\(\int \frac {(c^2-d^2 x^2)^{5/2} (A+B x+C x^2+D x^3)}{(c+d x)^4} \, dx\) [160]

Optimal result

Integrand size = 39, antiderivative size = 310 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^4} \, dx=-\frac {2 c \left (9 c^2 C d-7 B c d^2+5 A d^3-11 c^3 D\right ) \sqrt {c^2-d^2 x^2}}{d^4}+\frac {\left (47 c^2 C d-32 B c d^2+20 A d^3-60 c^3 D\right ) x \sqrt {c^2-d^2 x^2}}{8 d^3}+\frac {(C d-4 c D) x^3 \sqrt {c^2-d^2 x^2}}{4 d}+\frac {\left (4 c C d-B d^2-9 c^2 D\right ) \left (c^2-d^2 x^2\right )^{3/2}}{3 d^4}+\frac {D \left (c^2-d^2 x^2\right )^{5/2}}{5 d^4}-\frac {2 \left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \left (c^2-d^2 x^2\right )^{5/2}}{d^4 (c+d x)^3}-\frac {c^2 \left (95 c^2 C d-80 B c d^2+60 A d^3-108 c^3 D\right ) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{8 d^4} \] Output:

-2*c*(5*A*d^3-7*B*c*d^2+9*C*c^2*d-11*D*c^3)*(-d^2*x^2+c^2)^(1/2)/d^4+1/8*( 
20*A*d^3-32*B*c*d^2+47*C*c^2*d-60*D*c^3)*x*(-d^2*x^2+c^2)^(1/2)/d^3+1/4*(C 
*d-4*D*c)*x^3*(-d^2*x^2+c^2)^(1/2)/d+1/3*(-B*d^2+4*C*c*d-9*D*c^2)*(-d^2*x^ 
2+c^2)^(3/2)/d^4+1/5*D*(-d^2*x^2+c^2)^(5/2)/d^4-2*(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)^3-1/8*c^2*(60*A*d^3-80*B*c*d^2+95*C 
*c^2*d-108*D*c^3)*arctan(d*x/(-d^2*x^2+c^2)^(1/2))/d^4
 

Mathematica [A] (verified)

Time = 1.47 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.72 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^4} \, dx=\frac {\frac {\sqrt {c^2-d^2 x^2} \left (2544 c^5 D-28 c^4 d (80 C-33 D x)+c^3 d^2 (1880 B-x (815 C+348 D x))+2 d^5 x^2 (30 A+x (20 B+3 x (5 C+4 D x)))-2 c d^4 x (210 A+x (100 B+x (65 C+48 D x)))+c^2 d^3 (-1440 A+x (680 B+x (305 C+192 D x)))\right )}{c+d x}-30 c^2 \left (-95 c^2 C d+80 B c d^2-60 A d^3+108 c^3 D\right ) \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{120 d^4} \] Input:

Integrate[((c^2 - d^2*x^2)^(5/2)*(A + B*x + C*x^2 + D*x^3))/(c + d*x)^4,x]
 

Output:

((Sqrt[c^2 - d^2*x^2]*(2544*c^5*D - 28*c^4*d*(80*C - 33*D*x) + c^3*d^2*(18 
80*B - x*(815*C + 348*D*x)) + 2*d^5*x^2*(30*A + x*(20*B + 3*x*(5*C + 4*D*x 
))) - 2*c*d^4*x*(210*A + x*(100*B + x*(65*C + 48*D*x))) + c^2*d^3*(-1440*A 
 + x*(680*B + x*(305*C + 192*D*x)))))/(c + d*x) - 30*c^2*(-95*c^2*C*d + 80 
*B*c*d^2 - 60*A*d^3 + 108*c^3*D)*ArcTan[(d*x)/(Sqrt[c^2] - Sqrt[c^2 - d^2* 
x^2])])/(120*d^4)
 

Rubi [A] (verified)

Time = 1.10 (sec) , antiderivative size = 281, normalized size of antiderivative = 0.91, 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, 466, 211, 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.

