\(\int x^2 (c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2} \, dx\) [610]

Optimal result

Integrand size = 32, antiderivative size = 201 \[ \int x^2 (c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2} \, dx=-\frac {64 c^4 \left (b c^2-b d^2 x^2\right )^{3/2}}{385 b d^3 (c+d x)^{3/2}}-\frac {48 c^3 \left (b c^2-b d^2 x^2\right )^{3/2}}{385 b d^3 \sqrt {c+d x}}-\frac {6 c^2 \sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{3/2}}{77 b d^3}+\frac {4 c (c+d x)^{3/2} \left (b c^2-b d^2 x^2\right )^{3/2}}{33 b d^3}-\frac {2 (c+d x)^{5/2} \left (b c^2-b d^2 x^2\right )^{3/2}}{11 b d^3} \] Output:

-64/385*c^4*(-b*d^2*x^2+b*c^2)^(3/2)/b/d^3/(d*x+c)^(3/2)-48/385*c^3*(-b*d^ 
2*x^2+b*c^2)^(3/2)/b/d^3/(d*x+c)^(1/2)-6/77*c^2*(d*x+c)^(1/2)*(-b*d^2*x^2+ 
b*c^2)^(3/2)/b/d^3+4/33*c*(d*x+c)^(3/2)*(-b*d^2*x^2+b*c^2)^(3/2)/b/d^3-2/1 
1*(d*x+c)^(5/2)*(-b*d^2*x^2+b*c^2)^(3/2)/b/d^3
 

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.40 \[ \int x^2 (c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2} \, dx=-\frac {2 (c-d x) \sqrt {b \left (c^2-d^2 x^2\right )} \left (248 c^4+372 c^3 d x+465 c^2 d^2 x^2+350 c d^3 x^3+105 d^4 x^4\right )}{1155 d^3 \sqrt {c+d x}} \] Input:

Integrate[x^2*(c + d*x)^(3/2)*Sqrt[b*c^2 - b*d^2*x^2],x]
 

Output:

(-2*(c - d*x)*Sqrt[b*(c^2 - d^2*x^2)]*(248*c^4 + 372*c^3*d*x + 465*c^2*d^2 
*x^2 + 350*c*d^3*x^3 + 105*d^4*x^4))/(1155*d^3*Sqrt[c + d*x])
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {581, 27, 672, 459, 459, 458}

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.

\(\displaystyle \int x^2 (c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2} \, dx\)

\(\Big \downarrow \) 581

\(\displaystyle \frac {2 \int \frac {1}{2} c (5 c-6 d x) (c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2}dx}{11 d^2}-\frac {2 (c+d x)^{5/2} \left (b c^2-b d^2 x^2\right )^{3/2}}{11 b d^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {c \int (5 c-6 d x) (c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2}dx}{11 d^2}-\frac {2 (c+d x)^{5/2} \left (b c^2-b d^2 x^2\right )^{3/2}}{11 b d^3}\)

\(\Big \downarrow \) 672

\(\displaystyle \frac {c \left (3 c \int (c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2}dx+\frac {4 (c+d x)^{3/2} \left (b c^2-b d^2 x^2\right )^{3/2}}{3 b d}\right )}{11 d^2}-\frac {2 (c+d x)^{5/2} \left (b c^2-b d^2 x^2\right )^{3/2}}{11 b d^3}\)

\(\Big \downarrow \) 459

\(\displaystyle \frac {c \left (3 c \left (\frac {8}{7} c \int \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}dx-\frac {2 \sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{3/2}}{7 b d}\right )+\frac {4 (c+d x)^{3/2} \left (b c^2-b d^2 x^2\right )^{3/2}}{3 b d}\right )}{11 d^2}-\frac {2 (c+d x)^{5/2} \left (b c^2-b d^2 x^2\right )^{3/2}}{11 b d^3}\)

\(\Big \downarrow \) 459

\(\displaystyle \frac {c \left (3 c \left (\frac {8}{7} c \left (\frac {4}{5} c \int \frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {c+d x}}dx-\frac {2 \left (b c^2-b d^2 x^2\right )^{3/2}}{5 b d \sqrt {c+d x}}\right )-\frac {2 \sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{3/2}}{7 b d}\right )+\frac {4 (c+d x)^{3/2} \left (b c^2-b d^2 x^2\right )^{3/2}}{3 b d}\right )}{11 d^2}-\frac {2 (c+d x)^{5/2} \left (b c^2-b d^2 x^2\right )^{3/2}}{11 b d^3}\)

