\(\int \frac {(c^2-d^2 x^2)^{5/2} (A+B x+C x^2+D x^3)}{(c+d x)^7} \, dx\) [163]

Optimal result

Integrand size = 39, antiderivative size = 299 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^7} \, dx=\frac {(C d-7 c D) \sqrt {c^2-d^2 x^2}}{d^4}+\frac {D x \sqrt {c^2-d^2 x^2}}{2 d^3}+\frac {2 \left (6 c C d-B d^2-19 c^2 D\right ) \sqrt {c^2-d^2 x^2}}{d^4 (c+d x)}-\frac {2 \left (4 c C d-B d^2-9 c^2 D\right ) \left (c^2-d^2 x^2\right )^{3/2}}{3 d^4 (c+d x)^3}+\frac {2 \left (2 c C d-B d^2-3 c^2 D\right ) \left (c^2-d^2 x^2\right )^{5/2}}{5 d^4 (c+d x)^5}-\frac {\left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \left (c^2-d^2 x^2\right )^{7/2}}{7 c d^4 (c+d x)^7}+\frac {\left (14 c C d-2 B d^2-51 c^2 D\right ) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{2 d^4} \] Output:

(C*d-7*D*c)*(-d^2*x^2+c^2)^(1/2)/d^4+1/2*D*x*(-d^2*x^2+c^2)^(1/2)/d^3+2*(- 
B*d^2+6*C*c*d-19*D*c^2)*(-d^2*x^2+c^2)^(1/2)/d^4/(d*x+c)-2/3*(-B*d^2+4*C*c 
*d-9*D*c^2)*(-d^2*x^2+c^2)^(3/2)/d^4/(d*x+c)^3+2/5*(-B*d^2+2*C*c*d-3*D*c^2 
)*(-d^2*x^2+c^2)^(5/2)/d^4/(d*x+c)^5-1/7*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)*(-d 
^2*x^2+c^2)^(7/2)/c/d^4/(d*x+c)^7+1/2*(-2*B*d^2+14*C*c*d-51*D*c^2)*arctan( 
d*x/(-d^2*x^2+c^2)^(1/2))/d^4
 

Mathematica [A] (verified)

Time = 2.66 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.75 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^7} \, dx=-\frac {\sqrt {c^2-d^2 x^2} \left (8412 c^6 D-30 A d^6 x^3+c^5 (-2308 C d+28293 d D x)+2 c^4 d^2 (167 B+x (-3881 C+16629 D x))+c d^5 x^2 (90 A+x (674 B-105 x (2 C+D x)))+2 c^2 d^4 x (-45 A+x (613 B+x (-2059 C+525 D x)))+2 c^3 d^3 (15 A+x (563 B+x (-4561 C+7386 D x)))\right )}{210 c d^4 (c+d x)^4}+\frac {\left (-14 c C d+2 B d^2+51 c^2 D\right ) \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{d^4} \] Input:

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

Output:

-1/210*(Sqrt[c^2 - d^2*x^2]*(8412*c^6*D - 30*A*d^6*x^3 + c^5*(-2308*C*d + 
28293*d*D*x) + 2*c^4*d^2*(167*B + x*(-3881*C + 16629*D*x)) + c*d^5*x^2*(90 
*A + x*(674*B - 105*x*(2*C + D*x))) + 2*c^2*d^4*x*(-45*A + x*(613*B + x*(- 
2059*C + 525*D*x))) + 2*c^3*d^3*(15*A + x*(563*B + x*(-4561*C + 7386*D*x)) 
)))/(c*d^4*(c + d*x)^4) + ((-14*c*C*d + 2*B*d^2 + 51*c^2*D)*ArcTan[(d*x)/( 
Sqrt[c^2] - Sqrt[c^2 - d^2*x^2])])/d^4
 

Rubi [A] (verified)

Time = 1.10 (sec) , antiderivative size = 272, normalized size of antiderivative = 0.91, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.282, Rules used = {2170, 25, 2170, 27, 671, 465, 465, 463, 25, 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.

