\(\int \frac {A+B x}{x^{7/2} (b x+c x^2)^{3/2}} \, dx\) [220]

Optimal result

Integrand size = 24, antiderivative size = 213 \[ \int \frac {A+B x}{x^{7/2} \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 c^3 (b B-A c) \sqrt {x}}{b^5 \sqrt {b x+c x^2}}-\frac {A \sqrt {b x+c x^2}}{4 b^2 x^{9/2}}-\frac {(8 b B-15 A c) \sqrt {b x+c x^2}}{24 b^3 x^{7/2}}+\frac {c (88 b B-123 A c) \sqrt {b x+c x^2}}{96 b^4 x^{5/2}}-\frac {c^2 (152 b B-187 A c) \sqrt {b x+c x^2}}{64 b^5 x^{3/2}}+\frac {35 c^3 (8 b B-9 A c) \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{64 b^{11/2}} \] Output:

-2*c^3*(-A*c+B*b)*x^(1/2)/b^5/(c*x^2+b*x)^(1/2)-1/4*A*(c*x^2+b*x)^(1/2)/b^ 
2/x^(9/2)-1/24*(-15*A*c+8*B*b)*(c*x^2+b*x)^(1/2)/b^3/x^(7/2)+1/96*c*(-123* 
A*c+88*B*b)*(c*x^2+b*x)^(1/2)/b^4/x^(5/2)-1/64*c^2*(-187*A*c+152*B*b)*(c*x 
^2+b*x)^(1/2)/b^5/x^(3/2)+35/64*c^3*(-9*A*c+8*B*b)*arctanh((c*x^2+b*x)^(1/ 
2)/b^(1/2)/x^(1/2))/b^(11/2)
 

Mathematica [A] (verified)

Time = 0.29 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.72 \[ \int \frac {A+B x}{x^{7/2} \left (b x+c x^2\right )^{3/2}} \, dx=\frac {\sqrt {b} \left (-8 b B x \left (8 b^3-14 b^2 c x+35 b c^2 x^2+105 c^3 x^3\right )+A \left (-48 b^4+72 b^3 c x-126 b^2 c^2 x^2+315 b c^3 x^3+945 c^4 x^4\right )\right )+105 c^3 (8 b B-9 A c) x^4 \sqrt {b+c x} \text {arctanh}\left (\frac {\sqrt {b+c x}}{\sqrt {b}}\right )}{192 b^{11/2} x^{7/2} \sqrt {x (b+c x)}} \] Input:

Integrate[(A + B*x)/(x^(7/2)*(b*x + c*x^2)^(3/2)),x]
 

Output:

(Sqrt[b]*(-8*b*B*x*(8*b^3 - 14*b^2*c*x + 35*b*c^2*x^2 + 105*c^3*x^3) + A*( 
-48*b^4 + 72*b^3*c*x - 126*b^2*c^2*x^2 + 315*b*c^3*x^3 + 945*c^4*x^4)) + 1 
05*c^3*(8*b*B - 9*A*c)*x^4*Sqrt[b + c*x]*ArcTanh[Sqrt[b + c*x]/Sqrt[b]])/( 
192*b^(11/2)*x^(7/2)*Sqrt[x*(b + c*x)])
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.93, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1220, 1135, 1135, 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.

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

\(\Big \downarrow \) 1220

\(\displaystyle \frac {(8 b B-9 A c) \int \frac {1}{x^{5/2} \left (c x^2+b x\right )^{3/2}}dx}{8 b}-\frac {A}{4 b x^{7/2} \sqrt {b x+c x^2}}\)

\(\Big \downarrow \) 1135

\(\displaystyle \frac {(8 b B-9 A c) \left (-\frac {7 c \int \frac {1}{x^{3/2} \left (c x^2+b x\right )^{3/2}}dx}{6 b}-\frac {1}{3 b x^{5/2} \sqrt {b x+c x^2}}\right )}{8 b}-\frac {A}{4 b x^{7/2} \sqrt {b x+c x^2}}\)

