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

Optimal result

Integrand size = 36, antiderivative size = 228 \[ \int \frac {(c x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{\sqrt {a x+b x^2}} \, dx=-\frac {2 a c^2 \left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \sqrt {a x+b x^2}}{b^5 \sqrt {c x}}+\frac {2 c^3 \left (A b^3-a \left (2 b^2 B-3 a b C+4 a^2 D\right )\right ) \left (a x+b x^2\right )^{3/2}}{3 b^5 (c x)^{3/2}}+\frac {2 c^4 \left (b^2 B-3 a b C+6 a^2 D\right ) \left (a x+b x^2\right )^{5/2}}{5 b^5 (c x)^{5/2}}+\frac {2 c^5 (b C-4 a D) \left (a x+b x^2\right )^{7/2}}{7 b^5 (c x)^{7/2}}+\frac {2 c^6 D \left (a x+b x^2\right )^{9/2}}{9 b^5 (c x)^{9/2}} \] Output:

-2*a*c^2*(A*b^3-a*(B*b^2-C*a*b+D*a^2))*(b*x^2+a*x)^(1/2)/b^5/(c*x)^(1/2)+2 
/3*c^3*(A*b^3-a*(2*B*b^2-3*C*a*b+4*D*a^2))*(b*x^2+a*x)^(3/2)/b^5/(c*x)^(3/ 
2)+2/5*c^4*(B*b^2-3*C*a*b+6*D*a^2)*(b*x^2+a*x)^(5/2)/b^5/(c*x)^(5/2)+2/7*c 
^5*(C*b-4*D*a)*(b*x^2+a*x)^(7/2)/b^5/(c*x)^(7/2)+2/9*c^6*D*(b*x^2+a*x)^(9/ 
2)/b^5/(c*x)^(9/2)
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.54 \[ \int \frac {(c x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{\sqrt {a x+b x^2}} \, dx=\frac {2 c^2 \sqrt {x (a+b x)} \left (128 a^4 D-16 a^3 b (9 C+4 D x)+24 a^2 b^2 (7 B+x (3 C+2 D x))+b^4 x (105 A+x (63 B+5 x (9 C+7 D x)))-2 a b^3 (105 A+x (42 B+x (27 C+20 D x)))\right )}{315 b^5 \sqrt {c x}} \] Input:

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

Output:

(2*c^2*Sqrt[x*(a + b*x)]*(128*a^4*D - 16*a^3*b*(9*C + 4*D*x) + 24*a^2*b^2* 
(7*B + x*(3*C + 2*D*x)) + b^4*x*(105*A + x*(63*B + 5*x*(9*C + 7*D*x))) - 2 
*a*b^3*(105*A + x*(42*B + x*(27*C + 20*D*x)))))/(315*b^5*Sqrt[c*x])
 

Rubi [A] (verified)

Time = 0.99 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {2169, 27, 2169, 27, 1221, 1128, 1122}

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.

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

\(\Big \downarrow \) 2169

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2169

\(\displaystyle \frac {\frac {2 \int \frac {3 c^5 (c x)^{3/2} \left (21 A b^2+\left (16 D a^2-18 b C a+21 b^2 B\right ) x\right )}{2 \sqrt {b x^2+a x}}dx}{7 b c^2}+\frac {2 c^2 (c x)^{5/2} \sqrt {a x+b x^2} (9 b C-8 a D)}{7 b}}{9 b c^3}+\frac {2 D (c x)^{7/2} \sqrt {a x+b x^2}}{9 b c^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3 c^3 \int \frac {(c x)^{3/2} \left (21 A b^2+\left (16 D a^2-18 b C a+21 b^2 B\right ) x\right )}{\sqrt {b x^2+a x}}dx}{7 b}+\frac {2 c^2 (c x)^{5/2} \sqrt {a x+b x^2} (9 b C-8 a D)}{7 b}}{9 b c^3}+\frac {2 D (c x)^{7/2} \sqrt {a x+b x^2}}{9 b c^2}\)

