\(\int \frac {A+B x}{\sqrt {x} (b x+c x^2)^{3/2}} \, dx\) [217]

Optimal result

Integrand size = 24, antiderivative size = 97 \[ \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )^{3/2}} \, dx=\frac {2 (b B-A c) \sqrt {x}}{b^2 \sqrt {b x+c x^2}}-\frac {A \sqrt {b x+c x^2}}{b^2 x^{3/2}}-\frac {(2 b B-3 A c) \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{b^{5/2}} \] Output:

2*(-A*c+B*b)*x^(1/2)/b^2/(c*x^2+b*x)^(1/2)-A*(c*x^2+b*x)^(1/2)/b^2/x^(3/2) 
-(-3*A*c+2*B*b)*arctanh((c*x^2+b*x)^(1/2)/b^(1/2)/x^(1/2))/b^(5/2)
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.84 \[ \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )^{3/2}} \, dx=\frac {\sqrt {b} (2 b B x-A (b+3 c x))-(2 b B-3 A c) x \sqrt {b+c x} \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )}{b^{5/2} \sqrt {x} \sqrt {x (b+c x)}} \] Input:

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

Output:

(Sqrt[b]*(2*b*B*x - A*(b + 3*c*x)) - (2*b*B - 3*A*c)*x*Sqrt[b + c*x]*ArcTa 
nh[Sqrt[b + c*x]/Sqrt[b]])/(b^(5/2)*Sqrt[x]*Sqrt[x*(b + c*x)])
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1220, 1132, 1136, 220}

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.

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

\(\Big \downarrow \) 1220

\(\displaystyle \frac {(2 b B-3 A c) \int \frac {\sqrt {x}}{\left (c x^2+b x\right )^{3/2}}dx}{2 b}-\frac {A}{b \sqrt {x} \sqrt {b x+c x^2}}\)

\(\Big \downarrow \) 1132

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

\(\Big \downarrow \) 1136

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

\(\Big \downarrow \) 220

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

Input:

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

Output:

-(A/(b*Sqrt[x]*Sqrt[b*x + c*x^2])) + ((2*b*B - 3*A*c)*((2*Sqrt[x])/(b*Sqrt 
[b*x + c*x^2]) - (2*ArcTanh[Sqrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/b^(3/2)) 
)/(2*b)
 

Defintions of rubi rules used

rule 220

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 
1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && 
 (LtQ[a, 0] || GtQ[b, 0])
 

rule 1132

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(2*c*d - b*e)*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/(e*(p + 
 1)*(b^2 - 4*a*c))), x] - Simp[(2*c*d - b*e)*((m + 2*p + 2)/((p + 1)*(b^2 - 
 4*a*c)))   Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1), x], x] /; Free 
Q[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && LtQ 
[0, m, 1] && IntegerQ[2*p]
 

rule 1136

Int[1/(Sqrt[(d_.) + (e_.)*(x_)]*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x 
_Symbol] :> Simp[2*e   Subst[Int[1/(2*c*d - b*e + e^2*x^2), x], x, Sqrt[a + 
 b*x + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 
- b*d*e + a*e^2, 0]
 

rule 1220

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + b*x + c*x 
^2)^(p + 1)/((2*c*d - b*e)*(m + p + 1))), x] + Simp[(m*(g*(c*d - b*e) + c*e 
*f) + e*(p + 1)*(2*c*f - b*g))/(e*(2*c*d - b*e)*(m + p + 1))   Int[(d + e*x 
)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, 
 x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p + 1, 0 
]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p + 1, 0 
]
 

Maple [A] (verified)

Time = 1.06 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.97

method	result	size
default	\(\frac {\sqrt {x \left (c x +b \right )}\, \left (3 A \sqrt {c x +b}\, \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) c x -2 B \sqrt {c x +b}\, \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) b x +2 B \,b^{\frac {3}{2}} x -3 A \sqrt {b}\, c x -A \,b^{\frac {3}{2}}\right )}{x^{\frac {3}{2}} \left (c x +b \right ) b^{\frac {5}{2}}}\)	\(94\)
risch	\(-\frac {A \left (c x +b \right )}{b^{2} \sqrt {x}\, \sqrt {x \left (c x +b \right )}}-\frac {\left (-\frac {2 \left (-2 A c +2 B b \right )}{\sqrt {c x +b}}-\frac {2 \left (3 A c -2 B b \right ) \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )}{\sqrt {b}}\right ) \sqrt {c x +b}\, \sqrt {x}}{2 b^{2} \sqrt {x \left (c x +b \right )}}\)	\(94\)

Input:

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

Output:

1/x^(3/2)*(x*(c*x+b))^(1/2)*(3*A*(c*x+b)^(1/2)*arctanh((c*x+b)^(1/2)/b^(1/ 
2))*c*x-2*B*(c*x+b)^(1/2)*arctanh((c*x+b)^(1/2)/b^(1/2))*b*x+2*B*b^(3/2)*x 
-3*A*b^(1/2)*c*x-A*b^(3/2))/(c*x+b)/b^(5/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 251, normalized size of antiderivative = 2.59 \[ \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )^{3/2}} \, dx=\left [-\frac {{\left ({\left (2 \, B b c - 3 \, A c^{2}\right )} x^{3} + {\left (2 \, B b^{2} - 3 \, A b c\right )} x^{2}\right )} \sqrt {b} \log \left (-\frac {c x^{2} + 2 \, b x + 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, {\left (A b^{2} - {\left (2 \, B b^{2} - 3 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{2 \, {\left (b^{3} c x^{3} + b^{4} x^{2}\right )}}, \frac {{\left ({\left (2 \, B b c - 3 \, A c^{2}\right )} x^{3} + {\left (2 \, B b^{2} - 3 \, A b c\right )} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-b}}{b \sqrt {x}}\right ) - {\left (A b^{2} - {\left (2 \, B b^{2} - 3 \, A b c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{b^{3} c x^{3} + b^{4} x^{2}}\right ] \] Input:

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

Output:

[-1/2*(((2*B*b*c - 3*A*c^2)*x^3 + (2*B*b^2 - 3*A*b*c)*x^2)*sqrt(b)*log(-(c 
*x^2 + 2*b*x + 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*(A*b^2 - (2*B 
*b^2 - 3*A*b*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^3*c*x^3 + b^4*x^2), (((2* 
B*b*c - 3*A*c^2)*x^3 + (2*B*b^2 - 3*A*b*c)*x^2)*sqrt(-b)*arctan(sqrt(c*x^2 
 + b*x)*sqrt(-b)/(b*sqrt(x))) - (A*b^2 - (2*B*b^2 - 3*A*b*c)*x)*sqrt(c*x^2 
 + b*x)*sqrt(x))/(b^3*c*x^3 + b^4*x^2)]
 

Sympy [F]

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

integrate((B*x+A)/x**(1/2)/(c*x**2+b*x)**(3/2),x)
 

Output:

Integral((A + B*x)/(sqrt(x)*(x*(b + c*x))**(3/2)), x)
 

Maxima [F]

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

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

Output:

integrate((B*x + A)/((c*x^2 + b*x)^(3/2)*sqrt(x)), x)
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.90 \[ \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )^{3/2}} \, dx=\frac {{\left (2 \, B b - 3 \, A c\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{2}} + \frac {2 \, {\left (c x + b\right )} B b - 2 \, B b^{2} - 3 \, {\left (c x + b\right )} A c + 2 \, A b c}{{\left ({\left (c x + b\right )}^{\frac {3}{2}} - \sqrt {c x + b} b\right )} b^{2}} \] Input:

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

Output:

(2*B*b - 3*A*c)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^2) + (2*(c*x + 
b)*B*b - 2*B*b^2 - 3*(c*x + b)*A*c + 2*A*b*c)/(((c*x + b)^(3/2) - sqrt(c*x 
 + b)*b)*b^2)
 

Mupad [F(-1)]

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

int((A + B*x)/(x^(1/2)*(b*x + c*x^2)^(3/2)),x)
 

Output:

int((A + B*x)/(x^(1/2)*(b*x + c*x^2)^(3/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.37 \[ \int \frac {A+B x}{\sqrt {x} \left (b x+c x^2\right )^{3/2}} \, dx=\frac {-3 \sqrt {b}\, \sqrt {c x +b}\, \mathrm {log}\left (\sqrt {c x +b}-\sqrt {b}\right ) a c x +2 \sqrt {b}\, \sqrt {c x +b}\, \mathrm {log}\left (\sqrt {c x +b}-\sqrt {b}\right ) b^{2} x +3 \sqrt {b}\, \sqrt {c x +b}\, \mathrm {log}\left (\sqrt {c x +b}+\sqrt {b}\right ) a c x -2 \sqrt {b}\, \sqrt {c x +b}\, \mathrm {log}\left (\sqrt {c x +b}+\sqrt {b}\right ) b^{2} x -2 a \,b^{2}-6 a b c x +4 b^{3} x}{2 \sqrt {c x +b}\, b^{3} x} \] Input:

int((B*x+A)/x^(1/2)/(c*x^2+b*x)^(3/2),x)
 

Output:

( - 3*sqrt(b)*sqrt(b + c*x)*log(sqrt(b + c*x) - sqrt(b))*a*c*x + 2*sqrt(b) 
*sqrt(b + c*x)*log(sqrt(b + c*x) - sqrt(b))*b**2*x + 3*sqrt(b)*sqrt(b + c* 
x)*log(sqrt(b + c*x) + sqrt(b))*a*c*x - 2*sqrt(b)*sqrt(b + c*x)*log(sqrt(b 
 + c*x) + sqrt(b))*b**2*x - 2*a*b**2 - 6*a*b*c*x + 4*b**3*x)/(2*sqrt(b + c 
*x)*b**3*x)