\(\int \frac {\sqrt {c^2-d^2 x^2} (A+B x+C x^2+D x^3)}{(c+d x)^5} \, dx\) [136]

Optimal result

Integrand size = 39, antiderivative size = 234 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^5} \, dx=-\frac {2 D \sqrt {c^2-d^2 x^2}}{d^4 (c+d x)}-\frac {\left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \left (c^2-d^2 x^2\right )^{3/2}}{7 c d^4 (c+d x)^5}+\frac {\left (12 c^2 C d-5 B c d^2-2 A d^3-19 c^3 D\right ) \left (c^2-d^2 x^2\right )^{3/2}}{35 c^2 d^4 (c+d x)^4}-\frac {\left (23 c^2 C d+5 B c d^2+2 A d^3-86 c^3 D\right ) \left (c^2-d^2 x^2\right )^{3/2}}{105 c^3 d^4 (c+d x)^3}-\frac {D \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{d^4} \] Output:

-2*D*(-d^2*x^2+c^2)^(1/2)/d^4/(d*x+c)-1/7*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)*(- 
d^2*x^2+c^2)^(3/2)/c/d^4/(d*x+c)^5+1/35*(-2*A*d^3-5*B*c*d^2+12*C*c^2*d-19* 
D*c^3)*(-d^2*x^2+c^2)^(3/2)/c^2/d^4/(d*x+c)^4-1/105*(2*A*d^3+5*B*c*d^2+23* 
C*c^2*d-86*D*c^3)*(-d^2*x^2+c^2)^(3/2)/c^3/d^4/(d*x+c)^3-D*arctan(d*x/(-d^ 
2*x^2+c^2)^(1/2))/d^4
 

Mathematica [A] (verified)

Time = 1.58 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.80 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^5} \, dx=\frac {-\frac {\sqrt {c^2-d^2 x^2} \left (166 c^6 D-2 A d^6 x^3-c d^5 x^2 (8 A+5 B x)+c^5 d (2 C+559 D x)-c^2 d^4 x (13 A+x (20 B+23 C x))+c^4 d^2 (5 B+x (8 C+659 D x))+c^3 d^3 \left (23 A+x \left (20 B+13 C x+296 D x^2\right )\right )\right )}{c^3 (c+d x)^4}+210 D \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{105 d^4} \] Input:

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

Output:

(-((Sqrt[c^2 - d^2*x^2]*(166*c^6*D - 2*A*d^6*x^3 - c*d^5*x^2*(8*A + 5*B*x) 
 + c^5*d*(2*C + 559*D*x) - c^2*d^4*x*(13*A + x*(20*B + 23*C*x)) + c^4*d^2* 
(5*B + x*(8*C + 659*D*x)) + c^3*d^3*(23*A + x*(20*B + 13*C*x + 296*D*x^2)) 
))/(c^3*(c + d*x)^4)) + 210*D*ArcTan[(d*x)/(Sqrt[c^2] - Sqrt[c^2 - d^2*x^2 
])])/(105*d^4)
 

Rubi [A] (verified)

Time = 0.98 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.59, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {2168, 2009}

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.

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

\(\Big \downarrow \) 2168

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 \left (c^2-d^2 x^2\right )^{3/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{105 c^3 d^4 (c+d x)^3}-\frac {2 \left (c^2-d^2 x^2\right )^{3/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{35 c^2 d^4 (c+d x)^4}-\frac {\left (c^2-d^2 x^2\right )^{3/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{7 c d^4 (c+d x)^5}-\frac {D \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{d^4}+\frac {\left (c^2-d^2 x^2\right )^{3/2} \left (-B d^2-3 c^2 D+2 c C d\right )}{15 c^2 d^4 (c+d x)^3}+\frac {\left (c^2-d^2 x^2\right )^{3/2} \left (-B d^2-3 c^2 D+2 c C d\right )}{5 c d^4 (c+d x)^4}-\frac {\left (c^2-d^2 x^2\right )^{3/2} (C d-3 c D)}{3 c d^4 (c+d x)^3}-\frac {2 D \sqrt {c^2-d^2 x^2}}{d^4 (c+d x)}\)

Input:

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

Output:

(-2*D*Sqrt[c^2 - d^2*x^2])/(d^4*(c + d*x)) - ((c^2*C*d - B*c*d^2 + A*d^3 - 
 c^3*D)*(c^2 - d^2*x^2)^(3/2))/(7*c*d^4*(c + d*x)^5) + ((2*c*C*d - B*d^2 - 
 3*c^2*D)*(c^2 - d^2*x^2)^(3/2))/(5*c*d^4*(c + d*x)^4) - (2*(c^2*C*d - B*c 
*d^2 + A*d^3 - c^3*D)*(c^2 - d^2*x^2)^(3/2))/(35*c^2*d^4*(c + d*x)^4) - (( 
C*d - 3*c*D)*(c^2 - d^2*x^2)^(3/2))/(3*c*d^4*(c + d*x)^3) + ((2*c*C*d - B* 
d^2 - 3*c^2*D)*(c^2 - d^2*x^2)^(3/2))/(15*c^2*d^4*(c + d*x)^3) - (2*(c^2*C 
*d - B*c*d^2 + A*d^3 - c^3*D)*(c^2 - d^2*x^2)^(3/2))/(105*c^3*d^4*(c + d*x 
)^3) - (D*ArcTan[(d*x)/Sqrt[c^2 - d^2*x^2]])/d^4
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2168

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
 Int[ExpandIntegrand[(a + b*x^2)^p, (d + e*x)^m*Pq, x], x] /; FreeQ[{a, b, 
d, e}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 0] && EqQ[m + Expon[Pq, x] 
+ 2*p + 1, 0] && ILtQ[m, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(462\) vs. \(2(218)=436\).

Time = 0.52 (sec) , antiderivative size = 463, normalized size of antiderivative = 1.98

method	result	size
default	\(\frac {D \left (-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{c d \left (x +\frac {c}{d}\right )^{2}}-\frac {d \left (\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}+\frac {c d \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{\sqrt {d^{2}}}\right )}{c}\right )}{d^{5}}-\frac {\left (C d -3 D c \right ) \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{3 d^{7} c \left (x +\frac {c}{d}\right )^{3}}+\frac {\left (B \,d^{2}-2 C c d +3 D c^{2}\right ) \left (-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{5 c d \left (x +\frac {c}{d}\right )^{4}}-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{15 c^{2} \left (x +\frac {c}{d}\right )^{3}}\right )}{d^{7}}+\frac {\left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right ) \left (-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{7 c d \left (x +\frac {c}{d}\right )^{5}}+\frac {2 d \left (-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{5 c d \left (x +\frac {c}{d}\right )^{4}}-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{15 c^{2} \left (x +\frac {c}{d}\right )^{3}}\right )}{7 c}\right )}{d^{8}}\)	\(463\)

Input:

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

Output:

D/d^5*(-1/c/d/(x+c/d)^2*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(3/2)-d/c*((-d^2*(x 
+c/d)^2+2*c*d*(x+c/d))^(1/2)+c*d/(d^2)^(1/2)*arctan((d^2)^(1/2)*x/(-d^2*(x 
+c/d)^2+2*c*d*(x+c/d))^(1/2))))-1/3*(C*d-3*D*c)/d^7/c/(x+c/d)^3*(-d^2*(x+c 
/d)^2+2*c*d*(x+c/d))^(3/2)+(B*d^2-2*C*c*d+3*D*c^2)/d^7*(-1/5/c/d/(x+c/d)^4 
*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(3/2)-1/15/c^2/(x+c/d)^3*(-d^2*(x+c/d)^2+2 
*c*d*(x+c/d))^(3/2))+(A*d^3-B*c*d^2+C*c^2*d-D*c^3)/d^8*(-1/7/c/d/(x+c/d)^5 
*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(3/2)+2/7*d/c*(-1/5/c/d/(x+c/d)^4*(-d^2*(x 
+c/d)^2+2*c*d*(x+c/d))^(3/2)-1/15/c^2/(x+c/d)^3*(-d^2*(x+c/d)^2+2*c*d*(x+c 
/d))^(3/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 478 vs. \(2 (218) = 436\).

