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

Optimal result

Integrand size = 24, antiderivative size = 137 \[ \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 c (b B-A c) \sqrt {x}}{b^3 \sqrt {b x+c x^2}}-\frac {A \sqrt {b x+c x^2}}{2 b^2 x^{5/2}}-\frac {(4 b B-7 A c) \sqrt {b x+c x^2}}{4 b^3 x^{3/2}}+\frac {3 c (4 b B-5 A c) \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{4 b^{7/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.77 \[ \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^{3/2}} \, dx=\frac {\sqrt {b} \left (-4 b B x (b+3 c x)+A \left (-2 b^2+5 b c x+15 c^2 x^2\right )\right )+3 c (4 b B-5 A c) x^2 \sqrt {b+c x} \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )}{4 b^{7/2} x^{3/2} \sqrt {x (b+c x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.96, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1220, 1135, 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.

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

\(\Big \downarrow \) 1220

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

\(\Big \downarrow \) 1135

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

\(\Big \downarrow \) 1132

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

\(\Big \downarrow \) 1136

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

\(\Big \downarrow \) 220

\(\displaystyle \frac {(4 b B-5 A c) \left (-\frac {3 c \left (\frac {2 \sqrt {x}}{b \sqrt {b x+c x^2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{b^{3/2}}\right )}{2 b}-\frac {1}{b \sqrt {x} \sqrt {b x+c x^2}}\right )}{4 b}-\frac {A}{2 b x^{3/2} \sqrt {b x+c x^2}}\)

Input:

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

Output:

-1/2*A/(b*x^(3/2)*Sqrt[b*x + c*x^2]) + ((4*b*B - 5*A*c)*(-(1/(b*Sqrt[x]*Sq 
rt[b*x + c*x^2])) - (3*c*((2*Sqrt[x])/(b*Sqrt[b*x + c*x^2]) - (2*ArcTanh[S 
qrt[b*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/b^(3/2)))/(2*b)))/(4*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 1135

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[(-e)*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/((m + p + 1)*(2* 
c*d - b*e))), x] + Simp[c*((m + 2*p + 2)/((m + p + 1)*(2*c*d - b*e)))   Int 
[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, 
 x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, 0] && NeQ[m + p + 1, 0] && I 
ntegerQ[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 = 1.03 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.80

method	result	size
risch	\(-\frac {\left (c x +b \right ) \left (-7 A c x +4 B b x +2 A b \right )}{4 b^{3} x^{\frac {3}{2}} \sqrt {x \left (c x +b \right )}}+\frac {c \left (-\frac {2 \left (-8 A c +8 B b \right )}{\sqrt {c x +b}}-\frac {2 \left (15 A c -12 B b \right ) \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )}{\sqrt {b}}\right ) \sqrt {c x +b}\, \sqrt {x}}{8 b^{3} \sqrt {x \left (c x +b \right )}}\)	\(109\)
default	\(-\frac {\sqrt {x \left (c x +b \right )}\, \left (15 A \sqrt {c x +b}\, \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) c^{2} x^{2}-12 B \sqrt {c x +b}\, \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) b c \,x^{2}+4 B \,b^{\frac {5}{2}} x +12 B \,b^{\frac {3}{2}} c \,x^{2}+2 A \,b^{\frac {5}{2}}-5 A \,b^{\frac {3}{2}} c x -15 A \sqrt {b}\, c^{2} x^{2}\right )}{4 x^{\frac {5}{2}} \left (c x +b \right ) b^{\frac {7}{2}}}\)	\(124\)

Input:

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

Output:

-1/4*(c*x+b)*(-7*A*c*x+4*B*b*x+2*A*b)/b^3/x^(3/2)/(x*(c*x+b))^(1/2)+1/8/b^ 
3*c*(-2*(-8*A*c+8*B*b)/(c*x+b)^(1/2)-2*(15*A*c-12*B*b)/b^(1/2)*arctanh((c* 
x+b)^(1/2)/b^(1/2)))*(c*x+b)^(1/2)*x^(1/2)/(x*(c*x+b))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 307, normalized size of antiderivative = 2.24 \[ \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^{3/2}} \, dx=\left [-\frac {3 \, {\left ({\left (4 \, B b c^{2} - 5 \, A c^{3}\right )} x^{4} + {\left (4 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{3}\right )} \sqrt {b} \log \left (-\frac {c x^{2} + 2 \, b x - 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, {\left (2 \, A b^{3} + 3 \, {\left (4 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{2} + {\left (4 \, B b^{3} - 5 \, A b^{2} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{8 \, {\left (b^{4} c x^{4} + b^{5} x^{3}\right )}}, -\frac {3 \, {\left ({\left (4 \, B b c^{2} - 5 \, A c^{3}\right )} x^{4} + {\left (4 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (2 \, A b^{3} + 3 \, {\left (4 \, B b^{2} c - 5 \, A b c^{2}\right )} x^{2} + {\left (4 \, B b^{3} - 5 \, A b^{2} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{4 \, {\left (b^{4} c x^{4} + b^{5} x^{3}\right )}}\right ] \] Input:

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

Output:

[-1/8*(3*((4*B*b*c^2 - 5*A*c^3)*x^4 + (4*B*b^2*c - 5*A*b*c^2)*x^3)*sqrt(b) 
*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*(2*A* 
b^3 + 3*(4*B*b^2*c - 5*A*b*c^2)*x^2 + (4*B*b^3 - 5*A*b^2*c)*x)*sqrt(c*x^2 
+ b*x)*sqrt(x))/(b^4*c*x^4 + b^5*x^3), -1/4*(3*((4*B*b*c^2 - 5*A*c^3)*x^4 
+ (4*B*b^2*c - 5*A*b*c^2)*x^3)*sqrt(-b)*arctan(sqrt(c*x^2 + b*x)*sqrt(-b)/ 
(b*sqrt(x))) + (2*A*b^3 + 3*(4*B*b^2*c - 5*A*b*c^2)*x^2 + (4*B*b^3 - 5*A*b 
^2*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^4*c*x^4 + b^5*x^3)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.91 \[ \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {3 \, {\left (4 \, B b c - 5 \, A c^{2}\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{4 \, \sqrt {-b} b^{3}} - \frac {2 \, {\left (B b c - A c^{2}\right )}}{\sqrt {c x + b} b^{3}} - \frac {4 \, {\left (c x + b\right )}^{\frac {3}{2}} B b c - 4 \, \sqrt {c x + b} B b^{2} c - 7 \, {\left (c x + b\right )}^{\frac {3}{2}} A c^{2} + 9 \, \sqrt {c x + b} A b c^{2}}{4 \, b^{3} c^{2} x^{2}} \] Input:

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

Output:

-3/4*(4*B*b*c - 5*A*c^2)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^3) - 2 
*(B*b*c - A*c^2)/(sqrt(c*x + b)*b^3) - 1/4*(4*(c*x + b)^(3/2)*B*b*c - 4*sq 
rt(c*x + b)*B*b^2*c - 7*(c*x + b)^(3/2)*A*c^2 + 9*sqrt(c*x + b)*A*b*c^2)/( 
b^3*c^2*x^2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.23 \[ \int \frac {A+B x}{x^{3/2} \left (b x+c x^2\right )^{3/2}} \, dx=\frac {15 \sqrt {b}\, \sqrt {c x +b}\, \mathrm {log}\left (\sqrt {c x +b}-\sqrt {b}\right ) a \,c^{2} x^{2}-12 \sqrt {b}\, \sqrt {c x +b}\, \mathrm {log}\left (\sqrt {c x +b}-\sqrt {b}\right ) b^{2} c \,x^{2}-15 \sqrt {b}\, \sqrt {c x +b}\, \mathrm {log}\left (\sqrt {c x +b}+\sqrt {b}\right ) a \,c^{2} x^{2}+12 \sqrt {b}\, \sqrt {c x +b}\, \mathrm {log}\left (\sqrt {c x +b}+\sqrt {b}\right ) b^{2} c \,x^{2}-4 a \,b^{3}+10 a \,b^{2} c x +30 a b \,c^{2} x^{2}-8 b^{4} x -24 b^{3} c \,x^{2}}{8 \sqrt {c x +b}\, b^{4} x^{2}} \] Input:

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

Output:

(15*sqrt(b)*sqrt(b + c*x)*log(sqrt(b + c*x) - sqrt(b))*a*c**2*x**2 - 12*sq 
rt(b)*sqrt(b + c*x)*log(sqrt(b + c*x) - sqrt(b))*b**2*c*x**2 - 15*sqrt(b)* 
sqrt(b + c*x)*log(sqrt(b + c*x) + sqrt(b))*a*c**2*x**2 + 12*sqrt(b)*sqrt(b 
 + c*x)*log(sqrt(b + c*x) + sqrt(b))*b**2*c*x**2 - 4*a*b**3 + 10*a*b**2*c* 
x + 30*a*b*c**2*x**2 - 8*b**4*x - 24*b**3*c*x**2)/(8*sqrt(b + c*x)*b**4*x* 
*2)