\(\int (c+d x)^{5/2} \sqrt {c^2-d^2 x^2} (A+B x+C x^2+D x^3) \, dx\) [193]

Optimal result

Integrand size = 41, antiderivative size = 368 \[ \int (c+d x)^{5/2} \sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=-\frac {256 c^3 \left (47 c^2 C d+65 B c d^2+143 A d^3+33 c^3 D\right ) \left (c^2-d^2 x^2\right )^{3/2}}{45045 d^4 (c+d x)^{3/2}}-\frac {64 c^2 \left (47 c^2 C d+65 B c d^2+143 A d^3+33 c^3 D\right ) \left (c^2-d^2 x^2\right )^{3/2}}{15015 d^4 \sqrt {c+d x}}-\frac {8 c \left (47 c^2 C d+65 B c d^2+143 A d^3+33 c^3 D\right ) \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}{3003 d^4}-\frac {2 \left (47 c^2 C d+65 B c d^2+143 A d^3+33 c^3 D\right ) (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}}{1287 d^4}+\frac {2 \left (6 c C d-13 B d^2-11 c^2 D\right ) (c+d x)^{5/2} \left (c^2-d^2 x^2\right )^{3/2}}{143 d^4}-\frac {2 (5 C d-7 c D) (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{3/2}}{65 d^4}-\frac {2 D (c+d x)^{9/2} \left (c^2-d^2 x^2\right )^{3/2}}{15 d^4} \] Output:

-256/45045*c^3*(143*A*d^3+65*B*c*d^2+47*C*c^2*d+33*D*c^3)*(-d^2*x^2+c^2)^( 
3/2)/d^4/(d*x+c)^(3/2)-64/15015*c^2*(143*A*d^3+65*B*c*d^2+47*C*c^2*d+33*D* 
c^3)*(-d^2*x^2+c^2)^(3/2)/d^4/(d*x+c)^(1/2)-8/3003*c*(143*A*d^3+65*B*c*d^2 
+47*C*c^2*d+33*D*c^3)*(d*x+c)^(1/2)*(-d^2*x^2+c^2)^(3/2)/d^4-2/1287*(143*A 
*d^3+65*B*c*d^2+47*C*c^2*d+33*D*c^3)*(d*x+c)^(3/2)*(-d^2*x^2+c^2)^(3/2)/d^ 
4+2/143*(-13*B*d^2+6*C*c*d-11*D*c^2)*(d*x+c)^(5/2)*(-d^2*x^2+c^2)^(3/2)/d^ 
4-2/65*(5*C*d-7*D*c)*(d*x+c)^(7/2)*(-d^2*x^2+c^2)^(3/2)/d^4-2/15*D*(d*x+c) 
^(9/2)*(-d^2*x^2+c^2)^(3/2)/d^4
 

Mathematica [A] (verified)

Time = 0.70 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.54 \[ \int (c+d x)^{5/2} \sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=-\frac {2 (c-d x) \sqrt {c^2-d^2 x^2} \left (12144 c^6 D+8 c^5 d (2071 C+2277 D x)+7 d^6 x^3 \left (715 A+585 B x+495 C x^2+429 D x^3\right )+2 c^4 d^2 (12415 B+3 x (4142 C+3795 D x))+c^3 d^3 \left (45617 A+15 x \left (2483 B+2071 C x+1771 D x^2\right )\right )+c d^5 x^2 (23595 A+7 x (2665 B+9 x (245 C+209 D x)))+c^2 d^4 x (45903 A+5 x (7059 B+7 x (821 C+693 D x)))\right )}{45045 d^4 \sqrt {c+d x}} \] Input:

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

Output:

(-2*(c - d*x)*Sqrt[c^2 - d^2*x^2]*(12144*c^6*D + 8*c^5*d*(2071*C + 2277*D* 
x) + 7*d^6*x^3*(715*A + 585*B*x + 495*C*x^2 + 429*D*x^3) + 2*c^4*d^2*(1241 
5*B + 3*x*(4142*C + 3795*D*x)) + c^3*d^3*(45617*A + 15*x*(2483*B + 2071*C* 
x + 1771*D*x^2)) + c*d^5*x^2*(23595*A + 7*x*(2665*B + 9*x*(245*C + 209*D*x 
))) + c^2*d^4*x*(45903*A + 5*x*(7059*B + 7*x*(821*C + 693*D*x)))))/(45045* 
d^4*Sqrt[c + d*x])
 

Rubi [A] (verified)

Time = 1.20 (sec) , antiderivative size = 312, normalized size of antiderivative = 0.85, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.220, Rules used = {2170, 27, 2170, 27, 672, 459, 459, 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.

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

\(\Big \downarrow \) 2170

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2170

\(\displaystyle \frac {-\frac {2 \int -\frac {5}{2} d^6 (c+d x)^{5/2} \left (-2 D c^3+7 C d c^2+13 A d^3-d \left (-11 D c^2+6 C d c-13 B d^2\right ) x\right ) \sqrt {c^2-d^2 x^2}dx}{13 d^4}-\frac {2}{13} d \left (c^2-d^2 x^2\right )^{3/2} (c+d x)^{7/2} (5 C d-7 c D)}{5 d^5}-\frac {2 D (c+d x)^{9/2} \left (c^2-d^2 x^2\right )^{3/2}}{15 d^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {5}{13} d^2 \int (c+d x)^{5/2} \left (-2 D c^3+7 C d c^2+13 A d^3-d \left (-11 D c^2+6 C d c-13 B d^2\right ) x\right ) \sqrt {c^2-d^2 x^2}dx-\frac {2}{13} d (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{3/2} (5 C d-7 c D)}{5 d^5}-\frac {2 D (c+d x)^{9/2} \left (c^2-d^2 x^2\right )^{3/2}}{15 d^4}\)

\(\Big \downarrow \) 672

\(\displaystyle \frac {\frac {5}{13} d^2 \left (\frac {1}{11} \left (143 A d^3+65 B c d^2+33 c^3 D+47 c^2 C d\right ) \int (c+d x)^{5/2} \sqrt {c^2-d^2 x^2}dx+\frac {2 \left (c^2-d^2 x^2\right )^{3/2} (c+d x)^{5/2} \left (-13 B d^2-11 c^2 D+6 c C d\right )}{11 d}\right )-\frac {2}{13} d (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{3/2} (5 C d-7 c D)}{5 d^5}-\frac {2 D (c+d x)^{9/2} \left (c^2-d^2 x^2\right )^{3/2}}{15 d^4}\)

\(\Big \downarrow \) 459

\(\displaystyle \frac {\frac {5}{13} d^2 \left (\frac {1}{11} \left (143 A d^3+65 B c d^2+33 c^3 D+47 c^2 C d\right ) \left (\frac {4}{3} c \int (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}dx-\frac {2 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}}{9 d}\right )+\frac {2 \left (c^2-d^2 x^2\right )^{3/2} (c+d x)^{5/2} \left (-13 B d^2-11 c^2 D+6 c C d\right )}{11 d}\right )-\frac {2}{13} d (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{3/2} (5 C d-7 c D)}{5 d^5}-\frac {2 D (c+d x)^{9/2} \left (c^2-d^2 x^2\right )^{3/2}}{15 d^4}\)

