\(\int \frac {1}{x (c+d x)^2 \sqrt {a x^2+b x^3}} \, dx\) [252]

Optimal result

Integrand size = 26, antiderivative size = 195 \[ \int \frac {1}{x (c+d x)^2 \sqrt {a x^2+b x^3}} \, dx=-\frac {\sqrt {a x^2+b x^3}}{a c x^2 (c+d x)}-\frac {d (b c-2 a d) \sqrt {a x^2+b x^3}}{a c^2 (b c-a d) x (c+d x)}+\frac {d^{3/2} (5 b c-4 a d) \arctan \left (\frac {\sqrt {d} \sqrt {a x^2+b x^3}}{\sqrt {b c-a d} x}\right )}{c^3 (b c-a d)^{3/2}}+\frac {(b c+4 a d) \text {arctanh}\left (\frac {\sqrt {a x^2+b x^3}}{\sqrt {a} x}\right )}{a^{3/2} c^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.87 \[ \int \frac {1}{x (c+d x)^2 \sqrt {a x^2+b x^3}} \, dx=\frac {\frac {c (a+b x) (b c (c+d x)-a d (c+2 d x))}{a (-b c+a d) (c+d x)}+\frac {d^{3/2} (5 b c-4 a d) x \sqrt {a+b x} \arctan \left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b c-a d}}\right )}{(b c-a d)^{3/2}}+\frac {(b c+4 a d) x \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{3/2}}}{c^3 \sqrt {x^2 (a+b x)}} \] Input:

Integrate[1/(x*(c + d*x)^2*Sqrt[a*x^2 + b*x^3]),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.14, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {1948, 114, 27, 168, 25, 174, 73, 218, 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.

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

\(\Big \downarrow \) 1948

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

\(\Big \downarrow \) 114

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 168

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 174

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 221

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

Input:

Int[1/(x*(c + d*x)^2*Sqrt[a*x^2 + b*x^3]),x]
 

Output:

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

Defintions of rubi rules used

rule 25

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

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 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 114

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

rule 168

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + 
 d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S 
imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n 
*(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* 
h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]
 

rule 174

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

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

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 1948

Int[((e_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(jn_.))^(p_)*((c_) + 
(d_.)*(x_)^(n_.))^(q_.), x_Symbol] :> Simp[e^IntPart[m]*(e*x)^FracPart[m]*( 
(a*x^j + b*x^(j + n))^FracPart[p]/(x^(FracPart[m] + j*FracPart[p])*(a + b*x 
^n)^FracPart[p]))   Int[x^(m + j*p)*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; 
FreeQ[{a, b, c, d, e, j, m, n, p, q}, x] && EqQ[jn, j + n] &&  !IntegerQ[p] 
 && NeQ[b*c - a*d, 0] &&  !(EqQ[n, 1] && EqQ[j, 1])
 

Maple [A] (verified)