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

\(\Big \downarrow \) 2170

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2170

\(\displaystyle \frac {-\frac {\int \frac {d^6 \left (-28 D c^3+15 C d c^2-20 A d^3+d \left (-48 D c^2+35 C d c-20 B d^2\right ) x\right ) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^4}dx}{4 d^4}-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (5 C d-12 c D)}{4 (c+d x)^3}}{5 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{5 d^4 (c+d x)^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {1}{4} d^2 \int \frac {\left (-28 D c^3+15 C d c^2-20 A d^3+d \left (-48 D c^2+35 C d c-20 B d^2\right ) x\right ) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^4}dx-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (5 C d-12 c D)}{4 (c+d x)^3}}{5 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{5 d^4 (c+d x)^2}\)

\(\Big \downarrow \) 671

\(\displaystyle \frac {-\frac {1}{4} d^2 \left (\frac {\left (60 A d^3-80 B c d^2-108 c^3 D+95 c^2 C d\right ) \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^3}dx}{c}+\frac {20 \left (c^2-d^2 x^2\right )^{7/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^4}\right )-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (5 C d-12 c D)}{4 (c+d x)^3}}{5 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{5 d^4 (c+d x)^2}\)

\(\Big \downarrow \) 465

\(\displaystyle \frac {-\frac {1}{4} d^2 \left (\frac {\left (60 A d^3-80 B c d^2-108 c^3 D+95 c^2 C d\right ) \left (5 \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{c+d x}dx+\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{d (c+d x)^2}\right )}{c}+\frac {20 \left (c^2-d^2 x^2\right )^{7/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^4}\right )-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (5 C d-12 c D)}{4 (c+d x)^3}}{5 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{5 d^4 (c+d x)^2}\)

\(\Big \downarrow \) 466

\(\displaystyle \frac {-\frac {1}{4} d^2 \left (\frac {\left (60 A d^3-80 B c d^2-108 c^3 D+95 c^2 C d\right ) \left (5 \left (c \int \sqrt {c^2-d^2 x^2}dx+\frac {\left (c^2-d^2 x^2\right )^{3/2}}{3 d}\right )+\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{d (c+d x)^2}\right )}{c}+\frac {20 \left (c^2-d^2 x^2\right )^{7/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^4}\right )-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (5 C d-12 c D)}{4 (c+d x)^3}}{5 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{5 d^4 (c+d x)^2}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {-\frac {1}{4} d^2 \left (\frac {\left (60 A d^3-80 B c d^2-108 c^3 D+95 c^2 C d\right ) \left (5 \left (c \left (\frac {1}{2} c^2 \int \frac {1}{\sqrt {c^2-d^2 x^2}}dx+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )+\frac {\left (c^2-d^2 x^2\right )^{3/2}}{3 d}\right )+\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{d (c+d x)^2}\right )}{c}+\frac {20 \left (c^2-d^2 x^2\right )^{7/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^4}\right )-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (5 C d-12 c D)}{4 (c+d x)^3}}{5 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{5 d^4 (c+d x)^2}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {-\frac {1}{4} d^2 \left (\frac {\left (60 A d^3-80 B c d^2-108 c^3 D+95 c^2 C d\right ) \left (5 \left (c \left (\frac {1}{2} c^2 \int \frac {1}{\frac {d^2 x^2}{c^2-d^2 x^2}+1}d\frac {x}{\sqrt {c^2-d^2 x^2}}+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )+\frac {\left (c^2-d^2 x^2\right )^{3/2}}{3 d}\right )+\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{d (c+d x)^2}\right )}{c}+\frac {20 \left (c^2-d^2 x^2\right )^{7/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^4}\right )-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (5 C d-12 c D)}{4 (c+d x)^3}}{5 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{5 d^4 (c+d x)^2}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {-\frac {1}{4} d^2 \left (\frac {\left (5 \left (c \left (\frac {c^2 \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{2 d}+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )+\frac {\left (c^2-d^2 x^2\right )^{3/2}}{3 d}\right )+\frac {2 \left (c^2-d^2 x^2\right )^{5/2}}{d (c+d x)^2}\right ) \left (60 A d^3-80 B c d^2-108 c^3 D+95 c^2 C d\right )}{c}+\frac {20 \left (c^2-d^2 x^2\right )^{7/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^4}\right )-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (5 C d-12 c D)}{4 (c+d x)^3}}{5 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{5 d^4 (c+d x)^2}\)

Input:

Int[((c^2 - d^2*x^2)^(5/2)*(A + B*x + C*x^2 + D*x^3))/(c + d*x)^4,x]
 

Output:

-1/5*(D*(c^2 - d^2*x^2)^(7/2))/(d^4*(c + d*x)^2) + (-1/4*(d*(5*C*d - 12*c* 
D)*(c^2 - d^2*x^2)^(7/2))/(c + d*x)^3 - (d^2*((20*(c^2*C*d - B*c*d^2 + A*d 
^3 - c^3*D)*(c^2 - d^2*x^2)^(7/2))/(c*d*(c + d*x)^4) + ((95*c^2*C*d - 80*B 
*c*d^2 + 60*A*d^3 - 108*c^3*D)*((2*(c^2 - d^2*x^2)^(5/2))/(d*(c + d*x)^2) 
+ 5*((c^2 - d^2*x^2)^(3/2)/(3*d) + c*((x*Sqrt[c^2 - d^2*x^2])/2 + (c^2*Arc 
Tan[(d*x)/Sqrt[c^2 - d^2*x^2]])/(2*d)))))/c))/4)/(5*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 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

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 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 466

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 + 2*p + 1))), x] - Simp[2*b*c*(p/(d^ 
2*(n + 2*p + 1)))   Int[(c + d*x)^(n + 1)*(a + b*x^2)^(p - 1), x], x] /; Fr 
eeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0] && (LeQ[-2, n, 0 
] || EqQ[n + p + 1, 0]) && NeQ[n + 2*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. \(1126\) vs. \(2(286)=572\).

Time = 0.48 (sec) , antiderivative size = 1127, normalized size of antiderivative = 3.64

method	result	size
default	\(\text {Expression too large to display}\)	\(1127\)

Input:

int((-d^2*x^2+c^2)^(5/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^4,x,method=_RETURNVER 
BOSE)
 

Output:

D/d^4*(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)*ar 
ctan((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^ 
5*(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^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)))))))+(A*d^3-B*c*d^2+C*c^2*d-D*c^3)/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))^(...
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.15 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^4} \, dx=\frac {2544 \, D c^{6} - 2240 \, C c^{5} d + 1880 \, B c^{4} d^{2} - 1440 \, A c^{3} d^{3} + 8 \, {\left (318 \, D c^{5} d - 280 \, C c^{4} d^{2} + 235 \, B c^{3} d^{3} - 180 \, A c^{2} d^{4}\right )} x - 30 \, {\left (108 \, D c^{6} - 95 \, C c^{5} d + 80 \, B c^{4} d^{2} - 60 \, A c^{3} d^{3} + {\left (108 \, D c^{5} d - 95 \, C c^{4} d^{2} + 80 \, B c^{3} d^{3} - 60 \, A c^{2} d^{4}\right )} x\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + {\left (24 \, D d^{5} x^{5} + 2544 \, D c^{5} - 2240 \, C c^{4} d + 1880 \, B c^{3} d^{2} - 1440 \, A c^{2} d^{3} - 6 \, {\left (16 \, D c d^{4} - 5 \, C d^{5}\right )} x^{4} + 2 \, {\left (96 \, D c^{2} d^{3} - 65 \, C c d^{4} + 20 \, B d^{5}\right )} x^{3} - {\left (348 \, D c^{3} d^{2} - 305 \, C c^{2} d^{3} + 200 \, B c d^{4} - 60 \, A d^{5}\right )} x^{2} + {\left (924 \, D c^{4} d - 815 \, C c^{3} d^{2} + 680 \, B c^{2} d^{3} - 420 \, A c d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{120 \, {\left (d^{5} x + c d^{4}\right )}} \] Input:

integrate((-d^2*x^2+c^2)^(5/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^4,x, algorithm= 
"fricas")
 

Output:

1/120*(2544*D*c^6 - 2240*C*c^5*d + 1880*B*c^4*d^2 - 1440*A*c^3*d^3 + 8*(31 
8*D*c^5*d - 280*C*c^4*d^2 + 235*B*c^3*d^3 - 180*A*c^2*d^4)*x - 30*(108*D*c 
^6 - 95*C*c^5*d + 80*B*c^4*d^2 - 60*A*c^3*d^3 + (108*D*c^5*d - 95*C*c^4*d^ 
2 + 80*B*c^3*d^3 - 60*A*c^2*d^4)*x)*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d* 
x)) + (24*D*d^5*x^5 + 2544*D*c^5 - 2240*C*c^4*d + 1880*B*c^3*d^2 - 1440*A* 
c^2*d^3 - 6*(16*D*c*d^4 - 5*C*d^5)*x^4 + 2*(96*D*c^2*d^3 - 65*C*c*d^4 + 20 
*B*d^5)*x^3 - (348*D*c^3*d^2 - 305*C*c^2*d^3 + 200*B*c*d^4 - 60*A*d^5)*x^2 
 + (924*D*c^4*d - 815*C*c^3*d^2 + 680*B*c^2*d^3 - 420*A*c*d^4)*x)*sqrt(-d^ 
2*x^2 + c^2))/(d^5*x + c*d^4)
 

Sympy [F]

\[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^4} \, dx=\int \frac {\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {5}{2}} \left (A + B x + C x^{2} + D x^{3}\right )}{\left (c + d x\right )^{4}}\, dx \] Input:

integrate((-d**2*x**2+c**2)**(5/2)*(D*x**3+C*x**2+B*x+A)/(d*x+c)**4,x)
 

Output:

Integral((-(-c + d*x)*(c + d*x))**(5/2)*(A + B*x + C*x**2 + D*x**3)/(c + d 
*x)**4, x)
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 1173, normalized size of antiderivative = 3.78 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^4} \, dx=\text {Too large to display} \] Input:

integrate((-d^2*x^2+c^2)^(5/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^4,x, algorithm= 
"maxima")
 

Output:

-1/2*(-d^2*x^2 + c^2)^(5/2)*D*c^3/(d^7*x^3 + 3*c*d^6*x^2 + 3*c^2*d^5*x + c 
^3*d^4) - 5/2*(-d^2*x^2 + c^2)^(3/2)*D*c^4/(d^6*x^2 + 2*c*d^5*x + c^2*d^4) 
 + 15*sqrt(-d^2*x^2 + c^2)*D*c^5/(d^5*x + c*d^4) + 1/2*(-d^2*x^2 + c^2)^(5 
/2)*C*c^2/(d^6*x^3 + 3*c*d^5*x^2 + 3*c^2*d^4*x + c^3*d^3) + (-d^2*x^2 + c^ 
2)^(5/2)*D*c^2/(d^6*x^2 + 2*c*d^5*x + c^2*d^4) + 5/2*(-d^2*x^2 + c^2)^(3/2 
)*C*c^3/(d^5*x^2 + 2*c*d^4*x + c^2*d^3) + 5/2*(-d^2*x^2 + c^2)^(3/2)*D*c^3 
/(d^5*x + c*d^4) - 15*sqrt(-d^2*x^2 + c^2)*C*c^4/(d^4*x + c*d^3) - 1/2*(-d 
^2*x^2 + c^2)^(5/2)*B*c/(d^5*x^3 + 3*c*d^4*x^2 + 3*c^2*d^3*x + c^3*d^2) - 
2/3*(-d^2*x^2 + c^2)^(5/2)*C*c/(d^5*x^2 + 2*c*d^4*x + c^2*d^3) - 3/4*(-d^2 
*x^2 + c^2)^(5/2)*D*c/(d^5*x + c*d^4) - 5/2*(-d^2*x^2 + c^2)^(3/2)*B*c^2/( 
d^4*x^2 + 2*c*d^3*x + c^2*d^2) - 5/3*(-d^2*x^2 + c^2)^(3/2)*C*c^2/(d^4*x + 
 c*d^3) + 15*sqrt(-d^2*x^2 + c^2)*B*c^3/(d^3*x + c*d^2) + 3/2*I*D*c^5*arcs 
in(d*x/c + 2)/d^4 - 5/8*I*C*c^4*arcsin(d*x/c + 2)/d^3 + 15*D*c^5*arcsin(d* 
x/c)/d^4 - 25/2*C*c^4*arcsin(d*x/c)/d^3 + 10*B*c^3*arcsin(d*x/c)/d^2 - 15/ 
2*A*c^2*arcsin(d*x/c)/d + 1/2*(-d^2*x^2 + c^2)^(5/2)*A/(d^4*x^3 + 3*c*d^3* 
x^2 + 3*c^2*d^2*x + c^3*d) + 1/3*(-d^2*x^2 + c^2)^(5/2)*B/(d^4*x^2 + 2*c*d 
^3*x + c^2*d^2) + 1/4*(-d^2*x^2 + c^2)^(5/2)*C/(d^4*x + c*d^3) + 5/2*(-d^2 
*x^2 + c^2)^(3/2)*A*c/(d^3*x^2 + 2*c*d^2*x + c^2*d) + 5/6*(-d^2*x^2 + c^2) 
^(3/2)*B*c/(d^3*x + c*d^2) - 15*sqrt(-d^2*x^2 + c^2)*A*c^2/(d^2*x + c*d) - 
 3/2*sqrt(d^2*x^2 + 4*c*d*x + 3*c^2)*D*c^3*x/d^3 + 5/8*sqrt(d^2*x^2 + 4...
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 280, normalized size of antiderivative = 0.90 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^4} \, dx=\frac {1}{120} \, \sqrt {-d^{2} x^{2} + c^{2}} {\left ({\left (2 \, {\left (3 \, {\left (4 \, D x - \frac {5 \, {\left (4 \, D c d^{12} - C d^{13}\right )}}{d^{13}}\right )} x + \frac {4 \, {\left (39 \, D c^{2} d^{11} - 20 \, C c d^{12} + 5 \, B d^{13}\right )}}{d^{13}}\right )} x - \frac {15 \, {\left (44 \, D c^{3} d^{10} - 31 \, C c^{2} d^{11} + 16 \, B c d^{12} - 4 \, A d^{13}\right )}}{d^{13}}\right )} x + \frac {8 \, {\left (198 \, D c^{4} d^{9} - 160 \, C c^{3} d^{10} + 115 \, B c^{2} d^{11} - 60 \, A c d^{12}\right )}}{d^{13}}\right )} + \frac {{\left (108 \, D c^{5} - 95 \, C c^{4} d + 80 \, B c^{3} d^{2} - 60 \, A c^{2} d^{3}\right )} \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{8 \, d^{3} {\left | d \right |}} - \frac {16 \, {\left (D c^{5} - C c^{4} d + B c^{3} d^{2} - A c^{2} d^{3}\right )}}{d^{3} {\left (\frac {c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}}{d^{2} x} + 1\right )} {\left | d \right |}} \] Input:

