\(\int x^2 \sqrt {a+\frac {b}{(c+d x)^2}} \, dx\) [64]

Optimal result

Integrand size = 19, antiderivative size = 149 \[ \int x^2 \sqrt {a+\frac {b}{(c+d x)^2}} \, dx=\frac {(c+d x)^3 \left (a+\frac {b}{(c+d x)^2}\right )^{3/2}}{3 a d^3}-\frac {c (c+d x)^2 \sqrt {a+\frac {b}{(c+d x)^2}} \left (1-\frac {c}{c+d x}\right )}{d^3}-\frac {\sqrt {b} c^2 \text {arctanh}\left (\frac {\sqrt {b}}{(c+d x) \sqrt {a+\frac {b}{(c+d x)^2}}}\right )}{d^3}-\frac {b c \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{(c+d x)^2}}}{\sqrt {a}}\right )}{\sqrt {a} d^3} \] Output:

1/3*(d*x+c)^3*(a+b/(d*x+c)^2)^(3/2)/a/d^3-c*(d*x+c)^2*(a+b/(d*x+c)^2)^(1/2 
)*(1-c/(d*x+c))/d^3-b^(1/2)*c^2*arctanh(b^(1/2)/(d*x+c)/(a+b/(d*x+c)^2)^(1 
/2))/d^3-b*c*arctanh((a+b/(d*x+c)^2)^(1/2)/a^(1/2))/a^(1/2)/d^3
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 10.22 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.58 \[ \int x^2 \sqrt {a+\frac {b}{(c+d x)^2}} \, dx=\frac {b c \sqrt {\frac {b+a (c+d x)^2}{(c+d x)^2}}+a c^3 \sqrt {\frac {b+a (c+d x)^2}{(c+d x)^2}}+b d x \sqrt {\frac {b+a (c+d x)^2}{(c+d x)^2}}+a d^3 x^3 \sqrt {\frac {b+a (c+d x)^2}{(c+d x)^2}}+3 a \sqrt {b} c^2 \log (c+d x)-3 \sqrt {a} b c \log \left ((c+d x) \left (a+\sqrt {a} \sqrt {\frac {b+a (c+d x)^2}{(c+d x)^2}}\right )\right )-3 a \sqrt {b} c^2 \log \left (b+\sqrt {b} (c+d x) \sqrt {\frac {b+a (c+d x)^2}{(c+d x)^2}}\right )}{3 a d^3} \] Input:

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

Output:

(b*c*Sqrt[(b + a*(c + d*x)^2)/(c + d*x)^2] + a*c^3*Sqrt[(b + a*(c + d*x)^2 
)/(c + d*x)^2] + b*d*x*Sqrt[(b + a*(c + d*x)^2)/(c + d*x)^2] + a*d^3*x^3*S 
qrt[(b + a*(c + d*x)^2)/(c + d*x)^2] + 3*a*Sqrt[b]*c^2*Log[c + d*x] - 3*Sq 
rt[a]*b*c*Log[(c + d*x)*(a + Sqrt[a]*Sqrt[(b + a*(c + d*x)^2)/(c + d*x)^2] 
)] - 3*a*Sqrt[b]*c^2*Log[b + Sqrt[b]*(c + d*x)*Sqrt[(b + a*(c + d*x)^2)/(c 
 + d*x)^2]])/(3*a*d^3)
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.97, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.684, Rules used = {896, 1774, 1803, 540, 27, 537, 27, 538, 224, 219, 243, 73, 221}

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.

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

\(\Big \downarrow \) 896

\(\displaystyle \frac {\int d^2 x^2 \sqrt {a+\frac {b}{(c+d x)^2}}d(c+d x)}{d^3}\)

\(\Big \downarrow \) 1774

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

\(\Big \downarrow \) 1803

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

\(\Big \downarrow \) 540

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 537

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 538

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

-((-1/3*((c + d*x)^3*(a + b/(c + d*x)^2)^(3/2))/a - c*(-((c + d*x)^2*Sqrt[ 
a + b/(c + d*x)^2]*(1 - c/(c + d*x))) + b*(-((c*ArcTanh[Sqrt[b]/((c + d*x) 
*Sqrt[a + b/(c + d*x)^2])])/Sqrt[b]) - ArcTanh[Sqrt[a + b/(c + d*x)^2]/Sqr 
t[a]]/Sqrt[a])))/d^3)
 

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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

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 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 537

Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[x^(m + 1)*(c*(m + 2) + d*(m + 1)*x)*((a + b*x^2)^p/((m + 1)*(m + 2))), 
 x] - Simp[2*b*(p/((m + 1)*(m + 2)))   Int[x^(m + 2)*(c*(m + 2) + d*(m + 1) 
*x)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, -2] && 
 GtQ[p, 0] &&  !ILtQ[m + 2*p + 3, 0] && IntegerQ[2*p]
 

rule 538

Int[((c_) + (d_.)*(x_))/((x_)*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Simp 
[c   Int[1/(x*Sqrt[a + b*x^2]), x], x] + Simp[d   Int[1/Sqrt[a + b*x^2], x] 
, x] /; FreeQ[{a, b, c, d}, x]
 

rule 540

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol 
] :> With[{Qx = PolynomialQuotient[(c + d*x)^n, x, x], R = PolynomialRemain 
der[(c + d*x)^n, x, x]}, Simp[R*x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))) 
, x] + Simp[1/(a*(m + 1))   Int[x^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*(m + 
1)*Qx - b*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, p}, x] && IG 
tQ[n, 1] && ILtQ[m, -1] && GtQ[p, -1] && IntegerQ[2*p]
 

rule 896

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coeff 
icient[v, x, 0], d = Coefficient[v, x, 1]}, Simp[1/d^(m + 1)   Subst[Int[Si 
mplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]] /; 
 FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]
 

rule 1774

Int[((d_) + (e_.)*(x_)^(mn_.))^(q_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.), x_Sy 
mbol] :> Int[x^(mn*q)*(e + d/x^mn)^q*(a + c*x^n2)^p, x] /; FreeQ[{a, c, d, 
e, mn, p}, x] && EqQ[n2, -2*mn] && IntegerQ[q] && (PosQ[n2] ||  !IntegerQ[p 
])
 

rule 1803

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q 
_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(d + e*x 
)^q*(a + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, c, d, e, m, n, p, q}, x] && 
 EqQ[n2, 2*n] && IntegerQ[Simplify[(m + 1)/n]]
 

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.58

method	result	size
risch	\(\frac {\left (a \,d^{2} x^{2}-a d x c +a \,c^{2}+b \right ) \sqrt {\frac {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}{\left (d x +c \right )^{2}}}\, \left (d x +c \right )}{3 a \,d^{3}}+\frac {\left (-\frac {b c \ln \left (\frac {a \,d^{2} x +a c d}{\sqrt {a \,d^{2}}}+\sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}\right )}{d^{2} \sqrt {a \,d^{2}}}-\frac {\sqrt {b}\, c^{2} \ln \left (\frac {2 b +2 \sqrt {b}\, \sqrt {a \,d^{2} \left (x +\frac {c}{d}\right )^{2}+b}}{x +\frac {c}{d}}\right )}{d^{3}}\right ) \sqrt {\frac {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}{\left (d x +c \right )^{2}}}\, \left (d x +c \right )}{\sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}}\)	\(235\)
default	\(\frac {\sqrt {\frac {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}{\left (d x +c \right )^{2}}}\, \left (d x +c \right ) \left (-3 \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}\, \sqrt {a \,d^{2}}\, a c d x -3 \sqrt {a \,d^{2}}\, \sqrt {b}\, \ln \left (\frac {2 \left (\sqrt {b}\, \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}+b \right ) d}{d x +c}\right ) a \,c^{2}-3 \ln \left (\frac {a \,d^{2} x +a c d +\sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}\, \sqrt {a \,d^{2}}}{\sqrt {a \,d^{2}}}\right ) a b c d +\left (a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b \right )^{\frac {3}{2}} \sqrt {a \,d^{2}}\right )}{3 d^{3} \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b}\, a \sqrt {a \,d^{2}}}\)	\(256\)

Input:

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

Output:

