\(\int \frac {(c^2-d^2 x^2)^{5/2} (A+B x+C x^2+D x^3)}{(c+d x)^6} \, dx\) [162]

Optimal result

Integrand size = 39, antiderivative size = 315 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^6} \, dx=-\frac {\left (6 c C d-B d^2-19 c^2 D\right ) \sqrt {c^2-d^2 x^2}}{d^4}+\frac {(C d-6 c D) x \sqrt {c^2-d^2 x^2}}{2 d^3}-\frac {2 \left (13 c^2 C d-5 B c d^2+A d^3-25 c^3 D\right ) \sqrt {c^2-d^2 x^2}}{d^4 (c+d x)}-\frac {D \left (c^2-d^2 x^2\right )^{3/2}}{3 d^4}+\frac {2 \left (5 c^2 C d-3 B c d^2+A d^3-7 c^3 D\right ) \left (c^2-d^2 x^2\right )^{3/2}}{3 d^4 (c+d x)^3}-\frac {2 \left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \left (c^2-d^2 x^2\right )^{5/2}}{5 d^4 (c+d x)^5}-\frac {\left (37 c^2 C d-12 B c d^2+2 A d^3-82 c^3 D\right ) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{2 d^4} \] Output:

-(-B*d^2+6*C*c*d-19*D*c^2)*(-d^2*x^2+c^2)^(1/2)/d^4+1/2*(C*d-6*D*c)*x*(-d^ 
2*x^2+c^2)^(1/2)/d^3-2*(A*d^3-5*B*c*d^2+13*C*c^2*d-25*D*c^3)*(-d^2*x^2+c^2 
)^(1/2)/d^4/(d*x+c)-1/3*D*(-d^2*x^2+c^2)^(3/2)/d^4+2/3*(A*d^3-3*B*c*d^2+5* 
C*c^2*d-7*D*c^3)*(-d^2*x^2+c^2)^(3/2)/d^4/(d*x+c)^3-2/5*(A*d^3-B*c*d^2+C*c 
^2*d-D*c^3)*(-d^2*x^2+c^2)^(5/2)/d^4/(d*x+c)^5-1/2*(2*A*d^3-12*B*c*d^2+37* 
C*c^2*d-82*D*c^3)*arctan(d*x/(-d^2*x^2+c^2)^(1/2))/d^4
 

Mathematica [A] (verified)

Time = 2.32 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.70 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^6} \, dx=\frac {\frac {\sqrt {c^2-d^2 x^2} \left (1932 c^5 D+c^4 (-872 C d+4566 d D x)+3 c^3 d^2 (94 B+x (-687 C+1024 D x))+d^5 x^2 (-92 A+5 x (6 B+x (3 C+2 D x)))-3 c d^4 x (32 A+x (-154 B+5 x (9 C+4 D x)))+c^2 d^3 (-52 A+x (666 B+x (-1387 C+320 D x)))\right )}{(c+d x)^3}+30 \left (37 c^2 C d-12 B c d^2+2 A d^3-82 c^3 D\right ) \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{30 d^4} \] Input:

Integrate[((c^2 - d^2*x^2)^(5/2)*(A + B*x + C*x^2 + D*x^3))/(c + d*x)^6,x]
 

Output:

((Sqrt[c^2 - d^2*x^2]*(1932*c^5*D + c^4*(-872*C*d + 4566*d*D*x) + 3*c^3*d^ 
2*(94*B + x*(-687*C + 1024*D*x)) + d^5*x^2*(-92*A + 5*x*(6*B + x*(3*C + 2* 
D*x))) - 3*c*d^4*x*(32*A + x*(-154*B + 5*x*(9*C + 4*D*x))) + c^2*d^3*(-52* 
A + x*(666*B + x*(-1387*C + 320*D*x)))))/(c + d*x)^3 + 30*(37*c^2*C*d - 12 
*B*c*d^2 + 2*A*d^3 - 82*c^3*D)*ArcTan[(d*x)/(Sqrt[c^2] - Sqrt[c^2 - d^2*x^ 
2])])/(30*d^4)
 

Rubi [A] (verified)

Time = 1.16 (sec) , antiderivative size = 289, normalized size of antiderivative = 0.92, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.256, Rules used = {2170, 25, 2170, 27, 671, 465, 463, 455, 224, 216}

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.