\(\Big \downarrow \) 1135

\(\displaystyle \frac {(8 b B-9 A c) \left (-\frac {7 c \left (-\frac {5 c \int \frac {1}{\sqrt {x} \left (c x^2+b x\right )^{3/2}}dx}{4 b}-\frac {1}{2 b x^{3/2} \sqrt {b x+c x^2}}\right )}{6 b}-\frac {1}{3 b x^{5/2} \sqrt {b x+c x^2}}\right )}{8 b}-\frac {A}{4 b x^{7/2} \sqrt {b x+c x^2}}\)

\(\Big \downarrow \) 1135

\(\displaystyle \frac {(8 b B-9 A c) \left (-\frac {7 c \left (-\frac {5 c \left (-\frac {3 c \int \frac {\sqrt {x}}{\left (c x^2+b x\right )^{3/2}}dx}{2 b}-\frac {1}{b \sqrt {x} \sqrt {b x+c x^2}}\right )}{4 b}-\frac {1}{2 b x^{3/2} \sqrt {b x+c x^2}}\right )}{6 b}-\frac {1}{3 b x^{5/2} \sqrt {b x+c x^2}}\right )}{8 b}-\frac {A}{4 b x^{7/2} \sqrt {b x+c x^2}}\)

\(\Big \downarrow \) 1132

\(\displaystyle \frac {(8 b B-9 A c) \left (-\frac {7 c \left (-\frac {5 c \left (-\frac {3 c \left (\frac {\int \frac {1}{\sqrt {x} \sqrt {c x^2+b x}}dx}{b}+\frac {2 \sqrt {x}}{b \sqrt {b x+c x^2}}\right )}{2 b}-\frac {1}{b \sqrt {x} \sqrt {b x+c x^2}}\right )}{4 b}-\frac {1}{2 b x^{3/2} \sqrt {b x+c x^2}}\right )}{6 b}-\frac {1}{3 b x^{5/2} \sqrt {b x+c x^2}}\right )}{8 b}-\frac {A}{4 b x^{7/2} \sqrt {b x+c x^2}}\)

\(\Big \downarrow \) 1136

\(\displaystyle \frac {(8 b B-9 A c) \left (-\frac {7 c \left (-\frac {5 c \left (-\frac {3 c \left (\frac {2 \int \frac {1}{\frac {c x^2+b x}{x}-b}d\frac {\sqrt {c x^2+b x}}{\sqrt {x}}}{b}+\frac {2 \sqrt {x}}{b \sqrt {b x+c x^2}}\right )}{2 b}-\frac {1}{b \sqrt {x} \sqrt {b x+c x^2}}\right )}{4 b}-\frac {1}{2 b x^{3/2} \sqrt {b x+c x^2}}\right )}{6 b}-\frac {1}{3 b x^{5/2} \sqrt {b x+c x^2}}\right )}{8 b}-\frac {A}{4 b x^{7/2} \sqrt {b x+c x^2}}\)

\(\Big \downarrow \) 220

\(\displaystyle \frac {(8 b B-9 A c) \left (-\frac {7 c \left (-\frac {5 c \left (-\frac {3 c \left (\frac {2 \sqrt {x}}{b \sqrt {b x+c x^2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {b x+c x^2}}{\sqrt {b} \sqrt {x}}\right )}{b^{3/2}}\right )}{2 b}-\frac {1}{b \sqrt {x} \sqrt {b x+c x^2}}\right )}{4 b}-\frac {1}{2 b x^{3/2} \sqrt {b x+c x^2}}\right )}{6 b}-\frac {1}{3 b x^{5/2} \sqrt {b x+c x^2}}\right )}{8 b}-\frac {A}{4 b x^{7/2} \sqrt {b x+c x^2}}\)

Input:

Int[(A + B*x)/(x^(7/2)*(b*x + c*x^2)^(3/2)),x]
 

Output:

-1/4*A/(b*x^(7/2)*Sqrt[b*x + c*x^2]) + ((8*b*B - 9*A*c)*(-1/3*1/(b*x^(5/2) 
*Sqrt[b*x + c*x^2]) - (7*c*(-1/2*1/(b*x^(3/2)*Sqrt[b*x + c*x^2]) - (5*c*(- 
(1/(b*Sqrt[x]*Sqrt[b*x + c*x^2])) - (3*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)))/(4* 
b)))/(6*b)))/(8*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 = 157, normalized size of antiderivative = 0.74

method	result	size
risch	\(-\frac {\left (c x +b \right ) \left (-561 A \,c^{3} x^{3}+456 x^{3} B b \,c^{2}+246 A b \,c^{2} x^{2}-176 x^{2} B \,b^{2} c -120 A \,b^{2} c x +64 x B \,b^{3}+48 A \,b^{3}\right )}{192 b^{5} x^{\frac {7}{2}} \sqrt {x \left (c x +b \right )}}+\frac {c^{3} \left (-\frac {2 \left (-128 A c +128 B b \right )}{\sqrt {c x +b}}-\frac {2 \left (315 A c -280 B b \right ) \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right )}{\sqrt {b}}\right ) \sqrt {c x +b}\, \sqrt {x}}{128 b^{5} \sqrt {x \left (c x +b \right )}}\)	\(157\)
default	\(-\frac {\sqrt {x \left (c x +b \right )}\, \left (945 A \sqrt {c x +b}\, \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) c^{4} x^{4}-840 B \sqrt {c x +b}\, \operatorname {arctanh}\left (\frac {\sqrt {c x +b}}{\sqrt {b}}\right ) b \,c^{3} x^{4}+64 B \,b^{\frac {9}{2}} x -112 B \,b^{\frac {7}{2}} c \,x^{2}+280 B \,b^{\frac {5}{2}} c^{2} x^{3}+840 B \,b^{\frac {3}{2}} c^{3} x^{4}+48 A \,b^{\frac {9}{2}}-72 A \,b^{\frac {7}{2}} c x +126 A \,b^{\frac {5}{2}} c^{2} x^{2}-315 A \,b^{\frac {3}{2}} c^{3} x^{3}-945 A \sqrt {b}\, c^{4} x^{4}\right )}{192 x^{\frac {9}{2}} \left (c x +b \right ) b^{\frac {11}{2}}}\)	\(174\)

Input:

int((B*x+A)/x^(7/2)/(c*x^2+b*x)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/192*(c*x+b)*(-561*A*c^3*x^3+456*B*b*c^2*x^3+246*A*b*c^2*x^2-176*B*b^2*c 
*x^2-120*A*b^2*c*x+64*B*b^3*x+48*A*b^3)/b^5/x^(7/2)/(x*(c*x+b))^(1/2)+1/12 
8*c^3/b^5*(-2*(-128*A*c+128*B*b)/(c*x+b)^(1/2)-2*(315*A*c-280*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 = 409, normalized size of antiderivative = 1.92 \[ \int \frac {A+B x}{x^{7/2} \left (b x+c x^2\right )^{3/2}} \, dx=\left [-\frac {105 \, {\left ({\left (8 \, B b c^{4} - 9 \, A c^{5}\right )} x^{6} + {\left (8 \, B b^{2} c^{3} - 9 \, A b c^{4}\right )} x^{5}\right )} \sqrt {b} \log \left (-\frac {c x^{2} + 2 \, b x - 2 \, \sqrt {c x^{2} + b x} \sqrt {b} \sqrt {x}}{x^{2}}\right ) + 2 \, {\left (48 \, A b^{5} + 105 \, {\left (8 \, B b^{2} c^{3} - 9 \, A b c^{4}\right )} x^{4} + 35 \, {\left (8 \, B b^{3} c^{2} - 9 \, A b^{2} c^{3}\right )} x^{3} - 14 \, {\left (8 \, B b^{4} c - 9 \, A b^{3} c^{2}\right )} x^{2} + 8 \, {\left (8 \, B b^{5} - 9 \, A b^{4} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{384 \, {\left (b^{6} c x^{6} + b^{7} x^{5}\right )}}, -\frac {105 \, {\left ({\left (8 \, B b c^{4} - 9 \, A c^{5}\right )} x^{6} + {\left (8 \, B b^{2} c^{3} - 9 \, A b c^{4}\right )} x^{5}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-b}}{b \sqrt {x}}\right ) + {\left (48 \, A b^{5} + 105 \, {\left (8 \, B b^{2} c^{3} - 9 \, A b c^{4}\right )} x^{4} + 35 \, {\left (8 \, B b^{3} c^{2} - 9 \, A b^{2} c^{3}\right )} x^{3} - 14 \, {\left (8 \, B b^{4} c - 9 \, A b^{3} c^{2}\right )} x^{2} + 8 \, {\left (8 \, B b^{5} - 9 \, A b^{4} c\right )} x\right )} \sqrt {c x^{2} + b x} \sqrt {x}}{192 \, {\left (b^{6} c x^{6} + b^{7} x^{5}\right )}}\right ] \] Input:

integrate((B*x+A)/x^(7/2)/(c*x^2+b*x)^(3/2),x, algorithm="fricas")
 

Output:

[-1/384*(105*((8*B*b*c^4 - 9*A*c^5)*x^6 + (8*B*b^2*c^3 - 9*A*b*c^4)*x^5)*s 
qrt(b)*log(-(c*x^2 + 2*b*x - 2*sqrt(c*x^2 + b*x)*sqrt(b)*sqrt(x))/x^2) + 2 
*(48*A*b^5 + 105*(8*B*b^2*c^3 - 9*A*b*c^4)*x^4 + 35*(8*B*b^3*c^2 - 9*A*b^2 
*c^3)*x^3 - 14*(8*B*b^4*c - 9*A*b^3*c^2)*x^2 + 8*(8*B*b^5 - 9*A*b^4*c)*x)* 
sqrt(c*x^2 + b*x)*sqrt(x))/(b^6*c*x^6 + b^7*x^5), -1/192*(105*((8*B*b*c^4 
- 9*A*c^5)*x^6 + (8*B*b^2*c^3 - 9*A*b*c^4)*x^5)*sqrt(-b)*arctan(sqrt(c*x^2 
 + b*x)*sqrt(-b)/(b*sqrt(x))) + (48*A*b^5 + 105*(8*B*b^2*c^3 - 9*A*b*c^4)* 
x^4 + 35*(8*B*b^3*c^2 - 9*A*b^2*c^3)*x^3 - 14*(8*B*b^4*c - 9*A*b^3*c^2)*x^ 
2 + 8*(8*B*b^5 - 9*A*b^4*c)*x)*sqrt(c*x^2 + b*x)*sqrt(x))/(b^6*c*x^6 + b^7 
*x^5)]
 

Sympy [F]

\[ \int \frac {A+B x}{x^{7/2} \left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {A + B x}{x^{\frac {7}{2}} \left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate((B*x+A)/x**(7/2)/(c*x**2+b*x)**(3/2),x)
 

Output:

Integral((A + B*x)/(x**(7/2)*(x*(b + c*x))**(3/2)), x)
 

Maxima [F]

\[ \int \frac {A+B x}{x^{7/2} \left (b x+c x^2\right )^{3/2}} \, dx=\int { \frac {B x + A}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} x^{\frac {7}{2}}} \,d x } \] Input:

integrate((B*x+A)/x^(7/2)/(c*x^2+b*x)^(3/2),x, algorithm="maxima")
 

Output:

integrate((B*x + A)/((c*x^2 + b*x)^(3/2)*x^(7/2)), x)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x}{x^{7/2} \left (b x+c x^2\right )^{3/2}} \, dx=-\frac {35 \, {\left (8 \, B b c^{3} - 9 \, A c^{4}\right )} \arctan \left (\frac {\sqrt {c x + b}}{\sqrt {-b}}\right )}{64 \, \sqrt {-b} b^{5}} - \frac {2 \, {\left (B b c^{3} - A c^{4}\right )}}{\sqrt {c x + b} b^{5}} - \frac {456 \, {\left (c x + b\right )}^{\frac {7}{2}} B b c^{3} - 1544 \, {\left (c x + b\right )}^{\frac {5}{2}} B b^{2} c^{3} + 1784 \, {\left (c x + b\right )}^{\frac {3}{2}} B b^{3} c^{3} - 696 \, \sqrt {c x + b} B b^{4} c^{3} - 561 \, {\left (c x + b\right )}^{\frac {7}{2}} A c^{4} + 1929 \, {\left (c x + b\right )}^{\frac {5}{2}} A b c^{4} - 2295 \, {\left (c x + b\right )}^{\frac {3}{2}} A b^{2} c^{4} + 975 \, \sqrt {c x + b} A b^{3} c^{4}}{192 \, b^{5} c^{4} x^{4}} \] Input:

integrate((B*x+A)/x^(7/2)/(c*x^2+b*x)^(3/2),x, algorithm="giac")
 

