\(\int \frac {x^2 (c+d x)^3}{(b c^2-b d^2 x^2)^{5/2}} \, dx\) [432]

Optimal result

Integrand size = 30, antiderivative size = 142 \[ \int \frac {x^2 (c+d x)^3}{\left (b c^2-b d^2 x^2\right )^{5/2}} \, dx=\frac {4 c^3 (c+d x)}{3 b d^3 \left (b c^2-b d^2 x^2\right )^{3/2}}-\frac {c (15 c+13 d x)}{3 b^2 d^3 \sqrt {b c^2-b d^2 x^2}}-\frac {\sqrt {b c^2-b d^2 x^2}}{b^3 d^3}+\frac {3 c \arctan \left (\frac {\sqrt {b} d x}{\sqrt {b c^2-b d^2 x^2}}\right )}{b^{5/2} d^3} \] Output:

4/3*c^3*(d*x+c)/b/d^3/(-b*d^2*x^2+b*c^2)^(3/2)-1/3*c*(13*d*x+15*c)/b^2/d^3 
/(-b*d^2*x^2+b*c^2)^(1/2)-(-b*d^2*x^2+b*c^2)^(1/2)/b^3/d^3+3*c*arctan(b^(1 
/2)*d*x/(-b*d^2*x^2+b*c^2)^(1/2))/b^(5/2)/d^3
 

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.87 \[ \int \frac {x^2 (c+d x)^3}{\left (b c^2-b d^2 x^2\right )^{5/2}} \, dx=\frac {14 c^3-5 c^2 d x-16 c d^2 x^2+3 d^3 x^3+18 c (c-d x) \sqrt {c^2-d^2 x^2} \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{3 b^2 d^3 (-c+d x) \sqrt {b \left (c^2-d^2 x^2\right )}} \] Input:

Integrate[(x^2*(c + d*x)^3)/(b*c^2 - b*d^2*x^2)^(5/2),x]
 

Output:

(14*c^3 - 5*c^2*d*x - 16*c*d^2*x^2 + 3*d^3*x^3 + 18*c*(c - d*x)*Sqrt[c^2 - 
 d^2*x^2]*ArcTan[(d*x)/(Sqrt[c^2] - Sqrt[c^2 - d^2*x^2])])/(3*b^2*d^3*(-c 
+ d*x)*Sqrt[b*(c^2 - d^2*x^2)])
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {529, 27, 462, 455, 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.

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

\(\Big \downarrow \) 529

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 462

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

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

Input:

Int[(x^2*(c + d*x)^3)/(b*c^2 - b*d^2*x^2)^(5/2),x]
 

Output:

(c*(c + d*x)^3)/(3*b*d^3*(b*c^2 - b*d^2*x^2)^(3/2)) - ((4*c*(c + d*x))/(b* 
d*Sqrt[b*c^2 - b*d^2*x^2]) - (-(Sqrt[b*c^2 - b*d^2*x^2]/(b*d)) + (3*c*ArcT 
an[(Sqrt[b]*d*x)/Sqrt[b*c^2 - b*d^2*x^2]])/(Sqrt[b]*d))/b)/(b*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 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 462

Int[((c_) + (d_.)*(x_))^(n_)/((a_) + (b_.)*(x_)^2)^(3/2), x_Symbol] :> Simp 
[(-2^(n - 1))*d*c^(n - 2)*((c + d*x)/(b*Sqrt[a + b*x^2])), x] + Simp[d^2/b 
  Int[(1/Sqrt[a + b*x^2])*ExpandToSum[(2^(n - 1)*c^(n - 1) - (c + d*x)^(n - 
 1))/(c - d*x), x], x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 
0] && IGtQ[n, 2]
 

rule 529

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{Qx = PolynomialQuotient[x^m, a*d + b*c*x, x], R = PolynomialRem 
ainder[x^m, a*d + b*c*x, x]}, Simp[(-c)*R*(c + d*x)^n*((a + b*x^2)^(p + 1)/ 
(2*a*d*(p + 1))), x] + Simp[c/(2*a*(p + 1))   Int[(c + d*x)^(n - 1)*(a + b* 
x^2)^(p + 1)*ExpandToSum[2*a*d*(p + 1)*Qx + R*(n + 2*p + 2), x], x], x]] /; 
 FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && IGtQ[m, 1] && LtQ[p, -1] && EqQ[b* 
c^2 + a*d^2, 0]
 

