\(\int \frac {A+B x+C x^2+D x^3}{(c x)^{9/2} \sqrt {a x+b x^2}} \, dx\) [9]

Optimal result

Integrand size = 36, antiderivative size = 249 \[ \int \frac {A+B x+C x^2+D x^3}{(c x)^{9/2} \sqrt {a x+b x^2}} \, dx=-\frac {A \sqrt {a x+b x^2}}{4 a (c x)^{9/2}}+\frac {(7 A b-8 a B) \sqrt {a x+b x^2}}{24 a^2 c (c x)^{7/2}}-\frac {\left (35 A b^2-40 a b B+48 a^2 C\right ) \sqrt {a x+b x^2}}{96 a^3 c^2 (c x)^{5/2}}+\frac {\left (35 A b^3-8 a \left (5 b^2 B-6 a b C+8 a^2 D\right )\right ) \sqrt {a x+b x^2}}{64 a^4 c^3 (c x)^{3/2}}-\frac {b \left (35 A b^3-8 a \left (5 b^2 B-6 a b C+8 a^2 D\right )\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a x+b x^2}}{\sqrt {a} \sqrt {c x}}\right )}{64 a^{9/2} c^{9/2}} \] Output:

-1/4*A*(b*x^2+a*x)^(1/2)/a/(c*x)^(9/2)+1/24*(7*A*b-8*B*a)*(b*x^2+a*x)^(1/2 
)/a^2/c/(c*x)^(7/2)-1/96*(35*A*b^2-40*B*a*b+48*C*a^2)*(b*x^2+a*x)^(1/2)/a^ 
3/c^2/(c*x)^(5/2)+1/64*(35*A*b^3-8*a*(5*B*b^2-6*C*a*b+8*D*a^2))*(b*x^2+a*x 
)^(1/2)/a^4/c^3/(c*x)^(3/2)-1/64*b*(35*A*b^3-8*a*(5*B*b^2-6*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^(9/2)/c^(9/2)
 

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.71 \[ \int \frac {A+B x+C x^2+D x^3}{(c x)^{9/2} \sqrt {a x+b x^2}} \, dx=-\frac {\sqrt {c x} \left (\sqrt {a} (a+b x) \left (-105 A b^3 x^3+10 a b^2 x^2 (7 A+12 B x)-8 a^2 b x (7 A+2 x (5 B+9 C x))+16 a^3 \left (3 A+4 B x+6 x^2 (C+2 D x)\right )\right )+3 b \left (35 A b^3-8 a \left (5 b^2 B-6 a b C+8 a^2 D\right )\right ) x^4 \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{192 a^{9/2} c^5 x^4 \sqrt {x (a+b x)}} \] Input:

Integrate[(A + B*x + C*x^2 + D*x^3)/((c*x)^(9/2)*Sqrt[a*x + b*x^2]),x]
 

Output:

-1/192*(Sqrt[c*x]*(Sqrt[a]*(a + b*x)*(-105*A*b^3*x^3 + 10*a*b^2*x^2*(7*A + 
 12*B*x) - 8*a^2*b*x*(7*A + 2*x*(5*B + 9*C*x)) + 16*a^3*(3*A + 4*B*x + 6*x 
^2*(C + 2*D*x))) + 3*b*(35*A*b^3 - 8*a*(5*b^2*B - 6*a*b*C + 8*a^2*D))*x^4* 
Sqrt[a + b*x]*ArcTanh[Sqrt[a + b*x]/Sqrt[a]]))/(a^(9/2)*c^5*x^4*Sqrt[x*(a 
+ b*x)])
 

Rubi [A] (verified)

Time = 1.18 (sec) , antiderivative size = 309, normalized size of antiderivative = 1.24, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {2169, 27, 2169, 27, 1220, 1135, 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.

\(\displaystyle \int \frac {A+B x+C x^2+D x^3}{(c x)^{9/2} \sqrt {a x+b x^2}} \, dx\)

\(\Big \downarrow \) 2169

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2169

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1220

\(\displaystyle \frac {\frac {3 c^3 \left (-\frac {\left (-64 a^3 D+48 a^2 b C-40 a b^2 B+35 A b^3\right ) \int \frac {1}{(c x)^{7/2} \sqrt {b x^2+a x}}dx}{8 a c}-\frac {5 A b^2 \sqrt {a x+b x^2}}{4 a (c x)^{9/2}}\right )}{5 b}-\frac {2 c^2 \sqrt {a x+b x^2} (3 b C-4 a D)}{5 b (c x)^{7/2}}}{3 b c^3}-\frac {2 D \sqrt {a x+b x^2}}{3 b c^2 (c x)^{5/2}}\)