\(\Big \downarrow \) 458

\(\displaystyle \frac {c \left (\frac {4 (c+d x)^{3/2} \left (b c^2-b d^2 x^2\right )^{3/2}}{3 b d}+3 c \left (\frac {8}{7} c \left (-\frac {2 \left (b c^2-b d^2 x^2\right )^{3/2}}{5 b d \sqrt {c+d x}}-\frac {8 c \left (b c^2-b d^2 x^2\right )^{3/2}}{15 b d (c+d x)^{3/2}}\right )-\frac {2 \sqrt {c+d x} \left (b c^2-b d^2 x^2\right )^{3/2}}{7 b d}\right )\right )}{11 d^2}-\frac {2 (c+d x)^{5/2} \left (b c^2-b d^2 x^2\right )^{3/2}}{11 b d^3}\)

Input:

Int[x^2*(c + d*x)^(3/2)*Sqrt[b*c^2 - b*d^2*x^2],x]
 

Output:

(-2*(c + d*x)^(5/2)*(b*c^2 - b*d^2*x^2)^(3/2))/(11*b*d^3) + (c*((4*(c + d* 
x)^(3/2)*(b*c^2 - b*d^2*x^2)^(3/2))/(3*b*d) + 3*c*((-2*Sqrt[c + d*x]*(b*c^ 
2 - b*d^2*x^2)^(3/2))/(7*b*d) + (8*c*((-8*c*(b*c^2 - b*d^2*x^2)^(3/2))/(15 
*b*d*(c + d*x)^(3/2)) - (2*(b*c^2 - b*d^2*x^2)^(3/2))/(5*b*d*Sqrt[c + d*x] 
)))/7)))/(11*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 458

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

rule 459

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

rule 581

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

rule 672

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ 
), x_Symbol] :> Simp[g*(d + e*x)^m*((a + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), 
 x] + Simp[(m*(d*g + e*f) + 2*e*f*(p + 1))/(e*(m + 2*p + 2))   Int[(d + e*x 
)^m*(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] && NeQ[m + 2*p + 2, 0]
 

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.38

method	result	size
default	\(-\frac {2 \sqrt {b \left (-d^{2} x^{2}+c^{2}\right )}\, \left (-d x +c \right ) \left (105 d^{4} x^{4}+350 c \,d^{3} x^{3}+465 d^{2} c^{2} x^{2}+372 c^{3} d x +248 c^{4}\right )}{1155 \sqrt {d x +c}\, d^{3}}\)	\(76\)
gosper	\(-\frac {2 \left (-d x +c \right ) \left (105 d^{4} x^{4}+350 c \,d^{3} x^{3}+465 d^{2} c^{2} x^{2}+372 c^{3} d x +248 c^{4}\right ) \sqrt {-b \,x^{2} d^{2}+b \,c^{2}}}{1155 d^{3} \sqrt {d x +c}}\)	\(77\)
orering	\(-\frac {2 \left (-d x +c \right ) \left (105 d^{4} x^{4}+350 c \,d^{3} x^{3}+465 d^{2} c^{2} x^{2}+372 c^{3} d x +248 c^{4}\right ) \sqrt {-b \,x^{2} d^{2}+b \,c^{2}}}{1155 d^{3} \sqrt {d x +c}}\)	\(77\)
risch	\(-\frac {2 \sqrt {-\frac {b \left (d^{2} x^{2}-c^{2}\right )}{d x +c}}\, \sqrt {d x +c}\, b \left (-105 x^{5} d^{5}-245 x^{4} c \,d^{4}-115 c^{2} d^{3} x^{3}+93 c^{3} d^{2} x^{2}+124 c^{4} d x +248 c^{5}\right ) \left (-d x +c \right )}{1155 \sqrt {-b \left (d^{2} x^{2}-c^{2}\right )}\, d^{3} \sqrt {-b \left (d x -c \right )}}\)	\(127\)

Input:

int(x^2*(d*x+c)^(3/2)*(-b*d^2*x^2+b*c^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-2/1155/(d*x+c)^(1/2)*(b*(-d^2*x^2+c^2))^(1/2)*(-d*x+c)*(105*d^4*x^4+350*c 
*d^3*x^3+465*c^2*d^2*x^2+372*c^3*d*x+248*c^4)/d^3
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.45 \[ \int x^2 (c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2} \, dx=\frac {2 \, {\left (105 \, d^{5} x^{5} + 245 \, c d^{4} x^{4} + 115 \, c^{2} d^{3} x^{3} - 93 \, c^{3} d^{2} x^{2} - 124 \, c^{4} d x - 248 \, c^{5}\right )} \sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c}}{1155 \, {\left (d^{4} x + c d^{3}\right )}} \] Input:

integrate(x^2*(d*x+c)^(3/2)*(-b*d^2*x^2+b*c^2)^(1/2),x, algorithm="fricas" 
)
 

