\(\int x (a+\frac {b}{c+d x})^{3/2} \, dx\) [22]

Optimal result

Integrand size = 17, antiderivative size = 129 \[ \int x \left (a+\frac {b}{c+d x}\right )^{3/2} \, dx=\frac {2 b c \sqrt {a+\frac {b}{c+d x}}}{d^2}+\frac {(5 b-4 a c) (c+d x) \sqrt {a+\frac {b}{c+d x}}}{4 d^2}+\frac {a (c+d x)^2 \sqrt {a+\frac {b}{c+d x}}}{2 d^2}+\frac {3 b (b-4 a c) \text {arctanh}\left (\frac {\sqrt {a+\frac {b}{c+d x}}}{\sqrt {a}}\right )}{4 \sqrt {a} d^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.80 \[ \int x \left (a+\frac {b}{c+d x}\right )^{3/2} \, dx=\frac {\sqrt {\frac {b+a c+a d x}{c+d x}} \left (13 b c-2 a c^2+5 b d x+2 a d^2 x^2\right )}{4 d^2}-\frac {3 b (-b+4 a c) \text {arctanh}\left (\frac {\sqrt {\frac {b+a c+a d x}{c+d x}}}{\sqrt {a}}\right )}{4 \sqrt {a} d^2} \] Input:

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

Output:

(Sqrt[(b + a*c + a*d*x)/(c + d*x)]*(13*b*c - 2*a*c^2 + 5*b*d*x + 2*a*d^2*x 
^2))/(4*d^2) - (3*b*(-b + 4*a*c)*ArcTanh[Sqrt[(b + a*c + a*d*x)/(c + d*x)] 
/Sqrt[a]])/(4*Sqrt[a]*d^2)
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.95, 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, 51, 60, 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.

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

\(\Big \downarrow \) 896

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 941

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

\(\Big \downarrow \) 948

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(((c + d*x)^2*(a + b/(c + d*x))^(5/2))/(2*a) - ((b - 4*a*c)*(-((c + d*x)*( 
a + b/(c + d*x))^(3/2)) + (3*b*(2*Sqrt[a + b/(c + d*x)] - 2*Sqrt[a]*ArcTan 
h[Sqrt[a + b/(c + d*x)]/Sqrt[a]]))/2))/(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 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[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. \(1172\) vs. \(2(111)=222\).

Time = 0.13 (sec) , antiderivative size = 1173, normalized size of antiderivative = 9.09

method	result	size
default	\(\text {Expression too large to display}\)	\(1173\)

Input:

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

Output:

1/8*(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^3*x^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*b*c*d^3*x^2-24*((a*d*x+a*c+b)*(d*x+c))^(1/2)*(a*d^ 
2)^(1/2)*a*c*d^2*x^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*c*d^2*x^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))*b^2*d^3*x^2-24*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*b*c^2*d^2*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))*b^2*d^3*x^2-48*((a*d*x+a*c+b) 
*(d*x+c))^(1/2)*(a*d^2)^(1/2)*a*c^2*d*x+8*((a*d*x+a*c+b)*(d*x+c))^(1/2)*(a 
*d^2)^(1/2)*b*d^2*x^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*c^2*d*x+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^2*x^2-2*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))*b^2*c*d^2*x-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*b*c^3*d+8*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))*b^2*c*d^2*x-24*((a*d*x+a*c+b)*(d*x+c 
))^(1/2)*(a*d^2)^(1/2)*a*c^3+16*((a*d*x+a*c+b)*(d*x+c))^(1/2)*(a*d^2)^(1/2 
)*b*c*d*x+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^ 
3+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)*b*c*d*x-l...
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.85 \[ \int x \left (a+\frac {b}{c+d x}\right )^{3/2} \, dx=\left [\frac {3 \, {\left (4 \, a b c - b^{2}\right )} \sqrt {a} \log \left (2 \, a d x + 2 \, a c - 2 \, {\left (d x + c\right )} \sqrt {a} \sqrt {\frac {a d x + a c + b}{d x + c}} + b\right ) + 2 \, {\left (2 \, a^{2} d^{2} x^{2} - 2 \, a^{2} c^{2} + 5 \, a b d x + 13 \, a b c\right )} \sqrt {\frac {a d x + a c + b}{d x + c}}}{8 \, a d^{2}}, \frac {3 \, {\left (4 \, a b c - b^{2}\right )} \sqrt {-a} \arctan \left (\frac {{\left (d x + c\right )} \sqrt {-a} \sqrt {\frac {a d x + a c + b}{d x + c}}}{a d x + a c + b}\right ) + {\left (2 \, a^{2} d^{2} x^{2} - 2 \, a^{2} c^{2} + 5 \, a b d x + 13 \, a b c\right )} \sqrt {\frac {a d x + a c + b}{d x + c}}}{4 \, a d^{2}}\right ] \] Input:

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

Output:

[1/8*(3*(4*a*b*c - b^2)*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^2*d^2*x^2 - 2*a^2*c^2 + 5* 
a*b*d*x + 13*a*b*c)*sqrt((a*d*x + a*c + b)/(d*x + c)))/(a*d^2), 1/4*(3*(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 + 5*a*b*d*x + 13*a 
*b*c)*sqrt((a*d*x + a*c + b)/(d*x + c)))/(a*d^2)]
 

Sympy [F]

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

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 425 vs. \(2 (113) = 226\).

