\(\int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{9/2}} \, dx\) [179]

Optimal result

Integrand size = 24, antiderivative size = 138 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{9/2}} \, dx=-\frac {A \sqrt {b x+c x^2}}{3 x^{7/2}}-\frac {(6 b B+A c) \sqrt {b x+c x^2}}{12 b x^{5/2}}-\frac {c (2 b B-A c) \sqrt {b x+c x^2}}{8 b^2 x^{3/2}}+\frac {c^2 (2 b B-A c) \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{8 b^{5/2}} \] Output:

-1/3*A*(c*x^2+b*x)^(1/2)/x^(7/2)-1/12*(A*c+6*B*b)*(c*x^2+b*x)^(1/2)/b/x^(5 
/2)-1/8*c*(-A*c+2*B*b)*(c*x^2+b*x)^(1/2)/b^2/x^(3/2)+1/8*c^2*(-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 = 0.23 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.84 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{9/2}} \, dx=-\frac {\sqrt {x (b+c x)} \left (\sqrt {b} \sqrt {b+c x} \left (6 b B x (2 b+c x)+A \left (8 b^2+2 b c x-3 c^2 x^2\right )\right )+3 c^2 (-2 b B+A c) x^3 \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )\right )}{24 b^{5/2} x^{7/2} \sqrt {b+c x}} \] Input:

Integrate[((A + B*x)*Sqrt[b*x + c*x^2])/x^(9/2),x]
 

Output:

-1/24*(Sqrt[x*(b + c*x)]*(Sqrt[b]*Sqrt[b + c*x]*(6*b*B*x*(2*b + c*x) + A*( 
8*b^2 + 2*b*c*x - 3*c^2*x^2)) + 3*c^2*(-2*b*B + A*c)*x^3*ArcTanh[Sqrt[b + 
c*x]/Sqrt[b]]))/(b^(5/2)*x^(7/2)*Sqrt[b + c*x])
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {1220, 1130, 1135, 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.

\(\displaystyle \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{9/2}} \, dx\)

\(\Big \downarrow \) 1220

\(\displaystyle \frac {(2 b B-A c) \int \frac {\sqrt {c x^2+b x}}{x^{7/2}}dx}{2 b}-\frac {A \left (b x+c x^2\right )^{3/2}}{3 b x^{9/2}}\)

\(\Big \downarrow \) 1130

\(\displaystyle \frac {(2 b B-A c) \left (\frac {1}{4} c \int \frac {1}{x^{3/2} \sqrt {c x^2+b x}}dx-\frac {\sqrt {b x+c x^2}}{2 x^{5/2}}\right )}{2 b}-\frac {A \left (b x+c x^2\right )^{3/2}}{3 b x^{9/2}}\)

\(\Big \downarrow \) 1135

\(\displaystyle \frac {(2 b B-A c) \left (\frac {1}{4} c \left (-\frac {c \int \frac {1}{\sqrt {x} \sqrt {c x^2+b x}}dx}{2 b}-\frac {\sqrt {b x+c x^2}}{b x^{3/2}}\right )-\frac {\sqrt {b x+c x^2}}{2 x^{5/2}}\right )}{2 b}-\frac {A \left (b x+c x^2\right )^{3/2}}{3 b x^{9/2}}\)

\(\Big \downarrow \) 1136

\(\displaystyle \frac {(2 b B-A c) \left (\frac {1}{4} c \left (-\frac {c \int \frac {1}{\frac {c x^2+b x}{x}-b}d\frac {\sqrt {c x^2+b x}}{\sqrt {x}}}{b}-\frac {\sqrt {b x+c x^2}}{b x^{3/2}}\right )-\frac {\sqrt {b x+c x^2}}{2 x^{5/2}}\right )}{2 b}-\frac {A \left (b x+c x^2\right )^{3/2}}{3 b x^{9/2}}\)

\(\Big \downarrow \) 220

\(\displaystyle \frac {(2 b B-A c) \left (\frac {1}{4} c \left (\frac {c \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{b^{3/2}}-\frac {\sqrt {b x+c x^2}}{b x^{3/2}}\right )-\frac {\sqrt {b x+c x^2}}{2 x^{5/2}}\right )}{2 b}-\frac {A \left (b x+c x^2\right )^{3/2}}{3 b x^{9/2}}\)

Input:

Int[((A + B*x)*Sqrt[b*x + c*x^2])/x^(9/2),x]
 

