antiderivative size = 394, normalized size of antiderivative = 2.33

method	result	size
default	\(A \,c^{2} \left (\frac {x}{3 c^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{4} \sqrt {-d^{2} x^{2}+c^{2}}}\right )+\frac {c \left (2 A d +B c \right )}{3 d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+d \left (C d +2 D c \right ) \left (\frac {x^{3}}{3 d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}-\frac {\frac {x}{\sqrt {-d^{2} x^{2}+c^{2}}\, d^{2}}-\frac {\arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{d^{2} \sqrt {d^{2}}}}{d^{2}}\right )+\left (A \,d^{2}+2 B c d +C \,c^{2}\right ) \left (\frac {x}{2 d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}-\frac {c^{2} \left (\frac {x}{3 c^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{4} \sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 d^{2}}\right )+\left (B \,d^{2}+2 C c d +D c^{2}\right ) \left (\frac {x^{2}}{d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}-\frac {2 c^{2}}{3 d^{4} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}\right )+D d^{2} \left (-\frac {x^{4}}{d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+\frac {4 c^{2} \left (\frac {x^{2}}{d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}-\frac {2 c^{2}}{3 d^{4} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}\right )}{d^{2}}\right )\)	\(394\)

Input:

int((d*x+c)^2*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),x,method=_RETURNVER 
BOSE)
 

Output:

A*c^2*(1/3*x/c^2/(-d^2*x^2+c^2)^(3/2)+2/3*x/c^4/(-d^2*x^2+c^2)^(1/2))+1/3* 
c*(2*A*d+B*c)/d^2/(-d^2*x^2+c^2)^(3/2)+d*(C*d+2*D*c)*(1/3*x^3/d^2/(-d^2*x^ 
2+c^2)^(3/2)-1/d^2*(1/(-d^2*x^2+c^2)^(1/2)/d^2*x-1/d^2/(d^2)^(1/2)*arctan( 
(d^2)^(1/2)*x/(-d^2*x^2+c^2)^(1/2))))+(A*d^2+2*B*c*d+C*c^2)*(1/2*x/d^2/(-d 
^2*x^2+c^2)^(3/2)-1/2*c^2/d^2*(1/3*x/c^2/(-d^2*x^2+c^2)^(3/2)+2/3*x/c^4/(- 
d^2*x^2+c^2)^(1/2)))+(B*d^2+2*C*c*d+D*c^2)*(1/d^2*x^2/(-d^2*x^2+c^2)^(3/2) 
-2/3*c^2/d^4/(-d^2*x^2+c^2)^(3/2))+D*d^2*(-x^4/d^2/(-d^2*x^2+c^2)^(3/2)+4* 
c^2/d^2*(1/d^2*x^2/(-d^2*x^2+c^2)^(3/2)-2/3*c^2/d^4/(-d^2*x^2+c^2)^(3/2)))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.80 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {10 \, D c^{5} + 4 \, C c^{4} d + B c^{3} d^{2} - 2 \, A c^{2} d^{3} + {\left (10 \, D c^{3} d^{2} + 4 \, C c^{2} d^{3} + B c d^{4} - 2 \, A d^{5}\right )} x^{2} - 2 \, {\left (10 \, D c^{4} d + 4 \, C c^{3} d^{2} + B c^{2} d^{3} - 2 \, A c d^{4}\right )} x + 6 \, {\left (2 \, D c^{5} + C c^{4} d + {\left (2 \, D c^{3} d^{2} + C c^{2} d^{3}\right )} x^{2} - 2 \, {\left (2 \, D c^{4} d + C c^{3} d^{2}\right )} x\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + {\left (3 \, D c^{2} d^{2} x^{2} + 10 \, D c^{4} + 4 \, C c^{3} d + B c^{2} d^{2} - 2 \, A c d^{3} - {\left (14 \, D c^{3} d + 5 \, C c^{2} d^{2} + 2 \, B c d^{3} - A d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{3 \, {\left (c^{2} d^{6} x^{2} - 2 \, c^{3} d^{5} x + c^{4} d^{4}\right )}} \] Input:

integrate((d*x+c)^2*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),x, algorithm= 
"fricas")
 

Output:

