\(\int \frac {(c^2-d^2 x^2)^{3/2} (A+B x+C x^2+D x^3)}{(c+d x)^{9/2}} \, dx\) [209]

Optimal result

Integrand size = 41, antiderivative size = 321 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{9/2}} \, dx=\frac {\left (19 c^2 C d-11 B c d^2+3 A d^3-27 c^3 D\right ) \sqrt {c^2-d^2 x^2}}{4 d^4 (c+d x)^{3/2}}+\frac {2 \left (14 c C d-3 B d^2-33 c^2 D\right ) \sqrt {c^2-d^2 x^2}}{3 d^4 \sqrt {c+d x}}-\frac {2 (C d-3 c D) \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{3 d^4}-\frac {\left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \left (c^2-d^2 x^2\right )^{3/2}}{2 d^4 (c+d x)^{7/2}}-\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{5 d^4 (c+d x)^{5/2}}-\frac {\left (83 c^2 C d-27 B c d^2+3 A d^3-171 c^3 D\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{4 \sqrt {2} \sqrt {c} d^4} \] Output:

1/4*(3*A*d^3-11*B*c*d^2+19*C*c^2*d-27*D*c^3)*(-d^2*x^2+c^2)^(1/2)/d^4/(d*x 
+c)^(3/2)+2/3*(-3*B*d^2+14*C*c*d-33*D*c^2)*(-d^2*x^2+c^2)^(1/2)/d^4/(d*x+c 
)^(1/2)-2/3*(C*d-3*D*c)*(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(1/2)/d^4-1/2*(A*d^3- 
B*c*d^2+C*c^2*d-D*c^3)*(-d^2*x^2+c^2)^(3/2)/d^4/(d*x+c)^(7/2)-2/5*D*(-d^2* 
x^2+c^2)^(5/2)/d^4/(d*x+c)^(5/2)-1/8*(3*A*d^3-27*B*c*d^2+83*C*c^2*d-171*D* 
c^3)*arctanh(2^(1/2)*c^(1/2)*(d*x+c)^(1/2)/(-d^2*x^2+c^2)^(1/2))*2^(1/2)/c 
^(1/2)/d^4
 

Mathematica [A] (verified)

Time = 3.27 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.64 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{9/2}} \, dx=\frac {-\frac {2 \sqrt {c^2-d^2 x^2} \left (1599 c^4 D+c^3 (-775 C d+2715 d D x)+c^2 d^2 (255 B+x (-1315 C+912 D x))+d^4 x \left (-75 A+8 x \left (15 B+5 C x+3 D x^2\right )\right )-5 c d^3 \left (3 A+x \left (-87 B+88 C x+24 D x^2\right )\right )\right )}{(c+d x)^{5/2}}+\frac {15 \sqrt {2} \left (-83 c^2 C d+27 B c d^2-3 A d^3+171 c^3 D\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{\sqrt {c}}}{120 d^4} \] Input:

Integrate[((c^2 - d^2*x^2)^(3/2)*(A + B*x + C*x^2 + D*x^3))/(c + d*x)^(9/2 
),x]
 

Output:

((-2*Sqrt[c^2 - d^2*x^2]*(1599*c^4*D + c^3*(-775*C*d + 2715*d*D*x) + c^2*d 
^2*(255*B + x*(-1315*C + 912*D*x)) + d^4*x*(-75*A + 8*x*(15*B + 5*C*x + 3* 
D*x^2)) - 5*c*d^3*(3*A + x*(-87*B + 88*C*x + 24*D*x^2))))/(c + d*x)^(5/2) 
+ (15*Sqrt[2]*(-83*c^2*C*d + 27*B*c*d^2 - 3*A*d^3 + 171*c^3*D)*ArcTanh[(Sq 
rt[2]*Sqrt[c]*Sqrt[c + d*x])/Sqrt[c^2 - d^2*x^2]])/Sqrt[c])/(120*d^4)
 

Rubi [A] (verified)

Time = 1.19 (sec) , antiderivative size = 297, normalized size of antiderivative = 0.93, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.220, Rules used = {2170, 27, 2170, 27, 671, 465, 466, 471, 221}

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.

