\(\int \frac {\sqrt {c^2-d^2 x^2} (A+B x+C x^2+D x^3)}{(c+d x)^{5/2}} \, dx\) [198]

Optimal result

Integrand size = 41, antiderivative size = 256 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx=-\frac {\left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \sqrt {c^2-d^2 x^2}}{d^4 (c+d x)^{3/2}}-\frac {2 \left (8 c C d-3 B d^2-11 c^2 D\right ) \sqrt {c^2-d^2 x^2}}{3 d^4 \sqrt {c+d x}}+\frac {2 (C d-c D) \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{3 d^4}+\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{5 d^4 (c+d x)^{5/2}}+\frac {\left (9 c^2 C d-5 B c d^2+A d^3-13 c^3 D\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{\sqrt {2} \sqrt {c} d^4} \] Output:

-(A*d^3-B*c*d^2+C*c^2*d-D*c^3)*(-d^2*x^2+c^2)^(1/2)/d^4/(d*x+c)^(3/2)-2/3* 
(-3*B*d^2+8*C*c*d-11*D*c^2)*(-d^2*x^2+c^2)^(1/2)/d^4/(d*x+c)^(1/2)+2/3*(C* 
d-D*c)*(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(1/2)/d^4+2/5*D*(-d^2*x^2+c^2)^(5/2)/d 
^4/(d*x+c)^(5/2)+1/2*(A*d^3-5*B*c*d^2+9*C*c^2*d-13*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 = 0.91 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.69 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx=\frac {\frac {2 \sqrt {c^2-d^2 x^2} \left (121 c^3 D+c^2 (-85 C d+84 d D x)+c d^2 (45 B-4 x (15 C+4 D x))+d^3 \left (-15 A+2 x \left (15 B+5 C x+3 D x^2\right )\right )\right )}{(c+d x)^{3/2}}-\frac {15 \sqrt {2} \left (-9 c^2 C d+5 B c d^2-A d^3+13 c^3 D\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{\sqrt {c}}}{30 d^4} \] Input:

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

Output:

((2*Sqrt[c^2 - d^2*x^2]*(121*c^3*D + c^2*(-85*C*d + 84*d*D*x) + c*d^2*(45* 
B - 4*x*(15*C + 4*D*x)) + d^3*(-15*A + 2*x*(15*B + 5*C*x + 3*D*x^2))))/(c 
+ d*x)^(3/2) - (15*Sqrt[2]*(-9*c^2*C*d + 5*B*c*d^2 - A*d^3 + 13*c^3*D)*Arc 
Tanh[(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x])/Sqrt[c^2 - d^2*x^2]])/Sqrt[c])/(30*d^ 
4)
 

Rubi [A] (verified)

Time = 1.12 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {2170, 27, 2170, 27, 671, 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.

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

\(\Big \downarrow \) 2170

\(\displaystyle -\frac {2 \int -\frac {\sqrt {c^2-d^2 x^2} \left ((5 C d-11 c D) x^2 d^4+\left (5 B d^2-7 c^2 D\right ) x d^3+\left (5 A d^3-c^3 D\right ) d^2\right )}{2 (c+d x)^{5/2}}dx}{5 d^5}-\frac {2 D \left (c^2-d^2 x^2\right )^{3/2}}{5 d^4 \sqrt {c+d x}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sqrt {c^2-d^2 x^2} \left ((5 C d-11 c D) x^2 d^4+\left (5 B d^2-7 c^2 D\right ) x d^3+\left (5 A d^3-c^3 D\right ) d^2\right )}{(c+d x)^{5/2}}dx}{5 d^5}-\frac {2 D \left (c^2-d^2 x^2\right )^{3/2}}{5 d^4 \sqrt {c+d x}}\)

