\(\int \frac {x}{(a+\frac {b}{c+d x})^{3/2}} \, dx\) [46]

Optimal result

Integrand size = 17, antiderivative size = 143 \[ \int \frac {x}{\left (a+\frac {b}{c+d x}\right )^{3/2}} \, dx=-\frac {2 b (b+a c)}{a^3 d^2 \sqrt {a+\frac {b}{c+d x}}}-\frac {(7 b+4 a c) (c+d x) \sqrt {a+\frac {b}{c+d x}}}{4 a^3 d^2}+\frac {(c+d x)^2 \sqrt {a+\frac {b}{c+d x}}}{2 a^2 d^2}+\frac {3 b (5 b+4 a c) \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{c+d x}}}{\sqrt {a}}\right )}{4 a^{7/2} d^2} \] Output:

-2*b*(a*c+b)/a^3/d^2/(a+b/(d*x+c))^(1/2)-1/4*(4*a*c+7*b)*(d*x+c)*(a+b/(d*x 
+c))^(1/2)/a^3/d^2+1/2*(d*x+c)^2*(a+b/(d*x+c))^(1/2)/a^2/d^2+3/4*b*(4*a*c+ 
5*b)*arctanh((a+b/(d*x+c))^(1/2)/a^(1/2))/a^(7/2)/d^2
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.91 \[ \int \frac {x}{\left (a+\frac {b}{c+d x}\right )^{3/2}} \, dx=\frac {-\frac {\sqrt {a} (c+d x) \sqrt {\frac {b+a c+a d x}{c+d x}} \left (15 b^2+a b (17 c+5 d x)+2 a^2 \left (c^2-d^2 x^2\right )\right )}{b+a (c+d x)}+3 b (5 b+4 a c) \text {arctanh}\left (\frac {\sqrt {\frac {b+a c+a d x}{c+d x}}}{\sqrt {a}}\right )}{4 a^{7/2} d^2} \] Input:

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

Output:

(-((Sqrt[a]*(c + d*x)*Sqrt[(b + a*c + a*d*x)/(c + d*x)]*(15*b^2 + a*b*(17* 
c + 5*d*x) + 2*a^2*(c^2 - d^2*x^2)))/(b + a*(c + d*x))) + 3*b*(5*b + 4*a*c 
)*ArcTanh[Sqrt[(b + a*c + a*d*x)/(c + d*x)]/Sqrt[a]])/(4*a^(7/2)*d^2)
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.93, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.588, Rules used = {896, 25, 941, 948, 25, 87, 52, 61, 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.

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

\(\Big \downarrow \) 896

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 941

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 87

\(\displaystyle \frac {\frac {(4 a c+5 b) \int \frac {(c+d x)^2}{\left (a+\frac {b}{c+d x}\right )^{3/2}}d\frac {1}{c+d x}}{4 a}+\frac {(c+d x)^2}{2 a \sqrt {a+\frac {b}{c+d x}}}}{d^2}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {\frac {(4 a c+5 b) \left (-\frac {3 b \int \frac {c+d x}{\left (a+\frac {b}{c+d x}\right )^{3/2}}d\frac {1}{c+d x}}{2 a}-\frac {c+d x}{a \sqrt {a+\frac {b}{c+d x}}}\right )}{4 a}+\frac {(c+d x)^2}{2 a \sqrt {a+\frac {b}{c+d x}}}}{d^2}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {\frac {(4 a c+5 b) \left (-\frac {3 b \left (\frac {\int \frac {c+d x}{\sqrt {a+\frac {b}{c+d x}}}d\frac {1}{c+d x}}{a}+\frac {2}{a \sqrt {a+\frac {b}{c+d x}}}\right )}{2 a}-\frac {c+d x}{a \sqrt {a+\frac {b}{c+d x}}}\right )}{4 a}+\frac {(c+d x)^2}{2 a \sqrt {a+\frac {b}{c+d x}}}}{d^2}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {(4 a c+5 b) \left (-\frac {3 b \left (\frac {2 \int \frac {1}{\frac {1}{b (c+d x)^2}-\frac {a}{b}}d\sqrt {a+\frac {b}{c+d x}}}{a b}+\frac {2}{a \sqrt {a+\frac {b}{c+d x}}}\right )}{2 a}-\frac {c+d x}{a \sqrt {a+\frac {b}{c+d x}}}\right )}{4 a}+\frac {(c+d x)^2}{2 a \sqrt {a+\frac {b}{c+d x}}}}{d^2}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {(4 a c+5 b) \left (-\frac {3 b \left (\frac {2}{a \sqrt {a+\frac {b}{c+d x}}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{c+d x}}}{\sqrt {a}}\right )}{a^{3/2}}\right )}{2 a}-\frac {c+d x}{a \sqrt {a+\frac {b}{c+d x}}}\right )}{4 a}+\frac {(c+d x)^2}{2 a \sqrt {a+\frac {b}{c+d x}}}}{d^2}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 52