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.33

method	result	size
risch	\(-\frac {-d^{2} x^{2}+c^{2}}{d^{3} \sqrt {-b \left (d^{2} x^{2}-c^{2}\right )}\, b^{2}}+\frac {\frac {3 c \arctan \left (\frac {\sqrt {b \,d^{2}}\, x}{\sqrt {-b \,x^{2} d^{2}+b \,c^{2}}}\right )}{d^{2} \sqrt {b \,d^{2}}}+\frac {13 c \sqrt {-b \,d^{2} \left (x -\frac {c}{d}\right )^{2}-2 d b c \left (x -\frac {c}{d}\right )}}{3 d^{4} b \left (x -\frac {c}{d}\right )}+\frac {2 c^{2} \sqrt {-b \,d^{2} \left (x -\frac {c}{d}\right )^{2}-2 d b c \left (x -\frac {c}{d}\right )}}{3 d^{5} b \left (x -\frac {c}{d}\right )^{2}}}{b^{2}}\)	\(189\)
default	\(c^{3} \left (\frac {x}{2 b \,d^{2} \left (-b \,x^{2} d^{2}+b \,c^{2}\right )^{\frac {3}{2}}}-\frac {c^{2} \left (\frac {x}{3 b \,c^{2} \left (-b \,x^{2} d^{2}+b \,c^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 b^{2} c^{4} \sqrt {-b \,x^{2} d^{2}+b \,c^{2}}}\right )}{2 d^{2}}\right )+d^{3} \left (-\frac {x^{4}}{b \,d^{2} \left (-b \,x^{2} d^{2}+b \,c^{2}\right )^{\frac {3}{2}}}+\frac {4 c^{2} \left (\frac {x^{2}}{b \,d^{2} \left (-b \,x^{2} d^{2}+b \,c^{2}\right )^{\frac {3}{2}}}-\frac {2 c^{2}}{3 b \,d^{4} \left (-b \,x^{2} d^{2}+b \,c^{2}\right )^{\frac {3}{2}}}\right )}{d^{2}}\right )+3 c \,d^{2} \left (\frac {x^{3}}{3 b \,d^{2} \left (-b \,x^{2} d^{2}+b \,c^{2}\right )^{\frac {3}{2}}}-\frac {\frac {x}{b \,d^{2} \sqrt {-b \,x^{2} d^{2}+b \,c^{2}}}-\frac {\arctan \left (\frac {\sqrt {b \,d^{2}}\, x}{\sqrt {-b \,x^{2} d^{2}+b \,c^{2}}}\right )}{b \,d^{2} \sqrt {b \,d^{2}}}}{b \,d^{2}}\right )+3 c^{2} d \left (\frac {x^{2}}{b \,d^{2} \left (-b \,x^{2} d^{2}+b \,c^{2}\right )^{\frac {3}{2}}}-\frac {2 c^{2}}{3 b \,d^{4} \left (-b \,x^{2} d^{2}+b \,c^{2}\right )^{\frac {3}{2}}}\right )\)	\(364\)