-1/3*(10*D*c^5 + 4*C*c^4*d + B*c^3*d^2 - 2*A*c^2*d^3 + (10*D*c^3*d^2 + 4*C 
*c^2*d^3 + B*c*d^4 - 2*A*d^5)*x^2 - 2*(10*D*c^4*d + 4*C*c^3*d^2 + B*c^2*d^ 
3 - 2*A*c*d^4)*x + 6*(2*D*c^5 + C*c^4*d + (2*D*c^3*d^2 + C*c^2*d^3)*x^2 - 
2*(2*D*c^4*d + C*c^3*d^2)*x)*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x)) + ( 
3*D*c^2*d^2*x^2 + 10*D*c^4 + 4*C*c^3*d + B*c^2*d^2 - 2*A*c*d^3 - (14*D*c^3 
*d + 5*C*c^2*d^2 + 2*B*c*d^3 - A*d^4)*x)*sqrt(-d^2*x^2 + c^2))/(c^2*d^6*x^ 
2 - 2*c^3*d^5*x + c^4*d^4)
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {(c+d x)^2 \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (c + d x\right )^{2} \left (A + B x + C x^{2} + D x^{3}\right )}{\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((d*x+c)**2*(D*x**3+C*x**2+B*x+A)/(-d**2*x**2+c**2)**(5/2),x)
 

Output:

Integral((c + d*x)**2*(A + B*x + C*x**2 + D*x**3)/(-(-c + d*x)*(c + d*x))* 
*(5/2), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 415 vs. \(2 (157) = 314\).

Time = 0.12 (sec) , antiderivative size = 415, normalized size of antiderivative = 2.46 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {D x^{4}}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}}} + \frac {1}{3} \, {\left (2 \, D c d + C d^{2}\right )} x {\left (\frac {3 \, x^{2}}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{2}} - \frac {2 \, c^{2}}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{4}}\right )} + \frac {4 \, D c^{2} x^{2}}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{2}} + \frac {A x}{3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}}} - \frac {8 \, D c^{4}}{3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{4}} + \frac {B c^{2}}{3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{2}} + \frac {2 \, A c}{3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d} + \frac {2 \, A x}{3 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{2}} + \frac {{\left (D c^{2} + 2 \, C c d + B d^{2}\right )} x^{2}}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{2}} + \frac {{\left (C c^{2} + 2 \, B c d + A d^{2}\right )} x}{3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{2}} - \frac {2 \, {\left (D c^{2} + 2 \, C c d + B d^{2}\right )} c^{2}}{3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{4}} - \frac {{\left (2 \, D c d + C d^{2}\right )} x}{3 \, \sqrt {-d^{2} x^{2} + c^{2}} d^{4}} - \frac {{\left (C c^{2} + 2 \, B c d + A d^{2}\right )} x}{3 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{2} d^{2}} + \frac {{\left (2 \, D c d + C d^{2}\right )} \arcsin \left (\frac {d x}{c}\right )}{d^{5}} \] Input:

integrate((d*x+c)^2*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),x, algorithm= 
"maxima")
 

Output:

-D*x^4/(-d^2*x^2 + c^2)^(3/2) + 1/3*(2*D*c*d + C*d^2)*x*(3*x^2/((-d^2*x^2 
+ c^2)^(3/2)*d^2) - 2*c^2/((-d^2*x^2 + c^2)^(3/2)*d^4)) + 4*D*c^2*x^2/((-d 
^2*x^2 + c^2)^(3/2)*d^2) + 1/3*A*x/(-d^2*x^2 + c^2)^(3/2) - 8/3*D*c^4/((-d 
^2*x^2 + c^2)^(3/2)*d^4) + 1/3*B*c^2/((-d^2*x^2 + c^2)^(3/2)*d^2) + 2/3*A* 
c/((-d^2*x^2 + c^2)^(3/2)*d) + 2/3*A*x/(sqrt(-d^2*x^2 + c^2)*c^2) + (D*c^2 
 + 2*C*c*d + B*d^2)*x^2/((-d^2*x^2 + c^2)^(3/2)*d^2) + 1/3*(C*c^2 + 2*B*c* 
d + A*d^2)*x/((-d^2*x^2 + c^2)^(3/2)*d^2) - 2/3*(D*c^2 + 2*C*c*d + B*d^2)* 
c^2/((-d^2*x^2 + c^2)^(3/2)*d^4) - 1/3*(2*D*c*d + C*d^2)*x/(sqrt(-d^2*x^2 
+ c^2)*d^4) - 1/3*(C*c^2 + 2*B*c*d + A*d^2)*x/(sqrt(-d^2*x^2 + c^2)*c^2*d^ 
2) + (2*D*c*d + C*d^2)*arcsin(d*x/c)/d^5
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 342 vs. \(2 (157) = 314\).

