\(\int x \sqrt {a+\frac {b}{c+d x}} \, dx\) [14]

Optimal result

Integrand size = 17, antiderivative size = 107 \[ \int x \sqrt {a+\frac {b}{c+d x}} \, dx=\frac {(b-4 a c) (c+d x) \sqrt {a+\frac {b}{c+d x}}}{4 a d^2}+\frac {(c+d x)^2 \sqrt {a+\frac {b}{c+d x}}}{2 d^2}-\frac {b (b+4 a c) \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{c+d x}}}{\sqrt {a}}\right )}{4 a^{3/2} d^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.89 \[ \int x \sqrt {a+\frac {b}{c+d x}} \, dx=\frac {(c+d x) \sqrt {\frac {b+a c+a d x}{c+d x}} (b-2 a c+2 a d x)}{4 a d^2}-\frac {b (b+4 a c) \text {arctanh}\left (\frac {\sqrt {\frac {b+a c+a d x}{c+d x}}}{\sqrt {a}}\right )}{4 a^{3/2} d^2} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.93, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {896, 25, 941, 948, 25, 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.

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

\(\Big \downarrow \) 896

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 941

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(((c + d*x)^2*(a + b/(c + d*x))^(3/2))/(2*a) + ((b + 4*a*c)*(-((c + d*x)*S 
qrt[a + b/(c + d*x)]) - (b*ArcTanh[Sqrt[a + b/(c + d*x)]/Sqrt[a]])/Sqrt[a] 
))/(4*a))/d^2
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 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. \(332\) vs. \(2(91)=182\).

Time = 0.12 (sec) , antiderivative size = 333, normalized size of antiderivative = 3.11

method	result	size
default	\(-\frac {\sqrt {\frac {a d x +a c +b}{d x +c}}\, \left (d x +c \right ) \left (-4 \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b d x +b c}\, \sqrt {a \,d^{2}}\, a d x +4 \ln \left (\frac {2 a \,d^{2} x +2 a c d +2 \sqrt {\left (a d x +a c +b \right ) \left (d x +c \right )}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right ) a b c d +8 \sqrt {\left (a d x +a c +b \right ) \left (d x +c \right )}\, \sqrt {a \,d^{2}}\, a c -4 \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b d x +b c}\, \sqrt {a \,d^{2}}\, a c +b^{2} d \ln \left (\frac {2 a \,d^{2} x +2 a c d +2 \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b d x +b c}\, \sqrt {a \,d^{2}}+b d}{2 \sqrt {a \,d^{2}}}\right )-2 \sqrt {a \,d^{2} x^{2}+2 a d x c +a \,c^{2}+b d x +b c}\, \sqrt {a \,d^{2}}\, b \right )}{8 d^{2} \sqrt {\left (a d x +a c +b \right ) \left (d x +c \right )}\, a \sqrt {a \,d^{2}}}\)	\(333\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 229, normalized size of antiderivative = 2.14 \[ \int x \sqrt {a+\frac {b}{c+d x}} \, dx=\left [\frac {{\left (4 \, a b c + b^{2}\right )} \sqrt {a} \log \left (2 \, a d x + 2 \, a c - 2 \, {\left (d x + c\right )} \sqrt {a} \sqrt {\frac {a d x + a c + b}{d x + c}} + b\right ) + 2 \, {\left (2 \, a^{2} d^{2} x^{2} - 2 \, a^{2} c^{2} + a b d x + a b c\right )} \sqrt {\frac {a d x + a c + b}{d x + c}}}{8 \, a^{2} d^{2}}, \frac {{\left (4 \, a b c + b^{2}\right )} \sqrt {-a} \arctan \left (\frac {{\left (d x + c\right )} \sqrt {-a} \sqrt {\frac {a d x + a c + b}{d x + c}}}{a d x + a c + b}\right ) + {\left (2 \, a^{2} d^{2} x^{2} - 2 \, a^{2} c^{2} + a b d x + a b c\right )} \sqrt {\frac {a d x + a c + b}{d x + c}}}{4 \, a^{2} d^{2}}\right ] \] Input:

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

Output:

[1/8*((4*a*b*c + b^2)*sqrt(a)*log(2*a*d*x + 2*a*c - 2*(d*x + c)*sqrt(a)*sq 
rt((a*d*x + a*c + b)/(d*x + c)) + b) + 2*(2*a^2*d^2*x^2 - 2*a^2*c^2 + a*b* 
d*x + a*b*c)*sqrt((a*d*x + a*c + b)/(d*x + c)))/(a^2*d^2), 1/4*((4*a*b*c + 
 b^2)*sqrt(-a)*arctan((d*x + c)*sqrt(-a)*sqrt((a*d*x + a*c + b)/(d*x + c)) 
/(a*d*x + a*c + b)) + (2*a^2*d^2*x^2 - 2*a^2*c^2 + a*b*d*x + a*b*c)*sqrt(( 
a*d*x + a*c + b)/(d*x + c)))/(a^2*d^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.57 \[ \int x \sqrt {a+\frac {b}{c+d x}} \, dx=\frac {1}{4} \, \sqrt {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b d x + b c} {\left (\frac {2 \, x \mathrm {sgn}\left (d x + c\right )}{d} - \frac {2 \, a c d \mathrm {sgn}\left (d x + c\right ) - b d \mathrm {sgn}\left (d x + c\right )}{a d^{3}}\right )} + \frac {{\left (4 \, a b c \mathrm {sgn}\left (d x + c\right ) + b^{2} \mathrm {sgn}\left (d x + c\right )\right )} \log \left ({\left | 2 \, a c d + 2 \, {\left (\sqrt {a d^{2}} x - \sqrt {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b d x + b c}\right )} \sqrt {a} {\left | d \right |} + b d \right |}\right )}{8 \, a^{\frac {3}{2}} d {\left | d \right |}} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.30 \[ \int x \sqrt {a+\frac {b}{c+d x}} \, dx=\frac {-2 \sqrt {d x +c}\, \sqrt {a d x +a c +b}\, a^{2} c +2 \sqrt {d x +c}\, \sqrt {a d x +a c +b}\, a^{2} d x +\sqrt {d x +c}\, \sqrt {a d x +a c +b}\, a b -4 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {a d x +a c +b}+\sqrt {a}\, \sqrt {d x +c}}{\sqrt {b}}\right ) a b c -\sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {a d x +a c +b}+\sqrt {a}\, \sqrt {d x +c}}{\sqrt {b}}\right ) b^{2}}{4 a^{2} d^{2}} \] Input:

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

Output:

( - 2*sqrt(c + d*x)*sqrt(a*c + a*d*x + b)*a**2*c + 2*sqrt(c + d*x)*sqrt(a* 
c + a*d*x + b)*a**2*d*x + sqrt(c + d*x)*sqrt(a*c + a*d*x + b)*a*b - 4*sqrt 
(a)*log((sqrt(a*c + a*d*x + b) + sqrt(a)*sqrt(c + d*x))/sqrt(b))*a*b*c - s 
qrt(a)*log((sqrt(a*c + a*d*x + b) + sqrt(a)*sqrt(c + d*x))/sqrt(b))*b**2)/ 
(4*a**2*d**2)