\(\int \frac {(c^2-d^2 x^2)^{3/2} (A+B x+C x^2+D x^3)}{(c+d x)^{15/2}} \, dx\) [212]

Optimal result

Integrand size = 41, antiderivative size = 384 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{15/2}} \, dx=\frac {\left (43 c^2 C d-23 B c d^2+3 A d^3-63 c^3 D\right ) \sqrt {c^2-d^2 x^2}}{40 d^4 (c+d x)^{9/2}}-\frac {\left (683 c^2 C d-183 B c d^2+3 A d^3-1503 c^3 D\right ) \sqrt {c^2-d^2 x^2}}{480 c d^4 (c+d x)^{7/2}}+\frac {\left (469 c^2 C d-9 B c d^2-3 A d^3-2145 c^3 D\right ) \sqrt {c^2-d^2 x^2}}{768 c^2 d^4 (c+d x)^{5/2}}-\frac {\left (43 c^2 C d+9 B c d^2+3 A d^3-1439 c^3 D\right ) \sqrt {c^2-d^2 x^2}}{1024 c^3 d^4 (c+d x)^{3/2}}-\frac {\left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \left (c^2-d^2 x^2\right )^{3/2}}{5 d^4 (c+d x)^{13/2}}-\frac {\left (43 c^2 C d+9 B c d^2+3 A d^3+609 c^3 D\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{1024 \sqrt {2} c^{7/2} d^4} \] Output:

1/40*(3*A*d^3-23*B*c*d^2+43*C*c^2*d-63*D*c^3)*(-d^2*x^2+c^2)^(1/2)/d^4/(d* 
x+c)^(9/2)-1/480*(3*A*d^3-183*B*c*d^2+683*C*c^2*d-1503*D*c^3)*(-d^2*x^2+c^ 
2)^(1/2)/c/d^4/(d*x+c)^(7/2)+1/768*(-3*A*d^3-9*B*c*d^2+469*C*c^2*d-2145*D* 
c^3)*(-d^2*x^2+c^2)^(1/2)/c^2/d^4/(d*x+c)^(5/2)-1/1024*(3*A*d^3+9*B*c*d^2+ 
43*C*c^2*d-1439*D*c^3)*(-d^2*x^2+c^2)^(1/2)/c^3/d^4/(d*x+c)^(3/2)-1/5*(A*d 
^3-B*c*d^2+C*c^2*d-D*c^3)*(-d^2*x^2+c^2)^(3/2)/d^4/(d*x+c)^(13/2)-1/2048*( 
3*A*d^3+9*B*c*d^2+43*C*c^2*d+609*D*c^3)*arctanh(2^(1/2)*c^(1/2)*(d*x+c)^(1 
/2)/(-d^2*x^2+c^2)^(1/2))*2^(1/2)/c^(7/2)/d^4
 

Mathematica [A] (verified)

Time = 3.73 (sec) , antiderivative size = 266, normalized size of antiderivative = 0.69 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{15/2}} \, dx=\frac {\frac {2 \sqrt {c} \sqrt {c^2-d^2 x^2} \left (5661 c^7 D-45 A d^7 x^4-15 c d^6 x^3 (16 A+9 B x)+c^6 d (319 C+26568 D x)-3 c^2 d^5 x^2 (182 A+5 x (48 B+43 C x))+c^5 d^2 (-219 B+2 x (716 C+24453 D x))+c^3 d^4 x \left (3672 A+x \left (4506 B+6800 C x+21585 D x^2\right )\right )+c^4 d^3 \left (-2121 A+2 x \left (-636 B+1207 C x+21720 D x^2\right )\right )\right )}{(c+d x)^{11/2}}-15 \sqrt {2} \left (43 c^2 C d+9 B c d^2+3 A d^3+609 c^3 D\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )}{30720 c^{7/2} d^4} \] Input:

Integrate[((c^2 - d^2*x^2)^(3/2)*(A + B*x + C*x^2 + D*x^3))/(c + d*x)^(15/ 
2),x]
 

Output:

((2*Sqrt[c]*Sqrt[c^2 - d^2*x^2]*(5661*c^7*D - 45*A*d^7*x^4 - 15*c*d^6*x^3* 
(16*A + 9*B*x) + c^6*d*(319*C + 26568*D*x) - 3*c^2*d^5*x^2*(182*A + 5*x*(4 
8*B + 43*C*x)) + c^5*d^2*(-219*B + 2*x*(716*C + 24453*D*x)) + c^3*d^4*x*(3 
672*A + x*(4506*B + 6800*C*x + 21585*D*x^2)) + c^4*d^3*(-2121*A + 2*x*(-63 
6*B + 1207*C*x + 21720*D*x^2))))/(c + d*x)^(11/2) - 15*Sqrt[2]*(43*c^2*C*d 
 + 9*B*c*d^2 + 3*A*d^3 + 609*c^3*D)*ArcTanh[(Sqrt[2]*Sqrt[c]*Sqrt[c + d*x] 
)/Sqrt[c^2 - d^2*x^2]])/(30720*c^(7/2)*d^4)
 

Rubi [A] (verified)

Time = 1.33 (sec) , antiderivative size = 384, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.268, Rules used = {2170, 27, 2170, 27, 671, 465, 465, 470, 470, 471, 221}

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.

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

\(\Big \downarrow \) 2170

\(\displaystyle \frac {2 \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left ((C d+9 c D) x^2 d^4+\left (21 D c^2+B d^2\right ) x d^3+\left (11 D c^3+A d^3\right ) d^2\right )}{2 (c+d x)^{15/2}}dx}{d^5}+\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{d^4 (c+d x)^{11/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left ((C d+9 c D) x^2 d^4+\left (21 D c^2+B d^2\right ) x d^3+\left (11 D c^3+A d^3\right ) d^2\right )}{(c+d x)^{15/2}}dx}{d^5}+\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{d^4 (c+d x)^{11/2}}\)

\(\Big \downarrow \) 2170

\(\displaystyle \frac {\frac {2 \int \frac {d^6 \left (150 D c^3+13 C d c^2+3 A d^3+d \left (153 D c^2+10 C d c+3 B d^2\right ) x\right ) \left (c^2-d^2 x^2\right )^{3/2}}{2 (c+d x)^{15/2}}dx}{3 d^4}+\frac {2 d \left (c^2-d^2 x^2\right )^{5/2} (9 c D+C d)}{3 (c+d x)^{13/2}}}{d^5}+\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{d^4 (c+d x)^{11/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{3} d^2 \int \frac {\left (150 D c^3+13 C d c^2+3 A d^3+d \left (153 D c^2+10 C d c+3 B d^2\right ) x\right ) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{15/2}}dx+\frac {2 d \left (c^2-d^2 x^2\right )^{5/2} (9 c D+C d)}{3 (c+d x)^{13/2}}}{d^5}+\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{d^4 (c+d x)^{11/2}}\)

\(\Big \downarrow \) 671

\(\displaystyle \frac {\frac {1}{3} d^2 \left (\frac {\left (3 A d^3+9 B c d^2+609 c^3 D+43 c^2 C d\right ) \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{13/2}}dx}{4 c}-\frac {3 \left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{10 c d (c+d x)^{15/2}}\right )+\frac {2 d \left (c^2-d^2 x^2\right )^{5/2} (9 c D+C d)}{3 (c+d x)^{13/2}}}{d^5}+\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{d^4 (c+d x)^{11/2}}\)

\(\Big \downarrow \) 465

\(\displaystyle \frac {\frac {1}{3} d^2 \left (\frac {\left (3 A d^3+9 B c d^2+609 c^3 D+43 c^2 C d\right ) \left (-\frac {3}{8} \int \frac {\sqrt {c^2-d^2 x^2}}{(c+d x)^{9/2}}dx-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{4 d (c+d x)^{11/2}}\right )}{4 c}-\frac {3 \left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{10 c d (c+d x)^{15/2}}\right )+\frac {2 d \left (c^2-d^2 x^2\right )^{5/2} (9 c D+C d)}{3 (c+d x)^{13/2}}}{d^5}+\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{d^4 (c+d x)^{11/2}}\)

