\(\int \sqrt {c^2-d^2 x^2} (A+B x+C x^2+D x^3) \, dx\) [131]

Optimal result

Integrand size = 32, antiderivative size = 159 \[ \int \sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{8} \left (4 A+\frac {c^2 C}{d^2}\right ) x \sqrt {c^2-d^2 x^2}-\frac {\left (B d^2+c^2 D\right ) \left (c^2-d^2 x^2\right )^{3/2}}{3 d^4}-\frac {C x \left (c^2-d^2 x^2\right )^{3/2}}{4 d^2}+\frac {D \left (c^2-d^2 x^2\right )^{5/2}}{5 d^4}+\frac {c^2 \left (c^2 C+4 A d^2\right ) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{8 d^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.98 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.88 \[ \int \sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {\sqrt {c^2-d^2 x^2} \left (-16 c^4 D-c^2 d^2 x (15 C+8 D x)+40 B d^2 \left (-c^2+d^2 x^2\right )+6 d^4 x \left (10 A+5 C x^2+4 D x^3\right )\right )-30 c^2 d \left (c^2 C+4 A d^2\right ) \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{120 d^4} \] Input:

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

Output:

(Sqrt[c^2 - d^2*x^2]*(-16*c^4*D - c^2*d^2*x*(15*C + 8*D*x) + 40*B*d^2*(-c^ 
2 + d^2*x^2) + 6*d^4*x*(10*A + 5*C*x^2 + 4*D*x^3)) - 30*c^2*d*(c^2*C + 4*A 
*d^2)*ArcTan[(d*x)/(Sqrt[c^2] - Sqrt[c^2 - d^2*x^2])])/(120*d^4)
 

Rubi [A] (verified)

Time = 0.64 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.02, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.281, Rules used = {2346, 25, 2346, 25, 27, 455, 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.

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

\(\Big \downarrow \) 2346

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2346

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {1}{4} \left (5 \left (4 A d^2+c^2 C\right ) \left (\frac {c^2 \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{2 d}+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )-\frac {4}{3} \left (c^2-d^2 x^2\right )^{3/2} \left (5 B+\frac {2 c^2 D}{d^2}\right )\right )-\frac {5}{4} C x \left (c^2-d^2 x^2\right )^{3/2}}{5 d^2}-\frac {D x^2 \left (c^2-d^2 x^2\right )^{3/2}}{5 d^2}\)

Input:

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

Output:

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

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 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 2346

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], 
e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*( 
q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1))   Int[(a + b*x^2)^p*ExpandToS 
um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], 
x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] &&  !LeQ[p, -1]
 

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.27

method	result	size
default	\(A \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )-\frac {B \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{3 d^{2}}+C \left (-\frac {x \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{4 d^{2}}+\frac {c^{2} \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )}{4 d^{2}}\right )+D \left (-\frac {x^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{5 d^{2}}-\frac {2 c^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{15 d^{4}}\right )\)	\(202\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.87 \[ \int \sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=-\frac {30 \, {\left (C c^{4} d + 4 \, A c^{2} d^{3}\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) - {\left (24 \, D d^{4} x^{4} + 30 \, C d^{4} x^{3} - 16 \, D c^{4} - 40 \, B c^{2} d^{2} - 8 \, {\left (D c^{2} d^{2} - 5 \, B d^{4}\right )} x^{2} - 15 \, {\left (C c^{2} d^{2} - 4 \, A d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{120 \, d^{4}} \] Input:

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

Output:

-1/120*(30*(C*c^4*d + 4*A*c^2*d^3)*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x 
)) - (24*D*d^4*x^4 + 30*C*d^4*x^3 - 16*D*c^4 - 40*B*c^2*d^2 - 8*(D*c^2*d^2 
 - 5*B*d^4)*x^2 - 15*(C*c^2*d^2 - 4*A*d^4)*x)*sqrt(-d^2*x^2 + c^2))/d^4
 

Sympy [A] (verification not implemented)