Output:

2/1155*(105*d^5*x^5 + 245*c*d^4*x^4 + 115*c^2*d^3*x^3 - 93*c^3*d^2*x^2 - 1 
24*c^4*d*x - 248*c^5)*sqrt(-b*d^2*x^2 + b*c^2)*sqrt(d*x + c)/(d^4*x + c*d^ 
3)
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int x^2 (c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2} \, dx=\int x^{2} \sqrt {- b \left (- c + d x\right ) \left (c + d x\right )} \left (c + d x\right )^{\frac {3}{2}}\, dx \] Input:

integrate(x**2*(d*x+c)**(3/2)*(-b*d**2*x**2+b*c**2)**(1/2),x)
 

Output:

Integral(x**2*sqrt(-b*(-c + d*x)*(c + d*x))*(c + d*x)**(3/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.49 \[ \int x^2 (c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2} \, dx=\frac {2 \, {\left (105 \, \sqrt {b} d^{5} x^{5} + 245 \, \sqrt {b} c d^{4} x^{4} + 115 \, \sqrt {b} c^{2} d^{3} x^{3} - 93 \, \sqrt {b} c^{3} d^{2} x^{2} - 124 \, \sqrt {b} c^{4} d x - 248 \, \sqrt {b} c^{5}\right )} {\left (d x + c\right )} \sqrt {-d x + c}}{1155 \, {\left (d^{4} x + c d^{3}\right )}} \] Input:

integrate(x^2*(d*x+c)^(3/2)*(-b*d^2*x^2+b*c^2)^(1/2),x, algorithm="maxima" 
)
 

Output:

2/1155*(105*sqrt(b)*d^5*x^5 + 245*sqrt(b)*c*d^4*x^4 + 115*sqrt(b)*c^2*d^3* 
x^3 - 93*sqrt(b)*c^3*d^2*x^2 - 124*sqrt(b)*c^4*d*x - 248*sqrt(b)*c^5)*(d*x 
 + c)*sqrt(-d*x + c)/(d^4*x + c*d^3)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 449 vs. \(2 (171) = 342\).

Time = 0.12 (sec) , antiderivative size = 449, normalized size of antiderivative = 2.23 \[ \int x^2 (c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2} \, dx=-\frac {2 \, {\left (\frac {231 \, {\left (15 \, \sqrt {-b d x + b c} b^{2} c^{2} - 10 \, {\left (-b d x + b c\right )}^{\frac {3}{2}} b c + 3 \, {\left (b d x - b c\right )}^{2} \sqrt {-b d x + b c}\right )} c^{3}}{b^{2}} + \frac {99 \, {\left (35 \, \sqrt {-b d x + b c} b^{3} c^{3} - 35 \, {\left (-b d x + b c\right )}^{\frac {3}{2}} b^{2} c^{2} + 21 \, {\left (b d x - b c\right )}^{2} \sqrt {-b d x + b c} b c + 5 \, {\left (b d x - b c\right )}^{3} \sqrt {-b d x + b c}\right )} c^{2}}{b^{3}} - \frac {11 \, {\left (315 \, \sqrt {-b d x + b c} b^{4} c^{4} - 420 \, {\left (-b d x + b c\right )}^{\frac {3}{2}} b^{3} c^{3} + 378 \, {\left (b d x - b c\right )}^{2} \sqrt {-b d x + b c} b^{2} c^{2} + 180 \, {\left (b d x - b c\right )}^{3} \sqrt {-b d x + b c} b c + 35 \, {\left (b d x - b c\right )}^{4} \sqrt {-b d x + b c}\right )} c}{b^{4}} - \frac {5 \, {\left (693 \, \sqrt {-b d x + b c} b^{5} c^{5} - 1155 \, {\left (-b d x + b c\right )}^{\frac {3}{2}} b^{4} c^{4} + 1386 \, {\left (b d x - b c\right )}^{2} \sqrt {-b d x + b c} b^{3} c^{3} + 990 \, {\left (b d x - b c\right )}^{3} \sqrt {-b d x + b c} b^{2} c^{2} + 385 \, {\left (b d x - b c\right )}^{4} \sqrt {-b d x + b c} b c + 63 \, {\left (b d x - b c\right )}^{5} \sqrt {-b d x + b c}\right )}}{b^{5}}\right )}}{3465 \, d^{3}} \] Input:

integrate(x^2*(d*x+c)^(3/2)*(-b*d^2*x^2+b*c^2)^(1/2),x, algorithm="giac")
 