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

\(\Big \downarrow \) 2170

\(\displaystyle -\frac {\int -\frac {\left (c^2-d^2 x^2\right )^{5/2} \left ((3 C d-10 c D) x^2 d^4+\left (3 B d^2-11 c^2 D\right ) x d^3+\left (3 A d^3-4 c^3 D\right ) d^2\right )}{(c+d x)^6}dx}{3 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{3 d^4 (c+d x)^4}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left ((3 C d-10 c D) x^2 d^4+\left (3 B d^2-11 c^2 D\right ) x d^3+\left (3 A d^3-4 c^3 D\right ) d^2\right )}{(c+d x)^6}dx}{3 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{3 d^4 (c+d x)^4}\)

\(\Big \downarrow \) 2170

\(\displaystyle \frac {-\frac {\int \frac {3 d^6 \left (-14 D c^3+5 C d c^2-2 A d^3+d \left (-16 D c^2+7 C d c-2 B d^2\right ) x\right ) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^6}dx}{2 d^4}-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (3 C d-10 c D)}{2 (c+d x)^5}}{3 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{3 d^4 (c+d x)^4}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {3}{2} d^2 \int \frac {\left (-14 D c^3+5 C d c^2-2 A d^3+d \left (-16 D c^2+7 C d c-2 B d^2\right ) x\right ) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^6}dx-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (3 C d-10 c D)}{2 (c+d x)^5}}{3 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{3 d^4 (c+d x)^4}\)

\(\Big \downarrow \) 671

\(\displaystyle \frac {-\frac {3}{2} d^2 \left (\frac {\left (2 A d^3-12 B c d^2-82 c^3 D+37 c^2 C d\right ) \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^5}dx}{5 c}+\frac {2 \left (c^2-d^2 x^2\right )^{7/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{5 c d (c+d x)^6}\right )-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (3 C d-10 c D)}{2 (c+d x)^5}}{3 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{3 d^4 (c+d x)^4}\)

\(\Big \downarrow \) 465

\(\displaystyle \frac {-\frac {3}{2} d^2 \left (\frac {\left (2 A d^3-12 B c d^2-82 c^3 D+37 c^2 C d\right ) \left (-\frac {5}{3} \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^3}dx-\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{3 d (c+d x)^4}\right )}{5 c}+\frac {2 \left (c^2-d^2 x^2\right )^{7/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{5 c d (c+d x)^6}\right )-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (3 C d-10 c D)}{2 (c+d x)^5}}{3 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{3 d^4 (c+d x)^4}\)

\(\Big \downarrow \) 463

\(\displaystyle \frac {-\frac {3}{2} d^2 \left (\frac {\left (2 A d^3-12 B c d^2-82 c^3 D+37 c^2 C d\right ) \left (-\frac {5}{3} \left (-\int \frac {3 c-d x}{\sqrt {c^2-d^2 x^2}}dx-\frac {4 c \sqrt {c^2-d^2 x^2}}{d (c+d x)}\right )-\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{3 d (c+d x)^4}\right )}{5 c}+\frac {2 \left (c^2-d^2 x^2\right )^{7/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{5 c d (c+d x)^6}\right )-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (3 C d-10 c D)}{2 (c+d x)^5}}{3 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{3 d^4 (c+d x)^4}\)

\(\Big \downarrow \) 455

\(\displaystyle \frac {-\frac {3}{2} d^2 \left (\frac {\left (2 A d^3-12 B c d^2-82 c^3 D+37 c^2 C d\right ) \left (-\frac {5}{3} \left (-3 c \int \frac {1}{\sqrt {c^2-d^2 x^2}}dx-\frac {4 c \sqrt {c^2-d^2 x^2}}{d (c+d x)}-\frac {\sqrt {c^2-d^2 x^2}}{d}\right )-\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{3 d (c+d x)^4}\right )}{5 c}+\frac {2 \left (c^2-d^2 x^2\right )^{7/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{5 c d (c+d x)^6}\right )-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (3 C d-10 c D)}{2 (c+d x)^5}}{3 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{3 d^4 (c+d x)^4}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {-\frac {3}{2} d^2 \left (\frac {\left (2 A d^3-12 B c d^2-82 c^3 D+37 c^2 C d\right ) \left (-\frac {5}{3} \left (-3 c \int \frac {1}{\frac {d^2 x^2}{c^2-d^2 x^2}+1}d\frac {x}{\sqrt {c^2-d^2 x^2}}-\frac {4 c \sqrt {c^2-d^2 x^2}}{d (c+d x)}-\frac {\sqrt {c^2-d^2 x^2}}{d}\right )-\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{3 d (c+d x)^4}\right )}{5 c}+\frac {2 \left (c^2-d^2 x^2\right )^{7/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{5 c d (c+d x)^6}\right )-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (3 C d-10 c D)}{2 (c+d x)^5}}{3 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{3 d^4 (c+d x)^4}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {-\frac {3}{2} d^2 \left (\frac {\left (-\frac {5}{3} \left (-\frac {3 c \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{d}-\frac {4 c \sqrt {c^2-d^2 x^2}}{d (c+d x)}-\frac {\sqrt {c^2-d^2 x^2}}{d}\right )-\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{3 d (c+d x)^4}\right ) \left (2 A d^3-12 B c d^2-82 c^3 D+37 c^2 C d\right )}{5 c}+\frac {2 \left (c^2-d^2 x^2\right )^{7/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{5 c d (c+d x)^6}\right )-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (3 C d-10 c D)}{2 (c+d x)^5}}{3 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{3 d^4 (c+d x)^4}\)