\(\Big \downarrow \) 1135

\(\displaystyle \frac {\frac {3 c^3 \left (-\frac {\left (-64 a^3 D+48 a^2 b C-40 a b^2 B+35 A b^3\right ) \left (-\frac {5 b \int \frac {1}{(c x)^{5/2} \sqrt {b x^2+a x}}dx}{6 a c}-\frac {\sqrt {a x+b x^2}}{3 a (c x)^{7/2}}\right )}{8 a c}-\frac {5 A b^2 \sqrt {a x+b x^2}}{4 a (c x)^{9/2}}\right )}{5 b}-\frac {2 c^2 \sqrt {a x+b x^2} (3 b C-4 a D)}{5 b (c x)^{7/2}}}{3 b c^3}-\frac {2 D \sqrt {a x+b x^2}}{3 b c^2 (c x)^{5/2}}\)

\(\Big \downarrow \) 1135

\(\displaystyle \frac {\frac {3 c^3 \left (-\frac {\left (-64 a^3 D+48 a^2 b C-40 a b^2 B+35 A b^3\right ) \left (-\frac {5 b \left (-\frac {3 b \int \frac {1}{(c x)^{3/2} \sqrt {b x^2+a x}}dx}{4 a c}-\frac {\sqrt {a x+b x^2}}{2 a (c x)^{5/2}}\right )}{6 a c}-\frac {\sqrt {a x+b x^2}}{3 a (c x)^{7/2}}\right )}{8 a c}-\frac {5 A b^2 \sqrt {a x+b x^2}}{4 a (c x)^{9/2}}\right )}{5 b}-\frac {2 c^2 \sqrt {a x+b x^2} (3 b C-4 a D)}{5 b (c x)^{7/2}}}{3 b c^3}-\frac {2 D \sqrt {a x+b x^2}}{3 b c^2 (c x)^{5/2}}\)

\(\Big \downarrow \) 1135

\(\displaystyle \frac {\frac {3 c^3 \left (-\frac {\left (-64 a^3 D+48 a^2 b C-40 a b^2 B+35 A b^3\right ) \left (-\frac {5 b \left (-\frac {3 b \left (-\frac {b \int \frac {1}{\sqrt {c x} \sqrt {b x^2+a x}}dx}{2 a c}-\frac {\sqrt {a x+b x^2}}{a (c x)^{3/2}}\right )}{4 a c}-\frac {\sqrt {a x+b x^2}}{2 a (c x)^{5/2}}\right )}{6 a c}-\frac {\sqrt {a x+b x^2}}{3 a (c x)^{7/2}}\right )}{8 a c}-\frac {5 A b^2 \sqrt {a x+b x^2}}{4 a (c x)^{9/2}}\right )}{5 b}-\frac {2 c^2 \sqrt {a x+b x^2} (3 b C-4 a D)}{5 b (c x)^{7/2}}}{3 b c^3}-\frac {2 D \sqrt {a x+b x^2}}{3 b c^2 (c x)^{5/2}}\)

\(\Big \downarrow \) 1136

\(\displaystyle \frac {\frac {3 c^3 \left (-\frac {\left (-64 a^3 D+48 a^2 b C-40 a b^2 B+35 A b^3\right ) \left (-\frac {5 b \left (-\frac {3 b \left (-\frac {b \int \frac {1}{\frac {c \left (b x^2+a x\right )}{x}-a c}d\frac {\sqrt {b x^2+a x}}{\sqrt {c x}}}{a}-\frac {\sqrt {a x+b x^2}}{a (c x)^{3/2}}\right )}{4 a c}-\frac {\sqrt {a x+b x^2}}{2 a (c x)^{5/2}}\right )}{6 a c}-\frac {\sqrt {a x+b x^2}}{3 a (c x)^{7/2}}\right )}{8 a c}-\frac {5 A b^2 \sqrt {a x+b x^2}}{4 a (c x)^{9/2}}\right )}{5 b}-\frac {2 c^2 \sqrt {a x+b x^2} (3 b C-4 a D)}{5 b (c x)^{7/2}}}{3 b c^3}-\frac {2 D \sqrt {a x+b x^2}}{3 b c^2 (c x)^{5/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {3 c^3 \left (-\frac {\left (-\frac {5 b \left (-\frac {3 b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a x+b x^2}}{\sqrt {a} \sqrt {c x}}\right )}{a^{3/2} c^{3/2}}-\frac {\sqrt {a x+b x^2}}{a (c x)^{3/2}}\right )}{4 a c}-\frac {\sqrt {a x+b x^2}}{2 a (c x)^{5/2}}\right )}{6 a c}-\frac {\sqrt {a x+b x^2}}{3 a (c x)^{7/2}}\right ) \left (-64 a^3 D+48 a^2 b C-40 a b^2 B+35 A b^3\right )}{8 a c}-\frac {5 A b^2 \sqrt {a x+b x^2}}{4 a (c x)^{9/2}}\right )}{5 b}-\frac {2 c^2 \sqrt {a x+b x^2} (3 b C-4 a D)}{5 b (c x)^{7/2}}}{3 b c^3}-\frac {2 D \sqrt {a x+b x^2}}{3 b c^2 (c x)^{5/2}}\)

