\(\int x^2 \sqrt {a+\frac {b}{c+d x}} \, dx\) [13]

Optimal result

Integrand size = 19, antiderivative size = 167 \[ \int x^2 \sqrt {a+\frac {b}{c+d x}} \, dx=-\frac {\left (b^2+4 a b c-8 a^2 c^2\right ) (c+d x) \sqrt {a+\frac {b}{c+d x}}}{8 a^2 d^3}+\frac {(b-12 a c) (c+d x)^2 \sqrt {a+\frac {b}{c+d x}}}{12 a d^3}+\frac {(c+d x)^3 \sqrt {a+\frac {b}{c+d x}}}{3 d^3}+\frac {b \left (b^2+4 a b c+8 a^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{c+d x}}}{\sqrt {a}}\right )}{8 a^{5/2} d^3} \] Output:

-1/8*(-8*a^2*c^2+4*a*b*c+b^2)*(d*x+c)*(a+b/(d*x+c))^(1/2)/a^2/d^3+1/12*(-1 
2*a*c+b)*(d*x+c)^2*(a+b/(d*x+c))^(1/2)/a/d^3+1/3*(d*x+c)^3*(a+b/(d*x+c))^( 
1/2)/d^3+1/8*b*(8*a^2*c^2+4*a*b*c+b^2)*arctanh((a+b/(d*x+c))^(1/2)/a^(1/2) 
)/a^(5/2)/d^3
 

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.78 \[ \int x^2 \sqrt {a+\frac {b}{c+d x}} \, dx=\frac {\sqrt {a} (c+d x) \sqrt {\frac {b+a c+a d x}{c+d x}} \left (-3 b^2+2 a b (-5 c+d x)+8 a^2 \left (c^2-c d x+d^2 x^2\right )\right )+3 b \left (b^2+4 a b c+8 a^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {\frac {b+a c+a d x}{c+d x}}}{\sqrt {a}}\right )}{24 a^{5/2} d^3} \] Input:

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

Output:

(Sqrt[a]*(c + d*x)*Sqrt[(b + a*c + a*d*x)/(c + d*x)]*(-3*b^2 + 2*a*b*(-5*c 
 + d*x) + 8*a^2*(c^2 - c*d*x + d^2*x^2)) + 3*b*(b^2 + 4*a*b*c + 8*a^2*c^2) 
*ArcTanh[Sqrt[(b + a*c + a*d*x)/(c + d*x)]/Sqrt[a]])/(24*a^(5/2)*d^3)
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.93, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.474, Rules used = {896, 941, 948, 100, 27, 87, 51, 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.

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

\(\Big \downarrow \) 896

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

\(\Big \downarrow \) 941

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 100

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 87

\(\displaystyle -\frac {-\frac {-\frac {\left (8 a^2 c^2+4 a b c+b^2\right ) \int (c+d x)^2 \sqrt {a+\frac {b}{c+d x}}d\frac {1}{c+d x}}{4 a}-\frac {(4 a c+b) (c+d x)^2 \left (a+\frac {b}{c+d x}\right )^{3/2}}{2 a}}{2 a}-\frac {(c+d x)^3 \left (a+\frac {b}{c+d x}\right )^{3/2}}{3 a}}{d^3}\)

\(\Big \downarrow \) 51

\(\displaystyle -\frac {-\frac {-\frac {\left (8 a^2 c^2+4 a b c+b^2\right ) \left (\frac {1}{2} b \int \frac {c+d x}{\sqrt {a+\frac {b}{c+d x}}}d\frac {1}{c+d x}-(c+d x) \sqrt {a+\frac {b}{c+d x}}\right )}{4 a}-\frac {(4 a c+b) (c+d x)^2 \left (a+\frac {b}{c+d x}\right )^{3/2}}{2 a}}{2 a}-\frac {(c+d x)^3 \left (a+\frac {b}{c+d x}\right )^{3/2}}{3 a}}{d^3}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {-\frac {-\frac {\left (8 a^2 c^2+4 a b c+b^2\right ) \left (\int \frac {1}{\frac {1}{b (c+d x)^2}-\frac {a}{b}}d\sqrt {a+\frac {b}{c+d x}}-(c+d x) \sqrt {a+\frac {b}{c+d x}}\right )}{4 a}-\frac {(4 a c+b) (c+d x)^2 \left (a+\frac {b}{c+d x}\right )^{3/2}}{2 a}}{2 a}-\frac {(c+d x)^3 \left (a+\frac {b}{c+d x}\right )^{3/2}}{3 a}}{d^3}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {-\frac {-\frac {\left (8 a^2 c^2+4 a b c+b^2\right ) \left ((c+d x) \left (-\sqrt {a+\frac {b}{c+d x}}\right )-\frac {b \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{c+d x}}}{\sqrt {a}}\right )}{\sqrt {a}}\right )}{4 a}-\frac {(4 a c+b) (c+d x)^2 \left (a+\frac {b}{c+d x}\right )^{3/2}}{2 a}}{2 a}-\frac {(c+d x)^3 \left (a+\frac {b}{c+d x}\right )^{3/2}}{3 a}}{d^3}\)