1/3*(a*d^2*x^2-a*c*d*x+a*c^2+b)/a/d^3*((a*d^2*x^2+2*a*c*d*x+a*c^2+b)/(d*x+ 
c)^2)^(1/2)*(d*x+c)+(-b*c/d^2*ln((a*d^2*x+a*c*d)/(a*d^2)^(1/2)+(a*d^2*x^2+ 
2*a*c*d*x+a*c^2+b)^(1/2))/(a*d^2)^(1/2)-b^(1/2)*c^2/d^3*ln((2*b+2*b^(1/2)* 
(a*d^2*(x+c/d)^2+b)^(1/2))/(x+c/d)))*((a*d^2*x^2+2*a*c*d*x+a*c^2+b)/(d*x+c 
)^2)^(1/2)*(d*x+c)/(a*d^2*x^2+2*a*c*d*x+a*c^2+b)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 1028, normalized size of antiderivative = 6.90 \[ \int x^2 \sqrt {a+\frac {b}{(c+d x)^2}} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/6*(3*a*sqrt(b)*c^2*log(-(a*d^2*x^2 + 2*a*c*d*x + a*c^2 - 2*(d*x + c)*sq 
rt(b)*sqrt((a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b)/(d^2*x^2 + 2*c*d*x + c^2)) 
+ 2*b)/(d^2*x^2 + 2*c*d*x + c^2)) + 3*sqrt(a)*b*c*log(-2*a*d^2*x^2 - 4*a*c 
*d*x - 2*a*c^2 + 2*(d^2*x^2 + 2*c*d*x + c^2)*sqrt(a)*sqrt((a*d^2*x^2 + 2*a 
*c*d*x + a*c^2 + b)/(d^2*x^2 + 2*c*d*x + c^2)) - b) + 2*(a*d^3*x^3 + a*c^3 
 + b*d*x + b*c)*sqrt((a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b)/(d^2*x^2 + 2*c*d* 
x + c^2)))/(a*d^3), 1/6*(3*a*sqrt(b)*c^2*log(-(a*d^2*x^2 + 2*a*c*d*x + a*c 
^2 - 2*(d*x + c)*sqrt(b)*sqrt((a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b)/(d^2*x^2 
 + 2*c*d*x + c^2)) + 2*b)/(d^2*x^2 + 2*c*d*x + c^2)) + 6*sqrt(-a)*b*c*arct 
an((d^2*x^2 + 2*c*d*x + c^2)*sqrt(-a)*sqrt((a*d^2*x^2 + 2*a*c*d*x + a*c^2 
+ b)/(d^2*x^2 + 2*c*d*x + c^2))/(a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b)) + 2*( 
a*d^3*x^3 + a*c^3 + b*d*x + b*c)*sqrt((a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b)/ 
(d^2*x^2 + 2*c*d*x + c^2)))/(a*d^3), 1/6*(6*a*sqrt(-b)*c^2*arctan((d*x + c 
)*sqrt(-b)*sqrt((a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b)/(d^2*x^2 + 2*c*d*x + c 
^2))/b) + 3*sqrt(a)*b*c*log(-2*a*d^2*x^2 - 4*a*c*d*x - 2*a*c^2 + 2*(d^2*x^ 
2 + 2*c*d*x + c^2)*sqrt(a)*sqrt((a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b)/(d^2*x 
^2 + 2*c*d*x + c^2)) - b) + 2*(a*d^3*x^3 + a*c^3 + b*d*x + b*c)*sqrt((a*d^ 
2*x^2 + 2*a*c*d*x + a*c^2 + b)/(d^2*x^2 + 2*c*d*x + c^2)))/(a*d^3), 1/3*(3 
*a*sqrt(-b)*c^2*arctan((d*x + c)*sqrt(-b)*sqrt((a*d^2*x^2 + 2*a*c*d*x + a* 
c^2 + b)/(d^2*x^2 + 2*c*d*x + c^2))/b) + 3*sqrt(-a)*b*c*arctan((d^2*x^2...
 

Sympy [F]

\[ \int x^2 \sqrt {a+\frac {b}{(c+d x)^2}} \, dx=\int x^{2} \sqrt {\frac {a c^{2} + 2 a c d x + a d^{2} x^{2} + b}{c^{2} + 2 c d x + d^{2} x^{2}}}\, dx \] Input:

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

Output:

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

Maxima [F]

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

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

Output:

integrate(sqrt(a + b/(d*x + c)^2)*x^2, x)
 

Giac [F(-1)]

Timed out. \[ \int x^2 \sqrt {a+\frac {b}{(c+d x)^2}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.74 \[ \int x^2 \sqrt {a+\frac {b}{(c+d x)^2}} \, dx=\frac {\sqrt {a \,d^{2} x^{2}+2 a c d x +a \,c^{2}+b}\, a \,c^{2}-\sqrt {a \,d^{2} x^{2}+2 a c d x +a \,c^{2}+b}\, a c d x +\sqrt {a \,d^{2} x^{2}+2 a c d x +a \,c^{2}+b}\, a \,d^{2} x^{2}+\sqrt {a \,d^{2} x^{2}+2 a c d x +a \,c^{2}+b}\, b -3 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {a \,d^{2} x^{2}+2 a c d x +a \,c^{2}+b}+\sqrt {a}\, c +\sqrt {a}\, d x}{\sqrt {b}}\right ) b c +3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {a \,d^{2} x^{2}+2 a c d x +a \,c^{2}+b}+\sqrt {a}\, c +\sqrt {a}\, d x -\sqrt {b}}{\sqrt {b}}\right ) a \,c^{2}-3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {a \,d^{2} x^{2}+2 a c d x +a \,c^{2}+b}+\sqrt {a}\, c +\sqrt {a}\, d x +\sqrt {b}}{\sqrt {b}}\right ) a \,c^{2}}{3 a \,d^{3}} \] Input:

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

Output:

(sqrt(a*c**2 + 2*a*c*d*x + a*d**2*x**2 + b)*a*c**2 - sqrt(a*c**2 + 2*a*c*d 
*x + a*d**2*x**2 + b)*a*c*d*x + sqrt(a*c**2 + 2*a*c*d*x + a*d**2*x**2 + b) 
*a*d**2*x**2 + sqrt(a*c**2 + 2*a*c*d*x + a*d**2*x**2 + b)*b - 3*sqrt(a)*lo 
g((sqrt(a*c**2 + 2*a*c*d*x + a*d**2*x**2 + b) + sqrt(a)*c + sqrt(a)*d*x)/s 
qrt(b))*b*c + 3*sqrt(b)*log((sqrt(a*c**2 + 2*a*c*d*x + a*d**2*x**2 + b) + 
sqrt(a)*c + sqrt(a)*d*x - sqrt(b))/sqrt(b))*a*c**2 - 3*sqrt(b)*log((sqrt(a 
*c**2 + 2*a*c*d*x + a*d**2*x**2 + b) + sqrt(a)*c + sqrt(a)*d*x + sqrt(b))/ 
sqrt(b))*a*c**2)/(3*a*d**3)