\(\Big \downarrow \) 2170

\(\displaystyle \frac {-\frac {2 \int \frac {15 d^6 \left (-2 D c^3+C d c^2-A d^3+d \left (-3 D c^2+2 C d c-B d^2\right ) x\right ) \sqrt {c^2-d^2 x^2}}{2 (c+d x)^{5/2}}dx}{3 d^4}-\frac {2 d \left (c^2-d^2 x^2\right )^{3/2} (5 C d-11 c D)}{3 (c+d x)^{3/2}}}{5 d^5}-\frac {2 D \left (c^2-d^2 x^2\right )^{3/2}}{5 d^4 \sqrt {c+d x}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-5 d^2 \int \frac {\left (-2 D c^3+C d c^2-A d^3+d \left (-3 D c^2+2 C d c-B d^2\right ) x\right ) \sqrt {c^2-d^2 x^2}}{(c+d x)^{5/2}}dx-\frac {2 d \left (c^2-d^2 x^2\right )^{3/2} (5 C d-11 c D)}{3 (c+d x)^{3/2}}}{5 d^5}-\frac {2 D \left (c^2-d^2 x^2\right )^{3/2}}{5 d^4 \sqrt {c+d x}}\)

\(\Big \downarrow \) 671

\(\displaystyle \frac {-5 d^2 \left (\frac {\left (A d^3-5 B c d^2-13 c^3 D+9 c^2 C d\right ) \int \frac {\sqrt {c^2-d^2 x^2}}{(c+d x)^{3/2}}dx}{4 c}+\frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{2 c d (c+d x)^{5/2}}\right )-\frac {2 d \left (c^2-d^2 x^2\right )^{3/2} (5 C d-11 c D)}{3 (c+d x)^{3/2}}}{5 d^5}-\frac {2 D \left (c^2-d^2 x^2\right )^{3/2}}{5 d^4 \sqrt {c+d x}}\)

\(\Big \downarrow \) 466

\(\displaystyle \frac {-5 d^2 \left (\frac {\left (A d^3-5 B c d^2-13 c^3 D+9 c^2 C d\right ) \left (2 c \int \frac {1}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx+\frac {2 \sqrt {c^2-d^2 x^2}}{d \sqrt {c+d x}}\right )}{4 c}+\frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{2 c d (c+d x)^{5/2}}\right )-\frac {2 d \left (c^2-d^2 x^2\right )^{3/2} (5 C d-11 c D)}{3 (c+d x)^{3/2}}}{5 d^5}-\frac {2 D \left (c^2-d^2 x^2\right )^{3/2}}{5 d^4 \sqrt {c+d x}}\)

\(\Big \downarrow \) 471

\(\displaystyle \frac {-5 d^2 \left (\frac {\left (A d^3-5 B c d^2-13 c^3 D+9 c^2 C d\right ) \left (4 c d \int \frac {1}{\frac {d^2 \left (c^2-d^2 x^2\right )}{c+d x}-2 c d^2}d\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}+\frac {2 \sqrt {c^2-d^2 x^2}}{d \sqrt {c+d x}}\right )}{4 c}+\frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{2 c d (c+d x)^{5/2}}\right )-\frac {2 d \left (c^2-d^2 x^2\right )^{3/2} (5 C d-11 c D)}{3 (c+d x)^{3/2}}}{5 d^5}-\frac {2 D \left (c^2-d^2 x^2\right )^{3/2}}{5 d^4 \sqrt {c+d x}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {-5 d^2 \left (\frac {\left (\frac {2 \sqrt {c^2-d^2 x^2}}{d \sqrt {c+d x}}-\frac {2 \sqrt {2} \sqrt {c} \text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {c+d x}}\right )}{d}\right ) \left (A d^3-5 B c d^2-13 c^3 D+9 c^2 C d\right )}{4 c}+\frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{2 c d (c+d x)^{5/2}}\right )-\frac {2 d \left (c^2-d^2 x^2\right )^{3/2} (5 C d-11 c D)}{3 (c+d x)^{3/2}}}{5 d^5}-\frac {2 D \left (c^2-d^2 x^2\right )^{3/2}}{5 d^4 \sqrt {c+d x}}\)

Input:

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

Output:

(-2*D*(c^2 - d^2*x^2)^(3/2))/(5*d^4*Sqrt[c + d*x]) + ((-2*d*(5*C*d - 11*c* 
D)*(c^2 - d^2*x^2)^(3/2))/(3*(c + d*x)^(3/2)) - 5*d^2*(((c^2*C*d - B*c*d^2 
 + A*d^3 - c^3*D)*(c^2 - d^2*x^2)^(3/2))/(2*c*d*(c + d*x)^(5/2)) + ((9*c^2 
*C*d - 5*B*c*d^2 + A*d^3 - 13*c^3*D)*((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]*Sq 
rt[c + d*x])])/d))/(4*c)))/(5*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 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 [A] (verified)