Input:

int(x^2*(d*x+c)^3/(-b*d^2*x^2+b*c^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-(-d^2*x^2+c^2)/d^3/(-b*(d^2*x^2-c^2))^(1/2)/b^2+(3*c/d^2/(b*d^2)^(1/2)*ar 
ctan((b*d^2)^(1/2)*x/(-b*d^2*x^2+b*c^2)^(1/2))+13/3*c/d^4/b/(x-c/d)*(-b*d^ 
2*(x-c/d)^2-2*d*b*c*(x-c/d))^(1/2)+2/3*c^2/d^5/b/(x-c/d)^2*(-b*d^2*(x-c/d) 
^2-2*d*b*c*(x-c/d))^(1/2))/b^2
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.99 \[ \int \frac {x^2 (c+d x)^3}{\left (b c^2-b d^2 x^2\right )^{5/2}} \, dx=\left [-\frac {9 \, {\left (c d^{2} x^{2} - 2 \, c^{2} d x + c^{3}\right )} \sqrt {-b} \log \left (2 \, b d^{2} x^{2} - b c^{2} - 2 \, \sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {-b} d x\right ) + 2 \, \sqrt {-b d^{2} x^{2} + b c^{2}} {\left (3 \, d^{2} x^{2} - 19 \, c d x + 14 \, c^{2}\right )}}{6 \, {\left (b^{3} d^{5} x^{2} - 2 \, b^{3} c d^{4} x + b^{3} c^{2} d^{3}\right )}}, -\frac {9 \, {\left (c d^{2} x^{2} - 2 \, c^{2} d x + c^{3}\right )} \sqrt {b} \arctan \left (\frac {\sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {b} d x}{b d^{2} x^{2} - b c^{2}}\right ) + \sqrt {-b d^{2} x^{2} + b c^{2}} {\left (3 \, d^{2} x^{2} - 19 \, c d x + 14 \, c^{2}\right )}}{3 \, {\left (b^{3} d^{5} x^{2} - 2 \, b^{3} c d^{4} x + b^{3} c^{2} d^{3}\right )}}\right ] \] Input:

integrate(x^2*(d*x+c)^3/(-b*d^2*x^2+b*c^2)^(5/2),x, algorithm="fricas")
 

Output:

[-1/6*(9*(c*d^2*x^2 - 2*c^2*d*x + c^3)*sqrt(-b)*log(2*b*d^2*x^2 - b*c^2 - 
2*sqrt(-b*d^2*x^2 + b*c^2)*sqrt(-b)*d*x) + 2*sqrt(-b*d^2*x^2 + b*c^2)*(3*d 
^2*x^2 - 19*c*d*x + 14*c^2))/(b^3*d^5*x^2 - 2*b^3*c*d^4*x + b^3*c^2*d^3), 
-1/3*(9*(c*d^2*x^2 - 2*c^2*d*x + c^3)*sqrt(b)*arctan(sqrt(-b*d^2*x^2 + b*c 
^2)*sqrt(b)*d*x/(b*d^2*x^2 - b*c^2)) + sqrt(-b*d^2*x^2 + b*c^2)*(3*d^2*x^2 
 - 19*c*d*x + 14*c^2))/(b^3*d^5*x^2 - 2*b^3*c*d^4*x + b^3*c^2*d^3)]
 

Sympy [F]

\[ \int \frac {x^2 (c+d x)^3}{\left (b c^2-b d^2 x^2\right )^{5/2}} \, dx=\int \frac {x^{2} \left (c + d x\right )^{3}}{\left (- b \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(x**2*(d*x+c)**3/(-b*d**2*x**2+b*c**2)**(5/2),x)
 

Output:

Integral(x**2*(c + d*x)**3/(-b*(-c + d*x)*(c + d*x))**(5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.56 \[ \int \frac {x^2 (c+d x)^3}{\left (b c^2-b d^2 x^2\right )^{5/2}} \, dx=c d^{2} x {\left (\frac {3 \, x^{2}}{{\left (-b d^{2} x^{2} + b c^{2}\right )}^{\frac {3}{2}} b d^{2}} - \frac {2 \, c^{2}}{{\left (-b d^{2} x^{2} + b c^{2}\right )}^{\frac {3}{2}} b d^{4}}\right )} - \frac {d x^{4}}{{\left (-b d^{2} x^{2} + b c^{2}\right )}^{\frac {3}{2}} b} + \frac {7 \, c^{2} x^{2}}{{\left (-b d^{2} x^{2} + b c^{2}\right )}^{\frac {3}{2}} b d} + \frac {c^{3} x}{3 \, {\left (-b d^{2} x^{2} + b c^{2}\right )}^{\frac {3}{2}} b d^{2}} - \frac {14 \, c^{4}}{3 \, {\left (-b d^{2} x^{2} + b c^{2}\right )}^{\frac {3}{2}} b d^{3}} - \frac {4 \, c x}{3 \, \sqrt {-b d^{2} x^{2} + b c^{2}} b^{2} d^{2}} + \frac {3 \, c \arcsin \left (\frac {d x}{c}\right )}{b^{\frac {5}{2}} d^{3}} \] Input:

integrate(x^2*(d*x+c)^3/(-b*d^2*x^2+b*c^2)^(5/2),x, algorithm="maxima")
 

Output:

c*d^2*x*(3*x^2/((-b*d^2*x^2 + b*c^2)^(3/2)*b*d^2) - 2*c^2/((-b*d^2*x^2 + b 
*c^2)^(3/2)*b*d^4)) - d*x^4/((-b*d^2*x^2 + b*c^2)^(3/2)*b) + 7*c^2*x^2/((- 
b*d^2*x^2 + b*c^2)^(3/2)*b*d) + 1/3*c^3*x/((-b*d^2*x^2 + b*c^2)^(3/2)*b*d^ 
2) - 14/3*c^4/((-b*d^2*x^2 + b*c^2)^(3/2)*b*d^3) - 4/3*c*x/(sqrt(-b*d^2*x^ 
2 + b*c^2)*b^2*d^2) + 3*c*arcsin(d*x/c)/(b^(5/2)*d^3)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {x^2 (c+d x)^3}{\left (b c^2-b d^2 x^2\right )^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(x^2*(d*x+c)^3/(-b*d^2*x^2+b*c^2)^(5/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E 
rror: Bad Argument Value
 

Mupad [F(-1)]

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

int((x^2*(c + d*x)^3)/(b*c^2 - b*d^2*x^2)^(5/2),x)
 

Output:

int((x^2*(c + d*x)^3)/(b*c^2 - b*d^2*x^2)^(5/2), x)
                                                                                    
                                                                                    
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.70 \[ \int \frac {x^2 (c+d x)^3}{\left (b c^2-b d^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {b}\, \left (9 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) c^{2}-9 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) c d x -9 \mathit {asin} \left (\frac {d x}{c}\right ) c^{3}+18 \mathit {asin} \left (\frac {d x}{c}\right ) c^{2} d x -9 \mathit {asin} \left (\frac {d x}{c}\right ) c \,d^{2} x^{2}+6 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{2}-11 \sqrt {-d^{2} x^{2}+c^{2}}\, c d x +3 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{2} x^{2}-6 c^{3}-11 c^{2} d x +24 c \,d^{2} x^{2}-3 d^{3} x^{3}\right )}{3 b^{3} d^{3} \left (\sqrt {-d^{2} x^{2}+c^{2}}\, c -\sqrt {-d^{2} x^{2}+c^{2}}\, d x -c^{2}+2 c d x -d^{2} x^{2}\right )} \] Input:

int(x^2*(d*x+c)^3/(-b*d^2*x^2+b*c^2)^(5/2),x)
 

Output:

(sqrt(b)*(9*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*c**2 - 9*sqrt(c**2 - d**2 
*x**2)*asin((d*x)/c)*c*d*x - 9*asin((d*x)/c)*c**3 + 18*asin((d*x)/c)*c**2* 
d*x - 9*asin((d*x)/c)*c*d**2*x**2 + 6*sqrt(c**2 - d**2*x**2)*c**2 - 11*sqr 
t(c**2 - d**2*x**2)*c*d*x + 3*sqrt(c**2 - d**2*x**2)*d**2*x**2 - 6*c**3 - 
11*c**2*d*x + 24*c*d**2*x**2 - 3*d**3*x**3))/(3*b**3*d**3*(sqrt(c**2 - d** 
2*x**2)*c - sqrt(c**2 - d**2*x**2)*d*x - c**2 + 2*c*d*x - d**2*x**2))