Time = 0.19 (sec) , antiderivative size = 425, normalized size of antiderivative = 3.29 \[ \int x \left (a+\frac {b}{c+d x}\right )^{3/2} \, dx=\frac {1}{4} \, \sqrt {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b d x + b c} {\left (\frac {2 \, a x \mathrm {sgn}\left (d x + c\right )}{d} - \frac {2 \, a^{2} c d^{2} \mathrm {sgn}\left (d x + c\right ) - 5 \, a b d^{2} \mathrm {sgn}\left (d x + c\right )}{a d^{4}}\right )} + \frac {{\left (4 \, a b c \mathrm {sgn}\left (d x + c\right ) - b^{2} \mathrm {sgn}\left (d x + c\right )\right )} \log \left ({\left | 2 \, a^{2} c^{3} d + 6 \, {\left (\sqrt {a d^{2}} x - \sqrt {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b d x + b c}\right )} a^{\frac {3}{2}} c^{2} {\left | d \right |} + 6 \, {\left (\sqrt {a d^{2}} x - \sqrt {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b d x + b c}\right )}^{2} a c d + a b c^{2} d + 2 \, {\left (\sqrt {a d^{2}} x - \sqrt {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b d x + b c}\right )}^{3} \sqrt {a} {\left | d \right |} + 2 \, {\left (\sqrt {a d^{2}} x - \sqrt {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b d x + b c}\right )} \sqrt {a} b c {\left | d \right |} + {\left (\sqrt {a d^{2}} x - \sqrt {a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b d x + b c}\right )}^{2} b d \right |}\right )}{8 \, \sqrt {a} d {\left | d \right |}} + \frac {{\left (4 \, a b c d^{2} {\left | d \right |} \mathrm {sgn}\left (d x + c\right ) - b^{2} d^{2} {\left | d \right |} \mathrm {sgn}\left (d x + c\right )\right )} \log \left (48\right )}{4 \, \sqrt {a} d^{5}} \] 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*a*x*sgn(d*x + c)/ 
d - (2*a^2*c*d^2*sgn(d*x + c) - 5*a*b*d^2*sgn(d*x + c))/(a*d^4)) + 1/8*(4* 
a*b*c*sgn(d*x + c) - b^2*sgn(d*x + c))*log(abs(2*a^2*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^(3/2)*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*c*d + a*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*sqrt(a)*abs(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)*b*c*abs(d) + (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))^2*b*d))/(sqrt(a)*d*abs(d)) 
 + 1/4*(4*a*b*c*d^2*abs(d)*sgn(d*x + c) - b^2*d^2*abs(d)*sgn(d*x + c))*log 
(48)/(sqrt(a)*d^5)
 

Mupad [F(-1)]

Timed out. \[ \int x \left (a+\frac {b}{c+d x}\right )^{3/2} \, dx=\int x\,{\left (a+\frac {b}{c+d\,x}\right )}^{3/2} \,d x \] 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 = 0.21 (sec) , antiderivative size = 284, normalized size of antiderivative = 2.20 \[ \int x \left (a+\frac {b}{c+d x}\right )^{3/2} \, dx=\frac {-2 \sqrt {d x +c}\, \sqrt {a d x +a c +b}\, a^{2} c^{2}+2 \sqrt {d x +c}\, \sqrt {a d x +a c +b}\, a^{2} d^{2} x^{2}+13 \sqrt {d x +c}\, \sqrt {a d x +a c +b}\, a b c +5 \sqrt {d x +c}\, \sqrt {a d x +a c +b}\, a b d x -12 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {a d x +a c +b}+\sqrt {a}\, \sqrt {d x +c}}{\sqrt {b}}\right ) a b \,c^{2}-12 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {a d x +a c +b}+\sqrt {a}\, \sqrt {d x +c}}{\sqrt {b}}\right ) a b c d x +3 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {a d x +a c +b}+\sqrt {a}\, \sqrt {d x +c}}{\sqrt {b}}\right ) b^{2} c +3 \sqrt {a}\, \mathrm {log}\left (\frac {\sqrt {a d x +a c +b}+\sqrt {a}\, \sqrt {d x +c}}{\sqrt {b}}\right ) b^{2} d x +9 \sqrt {a}\, a b \,c^{2}+9 \sqrt {a}\, a b c d x -\sqrt {a}\, b^{2} c -\sqrt {a}\, b^{2} d x}{4 a \,d^{2} \left (d x +c \right )} \] Input:

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

Output:

( - 2*sqrt(c + d*x)*sqrt(a*c + a*d*x + b)*a**2*c**2 + 2*sqrt(c + d*x)*sqrt 
(a*c + a*d*x + b)*a**2*d**2*x**2 + 13*sqrt(c + d*x)*sqrt(a*c + a*d*x + b)* 
a*b*c + 5*sqrt(c + d*x)*sqrt(a*c + a*d*x + b)*a*b*d*x - 12*sqrt(a)*log((sq 
rt(a*c + a*d*x + b) + sqrt(a)*sqrt(c + d*x))/sqrt(b))*a*b*c**2 - 12*sqrt(a 
)*log((sqrt(a*c + a*d*x + b) + sqrt(a)*sqrt(c + d*x))/sqrt(b))*a*b*c*d*x + 
 3*sqrt(a)*log((sqrt(a*c + a*d*x + b) + sqrt(a)*sqrt(c + d*x))/sqrt(b))*b* 
*2*c + 3*sqrt(a)*log((sqrt(a*c + a*d*x + b) + sqrt(a)*sqrt(c + d*x))/sqrt( 
b))*b**2*d*x + 9*sqrt(a)*a*b*c**2 + 9*sqrt(a)*a*b*c*d*x - sqrt(a)*b**2*c - 
 sqrt(a)*b**2*d*x)/(4*a*d**2*(c + d*x))