Time = 0.71 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.51

method	result	size
pseudoelliptic	\(\frac {-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {d \left (-\frac {c \sqrt {b x +a}}{d x +c}-\frac {\left (2 a d -3 b c \right ) \operatorname {arctanh}\left (\frac {d \sqrt {b x +a}}{\sqrt {d \left (a d -b c \right )}}\right )}{\sqrt {d \left (a d -b c \right )}}\right )}{a d -b c}}{c^{2}}\)	\(99\)
risch	\(-\frac {b x +a}{a \,c^{2} \sqrt {x^{2} \left (b x +a \right )}}-\frac {b \left (-\frac {\left (4 a d +b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{b c \sqrt {a}}-\frac {2 a \,d^{2} \left (-\frac {b c \sqrt {b x +a}}{2 \left (a d -b c \right ) \left (d \left (b x +a \right )-a d +b c \right )}-\frac {\left (4 a d -5 b c \right ) \operatorname {arctanh}\left (\frac {d \sqrt {b x +a}}{\sqrt {d \left (a d -b c \right )}}\right )}{2 \left (a d -b c \right ) \sqrt {d \left (a d -b c \right )}}\right )}{b c}\right ) \sqrt {b x +a}\, x}{c^{2} a \sqrt {x^{2} \left (b x +a \right )}}\)	\(192\)
default	\(-\frac {\sqrt {b x +a}\, \left (4 a^{\frac {7}{2}} \operatorname {arctanh}\left (\frac {d \sqrt {b x +a}}{\sqrt {d \left (a d -b c \right )}}\right ) d^{4} x^{2}-5 a^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {d \sqrt {b x +a}}{\sqrt {d \left (a d -b c \right )}}\right ) b c \,d^{3} x^{2}+4 a^{\frac {7}{2}} \operatorname {arctanh}\left (\frac {d \sqrt {b x +a}}{\sqrt {d \left (a d -b c \right )}}\right ) c \,d^{3} x -5 a^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {d \sqrt {b x +a}}{\sqrt {d \left (a d -b c \right )}}\right ) b \,c^{2} d^{2} x +2 a^{\frac {5}{2}} \sqrt {d \left (a d -b c \right )}\, \sqrt {b x +a}\, c \,d^{2} x -a^{\frac {3}{2}} \sqrt {d \left (a d -b c \right )}\, \sqrt {b x +a}\, b \,c^{2} d x -4 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {d \left (a d -b c \right )}\, a^{3} d^{3} x^{2}+3 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {d \left (a d -b c \right )}\, a^{2} b c \,d^{2} x^{2}+\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {d \left (a d -b c \right )}\, a \,b^{2} c^{2} d \,x^{2}+a^{\frac {5}{2}} \sqrt {d \left (a d -b c \right )}\, \sqrt {b x +a}\, c^{2} d -a^{\frac {3}{2}} \sqrt {d \left (a d -b c \right )}\, \sqrt {b x +a}\, b \,c^{3}-4 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {d \left (a d -b c \right )}\, a^{3} c \,d^{2} x +3 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {d \left (a d -b c \right )}\, a^{2} b \,c^{2} d x +\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right ) \sqrt {d \left (a d -b c \right )}\, a \,b^{2} c^{3} x \right )}{\sqrt {b \,x^{3}+a \,x^{2}}\, c^{3} a^{\frac {5}{2}} \left (a d -b c \right ) \left (d x +c \right ) \sqrt {d \left (a d -b c \right )}}\)	\(518\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 1256, normalized size of antiderivative = 6.44 \[ \int \frac {1}{x (c+d x)^2 \sqrt {a x^2+b x^3}} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/2*(((5*a^2*b*c*d^2 - 4*a^3*d^3)*x^3 + (5*a^2*b*c^2*d - 4*a^3*c*d^2)*x^2 
)*sqrt(-d/(b*c - a*d))*log((b*d*x^2 - (b*c - 2*a*d)*x + 2*sqrt(b*x^3 + a*x 
^2)*(b*c - a*d)*sqrt(-d/(b*c - a*d)))/(d*x^2 + c*x)) + ((b^2*c^2*d + 3*a*b 
*c*d^2 - 4*a^2*d^3)*x^3 + (b^2*c^3 + 3*a*b*c^2*d - 4*a^2*c*d^2)*x^2)*sqrt( 
a)*log((b*x^2 + 2*a*x + 2*sqrt(b*x^3 + a*x^2)*sqrt(a))/x^2) - 2*(a*b*c^3 - 
 a^2*c^2*d + (a*b*c^2*d - 2*a^2*c*d^2)*x)*sqrt(b*x^3 + a*x^2))/((a^2*b*c^4 
*d - a^3*c^3*d^2)*x^3 + (a^2*b*c^5 - a^3*c^4*d)*x^2), 1/2*(2*((5*a^2*b*c*d 
^2 - 4*a^3*d^3)*x^3 + (5*a^2*b*c^2*d - 4*a^3*c*d^2)*x^2)*sqrt(d/(b*c - a*d 
))*arctan(sqrt(b*x^3 + a*x^2)*sqrt(d/(b*c - a*d))/x) + ((b^2*c^2*d + 3*a*b 
*c*d^2 - 4*a^2*d^3)*x^3 + (b^2*c^3 + 3*a*b*c^2*d - 4*a^2*c*d^2)*x^2)*sqrt( 
a)*log((b*x^2 + 2*a*x + 2*sqrt(b*x^3 + a*x^2)*sqrt(a))/x^2) - 2*(a*b*c^3 - 
 a^2*c^2*d + (a*b*c^2*d - 2*a^2*c*d^2)*x)*sqrt(b*x^3 + a*x^2))/((a^2*b*c^4 
*d - a^3*c^3*d^2)*x^3 + (a^2*b*c^5 - a^3*c^4*d)*x^2), -1/2*(2*((b^2*c^2*d 
+ 3*a*b*c*d^2 - 4*a^2*d^3)*x^3 + (b^2*c^3 + 3*a*b*c^2*d - 4*a^2*c*d^2)*x^2 
)*sqrt(-a)*arctan(sqrt(b*x^3 + a*x^2)*sqrt(-a)/(b*x^2 + a*x)) - ((5*a^2*b* 
c*d^2 - 4*a^3*d^3)*x^3 + (5*a^2*b*c^2*d - 4*a^3*c*d^2)*x^2)*sqrt(-d/(b*c - 
 a*d))*log((b*d*x^2 - (b*c - 2*a*d)*x + 2*sqrt(b*x^3 + a*x^2)*(b*c - a*d)* 
sqrt(-d/(b*c - a*d)))/(d*x^2 + c*x)) + 2*(a*b*c^3 - a^2*c^2*d + (a*b*c^2*d 
 - 2*a^2*c*d^2)*x)*sqrt(b*x^3 + a*x^2))/((a^2*b*c^4*d - a^3*c^3*d^2)*x^3 + 
 (a^2*b*c^5 - a^3*c^4*d)*x^2), -(((b^2*c^2*d + 3*a*b*c*d^2 - 4*a^2*d^3)...
 

