\(\int \frac {(c+d x)^{3/2} (A+B x+C x^2+D x^3)}{\sqrt {c^2-d^2 x^2}} \, dx\) [214]

Optimal result

Integrand size = 41, antiderivative size = 244 \[ \int \frac {(c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c^2-d^2 x^2}} \, dx=-\frac {8 c \left (57 c^2 C d+63 B c d^2+105 A d^3+47 c^3 D\right ) \sqrt {c^2-d^2 x^2}}{315 d^4 \sqrt {c+d x}}-\frac {2 \left (57 c^2 C d+63 B c d^2+105 A d^3+47 c^3 D\right ) \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{315 d^4}+\frac {2 \left (6 c C d-21 B d^2-19 c^2 D\right ) (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{105 d^4}-\frac {2 (9 C d-11 c D) (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}{63 d^4}-\frac {2 D (c+d x)^{7/2} \sqrt {c^2-d^2 x^2}}{9 d^4} \] Output:

-8/315*c*(105*A*d^3+63*B*c*d^2+57*C*c^2*d+47*D*c^3)*(-d^2*x^2+c^2)^(1/2)/d 
^4/(d*x+c)^(1/2)-2/315*(105*A*d^3+63*B*c*d^2+57*C*c^2*d+47*D*c^3)*(d*x+c)^ 
(1/2)*(-d^2*x^2+c^2)^(1/2)/d^4+2/105*(-21*B*d^2+6*C*c*d-19*D*c^2)*(d*x+c)^ 
(3/2)*(-d^2*x^2+c^2)^(1/2)/d^4-2/63*(9*C*d-11*D*c)*(d*x+c)^(5/2)*(-d^2*x^2 
+c^2)^(1/2)/d^4-2/9*D*(d*x+c)^(7/2)*(-d^2*x^2+c^2)^(1/2)/d^4
 

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.52 \[ \int \frac {(c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c^2-d^2 x^2}} \, dx=-\frac {2 \sqrt {c^2-d^2 x^2} \left (272 c^4 D+8 c^3 d (39 C+17 D x)+6 c^2 d^2 (63 B+x (26 C+17 D x))+d^4 x (105 A+x (63 B+5 x (9 C+7 D x)))+c d^3 (525 A+x (189 B+x (117 C+85 D x)))\right )}{315 d^4 \sqrt {c+d x}} \] Input:

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

Output:

(-2*Sqrt[c^2 - d^2*x^2]*(272*c^4*D + 8*c^3*d*(39*C + 17*D*x) + 6*c^2*d^2*( 
63*B + x*(26*C + 17*D*x)) + d^4*x*(105*A + x*(63*B + 5*x*(9*C + 7*D*x))) + 
 c*d^3*(525*A + x*(189*B + x*(117*C + 85*D*x)))))/(315*d^4*Sqrt[c + d*x])
 

Rubi [A] (verified)

Time = 1.06 (sec) , antiderivative size = 236, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {2170, 27, 2170, 27, 672, 459, 458}

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.

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

\(\Big \downarrow \) 2170

\(\displaystyle -\frac {2 \int -\frac {(c+d x)^{3/2} \left ((9 C d-11 c D) x^2 d^4+\left (5 D c^2+9 B d^2\right ) x d^3+\left (7 D c^3+9 A d^3\right ) d^2\right )}{2 \sqrt {c^2-d^2 x^2}}dx}{9 d^5}-\frac {2 D \sqrt {c^2-d^2 x^2} (c+d x)^{7/2}}{9 d^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(c+d x)^{3/2} \left ((9 C d-11 c D) x^2 d^4+\left (5 D c^2+9 B d^2\right ) x d^3+\left (7 D c^3+9 A d^3\right ) d^2\right )}{\sqrt {c^2-d^2 x^2}}dx}{9 d^5}-\frac {2 D (c+d x)^{7/2} \sqrt {c^2-d^2 x^2}}{9 d^4}\)

\(\Big \downarrow \) 2170

\(\displaystyle \frac {-\frac {2 \int -\frac {3 d^6 (c+d x)^{3/2} \left (-2 D c^3+15 C d c^2+21 A d^3-d \left (-19 D c^2+6 C d c-21 B d^2\right ) x\right )}{2 \sqrt {c^2-d^2 x^2}}dx}{7 d^4}-\frac {2}{7} d \sqrt {c^2-d^2 x^2} (c+d x)^{5/2} (9 C d-11 c D)}{9 d^5}-\frac {2 D (c+d x)^{7/2} \sqrt {c^2-d^2 x^2}}{9 d^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3}{7} d^2 \int \frac {(c+d x)^{3/2} \left (-2 D c^3+15 C d c^2+21 A d^3-d \left (-19 D c^2+6 C d c-21 B d^2\right ) x\right )}{\sqrt {c^2-d^2 x^2}}dx-\frac {2}{7} d (c+d x)^{5/2} \sqrt {c^2-d^2 x^2} (9 C d-11 c D)}{9 d^5}-\frac {2 D (c+d x)^{7/2} \sqrt {c^2-d^2 x^2}}{9 d^4}\)