Output:

-2/3465*(231*(15*sqrt(-b*d*x + b*c)*b^2*c^2 - 10*(-b*d*x + b*c)^(3/2)*b*c 
+ 3*(b*d*x - b*c)^2*sqrt(-b*d*x + b*c))*c^3/b^2 + 99*(35*sqrt(-b*d*x + b*c 
)*b^3*c^3 - 35*(-b*d*x + b*c)^(3/2)*b^2*c^2 + 21*(b*d*x - b*c)^2*sqrt(-b*d 
*x + b*c)*b*c + 5*(b*d*x - b*c)^3*sqrt(-b*d*x + b*c))*c^2/b^3 - 11*(315*sq 
rt(-b*d*x + b*c)*b^4*c^4 - 420*(-b*d*x + b*c)^(3/2)*b^3*c^3 + 378*(b*d*x - 
 b*c)^2*sqrt(-b*d*x + b*c)*b^2*c^2 + 180*(b*d*x - b*c)^3*sqrt(-b*d*x + b*c 
)*b*c + 35*(b*d*x - b*c)^4*sqrt(-b*d*x + b*c))*c/b^4 - 5*(693*sqrt(-b*d*x 
+ b*c)*b^5*c^5 - 1155*(-b*d*x + b*c)^(3/2)*b^4*c^4 + 1386*(b*d*x - b*c)^2* 
sqrt(-b*d*x + b*c)*b^3*c^3 + 990*(b*d*x - b*c)^3*sqrt(-b*d*x + b*c)*b^2*c^ 
2 + 385*(b*d*x - b*c)^4*sqrt(-b*d*x + b*c)*b*c + 63*(b*d*x - b*c)^5*sqrt(- 
b*d*x + b*c))/b^5)/d^3
 

Mupad [B] (verification not implemented)

Time = 7.09 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.61 \[ \int x^2 (c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2} \, dx=-\frac {\sqrt {b\,c^2-b\,d^2\,x^2}\,\left (\frac {496\,c^5\,\sqrt {c+d\,x}}{1155\,d^4}-\frac {14\,c\,x^4\,\sqrt {c+d\,x}}{33}-\frac {2\,d\,x^5\,\sqrt {c+d\,x}}{11}-\frac {46\,c^2\,x^3\,\sqrt {c+d\,x}}{231\,d}+\frac {62\,c^3\,x^2\,\sqrt {c+d\,x}}{385\,d^2}+\frac {248\,c^4\,x\,\sqrt {c+d\,x}}{1155\,d^3}\right )}{x+\frac {c}{d}} \] Input:

int(x^2*(b*c^2 - b*d^2*x^2)^(1/2)*(c + d*x)^(3/2),x)
 

Output:

-((b*c^2 - b*d^2*x^2)^(1/2)*((496*c^5*(c + d*x)^(1/2))/(1155*d^4) - (14*c* 
x^4*(c + d*x)^(1/2))/33 - (2*d*x^5*(c + d*x)^(1/2))/11 - (46*c^2*x^3*(c + 
d*x)^(1/2))/(231*d) + (62*c^3*x^2*(c + d*x)^(1/2))/(385*d^2) + (248*c^4*x* 
(c + d*x)^(1/2))/(1155*d^3)))/(x + c/d)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.33 \[ \int x^2 (c+d x)^{3/2} \sqrt {b c^2-b d^2 x^2} \, dx=\frac {2 \sqrt {b}\, \sqrt {-d x +c}\, \left (105 d^{5} x^{5}+245 c \,d^{4} x^{4}+115 c^{2} d^{3} x^{3}-93 c^{3} d^{2} x^{2}-124 c^{4} d x -248 c^{5}\right )}{1155 d^{3}} \] Input:

int(x^2*(d*x+c)^(3/2)*(-b*d^2*x^2+b*c^2)^(1/2),x)
 

Output:

(2*sqrt(b)*sqrt(c - d*x)*( - 248*c**5 - 124*c**4*d*x - 93*c**3*d**2*x**2 + 
 115*c**2*d**3*x**3 + 245*c*d**4*x**4 + 105*d**5*x**5))/(1155*d**3)