\(\int \frac {(c+d x)^3 (A+B x+C x^2+D x^3)}{(c^2-d^2 x^2)^{5/2}} \, dx\) [186]

Optimal result

Integrand size = 39, antiderivative size = 201 \[ \int \frac {(c+d x)^3 \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {\left (c^2 C d+B c d^2+A d^3+c^3 D\right ) (c+d x)^3}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\left (2 c C d+B d^2+3 c^2 D\right ) (c+d x)^2}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\left (2 \left (3 c C d+B d^2+6 c^2 D\right )+c d D x\right ) \sqrt {c^2-d^2 x^2}}{2 c d^4}+\frac {\left (6 c C d+2 B d^2+11 c^2 D\right ) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{2 d^4} \] Output:

1/3*(A*d^3+B*c*d^2+C*c^2*d+D*c^3)*(d*x+c)^3/c/d^4/(-d^2*x^2+c^2)^(3/2)-(B* 
d^2+2*C*c*d+3*D*c^2)*(d*x+c)^2/c/d^4/(-d^2*x^2+c^2)^(1/2)-1/2*(D*c*d*x+2*B 
*d^2+6*C*c*d+12*D*c^2)*(-d^2*x^2+c^2)^(1/2)/c/d^4+1/2*(2*B*d^2+6*C*c*d+11* 
D*c^2)*arctan(d*x/(-d^2*x^2+c^2)^(1/2))/d^4
 

Mathematica [A] (verified)

Time = 1.36 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.81 \[ \int \frac {(c+d x)^3 \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {\frac {\sqrt {c^2-d^2 x^2} \left (52 c^4 D-2 A d^4 x+c^3 d (28 C-71 D x)+2 c^2 d^2 (5 B+x (-19 C+6 D x))+c d^3 \left (-2 A+x \left (-14 B+6 C x+3 D x^2\right )\right )\right )}{c (c-d x)^2}+6 \left (6 c C d+2 B d^2+11 c^2 D\right ) \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{6 d^4} \] Input:

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

Output:

-1/6*((Sqrt[c^2 - d^2*x^2]*(52*c^4*D - 2*A*d^4*x + c^3*d*(28*C - 71*D*x) + 
 2*c^2*d^2*(5*B + x*(-19*C + 6*D*x)) + c*d^3*(-2*A + x*(-14*B + 6*C*x + 3* 
D*x^2))))/(c*(c - d*x)^2) + 6*(6*c*C*d + 2*B*d^2 + 11*c^2*D)*ArcTan[(d*x)/ 
(Sqrt[c^2] - Sqrt[c^2 - d^2*x^2])])/d^4
 

Rubi [A] (verified)

Time = 1.07 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {2166, 27, 2166, 27, 676, 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.

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

\(\Big \downarrow \) 2166

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2166

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 676

\(\displaystyle \frac {(c+d x)^3 \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {(c+d x)^2 \left (B d^2+3 c^2 D+2 c C d\right )}{d^4 \sqrt {c^2-d^2 x^2}}-\frac {\frac {1}{2} c \left (2 B d^2+11 c^2 D+6 c C d\right ) \int \frac {1}{\sqrt {c^2-d^2 x^2}}dx-\frac {\sqrt {c^2-d^2 x^2} \left (B d^2+6 c^2 D+3 c C d\right )}{d}-\frac {1}{2} c D x \sqrt {c^2-d^2 x^2}}{d^3}}{c}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {(c+d x)^3 \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {(c+d x)^2 \left (B d^2+3 c^2 D+2 c C d\right )}{d^4 \sqrt {c^2-d^2 x^2}}-\frac {\frac {1}{2} c \left (2 B d^2+11 c^2 D+6 c C d\right ) \int \frac {1}{\frac {d^2 x^2}{c^2-d^2 x^2}+1}d\frac {x}{\sqrt {c^2-d^2 x^2}}-\frac {\sqrt {c^2-d^2 x^2} \left (B d^2+6 c^2 D+3 c C d\right )}{d}-\frac {1}{2} c D x \sqrt {c^2-d^2 x^2}}{d^3}}{c}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {(c+d x)^3 \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {(c+d x)^2 \left (B d^2+3 c^2 D+2 c C d\right )}{d^4 \sqrt {c^2-d^2 x^2}}-\frac {\frac {c \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right ) \left (2 B d^2+11 c^2 D+6 c C d\right )}{2 d}-\frac {\sqrt {c^2-d^2 x^2} \left (B d^2+6 c^2 D+3 c C d\right )}{d}-\frac {1}{2} c D x \sqrt {c^2-d^2 x^2}}{d^3}}{c}\)