Input:

Int[(A + B*x + C*x^2 + D*x^3)/((c*x)^(9/2)*Sqrt[a*x + b*x^2]),x]
 

Output:

(-2*D*Sqrt[a*x + b*x^2])/(3*b*c^2*(c*x)^(5/2)) + ((-2*c^2*(3*b*C - 4*a*D)* 
Sqrt[a*x + b*x^2])/(5*b*(c*x)^(7/2)) + (3*c^3*((-5*A*b^2*Sqrt[a*x + b*x^2] 
)/(4*a*(c*x)^(9/2)) - ((35*A*b^3 - 40*a*b^2*B + 48*a^2*b*C - 64*a^3*D)*(-1 
/3*Sqrt[a*x + b*x^2]/(a*(c*x)^(7/2)) - (5*b*(-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)))/ 
(6*a*c)))/(8*a*c)))/(5*b))/(3*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 = 0.42 (sec) , antiderivative size = 378, normalized size of antiderivative = 1.52

method	result	size
default	\(-\frac {\sqrt {x \left (b x +a \right )}\, \left (105 A \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (b x +a \right )}}{\sqrt {a c}}\right ) b^{4} c \,x^{4}-120 B \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (b x +a \right )}}{\sqrt {a c}}\right ) a \,b^{3} c \,x^{4}+144 C \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (b x +a \right )}}{\sqrt {a c}}\right ) a^{2} b^{2} c \,x^{4}-192 D \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (b x +a \right )}}{\sqrt {a c}}\right ) a^{3} b c \,x^{4}-105 A \,b^{3} x^{3} \sqrt {c \left (b x +a \right )}\, \sqrt {a c}+120 B a \,b^{2} x^{3} \sqrt {c \left (b x +a \right )}\, \sqrt {a c}-144 C \,a^{2} b \,x^{3} \sqrt {c \left (b x +a \right )}\, \sqrt {a c}+192 D a^{3} x^{3} \sqrt {c \left (b x +a \right )}\, \sqrt {a c}+70 A a \,b^{2} x^{2} \sqrt {c \left (b x +a \right )}\, \sqrt {a c}-80 B \,a^{2} b \,x^{2} \sqrt {c \left (b x +a \right )}\, \sqrt {a c}+96 C \,a^{3} x^{2} \sqrt {c \left (b x +a \right )}\, \sqrt {a c}-56 A \,a^{2} b x \sqrt {c \left (b x +a \right )}\, \sqrt {a c}+64 B \,a^{3} x \sqrt {c \left (b x +a \right )}\, \sqrt {a c}+48 A \,a^{3} \sqrt {c \left (b x +a \right )}\, \sqrt {a c}\right )}{192 c^{4} x^{4} \sqrt {c x}\, \sqrt {c \left (b x +a \right )}\, a^{4} \sqrt {a c}}\)	\(378\)

Input:

int((D*x^3+C*x^2+B*x+A)/(c*x)^(9/2)/(b*x^2+a*x)^(1/2),x,method=_RETURNVERB 
OSE)
 

Output:

-1/192*(x*(b*x+a))^(1/2)/c^4*(105*A*arctanh((c*(b*x+a))^(1/2)/(a*c)^(1/2)) 
*b^4*c*x^4-120*B*arctanh((c*(b*x+a))^(1/2)/(a*c)^(1/2))*a*b^3*c*x^4+144*C* 
arctanh((c*(b*x+a))^(1/2)/(a*c)^(1/2))*a^2*b^2*c*x^4-192*D*arctanh((c*(b*x 
+a))^(1/2)/(a*c)^(1/2))*a^3*b*c*x^4-105*A*b^3*x^3*(c*(b*x+a))^(1/2)*(a*c)^ 
(1/2)+120*B*a*b^2*x^3*(c*(b*x+a))^(1/2)*(a*c)^(1/2)-144*C*a^2*b*x^3*(c*(b* 
x+a))^(1/2)*(a*c)^(1/2)+192*D*a^3*x^3*(c*(b*x+a))^(1/2)*(a*c)^(1/2)+70*A*a 
*b^2*x^2*(c*(b*x+a))^(1/2)*(a*c)^(1/2)-80*B*a^2*b*x^2*(c*(b*x+a))^(1/2)*(a 
*c)^(1/2)+96*C*a^3*x^2*(c*(b*x+a))^(1/2)*(a*c)^(1/2)-56*A*a^2*b*x*(c*(b*x+ 
a))^(1/2)*(a*c)^(1/2)+64*B*a^3*x*(c*(b*x+a))^(1/2)*(a*c)^(1/2)+48*A*a^3*(c 
*(b*x+a))^(1/2)*(a*c)^(1/2))/x^4/(c*x)^(1/2)/(c*(b*x+a))^(1/2)/a^4/(a*c)^( 
1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 389, normalized size of antiderivative = 1.56 \[ \int \frac {A+B x+C x^2+D x^3}{(c x)^{9/2} \sqrt {a x+b x^2}} \, dx=\left [-\frac {3 \, {\left (64 \, D a^{3} b - 48 \, C a^{2} b^{2} + 40 \, B a b^{3} - 35 \, A b^{4}\right )} \sqrt {a c} x^{5} \log \left (-\frac {b c x^{2} + 2 \, a c x - 2 \, \sqrt {b x^{2} + a x} \sqrt {a c} \sqrt {c x}}{x^{2}}\right ) + 2 \, {\left (48 \, A a^{4} + 3 \, {\left (64 \, D a^{4} - 48 \, C a^{3} b + 40 \, B a^{2} b^{2} - 35 \, A a b^{3}\right )} x^{3} + 2 \, {\left (48 \, C a^{4} - 40 \, B a^{3} b + 35 \, A a^{2} b^{2}\right )} x^{2} + 8 \, {\left (8 \, B a^{4} - 7 \, A a^{3} b\right )} x\right )} \sqrt {b x^{2} + a x} \sqrt {c x}}{384 \, a^{5} c^{5} x^{5}}, -\frac {3 \, {\left (64 \, D a^{3} b - 48 \, C a^{2} b^{2} + 40 \, B a b^{3} - 35 \, A b^{4}\right )} \sqrt {-a c} x^{5} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-a c} \sqrt {c x}}{a c x}\right ) + {\left (48 \, A a^{4} + 3 \, {\left (64 \, D a^{4} - 48 \, C a^{3} b + 40 \, B a^{2} b^{2} - 35 \, A a b^{3}\right )} x^{3} + 2 \, {\left (48 \, C a^{4} - 40 \, B a^{3} b + 35 \, A a^{2} b^{2}\right )} x^{2} + 8 \, {\left (8 \, B a^{4} - 7 \, A a^{3} b\right )} x\right )} \sqrt {b x^{2} + a x} \sqrt {c x}}{192 \, a^{5} c^{5} x^{5}}\right ] \] Input:

integrate((D*x^3+C*x^2+B*x+A)/(c*x)^(9/2)/(b*x^2+a*x)^(1/2),x, algorithm=" 
fricas")
                                                                                    
                                                                                    
 

Output:

[-1/384*(3*(64*D*a^3*b - 48*C*a^2*b^2 + 40*B*a*b^3 - 35*A*b^4)*sqrt(a*c)*x 
^5*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*(48*A*a^4 + 3*(64*D*a^4 - 48*C*a^3*b + 40*B*a^2*b^2 - 35*A*a*b^3)*x^3 
 + 2*(48*C*a^4 - 40*B*a^3*b + 35*A*a^2*b^2)*x^2 + 8*(8*B*a^4 - 7*A*a^3*b)* 
x)*sqrt(b*x^2 + a*x)*sqrt(c*x))/(a^5*c^5*x^5), -1/192*(3*(64*D*a^3*b - 48* 
C*a^2*b^2 + 40*B*a*b^3 - 35*A*b^4)*sqrt(-a*c)*x^5*arctan(sqrt(b*x^2 + a*x) 
*sqrt(-a*c)*sqrt(c*x)/(a*c*x)) + (48*A*a^4 + 3*(64*D*a^4 - 48*C*a^3*b + 40 
*B*a^2*b^2 - 35*A*a*b^3)*x^3 + 2*(48*C*a^4 - 40*B*a^3*b + 35*A*a^2*b^2)*x^ 
2 + 8*(8*B*a^4 - 7*A*a^3*b)*x)*sqrt(b*x^2 + a*x)*sqrt(c*x))/(a^5*c^5*x^5)]
 

