\(\int \frac {\sqrt {c x} (A+B x+C x^2+D x^3)}{\sqrt {a x+b x^2}} \, dx\) [4]

Optimal result

Integrand size = 36, antiderivative size = 168 \[ \int \frac {\sqrt {c x} \left (A+B x+C x^2+D x^3\right )}{\sqrt {a x+b x^2}} \, dx=\frac {2 c \left (A b^3-a \left (b^2 B-a b C+a^2 D\right )\right ) \sqrt {a x+b x^2}}{b^4 \sqrt {c x}}+\frac {2 c^2 \left (b^2 B-2 a b C+3 a^2 D\right ) \left (a x+b x^2\right )^{3/2}}{3 b^4 (c x)^{3/2}}+\frac {2 c^3 (b C-3 a D) \left (a x+b x^2\right )^{5/2}}{5 b^4 (c x)^{5/2}}+\frac {2 c^4 D \left (a x+b x^2\right )^{7/2}}{7 b^4 (c x)^{7/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.55 \[ \int \frac {\sqrt {c x} \left (A+B x+C x^2+D x^3\right )}{\sqrt {a x+b x^2}} \, dx=\frac {2 c \sqrt {x (a+b x)} \left (105 A b^3-48 a^3 D+8 a^2 b (7 C+3 D x)+b^3 x (35 B+3 x (7 C+5 D x))-2 a b^2 (35 B+x (14 C+9 D x))\right )}{105 b^4 \sqrt {c x}} \] Input:

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

Output:

(2*c*Sqrt[x*(a + b*x)]*(105*A*b^3 - 48*a^3*D + 8*a^2*b*(7*C + 3*D*x) + b^3 
*x*(35*B + 3*x*(7*C + 5*D*x)) - 2*a*b^2*(35*B + x*(14*C + 9*D*x))))/(105*b 
^4*Sqrt[c*x])
 

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.15, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2169, 27, 2169, 27, 1221, 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.

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

\(\Big \downarrow \) 2169

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2169

\(\displaystyle \frac {\frac {2 \int \frac {c^5 \sqrt {c x} \left (35 A b^2+\left (24 D a^2-28 b C a+35 b^2 B\right ) x\right )}{2 \sqrt {b x^2+a x}}dx}{5 b c^2}+\frac {2 c^2 (c x)^{3/2} \sqrt {a x+b x^2} (7 b C-6 a D)}{5 b}}{7 b c^3}+\frac {2 D (c x)^{5/2} \sqrt {a x+b x^2}}{7 b c^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {c^3 \int \frac {\sqrt {c x} \left (35 A b^2+\left (24 D a^2-28 b C a+35 b^2 B\right ) x\right )}{\sqrt {b x^2+a x}}dx}{5 b}+\frac {2 c^2 (c x)^{3/2} \sqrt {a x+b x^2} (7 b C-6 a D)}{5 b}}{7 b c^3}+\frac {2 D (c x)^{5/2} \sqrt {a x+b x^2}}{7 b c^2}\)

\(\Big \downarrow \) 1221

\(\displaystyle \frac {\frac {c^3 \left (\frac {\left (-48 a^3 D+56 a^2 b C-70 a b^2 B+105 A b^3\right ) \int \frac {\sqrt {c x}}{\sqrt {b x^2+a x}}dx}{3 b}+\frac {2 \sqrt {c x} \sqrt {a x+b x^2} \left (24 a^2 D-28 a b C+35 b^2 B\right )}{3 b}\right )}{5 b}+\frac {2 c^2 (c x)^{3/2} \sqrt {a x+b x^2} (7 b C-6 a D)}{5 b}}{7 b c^3}+\frac {2 D (c x)^{5/2} \sqrt {a x+b x^2}}{7 b c^2}\)

\(\Big \downarrow \) 1122

\(\displaystyle \frac {\frac {c^3 \left (\frac {2 \sqrt {c x} \sqrt {a x+b x^2} \left (24 a^2 D-28 a b C+35 b^2 B\right )}{3 b}+\frac {2 c \sqrt {a x+b x^2} \left (-48 a^3 D+56 a^2 b C-70 a b^2 B+105 A b^3\right )}{3 b^2 \sqrt {c x}}\right )}{5 b}+\frac {2 c^2 (c x)^{3/2} \sqrt {a x+b x^2} (7 b C-6 a D)}{5 b}}{7 b c^3}+\frac {2 D (c x)^{5/2} \sqrt {a x+b x^2}}{7 b c^2}\)