\(\Big \downarrow \) 1221

\(\displaystyle \frac {\frac {3 c^3 \left (\frac {\left (-64 a^3 D+72 a^2 b C-84 a b^2 B+105 A b^3\right ) \int \frac {(c x)^{3/2}}{\sqrt {b x^2+a x}}dx}{5 b}+\frac {2 (c x)^{3/2} \sqrt {a x+b x^2} \left (16 a^2 D-18 a b C+21 b^2 B\right )}{5 b}\right )}{7 b}+\frac {2 c^2 (c x)^{5/2} \sqrt {a x+b x^2} (9 b C-8 a D)}{7 b}}{9 b c^3}+\frac {2 D (c x)^{7/2} \sqrt {a x+b x^2}}{9 b c^2}\)

\(\Big \downarrow \) 1128

\(\displaystyle \frac {\frac {3 c^3 \left (\frac {\left (-64 a^3 D+72 a^2 b C-84 a b^2 B+105 A b^3\right ) \left (\frac {2 c \sqrt {c x} \sqrt {a x+b x^2}}{3 b}-\frac {2 a c \int \frac {\sqrt {c x}}{\sqrt {b x^2+a x}}dx}{3 b}\right )}{5 b}+\frac {2 (c x)^{3/2} \sqrt {a x+b x^2} \left (16 a^2 D-18 a b C+21 b^2 B\right )}{5 b}\right )}{7 b}+\frac {2 c^2 (c x)^{5/2} \sqrt {a x+b x^2} (9 b C-8 a D)}{7 b}}{9 b c^3}+\frac {2 D (c x)^{7/2} \sqrt {a x+b x^2}}{9 b c^2}\)

\(\Big \downarrow \) 1122

\(\displaystyle \frac {\frac {3 c^3 \left (\frac {2 (c x)^{3/2} \sqrt {a x+b x^2} \left (16 a^2 D-18 a b C+21 b^2 B\right )}{5 b}+\frac {\left (\frac {2 c \sqrt {c x} \sqrt {a x+b x^2}}{3 b}-\frac {4 a c^2 \sqrt {a x+b x^2}}{3 b^2 \sqrt {c x}}\right ) \left (-64 a^3 D+72 a^2 b C-84 a b^2 B+105 A b^3\right )}{5 b}\right )}{7 b}+\frac {2 c^2 (c x)^{5/2} \sqrt {a x+b x^2} (9 b C-8 a D)}{7 b}}{9 b c^3}+\frac {2 D (c x)^{7/2} \sqrt {a x+b x^2}}{9 b c^2}\)

Input:

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

Output:

(2*D*(c*x)^(7/2)*Sqrt[a*x + b*x^2])/(9*b*c^2) + ((2*c^2*(9*b*C - 8*a*D)*(c 
*x)^(5/2)*Sqrt[a*x + b*x^2])/(7*b) + (3*c^3*((2*(21*b^2*B - 18*a*b*C + 16* 
a^2*D)*(c*x)^(3/2)*Sqrt[a*x + b*x^2])/(5*b) + ((105*A*b^3 - 84*a*b^2*B + 7 
2*a^2*b*C - 64*a^3*D)*((-4*a*c^2*Sqrt[a*x + b*x^2])/(3*b^2*Sqrt[c*x]) + (2 
*c*Sqrt[c*x]*Sqrt[a*x + b*x^2])/(3*b)))/(5*b)))/(7*b))/(9*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 1122