\(\displaystyle \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{9/2}} \, dx\)

\(\Big \downarrow \) 2170

\(\displaystyle -\frac {2 \int -\frac {5 \left (c^2-d^2 x^2\right )^{3/2} \left ((C d-3 c D) x^2 d^4+\left (B d^2-3 c^2 D\right ) x d^3+\left (A d^3-c^3 D\right ) d^2\right )}{2 (c+d x)^{9/2}}dx}{5 d^5}-\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{5 d^4 (c+d x)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left ((C d-3 c D) x^2 d^4+\left (B d^2-3 c^2 D\right ) x d^3+\left (A d^3-c^3 D\right ) d^2\right )}{(c+d x)^{9/2}}dx}{d^5}-\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{5 d^4 (c+d x)^{5/2}}\)

\(\Big \downarrow \) 2170

\(\displaystyle \frac {-\frac {2 \int \frac {d^6 \left (-18 D c^3+7 C d c^2-3 A d^3+d \left (-21 D c^2+10 C d c-3 B d^2\right ) x\right ) \left (c^2-d^2 x^2\right )^{3/2}}{2 (c+d x)^{9/2}}dx}{3 d^4}-\frac {2 d \left (c^2-d^2 x^2\right )^{5/2} (C d-3 c D)}{3 (c+d x)^{7/2}}}{d^5}-\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{5 d^4 (c+d x)^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {1}{3} d^2 \int \frac {\left (-18 D c^3+7 C d c^2-3 A d^3+d \left (-21 D c^2+10 C d c-3 B d^2\right ) x\right ) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{9/2}}dx-\frac {2 d \left (c^2-d^2 x^2\right )^{5/2} (C d-3 c D)}{3 (c+d x)^{7/2}}}{d^5}-\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{5 d^4 (c+d x)^{5/2}}\)

\(\Big \downarrow \) 671

\(\displaystyle \frac {-\frac {1}{3} d^2 \left (\frac {\left (3 A d^3-27 B c d^2-171 c^3 D+83 c^2 C d\right ) \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{7/2}}dx}{8 c}+\frac {3 \left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{4 c d (c+d x)^{9/2}}\right )-\frac {2 d \left (c^2-d^2 x^2\right )^{5/2} (C d-3 c D)}{3 (c+d x)^{7/2}}}{d^5}-\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{5 d^4 (c+d x)^{5/2}}\)

\(\Big \downarrow \) 465

\(\displaystyle \frac {-\frac {1}{3} d^2 \left (\frac {\left (3 A d^3-27 B c d^2-171 c^3 D+83 c^2 C d\right ) \left (-\frac {3}{2} \int \frac {\sqrt {c^2-d^2 x^2}}{(c+d x)^{3/2}}dx-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{d (c+d x)^{5/2}}\right )}{8 c}+\frac {3 \left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{4 c d (c+d x)^{9/2}}\right )-\frac {2 d \left (c^2-d^2 x^2\right )^{5/2} (C d-3 c D)}{3 (c+d x)^{7/2}}}{d^5}-\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{5 d^4 (c+d x)^{5/2}}\)

\(\Big \downarrow \) 466

\(\displaystyle \frac {-\frac {1}{3} d^2 \left (\frac {\left (3 A d^3-27 B c d^2-171 c^3 D+83 c^2 C d\right ) \left (-\frac {3}{2} \left (2 c \int \frac {1}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx+\frac {2 \sqrt {c^2-d^2 x^2}}{d \sqrt {c+d x}}\right )-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{d (c+d x)^{5/2}}\right )}{8 c}+\frac {3 \left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{4 c d (c+d x)^{9/2}}\right )-\frac {2 d \left (c^2-d^2 x^2\right )^{5/2} (C d-3 c D)}{3 (c+d x)^{7/2}}}{d^5}-\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{5 d^4 (c+d x)^{5/2}}\)