\(\Big \downarrow \) 465

\(\displaystyle \frac {\frac {1}{3} d^2 \left (\frac {\left (3 A d^3+9 B c d^2+609 c^3 D+43 c^2 C d\right ) \left (-\frac {3}{8} \left (-\frac {1}{6} \int \frac {1}{(c+d x)^{5/2} \sqrt {c^2-d^2 x^2}}dx-\frac {\sqrt {c^2-d^2 x^2}}{3 d (c+d x)^{7/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{4 d (c+d x)^{11/2}}\right )}{4 c}-\frac {3 \left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{10 c d (c+d x)^{15/2}}\right )+\frac {2 d \left (c^2-d^2 x^2\right )^{5/2} (9 c D+C d)}{3 (c+d x)^{13/2}}}{d^5}+\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{d^4 (c+d x)^{11/2}}\)

\(\Big \downarrow \) 470

\(\displaystyle \frac {\frac {1}{3} d^2 \left (\frac {\left (3 A d^3+9 B c d^2+609 c^3 D+43 c^2 C d\right ) \left (-\frac {3}{8} \left (\frac {1}{6} \left (\frac {\sqrt {c^2-d^2 x^2}}{4 c d (c+d x)^{5/2}}-\frac {3 \int \frac {1}{(c+d x)^{3/2} \sqrt {c^2-d^2 x^2}}dx}{8 c}\right )-\frac {\sqrt {c^2-d^2 x^2}}{3 d (c+d x)^{7/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{4 d (c+d x)^{11/2}}\right )}{4 c}-\frac {3 \left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{10 c d (c+d x)^{15/2}}\right )+\frac {2 d \left (c^2-d^2 x^2\right )^{5/2} (9 c D+C d)}{3 (c+d x)^{13/2}}}{d^5}+\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{d^4 (c+d x)^{11/2}}\)

\(\Big \downarrow \) 470

\(\displaystyle \frac {\frac {1}{3} d^2 \left (\frac {\left (3 A d^3+9 B c d^2+609 c^3 D+43 c^2 C d\right ) \left (-\frac {3}{8} \left (\frac {1}{6} \left (\frac {\sqrt {c^2-d^2 x^2}}{4 c d (c+d x)^{5/2}}-\frac {3 \left (\frac {\int \frac {1}{\sqrt {c+d x} \sqrt {c^2-d^2 x^2}}dx}{4 c}-\frac {\sqrt {c^2-d^2 x^2}}{2 c d (c+d x)^{3/2}}\right )}{8 c}\right )-\frac {\sqrt {c^2-d^2 x^2}}{3 d (c+d x)^{7/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{4 d (c+d x)^{11/2}}\right )}{4 c}-\frac {3 \left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{10 c d (c+d x)^{15/2}}\right )+\frac {2 d \left (c^2-d^2 x^2\right )^{5/2} (9 c D+C d)}{3 (c+d x)^{13/2}}}{d^5}+\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{d^4 (c+d x)^{11/2}}\)

\(\Big \downarrow \) 471

\(\displaystyle \frac {\frac {1}{3} d^2 \left (\frac {\left (3 A d^3+9 B c d^2+609 c^3 D+43 c^2 C d\right ) \left (-\frac {3}{8} \left (\frac {1}{6} \left (\frac {\sqrt {c^2-d^2 x^2}}{4 c d (c+d x)^{5/2}}-\frac {3 \left (\frac {d \int \frac {1}{\frac {d^2 \left (c^2-d^2 x^2\right )}{c+d x}-2 c d^2}d\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {c+d x}}}{2 c}-\frac {\sqrt {c^2-d^2 x^2}}{2 c d (c+d x)^{3/2}}\right )}{8 c}\right )-\frac {\sqrt {c^2-d^2 x^2}}{3 d (c+d x)^{7/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{4 d (c+d x)^{11/2}}\right )}{4 c}-\frac {3 \left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{10 c d (c+d x)^{15/2}}\right )+\frac {2 d \left (c^2-d^2 x^2\right )^{5/2} (9 c D+C d)}{3 (c+d x)^{13/2}}}{d^5}+\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{d^4 (c+d x)^{11/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {1}{3} d^2 \left (\frac {\left (-\frac {3}{8} \left (\frac {1}{6} \left (\frac {\sqrt {c^2-d^2 x^2}}{4 c d (c+d x)^{5/2}}-\frac {3 \left (-\frac {\text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {c+d x}}\right )}{2 \sqrt {2} c^{3/2} d}-\frac {\sqrt {c^2-d^2 x^2}}{2 c d (c+d x)^{3/2}}\right )}{8 c}\right )-\frac {\sqrt {c^2-d^2 x^2}}{3 d (c+d x)^{7/2}}\right )-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{4 d (c+d x)^{11/2}}\right ) \left (3 A d^3+9 B c d^2+609 c^3 D+43 c^2 C d\right )}{4 c}-\frac {3 \left (c^2-d^2 x^2\right )^{5/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{10 c d (c+d x)^{15/2}}\right )+\frac {2 d \left (c^2-d^2 x^2\right )^{5/2} (9 c D+C d)}{3 (c+d x)^{13/2}}}{d^5}+\frac {2 D \left (c^2-d^2 x^2\right )^{5/2}}{d^4 (c+d x)^{11/2}}\)