Time = 0.36 (sec) , antiderivative size = 436, normalized size of antiderivative = 1.70

method	result	size
default	\(\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, \left (12 D d^{3} x^{3} \sqrt {-d x +c}\, \sqrt {c}+15 A \,\operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, d^{4} x -75 B \,\operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c \,d^{3} x +135 C \,\operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{2} d^{2} x +20 C \,d^{3} x^{2} \sqrt {-d x +c}\, \sqrt {c}-195 D \,\operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{3} d x -32 D c^{\frac {3}{2}} d^{2} x^{2} \sqrt {-d x +c}+15 A \,\operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c \,d^{3}-75 B \,\operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{2} d^{2}+60 B \sqrt {-d x +c}\, \sqrt {c}\, d^{3} x +135 C \,\operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{3} d -120 C \sqrt {-d x +c}\, c^{\frac {3}{2}} d^{2} x -195 D \,\operatorname {arctanh}\left (\frac {\sqrt {-d x +c}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sqrt {2}\, c^{4}+168 D \sqrt {-d x +c}\, c^{\frac {5}{2}} d x -30 A \sqrt {-d x +c}\, \sqrt {c}\, d^{3}+90 B \sqrt {-d x +c}\, c^{\frac {3}{2}} d^{2}-170 C \sqrt {-d x +c}\, c^{\frac {5}{2}} d +242 D \sqrt {-d x +c}\, c^{\frac {7}{2}}\right )}{30 \left (d x +c \right )^{\frac {3}{2}} \sqrt {-d x +c}\, d^{4} \sqrt {c}}\)	\(436\)

Input:

int((-d^2*x^2+c^2)^(1/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(5/2),x,method=_RETUR 
NVERBOSE)
 

Output:

1/30*(-d^2*x^2+c^2)^(1/2)*(12*D*d^3*x^3*(-d*x+c)^(1/2)*c^(1/2)+15*A*arctan 
h(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*d^4*x-75*B*arctanh(1/2*(-d*x 
+c)^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c*d^3*x+135*C*arctanh(1/2*(-d*x+c)^(1/2 
)*2^(1/2)/c^(1/2))*2^(1/2)*c^2*d^2*x+20*C*d^3*x^2*(-d*x+c)^(1/2)*c^(1/2)-1 
95*D*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c^3*d*x-32*D*c^(3 
/2)*d^2*x^2*(-d*x+c)^(1/2)+15*A*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2) 
)*2^(1/2)*c*d^3-75*B*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c 
^2*d^2+60*B*(-d*x+c)^(1/2)*c^(1/2)*d^3*x+135*C*arctanh(1/2*(-d*x+c)^(1/2)* 
2^(1/2)/c^(1/2))*2^(1/2)*c^3*d-120*C*(-d*x+c)^(1/2)*c^(3/2)*d^2*x-195*D*ar 
ctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)*c^4+168*D*(-d*x+c)^(1/2) 
*c^(5/2)*d*x-30*A*(-d*x+c)^(1/2)*c^(1/2)*d^3+90*B*(-d*x+c)^(1/2)*c^(3/2)*d 
^2-170*C*(-d*x+c)^(1/2)*c^(5/2)*d+242*D*(-d*x+c)^(1/2)*c^(7/2))/(d*x+c)^(3 
/2)/(-d*x+c)^(1/2)/d^4/c^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 621, normalized size of antiderivative = 2.43 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx=\left [-\frac {15 \, \sqrt {2} {\left (13 \, D c^{5} - 9 \, C c^{4} d + 5 \, B c^{3} d^{2} - A c^{2} d^{3} + {\left (13 \, D c^{3} d^{2} - 9 \, C c^{2} d^{3} + 5 \, B c d^{4} - A d^{5}\right )} x^{2} + 2 \, {\left (13 \, D c^{4} d - 9 \, C c^{3} d^{2} + 5 \, B c^{2} d^{3} - A c d^{4}\right )} x\right )} \sqrt {c} \log \left (-\frac {d^{2} x^{2} - 2 \, c d x - 2 \, \sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {c} - 3 \, c^{2}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) - 4 \, {\left (6 \, D c d^{3} x^{3} + 121 \, D c^{4} - 85 \, C c^{3} d + 45 \, B c^{2} d^{2} - 15 \, A c d^{3} - 2 \, {\left (8 \, D c^{2} d^{2} - 5 \, C c d^{3}\right )} x^{2} + 6 \, {\left (14 \, D c^{3} d - 10 \, C c^{2} d^{2} + 5 \, B c d^{3}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{60 \, {\left (c d^{6} x^{2} + 2 \, c^{2} d^{5} x + c^{3} d^{4}\right )}}, \frac {15 \, \sqrt {2} {\left (13 \, D c^{5} - 9 \, C c^{4} d + 5 \, B c^{3} d^{2} - A c^{2} d^{3} + {\left (13 \, D c^{3} d^{2} - 9 \, C c^{2} d^{3} + 5 \, B c d^{4} - A d^{5}\right )} x^{2} + 2 \, {\left (13 \, D c^{4} d - 9 \, C c^{3} d^{2} + 5 \, B c^{2} d^{3} - A c d^{4}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {-c}}{2 \, {\left (c d x + c^{2}\right )}}\right ) + 2 \, {\left (6 \, D c d^{3} x^{3} + 121 \, D c^{4} - 85 \, C c^{3} d + 45 \, B c^{2} d^{2} - 15 \, A c d^{3} - 2 \, {\left (8 \, D c^{2} d^{2} - 5 \, C c d^{3}\right )} x^{2} + 6 \, {\left (14 \, D c^{3} d - 10 \, C c^{2} d^{2} + 5 \, B c d^{3}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{30 \, {\left (c d^{6} x^{2} + 2 \, c^{2} d^{5} x + c^{3} d^{4}\right )}}\right ] \] Input:

integrate((-d^2*x^2+c^2)^(1/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(5/2),x, algori 
thm="fricas")
                                                                                    
                                                                                    
 

Output:

[-1/60*(15*sqrt(2)*(13*D*c^5 - 9*C*c^4*d + 5*B*c^3*d^2 - A*c^2*d^3 + (13*D 
*c^3*d^2 - 9*C*c^2*d^3 + 5*B*c*d^4 - A*d^5)*x^2 + 2*(13*D*c^4*d - 9*C*c^3* 
d^2 + 5*B*c^2*d^3 - A*c*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*(6*D*c*d^3*x^3 + 121*D*c^4 - 85*C*c^3*d + 45*B*c^2*d^2 - 15*A*c 
*d^3 - 2*(8*D*c^2*d^2 - 5*C*c*d^3)*x^2 + 6*(14*D*c^3*d - 10*C*c^2*d^2 + 5* 
B*c*d^3)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c))/(c*d^6*x^2 + 2*c^2*d^5*x + 
 c^3*d^4), 1/30*(15*sqrt(2)*(13*D*c^5 - 9*C*c^4*d + 5*B*c^3*d^2 - A*c^2*d^ 
3 + (13*D*c^3*d^2 - 9*C*c^2*d^3 + 5*B*c*d^4 - A*d^5)*x^2 + 2*(13*D*c^4*d - 
 9*C*c^3*d^2 + 5*B*c^2*d^3 - A*c*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*(6*D*c*d^3*x^3 + 
 121*D*c^4 - 85*C*c^3*d + 45*B*c^2*d^2 - 15*A*c*d^3 - 2*(8*D*c^2*d^2 - 5*C 
*c*d^3)*x^2 + 6*(14*D*c^3*d - 10*C*c^2*d^2 + 5*B*c*d^3)*x)*sqrt(-d^2*x^2 + 
 c^2)*sqrt(d*x + c))/(c*d^6*x^2 + 2*c^2*d^5*x + c^3*d^4)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate((-d^2*x^2+c^2)^(1/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(5/2),x, algori 
thm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 213, normalized size of antiderivative = 0.83 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx=\frac {12 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} D + 20 \, {\left (-d x + c\right )}^{\frac {3}{2}} D c + 180 \, \sqrt {-d x + c} D c^{2} - 20 \, {\left (-d x + c\right )}^{\frac {3}{2}} C d - 120 \, \sqrt {-d x + c} C c d + 60 \, \sqrt {-d x + c} B d^{2} + \frac {15 \, \sqrt {2} {\left (13 \, D c^{3} - 9 \, C c^{2} d + 5 \, B c d^{2} - A d^{3}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c}} + \frac {30 \, {\left (\sqrt {-d x + c} D c^{3} - \sqrt {-d x + c} C c^{2} d + \sqrt {-d x + c} B c d^{2} - \sqrt {-d x + c} A d^{3}\right )}}{d x + c}}{30 \, d^{4}} \] Input:

integrate((-d^2*x^2+c^2)^(1/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(5/2),x, algori 
thm="giac")
 

Output:

1/30*(12*(d*x - c)^2*sqrt(-d*x + c)*D + 20*(-d*x + c)^(3/2)*D*c + 180*sqrt 
(-d*x + c)*D*c^2 - 20*(-d*x + c)^(3/2)*C*d - 120*sqrt(-d*x + c)*C*c*d + 60 
*sqrt(-d*x + c)*B*d^2 + 15*sqrt(2)*(13*D*c^3 - 9*C*c^2*d + 5*B*c*d^2 - A*d 
^3)*arctan(1/2*sqrt(2)*sqrt(-d*x + c)/sqrt(-c))/sqrt(-c) + 30*(sqrt(-d*x + 
 c)*D*c^3 - sqrt(-d*x + c)*C*c^2*d + sqrt(-d*x + c)*B*c*d^2 - sqrt(-d*x + 
c)*A*d^3)/(d*x + c))/d^4
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.44 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{5/2}} \, dx=\frac {-40 \sqrt {-d x +c}\, a c \,d^{2}+120 \sqrt {-d x +c}\, b \,c^{2} d +80 \sqrt {-d x +c}\, b c \,d^{2} x +96 \sqrt {-d x +c}\, c^{4}+64 \sqrt {-d x +c}\, c^{3} d x -16 \sqrt {-d x +c}\, c^{2} d^{2} x^{2}+16 \sqrt {-d x +c}\, c \,d^{3} x^{3}-20 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) a c \,d^{2}-20 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) a \,d^{3} x +100 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) b \,c^{2} d +100 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) b c \,d^{2} x +80 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) c^{4}+80 \sqrt {c}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {d x +c}}{\sqrt {c}\, \sqrt {2}}\right )}{2}\right )\right ) c^{3} d x +25 \sqrt {c}\, \sqrt {2}\, a c \,d^{2}+25 \sqrt {c}\, \sqrt {2}\, a \,d^{3} x -105 \sqrt {c}\, \sqrt {2}\, b \,c^{2} d -105 \sqrt {c}\, \sqrt {2}\, b c \,d^{2} x -144 \sqrt {c}\, \sqrt {2}\, c^{4}-144 \sqrt {c}\, \sqrt {2}\, c^{3} d x}{40 c \,d^{3} \left (d x +c \right )} \] Input:

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

Output:

( - 40*sqrt(c - d*x)*a*c*d**2 + 120*sqrt(c - d*x)*b*c**2*d + 80*sqrt(c - d 
*x)*b*c*d**2*x + 96*sqrt(c - d*x)*c**4 + 64*sqrt(c - d*x)*c**3*d*x - 16*sq 
rt(c - d*x)*c**2*d**2*x**2 + 16*sqrt(c - d*x)*c*d**3*x**3 - 20*sqrt(c)*sqr 
t(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*c*d**2 - 20*sqrt( 
c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*d**3*x + 10 
0*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*b*c**2 
*d + 100*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2)) 
*b*c*d**2*x + 80*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt( 
2)))/2))*c**4 + 80*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqr 
t(2)))/2))*c**3*d*x + 25*sqrt(c)*sqrt(2)*a*c*d**2 + 25*sqrt(c)*sqrt(2)*a*d 
**3*x - 105*sqrt(c)*sqrt(2)*b*c**2*d - 105*sqrt(c)*sqrt(2)*b*c*d**2*x - 14 
4*sqrt(c)*sqrt(2)*c**4 - 144*sqrt(c)*sqrt(2)*c**3*d*x)/(40*c*d**3*(c + d*x 
))