\(\Big \downarrow \) 459

\(\displaystyle \frac {\frac {5}{13} d^2 \left (\frac {1}{11} \left (143 A d^3+65 B c d^2+33 c^3 D+47 c^2 C d\right ) \left (\frac {4}{3} c \left (\frac {8}{7} c \int \sqrt {c+d x} \sqrt {c^2-d^2 x^2}dx-\frac {2 \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}{7 d}\right )-\frac {2 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}}{9 d}\right )+\frac {2 \left (c^2-d^2 x^2\right )^{3/2} (c+d x)^{5/2} \left (-13 B d^2-11 c^2 D+6 c C d\right )}{11 d}\right )-\frac {2}{13} d (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{3/2} (5 C d-7 c D)}{5 d^5}-\frac {2 D (c+d x)^{9/2} \left (c^2-d^2 x^2\right )^{3/2}}{15 d^4}\)

\(\Big \downarrow \) 459

\(\displaystyle \frac {\frac {5}{13} d^2 \left (\frac {1}{11} \left (143 A d^3+65 B c d^2+33 c^3 D+47 c^2 C d\right ) \left (\frac {4}{3} c \left (\frac {8}{7} c \left (\frac {4}{5} c \int \frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}dx-\frac {2 \left (c^2-d^2 x^2\right )^{3/2}}{5 d \sqrt {c+d x}}\right )-\frac {2 \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}{7 d}\right )-\frac {2 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}}{9 d}\right )+\frac {2 \left (c^2-d^2 x^2\right )^{3/2} (c+d x)^{5/2} \left (-13 B d^2-11 c^2 D+6 c C d\right )}{11 d}\right )-\frac {2}{13} d (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{3/2} (5 C d-7 c D)}{5 d^5}-\frac {2 D (c+d x)^{9/2} \left (c^2-d^2 x^2\right )^{3/2}}{15 d^4}\)

\(\Big \downarrow \) 458

\(\displaystyle \frac {\frac {5}{13} d^2 \left (\frac {1}{11} \left (\frac {4}{3} c \left (\frac {8}{7} c \left (-\frac {2 \left (c^2-d^2 x^2\right )^{3/2}}{5 d \sqrt {c+d x}}-\frac {8 c \left (c^2-d^2 x^2\right )^{3/2}}{15 d (c+d x)^{3/2}}\right )-\frac {2 \sqrt {c+d x} \left (c^2-d^2 x^2\right )^{3/2}}{7 d}\right )-\frac {2 (c+d x)^{3/2} \left (c^2-d^2 x^2\right )^{3/2}}{9 d}\right ) \left (143 A d^3+65 B c d^2+33 c^3 D+47 c^2 C d\right )+\frac {2 \left (c^2-d^2 x^2\right )^{3/2} (c+d x)^{5/2} \left (-13 B d^2-11 c^2 D+6 c C d\right )}{11 d}\right )-\frac {2}{13} d (c+d x)^{7/2} \left (c^2-d^2 x^2\right )^{3/2} (5 C d-7 c D)}{5 d^5}-\frac {2 D (c+d x)^{9/2} \left (c^2-d^2 x^2\right )^{3/2}}{15 d^4}\)

Input:

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

Output:

(-2*D*(c + d*x)^(9/2)*(c^2 - d^2*x^2)^(3/2))/(15*d^4) + ((-2*d*(5*C*d - 7* 
c*D)*(c + d*x)^(7/2)*(c^2 - d^2*x^2)^(3/2))/13 + (5*d^2*((2*(6*c*C*d - 13* 
B*d^2 - 11*c^2*D)*(c + d*x)^(5/2)*(c^2 - d^2*x^2)^(3/2))/(11*d) + ((47*c^2 
*C*d + 65*B*c*d^2 + 143*A*d^3 + 33*c^3*D)*((-2*(c + d*x)^(3/2)*(c^2 - d^2* 
x^2)^(3/2))/(9*d) + (4*c*((-2*Sqrt[c + d*x]*(c^2 - d^2*x^2)^(3/2))/(7*d) + 
 (8*c*((-8*c*(c^2 - d^2*x^2)^(3/2))/(15*d*(c + d*x)^(3/2)) - (2*(c^2 - d^2 
*x^2)^(3/2))/(5*d*Sqrt[c + d*x])))/7))/3))/11))/13)/(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 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 = 251, normalized size of antiderivative = 0.68

