\(\int x^2 (c+d x)^2 \sqrt {c^2-d^2 x^2} \, dx\) [93]

Optimal result

Integrand size = 27, antiderivative size = 168 \[ \int x^2 (c+d x)^2 \sqrt {c^2-d^2 x^2} \, dx=\frac {3 c^4 x \sqrt {c^2-d^2 x^2}}{16 d^2}+\frac {c^2 x \left (c^2-d^2 x^2\right )^{3/2}}{8 d^2}-\frac {c (c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2}}{3 d^3}-\frac {c \left (c^2-d^2 x^2\right )^{5/2}}{10 d^3}+\frac {(c+d x) \left (c^2-d^2 x^2\right )^{5/2}}{6 d^3}+\frac {3 c^6 \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{16 d^3} \] Output:

3/16*c^4*x*(-d^2*x^2+c^2)^(1/2)/d^2+1/8*c^2*x*(-d^2*x^2+c^2)^(3/2)/d^2-1/3 
*c*(d*x+c)^2*(-d^2*x^2+c^2)^(3/2)/d^3-1/10*c*(-d^2*x^2+c^2)^(5/2)/d^3+1/6* 
(d*x+c)*(-d^2*x^2+c^2)^(5/2)/d^3+3/16*c^6*arctan(d*x/(-d^2*x^2+c^2)^(1/2)) 
/d^3
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.68 \[ \int x^2 (c+d x)^2 \sqrt {c^2-d^2 x^2} \, dx=\frac {\sqrt {c^2-d^2 x^2} \left (-64 c^5-45 c^4 d x-32 c^3 d^2 x^2+50 c^2 d^3 x^3+96 c d^4 x^4+40 d^5 x^5\right )-90 c^6 \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{240 d^3} \] Input:

Integrate[x^2*(c + d*x)^2*Sqrt[c^2 - d^2*x^2],x]
 

Output:

(Sqrt[c^2 - d^2*x^2]*(-64*c^5 - 45*c^4*d*x - 32*c^3*d^2*x^2 + 50*c^2*d^3*x 
^3 + 96*c*d^4*x^4 + 40*d^5*x^5) - 90*c^6*ArcTan[(d*x)/(Sqrt[c^2] - Sqrt[c^ 
2 - d^2*x^2])])/(240*d^3)
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.05, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {541, 27, 533, 27, 533, 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.

\(\displaystyle \int x^2 (c+d x)^2 \sqrt {c^2-d^2 x^2} \, dx\)

\(\Big \downarrow \) 541

\(\displaystyle -\frac {\int -3 c d^2 x^2 (3 c+4 d x) \sqrt {c^2-d^2 x^2}dx}{6 d^2}-\frac {1}{6} x^3 \left (c^2-d^2 x^2\right )^{3/2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} c \int x^2 (3 c+4 d x) \sqrt {c^2-d^2 x^2}dx-\frac {1}{6} x^3 \left (c^2-d^2 x^2\right )^{3/2}\)

\(\Big \downarrow \) 533

\(\displaystyle \frac {1}{2} c \left (\frac {\int c d x (8 c+15 d x) \sqrt {c^2-d^2 x^2}dx}{5 d^2}-\frac {4 x^2 \left (c^2-d^2 x^2\right )^{3/2}}{5 d}\right )-\frac {1}{6} x^3 \left (c^2-d^2 x^2\right )^{3/2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} c \left (\frac {c \int x (8 c+15 d x) \sqrt {c^2-d^2 x^2}dx}{5 d}-\frac {4 x^2 \left (c^2-d^2 x^2\right )^{3/2}}{5 d}\right )-\frac {1}{6} x^3 \left (c^2-d^2 x^2\right )^{3/2}\)

\(\Big \downarrow \) 533

\(\displaystyle \frac {1}{2} c \left (\frac {c \left (\frac {\int c d (15 c+32 d x) \sqrt {c^2-d^2 x^2}dx}{4 d^2}-\frac {15 x \left (c^2-d^2 x^2\right )^{3/2}}{4 d}\right )}{5 d}-\frac {4 x^2 \left (c^2-d^2 x^2\right )^{3/2}}{5 d}\right )-\frac {1}{6} x^3 \left (c^2-d^2 x^2\right )^{3/2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} c \left (\frac {c \left (\frac {c \int (15 c+32 d x) \sqrt {c^2-d^2 x^2}dx}{4 d}-\frac {15 x \left (c^2-d^2 x^2\right )^{3/2}}{4 d}\right )}{5 d}-\frac {4 x^2 \left (c^2-d^2 x^2\right )^{3/2}}{5 d}\right )-\frac {1}{6} x^3 \left (c^2-d^2 x^2\right )^{3/2}\)

