\(\int \frac {(c+d x)^{5/2} (A+B x+C x^2+D x^3)}{(c^2-d^2 x^2)^{3/2}} \, dx\) [221]

Optimal result

Integrand size = 41, antiderivative size = 251 \[ \int \frac {(c+d x)^{5/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {\left (c^2 C d+B c d^2+A d^3+c^3 D\right ) (c+d x)^{5/2}}{c d^4 \sqrt {c^2-d^2 x^2}}+\frac {4 \left (217 c^2 C d+175 B c d^2+105 A d^3+255 c^3 D\right ) \sqrt {c^2-d^2 x^2}}{105 d^4 \sqrt {c+d x}}+\frac {\left (217 c^2 C d+175 B c d^2+105 A d^3+255 c^3 D\right ) \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{105 c d^4}+\frac {2 (7 C d+5 c D) (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{35 d^4}+\frac {2 D (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}{7 d^4} \] Output:

(A*d^3+B*c*d^2+C*c^2*d+D*c^3)*(d*x+c)^(5/2)/c/d^4/(-d^2*x^2+c^2)^(1/2)+4/1 
05*(105*A*d^3+175*B*c*d^2+217*C*c^2*d+255*D*c^3)*(-d^2*x^2+c^2)^(1/2)/d^4/ 
(d*x+c)^(1/2)+1/105*(105*A*d^3+175*B*c*d^2+217*C*c^2*d+255*D*c^3)*(d*x+c)^ 
(1/2)*(-d^2*x^2+c^2)^(1/2)/c/d^4+2/35*(7*C*d+5*D*c)*(d*x+c)^(3/2)*(-d^2*x^ 
2+c^2)^(1/2)/d^4+2/7*D*(d*x+c)^(5/2)*(-d^2*x^2+c^2)^(1/2)/d^4
 

Mathematica [A] (verified)

Time = 2.64 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.54 \[ \int \frac {(c+d x)^{5/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {2 \sqrt {c^2-d^2 x^2} \left (-720 c^4 D-8 c^3 d (77 C-45 D x)+2 c^2 d^2 (-245 B+x (154 C+45 D x))+d^4 x (105 A+x (35 B+3 x (7 C+5 D x)))+c d^3 (-315 A+x (245 B+x (77 C+45 D x)))\right )}{105 d^4 (-c+d x) \sqrt {c+d x}} \] Input:

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

Output:

(2*Sqrt[c^2 - d^2*x^2]*(-720*c^4*D - 8*c^3*d*(77*C - 45*D*x) + 2*c^2*d^2*( 
-245*B + x*(154*C + 45*D*x)) + d^4*x*(105*A + x*(35*B + 3*x*(7*C + 5*D*x)) 
) + c*d^3*(-315*A + x*(245*B + x*(77*C + 45*D*x)))))/(105*d^4*(-c + d*x)*S 
qrt[c + d*x])
 

Rubi [A] (verified)

Time = 1.18 (sec) , antiderivative size = 245, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {2166, 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.

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

\(\Big \downarrow \) 2166

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2170

\(\displaystyle \frac {(c+d x)^{5/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {-\frac {2 \int -\frac {d (c+d x)^{3/2} \left (45 D c^3+35 C d c^2+35 B d^2 c+2 d (7 C d+5 c D) x c+21 A d^3\right )}{2 \sqrt {c^2-d^2 x^2}}dx}{7 d^4}-\frac {4 c D \sqrt {c^2-d^2 x^2} (c+d x)^{5/2}}{7 d^4}}{2 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(c+d x)^{5/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\frac {\int \frac {(c+d x)^{3/2} \left (45 D c^3+35 C d c^2+35 B d^2 c+2 d (7 C d+5 c D) x c+21 A d^3\right )}{\sqrt {c^2-d^2 x^2}}dx}{7 d^3}-\frac {4 c D (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}{7 d^4}}{2 c}\)

\(\Big \downarrow \) 672

\(\displaystyle \frac {(c+d x)^{5/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\frac {\frac {1}{5} \left (105 A d^3+175 B c d^2+255 c^3 D+217 c^2 C d\right ) \int \frac {(c+d x)^{3/2}}{\sqrt {c^2-d^2 x^2}}dx-\frac {4 c (c+d x)^{3/2} \sqrt {c^2-d^2 x^2} (5 c D+7 C d)}{5 d}}{7 d^3}-\frac {4 c D (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}{7 d^4}}{2 c}\)

\(\Big \downarrow \) 459

\(\displaystyle \frac {(c+d x)^{5/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\frac {\frac {1}{5} \left (105 A d^3+175 B c d^2+255 c^3 D+217 c^2 C d\right ) \left (\frac {4}{3} c \int \frac {\sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}dx-\frac {2 \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{3 d}\right )-\frac {4 c (c+d x)^{3/2} \sqrt {c^2-d^2 x^2} (5 c D+7 C d)}{5 d}}{7 d^3}-\frac {4 c D (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}{7 d^4}}{2 c}\)