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

rule 1128

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

rule 1221

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c 
_.)*(x_)^2)^(p_), x_Symbol] :> Simp[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1 
)/(c*(m + 2*p + 2))), x] + Simp[(m*(g*(c*d - b*e) + c*e*f) + e*(p + 1)*(2*c 
*f - b*g))/(c*e*(m + 2*p + 2))   Int[(d + e*x)^m*(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] 
 && NeQ[m + 2*p + 2, 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 = 0.49 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.64

method	result	size
default	\(-\frac {2 c \sqrt {c x}\, \sqrt {x \left (b x +a \right )}\, \left (-35 D x^{4} b^{4}-45 C \,b^{4} x^{3}+40 D a \,b^{3} x^{3}-63 B \,b^{4} x^{2}+54 C a \,b^{3} x^{2}-48 D a^{2} b^{2} x^{2}-105 A \,b^{4} x +84 B a \,b^{3} x -72 C \,a^{2} b^{2} x +64 D a^{3} b x +210 A a \,b^{3}-168 B \,a^{2} b^{2}+144 C \,a^{3} b -128 D a^{4}\right )}{315 x \,b^{5}}\)	\(146\)
gosper	\(-\frac {2 \left (b x +a \right ) \left (-35 D x^{4} b^{4}-45 C \,b^{4} x^{3}+40 D a \,b^{3} x^{3}-63 B \,b^{4} x^{2}+54 C a \,b^{3} x^{2}-48 D a^{2} b^{2} x^{2}-105 A \,b^{4} x +84 B a \,b^{3} x -72 C \,a^{2} b^{2} x +64 D a^{3} b x +210 A a \,b^{3}-168 B \,a^{2} b^{2}+144 C \,a^{3} b -128 D a^{4}\right ) \left (c x \right )^{\frac {3}{2}}}{315 b^{5} x \sqrt {b \,x^{2}+a x}}\)	\(152\)
orering	\(-\frac {2 \left (b x +a \right ) \left (-35 D x^{4} b^{4}-45 C \,b^{4} x^{3}+40 D a \,b^{3} x^{3}-63 B \,b^{4} x^{2}+54 C a \,b^{3} x^{2}-48 D a^{2} b^{2} x^{2}-105 A \,b^{4} x +84 B a \,b^{3} x -72 C \,a^{2} b^{2} x +64 D a^{3} b x +210 A a \,b^{3}-168 B \,a^{2} b^{2}+144 C \,a^{3} b -128 D a^{4}\right ) \left (c x \right )^{\frac {3}{2}}}{315 b^{5} x \sqrt {b \,x^{2}+a x}}\)	\(152\)

Input:

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

Output:

-2/315*c/x*(c*x)^(1/2)*(x*(b*x+a))^(1/2)*(-35*D*b^4*x^4-45*C*b^4*x^3+40*D* 
a*b^3*x^3-63*B*b^4*x^2+54*C*a*b^3*x^2-48*D*a^2*b^2*x^2-105*A*b^4*x+84*B*a* 
b^3*x-72*C*a^2*b^2*x+64*D*a^3*b*x+210*A*a*b^3-168*B*a^2*b^2+144*C*a^3*b-12 
8*D*a^4)/b^5
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.66 \[ \int \frac {(c x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{\sqrt {a x+b x^2}} \, dx=\frac {2 \, {\left (35 \, D b^{4} c x^{4} - 5 \, {\left (8 \, D a b^{3} - 9 \, C b^{4}\right )} c x^{3} + 3 \, {\left (16 \, D a^{2} b^{2} - 18 \, C a b^{3} + 21 \, B b^{4}\right )} c x^{2} - {\left (64 \, D a^{3} b - 72 \, C a^{2} b^{2} + 84 \, B a b^{3} - 105 \, A b^{4}\right )} c x + 2 \, {\left (64 \, D a^{4} - 72 \, C a^{3} b + 84 \, B a^{2} b^{2} - 105 \, A a b^{3}\right )} c\right )} \sqrt {b x^{2} + a x} \sqrt {c x}}{315 \, b^{5} x} \] Input:

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

Output:

2/315*(35*D*b^4*c*x^4 - 5*(8*D*a*b^3 - 9*C*b^4)*c*x^3 + 3*(16*D*a^2*b^2 - 
18*C*a*b^3 + 21*B*b^4)*c*x^2 - (64*D*a^3*b - 72*C*a^2*b^2 + 84*B*a*b^3 - 1 
05*A*b^4)*c*x + 2*(64*D*a^4 - 72*C*a^3*b + 84*B*a^2*b^2 - 105*A*a*b^3)*c)* 
sqrt(b*x^2 + a*x)*sqrt(c*x)/(b^5*x)
                                                                                    
                                                                                    
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.09 \[ \int \frac {(c x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{\sqrt {a x+b x^2}} \, dx=\frac {2 \, {\left (b^{2} c^{\frac {3}{2}} x^{2} - a b c^{\frac {3}{2}} x - 2 \, a^{2} c^{\frac {3}{2}}\right )} A}{3 \, \sqrt {b x + a} b^{2}} + \frac {2 \, {\left (3 \, b^{3} c^{\frac {3}{2}} x^{3} - a b^{2} c^{\frac {3}{2}} x^{2} + 4 \, a^{2} b c^{\frac {3}{2}} x + 8 \, a^{3} c^{\frac {3}{2}}\right )} B}{15 \, \sqrt {b x + a} b^{3}} + \frac {2 \, {\left (5 \, b^{4} c^{\frac {3}{2}} x^{4} - a b^{3} c^{\frac {3}{2}} x^{3} + 2 \, a^{2} b^{2} c^{\frac {3}{2}} x^{2} - 8 \, a^{3} b c^{\frac {3}{2}} x - 16 \, a^{4} c^{\frac {3}{2}}\right )} C}{35 \, \sqrt {b x + a} b^{4}} + \frac {2 \, {\left (35 \, b^{5} c^{\frac {3}{2}} x^{5} - 5 \, a b^{4} c^{\frac {3}{2}} x^{4} + 8 \, a^{2} b^{3} c^{\frac {3}{2}} x^{3} - 16 \, a^{3} b^{2} c^{\frac {3}{2}} x^{2} + 64 \, a^{4} b c^{\frac {3}{2}} x + 128 \, a^{5} c^{\frac {3}{2}}\right )} D}{315 \, \sqrt {b x + a} b^{5}} \] Input:

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

Output:

2/3*(b^2*c^(3/2)*x^2 - a*b*c^(3/2)*x - 2*a^2*c^(3/2))*A/(sqrt(b*x + a)*b^2 
) + 2/15*(3*b^3*c^(3/2)*x^3 - a*b^2*c^(3/2)*x^2 + 4*a^2*b*c^(3/2)*x + 8*a^ 
3*c^(3/2))*B/(sqrt(b*x + a)*b^3) + 2/35*(5*b^4*c^(3/2)*x^4 - a*b^3*c^(3/2) 
*x^3 + 2*a^2*b^2*c^(3/2)*x^2 - 8*a^3*b*c^(3/2)*x - 16*a^4*c^(3/2))*C/(sqrt 
(b*x + a)*b^4) + 2/315*(35*b^5*c^(3/2)*x^5 - 5*a*b^4*c^(3/2)*x^4 + 8*a^2*b 
^3*c^(3/2)*x^3 - 16*a^3*b^2*c^(3/2)*x^2 + 64*a^4*b*c^(3/2)*x + 128*a^5*c^( 
3/2))*D/(sqrt(b*x + a)*b^5)
 