Time = 0.44 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.33 \[ \int \sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\begin {cases} \sqrt {c^{2} - d^{2} x^{2}} \left (\frac {C x^{3}}{4} + \frac {D x^{4}}{5} - \frac {x^{2} \left (- B d^{2} + \frac {D c^{2}}{5}\right )}{3 d^{2}} - \frac {x \left (- A d^{2} + \frac {C c^{2}}{4}\right )}{2 d^{2}} - \frac {B c^{2} + \frac {2 c^{2} \left (- B d^{2} + \frac {D c^{2}}{5}\right )}{3 d^{2}}}{d^{2}}\right ) + \left (A c^{2} + \frac {c^{2} \left (- A d^{2} + \frac {C c^{2}}{4}\right )}{2 d^{2}}\right ) \left (\begin {cases} \frac {\log {\left (- 2 d^{2} x + 2 \sqrt {- d^{2}} \sqrt {c^{2} - d^{2} x^{2}} \right )}}{\sqrt {- d^{2}}} & \text {for}\: c^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- d^{2} x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: d^{2} \neq 0 \\\left (A x + \frac {B x^{2}}{2} + \frac {C x^{3}}{3} + \frac {D x^{4}}{4}\right ) \sqrt {c^{2}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((sqrt(c**2 - d**2*x**2)*(C*x**3/4 + D*x**4/5 - x**2*(-B*d**2 + D 
*c**2/5)/(3*d**2) - x*(-A*d**2 + C*c**2/4)/(2*d**2) - (B*c**2 + 2*c**2*(-B 
*d**2 + D*c**2/5)/(3*d**2))/d**2) + (A*c**2 + c**2*(-A*d**2 + C*c**2/4)/(2 
*d**2))*Piecewise((log(-2*d**2*x + 2*sqrt(-d**2)*sqrt(c**2 - d**2*x**2))/s 
qrt(-d**2), Ne(c**2, 0)), (x*log(x)/sqrt(-d**2*x**2), True)), Ne(d**2, 0)) 
, ((A*x + B*x**2/2 + C*x**3/3 + D*x**4/4)*sqrt(c**2), True))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.02 \[ \int \sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {C c^{4} \arcsin \left (\frac {d x}{c}\right )}{8 \, d^{3}} + \frac {A c^{2} \arcsin \left (\frac {d x}{c}\right )}{2 \, d} + \frac {1}{2} \, \sqrt {-d^{2} x^{2} + c^{2}} A x + \frac {\sqrt {-d^{2} x^{2} + c^{2}} C c^{2} x}{8 \, d^{2}} - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} D x^{2}}{5 \, d^{2}} - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} C x}{4 \, d^{2}} - \frac {2 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} D c^{2}}{15 \, d^{4}} - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} B}{3 \, d^{2}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.85 \[ \int \sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {1}{120} \, \sqrt {-d^{2} x^{2} + c^{2}} {\left ({\left (2 \, {\left (3 \, {\left (4 \, D x + 5 \, C\right )} x - \frac {4 \, {\left (D c^{2} d^{4} - 5 \, B d^{6}\right )}}{d^{6}}\right )} x - \frac {15 \, {\left (C c^{2} d^{4} - 4 \, A d^{6}\right )}}{d^{6}}\right )} x - \frac {8 \, {\left (2 \, D c^{4} d^{2} + 5 \, B c^{2} d^{4}\right )}}{d^{6}}\right )} + \frac {{\left (C c^{4} + 4 \, A c^{2} d^{2}\right )} \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{8 \, d^{2} {\left | d \right |}} \] Input:

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

Output:

1/120*sqrt(-d^2*x^2 + c^2)*((2*(3*(4*D*x + 5*C)*x - 4*(D*c^2*d^4 - 5*B*d^6 
)/d^6)*x - 15*(C*c^2*d^4 - 4*A*d^6)/d^6)*x - 8*(2*D*c^4*d^2 + 5*B*c^2*d^4) 
/d^6) + 1/8*(C*c^4 + 4*A*c^2*d^2)*arcsin(d*x/c)*sgn(c)*sgn(d)/(d^2*abs(d))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.34 \[ \int \sqrt {c^2-d^2 x^2} \left (A+B x+C x^2+D x^3\right ) \, dx=\frac {60 \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{2} d^{2}+15 \mathit {asin} \left (\frac {d x}{c}\right ) c^{5}+60 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{3} x -40 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d +40 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,d^{3} x^{2}-16 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{4}-15 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{3} d x -8 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{2} d^{2} x^{2}+30 \sqrt {-d^{2} x^{2}+c^{2}}\, c \,d^{3} x^{3}+24 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{4} x^{4}+40 b \,c^{3} d +16 c^{5}}{120 d^{3}} \] Input:

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

Output:

(60*asin((d*x)/c)*a*c**2*d**2 + 15*asin((d*x)/c)*c**5 + 60*sqrt(c**2 - d** 
2*x**2)*a*d**3*x - 40*sqrt(c**2 - d**2*x**2)*b*c**2*d + 40*sqrt(c**2 - d** 
2*x**2)*b*d**3*x**2 - 16*sqrt(c**2 - d**2*x**2)*c**4 - 15*sqrt(c**2 - d**2 
*x**2)*c**3*d*x - 8*sqrt(c**2 - d**2*x**2)*c**2*d**2*x**2 + 30*sqrt(c**2 - 
 d**2*x**2)*c*d**3*x**3 + 24*sqrt(c**2 - d**2*x**2)*d**4*x**4 + 40*b*c**3* 
d + 16*c**5)/(120*d**3)