method	result	size
gosper	\(-\frac {2 \left (-d x +c \right ) \left (3003 D d^{6} x^{6}+3465 C \,d^{6} x^{5}+13167 D c \,d^{5} x^{5}+4095 B \,d^{6} x^{4}+15435 C c \,d^{5} x^{4}+24255 D c^{2} d^{4} x^{4}+5005 A \,d^{6} x^{3}+18655 B c \,d^{5} x^{3}+28735 C \,c^{2} d^{4} x^{3}+26565 D c^{3} d^{3} x^{3}+23595 A c \,d^{5} x^{2}+35295 B \,c^{2} d^{4} x^{2}+31065 C \,c^{3} d^{3} x^{2}+22770 D c^{4} d^{2} x^{2}+45903 A \,c^{2} d^{4} x +37245 B \,c^{3} d^{3} x +24852 C \,c^{4} d^{2} x +18216 D c^{5} d x +45617 A \,c^{3} d^{3}+24830 B \,c^{4} d^{2}+16568 C \,c^{5} d +12144 D c^{6}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{45045 d^{4} \sqrt {d x +c}}\)	\(251\)
default	\(-\frac {2 \left (-d x +c \right ) \left (3003 D d^{6} x^{6}+3465 C \,d^{6} x^{5}+13167 D c \,d^{5} x^{5}+4095 B \,d^{6} x^{4}+15435 C c \,d^{5} x^{4}+24255 D c^{2} d^{4} x^{4}+5005 A \,d^{6} x^{3}+18655 B c \,d^{5} x^{3}+28735 C \,c^{2} d^{4} x^{3}+26565 D c^{3} d^{3} x^{3}+23595 A c \,d^{5} x^{2}+35295 B \,c^{2} d^{4} x^{2}+31065 C \,c^{3} d^{3} x^{2}+22770 D c^{4} d^{2} x^{2}+45903 A \,c^{2} d^{4} x +37245 B \,c^{3} d^{3} x +24852 C \,c^{4} d^{2} x +18216 D c^{5} d x +45617 A \,c^{3} d^{3}+24830 B \,c^{4} d^{2}+16568 C \,c^{5} d +12144 D c^{6}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{45045 d^{4} \sqrt {d x +c}}\)	\(251\)
orering	\(-\frac {2 \left (-d x +c \right ) \left (3003 D d^{6} x^{6}+3465 C \,d^{6} x^{5}+13167 D c \,d^{5} x^{5}+4095 B \,d^{6} x^{4}+15435 C c \,d^{5} x^{4}+24255 D c^{2} d^{4} x^{4}+5005 A \,d^{6} x^{3}+18655 B c \,d^{5} x^{3}+28735 C \,c^{2} d^{4} x^{3}+26565 D c^{3} d^{3} x^{3}+23595 A c \,d^{5} x^{2}+35295 B \,c^{2} d^{4} x^{2}+31065 C \,c^{3} d^{3} x^{2}+22770 D c^{4} d^{2} x^{2}+45903 A \,c^{2} d^{4} x +37245 B \,c^{3} d^{3} x +24852 C \,c^{4} d^{2} x +18216 D c^{5} d x +45617 A \,c^{3} d^{3}+24830 B \,c^{4} d^{2}+16568 C \,c^{5} d +12144 D c^{6}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{45045 d^{4} \sqrt {d x +c}}\)	\(251\)

Input:

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

Output:

-2/45045*(-d*x+c)*(3003*D*d^6*x^6+3465*C*d^6*x^5+13167*D*c*d^5*x^5+4095*B* 
d^6*x^4+15435*C*c*d^5*x^4+24255*D*c^2*d^4*x^4+5005*A*d^6*x^3+18655*B*c*d^5 
*x^3+28735*C*c^2*d^4*x^3+26565*D*c^3*d^3*x^3+23595*A*c*d^5*x^2+35295*B*c^2 
*d^4*x^2+31065*C*c^3*d^3*x^2+22770*D*c^4*d^2*x^2+45903*A*c^2*d^4*x+37245*B 
*c^3*d^3*x+24852*C*c^4*d^2*x+18216*D*c^5*d*x+45617*A*c^3*d^3+24830*B*c^4*d 
^2+16568*C*c^5*d+12144*D*c^6)*(-d^2*x^2+c^2)^(1/2)/d^4/(d*x+c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 281, normalized size of antiderivative = 0.76 \[ \int (c+d x)^{5/2} \sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2 \, {\left (3003 \, D d^{7} x^{7} - 12144 \, D c^{7} - 16568 \, C c^{6} d - 24830 \, B c^{5} d^{2} - 45617 \, A c^{4} d^{3} + 231 \, {\left (44 \, D c d^{6} + 15 \, C d^{7}\right )} x^{6} + 63 \, {\left (176 \, D c^{2} d^{5} + 190 \, C c d^{6} + 65 \, B d^{7}\right )} x^{5} + 35 \, {\left (66 \, D c^{3} d^{4} + 380 \, C c^{2} d^{5} + 416 \, B c d^{6} + 143 \, A d^{7}\right )} x^{4} - 5 \, {\left (759 \, D c^{4} d^{3} - 466 \, C c^{3} d^{4} - 3328 \, B c^{2} d^{5} - 3718 \, A c d^{6}\right )} x^{3} - 3 \, {\left (1518 \, D c^{5} d^{2} + 2071 \, C c^{4} d^{3} - 650 \, B c^{3} d^{4} - 7436 \, A c^{2} d^{5}\right )} x^{2} - {\left (6072 \, D c^{6} d + 8284 \, C c^{5} d^{2} + 12415 \, B c^{4} d^{3} + 286 \, A c^{3} d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c}}{45045 \, {\left (d^{5} x + c d^{4}\right )}} \] Input:

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

Output:

2/45045*(3003*D*d^7*x^7 - 12144*D*c^7 - 16568*C*c^6*d - 24830*B*c^5*d^2 - 
45617*A*c^4*d^3 + 231*(44*D*c*d^6 + 15*C*d^7)*x^6 + 63*(176*D*c^2*d^5 + 19 
0*C*c*d^6 + 65*B*d^7)*x^5 + 35*(66*D*c^3*d^4 + 380*C*c^2*d^5 + 416*B*c*d^6 
 + 143*A*d^7)*x^4 - 5*(759*D*c^4*d^3 - 466*C*c^3*d^4 - 3328*B*c^2*d^5 - 37 
18*A*c*d^6)*x^3 - 3*(1518*D*c^5*d^2 + 2071*C*c^4*d^3 - 650*B*c^3*d^4 - 743 
6*A*c^2*d^5)*x^2 - (6072*D*c^6*d + 8284*C*c^5*d^2 + 12415*B*c^4*d^3 + 286* 
A*c^3*d^4)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)/(d^5*x + c*d^4)
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 345, normalized size of antiderivative = 0.94 \[ \int (c+d x)^{5/2} \sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2 \, {\left (35 \, d^{4} x^{4} + 130 \, c d^{3} x^{3} + 156 \, c^{2} d^{2} x^{2} - 2 \, c^{3} d x - 319 \, c^{4}\right )} {\left (d x + c\right )} \sqrt {-d x + c} A}{315 \, {\left (d^{2} x + c d\right )}} + \frac {2 \, {\left (63 \, d^{5} x^{5} + 224 \, c d^{4} x^{4} + 256 \, c^{2} d^{3} x^{3} + 30 \, c^{3} d^{2} x^{2} - 191 \, c^{4} d x - 382 \, c^{5}\right )} {\left (d x + c\right )} \sqrt {-d x + c} B}{693 \, {\left (d^{3} x + c d^{2}\right )}} + \frac {2 \, {\left (3465 \, d^{6} x^{6} + 11970 \, c d^{5} x^{5} + 13300 \, c^{2} d^{4} x^{4} + 2330 \, c^{3} d^{3} x^{3} - 6213 \, c^{4} d^{2} x^{2} - 8284 \, c^{5} d x - 16568 \, c^{6}\right )} {\left (d x + c\right )} \sqrt {-d x + c} C}{45045 \, {\left (d^{4} x + c d^{3}\right )}} + \frac {2 \, {\left (91 \, d^{7} x^{7} + 308 \, c d^{6} x^{6} + 336 \, c^{2} d^{5} x^{5} + 70 \, c^{3} d^{4} x^{4} - 115 \, c^{4} d^{3} x^{3} - 138 \, c^{5} d^{2} x^{2} - 184 \, c^{6} d x - 368 \, c^{7}\right )} {\left (d x + c\right )} \sqrt {-d x + c} D}{1365 \, {\left (d^{5} x + c d^{4}\right )}} \] Input:

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

Output:

2/315*(35*d^4*x^4 + 130*c*d^3*x^3 + 156*c^2*d^2*x^2 - 2*c^3*d*x - 319*c^4) 
*(d*x + c)*sqrt(-d*x + c)*A/(d^2*x + c*d) + 2/693*(63*d^5*x^5 + 224*c*d^4* 
x^4 + 256*c^2*d^3*x^3 + 30*c^3*d^2*x^2 - 191*c^4*d*x - 382*c^5)*(d*x + c)* 
sqrt(-d*x + c)*B/(d^3*x + c*d^2) + 2/45045*(3465*d^6*x^6 + 11970*c*d^5*x^5 
 + 13300*c^2*d^4*x^4 + 2330*c^3*d^3*x^3 - 6213*c^4*d^2*x^2 - 8284*c^5*d*x 
- 16568*c^6)*(d*x + c)*sqrt(-d*x + c)*C/(d^4*x + c*d^3) + 2/1365*(91*d^7*x 
^7 + 308*c*d^6*x^6 + 336*c^2*d^5*x^5 + 70*c^3*d^4*x^4 - 115*c^4*d^3*x^3 - 
138*c^5*d^2*x^2 - 184*c^6*d*x - 368*c^7)*(d*x + c)*sqrt(-d*x + c)*D/(d^5*x 
 + c*d^4)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1346 vs. \(2 (326) = 652\).

Time = 0.12 (sec) , antiderivative size = 1346, normalized size of antiderivative = 3.66 \[ \int (c+d x)^{5/2} \sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\text {Too large to display} \] Input:

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

Output:

-2/45045*(45045*sqrt(-d*x + c)*A*c^4*d^3 - 15015*((-d*x + c)^(3/2) - 3*sqr 
t(-d*x + c)*c)*B*c^4*d^2 - 30030*((-d*x + c)^(3/2) - 3*sqrt(-d*x + c)*c)*A 
*c^3*d^3 + 3003*(3*(d*x - c)^2*sqrt(-d*x + c) - 10*(-d*x + c)^(3/2)*c + 15 
*sqrt(-d*x + c)*c^2)*C*c^4*d + 6006*(3*(d*x - c)^2*sqrt(-d*x + c) - 10*(-d 
*x + c)^(3/2)*c + 15*sqrt(-d*x + c)*c^2)*B*c^3*d^2 + 1287*(5*(d*x - c)^3*s 
qrt(-d*x + c) + 21*(d*x - c)^2*sqrt(-d*x + c)*c - 35*(-d*x + c)^(3/2)*c^2 
+ 35*sqrt(-d*x + c)*c^3)*D*c^4 + 2574*(5*(d*x - c)^3*sqrt(-d*x + c) + 21*( 
d*x - c)^2*sqrt(-d*x + c)*c - 35*(-d*x + c)^(3/2)*c^2 + 35*sqrt(-d*x + c)* 
c^3)*C*c^3*d - 2574*(5*(d*x - c)^3*sqrt(-d*x + c) + 21*(d*x - c)^2*sqrt(-d 
*x + c)*c - 35*(-d*x + c)^(3/2)*c^2 + 35*sqrt(-d*x + c)*c^3)*A*c*d^3 + 286 
*(35*(d*x - c)^4*sqrt(-d*x + c) + 180*(d*x - c)^3*sqrt(-d*x + c)*c + 378*( 
d*x - c)^2*sqrt(-d*x + c)*c^2 - 420*(-d*x + c)^(3/2)*c^3 + 315*sqrt(-d*x + 
 c)*c^4)*D*c^3 - 286*(35*(d*x - c)^4*sqrt(-d*x + c) + 180*(d*x - c)^3*sqrt 
(-d*x + c)*c + 378*(d*x - c)^2*sqrt(-d*x + c)*c^2 - 420*(-d*x + c)^(3/2)*c 
^3 + 315*sqrt(-d*x + c)*c^4)*B*c*d^2 - 143*(35*(d*x - c)^4*sqrt(-d*x + c) 
+ 180*(d*x - c)^3*sqrt(-d*x + c)*c + 378*(d*x - c)^2*sqrt(-d*x + c)*c^2 - 
420*(-d*x + c)^(3/2)*c^3 + 315*sqrt(-d*x + c)*c^4)*A*d^3 - 130*(63*(d*x - 
c)^5*sqrt(-d*x + c) + 385*(d*x - c)^4*sqrt(-d*x + c)*c + 990*(d*x - c)^3*s 
qrt(-d*x + c)*c^2 + 1386*(d*x - c)^2*sqrt(-d*x + c)*c^3 - 1155*(-d*x + c)^ 
(3/2)*c^4 + 693*sqrt(-d*x + c)*c^5)*C*c*d - 65*(63*(d*x - c)^5*sqrt(-d*...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.53 \[ \int (c+d x)^{5/2} \sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {2 \sqrt {-d x +c}\, \left (3003 d^{7} x^{7}+13629 c \,d^{6} x^{6}+4095 b \,d^{6} x^{5}+23058 c^{2} d^{5} x^{5}+5005 a \,d^{6} x^{4}+14560 b c \,d^{5} x^{4}+15610 c^{3} d^{4} x^{4}+18590 a c \,d^{5} x^{3}+16640 b \,c^{2} d^{4} x^{3}-1465 c^{4} d^{3} x^{3}+22308 a \,c^{2} d^{4} x^{2}+1950 b \,c^{3} d^{3} x^{2}-10767 c^{5} d^{2} x^{2}-286 a \,c^{3} d^{3} x -12415 b \,c^{4} d^{2} x -14356 c^{6} d x -45617 a \,c^{4} d^{2}-24830 b \,c^{5} d -28712 c^{7}\right )}{45045 d^{3}} \] Input:

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

Output:

(2*sqrt(c - d*x)*( - 45617*a*c**4*d**2 - 286*a*c**3*d**3*x + 22308*a*c**2* 
d**4*x**2 + 18590*a*c*d**5*x**3 + 5005*a*d**6*x**4 - 24830*b*c**5*d - 1241 
5*b*c**4*d**2*x + 1950*b*c**3*d**3*x**2 + 16640*b*c**2*d**4*x**3 + 14560*b 
*c*d**5*x**4 + 4095*b*d**6*x**5 - 28712*c**7 - 14356*c**6*d*x - 10767*c**5 
*d**2*x**2 - 1465*c**4*d**3*x**3 + 15610*c**3*d**4*x**4 + 23058*c**2*d**5* 
x**5 + 13629*c*d**6*x**6 + 3003*d**7*x**7))/(45045*d**3)