\(\int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x^2} \, dx\) [218]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [F(-2)]
Giac [B] (verification not implemented)
Mupad [F(-1)]
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 22, antiderivative size = 144 \[ \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x^2} \, dx=2 b \sqrt {a+b x} \sqrt {c+d x}-\frac {(a+b x)^{3/2} \sqrt {c+d x}}{x}-\frac {\sqrt {a} (3 b c+a d) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{\sqrt {c}}+\frac {\sqrt {b} (b c+3 a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {d}} \] Output:

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

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.94 \[ \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x^2} \, dx=\frac {\sqrt {c+d x} \left (-a^2+b^2 x^2\right )}{x \sqrt {a+b x}}-\frac {\sqrt {a} (3 b c+a d) \text {arctanh}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{\sqrt {c}}+\frac {\sqrt {b} (b c+3 a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{\sqrt {d}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {108, 27, 171, 27, 175, 66, 104, 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.

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

\(\Big \downarrow \) 108

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 171

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 175

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

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 66
Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 
2   Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre 
eQ[{a, b, c, d}, x] &&  !GtQ[c - a*(d/b), 0]
 

rule 104
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 108
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^p/(b*(m + 1))) 
, x] - Simp[1/(b*(m + 1))   Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f* 
x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c 
, d, e, f}, x] && LtQ[m, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2 
*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
 

rule 171
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*(( 
e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Simp[1/(d*f*(m + n + p + 2)) 
  Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2 
) - h*(b*c*e*m + a*(d*e*(n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) 
+ h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x], x], x] /; Fre 
eQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] 
 && IntegersQ[2*m, 2*n, 2*p]
 

rule 175
Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_ 
)))/((a_.) + (b_.)*(x_)), x_] :> Simp[h/b   Int[(c + d*x)^n*(e + f*x)^p, x] 
, x] + Simp[(b*g - a*h)/b   Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)), x], x] 
 /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]
 

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]
 
Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(297\) vs. \(2(112)=224\).

Time = 0.22 (sec) , antiderivative size = 298, normalized size of antiderivative = 2.07

method result size
default \(\frac {\sqrt {b x +a}\, \sqrt {x d +c}\, \left (3 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a b d x \sqrt {a c}+\ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) b^{2} c x \sqrt {a c}-\ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) a^{2} d x \sqrt {d b}-3 \ln \left (\frac {a d x +b c x +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}+2 a c}{x}\right ) a b c x \sqrt {d b}+2 b x \sqrt {d b}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}-2 a \sqrt {d b}\, \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}\right )}{2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, x \sqrt {d b}\, \sqrt {a c}}\) \(298\)

Input:

int((b*x+a)^(3/2)*(d*x+c)^(1/2)/x^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 893, normalized size of antiderivative = 6.20 \[ \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x^2} \, dx =\text {Too large to display} \] Input:

integrate((b*x+a)^(3/2)*(d*x+c)^(1/2)/x^2,x, algorithm="fricas")
 

Output:

[1/4*((b*c + 3*a*d)*x*sqrt(b/d)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + 
a^2*d^2 + 4*(2*b*d^2*x + b*c*d + a*d^2)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(b 
/d) + 8*(b^2*c*d + a*b*d^2)*x) + (3*b*c + a*d)*x*sqrt(a/c)*log((8*a^2*c^2 
+ (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a*c^2 + (b*c^2 + a*c*d)*x)*sq 
rt(b*x + a)*sqrt(d*x + c)*sqrt(a/c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) + 4*sq 
rt(b*x + a)*(b*x - a)*sqrt(d*x + c))/x, -1/4*(2*(b*c + 3*a*d)*x*sqrt(-b/d) 
*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-b/d)/( 
b^2*d*x^2 + a*b*c + (b^2*c + a*b*d)*x)) - (3*b*c + a*d)*x*sqrt(a/c)*log((8 
*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^2 - 4*(2*a*c^2 + (b*c^2 + a*c 
*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(a/c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^ 
2) - 4*sqrt(b*x + a)*(b*x - a)*sqrt(d*x + c))/x, 1/4*(2*(3*b*c + a*d)*x*sq 
rt(-a/c)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sq 
rt(-a/c)/(a*b*d*x^2 + a^2*c + (a*b*c + a^2*d)*x)) + (b*c + 3*a*d)*x*sqrt(b 
/d)*log(8*b^2*d^2*x^2 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b*d^2*x + b*c 
*d + a*d^2)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(b/d) + 8*(b^2*c*d + a*b*d^2)* 
x) + 4*sqrt(b*x + a)*(b*x - a)*sqrt(d*x + c))/x, 1/2*((3*b*c + a*d)*x*sqrt 
(-a/c)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt 
(-a/c)/(a*b*d*x^2 + a^2*c + (a*b*c + a^2*d)*x)) - (b*c + 3*a*d)*x*sqrt(-b/ 
d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(b*x + a)*sqrt(d*x + c)*sqrt(-b/d) 
/(b^2*d*x^2 + a*b*c + (b^2*c + a*b*d)*x)) + 2*sqrt(b*x + a)*(b*x - a)*s...
 

Sympy [F]

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

integrate((b*x+a)**(3/2)*(d*x+c)**(1/2)/x**2,x)
 

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x^2} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((b*x+a)^(3/2)*(d*x+c)^(1/2)/x^2,x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m 
ore detail
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 496 vs. \(2 (112) = 224\).

Time = 0.30 (sec) , antiderivative size = 496, normalized size of antiderivative = 3.44 \[ \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x^2} \, dx=-\frac {1}{2} \, b {\left (\frac {{\left (b c {\left | b \right |} + 3 \, a d {\left | b \right |}\right )} \log \left ({\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{\sqrt {b d} b} - \frac {2 \, \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} {\left | b \right |}}{b^{2}} - \frac {2 \, {\left (3 \, a b c d {\left | b \right |} + a^{2} d^{2} {\left | b \right |}\right )} \arctan \left (\frac {b^{2} c + a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}}{2 \, \sqrt {-a b c d} b}\right )}{\sqrt {-a b c d} \sqrt {b d} b} + \frac {4 \, {\left (a b^{3} c^{2} d {\left | b \right |} - 2 \, a^{2} b^{2} c d^{2} {\left | b \right |} + a^{3} b d^{3} {\left | b \right |} - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b c d {\left | b \right |} - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a^{2} d^{2} {\left | b \right |}\right )}}{{\left (b^{4} c^{2} - 2 \, a b^{3} c d + a^{2} b^{2} d^{2} - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} b^{2} c - 2 \, {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2} a b d + {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{4}\right )} \sqrt {b d}}\right )} \] Input:

integrate((b*x+a)^(3/2)*(d*x+c)^(1/2)/x^2,x, algorithm="giac")
 

Output:

-1/2*b*((b*c*abs(b) + 3*a*d*abs(b))*log((sqrt(b*d)*sqrt(b*x + a) - sqrt(b^ 
2*c + (b*x + a)*b*d - a*b*d))^2)/(sqrt(b*d)*b) - 2*sqrt(b^2*c + (b*x + a)* 
b*d - a*b*d)*sqrt(b*x + a)*abs(b)/b^2 - 2*(3*a*b*c*d*abs(b) + a^2*d^2*abs( 
b))*arctan(1/2*(b^2*c + a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b 
*x + a)*b*d - a*b*d))^2)/(sqrt(-a*b*c*d)*b))/(sqrt(-a*b*c*d)*sqrt(b*d)*b) 
+ 4*(a*b^3*c^2*d*abs(b) - 2*a^2*b^2*c*d^2*abs(b) + a^3*b*d^3*abs(b) - (sqr 
t(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b*c*d*abs( 
b) - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2 
*d^2*abs(b))/((b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2 - 2*(sqrt(b*d)*sqrt(b*x 
 + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^2*c - 2*(sqrt(b*d)*sqrt(b 
*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b*d + (sqrt(b*d)*sqrt(b 
*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4)*sqrt(b*d)))
 

