\(\int \frac {(c+d x)^{7/2} (A+B x+C x^2+D x^3)}{(c^2-d^2 x^2)^{5/2}} \, dx\) [228]

Optimal result

Integrand size = 41, antiderivative size = 277 \[ \int \frac {(c+d x)^{7/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {\left (c^2 C d+B c d^2+A d^3+c^3 D\right ) (c+d x)^{7/2}}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\left (13 c^2 C d+7 B c d^2+A d^3+19 c^3 D\right ) (c+d x)^{5/2}}{6 c^2 d^4 \sqrt {c^2-d^2 x^2}}-\frac {2 \left (85 c^2 C d+35 B c d^2+5 A d^3+147 c^3 D\right ) \sqrt {c^2-d^2 x^2}}{15 c d^4 \sqrt {c+d x}}-\frac {\left (85 c^2 C d+35 B c d^2+5 A d^3+147 c^3 D\right ) \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{30 c^2 d^4}-\frac {2 D (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{5 d^4} \] Output:

1/3*(A*d^3+B*c*d^2+C*c^2*d+D*c^3)*(d*x+c)^(7/2)/c/d^4/(-d^2*x^2+c^2)^(3/2) 
-1/6*(A*d^3+7*B*c*d^2+13*C*c^2*d+19*D*c^3)*(d*x+c)^(5/2)/c^2/d^4/(-d^2*x^2 
+c^2)^(1/2)-2/15*(5*A*d^3+35*B*c*d^2+85*C*c^2*d+147*D*c^3)*(-d^2*x^2+c^2)^ 
(1/2)/c/d^4/(d*x+c)^(1/2)-1/30*(5*A*d^3+35*B*c*d^2+85*C*c^2*d+147*D*c^3)*( 
d*x+c)^(1/2)*(-d^2*x^2+c^2)^(1/2)/c^2/d^4-2/5*D*(d*x+c)^(3/2)*(-d^2*x^2+c^ 
2)^(1/2)/d^4
 

Mathematica [A] (verified)

Time = 2.78 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.48 \[ \int \frac {(c+d x)^{7/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {2 \sqrt {c+d x} \left (208 c^4 D+24 c^3 d (5 C-13 D x)+2 c^2 d^2 \left (25 B-90 C x+39 D x^2\right )+d^4 x \left (-15 A+x \left (15 B+5 C x+3 D x^2\right )\right )+c d^3 \left (5 A+x \left (-75 B+45 C x+13 D x^2\right )\right )\right )}{15 d^4 (-c+d x) \sqrt {c^2-d^2 x^2}} \] Input:

Integrate[((c + d*x)^(7/2)*(A + B*x + C*x^2 + D*x^3))/(c^2 - d^2*x^2)^(5/2 
),x]
 

Output:

(2*Sqrt[c + d*x]*(208*c^4*D + 24*c^3*d*(5*C - 13*D*x) + 2*c^2*d^2*(25*B - 
90*C*x + 39*D*x^2) + d^4*x*(-15*A + x*(15*B + 5*C*x + 3*D*x^2)) + c*d^3*(5 
*A + x*(-75*B + 45*C*x + 13*D*x^2))))/(15*d^4*(-c + d*x)*Sqrt[c^2 - d^2*x^ 
2])
 

Rubi [A] (verified)

Time = 1.21 (sec) , antiderivative size = 269, 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 = {2166, 27, 2166, 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.

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

\(\Big \downarrow \) 2166

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2166

\(\displaystyle \frac {(c+d x)^{7/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {(c+d x)^{5/2} \left (A d^3+7 B c d^2+19 c^3 D+13 c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {\int \frac {3 (c+d x)^{3/2} \left (27 D c^3+17 C d c^2+4 d D x c^2+7 B d^2 c+A d^3\right )}{2 d^3 \sqrt {c^2-d^2 x^2}}dx}{c}}{6 c}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(c+d x)^{7/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {(c+d x)^{5/2} \left (A d^3+7 B c d^2+19 c^3 D+13 c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {3 \int \frac {(c+d x)^{3/2} \left (27 D c^3+17 C d c^2+4 d D x c^2+7 B d^2 c+A d^3\right )}{\sqrt {c^2-d^2 x^2}}dx}{2 c d^3}}{6 c}\)