Time = 0.15 (sec) , antiderivative size = 342, normalized size of antiderivative = 2.02 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {{\left (2 \, D c + C d\right )} \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{d^{3} {\left | d \right |}} - \frac {\sqrt {-d^{2} x^{2} + c^{2}} D}{d^{4}} - \frac {2 \, {\left (7 \, D c^{3} + 4 \, C c^{2} d + B c d^{2} - 2 \, A d^{3} - \frac {3 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} B c}{x} - \frac {15 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} D c^{3}}{d^{2} x} - \frac {9 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} C c^{2}}{d x} + \frac {3 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} A d}{x} + \frac {6 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} D c^{3}}{d^{4} x^{2}} + \frac {3 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} C c^{2}}{d^{3} x^{2}} - \frac {3 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} A}{d x^{2}}\right )}}{3 \, c^{2} d^{3} {\left (\frac {c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}}{d^{2} x} - 1\right )}^{3} {\left | d \right |}} \] Input:

integrate((d*x+c)^2*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),x, algorithm= 
"giac")
 

Output:

(2*D*c + C*d)*arcsin(d*x/c)*sgn(c)*sgn(d)/(d^3*abs(d)) - sqrt(-d^2*x^2 + c 
^2)*D/d^4 - 2/3*(7*D*c^3 + 4*C*c^2*d + B*c*d^2 - 2*A*d^3 - 3*(c*d + sqrt(- 
d^2*x^2 + c^2)*abs(d))*B*c/x - 15*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))*D*c^ 
3/(d^2*x) - 9*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))*C*c^2/(d*x) + 3*(c*d + s 
qrt(-d^2*x^2 + c^2)*abs(d))*A*d/x + 6*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^ 
2*D*c^3/(d^4*x^2) + 3*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^2*C*c^2/(d^3*x^2 
) - 3*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^2*A/(d*x^2))/(c^2*d^3*((c*d + sq 
rt(-d^2*x^2 + c^2)*abs(d))/(d^2*x) - 1)^3*abs(d))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\int \frac {{\left (c+d\,x\right )}^2\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (c^2-d^2\,x^2\right )}^{5/2}} \,d x \] Input:

int(((c + d*x)^2*(A + B*x + C*x^2 + x^3*D))/(c^2 - d^2*x^2)^(5/2),x)
 

Output:

int(((c + d*x)^2*(A + B*x + C*x^2 + x^3*D))/(c^2 - d^2*x^2)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 359, normalized size of antiderivative = 2.12 \[ \int \frac {(c+d x)^2 \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {9 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) c^{4}-9 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) c^{3} d x -9 \mathit {asin} \left (\frac {d x}{c}\right ) c^{5}+18 \mathit {asin} \left (\frac {d x}{c}\right ) c^{4} d x -9 \mathit {asin} \left (\frac {d x}{c}\right ) c^{3} d^{2} x^{2}-2 \sqrt {-d^{2} x^{2}+c^{2}}\, a c \,d^{2}+\sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{3} x -\sqrt {-d^{2} x^{2}+c^{2}}\, b c \,d^{2} x +6 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{4}-11 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{3} d x +3 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{2} d^{2} x^{2}+2 a \,c^{2} d^{2}+a c \,d^{3} x -a \,d^{4} x^{2}-b \,c^{2} d^{2} x +3 b c \,d^{3} x^{2}-6 c^{5}-11 c^{4} d x +24 c^{3} d^{2} x^{2}-3 c^{2} d^{3} x^{3}}{3 c^{2} d^{3} \left (\sqrt {-d^{2} x^{2}+c^{2}}\, c -\sqrt {-d^{2} x^{2}+c^{2}}\, d x -c^{2}+2 c d x -d^{2} x^{2}\right )} \] Input:

int((d*x+c)^2*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),x)
 

Output:

(9*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*c**4 - 9*sqrt(c**2 - d**2*x**2)*as 
in((d*x)/c)*c**3*d*x - 9*asin((d*x)/c)*c**5 + 18*asin((d*x)/c)*c**4*d*x - 
9*asin((d*x)/c)*c**3*d**2*x**2 - 2*sqrt(c**2 - d**2*x**2)*a*c*d**2 + sqrt( 
c**2 - d**2*x**2)*a*d**3*x - sqrt(c**2 - d**2*x**2)*b*c*d**2*x + 6*sqrt(c* 
*2 - d**2*x**2)*c**4 - 11*sqrt(c**2 - d**2*x**2)*c**3*d*x + 3*sqrt(c**2 - 
d**2*x**2)*c**2*d**2*x**2 + 2*a*c**2*d**2 + a*c*d**3*x - a*d**4*x**2 - b*c 
**2*d**2*x + 3*b*c*d**3*x**2 - 6*c**5 - 11*c**4*d*x + 24*c**3*d**2*x**2 - 
3*c**2*d**3*x**3)/(3*c**2*d**3*(sqrt(c**2 - d**2*x**2)*c - sqrt(c**2 - d** 
2*x**2)*d*x - c**2 + 2*c*d*x - d**2*x**2))