Input:

Int[((c^2 - d^2*x^2)^(5/2)*(A + B*x + C*x^2 + D*x^3))/(c + d*x)^6,x]
 

Output:

-1/3*(D*(c^2 - d^2*x^2)^(7/2))/(d^4*(c + d*x)^4) + (-1/2*(d*(3*C*d - 10*c* 
D)*(c^2 - d^2*x^2)^(7/2))/(c + d*x)^5 - (3*d^2*((2*(c^2*C*d - B*c*d^2 + A* 
d^3 - c^3*D)*(c^2 - d^2*x^2)^(7/2))/(5*c*d*(c + d*x)^6) + ((37*c^2*C*d - 1 
2*B*c*d^2 + 2*A*d^3 - 82*c^3*D)*((-2*(c^2 - d^2*x^2)^(5/2))/(3*d*(c + d*x) 
^4) - (5*(-(Sqrt[c^2 - d^2*x^2]/d) - (4*c*Sqrt[c^2 - d^2*x^2])/(d*(c + d*x 
)) - (3*c*ArcTan[(d*x)/Sqrt[c^2 - d^2*x^2]])/d))/3))/(5*c)))/2)/(3*d^5)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 455

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

rule 463

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(-(-c)^(-n - 2))*d^(2*n + 3)*(Sqrt[a + b*x^2]/(2^(n + 1)*b^(n + 2)*(c + d*x 
))), x] - Simp[d^(2*n + 2)/b^(n + 1)   Int[(1/Sqrt[a + b*x^2])*ExpandToSum[ 
(2^(-n - 1)*(-c)^(-n - 1) - (-c + d*x)^(-n - 1))/(c + d*x), x], x], x] /; F 
reeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && ILtQ[n, 0] && EqQ[n + p, 
-3/2]
 

rule 465

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 + p + 1))), x] - Simp[b*(p/(d^2*(n + 
 p + 1)))   Int[(c + d*x)^(n + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, 
b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0] && (LtQ[n, -2] || EqQ[n 
+ 2*p + 1, 0]) && NeQ[n + p + 1, 0] && IntegerQ[2*p]
 

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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1540\) vs. \(2(291)=582\).

Time = 0.70 (sec) , antiderivative size = 1541, normalized size of antiderivative = 4.89

method	result	size
default	\(\text {Expression too large to display}\)	\(1541\)

Input:

int((-d^2*x^2+c^2)^(5/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^6,x,method=_RETURNVER 
BOSE)
 

Output:

D/d^6*(1/c/d/(x+c/d)^3*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(7/2)+4*d/c*(1/3/c/d 
/(x+c/d)^2*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(7/2)+5/3*d/c*(1/5*(-d^2*(x+c/d) 
^2+2*c*d*(x+c/d))^(5/2)+c*d*(-1/8*(-2*d^2*(x+c/d)+2*c*d)/d^2*(-d^2*(x+c/d) 
^2+2*c*d*(x+c/d))^(3/2)+3/4*c^2*(-1/4*(-2*d^2*(x+c/d)+2*c*d)/d^2*(-d^2*(x+ 
c/d)^2+2*c*d*(x+c/d))^(1/2)+1/2*c^2/(d^2)^(1/2)*arctan((d^2)^(1/2)*x/(-d^2 
*(x+c/d)^2+2*c*d*(x+c/d))^(1/2)))))))+(C*d-3*D*c)/d^7*(-1/c/d/(x+c/d)^4*(- 
d^2*(x+c/d)^2+2*c*d*(x+c/d))^(7/2)-3*d/c*(1/c/d/(x+c/d)^3*(-d^2*(x+c/d)^2+ 
2*c*d*(x+c/d))^(7/2)+4*d/c*(1/3/c/d/(x+c/d)^2*(-d^2*(x+c/d)^2+2*c*d*(x+c/d 
))^(7/2)+5/3*d/c*(1/5*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(5/2)+c*d*(-1/8*(-2*d 
^2*(x+c/d)+2*c*d)/d^2*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(3/2)+3/4*c^2*(-1/4*( 
-2*d^2*(x+c/d)+2*c*d)/d^2*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(1/2)+1/2*c^2/(d^ 
2)^(1/2)*arctan((d^2)^(1/2)*x/(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(1/2))))))))+ 
(B*d^2-2*C*c*d+3*D*c^2)/d^8*(-1/3/c/d/(x+c/d)^5*(-d^2*(x+c/d)^2+2*c*d*(x+c 
/d))^(7/2)-2/3*d/c*(-1/c/d/(x+c/d)^4*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(7/2)- 
3*d/c*(1/c/d/(x+c/d)^3*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(7/2)+4*d/c*(1/3/c/d 
/(x+c/d)^2*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(7/2)+5/3*d/c*(1/5*(-d^2*(x+c/d) 
^2+2*c*d*(x+c/d))^(5/2)+c*d*(-1/8*(-2*d^2*(x+c/d)+2*c*d)/d^2*(-d^2*(x+c/d) 
^2+2*c*d*(x+c/d))^(3/2)+3/4*c^2*(-1/4*(-2*d^2*(x+c/d)+2*c*d)/d^2*(-d^2*(x+ 
c/d)^2+2*c*d*(x+c/d))^(1/2)+1/2*c^2/(d^2)^(1/2)*arctan((d^2)^(1/2)*x/(-d^2 
*(x+c/d)^2+2*c*d*(x+c/d))^(1/2)))))))))+(A*d^3-B*c*d^2+C*c^2*d-D*c^3)/d...
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 534, normalized size of antiderivative = 1.70 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^6} \, dx=\frac {1932 \, D c^{6} - 872 \, C c^{5} d + 282 \, B c^{4} d^{2} - 52 \, A c^{3} d^{3} + 2 \, {\left (966 \, D c^{3} d^{3} - 436 \, C c^{2} d^{4} + 141 \, B c d^{5} - 26 \, A d^{6}\right )} x^{3} + 6 \, {\left (966 \, D c^{4} d^{2} - 436 \, C c^{3} d^{3} + 141 \, B c^{2} d^{4} - 26 \, A c d^{5}\right )} x^{2} + 6 \, {\left (966 \, D c^{5} d - 436 \, C c^{4} d^{2} + 141 \, B c^{3} d^{3} - 26 \, A c^{2} d^{4}\right )} x - 30 \, {\left (82 \, D c^{6} - 37 \, C c^{5} d + 12 \, B c^{4} d^{2} - 2 \, A c^{3} d^{3} + {\left (82 \, D c^{3} d^{3} - 37 \, C c^{2} d^{4} + 12 \, B c d^{5} - 2 \, A d^{6}\right )} x^{3} + 3 \, {\left (82 \, D c^{4} d^{2} - 37 \, C c^{3} d^{3} + 12 \, B c^{2} d^{4} - 2 \, A c d^{5}\right )} x^{2} + 3 \, {\left (82 \, D c^{5} d - 37 \, C c^{4} d^{2} + 12 \, B c^{3} d^{3} - 2 \, A c^{2} d^{4}\right )} x\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + {\left (10 \, D d^{5} x^{5} + 1932 \, D c^{5} - 872 \, C c^{4} d + 282 \, B c^{3} d^{2} - 52 \, A c^{2} d^{3} - 15 \, {\left (4 \, D c d^{4} - C d^{5}\right )} x^{4} + 5 \, {\left (64 \, D c^{2} d^{3} - 27 \, C c d^{4} + 6 \, B d^{5}\right )} x^{3} + {\left (3072 \, D c^{3} d^{2} - 1387 \, C c^{2} d^{3} + 462 \, B c d^{4} - 92 \, A d^{5}\right )} x^{2} + 3 \, {\left (1522 \, D c^{4} d - 687 \, C c^{3} d^{2} + 222 \, B c^{2} d^{3} - 32 \, A c d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{30 \, {\left (d^{7} x^{3} + 3 \, c d^{6} x^{2} + 3 \, c^{2} d^{5} x + c^{3} d^{4}\right )}} \] Input:

integrate((-d^2*x^2+c^2)^(5/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^6,x, algorithm= 
"fricas")
 