\(\Big \downarrow \) 672

\(\displaystyle \frac {\frac {3}{7} d^2 \left (\frac {1}{5} \left (105 A d^3+63 B c d^2+47 c^3 D+57 c^2 C d\right ) \int \frac {(c+d x)^{3/2}}{\sqrt {c^2-d^2 x^2}}dx+\frac {2 \sqrt {c^2-d^2 x^2} (c+d x)^{3/2} \left (-21 B d^2-19 c^2 D+6 c C d\right )}{5 d}\right )-\frac {2}{7} d (c+d x)^{5/2} \sqrt {c^2-d^2 x^2} (9 C d-11 c D)}{9 d^5}-\frac {2 D (c+d x)^{7/2} \sqrt {c^2-d^2 x^2}}{9 d^4}\)

\(\Big \downarrow \) 459

\(\displaystyle \frac {\frac {3}{7} d^2 \left (\frac {1}{5} \left (105 A d^3+63 B c d^2+47 c^3 D+57 c^2 C d\right ) \left (\frac {4}{3} c \int \frac {\sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}dx-\frac {2 \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{3 d}\right )+\frac {2 \sqrt {c^2-d^2 x^2} (c+d x)^{3/2} \left (-21 B d^2-19 c^2 D+6 c C d\right )}{5 d}\right )-\frac {2}{7} d (c+d x)^{5/2} \sqrt {c^2-d^2 x^2} (9 C d-11 c D)}{9 d^5}-\frac {2 D (c+d x)^{7/2} \sqrt {c^2-d^2 x^2}}{9 d^4}\)

\(\Big \downarrow \) 458

\(\displaystyle \frac {\frac {3}{7} d^2 \left (\frac {1}{5} \left (-\frac {8 c \sqrt {c^2-d^2 x^2}}{3 d \sqrt {c+d x}}-\frac {2 \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{3 d}\right ) \left (105 A d^3+63 B c d^2+47 c^3 D+57 c^2 C d\right )+\frac {2 \sqrt {c^2-d^2 x^2} (c+d x)^{3/2} \left (-21 B d^2-19 c^2 D+6 c C d\right )}{5 d}\right )-\frac {2}{7} d (c+d x)^{5/2} \sqrt {c^2-d^2 x^2} (9 C d-11 c D)}{9 d^5}-\frac {2 D (c+d x)^{7/2} \sqrt {c^2-d^2 x^2}}{9 d^4}\)

Input:

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

Output:

(-2*D*(c + d*x)^(7/2)*Sqrt[c^2 - d^2*x^2])/(9*d^4) + ((-2*d*(9*C*d - 11*c* 
D)*(c + d*x)^(5/2)*Sqrt[c^2 - d^2*x^2])/7 + (3*d^2*((2*(6*c*C*d - 21*B*d^2 
 - 19*c^2*D)*(c + d*x)^(3/2)*Sqrt[c^2 - d^2*x^2])/(5*d) + ((57*c^2*C*d + 6 
3*B*c*d^2 + 105*A*d^3 + 47*c^3*D)*((-8*c*Sqrt[c^2 - d^2*x^2])/(3*d*Sqrt[c 
+ d*x]) - (2*Sqrt[c + d*x]*Sqrt[c^2 - d^2*x^2])/(3*d)))/5))/7)/(9*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 458