\(\Big \downarrow \) 471

\(\displaystyle \frac {-\frac {1}{3} d^2 \left (\frac {\left (3 A d^3-27 B c d^2-171 c^3 D+83 c^2 C d\right ) \left (-\frac {3}{2} \left (4 c d \int \frac {1}{\frac {d^2 \left (c^2-d^2 x^2\right )}{c+d x}-2 c d^2}d\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}+\frac {2 \sqrt {c^2-d^2 x^2}}{d \sqrt {c+d x}}\right )-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{d (c+d x)^{5/2}}\right )}{8 c}+\frac {3 \left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{4 c d (c+d x)^{9/2}}\right )-\frac {2 d \left (c^2-d^2 x^2\right )^{5/2} (C d-3 c D)}{3 (c+d x)^{7/2}}}{d^5}-\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{5 d^4 (c+d x)^{5/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {-\frac {1}{3} d^2 \left (\frac {\left (-\frac {3}{2} \left (\frac {2 \sqrt {c^2-d^2 x^2}}{d \sqrt {c+d x}}-\frac {2 \sqrt {2} \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {c+d x}}\right )}{d}\right )-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{d (c+d x)^{5/2}}\right ) \left (3 A d^3-27 B c d^2-171 c^3 D+83 c^2 C d\right )}{8 c}+\frac {3 \left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{4 c d (c+d x)^{9/2}}\right )-\frac {2 d \left (c^2-d^2 x^2\right )^{5/2} (C d-3 c D)}{3 (c+d x)^{7/2}}}{d^5}-\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{5 d^4 (c+d x)^{5/2}}\)

Input:

Int[((c^2 - d^2*x^2)^(3/2)*(A + B*x + C*x^2 + D*x^3))/(c + d*x)^(9/2),x]
 

Output:

(-2*D*(c^2 - d^2*x^2)^(5/2))/(5*d^4*(c + d*x)^(5/2)) + ((-2*d*(C*d - 3*c*D 
)*(c^2 - d^2*x^2)^(5/2))/(3*(c + d*x)^(7/2)) - (d^2*((3*(c^2*C*d - B*c*d^2 
 + A*d^3 - c^3*D)*(c^2 - d^2*x^2)^(5/2))/(4*c*d*(c + d*x)^(9/2)) + ((83*c^ 
2*C*d - 27*B*c*d^2 + 3*A*d^3 - 171*c^3*D)*(-((c^2 - d^2*x^2)^(3/2)/(d*(c + 
 d*x)^(5/2))) - (3*((2*Sqrt[c^2 - d^2*x^2])/(d*Sqrt[c + d*x]) - (2*Sqrt[2] 
*Sqrt[c]*ArcTanh[Sqrt[c^2 - d^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x])])/d)) 
/2))/(8*c)))/3)/d^5
 

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 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 465

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 + p + 1))), x] - Simp[b*(p/(d^2*(n + 
 p + 1)))   Int[(c + d*x)^(n + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, 
b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0] && (LtQ[n, -2] || EqQ[n 
+ 2*p + 1, 0]) && NeQ[n + p + 1, 0] && IntegerQ[2*p]
 

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 471

Int[1/(Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Sim 
p[2*d   Subst[Int[1/(2*b*c + d^2*x^2), x], x, Sqrt[a + b*x^2]/Sqrt[c + d*x] 
], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0]
 

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 2170

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x) 
^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*e^(q - 1)*(m + q + 2*p + 1))), x] + Si 
mp[1/(b*e^q*(m + q + 2*p + 1))   Int[(d + e*x)^m*(a + b*x^2)^p*ExpandToSum[ 
b*e^q*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*(d + e*x)^q - 2*e*f*(m + 
 p + q)*(d + e*x)^(q - 2)*(a*e - b*d*x), x], x], x] /; NeQ[m + q + 2*p + 1, 
 0]] /; FreeQ[{a, b, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 
0] &&  !IGtQ[m, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(641\) vs. \(2(277)=554\).

Time = 0.37 (sec) , antiderivative size = 642, normalized size of antiderivative = 2.00

method	result	size
default	\(-\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, \left (-2565 D \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{5}-30 A \sqrt {-d x +c}\, c^{\frac {3}{2}} d^{3}+510 B \sqrt {-d x +c}\, c^{\frac {5}{2}} d^{2}-1550 C \sqrt {-d x +c}\, c^{\frac {7}{2}} d +45 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d^{3}-150 A \,d^{4} x \sqrt {-d x +c}\, \sqrt {c}-405 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} d^{2}+870 B \,c^{\frac {3}{2}} d^{3} x \sqrt {-d x +c}-2565 D \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} d^{2} x^{2}+90 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{4} x -810 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d^{3} x +5430 D \sqrt {-d x +c}\, c^{\frac {7}{2}} d x +1245 C \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4} d -2630 C \sqrt {-d x +c}\, c^{\frac {5}{2}} d^{2} x +45 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) d^{5} x^{2}-240 D c^{\frac {3}{2}} d^{3} x^{3} \sqrt {-d x +c}-880 C \sqrt {-d x +c}\, c^{\frac {3}{2}} d^{3} x^{2}+1824 D \sqrt {-d x +c}\, c^{\frac {5}{2}} d^{2} x^{2}+2490 C \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{3} d^{2} x -5130 D \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{4} d x -405 B \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \,d^{4} x^{2}+1245 C \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2} d^{3} x^{2}+48 D d^{4} x^{4} \sqrt {-d x +c}\, \sqrt {c}+80 C \,d^{4} x^{3} \sqrt {-d x +c}\, \sqrt {c}+240 B \,d^{4} x^{2} \sqrt {-d x +c}\, \sqrt {c}+3198 D \sqrt {-d x +c}\, c^{\frac {9}{2}}\right )}{120 \left (d x +c \right )^{\frac {5}{2}} \sqrt {-d x +c}\, d^{4} \sqrt {c}}\)	\(642\)