Output:

-35/64*(8*B*b*c^3 - 9*A*c^4)*arctan(sqrt(c*x + b)/sqrt(-b))/(sqrt(-b)*b^5) 
 - 2*(B*b*c^3 - A*c^4)/(sqrt(c*x + b)*b^5) - 1/192*(456*(c*x + b)^(7/2)*B* 
b*c^3 - 1544*(c*x + b)^(5/2)*B*b^2*c^3 + 1784*(c*x + b)^(3/2)*B*b^3*c^3 - 
696*sqrt(c*x + b)*B*b^4*c^3 - 561*(c*x + b)^(7/2)*A*c^4 + 1929*(c*x + b)^( 
5/2)*A*b*c^4 - 2295*(c*x + b)^(3/2)*A*b^2*c^4 + 975*sqrt(c*x + b)*A*b^3*c^ 
4)/(b^5*c^4*x^4)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x}{x^{7/2} \left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {A+B\,x}{x^{7/2}\,{\left (c\,x^2+b\,x\right )}^{3/2}} \,d x \] Input:

int((A + B*x)/(x^(7/2)*(b*x + c*x^2)^(3/2)),x)
 

Output:

int((A + B*x)/(x^(7/2)*(b*x + c*x^2)^(3/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.02 \[ \int \frac {A+B x}{x^{7/2} \left (b x+c x^2\right )^{3/2}} \, dx=\frac {945 \sqrt {b}\, \sqrt {c x +b}\, \mathrm {log}\left (\sqrt {c x +b}-\sqrt {b}\right ) a \,c^{4} x^{4}-840 \sqrt {b}\, \sqrt {c x +b}\, \mathrm {log}\left (\sqrt {c x +b}-\sqrt {b}\right ) b^{2} c^{3} x^{4}-945 \sqrt {b}\, \sqrt {c x +b}\, \mathrm {log}\left (\sqrt {c x +b}+\sqrt {b}\right ) a \,c^{4} x^{4}+840 \sqrt {b}\, \sqrt {c x +b}\, \mathrm {log}\left (\sqrt {c x +b}+\sqrt {b}\right ) b^{2} c^{3} x^{4}-96 a \,b^{5}+144 a \,b^{4} c x -252 a \,b^{3} c^{2} x^{2}+630 a \,b^{2} c^{3} x^{3}+1890 a b \,c^{4} x^{4}-128 b^{6} x +224 b^{5} c \,x^{2}-560 b^{4} c^{2} x^{3}-1680 b^{3} c^{3} x^{4}}{384 \sqrt {c x +b}\, b^{6} x^{4}} \] Input:

int((B*x+A)/x^(7/2)/(c*x^2+b*x)^(3/2),x)
 

Output:

(945*sqrt(b)*sqrt(b + c*x)*log(sqrt(b + c*x) - sqrt(b))*a*c**4*x**4 - 840* 
sqrt(b)*sqrt(b + c*x)*log(sqrt(b + c*x) - sqrt(b))*b**2*c**3*x**4 - 945*sq 
rt(b)*sqrt(b + c*x)*log(sqrt(b + c*x) + sqrt(b))*a*c**4*x**4 + 840*sqrt(b) 
*sqrt(b + c*x)*log(sqrt(b + c*x) + sqrt(b))*b**2*c**3*x**4 - 96*a*b**5 + 1 
44*a*b**4*c*x - 252*a*b**3*c**2*x**2 + 630*a*b**2*c**3*x**3 + 1890*a*b*c** 
4*x**4 - 128*b**6*x + 224*b**5*c*x**2 - 560*b**4*c**2*x**3 - 1680*b**3*c** 
3*x**4)/(384*sqrt(b + c*x)*b**6*x**4)