Input:

Int[((c^2 - d^2*x^2)^(3/2)*(A + B*x + C*x^2 + D*x^3))/(c + d*x)^(15/2),x]
 

Output:

(2*D*(c^2 - d^2*x^2)^(5/2))/(d^4*(c + d*x)^(11/2)) + ((2*d*(C*d + 9*c*D)*( 
c^2 - d^2*x^2)^(5/2))/(3*(c + d*x)^(13/2)) + (d^2*((-3*(c^2*C*d - B*c*d^2 
+ A*d^3 - c^3*D)*(c^2 - d^2*x^2)^(5/2))/(10*c*d*(c + d*x)^(15/2)) + ((43*c 
^2*C*d + 9*B*c*d^2 + 3*A*d^3 + 609*c^3*D)*(-1/4*(c^2 - d^2*x^2)^(3/2)/(d*( 
c + d*x)^(11/2)) - (3*(-1/3*Sqrt[c^2 - d^2*x^2]/(d*(c + d*x)^(7/2)) + (Sqr 
t[c^2 - d^2*x^2]/(4*c*d*(c + d*x)^(5/2)) - (3*(-1/2*Sqrt[c^2 - d^2*x^2]/(c 
*d*(c + d*x)^(3/2)) - ArcTanh[Sqrt[c^2 - d^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[c 
+ d*x])]/(2*Sqrt[2]*c^(3/2)*d)))/(8*c))/6))/8))/(4*c)))/3)/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 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

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 470

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*c*(n + p + 1))), x] + Simp[(n + 
2*p + 2)/(2*c*(n + p + 1))   Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x], x] /; 
 FreeQ[{a, b, c, d, p}, x] && EqQ[b*c^2 + a*d^2, 0] && LtQ[n, 0] && NeQ[n + 
 p + 1, 0] && IntegerQ[2*p]
 

rule 471

Int[1/(Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Sim 
p[2*d   Subst[Int[1/(2*b*c + d^2*x^2), x], x, Sqrt[a + b*x^2]/Sqrt[c + d*x] 
], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0]
 

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. \(1145\) vs. \(2(340)=680\).

Time = 0.37 (sec) , antiderivative size = 1146, normalized size of antiderivative = 2.98

method	result	size
default	\(\text {Expression too large to display}\)	\(1146\)

Input:

int((-d^2*x^2+c^2)^(3/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(15/2),x,method=_RETU 
RNVERBOSE)
 

Output:

-1/30720*(-d^2*x^2+c^2)^(1/2)/c^(7/2)*(6450*C*2^(1/2)*arctanh(1/2*(-d*x+c) 
^(1/2)*2^(1/2)/c^(1/2))*c^4*d^4*x^3-97812*D*(-d*x+c)^(1/2)*c^(11/2)*d^2*x^ 
2+91350*D*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^5*d^3*x^3+ 
225*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^4*d^4*x+675*B* 
2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^5*d^3*x+3225*C*2^(1/ 
2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^6*d^2*x+225*A*2^(1/2)*arc 
tanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c*d^7*x^4-53136*D*(-d*x+c)^(1/2)* 
c^(13/2)*d*x+450*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^3 
*d^5*x^2+6450*C*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^5*d^ 
3*x^2+45*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^5*d^3+135 
*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^6*d^2+645*C*2^(1/ 
2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^7*d+1350*B*2^(1/2)*arctan 
h(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^4*d^4*x^2+45675*D*2^(1/2)*arctanh( 
1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^7*d*x+4242*A*(-d*x+c)^(1/2)*c^(9/2)* 
d^3+438*B*(-d*x+c)^(1/2)*c^(11/2)*d^2-638*C*(-d*x+c)^(1/2)*c^(13/2)*d+9135 
0*D*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^(1/2)/c^(1/2))*c^6*d^2*x^2-86880* 
D*(-d*x+c)^(1/2)*c^(9/2)*d^3*x^3+450*A*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)* 
2^(1/2)/c^(1/2))*c^2*d^6*x^3+1350*B*2^(1/2)*arctanh(1/2*(-d*x+c)^(1/2)*2^( 
1/2)/c^(1/2))*c^3*d^5*x^3+1290*C*c^(5/2)*d^5*x^4*(-d*x+c)^(1/2)-43170*D*c^ 
(7/2)*d^4*x^4*(-d*x+c)^(1/2)+480*A*c^(3/2)*d^6*x^3*(-d*x+c)^(1/2)+1092*...
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 1259, normalized size of antiderivative = 3.28 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{15/2}} \, dx=\text {Too large to display} \] Input:

integrate((-d^2*x^2+c^2)^(3/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(15/2),x, algor 
ithm="fricas")
 

Output:

[1/61440*(15*sqrt(2)*(609*D*c^9 + 43*C*c^8*d + 9*B*c^7*d^2 + 3*A*c^6*d^3 + 
 (609*D*c^3*d^6 + 43*C*c^2*d^7 + 9*B*c*d^8 + 3*A*d^9)*x^6 + 6*(609*D*c^4*d 
^5 + 43*C*c^3*d^6 + 9*B*c^2*d^7 + 3*A*c*d^8)*x^5 + 15*(609*D*c^5*d^4 + 43* 
C*c^4*d^5 + 9*B*c^3*d^6 + 3*A*c^2*d^7)*x^4 + 20*(609*D*c^6*d^3 + 43*C*c^5* 
d^4 + 9*B*c^4*d^5 + 3*A*c^3*d^6)*x^3 + 15*(609*D*c^7*d^2 + 43*C*c^6*d^3 + 
9*B*c^5*d^4 + 3*A*c^4*d^5)*x^2 + 6*(609*D*c^8*d + 43*C*c^7*d^2 + 9*B*c^6*d 
^3 + 3*A*c^5*d^4)*x)*sqrt(c)*log(-(d^2*x^2 - 2*c*d*x + 2*sqrt(2)*sqrt(-d^2 
*x^2 + c^2)*sqrt(d*x + c)*sqrt(c) - 3*c^2)/(d^2*x^2 + 2*c*d*x + c^2)) + 4* 
(5661*D*c^8 + 319*C*c^7*d - 219*B*c^6*d^2 - 2121*A*c^5*d^3 + 15*(1439*D*c^ 
4*d^4 - 43*C*c^3*d^5 - 9*B*c^2*d^6 - 3*A*c*d^7)*x^4 + 80*(543*D*c^5*d^3 + 
85*C*c^4*d^4 - 9*B*c^3*d^5 - 3*A*c^2*d^6)*x^3 + 2*(24453*D*c^6*d^2 + 1207* 
C*c^5*d^3 + 2253*B*c^4*d^4 - 273*A*c^3*d^5)*x^2 + 8*(3321*D*c^7*d + 179*C* 
c^6*d^2 - 159*B*c^5*d^3 + 459*A*c^4*d^4)*x)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x 
+ c))/(c^4*d^10*x^6 + 6*c^5*d^9*x^5 + 15*c^6*d^8*x^4 + 20*c^7*d^7*x^3 + 15 
*c^8*d^6*x^2 + 6*c^9*d^5*x + c^10*d^4), 1/30720*(15*sqrt(2)*(609*D*c^9 + 4 
3*C*c^8*d + 9*B*c^7*d^2 + 3*A*c^6*d^3 + (609*D*c^3*d^6 + 43*C*c^2*d^7 + 9* 
B*c*d^8 + 3*A*d^9)*x^6 + 6*(609*D*c^4*d^5 + 43*C*c^3*d^6 + 9*B*c^2*d^7 + 3 
*A*c*d^8)*x^5 + 15*(609*D*c^5*d^4 + 43*C*c^4*d^5 + 9*B*c^3*d^6 + 3*A*c^2*d 
^7)*x^4 + 20*(609*D*c^6*d^3 + 43*C*c^5*d^4 + 9*B*c^4*d^5 + 3*A*c^3*d^6)*x^ 
3 + 15*(609*D*c^7*d^2 + 43*C*c^6*d^3 + 9*B*c^5*d^4 + 3*A*c^4*d^5)*x^2 +...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{15/2}} \, dx=\text {Timed out} \] Input:

integrate((-d**2*x**2+c**2)**(3/2)*(D*x**3+C*x**2+B*x+A)/(d*x+c)**(15/2),x 
)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{15/2}} \, dx=\int { \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} {\left (D x^{3} + C x^{2} + B x + A\right )}}{{\left (d x + c\right )}^{\frac {15}{2}}} \,d x } \] Input:

integrate((-d^2*x^2+c^2)^(3/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(15/2),x, algor 
ithm="maxima")
 

Output:

integrate((-d^2*x^2 + c^2)^(3/2)*(D*x^3 + C*x^2 + B*x + A)/(d*x + c)^(15/2 
), x)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 494, normalized size of antiderivative = 1.29 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{15/2}} \, dx=\frac {\frac {15 \, \sqrt {2} {\left (609 \, D c^{3} + 43 \, C c^{2} d + 9 \, B c d^{2} + 3 \, A d^{3}\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-d x + c}}{2 \, \sqrt {-c}}\right )}{\sqrt {-c} c^{3}} + \frac {2 \, {\left (21585 \, {\left (d x - c\right )}^{4} \sqrt {-d x + c} D c^{3} + 129780 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} D c^{4} + 308736 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} D c^{5} - 341040 \, {\left (-d x + c\right )}^{\frac {3}{2}} D c^{6} + 146160 \, \sqrt {-d x + c} D c^{7} - 645 \, {\left (d x - c\right )}^{4} \sqrt {-d x + c} C c^{2} d + 4220 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} C c^{3} d + 18944 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} C c^{4} d - 24080 \, {\left (-d x + c\right )}^{\frac {3}{2}} C c^{5} d + 10320 \, \sqrt {-d x + c} C c^{6} d - 135 \, {\left (d x - c\right )}^{4} \sqrt {-d x + c} B c d^{2} - 1260 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} B c^{2} d^{2} + 1536 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} B c^{3} d^{2} - 5040 \, {\left (-d x + c\right )}^{\frac {3}{2}} B c^{4} d^{2} + 2160 \, \sqrt {-d x + c} B c^{5} d^{2} - 45 \, {\left (d x - c\right )}^{4} \sqrt {-d x + c} A d^{3} - 420 \, {\left (d x - c\right )}^{3} \sqrt {-d x + c} A c d^{3} - 1536 \, {\left (d x - c\right )}^{2} \sqrt {-d x + c} A c^{2} d^{3} - 1680 \, {\left (-d x + c\right )}^{\frac {3}{2}} A c^{3} d^{3} + 720 \, \sqrt {-d x + c} A c^{4} d^{3}\right )}}{{\left (d x + c\right )}^{5} c^{3}}}{30720 \, d^{4}} \] Input:

integrate((-d^2*x^2+c^2)^(3/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(15/2),x, algor 
ithm="giac")
 

Output:

1/30720*(15*sqrt(2)*(609*D*c^3 + 43*C*c^2*d + 9*B*c*d^2 + 3*A*d^3)*arctan( 
1/2*sqrt(2)*sqrt(-d*x + c)/sqrt(-c))/(sqrt(-c)*c^3) + 2*(21585*(d*x - c)^4 
*sqrt(-d*x + c)*D*c^3 + 129780*(d*x - c)^3*sqrt(-d*x + c)*D*c^4 + 308736*( 
d*x - c)^2*sqrt(-d*x + c)*D*c^5 - 341040*(-d*x + c)^(3/2)*D*c^6 + 146160*s 
qrt(-d*x + c)*D*c^7 - 645*(d*x - c)^4*sqrt(-d*x + c)*C*c^2*d + 4220*(d*x - 
 c)^3*sqrt(-d*x + c)*C*c^3*d + 18944*(d*x - c)^2*sqrt(-d*x + c)*C*c^4*d - 
24080*(-d*x + c)^(3/2)*C*c^5*d + 10320*sqrt(-d*x + c)*C*c^6*d - 135*(d*x - 
 c)^4*sqrt(-d*x + c)*B*c*d^2 - 1260*(d*x - c)^3*sqrt(-d*x + c)*B*c^2*d^2 + 
 1536*(d*x - c)^2*sqrt(-d*x + c)*B*c^3*d^2 - 5040*(-d*x + c)^(3/2)*B*c^4*d 
^2 + 2160*sqrt(-d*x + c)*B*c^5*d^2 - 45*(d*x - c)^4*sqrt(-d*x + c)*A*d^3 - 
 420*(d*x - c)^3*sqrt(-d*x + c)*A*c*d^3 - 1536*(d*x - c)^2*sqrt(-d*x + c)* 
A*c^2*d^3 - 1680*(-d*x + c)^(3/2)*A*c^3*d^3 + 720*sqrt(-d*x + c)*A*c^4*d^3 
)/((d*x + c)^5*c^3))/d^4
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{15/2}} \, dx=\int \frac {{\left (c^2-d^2\,x^2\right )}^{3/2}\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (c+d\,x\right )}^{15/2}} \,d x \] Input:

int(((c^2 - d^2*x^2)^(3/2)*(A + B*x + C*x^2 + x^3*D))/(c + d*x)^(15/2),x)
 

Output:

int(((c^2 - d^2*x^2)^(3/2)*(A + B*x + C*x^2 + x^3*D))/(c + d*x)^(15/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 929, normalized size of antiderivative = 2.42 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^{15/2}} \, dx =\text {Too large to display} \] Input:

int((-d^2*x^2+c^2)^(3/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^(15/2),x)
 

Output:

( - 4242*sqrt(c - d*x)*a*c**5*d**2 + 7344*sqrt(c - d*x)*a*c**4*d**3*x - 10 
92*sqrt(c - d*x)*a*c**3*d**4*x**2 - 480*sqrt(c - d*x)*a*c**2*d**5*x**3 - 9 
0*sqrt(c - d*x)*a*c*d**6*x**4 - 438*sqrt(c - d*x)*b*c**6*d - 2544*sqrt(c - 
 d*x)*b*c**5*d**2*x + 9012*sqrt(c - d*x)*b*c**4*d**3*x**2 - 1440*sqrt(c - 
d*x)*b*c**3*d**4*x**3 - 270*sqrt(c - d*x)*b*c**2*d**5*x**4 + 11960*sqrt(c 
- d*x)*c**8 + 56000*sqrt(c - d*x)*c**7*d*x + 102640*sqrt(c - d*x)*c**6*d** 
2*x**2 + 100480*sqrt(c - d*x)*c**5*d**3*x**3 + 41880*sqrt(c - d*x)*c**4*d* 
*4*x**4 + 45*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2))) 
/2))*a*c**5*d**2 + 225*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c) 
*sqrt(2)))/2))*a*c**4*d**3*x + 450*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d 
*x)/(sqrt(c)*sqrt(2)))/2))*a*c**3*d**4*x**2 + 450*sqrt(c)*sqrt(2)*log(tan( 
asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*c**2*d**5*x**3 + 225*sqrt(c)*s 
qrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*c*d**6*x**4 + 4 
5*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*a*d**7 
*x**5 + 135*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/ 
2))*b*c**6*d + 675*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqr 
t(2)))/2))*b*c**5*d**2*x + 1350*sqrt(c)*sqrt(2)*log(tan(asin(sqrt(c + d*x) 
/(sqrt(c)*sqrt(2)))/2))*b*c**4*d**3*x**2 + 1350*sqrt(c)*sqrt(2)*log(tan(as 
in(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*b*c**3*d**4*x**3 + 675*sqrt(c)*sqr 
t(2)*log(tan(asin(sqrt(c + d*x)/(sqrt(c)*sqrt(2)))/2))*b*c**2*d**5*x**4...