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

\(\Big \downarrow \) 2170

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2170

\(\displaystyle \frac {-\frac {\int \frac {d^6 \left (-49 D c^3+12 C d c^2-2 A d^3+d \left (-51 D c^2+14 C d c-2 B d^2\right ) x\right ) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^7}dx}{d^4}-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (2 C d-9 c D)}{(c+d x)^6}}{2 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{2 d^4 (c+d x)^5}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-d^2 \int \frac {\left (-49 D c^3+12 C d c^2-2 A d^3+d \left (-51 D c^2+14 C d c-2 B d^2\right ) x\right ) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^7}dx-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (2 C d-9 c D)}{(c+d x)^6}}{2 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{2 d^4 (c+d x)^5}\)

\(\Big \downarrow \) 671

\(\displaystyle \frac {-d^2 \left (\left (-2 B d^2-51 c^2 D+14 c C d\right ) \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^6}dx+\frac {2 \left (c^2-d^2 x^2\right )^{7/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{7 c d (c+d x)^7}\right )-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (2 C d-9 c D)}{(c+d x)^6}}{2 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{2 d^4 (c+d x)^5}\)

\(\Big \downarrow \) 465

\(\displaystyle \frac {-d^2 \left (\left (-2 B d^2-51 c^2 D+14 c C d\right ) \left (-\int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^4}dx-\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{5 d (c+d x)^5}\right )+\frac {2 \left (c^2-d^2 x^2\right )^{7/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{7 c d (c+d x)^7}\right )-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (2 C d-9 c D)}{(c+d x)^6}}{2 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{2 d^4 (c+d x)^5}\)

\(\Big \downarrow \) 465

\(\displaystyle \frac {-d^2 \left (\left (-2 B d^2-51 c^2 D+14 c C d\right ) \left (\int \frac {\sqrt {c^2-d^2 x^2}}{(c+d x)^2}dx-\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{5 d (c+d x)^5}+\frac {2 \left (c^2-d^2 x^2\right )^{3/2}}{3 d (c+d x)^3}\right )+\frac {2 \left (c^2-d^2 x^2\right )^{7/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{7 c d (c+d x)^7}\right )-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (2 C d-9 c D)}{(c+d x)^6}}{2 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{2 d^4 (c+d x)^5}\)

\(\Big \downarrow \) 463

\(\displaystyle \frac {-d^2 \left (\left (-2 B d^2-51 c^2 D+14 c C d\right ) \left (\int -\frac {1}{\sqrt {c^2-d^2 x^2}}dx-\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{5 d (c+d x)^5}+\frac {2 \left (c^2-d^2 x^2\right )^{3/2}}{3 d (c+d x)^3}-\frac {2 \sqrt {c^2-d^2 x^2}}{d (c+d x)}\right )+\frac {2 \left (c^2-d^2 x^2\right )^{7/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{7 c d (c+d x)^7}\right )-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (2 C d-9 c D)}{(c+d x)^6}}{2 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{2 d^4 (c+d x)^5}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {-d^2 \left (\left (-2 B d^2-51 c^2 D+14 c C d\right ) \left (-\int \frac {1}{\sqrt {c^2-d^2 x^2}}dx-\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{5 d (c+d x)^5}+\frac {2 \left (c^2-d^2 x^2\right )^{3/2}}{3 d (c+d x)^3}-\frac {2 \sqrt {c^2-d^2 x^2}}{d (c+d x)}\right )+\frac {2 \left (c^2-d^2 x^2\right )^{7/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{7 c d (c+d x)^7}\right )-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (2 C d-9 c D)}{(c+d x)^6}}{2 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{2 d^4 (c+d x)^5}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {-d^2 \left (\left (-2 B d^2-51 c^2 D+14 c C d\right ) \left (-\int \frac {1}{\frac {d^2 x^2}{c^2-d^2 x^2}+1}d\frac {x}{\sqrt {c^2-d^2 x^2}}-\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{5 d (c+d x)^5}+\frac {2 \left (c^2-d^2 x^2\right )^{3/2}}{3 d (c+d x)^3}-\frac {2 \sqrt {c^2-d^2 x^2}}{d (c+d x)}\right )+\frac {2 \left (c^2-d^2 x^2\right )^{7/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{7 c d (c+d x)^7}\right )-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (2 C d-9 c D)}{(c+d x)^6}}{2 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{2 d^4 (c+d x)^5}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {-d^2 \left (\frac {2 \left (c^2-d^2 x^2\right )^{7/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{7 c d (c+d x)^7}+\left (-\frac {\arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{d}-\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{5 d (c+d x)^5}+\frac {2 \left (c^2-d^2 x^2\right )^{3/2}}{3 d (c+d x)^3}-\frac {2 \sqrt {c^2-d^2 x^2}}{d (c+d x)}\right ) \left (-2 B d^2-51 c^2 D+14 c C d\right )\right )-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (2 C d-9 c D)}{(c+d x)^6}}{2 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{2 d^4 (c+d x)^5}\)