\(\Big \downarrow \) 455

\(\displaystyle \frac {1}{2} c \left (\frac {c \left (\frac {c \left (15 c \int \sqrt {c^2-d^2 x^2}dx-\frac {32 \left (c^2-d^2 x^2\right )^{3/2}}{3 d}\right )}{4 d}-\frac {15 x \left (c^2-d^2 x^2\right )^{3/2}}{4 d}\right )}{5 d}-\frac {4 x^2 \left (c^2-d^2 x^2\right )^{3/2}}{5 d}\right )-\frac {1}{6} x^3 \left (c^2-d^2 x^2\right )^{3/2}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {1}{2} c \left (\frac {c \left (\frac {c \left (15 c \left (\frac {1}{2} c^2 \int \frac {1}{\sqrt {c^2-d^2 x^2}}dx+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )-\frac {32 \left (c^2-d^2 x^2\right )^{3/2}}{3 d}\right )}{4 d}-\frac {15 x \left (c^2-d^2 x^2\right )^{3/2}}{4 d}\right )}{5 d}-\frac {4 x^2 \left (c^2-d^2 x^2\right )^{3/2}}{5 d}\right )-\frac {1}{6} x^3 \left (c^2-d^2 x^2\right )^{3/2}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {1}{2} c \left (\frac {c \left (\frac {c \left (15 c \left (\frac {1}{2} c^2 \int \frac {1}{\frac {d^2 x^2}{c^2-d^2 x^2}+1}d\frac {x}{\sqrt {c^2-d^2 x^2}}+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )-\frac {32 \left (c^2-d^2 x^2\right )^{3/2}}{3 d}\right )}{4 d}-\frac {15 x \left (c^2-d^2 x^2\right )^{3/2}}{4 d}\right )}{5 d}-\frac {4 x^2 \left (c^2-d^2 x^2\right )^{3/2}}{5 d}\right )-\frac {1}{6} x^3 \left (c^2-d^2 x^2\right )^{3/2}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{2} c \left (\frac {c \left (\frac {c \left (15 c \left (\frac {c^2 \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{2 d}+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )-\frac {32 \left (c^2-d^2 x^2\right )^{3/2}}{3 d}\right )}{4 d}-\frac {15 x \left (c^2-d^2 x^2\right )^{3/2}}{4 d}\right )}{5 d}-\frac {4 x^2 \left (c^2-d^2 x^2\right )^{3/2}}{5 d}\right )-\frac {1}{6} x^3 \left (c^2-d^2 x^2\right )^{3/2}\)

Input:

Int[x^2*(c + d*x)^2*Sqrt[c^2 - d^2*x^2],x]
 

Output:

-1/6*(x^3*(c^2 - d^2*x^2)^(3/2)) + (c*((-4*x^2*(c^2 - d^2*x^2)^(3/2))/(5*d 
) + (c*((-15*x*(c^2 - d^2*x^2)^(3/2))/(4*d) + (c*((-32*(c^2 - d^2*x^2)^(3/ 
2))/(3*d) + 15*c*((x*Sqrt[c^2 - d^2*x^2])/2 + (c^2*ArcTan[(d*x)/Sqrt[c^2 - 
 d^2*x^2]])/(2*d))))/(4*d)))/(5*d)))/2
 

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 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 533

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
 Simp[d*x^m*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 2))), x] - Simp[1/(b*(m + 2* 
p + 2))   Int[x^(m - 1)*(a + b*x^2)^p*Simp[a*d*m - b*c*(m + 2*p + 2)*x, x], 
 x], x] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[m, 0] && GtQ[p, -1] && Integer 
Q[2*p]
 

rule 541

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> Simp[d^n*x^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*(m + n + 2*p + 1))), x 
] + Simp[1/(b*(m + n + 2*p + 1))   Int[x^m*(a + b*x^2)^p*ExpandToSum[b*(m + 
 n + 2*p + 1)*(c + d*x)^n - b*d^n*(m + n + 2*p + 1)*x^n - a*d^n*(m + n - 1) 
*x^(n - 2), x], x], x] /; FreeQ[{a, b, c, d, m, p}, x] && IGtQ[n, 1] && IGt 
Q[m, -2] && GtQ[p, -1] && IntegerQ[2*p]
 