Output:

-1/3*(A*(b*x + c*x^2)^(3/2))/(b*x^(9/2)) + ((2*b*B - A*c)*(-1/2*Sqrt[b*x + 
 c*x^2]/x^(5/2) + (c*(-(Sqrt[b*x + c*x^2]/(b*x^(3/2))) + (c*ArcTanh[Sqrt[b 
*x + c*x^2]/(Sqrt[b]*Sqrt[x])])/b^(3/2)))/4))/(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 1130

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 + p + 1))), x] 
- Simp[c*(p/(e^2*(m + p + 1)))   Int[(d + e*x)^(m + 2)*(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] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] & 
& 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 = 0.96 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.78

method	result	size
risch	\(-\frac {\left (c x +b \right ) \left (-3 A \,c^{2} x^{2}+6 x^{2} B b c +2 A b c x +12 x B \,b^{2}+8 b^{2} A \right )}{24 x^{\frac {5}{2}} b^{2} \sqrt {x \left (c x +b \right )}}-\frac {c^{2} \left (A c -2 B b \right ) \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) \sqrt {c x +b}\, \sqrt {x}}{8 b^{\frac {5}{2}} \sqrt {x \left (c x +b \right )}}\)	\(108\)
default	\(-\frac {\sqrt {x \left (c x +b \right )}\, \left (3 A \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) c^{3} x^{3}-6 B \,\operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) b \,c^{2} x^{3}-3 A \,c^{2} x^{2} \sqrt {c x +b}\, \sqrt {b}+6 B \,b^{\frac {3}{2}} c \,x^{2} \sqrt {c x +b}+2 A \,b^{\frac {3}{2}} c x \sqrt {c x +b}+12 B \,b^{\frac {5}{2}} x \sqrt {c x +b}+8 A \,b^{\frac {5}{2}} \sqrt {c x +b}\right )}{24 b^{\frac {5}{2}} x^{\frac {7}{2}} \sqrt {c x +b}}\)	\(147\)

Input:

int((B*x+A)*(c*x^2+b*x)^(1/2)/x^(9/2),x,method=_RETURNVERBOSE)
 

Output:

-1/24*(c*x+b)*(-3*A*c^2*x^2+6*B*b*c*x^2+2*A*b*c*x+12*B*b^2*x+8*A*b^2)/x^(5 
/2)/b^2/(x*(c*x+b))^(1/2)-1/8*c^2*(A*c-2*B*b)/b^(5/2)*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 = 0.09 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.75 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{9/2}} \, dx=\left [-\frac {3 \, {\left (2 \, B b c^{2} - A c^{3}\right )} \sqrt {b} x^{4} \log \left (-\frac {c x^{2} + 2 \, b x - 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, {\left (8 \, A b^{3} + 3 \, {\left (2 \, B b^{2} c - A b c^{2}\right )} x^{2} + 2 \, {\left (6 \, B b^{3} + A b^{2} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{48 \, b^{3} x^{4}}, -\frac {3 \, {\left (2 \, B b c^{2} - A c^{3}\right )} \sqrt {-b} x^{4} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (8 \, A b^{3} + 3 \, {\left (2 \, B b^{2} c - A b c^{2}\right )} x^{2} + 2 \, {\left (6 \, B b^{3} + A b^{2} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{24 \, b^{3} x^{4}}\right ] \] Input:

integrate((B*x+A)*(c*x^2+b*x)^(1/2)/x^(9/2),x, algorithm="fricas")
 

Output:

[-1/48*(3*(2*B*b*c^2 - A*c^3)*sqrt(b)*x^4*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x 
^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2*(8*A*b^3 + 3*(2*B*b^2*c - A*b*c^2)*x^2 
 + 2*(6*B*b^3 + A*b^2*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^3*x^4), -1/24*(3 
*(2*B*b*c^2 - A*c^3)*sqrt(-b)*x^4*arctan(sqrt(c*x^2 + b*x)*sqrt(-b)/(b*sqr 
t(x))) + (8*A*b^3 + 3*(2*B*b^2*c - A*b*c^2)*x^2 + 2*(6*B*b^3 + A*b^2*c)*x) 
*sqrt(c*x^2 + b*x)*sqrt(x))/(b^3*x^4)]
 