integrate((-d^2*x^2+c^2)^(5/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^4,x, algorithm= 
"giac")
 

Output:

1/120*sqrt(-d^2*x^2 + c^2)*((2*(3*(4*D*x - 5*(4*D*c*d^12 - C*d^13)/d^13)*x 
 + 4*(39*D*c^2*d^11 - 20*C*c*d^12 + 5*B*d^13)/d^13)*x - 15*(44*D*c^3*d^10 
- 31*C*c^2*d^11 + 16*B*c*d^12 - 4*A*d^13)/d^13)*x + 8*(198*D*c^4*d^9 - 160 
*C*c^3*d^10 + 115*B*c^2*d^11 - 60*A*c*d^12)/d^13) + 1/8*(108*D*c^5 - 95*C* 
c^4*d + 80*B*c^3*d^2 - 60*A*c^2*d^3)*arcsin(d*x/c)*sgn(c)*sgn(d)/(d^3*abs( 
d)) - 16*(D*c^5 - C*c^4*d + B*c^3*d^2 - A*c^2*d^3)/(d^3*((c*d + sqrt(-d^2* 
x^2 + c^2)*abs(d))/(d^2*x) + 1)*abs(d))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^4} \, dx=\int \frac {{\left (c^2-d^2\,x^2\right )}^{5/2}\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (c+d\,x\right )}^4} \,d x \] Input:

int(((c^2 - d^2*x^2)^(5/2)*(A + B*x + C*x^2 + x^3*D))/(c + d*x)^4,x)
 