\(\Big \downarrow \) 672

\(\displaystyle \frac {(c+d x)^{7/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {(c+d x)^{5/2} \left (A d^3+7 B c d^2+19 c^3 D+13 c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {3 \left (\frac {1}{5} \left (5 A d^3+35 B c d^2+147 c^3 D+85 c^2 C d\right ) \int \frac {(c+d x)^{3/2}}{\sqrt {c^2-d^2 x^2}}dx-\frac {8 c^2 D (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{5 d}\right )}{2 c d^3}}{6 c}\)

\(\Big \downarrow \) 459

\(\displaystyle \frac {(c+d x)^{7/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {(c+d x)^{5/2} \left (A d^3+7 B c d^2+19 c^3 D+13 c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {3 \left (\frac {1}{5} \left (5 A d^3+35 B c d^2+147 c^3 D+85 c^2 C d\right ) \left (\frac {4}{3} c \int \frac {\sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}dx-\frac {2 \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{3 d}\right )-\frac {8 c^2 D (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{5 d}\right )}{2 c d^3}}{6 c}\)

\(\Big \downarrow \) 458

\(\displaystyle \frac {(c+d x)^{7/2} \left (A d^3+B c d^2+c^3 D+c^2 C d\right )}{3 c d^4 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {\frac {(c+d x)^{5/2} \left (A d^3+7 B c d^2+19 c^3 D+13 c^2 C d\right )}{c d^4 \sqrt {c^2-d^2 x^2}}-\frac {3 \left (\frac {1}{5} \left (-\frac {8 c \sqrt {c^2-d^2 x^2}}{3 d \sqrt {c+d x}}-\frac {2 \sqrt {c+d x} \sqrt {c^2-d^2 x^2}}{3 d}\right ) \left (5 A d^3+35 B c d^2+147 c^3 D+85 c^2 C d\right )-\frac {8 c^2 D (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}{5 d}\right )}{2 c d^3}}{6 c}\)

Input:

Int[((c + d*x)^(7/2)*(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)^(7/2))/(3*c*d^4*(c^2 - d^2* 
x^2)^(3/2)) - (((13*c^2*C*d + 7*B*c*d^2 + A*d^3 + 19*c^3*D)*(c + d*x)^(5/2 
))/(c*d^4*Sqrt[c^2 - d^2*x^2]) - (3*((-8*c^2*D*(c + d*x)^(3/2)*Sqrt[c^2 - 
d^2*x^2])/(5*d) + ((85*c^2*C*d + 35*B*c*d^2 + 5*A*d^3 + 147*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))/(2*c*d^3))/(6*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]
 

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.56

method	result	size
gosper	\(-\frac {2 \left (-d x +c \right ) \left (3 D x^{4} d^{4}+5 C \,d^{4} x^{3}+13 D c \,d^{3} x^{3}+15 B \,d^{4} x^{2}+45 C c \,d^{3} x^{2}+78 D c^{2} d^{2} x^{2}-15 A \,d^{4} x -75 B c \,d^{3} x -180 C \,c^{2} d^{2} x -312 D c^{3} d x +5 A c \,d^{3}+50 B \,c^{2} d^{2}+120 C \,c^{3} d +208 c^{4} D\right ) \left (d x +c \right )^{\frac {5}{2}}}{15 d^{4} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}\)	\(155\)
orering	\(-\frac {2 \left (-d x +c \right ) \left (3 D x^{4} d^{4}+5 C \,d^{4} x^{3}+13 D c \,d^{3} x^{3}+15 B \,d^{4} x^{2}+45 C c \,d^{3} x^{2}+78 D c^{2} d^{2} x^{2}-15 A \,d^{4} x -75 B c \,d^{3} x -180 C \,c^{2} d^{2} x -312 D c^{3} d x +5 A c \,d^{3}+50 B \,c^{2} d^{2}+120 C \,c^{3} d +208 c^{4} D\right ) \left (d x +c \right )^{\frac {5}{2}}}{15 d^{4} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}\)	\(155\)
default	\(-\frac {2 \sqrt {-d^{2} x^{2}+c^{2}}\, \left (3 D x^{4} d^{4}+5 C \,d^{4} x^{3}+13 D c \,d^{3} x^{3}+15 B \,d^{4} x^{2}+45 C c \,d^{3} x^{2}+78 D c^{2} d^{2} x^{2}-15 A \,d^{4} x -75 B c \,d^{3} x -180 C \,c^{2} d^{2} x -312 D c^{3} d x +5 A c \,d^{3}+50 B \,c^{2} d^{2}+120 C \,c^{3} d +208 c^{4} D\right )}{15 \sqrt {d x +c}\, \left (-d x +c \right )^{2} d^{4}}\)	\(157\)