Input:

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

Output:

-1/2*(D*(c^2 - d^2*x^2)^(7/2))/(d^4*(c + d*x)^5) + (-((d*(2*C*d - 9*c*D)*( 
c^2 - d^2*x^2)^(7/2))/(c + d*x)^6) - 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))/(7*c*d*(c + d*x)^7) + (14*c*C*d - 2*B*d^2 - 
51*c^2*D)*((-2*Sqrt[c^2 - d^2*x^2])/(d*(c + d*x)) + (2*(c^2 - d^2*x^2)^(3/ 
2))/(3*d*(c + d*x)^3) - (2*(c^2 - d^2*x^2)^(5/2))/(5*d*(c + d*x)^5) - ArcT 
an[(d*x)/Sqrt[c^2 - d^2*x^2]]/d)))/(2*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 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. \(1291\) vs. \(2(275)=550\).

Time = 0.86 (sec) , antiderivative size = 1292, normalized size of antiderivative = 4.32

method	result	size
default	\(\text {Expression too large to display}\)	\(1292\)

Input:

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

Output:

D/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))))))))+(C*d-3*D*c)/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)))))))))+(B*d^2-2*C*c*d+3*D*c 
^2)/d^9*(-1/5/c/d/(x+c/d)^6*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(7/2)-1/5*d/c*( 
-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/...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 612 vs. \(2 (274) = 548\).

Time = 0.15 (sec) , antiderivative size = 612, normalized size of antiderivative = 2.05 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^7} \, dx=-\frac {8412 \, D c^{7} - 2308 \, C c^{6} d + 334 \, B c^{5} d^{2} + 30 \, A c^{4} d^{3} + 2 \, {\left (4206 \, D c^{3} d^{4} - 1154 \, C c^{2} d^{5} + 167 \, B c d^{6} + 15 \, A d^{7}\right )} x^{4} + 8 \, {\left (4206 \, D c^{4} d^{3} - 1154 \, C c^{3} d^{4} + 167 \, B c^{2} d^{5} + 15 \, A c d^{6}\right )} x^{3} + 12 \, {\left (4206 \, D c^{5} d^{2} - 1154 \, C c^{4} d^{3} + 167 \, B c^{3} d^{4} + 15 \, A c^{2} d^{5}\right )} x^{2} + 8 \, {\left (4206 \, D c^{6} d - 1154 \, C c^{5} d^{2} + 167 \, B c^{4} d^{3} + 15 \, A c^{3} d^{4}\right )} x - 210 \, {\left (51 \, D c^{7} - 14 \, C c^{6} d + 2 \, B c^{5} d^{2} + {\left (51 \, D c^{3} d^{4} - 14 \, C c^{2} d^{5} + 2 \, B c d^{6}\right )} x^{4} + 4 \, {\left (51 \, D c^{4} d^{3} - 14 \, C c^{3} d^{4} + 2 \, B c^{2} d^{5}\right )} x^{3} + 6 \, {\left (51 \, D c^{5} d^{2} - 14 \, C c^{4} d^{3} + 2 \, B c^{3} d^{4}\right )} x^{2} + 4 \, {\left (51 \, D c^{6} d - 14 \, C c^{5} d^{2} + 2 \, B c^{4} d^{3}\right )} x\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) - {\left (105 \, D c d^{5} x^{5} - 8412 \, D c^{6} + 2308 \, C c^{5} d - 334 \, B c^{4} d^{2} - 30 \, A c^{3} d^{3} - 210 \, {\left (5 \, D c^{2} d^{4} - C c d^{5}\right )} x^{4} - 2 \, {\left (7386 \, D c^{3} d^{3} - 2059 \, C c^{2} d^{4} + 337 \, B c d^{5} - 15 \, A d^{6}\right )} x^{3} - 2 \, {\left (16629 \, D c^{4} d^{2} - 4561 \, C c^{3} d^{3} + 613 \, B c^{2} d^{4} + 45 \, A c d^{5}\right )} x^{2} - {\left (28293 \, D c^{5} d - 7762 \, C c^{4} d^{2} + 1126 \, B c^{3} d^{3} - 90 \, A c^{2} d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{210 \, {\left (c d^{8} x^{4} + 4 \, c^{2} d^{7} x^{3} + 6 \, c^{3} d^{6} x^{2} + 4 \, c^{4} d^{5} x + c^{5} d^{4}\right )}} \] Input:

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

Output:

-1/210*(8412*D*c^7 - 2308*C*c^6*d + 334*B*c^5*d^2 + 30*A*c^4*d^3 + 2*(4206 
*D*c^3*d^4 - 1154*C*c^2*d^5 + 167*B*c*d^6 + 15*A*d^7)*x^4 + 8*(4206*D*c^4* 
d^3 - 1154*C*c^3*d^4 + 167*B*c^2*d^5 + 15*A*c*d^6)*x^3 + 12*(4206*D*c^5*d^ 
2 - 1154*C*c^4*d^3 + 167*B*c^3*d^4 + 15*A*c^2*d^5)*x^2 + 8*(4206*D*c^6*d - 
 1154*C*c^5*d^2 + 167*B*c^4*d^3 + 15*A*c^3*d^4)*x - 210*(51*D*c^7 - 14*C*c 
^6*d + 2*B*c^5*d^2 + (51*D*c^3*d^4 - 14*C*c^2*d^5 + 2*B*c*d^6)*x^4 + 4*(51 
*D*c^4*d^3 - 14*C*c^3*d^4 + 2*B*c^2*d^5)*x^3 + 6*(51*D*c^5*d^2 - 14*C*c^4* 
d^3 + 2*B*c^3*d^4)*x^2 + 4*(51*D*c^6*d - 14*C*c^5*d^2 + 2*B*c^4*d^3)*x)*ar 
ctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x)) - (105*D*c*d^5*x^5 - 8412*D*c^6 + 
2308*C*c^5*d - 334*B*c^4*d^2 - 30*A*c^3*d^3 - 210*(5*D*c^2*d^4 - C*c*d^5)* 
x^4 - 2*(7386*D*c^3*d^3 - 2059*C*c^2*d^4 + 337*B*c*d^5 - 15*A*d^6)*x^3 - 2 
*(16629*D*c^4*d^2 - 4561*C*c^3*d^3 + 613*B*c^2*d^4 + 45*A*c*d^5)*x^2 - (28 
293*D*c^5*d - 7762*C*c^4*d^2 + 1126*B*c^3*d^3 - 90*A*c^2*d^4)*x)*sqrt(-d^2 
*x^2 + c^2))/(c*d^8*x^4 + 4*c^2*d^7*x^3 + 6*c^3*d^6*x^2 + 4*c^4*d^5*x + c^ 
5*d^4)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2371 vs. \(2 (274) = 548\).