Sympy [F]

\[ \int \frac {1}{x (c+d x)^2 \sqrt {a x^2+b x^3}} \, dx=\int \frac {1}{x \sqrt {x^{2} \left (a + b x\right )} \left (c + d x\right )^{2}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {1}{x (c+d x)^2 \sqrt {a x^2+b x^3}} \, dx=\int { \frac {1}{\sqrt {b x^{3} + a x^{2}} {\left (d x + c\right )}^{2} x} \,d x } \] Input:

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

Output:

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

Giac [F(-1)]

Timed out. \[ \int \frac {1}{x (c+d x)^2 \sqrt {a x^2+b x^3}} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x (c+d x)^2 \sqrt {a x^2+b x^3}} \, dx=\int \frac {1}{x\,\sqrt {b\,x^3+a\,x^2}\,{\left (c+d\,x\right )}^2} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 743, normalized size of antiderivative = 3.81 \[ \int \frac {1}{x (c+d x)^2 \sqrt {a x^2+b x^3}} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 8*sqrt(d)*sqrt( - a*d + b*c)*atan((sqrt(a + b*x)*d)/(sqrt(d)*sqrt( - a 
*d + b*c)))*a**3*c*d**2*x - 8*sqrt(d)*sqrt( - a*d + b*c)*atan((sqrt(a + b* 
x)*d)/(sqrt(d)*sqrt( - a*d + b*c)))*a**3*d**3*x**2 + 10*sqrt(d)*sqrt( - a* 
d + b*c)*atan((sqrt(a + b*x)*d)/(sqrt(d)*sqrt( - a*d + b*c)))*a**2*b*c**2* 
d*x + 10*sqrt(d)*sqrt( - a*d + b*c)*atan((sqrt(a + b*x)*d)/(sqrt(d)*sqrt( 
- a*d + b*c)))*a**2*b*c*d**2*x**2 - 2*sqrt(a + b*x)*a**3*c**2*d**2 - 4*sqr 
t(a + b*x)*a**3*c*d**3*x + 4*sqrt(a + b*x)*a**2*b*c**3*d + 6*sqrt(a + b*x) 
*a**2*b*c**2*d**2*x - 2*sqrt(a + b*x)*a*b**2*c**4 - 2*sqrt(a + b*x)*a*b**2 
*c**3*d*x - 4*sqrt(a)*log(sqrt(a + b*x) - sqrt(a))*a**3*c*d**3*x - 4*sqrt( 
a)*log(sqrt(a + b*x) - sqrt(a))*a**3*d**4*x**2 + 7*sqrt(a)*log(sqrt(a + b* 
x) - sqrt(a))*a**2*b*c**2*d**2*x + 7*sqrt(a)*log(sqrt(a + b*x) - sqrt(a))* 
a**2*b*c*d**3*x**2 - 2*sqrt(a)*log(sqrt(a + b*x) - sqrt(a))*a*b**2*c**3*d* 
x - 2*sqrt(a)*log(sqrt(a + b*x) - sqrt(a))*a*b**2*c**2*d**2*x**2 - sqrt(a) 
*log(sqrt(a + b*x) - sqrt(a))*b**3*c**4*x - sqrt(a)*log(sqrt(a + b*x) - sq 
rt(a))*b**3*c**3*d*x**2 + 4*sqrt(a)*log(sqrt(a + b*x) + sqrt(a))*a**3*c*d* 
*3*x + 4*sqrt(a)*log(sqrt(a + b*x) + sqrt(a))*a**3*d**4*x**2 - 7*sqrt(a)*l 
og(sqrt(a + b*x) + sqrt(a))*a**2*b*c**2*d**2*x - 7*sqrt(a)*log(sqrt(a + b* 
x) + sqrt(a))*a**2*b*c*d**3*x**2 + 2*sqrt(a)*log(sqrt(a + b*x) + sqrt(a))* 
a*b**2*c**3*d*x + 2*sqrt(a)*log(sqrt(a + b*x) + sqrt(a))*a*b**2*c**2*d**2* 
x**2 + sqrt(a)*log(sqrt(a + b*x) + sqrt(a))*b**3*c**4*x + sqrt(a)*log(s...