Giac [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 300, normalized size of antiderivative = 1.32 \[ \int \frac {(c x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{\sqrt {a x+b x^2}} \, dx=\frac {2 \, c^{3} {\left (\frac {315 \, {\left (D a^{4} - C a^{3} b + B a^{2} b^{2} - A a b^{3}\right )} \sqrt {b c x + a c}}{b^{5} c} - \frac {2 \, {\left (64 \, \sqrt {a c} D a^{4} - 72 \, \sqrt {a c} C a^{3} b + 84 \, \sqrt {a c} B a^{2} b^{2} - 105 \, \sqrt {a c} A a b^{3}\right )}}{b^{5} c} - \frac {420 \, {\left (b c x + a c\right )}^{\frac {3}{2}} D a^{3} c^{3} - 315 \, {\left (b c x + a c\right )}^{\frac {3}{2}} C a^{2} b c^{3} + 210 \, {\left (b c x + a c\right )}^{\frac {3}{2}} B a b^{2} c^{3} - 105 \, {\left (b c x + a c\right )}^{\frac {3}{2}} A b^{3} c^{3} - 378 \, {\left (b c x + a c\right )}^{\frac {5}{2}} D a^{2} c^{2} + 189 \, {\left (b c x + a c\right )}^{\frac {5}{2}} C a b c^{2} - 63 \, {\left (b c x + a c\right )}^{\frac {5}{2}} B b^{2} c^{2} + 180 \, {\left (b c x + a c\right )}^{\frac {7}{2}} D a c - 45 \, {\left (b c x + a c\right )}^{\frac {7}{2}} C b c - 35 \, {\left (b c x + a c\right )}^{\frac {9}{2}} D}{b^{5} c^{5}}\right )}}{315 \, {\left | c \right |}} \] Input:

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

Output:

2/315*c^3*(315*(D*a^4 - C*a^3*b + B*a^2*b^2 - A*a*b^3)*sqrt(b*c*x + a*c)/( 
b^5*c) - 2*(64*sqrt(a*c)*D*a^4 - 72*sqrt(a*c)*C*a^3*b + 84*sqrt(a*c)*B*a^2 
*b^2 - 105*sqrt(a*c)*A*a*b^3)/(b^5*c) - (420*(b*c*x + a*c)^(3/2)*D*a^3*c^3 
 - 315*(b*c*x + a*c)^(3/2)*C*a^2*b*c^3 + 210*(b*c*x + a*c)^(3/2)*B*a*b^2*c 
^3 - 105*(b*c*x + a*c)^(3/2)*A*b^3*c^3 - 378*(b*c*x + a*c)^(5/2)*D*a^2*c^2 
 + 189*(b*c*x + a*c)^(5/2)*C*a*b*c^2 - 63*(b*c*x + a*c)^(5/2)*B*b^2*c^2 + 
180*(b*c*x + a*c)^(7/2)*D*a*c - 45*(b*c*x + a*c)^(7/2)*C*b*c - 35*(b*c*x + 
 a*c)^(9/2)*D)/(b^5*c^5))/abs(c)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.52 \[ \int \frac {(c x)^{3/2} \left (A+B x+C x^2+D x^3\right )}{\sqrt {a x+b x^2}} \, dx=\frac {2 \sqrt {c}\, \sqrt {b x +a}\, c \left (35 b^{4} d \,x^{4}-40 a \,b^{3} d \,x^{3}+45 b^{4} c \,x^{3}+48 a^{2} b^{2} d \,x^{2}-54 a \,b^{3} c \,x^{2}+63 b^{5} x^{2}-64 a^{3} b d x +72 a^{2} b^{2} c x +21 a \,b^{4} x +128 a^{4} d -144 a^{3} b c -42 a^{2} b^{3}\right )}{315 b^{5}} \] Input:

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

Output:

(2*sqrt(c)*sqrt(a + b*x)*c*(128*a**4*d - 144*a**3*b*c - 64*a**3*b*d*x - 42 
*a**2*b**3 + 72*a**2*b**2*c*x + 48*a**2*b**2*d*x**2 + 21*a*b**4*x - 54*a*b 
**3*c*x**2 - 40*a*b**3*d*x**3 + 63*b**5*x**2 + 45*b**4*c*x**3 + 35*b**4*d* 
x**4))/(315*b**5)