Time = 0.13 (sec) , antiderivative size = 478, normalized size of antiderivative = 2.04 \[ \int \frac {\sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^5} \, dx=-\frac {166 \, D c^{7} + 2 \, C c^{6} d + 5 \, B c^{5} d^{2} + 23 \, A c^{4} d^{3} + {\left (166 \, D c^{3} d^{4} + 2 \, C c^{2} d^{5} + 5 \, B c d^{6} + 23 \, A d^{7}\right )} x^{4} + 4 \, {\left (166 \, D c^{4} d^{3} + 2 \, C c^{3} d^{4} + 5 \, B c^{2} d^{5} + 23 \, A c d^{6}\right )} x^{3} + 6 \, {\left (166 \, D c^{5} d^{2} + 2 \, C c^{4} d^{3} + 5 \, B c^{3} d^{4} + 23 \, A c^{2} d^{5}\right )} x^{2} + 4 \, {\left (166 \, D c^{6} d + 2 \, C c^{5} d^{2} + 5 \, B c^{4} d^{3} + 23 \, A c^{3} d^{4}\right )} x - 210 \, {\left (D c^{3} d^{4} x^{4} + 4 \, D c^{4} d^{3} x^{3} + 6 \, D c^{5} d^{2} x^{2} + 4 \, D c^{6} d x + D c^{7}\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + {\left (166 \, D c^{6} + 2 \, C c^{5} d + 5 \, B c^{4} d^{2} + 23 \, A c^{3} d^{3} + {\left (296 \, D c^{3} d^{3} - 23 \, C c^{2} d^{4} - 5 \, B c d^{5} - 2 \, A d^{6}\right )} x^{3} + {\left (659 \, D c^{4} d^{2} + 13 \, C c^{3} d^{3} - 20 \, B c^{2} d^{4} - 8 \, A c d^{5}\right )} x^{2} + {\left (559 \, D c^{5} d + 8 \, C c^{4} d^{2} + 20 \, B c^{3} d^{3} - 13 \, A c^{2} d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{105 \, {\left (c^{3} d^{8} x^{4} + 4 \, c^{4} d^{7} x^{3} + 6 \, c^{5} d^{6} x^{2} + 4 \, c^{6} d^{5} x + c^{7} d^{4}\right )}} \] Input:

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

Output:

-1/105*(166*D*c^7 + 2*C*c^6*d + 5*B*c^5*d^2 + 23*A*c^4*d^3 + (166*D*c^3*d^ 
4 + 2*C*c^2*d^5 + 5*B*c*d^6 + 23*A*d^7)*x^4 + 4*(166*D*c^4*d^3 + 2*C*c^3*d 
^4 + 5*B*c^2*d^5 + 23*A*c*d^6)*x^3 + 6*(166*D*c^5*d^2 + 2*C*c^4*d^3 + 5*B* 
c^3*d^4 + 23*A*c^2*d^5)*x^2 + 4*(166*D*c^6*d + 2*C*c^5*d^2 + 5*B*c^4*d^3 + 
 23*A*c^3*d^4)*x - 210*(D*c^3*d^4*x^4 + 4*D*c^4*d^3*x^3 + 6*D*c^5*d^2*x^2 
+ 4*D*c^6*d*x + D*c^7)*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x)) + (166*D* 
c^6 + 2*C*c^5*d + 5*B*c^4*d^2 + 23*A*c^3*d^3 + (296*D*c^3*d^3 - 23*C*c^2*d 
^4 - 5*B*c*d^5 - 2*A*d^6)*x^3 + (659*D*c^4*d^2 + 13*C*c^3*d^3 - 20*B*c^2*d 
^4 - 8*A*c*d^5)*x^2 + (559*D*c^5*d + 8*C*c^4*d^2 + 20*B*c^3*d^3 - 13*A*c^2 
*d^4)*x)*sqrt(-d^2*x^2 + c^2))/(c^3*d^8*x^4 + 4*c^4*d^7*x^3 + 6*c^5*d^6*x^ 
2 + 4*c^6*d^5*x + c^7*d^4)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

Output:

Exception raised: NotImplementedError >> unable to parse Giac output: abs( 
sageVARd)*(-(2*i*sageVARA*sageVARd^3+5*i*sageVARB*sageVARc*sageVARd^2+23*i 
*sageVARC*sageVARc^2*sageVARd+210*sageVARD*sageVARc^3*atan(i)+(-296*i)*sag 
eVARD*sageVARc^3)/1
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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