Time = 0.18 (sec) , antiderivative size = 2371, normalized size of antiderivative = 7.93 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^7} \, dx=\text {Too large to display} \] Input:

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

Output:

(-d^2*x^2 + c^2)^(5/2)*D*c^3/(d^10*x^6 + 6*c*d^9*x^5 + 15*c^2*d^8*x^4 + 20 
*c^3*d^7*x^3 + 15*c^4*d^6*x^2 + 6*c^5*d^5*x + c^6*d^4) - 5/2*(-d^2*x^2 + c 
^2)^(3/2)*D*c^4/(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) + 15/7*sqrt(-d^2*x^2 + c^2)*D*c^5/(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)^(5/2)*C 
*c^2/(d^9*x^6 + 6*c*d^8*x^5 + 15*c^2*d^7*x^4 + 20*c^3*d^6*x^3 + 15*c^4*d^5 
*x^2 + 6*c^5*d^4*x + c^6*d^3) - 3/5*(-d^2*x^2 + c^2)^(5/2)*D*c^2/(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) + 
 5/2*(-d^2*x^2 + c^2)^(3/2)*C*c^3/(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)^(3/2)*D*c^3 
/(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) - 15/7*sq 
rt(-d^2*x^2 + c^2)*C*c^4/(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) + 237/70*sqrt(-d^2*x^2 + c^2)*D*c^4/(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)^(5/2)*B*c/(d^8*x^6 + 6*c*d^7*x 
^5 + 15*c^2*d^6*x^4 + 20*c^3*d^5*x^3 + 15*c^4*d^4*x^2 + 6*c^5*d^3*x + c^6* 
d^2) + 2/5*(-d^2*x^2 + c^2)^(5/2)*C*c/(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/(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) - 5/2*( 
-d^2*x^2 + c^2)^(3/2)*B*c^2/(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)^(3/2)*C*c^2/(d...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 857 vs. \(2 (274) = 548\).

Time = 0.17 (sec) , antiderivative size = 857, normalized size of antiderivative = 2.87 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^7} \, dx =\text {Too large to display} \] Input:

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

Output:

1/2*sqrt(-d^2*x^2 + c^2)*(D*x/d^3 - 2*(7*D*c*d^6 - C*d^7)/d^10) - 1/2*(51* 
D*c^2 - 14*C*c*d + 2*B*d^2)*arcsin(d*x/c)*sgn(c)*sgn(d)/(d^3*abs(d)) + 2/1 
05*(3471*D*c^3 - 1049*C*c^2*d + 167*B*c*d^2 + 15*A*d^3 + 1064*(c*d + sqrt( 
-d^2*x^2 + c^2)*abs(d))*B*c/x + 21672*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))* 
D*c^3/(d^2*x) - 6608*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))*C*c^2/(d*x) + 552 
51*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^2*D*c^3/(d^4*x^2) - 16989*(c*d + sq 
rt(-d^2*x^2 + c^2)*abs(d))^2*C*c^2/(d^3*x^2) + 2667*(c*d + sqrt(-d^2*x^2 + 
 c^2)*abs(d))^2*B*c/(d^2*x^2) + 315*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^2* 
A/(d*x^2) + 72240*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^3*D*c^3/(d^6*x^3) - 
22400*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^3*C*c^2/(d^5*x^3) + 3920*(c*d + 
sqrt(-d^2*x^2 + c^2)*abs(d))^3*B*c/(d^4*x^3) + 49245*(c*d + sqrt(-d^2*x^2 
+ c^2)*abs(d))^4*D*c^3/(d^8*x^4) - 14315*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d 
))^4*C*c^2/(d^7*x^4) + 1925*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^4*B*c/(d^6 
*x^4) + 525*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^4*A/(d^5*x^4) + 17640*(c*d 
 + sqrt(-d^2*x^2 + c^2)*abs(d))^5*D*c^3/(d^10*x^5) - 5040*(c*d + sqrt(-d^2 
*x^2 + c^2)*abs(d))^5*C*c^2/(d^9*x^5) + 840*(c*d + sqrt(-d^2*x^2 + c^2)*ab 
s(d))^5*B*c/(d^8*x^5) + 2625*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^6*D*c^3/( 
d^12*x^6) - 735*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^6*C*c^2/(d^11*x^6) + 1 
05*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^6*B*c/(d^10*x^6) + 105*(c*d + sqrt( 
-d^2*x^2 + c^2)*abs(d))^6*A/(d^9*x^6))/(c*d^3*((c*d + sqrt(-d^2*x^2 + c...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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