Output:

1/30*(1932*D*c^6 - 872*C*c^5*d + 282*B*c^4*d^2 - 52*A*c^3*d^3 + 2*(966*D*c 
^3*d^3 - 436*C*c^2*d^4 + 141*B*c*d^5 - 26*A*d^6)*x^3 + 6*(966*D*c^4*d^2 - 
436*C*c^3*d^3 + 141*B*c^2*d^4 - 26*A*c*d^5)*x^2 + 6*(966*D*c^5*d - 436*C*c 
^4*d^2 + 141*B*c^3*d^3 - 26*A*c^2*d^4)*x - 30*(82*D*c^6 - 37*C*c^5*d + 12* 
B*c^4*d^2 - 2*A*c^3*d^3 + (82*D*c^3*d^3 - 37*C*c^2*d^4 + 12*B*c*d^5 - 2*A* 
d^6)*x^3 + 3*(82*D*c^4*d^2 - 37*C*c^3*d^3 + 12*B*c^2*d^4 - 2*A*c*d^5)*x^2 
+ 3*(82*D*c^5*d - 37*C*c^4*d^2 + 12*B*c^3*d^3 - 2*A*c^2*d^4)*x)*arctan(-(c 
 - sqrt(-d^2*x^2 + c^2))/(d*x)) + (10*D*d^5*x^5 + 1932*D*c^5 - 872*C*c^4*d 
 + 282*B*c^3*d^2 - 52*A*c^2*d^3 - 15*(4*D*c*d^4 - C*d^5)*x^4 + 5*(64*D*c^2 
*d^3 - 27*C*c*d^4 + 6*B*d^5)*x^3 + (3072*D*c^3*d^2 - 1387*C*c^2*d^3 + 462* 
B*c*d^4 - 92*A*d^5)*x^2 + 3*(1522*D*c^4*d - 687*C*c^3*d^2 + 222*B*c^2*d^3 
- 32*A*c*d^4)*x)*sqrt(-d^2*x^2 + c^2))/(d^7*x^3 + 3*c*d^6*x^2 + 3*c^2*d^5* 
x + c^3*d^4)
 

Sympy [F]

\[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^6} \, dx=\int \frac {\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {5}{2}} \left (A + B x + C x^{2} + D x^{3}\right )}{\left (c + d x\right )^{6}}\, dx \] Input:

integrate((-d**2*x**2+c**2)**(5/2)*(D*x**3+C*x**2+B*x+A)/(d*x+c)**6,x)
 

Output:

Integral((-(-c + d*x)*(c + d*x))**(5/2)*(A + B*x + C*x**2 + D*x**3)/(c + d 
*x)**6, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1825 vs. \(2 (292) = 584\).

Time = 0.17 (sec) , antiderivative size = 1825, normalized size of antiderivative = 5.79 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^6} \, dx=\text {Too large to display} \] Input:

integrate((-d^2*x^2+c^2)^(5/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^6,x, algorithm= 
"maxima")
 

Output:

1/5*(-d^2*x^2 + c^2)^(5/2)*D*c^3/(d^9*x^5 + 5*c*d^8*x^4 + 10*c^2*d^7*x^3 + 
 10*c^3*d^6*x^2 + 5*c^4*d^5*x + c^5*d^4) + (-d^2*x^2 + c^2)^(3/2)*D*c^4/(d 
^8*x^4 + 4*c*d^7*x^3 + 6*c^2*d^6*x^2 + 4*c^3*d^5*x + c^4*d^4) - 6/5*sqrt(- 
d^2*x^2 + c^2)*D*c^5/(d^7*x^3 + 3*c*d^6*x^2 + 3*c^2*d^5*x + c^3*d^4) - 1/5 
*(-d^2*x^2 + c^2)^(5/2)*C*c^2/(d^8*x^5 + 5*c*d^7*x^4 + 10*c^2*d^6*x^3 + 10 
*c^3*d^5*x^2 + 5*c^4*d^4*x + c^5*d^3) + 3*(-d^2*x^2 + c^2)^(5/2)*D*c^2/(d^ 
8*x^4 + 4*c*d^7*x^3 + 6*c^2*d^6*x^2 + 4*c^3*d^5*x + c^4*d^4) - (-d^2*x^2 + 
 c^2)^(3/2)*C*c^3/(d^7*x^4 + 4*c*d^6*x^3 + 6*c^2*d^5*x^2 + 4*c^3*d^4*x + c 
^4*d^3) - 16/3*(-d^2*x^2 + c^2)^(3/2)*D*c^3/(d^7*x^3 + 3*c*d^6*x^2 + 3*c^2 
*d^5*x + c^3*d^4) + 6/5*sqrt(-d^2*x^2 + c^2)*C*c^4/(d^6*x^3 + 3*c*d^5*x^2 
+ 3*c^2*d^4*x + c^3*d^3) - 157/15*sqrt(-d^2*x^2 + c^2)*D*c^4/(d^6*x^2 + 2* 
c*d^5*x + c^2*d^4) + 1/5*(-d^2*x^2 + c^2)^(5/2)*B*c/(d^7*x^5 + 5*c*d^6*x^4 
 + 10*c^2*d^5*x^3 + 10*c^3*d^4*x^2 + 5*c^4*d^3*x + c^5*d^2) - 2*(-d^2*x^2 
+ c^2)^(5/2)*C*c/(d^7*x^4 + 4*c*d^6*x^3 + 6*c^2*d^5*x^2 + 4*c^3*d^4*x + c^ 
4*d^3) - 3/2*(-d^2*x^2 + c^2)^(5/2)*D*c/(d^7*x^3 + 3*c*d^6*x^2 + 3*c^2*d^5 
*x + c^3*d^4) + (-d^2*x^2 + c^2)^(3/2)*B*c^2/(d^6*x^4 + 4*c*d^5*x^3 + 6*c^ 
2*d^4*x^2 + 4*c^3*d^3*x + c^4*d^2) + 11/3*(-d^2*x^2 + c^2)^(3/2)*C*c^2/(d^ 
6*x^3 + 3*c*d^5*x^2 + 3*c^2*d^4*x + c^3*d^3) - 15/2*(-d^2*x^2 + c^2)^(3/2) 
*D*c^2/(d^6*x^2 + 2*c*d^5*x + c^2*d^4) - 6/5*sqrt(-d^2*x^2 + c^2)*B*c^3/(d 
^5*x^3 + 3*c*d^4*x^2 + 3*c^2*d^3*x + c^3*d^2) + 107/15*sqrt(-d^2*x^2 + ...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 717 vs. \(2 (292) = 584\).

Time = 0.15 (sec) , antiderivative size = 717, normalized size of antiderivative = 2.28 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^6} \, dx =\text {Too large to display} \] Input:

integrate((-d^2*x^2+c^2)^(5/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^6,x, algorithm= 
"giac")
 

Output:

1/6*sqrt(-d^2*x^2 + c^2)*(x*(2*D*x/d^2 - 3*(6*D*c*d^9 - C*d^10)/d^12) + 2* 
(56*D*c^2*d^8 - 18*C*c*d^9 + 3*B*d^10)/d^12) + 1/2*(82*D*c^3 - 37*C*c^2*d 
+ 12*B*c*d^2 - 2*A*d^3)*arcsin(d*x/c)*sgn(c)*sgn(d)/(d^3*abs(d)) - 4/15*(3 
43*D*c^3 - 173*C*c^2*d + 63*B*c*d^2 - 13*A*d^3 + 270*(c*d + sqrt(-d^2*x^2 
+ c^2)*abs(d))*B*c/x + 1430*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))*D*c^3/(d^2 
*x) - 730*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))*C*c^2/(d*x) - 50*(c*d + sqrt 
(-d^2*x^2 + c^2)*abs(d))*A*d/x + 2140*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^ 
2*D*c^3/(d^4*x^2) - 1100*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^2*C*c^2/(d^3* 
x^2) + 420*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^2*B*c/(d^2*x^2) - 100*(c*d 
+ sqrt(-d^2*x^2 + c^2)*abs(d))^2*A/(d*x^2) + 1290*(c*d + sqrt(-d^2*x^2 + c 
^2)*abs(d))^3*D*c^3/(d^6*x^3) - 630*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^3* 
C*c^2/(d^5*x^3) + 210*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^3*B*c/(d^4*x^3) 
- 30*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^3*A/(d^3*x^3) + 285*(c*d + sqrt(- 
d^2*x^2 + c^2)*abs(d))^4*D*c^3/(d^8*x^4) - 135*(c*d + sqrt(-d^2*x^2 + c^2) 
*abs(d))^4*C*c^2/(d^7*x^4) + 45*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^4*B*c/ 
(d^6*x^4) - 15*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^4*A/(d^5*x^4))/(d^3*((c 
*d + sqrt(-d^2*x^2 + c^2)*abs(d))/(d^2*x) + 1)^5*abs(d))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^6} \, dx=\int \frac {{\left (c^2-d^2\,x^2\right )}^{5/2}\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (c+d\,x\right )}^6} \,d x \] Input:

int(((c^2 - d^2*x^2)^(5/2)*(A + B*x + C*x^2 + x^3*D))/(c + d*x)^6,x)
 

Output:

int(((c^2 - d^2*x^2)^(5/2)*(A + B*x + C*x^2 + x^3*D))/(c + d*x)^6, x)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 1136, normalized size of antiderivative = 3.61 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^6} \, dx =\text {Too large to display} \] Input:

int((-d^2*x^2+c^2)^(5/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^6,x)
 

Output:

(30*atan((sqrt(c**2 - d**2*x**2)*c**2 - 2*sqrt(c**2 - d**2*x**2)*d**2*x**2 
)/(2*c**2*d*x - 2*d**3*x**3))*a*c**3*d**2 + 90*atan((sqrt(c**2 - d**2*x**2 
)*c**2 - 2*sqrt(c**2 - d**2*x**2)*d**2*x**2)/(2*c**2*d*x - 2*d**3*x**3))*a 
*c**2*d**3*x + 90*atan((sqrt(c**2 - d**2*x**2)*c**2 - 2*sqrt(c**2 - d**2*x 
**2)*d**2*x**2)/(2*c**2*d*x - 2*d**3*x**3))*a*c*d**4*x**2 + 30*atan((sqrt( 
c**2 - d**2*x**2)*c**2 - 2*sqrt(c**2 - d**2*x**2)*d**2*x**2)/(2*c**2*d*x - 
 2*d**3*x**3))*a*d**5*x**3 - 180*atan((sqrt(c**2 - d**2*x**2)*c**2 - 2*sqr 
t(c**2 - d**2*x**2)*d**2*x**2)/(2*c**2*d*x - 2*d**3*x**3))*b*c**4*d - 540* 
atan((sqrt(c**2 - d**2*x**2)*c**2 - 2*sqrt(c**2 - d**2*x**2)*d**2*x**2)/(2 
*c**2*d*x - 2*d**3*x**3))*b*c**3*d**2*x - 540*atan((sqrt(c**2 - d**2*x**2) 
*c**2 - 2*sqrt(c**2 - d**2*x**2)*d**2*x**2)/(2*c**2*d*x - 2*d**3*x**3))*b* 
c**2*d**3*x**2 - 180*atan((sqrt(c**2 - d**2*x**2)*c**2 - 2*sqrt(c**2 - d** 
2*x**2)*d**2*x**2)/(2*c**2*d*x - 2*d**3*x**3))*b*c*d**4*x**3 - 675*atan((s 
qrt(c**2 - d**2*x**2)*c**2 - 2*sqrt(c**2 - d**2*x**2)*d**2*x**2)/(2*c**2*d 
*x - 2*d**3*x**3))*c**6 - 2025*atan((sqrt(c**2 - d**2*x**2)*c**2 - 2*sqrt( 
c**2 - d**2*x**2)*d**2*x**2)/(2*c**2*d*x - 2*d**3*x**3))*c**5*d*x - 2025*a 
tan((sqrt(c**2 - d**2*x**2)*c**2 - 2*sqrt(c**2 - d**2*x**2)*d**2*x**2)/(2* 
c**2*d*x - 2*d**3*x**3))*c**4*d**2*x**2 - 675*atan((sqrt(c**2 - d**2*x**2) 
*c**2 - 2*sqrt(c**2 - d**2*x**2)*d**2*x**2)/(2*c**2*d*x - 2*d**3*x**3))*c* 
*3*d**3*x**3 - 104*sqrt(c**2 - d**2*x**2)*a*c**2*d**2 - 192*sqrt(c**2 -...