Sympy [F]

\[ \int \frac {A+B x+C x^2+D x^3}{(c x)^{9/2} \sqrt {a x+b x^2}} \, dx=\int \frac {A + B x + C x^{2} + D x^{3}}{\left (c x\right )^{\frac {9}{2}} \sqrt {x \left (a + b x\right )}}\, dx \] Input:

integrate((D*x**3+C*x**2+B*x+A)/(c*x)**(9/2)/(b*x**2+a*x)**(1/2),x)
 

Output:

Integral((A + B*x + C*x**2 + D*x**3)/((c*x)**(9/2)*sqrt(x*(a + b*x))), x)
 

Maxima [F]

\[ \int \frac {A+B x+C x^2+D x^3}{(c x)^{9/2} \sqrt {a x+b x^2}} \, dx=\int { \frac {D x^{3} + C x^{2} + B x + A}{\sqrt {b x^{2} + a x} \left (c x\right )^{\frac {9}{2}}} \,d x } \] Input:

integrate((D*x^3+C*x^2+B*x+A)/(c*x)^(9/2)/(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)^(9/2)), x)
 

Giac [A] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 417, normalized size of antiderivative = 1.67 \[ \int \frac {A+B x+C x^2+D x^3}{(c x)^{9/2} \sqrt {a x+b x^2}} \, dx=-\frac {\frac {3 \, {\left (64 \, D a^{3} b^{2} - 48 \, C a^{2} b^{3} + 40 \, B a b^{4} - 35 \, A b^{5}\right )} \arctan \left (\frac {\sqrt {b c x + a c}}{\sqrt {-a c}}\right )}{\sqrt {-a c} a^{4} c} - \frac {192 \, \sqrt {b c x + a c} D a^{6} b^{2} c^{3} - 240 \, \sqrt {b c x + a c} C a^{5} b^{3} c^{3} + 264 \, \sqrt {b c x + a c} B a^{4} b^{4} c^{3} - 279 \, \sqrt {b c x + a c} A a^{3} b^{5} c^{3} - 576 \, {\left (b c x + a c\right )}^{\frac {3}{2}} D a^{5} b^{2} c^{2} + 624 \, {\left (b c x + a c\right )}^{\frac {3}{2}} C a^{4} b^{3} c^{2} - 584 \, {\left (b c x + a c\right )}^{\frac {3}{2}} B a^{3} b^{4} c^{2} + 511 \, {\left (b c x + a c\right )}^{\frac {3}{2}} A a^{2} b^{5} c^{2} + 576 \, {\left (b c x + a c\right )}^{\frac {5}{2}} D a^{4} b^{2} c - 528 \, {\left (b c x + a c\right )}^{\frac {5}{2}} C a^{3} b^{3} c + 440 \, {\left (b c x + a c\right )}^{\frac {5}{2}} B a^{2} b^{4} c - 385 \, {\left (b c x + a c\right )}^{\frac {5}{2}} A a b^{5} c - 192 \, {\left (b c x + a c\right )}^{\frac {7}{2}} D a^{3} b^{2} + 144 \, {\left (b c x + a c\right )}^{\frac {7}{2}} C a^{2} b^{3} - 120 \, {\left (b c x + a c\right )}^{\frac {7}{2}} B a b^{4} + 105 \, {\left (b c x + a c\right )}^{\frac {7}{2}} A b^{5}}{a^{4} b^{4} c^{5} x^{4}}}{192 \, b c^{2} {\left | c \right |}} \] Input:

integrate((D*x^3+C*x^2+B*x+A)/(c*x)^(9/2)/(b*x^2+a*x)^(1/2),x, algorithm=" 
giac")
 

Output:

-1/192*(3*(64*D*a^3*b^2 - 48*C*a^2*b^3 + 40*B*a*b^4 - 35*A*b^5)*arctan(sqr 
t(b*c*x + a*c)/sqrt(-a*c))/(sqrt(-a*c)*a^4*c) - (192*sqrt(b*c*x + a*c)*D*a 
^6*b^2*c^3 - 240*sqrt(b*c*x + a*c)*C*a^5*b^3*c^3 + 264*sqrt(b*c*x + a*c)*B 
*a^4*b^4*c^3 - 279*sqrt(b*c*x + a*c)*A*a^3*b^5*c^3 - 576*(b*c*x + a*c)^(3/ 
2)*D*a^5*b^2*c^2 + 624*(b*c*x + a*c)^(3/2)*C*a^4*b^3*c^2 - 584*(b*c*x + a* 
c)^(3/2)*B*a^3*b^4*c^2 + 511*(b*c*x + a*c)^(3/2)*A*a^2*b^5*c^2 + 576*(b*c* 
x + a*c)^(5/2)*D*a^4*b^2*c - 528*(b*c*x + a*c)^(5/2)*C*a^3*b^3*c + 440*(b* 
c*x + a*c)^(5/2)*B*a^2*b^4*c - 385*(b*c*x + a*c)^(5/2)*A*a*b^5*c - 192*(b* 
c*x + a*c)^(7/2)*D*a^3*b^2 + 144*(b*c*x + a*c)^(7/2)*C*a^2*b^3 - 120*(b*c* 
x + a*c)^(7/2)*B*a*b^4 + 105*(b*c*x + a*c)^(7/2)*A*b^5)/(a^4*b^4*c^5*x^4)) 
/(b*c^2*abs(c))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{(c x)^{9/2} \sqrt {a x+b x^2}} \, dx=\int \frac {A+B\,x+C\,x^2+x^3\,D}{\sqrt {b\,x^2+a\,x}\,{\left (c\,x\right )}^{9/2}} \,d x \] Input:

int((A + B*x + C*x^2 + x^3*D)/((a*x + b*x^2)^(1/2)*(c*x)^(9/2)),x)
 

Output:

int((A + B*x + C*x^2 + x^3*D)/((a*x + b*x^2)^(1/2)*(c*x)^(9/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x+C x^2+D x^3}{(c x)^{9/2} \sqrt {a x+b x^2}} \, dx=\frac {\sqrt {c}\, \left (-96 \sqrt {b x +a}\, a^{4}-16 \sqrt {b x +a}\, a^{3} b x -192 \sqrt {b x +a}\, a^{3} c \,x^{2}-384 \sqrt {b x +a}\, a^{3} d \,x^{3}+20 \sqrt {b x +a}\, a^{2} b^{2} x^{2}+288 \sqrt {b x +a}\, a^{2} b c \,x^{3}-30 \sqrt {b x +a}\, a \,b^{3} x^{3}-192 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a^{2} b d \,x^{4}+144 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a \,b^{2} c \,x^{4}-15 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{4} x^{4}+192 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a^{2} b d \,x^{4}-144 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a \,b^{2} c \,x^{4}+15 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{4} x^{4}\right )}{384 a^{4} c^{5} x^{4}} \] Input:

int((D*x^3+C*x^2+B*x+A)/(c*x)^(9/2)/(b*x^2+a*x)^(1/2),x)
 

Output:

(sqrt(c)*( - 96*sqrt(a + b*x)*a**4 - 16*sqrt(a + b*x)*a**3*b*x - 192*sqrt( 
a + b*x)*a**3*c*x**2 - 384*sqrt(a + b*x)*a**3*d*x**3 + 20*sqrt(a + b*x)*a* 
*2*b**2*x**2 + 288*sqrt(a + b*x)*a**2*b*c*x**3 - 30*sqrt(a + b*x)*a*b**3*x 
**3 - 192*sqrt(a)*log(sqrt(a + b*x) - sqrt(a))*a**2*b*d*x**4 + 144*sqrt(a) 
*log(sqrt(a + b*x) - sqrt(a))*a*b**2*c*x**4 - 15*sqrt(a)*log(sqrt(a + b*x) 
 - sqrt(a))*b**4*x**4 + 192*sqrt(a)*log(sqrt(a + b*x) + sqrt(a))*a**2*b*d* 
x**4 - 144*sqrt(a)*log(sqrt(a + b*x) + sqrt(a))*a*b**2*c*x**4 + 15*sqrt(a) 
*log(sqrt(a + b*x) + sqrt(a))*b**4*x**4))/(384*a**4*c**5*x**4)