Input:

int((d*x+c)^(7/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),x,method=_RETUR 
NVERBOSE)
 

Output:

-2/15*(-d*x+c)*(3*D*d^4*x^4+5*C*d^4*x^3+13*D*c*d^3*x^3+15*B*d^4*x^2+45*C*c 
*d^3*x^2+78*D*c^2*d^2*x^2-15*A*d^4*x-75*B*c*d^3*x-180*C*c^2*d^2*x-312*D*c^ 
3*d*x+5*A*c*d^3+50*B*c^2*d^2+120*C*c^3*d+208*D*c^4)*(d*x+c)^(5/2)/d^4/(-d^ 
2*x^2+c^2)^(5/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.64 \[ \int \frac {(c+d x)^{7/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (3 \, D d^{4} x^{4} + 208 \, D c^{4} + 120 \, C c^{3} d + 50 \, B c^{2} d^{2} + 5 \, A c d^{3} + {\left (13 \, D c d^{3} + 5 \, C d^{4}\right )} x^{3} + 3 \, {\left (26 \, D c^{2} d^{2} + 15 \, C c d^{3} + 5 \, B d^{4}\right )} x^{2} - 3 \, {\left (104 \, D c^{3} d + 60 \, C c^{2} d^{2} + 25 \, B c d^{3} + 5 \, A d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{15 \, {\left (d^{7} x^{3} - c d^{6} x^{2} - c^{2} d^{5} x + c^{3} d^{4}\right )}} \] Input:

integrate((d*x+c)^(7/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),x, algori 
thm="fricas")
 

Output:

-2/15*(3*D*d^4*x^4 + 208*D*c^4 + 120*C*c^3*d + 50*B*c^2*d^2 + 5*A*c*d^3 + 
(13*D*c*d^3 + 5*C*d^4)*x^3 + 3*(26*D*c^2*d^2 + 15*C*c*d^3 + 5*B*d^4)*x^2 - 
 3*(104*D*c^3*d + 60*C*c^2*d^2 + 25*B*c*d^3 + 5*A*d^4)*x)*sqrt(-d^2*x^2 + 
c^2)*sqrt(d*x + c)/(d^7*x^3 - c*d^6*x^2 - c^2*d^5*x + c^3*d^4)
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {(c+d x)^{7/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (c + d x\right )^{\frac {7}{2}} \left (A + B x + C x^{2} + D x^{3}\right )}{\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((d*x+c)**(7/2)*(D*x**3+C*x**2+B*x+A)/(-d**2*x**2+c**2)**(5/2),x)
 

Output:

Integral((c + d*x)**(7/2)*(A + B*x + C*x**2 + D*x**3)/(-(-c + d*x)*(c + d* 
x))**(5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.71 \[ \int \frac {(c+d x)^{7/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (3 \, d x - c\right )} A}{3 \, {\left (d^{2} x - c d\right )} \sqrt {-d x + c}} + \frac {2 \, {\left (3 \, d^{2} x^{2} - 15 \, c d x + 10 \, c^{2}\right )} B}{3 \, {\left (d^{3} x - c d^{2}\right )} \sqrt {-d x + c}} + \frac {2 \, {\left (d^{3} x^{3} + 9 \, c d^{2} x^{2} - 36 \, c^{2} d x + 24 \, c^{3}\right )} C}{3 \, {\left (d^{4} x - c d^{3}\right )} \sqrt {-d x + c}} + \frac {2 \, {\left (3 \, d^{4} x^{4} + 13 \, c d^{3} x^{3} + 78 \, c^{2} d^{2} x^{2} - 312 \, c^{3} d x + 208 \, c^{4}\right )} D}{15 \, {\left (d^{5} x - c d^{4}\right )} \sqrt {-d x + c}} \] Input:

integrate((d*x+c)^(7/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),x, algori 
thm="maxima")
 

Output:

-2/3*(3*d*x - c)*A/((d^2*x - c*d)*sqrt(-d*x + c)) + 2/3*(3*d^2*x^2 - 15*c* 
d*x + 10*c^2)*B/((d^3*x - c*d^2)*sqrt(-d*x + c)) + 2/3*(d^3*x^3 + 9*c*d^2* 
x^2 - 36*c^2*d*x + 24*c^3)*C/((d^4*x - c*d^3)*sqrt(-d*x + c)) + 2/15*(3*d^ 
4*x^4 + 13*c*d^3*x^3 + 78*c^2*d^2*x^2 - 312*c^3*d*x + 208*c^4)*D/((d^5*x - 
 c*d^4)*sqrt(-d*x + c))
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.78 \[ \int \frac {(c+d x)^{7/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (\frac {5 \, {\left (21 \, {\left (d x - c\right )} D c^{3} + 2 \, D c^{4} + 15 \, {\left (d x - c\right )} C c^{2} d + 2 \, C c^{3} d + 9 \, {\left (d x - c\right )} B c d^{2} + 2 \, B c^{2} d^{2} + 3 \, {\left (d x - c\right )} A d^{3} + 2 \, A c d^{3}\right )}}{{\left (d x - c\right )} \sqrt {-d x + c} d^{3}} + \frac {3 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} D d^{12} - 25 \, {\left (-d x + c\right )}^{\frac {3}{2}} D c d^{12} + 135 \, \sqrt {-d x + c} D c^{2} d^{12} - 5 \, {\left (-d x + c\right )}^{\frac {3}{2}} C d^{13} + 60 \, \sqrt {-d x + c} C c d^{13} + 15 \, \sqrt {-d x + c} B d^{14}}{d^{15}}\right )}}{15 \, d} \] Input:

integrate((d*x+c)^(7/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),x, algori 
thm="giac")
 

Output:

-2/15*(5*(21*(d*x - c)*D*c^3 + 2*D*c^4 + 15*(d*x - c)*C*c^2*d + 2*C*c^3*d 
+ 9*(d*x - c)*B*c*d^2 + 2*B*c^2*d^2 + 3*(d*x - c)*A*d^3 + 2*A*c*d^3)/((d*x 
 - c)*sqrt(-d*x + c)*d^3) + (3*(d*x - c)^2*sqrt(-d*x + c)*D*d^12 - 25*(-d* 
x + c)^(3/2)*D*c*d^12 + 135*sqrt(-d*x + c)*D*c^2*d^12 - 5*(-d*x + c)^(3/2) 
*C*d^13 + 60*sqrt(-d*x + c)*C*c*d^13 + 15*sqrt(-d*x + c)*B*d^14)/d^15)/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^{7/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\int \frac {{\left (c+d\,x\right )}^{7/2}\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (c^2-d^2\,x^2\right )}^{5/2}} \,d x \] Input:

int(((c + d*x)^(7/2)*(A + B*x + C*x^2 + x^3*D))/(c^2 - d^2*x^2)^(5/2),x)
 

Output:

int(((c + d*x)^(7/2)*(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 = 101, normalized size of antiderivative = 0.36 \[ \int \frac {(c+d x)^{7/2} \left (A+B x+C x^2+D x^3\right )}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {-\frac {2}{5} d^{4} x^{4}-\frac {12}{5} c \,d^{3} x^{3}-2 b \,d^{3} x^{2}-\frac {82}{5} c^{2} d^{2} x^{2}+2 a \,d^{3} x +10 b c \,d^{2} x +\frac {328}{5} c^{3} d x -\frac {2}{3} a c \,d^{2}-\frac {20}{3} b \,c^{2} d -\frac {656}{15} c^{4}}{\sqrt {-d x +c}\, d^{3} \left (-d x +c \right )} \] Input:

int((d*x+c)^(7/2)*(D*x^3+C*x^2+B*x+A)/(-d^2*x^2+c^2)^(5/2),x)
 

Output:

(2*( - 5*a*c*d**2 + 15*a*d**3*x - 50*b*c**2*d + 75*b*c*d**2*x - 15*b*d**3* 
x**2 - 328*c**4 + 492*c**3*d*x - 123*c**2*d**2*x**2 - 18*c*d**3*x**3 - 3*d 
**4*x**4))/(15*sqrt(c - d*x)*d**3*(c - d*x))