Input:

int((-d^2*x^2+c^2)^(3/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(9/2),x,method=_RETUR 
NVERBOSE)
 

Output:

-1/120*(-d^2*x^2+c^2)^(1/2)*(-2565*D*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^ 
(1/2)/c^(1/2))*c^5-30*A*(-d*x+c)^(1/2)*c^(3/2)*d^3+510*B*(-d*x+c)^(1/2)*c^ 
(5/2)*d^2-1550*C*(-d*x+c)^(1/2)*c^(7/2)*d+45*A*2^(1/2)*arctanh(1/2*(-d*x+c 
)^(1/2)*2^(1/2)/c^(1/2))*c^2*d^3-150*A*d^4*x*(-d*x+c)^(1/2)*c^(1/2)-405*B* 
2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^3*d^2+870*B*c^(3/2)* 
d^3*x*(-d*x+c)^(1/2)-2565*D*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^( 
1/2))*c^3*d^2*x^2+90*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2)) 
*c*d^4*x-810*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^2*d^3 
*x+5430*D*(-d*x+c)^(1/2)*c^(7/2)*d*x+1245*C*2^(1/2)*arctanh(1/2*(-d*x+c)^( 
1/2)*2^(1/2)/c^(1/2))*c^4*d-2630*C*(-d*x+c)^(1/2)*c^(5/2)*d^2*x+45*A*2^(1/ 
2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*d^5*x^2-240*D*c^(3/2)*d^3*x 
^3*(-d*x+c)^(1/2)-880*C*(-d*x+c)^(1/2)*c^(3/2)*d^3*x^2+1824*D*(-d*x+c)^(1/ 
2)*c^(5/2)*d^2*x^2+2490*C*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/ 
2))*c^3*d^2*x-5130*D*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c 
^4*d*x-405*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c*d^4*x^2 
+1245*C*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^2*d^3*x^2+48 
*D*d^4*x^4*(-d*x+c)^(1/2)*c^(1/2)+80*C*d^4*x^3*(-d*x+c)^(1/2)*c^(1/2)+240* 
B*d^4*x^2*(-d*x+c)^(1/2)*c^(1/2)+3198*D*(-d*x+c)^(1/2)*c^(9/2))/(d*x+c)^(5 
/2)/(-d*x+c)^(1/2)/d^4/c^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 811, normalized size of antiderivative = 2.53 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{9/2}} \, dx =\text {Too large to display} \] Input:

integrate((-d^2*x^2+c^2)^(3/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(9/2),x, algori 
thm="fricas")
 

Output:

[-1/240*(15*sqrt(2)*(171*D*c^6 - 83*C*c^5*d + 27*B*c^4*d^2 - 3*A*c^3*d^3 + 
 (171*D*c^3*d^3 - 83*C*c^2*d^4 + 27*B*c*d^5 - 3*A*d^6)*x^3 + 3*(171*D*c^4* 
d^2 - 83*C*c^3*d^3 + 27*B*c^2*d^4 - 3*A*c*d^5)*x^2 + 3*(171*D*c^5*d - 83*C 
*c^4*d^2 + 27*B*c^3*d^3 - 3*A*c^2*d^4)*x)*sqrt(c)*log(-(d^2*x^2 - 2*c*d*x 
+ 2*sqrt(2)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)*sqrt(c) - 3*c^2)/(d^2*x^2 + 
 2*c*d*x + c^2)) + 4*(24*D*c*d^4*x^4 + 1599*D*c^5 - 775*C*c^4*d + 255*B*c^ 
3*d^2 - 15*A*c^2*d^3 - 40*(3*D*c^2*d^3 - C*c*d^4)*x^3 + 8*(114*D*c^3*d^2 - 
 55*C*c^2*d^3 + 15*B*c*d^4)*x^2 + 5*(543*D*c^4*d - 263*C*c^3*d^2 + 87*B*c^ 
2*d^3 - 15*A*c*d^4)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(c*d^7*x^3 + 3* 
c^2*d^6*x^2 + 3*c^3*d^5*x + c^4*d^4), -1/120*(15*sqrt(2)*(171*D*c^6 - 83*C 
*c^5*d + 27*B*c^4*d^2 - 3*A*c^3*d^3 + (171*D*c^3*d^3 - 83*C*c^2*d^4 + 27*B 
*c*d^5 - 3*A*d^6)*x^3 + 3*(171*D*c^4*d^2 - 83*C*c^3*d^3 + 27*B*c^2*d^4 - 3 
*A*c*d^5)*x^2 + 3*(171*D*c^5*d - 83*C*c^4*d^2 + 27*B*c^3*d^3 - 3*A*c^2*d^4 
)*x)*sqrt(-c)*arctan(1/2*sqrt(2)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)*sqrt(- 
c)/(c*d*x + c^2)) + 2*(24*D*c*d^4*x^4 + 1599*D*c^5 - 775*C*c^4*d + 255*B*c 
^3*d^2 - 15*A*c^2*d^3 - 40*(3*D*c^2*d^3 - C*c*d^4)*x^3 + 8*(114*D*c^3*d^2 
- 55*C*c^2*d^3 + 15*B*c*d^4)*x^2 + 5*(543*D*c^4*d - 263*C*c^3*d^2 + 87*B*c 
^2*d^3 - 15*A*c*d^4)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(c*d^7*x^3 + 3 
*c^2*d^6*x^2 + 3*c^3*d^5*x + c^4*d^4)]
 

Sympy [F]

\[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{9/2}} \, dx=\int \frac {\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {3}{2}} \left (A + B x + C x^{2} + D x^{3}\right )}{\left (c + d x\right )^{\frac {9}{2}}}\, dx \] Input:

integrate((-d**2*x**2+c**2)**(3/2)*(D*x**3+C*x**2+B*x+A)/(d*x+c)**(9/2),x)
 

Output:

Integral((-(-c + d*x)*(c + d*x))**(3/2)*(A + B*x + C*x**2 + D*x**3)/(c + d 
*x)**(9/2), x)
 

Maxima [F]

\[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{9/2}} \, dx=\int { \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} {\left (D x^{3} + C x^{2} + B x + A\right )}}{{\left (d x + c\right )}^{\frac {9}{2}}} \,d x } \] Input:

integrate((-d^2*x^2+c^2)^(3/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(9/2),x, algori 
thm="maxima")
 

Output:

integrate((-d^2*x^2 + c^2)^(3/2)*(D*x^3 + C*x^2 + B*x + A)/(d*x + c)^(9/2) 
, x)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 276, normalized size of antiderivative = 0.86 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{9/2}} \, dx=-\frac {48 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} D + 240 \, {\left (-d x + c\right )}^{\frac {3}{2}} D c + 2160 \, \sqrt {-d x + c} D c^{2} - 80 \, {\left (-d x + c\right )}^{\frac {3}{2}} C d - 960 \, \sqrt {-d x + c} C c d + 240 \, \sqrt {-d x + c} B d^{2} + \frac {15 \, \sqrt {2} {\left (171 \, D c^{3} - 83 \, C c^{2} d + 27 \, B c d^{2} - 3 \, A d^{3}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c}} - \frac {30 \, {\left (29 \, {\left (-d x + c\right )}^{\frac {3}{2}} D c^{3} - 54 \, \sqrt {-d x + c} D c^{4} - 21 \, {\left (-d x + c\right )}^{\frac {3}{2}} C c^{2} d + 38 \, \sqrt {-d x + c} C c^{3} d + 13 \, {\left (-d x + c\right )}^{\frac {3}{2}} B c d^{2} - 22 \, \sqrt {-d x + c} B c^{2} d^{2} - 5 \, {\left (-d x + c\right )}^{\frac {3}{2}} A d^{3} + 6 \, \sqrt {-d x + c} A c d^{3}\right )}}{{\left (d x + c\right )}^{2}}}{120 \, d^{4}} \] Input:

integrate((-d^2*x^2+c^2)^(3/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(9/2),x, algori 
thm="giac")
 