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.64

method	result	size
risch	\(-\frac {\left (-40 x^{5} d^{5}-96 x^{4} c \,d^{4}-50 c^{2} d^{3} x^{3}+32 c^{3} d^{2} x^{2}+45 c^{4} d x +64 c^{5}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{240 d^{3}}+\frac {3 c^{6} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{16 d^{2} \sqrt {d^{2}}}\)	\(108\)
default	\(c^{2} \left (-\frac {x \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{4 d^{2}}+\frac {c^{2} \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )}{4 d^{2}}\right )+d^{2} \left (-\frac {x^{3} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{6 d^{2}}+\frac {c^{2} \left (-\frac {x \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{4 d^{2}}+\frac {c^{2} \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )}{4 d^{2}}\right )}{2 d^{2}}\right )+2 c d \left (-\frac {x^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{5 d^{2}}-\frac {2 c^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{15 d^{4}}\right )\)	\(248\)

Input:

int(x^2*(d*x+c)^2*(-d^2*x^2+c^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/240*(-40*d^5*x^5-96*c*d^4*x^4-50*c^2*d^3*x^3+32*c^3*d^2*x^2+45*c^4*d*x+ 
64*c^5)/d^3*(-d^2*x^2+c^2)^(1/2)+3/16*c^6/d^2/(d^2)^(1/2)*arctan((d^2)^(1/ 
2)*x/(-d^2*x^2+c^2)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.63 \[ \int x^2 (c+d x)^2 \sqrt {c^2-d^2 x^2} \, dx=-\frac {90 \, c^{6} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) - {\left (40 \, d^{5} x^{5} + 96 \, c d^{4} x^{4} + 50 \, c^{2} d^{3} x^{3} - 32 \, c^{3} d^{2} x^{2} - 45 \, c^{4} d x - 64 \, c^{5}\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{240 \, d^{3}} \] Input:

integrate(x^2*(d*x+c)^2*(-d^2*x^2+c^2)^(1/2),x, algorithm="fricas")
 

Output:

-1/240*(90*c^6*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x)) - (40*d^5*x^5 + 9 
6*c*d^4*x^4 + 50*c^2*d^3*x^3 - 32*c^3*d^2*x^2 - 45*c^4*d*x - 64*c^5)*sqrt( 
-d^2*x^2 + c^2))/d^3
 

Sympy [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.06 \[ \int x^2 (c+d x)^2 \sqrt {c^2-d^2 x^2} \, dx=\begin {cases} \frac {3 c^{6} \left (\begin {cases} \frac {\log {\left (- 2 d^{2} x + 2 \sqrt {- d^{2}} \sqrt {c^{2} - d^{2} x^{2}} \right )}}{\sqrt {- d^{2}}} & \text {for}\: c^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- d^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{16 d^{2}} + \sqrt {c^{2} - d^{2} x^{2}} \left (- \frac {4 c^{5}}{15 d^{3}} - \frac {3 c^{4} x}{16 d^{2}} - \frac {2 c^{3} x^{2}}{15 d} + \frac {5 c^{2} x^{3}}{24} + \frac {2 c d x^{4}}{5} + \frac {d^{2} x^{5}}{6}\right ) & \text {for}\: d^{2} \neq 0 \\\left (\frac {c^{2} x^{3}}{3} + \frac {c d x^{4}}{2} + \frac {d^{2} x^{5}}{5}\right ) \sqrt {c^{2}} & \text {otherwise} \end {cases} \] Input:

integrate(x**2*(d*x+c)**2*(-d**2*x**2+c**2)**(1/2),x)
 

Output:

Piecewise((3*c**6*Piecewise((log(-2*d**2*x + 2*sqrt(-d**2)*sqrt(c**2 - d** 
2*x**2))/sqrt(-d**2), Ne(c**2, 0)), (x*log(x)/sqrt(-d**2*x**2), True))/(16 
*d**2) + sqrt(c**2 - d**2*x**2)*(-4*c**5/(15*d**3) - 3*c**4*x/(16*d**2) - 
2*c**3*x**2/(15*d) + 5*c**2*x**3/24 + 2*c*d*x**4/5 + d**2*x**5/6), Ne(d**2 
, 0)), ((c**2*x**3/3 + c*d*x**4/2 + d**2*x**5/5)*sqrt(c**2), True))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.75 \[ \int x^2 (c+d x)^2 \sqrt {c^2-d^2 x^2} \, dx=-\frac {1}{6} \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} x^{3} + \frac {3 \, c^{6} \arcsin \left (\frac {d x}{c}\right )}{16 \, d^{3}} + \frac {3 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{4} x}{16 \, d^{2}} - \frac {2 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c x^{2}}{5 \, d} - \frac {3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{2} x}{8 \, d^{2}} - \frac {4 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{3}}{15 \, d^{3}} \] Input:

integrate(x^2*(d*x+c)^2*(-d^2*x^2+c^2)^(1/2),x, algorithm="maxima")
 

Output:

-1/6*(-d^2*x^2 + c^2)^(3/2)*x^3 + 3/16*c^6*arcsin(d*x/c)/d^3 + 3/16*sqrt(- 
d^2*x^2 + c^2)*c^4*x/d^2 - 2/5*(-d^2*x^2 + c^2)^(3/2)*c*x^2/d - 3/8*(-d^2* 
x^2 + c^2)^(3/2)*c^2*x/d^2 - 4/15*(-d^2*x^2 + c^2)^(3/2)*c^3/d^3
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.57 \[ \int x^2 (c+d x)^2 \sqrt {c^2-d^2 x^2} \, dx=\frac {3 \, c^{6} \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{16 \, d^{2} {\left | d \right |}} - \frac {1}{240} \, \sqrt {-d^{2} x^{2} + c^{2}} {\left (\frac {64 \, c^{5}}{d^{3}} + {\left (\frac {45 \, c^{4}}{d^{2}} + 2 \, {\left (\frac {16 \, c^{3}}{d} - {\left (25 \, c^{2} + 4 \, {\left (5 \, d^{2} x + 12 \, c d\right )} x\right )} x\right )} x\right )} x\right )} \] Input:

integrate(x^2*(d*x+c)^2*(-d^2*x^2+c^2)^(1/2),x, algorithm="giac")
 

Output:

3/16*c^6*arcsin(d*x/c)*sgn(c)*sgn(d)/(d^2*abs(d)) - 1/240*sqrt(-d^2*x^2 + 
c^2)*(64*c^5/d^3 + (45*c^4/d^2 + 2*(16*c^3/d - (25*c^2 + 4*(5*d^2*x + 12*c 
*d)*x)*x)*x)*x)
 

Mupad [F(-1)]

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

int(x^2*(c^2 - d^2*x^2)^(1/2)*(c + d*x)^2,x)
 

Output:

int(x^2*(c^2 - d^2*x^2)^(1/2)*(c + d*x)^2, x)
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.90 \[ \int x^2 (c+d x)^2 \sqrt {c^2-d^2 x^2} \, dx=\frac {45 \mathit {asin} \left (\frac {d x}{c}\right ) c^{6}-64 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{5}-45 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{4} d x -32 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{3} d^{2} x^{2}+50 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{2} d^{3} x^{3}+96 \sqrt {-d^{2} x^{2}+c^{2}}\, c \,d^{4} x^{4}+40 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{5} x^{5}+64 c^{6}}{240 d^{3}} \] Input:

int(x^2*(d*x+c)^2*(-d^2*x^2+c^2)^(1/2),x)
 

Output:

(45*asin((d*x)/c)*c**6 - 64*sqrt(c**2 - d**2*x**2)*c**5 - 45*sqrt(c**2 - d 
**2*x**2)*c**4*d*x - 32*sqrt(c**2 - d**2*x**2)*c**3*d**2*x**2 + 50*sqrt(c* 
*2 - d**2*x**2)*c**2*d**3*x**3 + 96*sqrt(c**2 - d**2*x**2)*c*d**4*x**4 + 4 
0*sqrt(c**2 - d**2*x**2)*d**5*x**5 + 64*c**6)/(240*d**3)