\(\Big \downarrow \) 458

\(\displaystyle \frac {(c+d x)^{5/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\frac {\frac {1}{5} \left (-\frac {8 c \sqrt {c^2-d^2 x^2}}{3 d \sqrt {c+d x}}-\frac {2 \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{3 d}\right ) \left (105 A d^3+175 B c d^2+255 c^3 D+217 c^2 C d\right )-\frac {4 c (c+d x)^{3/2} \sqrt {c^2-d^2 x^2} (5 c D+7 C d)}{5 d}}{7 d^3}-\frac {4 c D (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}{7 d^4}}{2 c}\)

Input:

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

Output:

((c^2*C*d + B*c*d^2 + A*d^3 + c^3*D)*(c + d*x)^(5/2))/(c*d^4*Sqrt[c^2 - d^ 
2*x^2]) - ((-4*c*D*(c + d*x)^(5/2)*Sqrt[c^2 - d^2*x^2])/(7*d^4) + ((-4*c*( 
7*C*d + 5*c*D)*(c + d*x)^(3/2)*Sqrt[c^2 - d^2*x^2])/(5*d) + ((217*c^2*C*d 
+ 175*B*c*d^2 + 105*A*d^3 + 255*c^3*D)*((-8*c*Sqrt[c^2 - d^2*x^2])/(3*d*Sq 
rt[c + d*x]) - (2*Sqrt[c + d*x]*Sqrt[c^2 - d^2*x^2])/(3*d)))/5)/(7*d^3))/( 
2*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 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 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]
 

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 = 155, normalized size of antiderivative = 0.62

method	result	size
gosper	\(\frac {2 \left (-d x +c \right ) \left (-15 D x^{4} d^{4}-21 C \,d^{4} x^{3}-45 D c \,d^{3} x^{3}-35 B \,d^{4} x^{2}-77 C c \,d^{3} x^{2}-90 D c^{2} d^{2} x^{2}-105 A \,d^{4} x -245 B c \,d^{3} x -308 C \,c^{2} d^{2} x -360 D c^{3} d x +315 A c \,d^{3}+490 B \,c^{2} d^{2}+616 C \,c^{3} d +720 c^{4} D\right ) \left (d x +c \right )^{\frac {3}{2}}}{105 d^{4} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}\)	\(155\)
orering	\(\frac {2 \left (-d x +c \right ) \left (-15 D x^{4} d^{4}-21 C \,d^{4} x^{3}-45 D c \,d^{3} x^{3}-35 B \,d^{4} x^{2}-77 C c \,d^{3} x^{2}-90 D c^{2} d^{2} x^{2}-105 A \,d^{4} x -245 B c \,d^{3} x -308 C \,c^{2} d^{2} x -360 D c^{3} d x +315 A c \,d^{3}+490 B \,c^{2} d^{2}+616 C \,c^{3} d +720 c^{4} D\right ) \left (d x +c \right )^{\frac {3}{2}}}{105 d^{4} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}\)	\(155\)
default	\(\frac {2 \sqrt {-d^{2} x^{2}+c^{2}}\, \left (-15 D x^{4} d^{4}-21 C \,d^{4} x^{3}-45 D c \,d^{3} x^{3}-35 B \,d^{4} x^{2}-77 C c \,d^{3} x^{2}-90 D c^{2} d^{2} x^{2}-105 A \,d^{4} x -245 B c \,d^{3} x -308 C \,c^{2} d^{2} x -360 D c^{3} d x +315 A c \,d^{3}+490 B \,c^{2} d^{2}+616 C \,c^{3} d +720 c^{4} D\right )}{105 \sqrt {d x +c}\, \left (-d x +c \right ) d^{4}}\)	\(157\)

Input:

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

Output:

2/105*(-d*x+c)*(-15*D*d^4*x^4-21*C*d^4*x^3-45*D*c*d^3*x^3-35*B*d^4*x^2-77* 
C*c*d^3*x^2-90*D*c^2*d^2*x^2-105*A*d^4*x-245*B*c*d^3*x-308*C*c^2*d^2*x-360 
*D*c^3*d*x+315*A*c*d^3+490*B*c^2*d^2+616*C*c^3*d+720*D*c^4)*(d*x+c)^(3/2)/ 
d^4/(-d^2*x^2+c^2)^(3/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.63 \[ \int \frac {(c+d x)^{5/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (15 \, D d^{4} x^{4} - 720 \, D c^{4} - 616 \, C c^{3} d - 490 \, B c^{2} d^{2} - 315 \, A c d^{3} + 3 \, {\left (15 \, D c d^{3} + 7 \, C d^{4}\right )} x^{3} + {\left (90 \, D c^{2} d^{2} + 77 \, C c d^{3} + 35 \, B d^{4}\right )} x^{2} + {\left (360 \, D c^{3} d + 308 \, C c^{2} d^{2} + 245 \, B c d^{3} + 105 \, A d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{105 \, {\left (d^{6} x^{2} - c^{2} d^{4}\right )}} \] Input:

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

Output:

2/105*(15*D*d^4*x^4 - 720*D*c^4 - 616*C*c^3*d - 490*B*c^2*d^2 - 315*A*c*d^ 
3 + 3*(15*D*c*d^3 + 7*C*d^4)*x^3 + (90*D*c^2*d^2 + 77*C*c*d^3 + 35*B*d^4)* 
x^2 + (360*D*c^3*d + 308*C*c^2*d^2 + 245*B*c*d^3 + 105*A*d^4)*x)*sqrt(-d^2 
*x^2 + c^2)*sqrt(d*x + c)/(d^6*x^2 - c^2*d^4)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.61 \[ \int \frac {(c+d x)^{5/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=-\frac {2 \, {\left (d x - 3 \, c\right )} A}{\sqrt {-d x + c} d} - \frac {2 \, {\left (d^{2} x^{2} + 7 \, c d x - 14 \, c^{2}\right )} B}{3 \, \sqrt {-d x + c} d^{2}} - \frac {2 \, {\left (3 \, d^{3} x^{3} + 11 \, c d^{2} x^{2} + 44 \, c^{2} d x - 88 \, c^{3}\right )} C}{15 \, \sqrt {-d x + c} d^{3}} - \frac {2 \, {\left (d^{4} x^{4} + 3 \, c d^{3} x^{3} + 6 \, c^{2} d^{2} x^{2} + 24 \, c^{3} d x - 48 \, c^{4}\right )} D}{7 \, \sqrt {-d x + c} d^{4}} \] Input:

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

Output:

-2*(d*x - 3*c)*A/(sqrt(-d*x + c)*d) - 2/3*(d^2*x^2 + 7*c*d*x - 14*c^2)*B/( 
sqrt(-d*x + c)*d^2) - 2/15*(3*d^3*x^3 + 11*c*d^2*x^2 + 44*c^2*d*x - 88*c^3 
)*C/(sqrt(-d*x + c)*d^3) - 2/7*(d^4*x^4 + 3*c*d^3*x^3 + 6*c^2*d^2*x^2 + 24 
*c^3*d*x - 48*c^4)*D/(sqrt(-d*x + c)*d^4)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 229, normalized size of antiderivative = 0.91 \[ \int \frac {(c+d x)^{5/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (\frac {210 \, {\left (D c^{4} + C c^{3} d + B c^{2} d^{2} + A c d^{3}\right )}}{\sqrt {-d x + c} d^{3}} + \frac {15 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} D d^{18} + 105 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} D c d^{18} - 315 \, {\left (-d x + c\right )}^{\frac {3}{2}} D c^{2} d^{18} + 735 \, \sqrt {-d x + c} D c^{3} d^{18} + 21 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} C d^{19} - 140 \, {\left (-d x + c\right )}^{\frac {3}{2}} C c d^{19} + 525 \, \sqrt {-d x + c} C c^{2} d^{19} - 35 \, {\left (-d x + c\right )}^{\frac {3}{2}} B d^{20} + 315 \, \sqrt {-d x + c} B c d^{20} + 105 \, \sqrt {-d x + c} A d^{21}}{d^{21}}\right )}}{105 \, d} \] Input:

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

Output:

2/105*(210*(D*c^4 + C*c^3*d + B*c^2*d^2 + A*c*d^3)/(sqrt(-d*x + c)*d^3) + 
(15*(d*x - c)^3*sqrt(-d*x + c)*D*d^18 + 105*(d*x - c)^2*sqrt(-d*x + c)*D*c 
*d^18 - 315*(-d*x + c)^(3/2)*D*c^2*d^18 + 735*sqrt(-d*x + c)*D*c^3*d^18 + 
21*(d*x - c)^2*sqrt(-d*x + c)*C*d^19 - 140*(-d*x + c)^(3/2)*C*c*d^19 + 525 
*sqrt(-d*x + c)*C*c^2*d^19 - 35*(-d*x + c)^(3/2)*B*d^20 + 315*sqrt(-d*x + 
c)*B*c*d^20 + 105*sqrt(-d*x + c)*A*d^21)/d^21)/d
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.37 \[ \int \frac {(c+d x)^{5/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {-\frac {2}{7} d^{4} x^{4}-\frac {44}{35} c \,d^{3} x^{3}-\frac {2}{3} b \,d^{3} x^{2}-\frac {334}{105} c^{2} d^{2} x^{2}-2 a \,d^{3} x -\frac {14}{3} b c \,d^{2} x -\frac {1336}{105} c^{3} d x +6 a c \,d^{2}+\frac {28}{3} b \,c^{2} d +\frac {2672}{105} c^{4}}{\sqrt {-d x +c}\, d^{3}} \] Input:

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

Output:

(2*(315*a*c*d**2 - 105*a*d**3*x + 490*b*c**2*d - 245*b*c*d**2*x - 35*b*d** 
3*x**2 + 1336*c**4 - 668*c**3*d*x - 167*c**2*d**2*x**2 - 66*c*d**3*x**3 - 
15*d**4*x**4))/(105*sqrt(c - d*x)*d**3)