Output:

int(((c^2 - d^2*x^2)^(5/2)*(A + B*x + C*x^2 + x^3*D))/(c + d*x)^4, x)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 630, normalized size of antiderivative = 2.03 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^4} \, dx=\frac {-390 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{5}-2040 a \,c^{3} d^{2}+2400 b \,c^{4} d +420 \sqrt {-d^{2} x^{2}+c^{2}}\, a c \,d^{3} x -680 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d^{2} x +200 \sqrt {-d^{2} x^{2}+c^{2}}\, b c \,d^{3} x^{2}-900 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{2} d^{2}+1200 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{3} d +900 \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{2} d^{3} x -1200 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{3} d^{2} x -195 \mathit {asin} \left (\frac {d x}{c}\right ) c^{5} d x +420 a \,c^{2} d^{3} x +480 a c \,d^{4} x^{2}-680 b \,c^{3} d^{2} x -880 b \,c^{2} d^{3} x^{2}+240 b c \,d^{4} x^{3}+390 c^{6}-24 d^{6} x^{6}-195 \mathit {asin} \left (\frac {d x}{c}\right ) c^{6}-109 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{4} d x +43 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{3} d^{2} x^{2}-62 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{2} d^{3} x^{3}+66 \sqrt {-d^{2} x^{2}+c^{2}}\, c \,d^{4} x^{4}+900 \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{3} d^{2}-1200 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{4} d +2040 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{2} d^{2}-60 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{4} x^{2}-2400 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{3} d -40 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,d^{4} x^{3}-24 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{5} x^{5}+195 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) c^{5}-60 a \,d^{5} x^{3}-40 b \,d^{5} x^{4}-109 c^{5} d x -152 c^{4} d^{2} x^{2}+105 c^{3} d^{3} x^{3}-128 c^{2} d^{4} x^{4}+90 c \,d^{5} x^{5}}{120 d^{3} \left (\sqrt {-d^{2} x^{2}+c^{2}}-c -d x \right )} \] Input:

int((-d^2*x^2+c^2)^(5/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^4,x)
 

Output:

( - 900*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*a*c**2*d**2 + 1200*sqrt(c**2 
- d**2*x**2)*asin((d*x)/c)*b*c**3*d + 195*sqrt(c**2 - d**2*x**2)*asin((d*x 
)/c)*c**5 + 900*asin((d*x)/c)*a*c**3*d**2 + 900*asin((d*x)/c)*a*c**2*d**3* 
x - 1200*asin((d*x)/c)*b*c**4*d - 1200*asin((d*x)/c)*b*c**3*d**2*x - 195*a 
sin((d*x)/c)*c**6 - 195*asin((d*x)/c)*c**5*d*x + 2040*sqrt(c**2 - d**2*x** 
2)*a*c**2*d**2 + 420*sqrt(c**2 - d**2*x**2)*a*c*d**3*x - 60*sqrt(c**2 - d* 
*2*x**2)*a*d**4*x**2 - 2400*sqrt(c**2 - d**2*x**2)*b*c**3*d - 680*sqrt(c** 
2 - d**2*x**2)*b*c**2*d**2*x + 200*sqrt(c**2 - d**2*x**2)*b*c*d**3*x**2 - 
40*sqrt(c**2 - d**2*x**2)*b*d**4*x**3 - 390*sqrt(c**2 - d**2*x**2)*c**5 - 
109*sqrt(c**2 - d**2*x**2)*c**4*d*x + 43*sqrt(c**2 - d**2*x**2)*c**3*d**2* 
x**2 - 62*sqrt(c**2 - d**2*x**2)*c**2*d**3*x**3 + 66*sqrt(c**2 - d**2*x**2 
)*c*d**4*x**4 - 24*sqrt(c**2 - d**2*x**2)*d**5*x**5 - 2040*a*c**3*d**2 + 4 
20*a*c**2*d**3*x + 480*a*c*d**4*x**2 - 60*a*d**5*x**3 + 2400*b*c**4*d - 68 
0*b*c**3*d**2*x - 880*b*c**2*d**3*x**2 + 240*b*c*d**4*x**3 - 40*b*d**5*x** 
4 + 390*c**6 - 109*c**5*d*x - 152*c**4*d**2*x**2 + 105*c**3*d**3*x**3 - 12 
8*c**2*d**4*x**4 + 90*c*d**5*x**5 - 24*d**6*x**6)/(120*d**3*(sqrt(c**2 - d 
**2*x**2) - c - d*x))