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

rule 459

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(n + 2*p + 1))), x] + Simp[2*c* 
(Simplify[n + p]/(n + 2*p + 1))   Int[(c + d*x)^(n - 1)*(a + b*x^2)^p, x], 
x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && IGtQ[Simplif 
y[n + p], 0]
 

rule 672

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ 
), x_Symbol] :> Simp[g*(d + e*x)^m*((a + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), 
 x] + Simp[(m*(d*g + e*f) + 2*e*f*(p + 1))/(e*(m + 2*p + 2))   Int[(d + e*x 
)^m*(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] && NeQ[m + 2*p + 2, 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 = 0.37 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.61

method	result	size
default	\(-\frac {2 \sqrt {-d^{2} x^{2}+c^{2}}\, \left (35 D x^{4} d^{4}+45 C \,d^{4} x^{3}+85 D c \,d^{3} x^{3}+63 B \,d^{4} x^{2}+117 C c \,d^{3} x^{2}+102 D c^{2} d^{2} x^{2}+105 A \,d^{4} x +189 B c \,d^{3} x +156 C \,c^{2} d^{2} x +136 D c^{3} d x +525 A c \,d^{3}+378 B \,c^{2} d^{2}+312 C \,c^{3} d +272 c^{4} D\right )}{315 \sqrt {d x +c}\, d^{4}}\)	\(149\)
gosper	\(-\frac {2 \left (-d x +c \right ) \left (35 D x^{4} d^{4}+45 C \,d^{4} x^{3}+85 D c \,d^{3} x^{3}+63 B \,d^{4} x^{2}+117 C c \,d^{3} x^{2}+102 D c^{2} d^{2} x^{2}+105 A \,d^{4} x +189 B c \,d^{3} x +156 C \,c^{2} d^{2} x +136 D c^{3} d x +525 A c \,d^{3}+378 B \,c^{2} d^{2}+312 C \,c^{3} d +272 c^{4} D\right ) \sqrt {d x +c}}{315 d^{4} \sqrt {-d^{2} x^{2}+c^{2}}}\)	\(155\)
orering	\(-\frac {2 \left (-d x +c \right ) \left (35 D x^{4} d^{4}+45 C \,d^{4} x^{3}+85 D c \,d^{3} x^{3}+63 B \,d^{4} x^{2}+117 C c \,d^{3} x^{2}+102 D c^{2} d^{2} x^{2}+105 A \,d^{4} x +189 B c \,d^{3} x +156 C \,c^{2} d^{2} x +136 D c^{3} d x +525 A c \,d^{3}+378 B \,c^{2} d^{2}+312 C \,c^{3} d +272 c^{4} D\right ) \sqrt {d x +c}}{315 d^{4} \sqrt {-d^{2} x^{2}+c^{2}}}\)	\(155\)

Input:

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

Output:

-2/315/(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(1/2)*(35*D*d^4*x^4+45*C*d^4*x^3+85*D* 
c*d^3*x^3+63*B*d^4*x^2+117*C*c*d^3*x^2+102*D*c^2*d^2*x^2+105*A*d^4*x+189*B 
*c*d^3*x+156*C*c^2*d^2*x+136*D*c^3*d*x+525*A*c*d^3+378*B*c^2*d^2+312*C*c^3 
*d+272*D*c^4)/d^4
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.63 \[ \int \frac {(c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c^2-d^2 x^2}} \, dx=-\frac {2 \, {\left (35 \, D d^{4} x^{4} + 272 \, D c^{4} + 312 \, C c^{3} d + 378 \, B c^{2} d^{2} + 525 \, A c d^{3} + 5 \, {\left (17 \, D c d^{3} + 9 \, C d^{4}\right )} x^{3} + 3 \, {\left (34 \, D c^{2} d^{2} + 39 \, C c d^{3} + 21 \, B d^{4}\right )} x^{2} + {\left (136 \, D c^{3} d + 156 \, C c^{2} d^{2} + 189 \, B c d^{3} + 105 \, A d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{315 \, {\left (d^{5} x + c d^{4}\right )}} \] Input:

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

Output:

-2/315*(35*D*d^4*x^4 + 272*D*c^4 + 312*C*c^3*d + 378*B*c^2*d^2 + 525*A*c*d 
^3 + 5*(17*D*c*d^3 + 9*C*d^4)*x^3 + 3*(34*D*c^2*d^2 + 39*C*c*d^3 + 21*B*d^ 
4)*x^2 + (136*D*c^3*d + 156*C*c^2*d^2 + 189*B*c*d^3 + 105*A*d^4)*x)*sqrt(- 
d^2*x^2 + c^2)*sqrt(d*x + c)/(d^5*x + c*d^4)
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.81 \[ \int \frac {(c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c^2-d^2 x^2}} \, dx=\frac {2 \, {\left (d^{2} x^{2} + 4 \, c d x - 5 \, c^{2}\right )} A}{3 \, \sqrt {-d x + c} d} + \frac {2 \, {\left (d^{3} x^{3} + 2 \, c d^{2} x^{2} + 3 \, c^{2} d x - 6 \, c^{3}\right )} B}{5 \, \sqrt {-d x + c} d^{2}} + \frac {2 \, {\left (15 \, d^{4} x^{4} + 24 \, c d^{3} x^{3} + 13 \, c^{2} d^{2} x^{2} + 52 \, c^{3} d x - 104 \, c^{4}\right )} C}{105 \, \sqrt {-d x + c} d^{3}} + \frac {2 \, {\left (35 \, d^{5} x^{5} + 50 \, c d^{4} x^{4} + 17 \, c^{2} d^{3} x^{3} + 34 \, c^{3} d^{2} x^{2} + 136 \, c^{4} d x - 272 \, c^{5}\right )} D}{315 \, \sqrt {-d x + c} d^{4}} \] Input:

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

Output:

2/3*(d^2*x^2 + 4*c*d*x - 5*c^2)*A/(sqrt(-d*x + c)*d) + 2/5*(d^3*x^3 + 2*c* 
d^2*x^2 + 3*c^2*d*x - 6*c^3)*B/(sqrt(-d*x + c)*d^2) + 2/105*(15*d^4*x^4 + 
24*c*d^3*x^3 + 13*c^2*d^2*x^2 + 52*c^3*d*x - 104*c^4)*C/(sqrt(-d*x + c)*d^ 
3) + 2/315*(35*d^5*x^5 + 50*c*d^4*x^4 + 17*c^2*d^3*x^3 + 34*c^3*d^2*x^2 + 
136*c^4*d*x - 272*c^5)*D/(sqrt(-d*x + c)*d^4)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.05 \[ \int \frac {(c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c^2-d^2 x^2}} \, dx=-\frac {2 \, {\left (35 \, {\left (d x - c\right )}^{4} \sqrt {-d x + c} D + 225 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} D c + 567 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} D c^{2} - 735 \, {\left (-d x + c\right )}^{\frac {3}{2}} D c^{3} + 630 \, \sqrt {-d x + c} D c^{4} + 45 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} C d + 252 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} C c d - 525 \, {\left (-d x + c\right )}^{\frac {3}{2}} C c^{2} d + 630 \, \sqrt {-d x + c} C c^{3} d + 63 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} B d^{2} - 315 \, {\left (-d x + c\right )}^{\frac {3}{2}} B c d^{2} + 630 \, \sqrt {-d x + c} B c^{2} d^{2} - 105 \, {\left (-d x + c\right )}^{\frac {3}{2}} A d^{3} + 630 \, \sqrt {-d x + c} A c d^{3}\right )}}{315 \, d^{4}} \] Input:

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

Output:

-2/315*(35*(d*x - c)^4*sqrt(-d*x + c)*D + 225*(d*x - c)^3*sqrt(-d*x + c)*D 
*c + 567*(d*x - c)^2*sqrt(-d*x + c)*D*c^2 - 735*(-d*x + c)^(3/2)*D*c^3 + 6 
30*sqrt(-d*x + c)*D*c^4 + 45*(d*x - c)^3*sqrt(-d*x + c)*C*d + 252*(d*x - c 
)^2*sqrt(-d*x + c)*C*c*d - 525*(-d*x + c)^(3/2)*C*c^2*d + 630*sqrt(-d*x + 
c)*C*c^3*d + 63*(d*x - c)^2*sqrt(-d*x + c)*B*d^2 - 315*(-d*x + c)^(3/2)*B* 
c*d^2 + 630*sqrt(-d*x + c)*B*c^2*d^2 - 105*(-d*x + c)^(3/2)*A*d^3 + 630*sq 
rt(-d*x + c)*A*c*d^3)/d^4
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.37 \[ \int \frac {(c+d x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{\sqrt {c^2-d^2 x^2}} \, dx=\frac {2 \sqrt {-d x +c}\, \left (-35 d^{4} x^{4}-130 c \,d^{3} x^{3}-63 b \,d^{3} x^{2}-219 c^{2} d^{2} x^{2}-105 a \,d^{3} x -189 b c \,d^{2} x -292 c^{3} d x -525 a c \,d^{2}-378 b \,c^{2} d -584 c^{4}\right )}{315 d^{3}} \] Input:

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

Output:

(2*sqrt(c - d*x)*( - 525*a*c*d**2 - 105*a*d**3*x - 378*b*c**2*d - 189*b*c* 
d**2*x - 63*b*d**3*x**2 - 584*c**4 - 292*c**3*d*x - 219*c**2*d**2*x**2 - 1 
30*c*d**3*x**3 - 35*d**4*x**4))/(315*d**3)