\(\int \frac {(A+B x) (b x+c x^2)^{3/2}}{x^3} \, dx\) [119]

Optimal result

Integrand size = 22, antiderivative size = 105 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^3} \, dx=\frac {3}{4} (b B+4 A c) \sqrt {b x+c x^2}-\frac {2 A \left (b x+c x^2\right )^{3/2}}{x^2}+\frac {B \left (b x+c x^2\right )^{3/2}}{2 x}+\frac {3 b (b B+4 A c) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 \sqrt {c}} \] Output:

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

Mathematica [A] (verified)

Time = 0.38 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.96 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^3} \, dx=\frac {\sqrt {x (b+c x)} \left (B x (5 b+2 c x)+A (-8 b+4 c x)+\frac {6 b (b B+4 A c) \sqrt {x} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )}{\sqrt {c} \sqrt {b+c x}}\right )}{4 x} \] Input:

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

Output:

(Sqrt[x*(b + c*x)]*(B*x*(5*b + 2*c*x) + A*(-8*b + 4*c*x) + (6*b*(b*B + 4*A 
*c)*Sqrt[x]*ArcTanh[(Sqrt[c]*Sqrt[x])/(-Sqrt[b] + Sqrt[b + c*x])])/(Sqrt[c 
]*Sqrt[b + c*x])))/(4*x)
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1220, 1131, 1131, 1091, 219}

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.

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

\(\Big \downarrow \) 1220

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

\(\Big \downarrow \) 1131

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

\(\Big \downarrow \) 1131

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

\(\Big \downarrow \) 1091

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 219

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

rule 1091

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2   Subst[Int[1/(1 
 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
 

rule 1131

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

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 = 0.89 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.70

method	result	size
pseudoelliptic	\(\frac {3 b x \left (A c +\frac {B b}{4}\right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )-2 \left (-\frac {\left (\frac {B x}{2}+A \right ) x \,c^{\frac {3}{2}}}{2}+b \sqrt {c}\, \left (-\frac {5 B x}{8}+A \right )\right ) \sqrt {x \left (c x +b \right )}}{\sqrt {c}\, x}\)	\(73\)
risch	\(-\frac {\left (c x +b \right ) \left (-2 B c \,x^{2}-4 A c x -5 B b x +8 A b \right )}{4 \sqrt {x \left (c x +b \right )}}+\frac {3 \left (4 A c +B b \right ) b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 \sqrt {c}}\)	\(78\)
default	\(A \left (-\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{b \,x^{3}}+\frac {4 c \left (\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{b \,x^{2}}-\frac {6 c \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3}+\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2}\right )}{b}\right )}{b}\right )+B \left (\frac {2 \left (c \,x^{2}+b x \right )^{\frac {5}{2}}}{b \,x^{2}}-\frac {6 c \left (\frac {\left (c \,x^{2}+b x \right )^{\frac {3}{2}}}{3}+\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x}}{4 c}-\frac {b^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{8 c^{\frac {3}{2}}}\right )}{2}\right )}{b}\right )\)	\(228\)

Input:

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

Output:

3/c^(1/2)*(b*x*(A*c+1/4*B*b)*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2))-2/3*(-1/ 
2*(1/2*B*x+A)*x*c^(3/2)+b*c^(1/2)*(-5/8*B*x+A))*(x*(c*x+b))^(1/2))/x
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.78 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^3} \, dx=\left [\frac {3 \, {\left (B b^{2} + 4 \, A b c\right )} \sqrt {c} x \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (2 \, B c^{2} x^{2} - 8 \, A b c + {\left (5 \, B b c + 4 \, A c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{8 \, c x}, -\frac {3 \, {\left (B b^{2} + 4 \, A b c\right )} \sqrt {-c} x \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x + b}\right ) - {\left (2 \, B c^{2} x^{2} - 8 \, A b c + {\left (5 \, B b c + 4 \, A c^{2}\right )} x\right )} \sqrt {c x^{2} + b x}}{4 \, c x}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.23 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^3} \, dx=\frac {3 \, B b^{2} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, \sqrt {c}} + \frac {3}{2} \, A b \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + \frac {3}{4} \, \sqrt {c x^{2} + b x} B b + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} B}{2 \, x} - \frac {3 \, \sqrt {c x^{2} + b x} A b}{x} + \frac {{\left (c x^{2} + b x\right )}^{\frac {3}{2}} A}{x^{2}} \] Input:

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

Output:

3/8*B*b^2*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/sqrt(c) + 3/2*A*b*s 
qrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 3/4*sqrt(c*x^2 + b*x 
)*B*b + 1/2*(c*x^2 + b*x)^(3/2)*B/x - 3*sqrt(c*x^2 + b*x)*A*b/x + (c*x^2 + 
 b*x)^(3/2)*A/x^2
 

Giac [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.02 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^3} \, dx=\frac {2 \, A b^{2}}{\sqrt {c} x - \sqrt {c x^{2} + b x}} + \frac {1}{4} \, {\left (2 \, B c x + \frac {5 \, B b c + 4 \, A c^{2}}{c}\right )} \sqrt {c x^{2} + b x} - \frac {3 \, {\left (B b^{2} + 4 \, A b c\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{8 \, \sqrt {c}} \] Input:

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

Output:

2*A*b^2/(sqrt(c)*x - sqrt(c*x^2 + b*x)) + 1/4*(2*B*c*x + (5*B*b*c + 4*A*c^ 
2)/c)*sqrt(c*x^2 + b*x) - 3/8*(B*b^2 + 4*A*b*c)*log(abs(2*(sqrt(c)*x - sqr 
t(c*x^2 + b*x))*sqrt(c) + b))/sqrt(c)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.30 \[ \int \frac {(A+B x) \left (b x+c x^2\right )^{3/2}}{x^3} \, dx=\frac {-8 \sqrt {x}\, \sqrt {c x +b}\, a b c +4 \sqrt {x}\, \sqrt {c x +b}\, a \,c^{2} x +5 \sqrt {x}\, \sqrt {c x +b}\, b^{2} c x +2 \sqrt {x}\, \sqrt {c x +b}\, b \,c^{2} x^{2}+12 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) a b c x +3 \sqrt {c}\, \mathrm {log}\left (\frac {\sqrt {c x +b}+\sqrt {x}\, \sqrt {c}}{\sqrt {b}}\right ) b^{3} x -9 \sqrt {c}\, a b c x -\sqrt {c}\, b^{3} x}{4 c x} \] Input:

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

Output:

( - 8*sqrt(x)*sqrt(b + c*x)*a*b*c + 4*sqrt(x)*sqrt(b + c*x)*a*c**2*x + 5*s 
qrt(x)*sqrt(b + c*x)*b**2*c*x + 2*sqrt(x)*sqrt(b + c*x)*b*c**2*x**2 + 12*s 
qrt(c)*log((sqrt(b + c*x) + sqrt(x)*sqrt(c))/sqrt(b))*a*b*c*x + 3*sqrt(c)* 
log((sqrt(b + c*x) + sqrt(x)*sqrt(c))/sqrt(b))*b**3*x - 9*sqrt(c)*a*b*c*x 
- sqrt(c)*b**3*x)/(4*c*x)