Input:

Int[((c + d*x)^3*(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)^3)/(3*c*d^4*(c^2 - d^2*x^2) 
^(3/2)) - (((2*c*C*d + B*d^2 + 3*c^2*D)*(c + d*x)^2)/(d^4*Sqrt[c^2 - d^2*x 
^2]) - (-(((3*c*C*d + B*d^2 + 6*c^2*D)*Sqrt[c^2 - d^2*x^2])/d) - (c*D*x*Sq 
rt[c^2 - d^2*x^2])/2 + (c*(6*c*C*d + 2*B*d^2 + 11*c^2*D)*ArcTan[(d*x)/Sqrt 
[c^2 - d^2*x^2]])/(2*d))/d^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 676

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x 
_Symbol] :> Simp[(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + (Sim 
p[e*g*x*((a + c*x^2)^(p + 1)/(c*(2*p + 3))), x] - Simp[(a*e*g - c*d*f*(2*p 
+ 3))/(c*(2*p + 3))   Int[(a + c*x^2)^p, x], x]) /; FreeQ[{a, c, d, e, f, g 
, 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. \(538\) vs. \(2(185)=370\).

Time = 0.60 (sec) , antiderivative size = 539, normalized size of antiderivative = 2.68

method	result	size
default	\(A \,c^{3} \left (\frac {x}{3 c^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{4} \sqrt {-d^{2} x^{2}+c^{2}}}\right )+\frac {c^{2} \left (3 A d +B c \right )}{3 d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+d^{2} \left (C d +3 D c \right ) \left (-\frac {x^{4}}{d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+\frac {4 c^{2} \left (\frac {x^{2}}{d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}-\frac {2 c^{2}}{3 d^{4} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}\right )}{d^{2}}\right )+c \left (3 A \,d^{2}+3 B c d +C \,c^{2}\right ) \left (\frac {x}{2 d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}-\frac {c^{2} \left (\frac {x}{3 c^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{4} \sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 d^{2}}\right )+d \left (B \,d^{2}+3 C c d +3 D c^{2}\right ) \left (\frac {x^{3}}{3 d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}-\frac {\frac {x}{\sqrt {-d^{2} x^{2}+c^{2}}\, d^{2}}-\frac {\arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{d^{2} \sqrt {d^{2}}}}{d^{2}}\right )+\left (A \,d^{3}+3 B c \,d^{2}+3 C \,c^{2} d +D c^{3}\right ) \left (\frac {x^{2}}{d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}-\frac {2 c^{2}}{3 d^{4} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}\right )+d^{3} D \left (-\frac {x^{5}}{2 d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+\frac {5 c^{2} \left (\frac {x^{3}}{3 d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}-\frac {\frac {x}{\sqrt {-d^{2} x^{2}+c^{2}}\, d^{2}}-\frac {\arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{d^{2} \sqrt {d^{2}}}}{d^{2}}\right )}{2 d^{2}}\right )\)	\(539\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 354, normalized size of antiderivative = 1.76 \[ \int \frac {(c+d x)^3 \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {52 \, D c^{5} + 28 \, C c^{4} d + 10 \, B c^{3} d^{2} - 2 \, A c^{2} d^{3} + 2 \, {\left (26 \, D c^{3} d^{2} + 14 \, C c^{2} d^{3} + 5 \, B c d^{4} - A d^{5}\right )} x^{2} - 4 \, {\left (26 \, D c^{4} d + 14 \, C c^{3} d^{2} + 5 \, B c^{2} d^{3} - A c d^{4}\right )} x + 6 \, {\left (11 \, D c^{5} + 6 \, C c^{4} d + 2 \, B c^{3} d^{2} + {\left (11 \, D c^{3} d^{2} + 6 \, C c^{2} d^{3} + 2 \, B c d^{4}\right )} x^{2} - 2 \, {\left (11 \, D c^{4} d + 6 \, C c^{3} d^{2} + 2 \, B c^{2} d^{3}\right )} x\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + {\left (3 \, D c d^{3} x^{3} + 52 \, D c^{4} + 28 \, C c^{3} d + 10 \, B c^{2} d^{2} - 2 \, A c d^{3} + 6 \, {\left (2 \, D c^{2} d^{2} + C c d^{3}\right )} x^{2} - {\left (71 \, D c^{3} d + 38 \, C c^{2} d^{2} + 14 \, B c d^{3} + 2 \, A d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{6 \, {\left (c d^{6} x^{2} - 2 \, c^{2} d^{5} x + c^{3} d^{4}\right )}} \] Input:

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

Output:

-1/6*(52*D*c^5 + 28*C*c^4*d + 10*B*c^3*d^2 - 2*A*c^2*d^3 + 2*(26*D*c^3*d^2 
 + 14*C*c^2*d^3 + 5*B*c*d^4 - A*d^5)*x^2 - 4*(26*D*c^4*d + 14*C*c^3*d^2 + 
5*B*c^2*d^3 - A*c*d^4)*x + 6*(11*D*c^5 + 6*C*c^4*d + 2*B*c^3*d^2 + (11*D*c 
^3*d^2 + 6*C*c^2*d^3 + 2*B*c*d^4)*x^2 - 2*(11*D*c^4*d + 6*C*c^3*d^2 + 2*B* 
c^2*d^3)*x)*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x)) + (3*D*c*d^3*x^3 + 5 
2*D*c^4 + 28*C*c^3*d + 10*B*c^2*d^2 - 2*A*c*d^3 + 6*(2*D*c^2*d^2 + C*c*d^3 
)*x^2 - (71*D*c^3*d + 38*C*c^2*d^2 + 14*B*c*d^3 + 2*A*d^4)*x)*sqrt(-d^2*x^ 
2 + c^2))/(c*d^6*x^2 - 2*c^2*d^5*x + c^3*d^4)
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

Integral((c + d*x)**3*(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. 623 vs. \(2 (185) = 370\).

Time = 0.13 (sec) , antiderivative size = 623, normalized size of antiderivative = 3.10 \[ \int \frac {(c+d x)^3 \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

5/6*D*c^2*d*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)) - 1/2*D*d*x^5/(-d^2*x^2 + c^2)^(3/2) + 1/3*(3*D*c^2*d + 3*C* 
c*d^2 + B*d^3)*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)) + 1/3*A*c*x/(-d^2*x^2 + c^2)^(3/2) - (3*D*c*d^2 + C*d^3)* 
x^4/((-d^2*x^2 + c^2)^(3/2)*d^2) + 1/3*B*c^3/((-d^2*x^2 + c^2)^(3/2)*d^2) 
+ A*c^2/((-d^2*x^2 + c^2)^(3/2)*d) + 2/3*A*x/(sqrt(-d^2*x^2 + c^2)*c) - 5/ 
6*D*c^2*x/(sqrt(-d^2*x^2 + c^2)*d^3) + 5/2*D*c^2*arcsin(d*x/c)/d^4 + 4*(3* 
D*c*d^2 + C*d^3)*c^2*x^2/((-d^2*x^2 + c^2)^(3/2)*d^4) + (D*c^3 + 3*C*c^2*d 
 + 3*B*c*d^2 + A*d^3)*x^2/((-d^2*x^2 + c^2)^(3/2)*d^2) + 1/3*(C*c^3 + 3*B* 
c^2*d + 3*A*c*d^2)*x/((-d^2*x^2 + c^2)^(3/2)*d^2) - 8/3*(3*D*c*d^2 + C*d^3 
)*c^4/((-d^2*x^2 + c^2)^(3/2)*d^6) - 2/3*(D*c^3 + 3*C*c^2*d + 3*B*c*d^2 + 
A*d^3)*c^2/((-d^2*x^2 + c^2)^(3/2)*d^4) - 1/3*(3*D*c^2*d + 3*C*c*d^2 + B*d 
^3)*x/(sqrt(-d^2*x^2 + c^2)*d^4) - 1/3*(C*c^3 + 3*B*c^2*d + 3*A*c*d^2)*x/( 
sqrt(-d^2*x^2 + c^2)*c^2*d^2) + (3*D*c^2*d + 3*C*c*d^2 + B*d^3)*arcsin(d*x 
/c)/d^5
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 380 vs. \(2 (185) = 370\).

Time = 0.16 (sec) , antiderivative size = 380, normalized size of antiderivative = 1.89 \[ \int \frac {(c+d x)^3 \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {1}{2} \, \sqrt {-d^{2} x^{2} + c^{2}} {\left (\frac {D x}{d^{3}} + \frac {2 \, {\left (3 \, D c d^{6} + C d^{7}\right )}}{d^{10}}\right )} + \frac {{\left (11 \, D c^{2} + 6 \, C c d + 2 \, B d^{2}\right )} \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{2 \, d^{3} {\left | d \right |}} - \frac {2 \, {\left (17 \, D c^{3} + 11 \, C c^{2} d + 5 \, B c d^{2} - A d^{3} - \frac {12 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} B c}{x} - \frac {36 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} D c^{3}}{d^{2} x} - \frac {24 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} C c^{2}}{d x} + \frac {15 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} D c^{3}}{d^{4} x^{2}} + \frac {9 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} C c^{2}}{d^{3} x^{2}} + \frac {3 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} B c}{d^{2} x^{2}} - \frac {3 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} A}{d x^{2}}\right )}}{3 \, c d^{3} {\left (\frac {c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}}{d^{2} x} - 1\right )}^{3} {\left | d \right |}} \] Input:

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

Output:

-1/2*sqrt(-d^2*x^2 + c^2)*(D*x/d^3 + 2*(3*D*c*d^6 + C*d^7)/d^10) + 1/2*(11 
*D*c^2 + 6*C*c*d + 2*B*d^2)*arcsin(d*x/c)*sgn(c)*sgn(d)/(d^3*abs(d)) - 2/3 
*(17*D*c^3 + 11*C*c^2*d + 5*B*c*d^2 - A*d^3 - 12*(c*d + sqrt(-d^2*x^2 + c^ 
2)*abs(d))*B*c/x - 36*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))*D*c^3/(d^2*x) - 
24*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))*C*c^2/(d*x) + 15*(c*d + sqrt(-d^2*x 
^2 + c^2)*abs(d))^2*D*c^3/(d^4*x^2) + 9*(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*B*c/(d^2*x^2 
) - 3*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^2*A/(d*x^2))/(c*d^3*((c*d + sqrt 
(-d^2*x^2 + c^2)*abs(d))/(d^2*x) - 1)^3*abs(d))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 494, normalized size of antiderivative = 2.46 \[ \int \frac {(c+d x)^3 \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {-34 c^{5}+18 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{2} d^{2} x^{2}+34 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{4}+4 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d -6 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{3} d +102 \mathit {asin} \left (\frac {d x}{c}\right ) c^{4} d x -51 \mathit {asin} \left (\frac {d x}{c}\right ) c^{3} d^{2} x^{2}-4 \sqrt {-d^{2} x^{2}+c^{2}}\, a c \,d^{2}-8 b \,c^{2} d^{2} x +20 b c \,d^{3} x^{2}+51 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) c^{4}+4 a \,c^{2} d^{2}+4 a \,d^{4} x^{2}+6 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{2} d -51 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) c^{3} d x +12 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{2} d^{2} x -6 \mathit {asin} \left (\frac {d x}{c}\right ) b c \,d^{3} x^{2}-8 \sqrt {-d^{2} x^{2}+c^{2}}\, b c \,d^{2} x -63 c^{4} d x +137 c^{3} d^{2} x^{2}-21 c^{2} d^{3} x^{3}-3 c \,d^{4} x^{4}-63 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{3} d x +3 \sqrt {-d^{2} x^{2}+c^{2}}\, c \,d^{3} x^{3}-6 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) b c \,d^{2} x -51 \mathit {asin} \left (\frac {d x}{c}\right ) c^{5}-4 b \,c^{3} d}{6 c \,d^{3} \left (\sqrt {-d^{2} x^{2}+c^{2}}\, c -\sqrt {-d^{2} x^{2}+c^{2}}\, d x -c^{2}+2 c d x -d^{2} x^{2}\right )} \] Input:

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

Output:

(6*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*b*c**2*d - 6*sqrt(c**2 - d**2*x**2 
)*asin((d*x)/c)*b*c*d**2*x + 51*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*c**4 
- 51*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*c**3*d*x - 6*asin((d*x)/c)*b*c** 
3*d + 12*asin((d*x)/c)*b*c**2*d**2*x - 6*asin((d*x)/c)*b*c*d**3*x**2 - 51* 
asin((d*x)/c)*c**5 + 102*asin((d*x)/c)*c**4*d*x - 51*asin((d*x)/c)*c**3*d* 
*2*x**2 - 4*sqrt(c**2 - d**2*x**2)*a*c*d**2 + 4*sqrt(c**2 - d**2*x**2)*b*c 
**2*d - 8*sqrt(c**2 - d**2*x**2)*b*c*d**2*x + 34*sqrt(c**2 - d**2*x**2)*c* 
*4 - 63*sqrt(c**2 - d**2*x**2)*c**3*d*x + 18*sqrt(c**2 - d**2*x**2)*c**2*d 
**2*x**2 + 3*sqrt(c**2 - d**2*x**2)*c*d**3*x**3 + 4*a*c**2*d**2 + 4*a*d**4 
*x**2 - 4*b*c**3*d - 8*b*c**2*d**2*x + 20*b*c*d**3*x**2 - 34*c**5 - 63*c** 
4*d*x + 137*c**3*d**2*x**2 - 21*c**2*d**3*x**3 - 3*c*d**4*x**4)/(6*c*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))