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

rule 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, 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 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. \(1887\) vs. \(2(125)=250\).

Time = 0.23 (sec) , antiderivative size = 1888, normalized size of antiderivative = 13.20

method	result	size
default	\(\text {Expression too large to display}\)	\(1888\)

Input:

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

Output:

1/8*((a*d*x+a*c+b)/(d*x+c))^(1/2)*(d*x+c)/a^3/d^2*(20*(a*d^2)^(1/2)*(a*d^2 
*x^2+2*a*c*d*x+a*c^2+b*d*x+b*c)^(1/2)*a^2*b*c*d*x-112*(a*d^2)^(1/2)*((a*d* 
x+a*c+b)*(d*x+c))^(1/2)*a^2*b*c*d*x+12*(a*d^2)^(1/2)*(a*d^2*x^2+2*a*c*d*x+ 
a*c^2+b*d*x+b*c)^(1/2)*a^3*c^2*d*x+8*(a*d^2)^(1/2)*(a*d^2*x^2+2*a*c*d*x+a* 
c^2+b*d*x+b*c)^(1/2)*a*b^2*d*x+56*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^2*c*d^2*x-64*(a* 
d^2)^(1/2)*((a*d*x+a*c+b)*(d*x+c))^(1/2)*a*b^2*d*x+40*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^2*c^2*d-80*(a*d^2)^(1/2)*((a*d*x+a*c+b)*(d*x+c))^(1/2)*a^2*b*c^2+44*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^3*c*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))*a^2*b^2*c^2*d-2* 
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^3*c*d+10*(a*d^2)^(1/2)*(a*d^2*x^2+2*a* 
c*d*x+a*c^2+b*d*x+b*c)^(1/2)*a^2*b*c^2+8*(a*d^2)^(1/2)*(a*d^2*x^2+2*a*c*d* 
x+a*c^2+b*d*x+b*c)^(1/2)*a*b^2*c-24*((a*d*x+a*c+b)*(d*x+c))^(1/2)*(a*d^2)^ 
(1/2)*a^3*c^3+4*(a*d^2)^(1/2)*(a*d^2*x^2+2*a*c*d*x+a*c^2+b*d*x+b*c)^(1/2)* 
a^3*c^3+12*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^3*b*c*d^3*x^2+16*(a*d^2)^(1/2)*((a*d*x+a*c+ 
b)*(d*x+c))^(3/2)*a^2*c+16*(a*d^2)^(1/2)*((a*d*x+a*c+b)*(d*x+c))^(3/2)*...
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 438, normalized size of antiderivative = 3.06 \[ \int \frac {x}{\left (a+\frac {b}{c+d x}\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (4 \, a^{2} b c^{2} + 9 \, a b^{2} c + 5 \, b^{3} + {\left (4 \, a^{2} b c + 5 \, a b^{2}\right )} d x\right )} \sqrt {a} \log \left (2 \, a d x + 2 \, a c + 2 \, {\left (d x + c\right )} \sqrt {a} \sqrt {\frac {a d x + a c + b}{d x + c}} + b\right ) + 2 \, {\left (2 \, a^{3} d^{3} x^{3} - 2 \, a^{3} c^{3} - 17 \, a^{2} b c^{2} + {\left (2 \, a^{3} c - 5 \, a^{2} b\right )} d^{2} x^{2} - 15 \, a b^{2} c - {\left (2 \, a^{3} c^{2} + 22 \, a^{2} b c + 15 \, a b^{2}\right )} d x\right )} \sqrt {\frac {a d x + a c + b}{d x + c}}}{8 \, {\left (a^{5} d^{3} x + {\left (a^{5} c + a^{4} b\right )} d^{2}\right )}}, -\frac {3 \, {\left (4 \, a^{2} b c^{2} + 9 \, a b^{2} c + 5 \, b^{3} + {\left (4 \, a^{2} b c + 5 \, a b^{2}\right )} d x\right )} \sqrt {-a} \arctan \left (\frac {{\left (d x + c\right )} \sqrt {-a} \sqrt {\frac {a d x + a c + b}{d x + c}}}{a d x + a c + b}\right ) - {\left (2 \, a^{3} d^{3} x^{3} - 2 \, a^{3} c^{3} - 17 \, a^{2} b c^{2} + {\left (2 \, a^{3} c - 5 \, a^{2} b\right )} d^{2} x^{2} - 15 \, a b^{2} c - {\left (2 \, a^{3} c^{2} + 22 \, a^{2} b c + 15 \, a b^{2}\right )} d x\right )} \sqrt {\frac {a d x + a c + b}{d x + c}}}{4 \, {\left (a^{5} d^{3} x + {\left (a^{5} c + a^{4} b\right )} d^{2}\right )}}\right ] \] Input:

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

Output:

[1/8*(3*(4*a^2*b*c^2 + 9*a*b^2*c + 5*b^3 + (4*a^2*b*c + 5*a*b^2)*d*x)*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*(2*a^3*d^3*x^3 - 2*a^3*c^3 - 17*a^2*b*c^2 + (2*a^3*c - 5*a^ 
2*b)*d^2*x^2 - 15*a*b^2*c - (2*a^3*c^2 + 22*a^2*b*c + 15*a*b^2)*d*x)*sqrt( 
(a*d*x + a*c + b)/(d*x + c)))/(a^5*d^3*x + (a^5*c + a^4*b)*d^2), -1/4*(3*( 
4*a^2*b*c^2 + 9*a*b^2*c + 5*b^3 + (4*a^2*b*c + 5*a*b^2)*d*x)*sqrt(-a)*arct 
an((d*x + c)*sqrt(-a)*sqrt((a*d*x + a*c + b)/(d*x + c))/(a*d*x + a*c + b)) 
 - (2*a^3*d^3*x^3 - 2*a^3*c^3 - 17*a^2*b*c^2 + (2*a^3*c - 5*a^2*b)*d^2*x^2 
 - 15*a*b^2*c - (2*a^3*c^2 + 22*a^2*b*c + 15*a*b^2)*d*x)*sqrt((a*d*x + a*c 
 + b)/(d*x + c)))/(a^5*d^3*x + (a^5*c + a^4*b)*d^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

integrate(x/(a + b/(d*x + c))^(3/2), x)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 500 vs. \(2 (125) = 250\).

Time = 0.20 (sec) , antiderivative size = 500, normalized size of antiderivative = 3.50 \[ \int \frac {x}{\left (a+\frac {b}{c+d x}\right )^{3/2}} \, dx=\frac {1}{4} \, \sqrt {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b d x + b c} {\left (\frac {2 \, x}{a^{2} d \mathrm {sgn}\left (d x + c\right )} - \frac {2 \, a^{6} c d^{2} + 7 \, a^{5} b d^{2}}{a^{8} d^{4} \mathrm {sgn}\left (d x + c\right )}\right )} - \frac {{\left (4 \, a b c + 5 \, b^{2}\right )} \log \left ({\left | 2 \, a^{3} c^{3} d + 6 \, {\left (\sqrt {a d^{2}} x - \sqrt {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b d x + b c}\right )} a^{\frac {5}{2}} c^{2} {\left | d \right |} + 6 \, {\left (\sqrt {a d^{2}} x - \sqrt {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b d x + b c}\right )}^{2} a^{2} c d + 5 \, a^{2} b c^{2} d + 2 \, {\left (\sqrt {a d^{2}} x - \sqrt {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b d x + b c}\right )}^{3} a^{\frac {3}{2}} {\left | d \right |} + 10 \, {\left (\sqrt {a d^{2}} x - \sqrt {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b d x + b c}\right )} a^{\frac {3}{2}} b c {\left | d \right |} + 5 \, {\left (\sqrt {a d^{2}} x - \sqrt {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b d x + b c}\right )}^{2} a b d + 4 \, a b^{2} c d + 4 \, {\left (\sqrt {a d^{2}} x - \sqrt {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b d x + b c}\right )} \sqrt {a} b^{2} {\left | d \right |} + b^{3} d \right |}\right )}{8 \, a^{\frac {7}{2}} d {\left | d \right |} \mathrm {sgn}\left (d x + c\right )} - \frac {{\left (4 \, a^{\frac {9}{2}} b c d^{2} {\left | d \right |} \mathrm {sgn}\left (d x + c\right ) + 5 \, a^{\frac {7}{2}} b^{2} d^{2} {\left | d \right |} \mathrm {sgn}\left (d x + c\right )\right )} \log \left ({\left | a \right |}\right )}{8 \, a^{7} d^{5}} \] Input:

integrate(x/(a+b/(d*x+c))^(3/2),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/(a^2*d*sgn(d*x 
+ c)) - (2*a^6*c*d^2 + 7*a^5*b*d^2)/(a^8*d^4*sgn(d*x + c))) - 1/8*(4*a*b*c 
 + 5*b^2)*log(abs(2*a^3*c^3*d + 6*(sqrt(a*d^2)*x - sqrt(a*d^2*x^2 + 2*a*c* 
d*x + a*c^2 + b*d*x + b*c))*a^(5/2)*c^2*abs(d) + 6*(sqrt(a*d^2)*x - sqrt(a 
*d^2*x^2 + 2*a*c*d*x + a*c^2 + b*d*x + b*c))^2*a^2*c*d + 5*a^2*b*c^2*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))^3*a^( 
3/2)*abs(d) + 10*(sqrt(a*d^2)*x - sqrt(a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b*d 
*x + b*c))*a^(3/2)*b*c*abs(d) + 5*(sqrt(a*d^2)*x - sqrt(a*d^2*x^2 + 2*a*c* 
d*x + a*c^2 + b*d*x + b*c))^2*a*b*d + 4*a*b^2*c*d + 4*(sqrt(a*d^2)*x - sqr 
t(a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b*d*x + b*c))*sqrt(a)*b^2*abs(d) + b^3*d 
))/(a^(7/2)*d*abs(d)*sgn(d*x + c)) - 1/8*(4*a^(9/2)*b*c*d^2*abs(d)*sgn(d*x 
 + c) + 5*a^(7/2)*b^2*d^2*abs(d)*sgn(d*x + c))*log(abs(a))/(a^7*d^5)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.48 \[ \int \frac {x}{\left (a+\frac {b}{c+d x}\right )^{3/2}} \, dx=\frac {12 \sqrt {a}\, \sqrt {a d x +a c +b}\, \mathrm {log}\left (\frac {\sqrt {a d x +a c +b}+\sqrt {a}\, \sqrt {d x +c}}{\sqrt {b}}\right ) a b c +15 \sqrt {a}\, \sqrt {a d x +a c +b}\, \mathrm {log}\left (\frac {\sqrt {a d x +a c +b}+\sqrt {a}\, \sqrt {d x +c}}{\sqrt {b}}\right ) b^{2}-9 \sqrt {a}\, \sqrt {a d x +a c +b}\, a b c -10 \sqrt {a}\, \sqrt {a d x +a c +b}\, b^{2}-2 \sqrt {d x +c}\, a^{3} c^{2}+2 \sqrt {d x +c}\, a^{3} d^{2} x^{2}-17 \sqrt {d x +c}\, a^{2} b c -5 \sqrt {d x +c}\, a^{2} b d x -15 \sqrt {d x +c}\, a \,b^{2}}{4 \sqrt {a d x +a c +b}\, a^{4} d^{2}} \] Input:

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

Output:

(12*sqrt(a)*sqrt(a*c + a*d*x + b)*log((sqrt(a*c + a*d*x + b) + sqrt(a)*sqr 
t(c + d*x))/sqrt(b))*a*b*c + 15*sqrt(a)*sqrt(a*c + a*d*x + b)*log((sqrt(a* 
c + a*d*x + b) + sqrt(a)*sqrt(c + d*x))/sqrt(b))*b**2 - 9*sqrt(a)*sqrt(a*c 
 + a*d*x + b)*a*b*c - 10*sqrt(a)*sqrt(a*c + a*d*x + b)*b**2 - 2*sqrt(c + d 
*x)*a**3*c**2 + 2*sqrt(c + d*x)*a**3*d**2*x**2 - 17*sqrt(c + d*x)*a**2*b*c 
 - 5*sqrt(c + d*x)*a**2*b*d*x - 15*sqrt(c + d*x)*a*b**2)/(4*sqrt(a*c + a*d 
*x + b)*a**4*d**2)