Input:

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

Output:

-((-1/3*((c + d*x)^3*(a + b/(c + d*x))^(3/2))/a - (-1/2*((b + 4*a*c)*(c + 
d*x)^2*(a + b/(c + d*x))^(3/2))/a - ((b^2 + 4*a*b*c + 8*a^2*c^2)*(-((c + d 
*x)*Sqrt[a + b/(c + d*x)]) - (b*ArcTanh[Sqrt[a + b/(c + d*x)]/Sqrt[a]])/Sq 
rt[a]))/(4*a))/(2*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 51

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

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 87

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

rule 100

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d 
*e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1))   Int[(c + d*x)^ 
(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( 
p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x 
, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n 
 + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))
 

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 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 941

Int[((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Sym 
bol] :> Int[(a + b*x^n)^p*((d + c*x^n)^q/x^(n*q)), x] /; FreeQ[{a, b, c, d, 
 n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] ||  !IntegerQ[p])
 

rule 948

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(544\) vs. \(2(147)=294\).

Time = 0.12 (sec) , antiderivative size = 545, normalized size of antiderivative = 3.26

method	result	size
default	\(\frac {\sqrt {\frac {a d x +a c +b}{d x +c}}\, \left (d x +c \right ) \left (-48 \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b d x +b c}\, \sqrt {a \,d^{2}}\, a^{2} c d x +24 \ln \left (\frac {2 a \,d^{2} x +2 a c d +2 \sqrt {\left (a d x +a c +b \right ) \left (d x +c \right )}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) a^{2} b \,c^{2} d +48 \sqrt {\left (a d x +a c +b \right ) \left (d x +c \right )}\, \sqrt {a \,d^{2}}\, a^{2} c^{2}-48 \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b d x +b c}\, \sqrt {a \,d^{2}}\, a^{2} c^{2}-12 \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b d x +b c}\, \sqrt {a \,d^{2}}\, a b d x +12 \ln \left (\frac {2 a \,d^{2} x +2 a c d +2 \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b d x +b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) a \,b^{2} c d +16 \left (a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b d x +b c \right )^{\frac {3}{2}} a \sqrt {a \,d^{2}}-36 \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b d x +b c}\, \sqrt {a \,d^{2}}\, a b c +3 \ln \left (\frac {2 a \,d^{2} x +2 a c d +2 \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b d x +b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) b^{3} d -6 \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b d x +b c}\, \sqrt {a \,d^{2}}\, b^{2}\right )}{48 d^{3} \sqrt {\left (a d x +a c +b \right ) \left (d x +c \right )}\, a^{2} \sqrt {a \,d^{2}}}\)	\(545\)

Input:

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

Output:

1/48*((a*d*x+a*c+b)/(d*x+c))^(1/2)*(d*x+c)/d^3*(-48*(a*d^2*x^2+2*a*c*d*x+a 
*c^2+b*d*x+b*c)^(1/2)*(a*d^2)^(1/2)*a^2*c*d*x+24*ln(1/2*(2*a*d^2*x+2*a*c*d 
+2*((a*d*x+a*c+b)*(d*x+c))^(1/2)*(a*d^2)^(1/2)+b*d)/(a*d^2)^(1/2))*a^2*b*c 
^2*d+48*((a*d*x+a*c+b)*(d*x+c))^(1/2)*(a*d^2)^(1/2)*a^2*c^2-48*(a*d^2*x^2+ 
2*a*c*d*x+a*c^2+b*d*x+b*c)^(1/2)*(a*d^2)^(1/2)*a^2*c^2-12*(a*d^2*x^2+2*a*c 
*d*x+a*c^2+b*d*x+b*c)^(1/2)*(a*d^2)^(1/2)*a*b*d*x+12*ln(1/2*(2*a*d^2*x+2*a 
*c*d+2*(a*d^2*x^2+2*a*c*d*x+a*c^2+b*d*x+b*c)^(1/2)*(a*d^2)^(1/2)+b*d)/(a*d 
^2)^(1/2))*a*b^2*c*d+16*(a*d^2*x^2+2*a*c*d*x+a*c^2+b*d*x+b*c)^(3/2)*a*(a*d 
^2)^(1/2)-36*(a*d^2*x^2+2*a*c*d*x+a*c^2+b*d*x+b*c)^(1/2)*(a*d^2)^(1/2)*a*b 
*c+3*ln(1/2*(2*a*d^2*x+2*a*c*d+2*(a*d^2*x^2+2*a*c*d*x+a*c^2+b*d*x+b*c)^(1/ 
2)*(a*d^2)^(1/2)+b*d)/(a*d^2)^(1/2))*b^3*d-6*(a*d^2*x^2+2*a*c*d*x+a*c^2+b* 
d*x+b*c)^(1/2)*(a*d^2)^(1/2)*b^2)/((a*d*x+a*c+b)*(d*x+c))^(1/2)/a^2/(a*d^2 
)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.96 \[ \int x^2 \sqrt {a+\frac {b}{c+d x}} \, dx=\left [\frac {3 \, {\left (8 \, a^{2} b c^{2} + 4 \, a b^{2} c + b^{3}\right )} \sqrt {a} \log \left (2 \, a d x + 2 \, a c + 2 \, {\left (d x + c\right )} \sqrt {a} \sqrt {\frac {a d x + a c + b}{d x + c}} + b\right ) + 2 \, {\left (8 \, a^{3} d^{3} x^{3} + 2 \, a^{2} b d^{2} x^{2} + 8 \, a^{3} c^{3} - 10 \, a^{2} b c^{2} - 3 \, a b^{2} c - {\left (8 \, a^{2} b c + 3 \, a b^{2}\right )} d x\right )} \sqrt {\frac {a d x + a c + b}{d x + c}}}{48 \, a^{3} d^{3}}, -\frac {3 \, {\left (8 \, a^{2} b c^{2} + 4 \, a b^{2} c + b^{3}\right )} \sqrt {-a} \arctan \left (\frac {{\left (d x + c\right )} \sqrt {-a} \sqrt {\frac {a d x + a c + b}{d x + c}}}{a d x + a c + b}\right ) - {\left (8 \, a^{3} d^{3} x^{3} + 2 \, a^{2} b d^{2} x^{2} + 8 \, a^{3} c^{3} - 10 \, a^{2} b c^{2} - 3 \, a b^{2} c - {\left (8 \, a^{2} b c + 3 \, a b^{2}\right )} d x\right )} \sqrt {\frac {a d x + a c + b}{d x + c}}}{24 \, a^{3} d^{3}}\right ] \] Input:

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

Output:

[1/48*(3*(8*a^2*b*c^2 + 4*a*b^2*c + b^3)*sqrt(a)*log(2*a*d*x + 2*a*c + 2*( 
d*x + c)*sqrt(a)*sqrt((a*d*x + a*c + b)/(d*x + c)) + b) + 2*(8*a^3*d^3*x^3 
 + 2*a^2*b*d^2*x^2 + 8*a^3*c^3 - 10*a^2*b*c^2 - 3*a*b^2*c - (8*a^2*b*c + 3 
*a*b^2)*d*x)*sqrt((a*d*x + a*c + b)/(d*x + c)))/(a^3*d^3), -1/24*(3*(8*a^2 
*b*c^2 + 4*a*b^2*c + b^3)*sqrt(-a)*arctan((d*x + c)*sqrt(-a)*sqrt((a*d*x + 
 a*c + b)/(d*x + c))/(a*d*x + a*c + b)) - (8*a^3*d^3*x^3 + 2*a^2*b*d^2*x^2 
 + 8*a^3*c^3 - 10*a^2*b*c^2 - 3*a*b^2*c - (8*a^2*b*c + 3*a*b^2)*d*x)*sqrt( 
(a*d*x + a*c + b)/(d*x + c)))/(a^3*d^3)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.49 \[ \int x^2 \sqrt {a+\frac {b}{c+d x}} \, dx=\frac {1}{24} \, \sqrt {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b d x + b c} {\left (2 \, x {\left (\frac {4 \, x \mathrm {sgn}\left (d x + c\right )}{d} - \frac {4 \, a^{2} c d^{3} \mathrm {sgn}\left (d x + c\right ) - a b d^{3} \mathrm {sgn}\left (d x + c\right )}{a^{2} d^{5}}\right )} + \frac {8 \, a^{2} c^{2} d^{2} \mathrm {sgn}\left (d x + c\right ) - 10 \, a b c d^{2} \mathrm {sgn}\left (d x + c\right ) - 3 \, b^{2} d^{2} \mathrm {sgn}\left (d x + c\right )}{a^{2} d^{5}}\right )} - \frac {{\left (8 \, a^{2} b c^{2} \mathrm {sgn}\left (d x + c\right ) + 4 \, a b^{2} c \mathrm {sgn}\left (d x + c\right ) + b^{3} \mathrm {sgn}\left (d x + c\right )\right )} \log \left ({\left | 2 \, a c d + 2 \, {\left (\sqrt {a d^{2}} x - \sqrt {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b d x + b c}\right )} \sqrt {a} {\left | d \right |} + b d \right |}\right )}{16 \, a^{\frac {5}{2}} d^{2} {\left | d \right |}} \] Input:

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

Output:

1/24*sqrt(a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b*d*x + b*c)*(2*x*(4*x*sgn(d*x + 
 c)/d - (4*a^2*c*d^3*sgn(d*x + c) - a*b*d^3*sgn(d*x + c))/(a^2*d^5)) + (8* 
a^2*c^2*d^2*sgn(d*x + c) - 10*a*b*c*d^2*sgn(d*x + c) - 3*b^2*d^2*sgn(d*x + 
 c))/(a^2*d^5)) - 1/16*(8*a^2*b*c^2*sgn(d*x + c) + 4*a*b^2*c*sgn(d*x + c) 
+ b^3*sgn(d*x + c))*log(abs(2*a*c*d + 2*(sqrt(a*d^2)*x - sqrt(a*d^2*x^2 + 
2*a*c*d*x + a*c^2 + b*d*x + b*c))*sqrt(a)*abs(d) + b*d))/(a^(5/2)*d^2*abs( 
d))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.54 \[ \int x^2 \sqrt {a+\frac {b}{c+d x}} \, dx=\frac {8 \sqrt {d x +c}\, \sqrt {a d x +a c +b}\, a^{3} c^{2}-8 \sqrt {d x +c}\, \sqrt {a d x +a c +b}\, a^{3} c d x +8 \sqrt {d x +c}\, \sqrt {a d x +a c +b}\, a^{3} d^{2} x^{2}-10 \sqrt {d x +c}\, \sqrt {a d x +a c +b}\, a^{2} b c +2 \sqrt {d x +c}\, \sqrt {a d x +a c +b}\, a^{2} b d x -3 \sqrt {d x +c}\, \sqrt {a d x +a c +b}\, a \,b^{2}+24 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {a d x +a c +b}+\sqrt {a}\, \sqrt {d x +c}}{\sqrt {b}}\right ) a^{2} b \,c^{2}+12 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {a d x +a c +b}+\sqrt {a}\, \sqrt {d x +c}}{\sqrt {b}}\right ) a \,b^{2} c +3 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {a d x +a c +b}+\sqrt {a}\, \sqrt {d x +c}}{\sqrt {b}}\right ) b^{3}}{24 a^{3} d^{3}} \] Input:

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

Output:

(8*sqrt(c + d*x)*sqrt(a*c + a*d*x + b)*a**3*c**2 - 8*sqrt(c + d*x)*sqrt(a* 
c + a*d*x + b)*a**3*c*d*x + 8*sqrt(c + d*x)*sqrt(a*c + a*d*x + b)*a**3*d** 
2*x**2 - 10*sqrt(c + d*x)*sqrt(a*c + a*d*x + b)*a**2*b*c + 2*sqrt(c + d*x) 
*sqrt(a*c + a*d*x + b)*a**2*b*d*x - 3*sqrt(c + d*x)*sqrt(a*c + a*d*x + b)* 
a*b**2 + 24*sqrt(a)*log((sqrt(a*c + a*d*x + b) + sqrt(a)*sqrt(c + d*x))/sq 
rt(b))*a**2*b*c**2 + 12*sqrt(a)*log((sqrt(a*c + a*d*x + b) + sqrt(a)*sqrt( 
c + d*x))/sqrt(b))*a*b**2*c + 3*sqrt(a)*log((sqrt(a*c + a*d*x + b) + sqrt( 
a)*sqrt(c + d*x))/sqrt(b))*b**3)/(24*a**3*d**3)