Input:

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

Output:

(2*D*(c*x)^(5/2)*Sqrt[a*x + b*x^2])/(7*b*c^2) + ((2*c^2*(7*b*C - 6*a*D)*(c 
*x)^(3/2)*Sqrt[a*x + b*x^2])/(5*b) + (c^3*((2*c*(105*A*b^3 - 70*a*b^2*B + 
56*a^2*b*C - 48*a^3*D)*Sqrt[a*x + b*x^2])/(3*b^2*Sqrt[c*x]) + (2*(35*b^2*B 
 - 28*a*b*C + 24*a^2*D)*Sqrt[c*x]*Sqrt[a*x + b*x^2])/(3*b)))/(5*b))/(7*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 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 = 0.41 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.60

method	result	size
default	\(\frac {2 \sqrt {c x}\, \sqrt {x \left (b x +a \right )}\, \left (15 D x^{3} b^{3}+21 C \,b^{3} x^{2}-18 D a \,b^{2} x^{2}+35 x B \,b^{3}-28 C a \,b^{2} x +24 D a^{2} b x +105 A \,b^{3}-70 B a \,b^{2}+56 C \,a^{2} b -48 D a^{3}\right )}{105 x \,b^{4}}\)	\(101\)
gosper	\(\frac {2 \left (b x +a \right ) \left (15 D x^{3} b^{3}+21 C \,b^{3} x^{2}-18 D a \,b^{2} x^{2}+35 x B \,b^{3}-28 C a \,b^{2} x +24 D a^{2} b x +105 A \,b^{3}-70 B a \,b^{2}+56 C \,a^{2} b -48 D a^{3}\right ) \sqrt {c x}}{105 b^{4} \sqrt {b \,x^{2}+a x}}\)	\(105\)
orering	\(\frac {2 \left (b x +a \right ) \left (15 D x^{3} b^{3}+21 C \,b^{3} x^{2}-18 D a \,b^{2} x^{2}+35 x B \,b^{3}-28 C a \,b^{2} x +24 D a^{2} b x +105 A \,b^{3}-70 B a \,b^{2}+56 C \,a^{2} b -48 D a^{3}\right ) \sqrt {c x}}{105 b^{4} \sqrt {b \,x^{2}+a x}}\)	\(105\)

Input:

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

Output:

2/105*(c*x)^(1/2)*(x*(b*x+a))^(1/2)*(15*D*b^3*x^3+21*C*b^3*x^2-18*D*a*b^2* 
x^2+35*B*b^3*x-28*C*a*b^2*x+24*D*a^2*b*x+105*A*b^3-70*B*a*b^2+56*C*a^2*b-4 
8*D*a^3)/x/b^4
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.61 \[ \int \frac {\sqrt {c x} \left (A+B x+C x^2+D x^3\right )}{\sqrt {a x+b x^2}} \, dx=\frac {2 \, {\left (15 \, D b^{3} x^{3} - 48 \, D a^{3} + 56 \, C a^{2} b - 70 \, B a b^{2} + 105 \, A b^{3} - 3 \, {\left (6 \, D a b^{2} - 7 \, C b^{3}\right )} x^{2} + {\left (24 \, D a^{2} b - 28 \, C a b^{2} + 35 \, B b^{3}\right )} x\right )} \sqrt {b x^{2} + a x} \sqrt {c x}}{105 \, b^{4} x} \] Input:

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

Output:

2/105*(15*D*b^3*x^3 - 48*D*a^3 + 56*C*a^2*b - 70*B*a*b^2 + 105*A*b^3 - 3*( 
6*D*a*b^2 - 7*C*b^3)*x^2 + (24*D*a^2*b - 28*C*a*b^2 + 35*B*b^3)*x)*sqrt(b* 
x^2 + a*x)*sqrt(c*x)/(b^4*x)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.13 \[ \int \frac {\sqrt {c x} \left (A+B x+C x^2+D x^3\right )}{\sqrt {a x+b x^2}} \, dx=\frac {2 \, {\left (b \sqrt {c} x + a \sqrt {c}\right )} A}{\sqrt {b x + a} b} + \frac {2 \, {\left (b^{2} \sqrt {c} x^{2} - a b \sqrt {c} x - 2 \, a^{2} \sqrt {c}\right )} B}{3 \, \sqrt {b x + a} b^{2}} + \frac {2 \, {\left (3 \, b^{3} \sqrt {c} x^{3} - a b^{2} \sqrt {c} x^{2} + 4 \, a^{2} b \sqrt {c} x + 8 \, a^{3} \sqrt {c}\right )} C}{15 \, \sqrt {b x + a} b^{3}} + \frac {2 \, {\left (5 \, b^{4} \sqrt {c} x^{4} - a b^{3} \sqrt {c} x^{3} + 2 \, a^{2} b^{2} \sqrt {c} x^{2} - 8 \, a^{3} b \sqrt {c} x - 16 \, a^{4} \sqrt {c}\right )} D}{35 \, \sqrt {b x + a} b^{4}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.29 \[ \int \frac {\sqrt {c x} \left (A+B x+C x^2+D x^3\right )}{\sqrt {a x+b x^2}} \, dx=-\frac {2 \, c^{2} {\left (\frac {105 \, {\left (D a^{3} - C a^{2} b + B a b^{2} - A b^{3}\right )} \sqrt {b c x + a c}}{b^{4} c} - \frac {48 \, \sqrt {a c} D a^{3} - 56 \, \sqrt {a c} C a^{2} b + 70 \, \sqrt {a c} B a b^{2} - 105 \, \sqrt {a c} A b^{3}}{b^{4} c} - \frac {105 \, {\left (b c x + a c\right )}^{\frac {3}{2}} D a^{2} c^{2} - 70 \, {\left (b c x + a c\right )}^{\frac {3}{2}} C a b c^{2} + 35 \, {\left (b c x + a c\right )}^{\frac {3}{2}} B b^{2} c^{2} - 63 \, {\left (b c x + a c\right )}^{\frac {5}{2}} D a c + 21 \, {\left (b c x + a c\right )}^{\frac {5}{2}} C b c + 15 \, {\left (b c x + a c\right )}^{\frac {7}{2}} D}{b^{4} c^{4}}\right )}}{105 \, {\left | c \right |}} \] Input:

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

Output:

-2/105*c^2*(105*(D*a^3 - C*a^2*b + B*a*b^2 - A*b^3)*sqrt(b*c*x + a*c)/(b^4 
*c) - (48*sqrt(a*c)*D*a^3 - 56*sqrt(a*c)*C*a^2*b + 70*sqrt(a*c)*B*a*b^2 - 
105*sqrt(a*c)*A*b^3)/(b^4*c) - (105*(b*c*x + a*c)^(3/2)*D*a^2*c^2 - 70*(b* 
c*x + a*c)^(3/2)*C*a*b*c^2 + 35*(b*c*x + a*c)^(3/2)*B*b^2*c^2 - 63*(b*c*x 
+ a*c)^(5/2)*D*a*c + 21*(b*c*x + a*c)^(5/2)*C*b*c + 15*(b*c*x + a*c)^(7/2) 
*D)/(b^4*c^4))/abs(c)
 

Mupad [F(-1)]

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

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

Output:

int(((c*x)^(1/2)*(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 = 83, normalized size of antiderivative = 0.49 \[ \int \frac {\sqrt {c x} \left (A+B x+C x^2+D x^3\right )}{\sqrt {a x+b x^2}} \, dx=\frac {2 \sqrt {c}\, \sqrt {b x +a}\, \left (15 b^{3} d \,x^{3}-18 a \,b^{2} d \,x^{2}+21 b^{3} c \,x^{2}+24 a^{2} b d x -28 a \,b^{2} c x +35 b^{4} x -48 a^{3} d +56 a^{2} b c +35 a \,b^{3}\right )}{105 b^{4}} \] Input:

int((c*x)^(1/2)*(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)*( - 48*a**3*d + 56*a**2*b*c + 24*a**2*b*d*x + 35* 
a*b**3 - 28*a*b**2*c*x - 18*a*b**2*d*x**2 + 35*b**4*x + 21*b**3*c*x**2 + 1 
5*b**3*d*x**3))/(105*b**4)