\(\int \frac {A+B x+C x^2+D x^3}{(c x)^{7/2} \sqrt {a x+b x^2}} \, dx\) [8]

Optimal result

Integrand size = 36, antiderivative size = 190 \[ \int \frac {A+B x+C x^2+D x^3}{(c x)^{7/2} \sqrt {a x+b x^2}} \, dx=-\frac {A \sqrt {a x+b x^2}}{3 a (c x)^{7/2}}+\frac {(5 A b-6 a B) \sqrt {a x+b x^2}}{12 a^2 c (c x)^{5/2}}-\frac {\left (5 A b^2-6 a b B+8 a^2 C\right ) \sqrt {a x+b x^2}}{8 a^3 c^2 (c x)^{3/2}}+\frac {\left (5 A b^3-2 a \left (3 b^2 B-4 a b C+8 a^2 D\right )\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a x+b x^2}}{\sqrt {a} \sqrt {c x}}\right )}{8 a^{7/2} c^{7/2}} \] Output:

-1/3*A*(b*x^2+a*x)^(1/2)/a/(c*x)^(7/2)+1/12*(5*A*b-6*B*a)*(b*x^2+a*x)^(1/2 
)/a^2/c/(c*x)^(5/2)-1/8*(5*A*b^2-6*B*a*b+8*C*a^2)*(b*x^2+a*x)^(1/2)/a^3/c^ 
2/(c*x)^(3/2)+1/8*(5*A*b^3-2*a*(3*B*b^2-4*C*a*b+8*D*a^2))*arctanh(c^(1/2)* 
(b*x^2+a*x)^(1/2)/a^(1/2)/(c*x)^(1/2))/a^(7/2)/c^(7/2)
 

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.74 \[ \int \frac {A+B x+C x^2+D x^3}{(c x)^{7/2} \sqrt {a x+b x^2}} \, dx=\frac {x \left (-\sqrt {a} (a+b x) \left (15 A b^2 x^2-2 a b x (5 A+9 B x)+4 a^2 (2 A+3 x (B+2 C x))\right )+3 \left (5 A b^3-2 a \left (3 b^2 B-4 a b C+8 a^2 D\right )\right ) x^3 \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{24 a^{7/2} (c x)^{7/2} \sqrt {x (a+b x)}} \] Input:

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

Output:

(x*(-(Sqrt[a]*(a + b*x)*(15*A*b^2*x^2 - 2*a*b*x*(5*A + 9*B*x) + 4*a^2*(2*A 
 + 3*x*(B + 2*C*x)))) + 3*(5*A*b^3 - 2*a*(3*b^2*B - 4*a*b*C + 8*a^2*D))*x^ 
3*Sqrt[a + b*x]*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]))/(24*a^(7/2)*(c*x)^(7/2)*S 
qrt[x*(a + b*x)])
 

Rubi [A] (verified)

Time = 1.10 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.38, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2169, 27, 2169, 27, 1220, 1135, 1135, 1136, 221}

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.

\(\displaystyle \int \frac {A+B x+C x^2+D x^3}{(c x)^{7/2} \sqrt {a x+b x^2}} \, dx\)

\(\Big \downarrow \) 2169

\(\displaystyle -\frac {2 \int -\frac {(b C-2 a D) x^2 c^3+A b c^3+b B x c^3}{2 (c x)^{7/2} \sqrt {b x^2+a x}}dx}{b c^3}-\frac {2 D \sqrt {a x+b x^2}}{b c^2 (c x)^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {(b C-2 a D) x^2 c^3+A b c^3+b B x c^3}{(c x)^{7/2} \sqrt {b x^2+a x}}dx}{b c^3}-\frac {2 D \sqrt {a x+b x^2}}{b c^2 (c x)^{3/2}}\)

\(\Big \downarrow \) 2169

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1220

\(\displaystyle \frac {\frac {c^3 \left (-\frac {\left (-16 a^3 D+8 a^2 b C-6 a b^2 B+5 A b^3\right ) \int \frac {1}{(c x)^{5/2} \sqrt {b x^2+a x}}dx}{2 a c}-\frac {A b^2 \sqrt {a x+b x^2}}{a (c x)^{7/2}}\right )}{3 b}-\frac {2 c^2 \sqrt {a x+b x^2} (b C-2 a D)}{3 b (c x)^{5/2}}}{b c^3}-\frac {2 D \sqrt {a x+b x^2}}{b c^2 (c x)^{3/2}}\)