Sympy [F]

\[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{9/2}} \, dx=\int \frac {\sqrt {x \left (b + c x\right )} \left (A + B x\right )}{x^{\frac {9}{2}}}\, dx \] Input:

integrate((B*x+A)*(c*x**2+b*x)**(1/2)/x**(9/2),x)
 

Output:

Integral(sqrt(x*(b + c*x))*(A + B*x)/x**(9/2), x)
 

Maxima [F]

\[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{9/2}} \, dx=\int { \frac {\sqrt {c x^{2} + b x} {\left (B x + A\right )}}{x^{\frac {9}{2}}} \,d x } \] Input:

integrate((B*x+A)*(c*x^2+b*x)^(1/2)/x^(9/2),x, algorithm="maxima")
 

Output:

integrate(sqrt(c*x^2 + b*x)*(B*x + A)/x^(9/2), x)
 

Giac [A] (verification not implemented)

Time = 0.81 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.83 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{9/2}} \, dx=-\frac {1}{24} \, c^{3} {\left (\frac {3 \, {\left (2 \, B b - A c\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{\sqrt {-b} b^{2} c} + \frac {6 \, {\left (c x + b\right )}^{\frac {5}{2}} B b - 6 \, \sqrt {c x + b} B b^{3} - 3 \, {\left (c x + b\right )}^{\frac {5}{2}} A c + 8 \, {\left (c x + b\right )}^{\frac {3}{2}} A b c + 3 \, \sqrt {c x + b} A b^{2} c}{b^{2} c^{4} x^{3}}\right )} \] Input:

integrate((B*x+A)*(c*x^2+b*x)^(1/2)/x^(9/2),x, algorithm="giac")
 

Output:

-1/24*c^3*(3*(2*B*b - A*c)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^2*c) 
 + (6*(c*x + b)^(5/2)*B*b - 6*sqrt(c*x + b)*B*b^3 - 3*(c*x + b)^(5/2)*A*c 
+ 8*(c*x + b)^(3/2)*A*b*c + 3*sqrt(c*x + b)*A*b^2*c)/(b^2*c^4*x^3))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{9/2}} \, dx=\int \frac {\sqrt {c\,x^2+b\,x}\,\left (A+B\,x\right )}{x^{9/2}} \,d x \] Input:

int(((b*x + c*x^2)^(1/2)*(A + B*x))/x^(9/2),x)
 

Output:

int(((b*x + c*x^2)^(1/2)*(A + B*x))/x^(9/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.23 \[ \int \frac {(A+B x) \sqrt {b x+c x^2}}{x^{9/2}} \, dx=\frac {-16 \sqrt {c x +b}\, a \,b^{3}-4 \sqrt {c x +b}\, a \,b^{2} c x +6 \sqrt {c x +b}\, a b \,c^{2} x^{2}-24 \sqrt {c x +b}\, b^{4} x -12 \sqrt {c x +b}\, b^{3} c \,x^{2}+3 \sqrt {b}\, \mathrm {log}\left (\sqrt {c x +b}-\sqrt {b}\right ) a \,c^{3} x^{3}-6 \sqrt {b}\, \mathrm {log}\left (\sqrt {c x +b}-\sqrt {b}\right ) b^{2} c^{2} x^{3}-3 \sqrt {b}\, \mathrm {log}\left (\sqrt {c x +b}+\sqrt {b}\right ) a \,c^{3} x^{3}+6 \sqrt {b}\, \mathrm {log}\left (\sqrt {c x +b}+\sqrt {b}\right ) b^{2} c^{2} x^{3}}{48 b^{3} x^{3}} \] Input:

int((B*x+A)*(c*x^2+b*x)^(1/2)/x^(9/2),x)
 

Output:

( - 16*sqrt(b + c*x)*a*b**3 - 4*sqrt(b + c*x)*a*b**2*c*x + 6*sqrt(b + c*x) 
*a*b*c**2*x**2 - 24*sqrt(b + c*x)*b**4*x - 12*sqrt(b + c*x)*b**3*c*x**2 + 
3*sqrt(b)*log(sqrt(b + c*x) - sqrt(b))*a*c**3*x**3 - 6*sqrt(b)*log(sqrt(b 
+ c*x) - sqrt(b))*b**2*c**2*x**3 - 3*sqrt(b)*log(sqrt(b + c*x) + sqrt(b))* 
a*c**3*x**3 + 6*sqrt(b)*log(sqrt(b + c*x) + sqrt(b))*b**2*c**2*x**3)/(48*b 
**3*x**3)