Output:

-1/120*(48*(d*x - c)^2*sqrt(-d*x + c)*D + 240*(-d*x + c)^(3/2)*D*c + 2160* 
sqrt(-d*x + c)*D*c^2 - 80*(-d*x + c)^(3/2)*C*d - 960*sqrt(-d*x + c)*C*c*d 
+ 240*sqrt(-d*x + c)*B*d^2 + 15*sqrt(2)*(171*D*c^3 - 83*C*c^2*d + 27*B*c*d 
^2 - 3*A*d^3)*arctan(1/2*sqrt(2)*sqrt(-d*x + c)/sqrt(-c))/sqrt(-c) - 30*(2 
9*(-d*x + c)^(3/2)*D*c^3 - 54*sqrt(-d*x + c)*D*c^4 - 21*(-d*x + c)^(3/2)*C 
*c^2*d + 38*sqrt(-d*x + c)*C*c^3*d + 13*(-d*x + c)^(3/2)*B*c*d^2 - 22*sqrt 
(-d*x + c)*B*c^2*d^2 - 5*(-d*x + c)^(3/2)*A*d^3 + 6*sqrt(-d*x + c)*A*c*d^3 
)/(d*x + c)^2)/d^4
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{9/2}} \, dx=\int \frac {{\left (c^2-d^2\,x^2\right )}^{3/2}\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (c+d\,x\right )}^{9/2}} \,d x \] Input:

int(((c^2 - d^2*x^2)^(3/2)*(A + B*x + C*x^2 + x^3*D))/(c + d*x)^(9/2),x)
 

Output:

int(((c^2 - d^2*x^2)^(3/2)*(A + B*x + C*x^2 + x^3*D))/(c + d*x)^(9/2), x)
 

Reduce [B] (verification not implemented)

Time = 1.18 (sec) , antiderivative size = 588, normalized size of antiderivative = 1.83 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{9/2}} \, dx =\text {Too large to display} \] Input:

int((-d^2*x^2+c^2)^(3/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(9/2),x)
 

Output:

(120*sqrt(c - d*x)*a*c**2*d**2 + 600*sqrt(c - d*x)*a*c*d**3*x - 2040*sqrt( 
c - d*x)*b*c**3*d - 3480*sqrt(c - d*x)*b*c**2*d**2*x - 960*sqrt(c - d*x)*b 
*c*d**3*x**2 - 6592*sqrt(c - d*x)*c**5 - 11200*sqrt(c - d*x)*c**4*d*x - 37 
76*sqrt(c - d*x)*c**3*d**2*x**2 + 640*sqrt(c - d*x)*c**2*d**3*x**3 - 192*s 
qrt(c - d*x)*c*d**4*x**4 + 180*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/ 
(sqrt(c)*sqrt(2)))/2))*a*c**2*d**2 + 360*sqrt(c)*sqrt(2)*log(tan(asin(sqrt 
(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*c*d**3*x + 180*sqrt(c)*sqrt(2)*log(tan( 
asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*d**4*x**2 - 1620*sqrt(c)*sqrt( 
2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*b*c**3*d - 3240*sqrt( 
c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*b*c**2*d**2*x 
 - 1620*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))* 
b*c*d**3*x**2 - 5280*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*s 
qrt(2)))/2))*c**5 - 10560*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt 
(c)*sqrt(2)))/2))*c**4*d*x - 5280*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d* 
x)/(sqrt(c)*sqrt(2)))/2))*c**3*d**2*x**2 - 225*sqrt(c)*sqrt(2)*a*c**2*d**2 
 - 450*sqrt(c)*sqrt(2)*a*c*d**3*x - 225*sqrt(c)*sqrt(2)*a*d**4*x**2 + 1785 
*sqrt(c)*sqrt(2)*b*c**3*d + 3570*sqrt(c)*sqrt(2)*b*c**2*d**2*x + 1785*sqrt 
(c)*sqrt(2)*b*c*d**3*x**2 + 7448*sqrt(c)*sqrt(2)*c**5 + 14896*sqrt(c)*sqrt 
(2)*c**4*d*x + 7448*sqrt(c)*sqrt(2)*c**3*d**2*x**2)/(480*c*d**3*(c**2 + 2* 
c*d*x + d**2*x**2))