\(\Big \downarrow \) 1135

\(\displaystyle \frac {\frac {c^3 \left (-\frac {\left (-16 a^3 D+8 a^2 b C-6 a b^2 B+5 A b^3\right ) \left (-\frac {3 b \int \frac {1}{(c x)^{3/2} \sqrt {b x^2+a x}}dx}{4 a c}-\frac {\sqrt {a x+b x^2}}{2 a (c x)^{5/2}}\right )}{2 a c}-\frac {A b^2 \sqrt {a x+b x^2}}{a (c x)^{7/2}}\right )}{3 b}-\frac {2 c^2 \sqrt {a x+b x^2} (b C-2 a D)}{3 b (c x)^{5/2}}}{b c^3}-\frac {2 D \sqrt {a x+b x^2}}{b c^2 (c x)^{3/2}}\)

\(\Big \downarrow \) 1135

\(\displaystyle \frac {\frac {c^3 \left (-\frac {\left (-16 a^3 D+8 a^2 b C-6 a b^2 B+5 A b^3\right ) \left (-\frac {3 b \left (-\frac {b \int \frac {1}{\sqrt {c x} \sqrt {b x^2+a x}}dx}{2 a c}-\frac {\sqrt {a x+b x^2}}{a (c x)^{3/2}}\right )}{4 a c}-\frac {\sqrt {a x+b x^2}}{2 a (c x)^{5/2}}\right )}{2 a c}-\frac {A b^2 \sqrt {a x+b x^2}}{a (c x)^{7/2}}\right )}{3 b}-\frac {2 c^2 \sqrt {a x+b x^2} (b C-2 a D)}{3 b (c x)^{5/2}}}{b c^3}-\frac {2 D \sqrt {a x+b x^2}}{b c^2 (c x)^{3/2}}\)

\(\Big \downarrow \) 1136

\(\displaystyle \frac {\frac {c^3 \left (-\frac {\left (-16 a^3 D+8 a^2 b C-6 a b^2 B+5 A b^3\right ) \left (-\frac {3 b \left (-\frac {b \int \frac {1}{\frac {c \left (b x^2+a x\right )}{x}-a c}d\frac {\sqrt {b x^2+a x}}{\sqrt {c x}}}{a}-\frac {\sqrt {a x+b x^2}}{a (c x)^{3/2}}\right )}{4 a c}-\frac {\sqrt {a x+b x^2}}{2 a (c x)^{5/2}}\right )}{2 a c}-\frac {A b^2 \sqrt {a x+b x^2}}{a (c x)^{7/2}}\right )}{3 b}-\frac {2 c^2 \sqrt {a x+b x^2} (b C-2 a D)}{3 b (c x)^{5/2}}}{b c^3}-\frac {2 D \sqrt {a x+b x^2}}{b c^2 (c x)^{3/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {c^3 \left (-\frac {\left (-\frac {3 b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a x+b x^2}}{\sqrt {a} \sqrt {c x}}\right )}{a^{3/2} c^{3/2}}-\frac {\sqrt {a x+b x^2}}{a (c x)^{3/2}}\right )}{4 a c}-\frac {\sqrt {a x+b x^2}}{2 a (c x)^{5/2}}\right ) \left (-16 a^3 D+8 a^2 b C-6 a b^2 B+5 A b^3\right )}{2 a c}-\frac {A b^2 \sqrt {a x+b x^2}}{a (c x)^{7/2}}\right )}{3 b}-\frac {2 c^2 \sqrt {a x+b x^2} (b C-2 a D)}{3 b (c x)^{5/2}}}{b c^3}-\frac {2 D \sqrt {a x+b x^2}}{b c^2 (c x)^{3/2}}\)

Input:

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

Output:

(-2*D*Sqrt[a*x + b*x^2])/(b*c^2*(c*x)^(3/2)) + ((-2*c^2*(b*C - 2*a*D)*Sqrt 
[a*x + b*x^2])/(3*b*(c*x)^(5/2)) + (c^3*(-((A*b^2*Sqrt[a*x + b*x^2])/(a*(c 
*x)^(7/2))) - ((5*A*b^3 - 6*a*b^2*B + 8*a^2*b*C - 16*a^3*D)*(-1/2*Sqrt[a*x 
 + b*x^2]/(a*(c*x)^(5/2)) - (3*b*(-(Sqrt[a*x + b*x^2]/(a*(c*x)^(3/2))) + ( 
b*ArcTanh[(Sqrt[c]*Sqrt[a*x + b*x^2])/(Sqrt[a]*Sqrt[c*x])])/(a^(3/2)*c^(3/ 
2))))/(4*a*c)))/(2*a*c)))/(3*b))/(b*c^3)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 221

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

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 
]
 

rule 2169

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p 
_), x_Symbol] :> With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, S 
imp[f*(d + e*x)^(m + q - 1)*((a + b*x + c*x^2)^(p + 1)/(c*e^(q - 1)*(m + q 
+ 2*p + 1))), x] + Simp[1/(c*e^q*(m + q + 2*p + 1))   Int[(d + e*x)^m*(a + 
b*x + c*x^2)^p*ExpandToSum[c*e^q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 
1)*(d + e*x)^q + e*f*(m + p + q)*(d + e*x)^(q - 2)*(b*d - 2*a*e + (2*c*d - 
b*e)*x), x], x], x] /; NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, d, e, m, 
 p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2 
, 0]
 

Maple [A] (verified)

Time = 0.41 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.45

method	result	size
default	\(\frac {\sqrt {x \left (b x +a \right )}\, \left (15 A \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (b x +a \right )}}{\sqrt {a c}}\right ) b^{3} c \,x^{3}-18 B \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (b x +a \right )}}{\sqrt {a c}}\right ) a \,b^{2} c \,x^{3}+24 C \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (b x +a \right )}}{\sqrt {a c}}\right ) a^{2} b c \,x^{3}-48 D \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (b x +a \right )}}{\sqrt {a c}}\right ) a^{3} c \,x^{3}-15 A \,b^{2} x^{2} \sqrt {c \left (b x +a \right )}\, \sqrt {a c}+18 B a b \,x^{2} \sqrt {c \left (b x +a \right )}\, \sqrt {a c}-24 C \,a^{2} x^{2} \sqrt {c \left (b x +a \right )}\, \sqrt {a c}+10 A a b x \sqrt {c \left (b x +a \right )}\, \sqrt {a c}-12 B \,a^{2} x \sqrt {c \left (b x +a \right )}\, \sqrt {a c}-8 A \,a^{2} \sqrt {c \left (b x +a \right )}\, \sqrt {a c}\right )}{24 c^{3} x^{3} \sqrt {c x}\, \sqrt {c \left (b x +a \right )}\, a^{3} \sqrt {a c}}\)	\(276\)

Input:

int((D*x^3+C*x^2+B*x+A)/(c*x)^(7/2)/(b*x^2+a*x)^(1/2),x,method=_RETURNVERB 
OSE)
 

Output:

1/24*(x*(b*x+a))^(1/2)/c^3*(15*A*arctanh((c*(b*x+a))^(1/2)/(a*c)^(1/2))*b^ 
3*c*x^3-18*B*arctanh((c*(b*x+a))^(1/2)/(a*c)^(1/2))*a*b^2*c*x^3+24*C*arcta 
nh((c*(b*x+a))^(1/2)/(a*c)^(1/2))*a^2*b*c*x^3-48*D*arctanh((c*(b*x+a))^(1/ 
2)/(a*c)^(1/2))*a^3*c*x^3-15*A*b^2*x^2*(c*(b*x+a))^(1/2)*(a*c)^(1/2)+18*B* 
a*b*x^2*(c*(b*x+a))^(1/2)*(a*c)^(1/2)-24*C*a^2*x^2*(c*(b*x+a))^(1/2)*(a*c) 
^(1/2)+10*A*a*b*x*(c*(b*x+a))^(1/2)*(a*c)^(1/2)-12*B*a^2*x*(c*(b*x+a))^(1/ 
2)*(a*c)^(1/2)-8*A*a^2*(c*(b*x+a))^(1/2)*(a*c)^(1/2))/x^3/(c*x)^(1/2)/(c*( 
b*x+a))^(1/2)/a^3/(a*c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.63 \[ \int \frac {A+B x+C x^2+D x^3}{(c x)^{7/2} \sqrt {a x+b x^2}} \, dx=\left [-\frac {3 \, {\left (16 \, D a^{3} - 8 \, C a^{2} b + 6 \, B a b^{2} - 5 \, A b^{3}\right )} \sqrt {a c} x^{4} \log \left (-\frac {b c x^{2} + 2 \, a c x + 2 \, \sqrt {b x^{2} + a x} \sqrt {a c} \sqrt {c x}}{x^{2}}\right ) + 2 \, {\left (8 \, A a^{3} + 3 \, {\left (8 \, C a^{3} - 6 \, B a^{2} b + 5 \, A a b^{2}\right )} x^{2} + 2 \, {\left (6 \, B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {b x^{2} + a x} \sqrt {c x}}{48 \, a^{4} c^{4} x^{4}}, \frac {3 \, {\left (16 \, D a^{3} - 8 \, C a^{2} b + 6 \, B a b^{2} - 5 \, A b^{3}\right )} \sqrt {-a c} x^{4} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-a c} \sqrt {c x}}{a c x}\right ) - {\left (8 \, A a^{3} + 3 \, {\left (8 \, C a^{3} - 6 \, B a^{2} b + 5 \, A a b^{2}\right )} x^{2} + 2 \, {\left (6 \, B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {b x^{2} + a x} \sqrt {c x}}{24 \, a^{4} c^{4} x^{4}}\right ] \] Input:

integrate((D*x^3+C*x^2+B*x+A)/(c*x)^(7/2)/(b*x^2+a*x)^(1/2),x, algorithm=" 
fricas")
                                                                                    
                                                                                    
 

Output:

[-1/48*(3*(16*D*a^3 - 8*C*a^2*b + 6*B*a*b^2 - 5*A*b^3)*sqrt(a*c)*x^4*log(- 
(b*c*x^2 + 2*a*c*x + 2*sqrt(b*x^2 + a*x)*sqrt(a*c)*sqrt(c*x))/x^2) + 2*(8* 
A*a^3 + 3*(8*C*a^3 - 6*B*a^2*b + 5*A*a*b^2)*x^2 + 2*(6*B*a^3 - 5*A*a^2*b)* 
x)*sqrt(b*x^2 + a*x)*sqrt(c*x))/(a^4*c^4*x^4), 1/24*(3*(16*D*a^3 - 8*C*a^2 
*b + 6*B*a*b^2 - 5*A*b^3)*sqrt(-a*c)*x^4*arctan(sqrt(b*x^2 + a*x)*sqrt(-a* 
c)*sqrt(c*x)/(a*c*x)) - (8*A*a^3 + 3*(8*C*a^3 - 6*B*a^2*b + 5*A*a*b^2)*x^2 
 + 2*(6*B*a^3 - 5*A*a^2*b)*x)*sqrt(b*x^2 + a*x)*sqrt(c*x))/(a^4*c^4*x^4)]
 

Sympy [F]

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

integrate((D*x**3+C*x**2+B*x+A)/(c*x)**(7/2)/(b*x**2+a*x)**(1/2),x)
 

Output:

Integral((A + B*x + C*x**2 + D*x**3)/((c*x)**(7/2)*sqrt(x*(a + b*x))), x)
 

Maxima [F]

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

integrate((D*x^3+C*x^2+B*x+A)/(c*x)^(7/2)/(b*x^2+a*x)^(1/2),x, algorithm=" 
maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.31 \[ \int \frac {A+B x+C x^2+D x^3}{(c x)^{7/2} \sqrt {a x+b x^2}} \, dx=\frac {b^{3} {\left (\frac {3 \, {\left (16 \, D a^{3} - 8 \, C a^{2} b + 6 \, B a b^{2} - 5 \, A b^{3}\right )} \arctan \left (\frac {\sqrt {b c x + a c}}{\sqrt {-a c}}\right )}{\sqrt {-a c} a^{3} b^{3}} - \frac {24 \, \sqrt {b c x + a c} C a^{4} c^{2} - 30 \, \sqrt {b c x + a c} B a^{3} b c^{2} + 33 \, \sqrt {b c x + a c} A a^{2} b^{2} c^{2} - 48 \, {\left (b c x + a c\right )}^{\frac {3}{2}} C a^{3} c + 48 \, {\left (b c x + a c\right )}^{\frac {3}{2}} B a^{2} b c - 40 \, {\left (b c x + a c\right )}^{\frac {3}{2}} A a b^{2} c + 24 \, {\left (b c x + a c\right )}^{\frac {5}{2}} C a^{2} - 18 \, {\left (b c x + a c\right )}^{\frac {5}{2}} B a b + 15 \, {\left (b c x + a c\right )}^{\frac {5}{2}} A b^{2}}{a^{3} b^{5} c^{3} x^{3}}\right )}}{24 \, c^{2} {\left | c \right |}} \] Input:

integrate((D*x^3+C*x^2+B*x+A)/(c*x)^(7/2)/(b*x^2+a*x)^(1/2),x, algorithm=" 
giac")
 

Output:

1/24*b^3*(3*(16*D*a^3 - 8*C*a^2*b + 6*B*a*b^2 - 5*A*b^3)*arctan(sqrt(b*c*x 
 + a*c)/sqrt(-a*c))/(sqrt(-a*c)*a^3*b^3) - (24*sqrt(b*c*x + a*c)*C*a^4*c^2 
 - 30*sqrt(b*c*x + a*c)*B*a^3*b*c^2 + 33*sqrt(b*c*x + a*c)*A*a^2*b^2*c^2 - 
 48*(b*c*x + a*c)^(3/2)*C*a^3*c + 48*(b*c*x + a*c)^(3/2)*B*a^2*b*c - 40*(b 
*c*x + a*c)^(3/2)*A*a*b^2*c + 24*(b*c*x + a*c)^(5/2)*C*a^2 - 18*(b*c*x + a 
*c)^(5/2)*B*a*b + 15*(b*c*x + a*c)^(5/2)*A*b^2)/(a^3*b^5*c^3*x^3))/(c^2*ab 
s(c))
 

Mupad [F(-1)]

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

int((A + B*x + C*x^2 + x^3*D)/((a*x + b*x^2)^(1/2)*(c*x)^(7/2)),x)
 

Output:

int((A + B*x + C*x^2 + x^3*D)/((a*x + b*x^2)^(1/2)*(c*x)^(7/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.03 \[ \int \frac {A+B x+C x^2+D x^3}{(c x)^{7/2} \sqrt {a x+b x^2}} \, dx=\frac {\sqrt {c}\, \left (-16 \sqrt {b x +a}\, a^{3}-4 \sqrt {b x +a}\, a^{2} b x -48 \sqrt {b x +a}\, a^{2} c \,x^{2}+6 \sqrt {b x +a}\, a \,b^{2} x^{2}+48 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a^{2} d \,x^{3}-24 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a b c \,x^{3}+3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{3} x^{3}-48 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a^{2} d \,x^{3}+24 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a b c \,x^{3}-3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{3} x^{3}\right )}{48 a^{3} c^{4} x^{3}} \] Input:

int((D*x^3+C*x^2+B*x+A)/(c*x)^(7/2)/(b*x^2+a*x)^(1/2),x)
 

Output:

(sqrt(c)*( - 16*sqrt(a + b*x)*a**3 - 4*sqrt(a + b*x)*a**2*b*x - 48*sqrt(a 
+ b*x)*a**2*c*x**2 + 6*sqrt(a + b*x)*a*b**2*x**2 + 48*sqrt(a)*log(sqrt(a + 
 b*x) - sqrt(a))*a**2*d*x**3 - 24*sqrt(a)*log(sqrt(a + b*x) - sqrt(a))*a*b 
*c*x**3 + 3*sqrt(a)*log(sqrt(a + b*x) - sqrt(a))*b**3*x**3 - 48*sqrt(a)*lo 
g(sqrt(a + b*x) + sqrt(a))*a**2*d*x**3 + 24*sqrt(a)*log(sqrt(a + b*x) + sq 
rt(a))*a*b*c*x**3 - 3*sqrt(a)*log(sqrt(a + b*x) + sqrt(a))*b**3*x**3))/(48 
*a**3*c**4*x**3)