Mupad [F(-1)]

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

int(((a + b*x)^(3/2)*(c + d*x)^(1/2))/x^2,x)
                                                                                    
                                                                                    
 

Output:

int(((a + b*x)^(3/2)*(c + d*x)^(1/2))/x^2, x)
 

Reduce [B] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 419, normalized size of antiderivative = 2.91 \[ \int \frac {(a+b x)^{3/2} \sqrt {c+d x}}{x^2} \, dx=\frac {-2 \sqrt {d x +c}\, \sqrt {b x +a}\, a c d +2 \sqrt {d x +c}\, \sqrt {b x +a}\, b c d x +\sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (-\sqrt {2 \sqrt {d}\, \sqrt {c}\, \sqrt {b}\, \sqrt {a}+a d +b c}+\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}\right ) a \,d^{2} x +3 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (-\sqrt {2 \sqrt {d}\, \sqrt {c}\, \sqrt {b}\, \sqrt {a}+a d +b c}+\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}\right ) b c d x +\sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {2 \sqrt {d}\, \sqrt {c}\, \sqrt {b}\, \sqrt {a}+a d +b c}+\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}\right ) a \,d^{2} x +3 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (\sqrt {2 \sqrt {d}\, \sqrt {c}\, \sqrt {b}\, \sqrt {a}+a d +b c}+\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}\right ) b c d x -\sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (2 \sqrt {d}\, \sqrt {b}\, \sqrt {d x +c}\, \sqrt {b x +a}+2 \sqrt {d}\, \sqrt {c}\, \sqrt {b}\, \sqrt {a}+2 b d x \right ) a \,d^{2} x -3 \sqrt {c}\, \sqrt {a}\, \mathrm {log}\left (2 \sqrt {d}\, \sqrt {b}\, \sqrt {d x +c}\, \sqrt {b x +a}+2 \sqrt {d}\, \sqrt {c}\, \sqrt {b}\, \sqrt {a}+2 b d x \right ) b c d x +6 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a c d x +2 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) b \,c^{2} x}{2 c d x} \] Input:

int((b*x+a)^(3/2)*(d*x+c)^(1/2)/x^2,x)
 

Output:

( - 2*sqrt(c + d*x)*sqrt(a + b*x)*a*c*d + 2*sqrt(c + d*x)*sqrt(a + b*x)*b* 
c*d*x + sqrt(c)*sqrt(a)*log( - sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + a* 
d + b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))*a*d**2*x + 3*sqr 
t(c)*sqrt(a)*log( - sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + a*d + b*c) + 
sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))*b*c*d*x + sqrt(c)*sqrt(a)*l 
og(sqrt(2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + a*d + b*c) + sqrt(d)*sqrt(a + 
b*x) + sqrt(b)*sqrt(c + d*x))*a*d**2*x + 3*sqrt(c)*sqrt(a)*log(sqrt(2*sqrt 
(d)*sqrt(c)*sqrt(b)*sqrt(a) + a*d + b*c) + sqrt(d)*sqrt(a + b*x) + sqrt(b) 
*sqrt(c + d*x))*b*c*d*x - sqrt(c)*sqrt(a)*log(2*sqrt(d)*sqrt(b)*sqrt(c + d 
*x)*sqrt(a + b*x) + 2*sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + 2*b*d*x)*a*d**2*x 
- 3*sqrt(c)*sqrt(a)*log(2*sqrt(d)*sqrt(b)*sqrt(c + d*x)*sqrt(a + b*x) + 2* 
sqrt(d)*sqrt(c)*sqrt(b)*sqrt(a) + 2*b*d*x)*b*c*d*x + 6*sqrt(d)*sqrt(b)*log 
((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a*c*d*x 
+ 2*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sq 
rt(a*d - b*c))*b*c**2*x)/(2*c*d*x)