\(\int \frac {A+B x+C x^2+D x^3}{(c x)^{3/2} \sqrt {a x+b x^2}} \, dx\) [6]

Optimal result

Integrand size = 36, antiderivative size = 142 \[ \int \frac {A+B x+C x^2+D x^3}{(c x)^{3/2} \sqrt {a x+b x^2}} \, dx=-\frac {A \sqrt {a x+b x^2}}{a (c x)^{3/2}}+\frac {2 (b C-a D) \sqrt {a x+b x^2}}{b^2 c \sqrt {c x}}+\frac {2 D \left (a x+b x^2\right )^{3/2}}{3 b^2 (c x)^{3/2}}+\frac {(A b-2 a B) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a x+b x^2}}{\sqrt {a} \sqrt {c x}}\right )}{a^{3/2} c^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.75 \[ \int \frac {A+B x+C x^2+D x^3}{(c x)^{3/2} \sqrt {a x+b x^2}} \, dx=\frac {x \left (-\sqrt {a} (a+b x) \left (3 A b^2-2 a x (3 b C-2 a D+b D x)\right )+3 b^2 (A b-2 a B) x \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{3 a^{3/2} b^2 (c x)^{3/2} \sqrt {x (a+b x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.15, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {2169, 27, 2169, 27, 1220, 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.

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

\(\Big \downarrow \) 2169

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2169

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1220

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

\(\Big \downarrow \) 1136

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(2*D*Sqrt[c*x]*Sqrt[a*x + b*x^2])/(3*b*c^2) + ((2*c^2*(3*b*C - 2*a*D)*Sqrt 
[a*x + b*x^2])/(b*Sqrt[c*x]) + 3*b*c^3*(-((A*Sqrt[a*x + b*x^2])/(a*(c*x)^( 
3/2))) + ((A*b - 2*a*B)*ArcTanh[(Sqrt[c]*Sqrt[a*x + b*x^2])/(Sqrt[a]*Sqrt[ 
c*x])])/(a^(3/2)*c^(3/2))))/(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 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 = 176, normalized size of antiderivative = 1.24

method	result	size
default	\(\frac {\sqrt {x \left (b x +a \right )}\, \left (3 A \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (b x +a \right )}}{\sqrt {a c}}\right ) b^{3} c x -6 B \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (b x +a \right )}}{\sqrt {a c}}\right ) a \,b^{2} c x +2 D a b \,x^{2} \sqrt {c \left (b x +a \right )}\, \sqrt {a c}+6 C b \sqrt {c \left (b x +a \right )}\, a x \sqrt {a c}-4 D a^{2} \sqrt {c \left (b x +a \right )}\, x \sqrt {a c}-3 A \sqrt {c \left (b x +a \right )}\, \sqrt {a c}\, b^{2}\right )}{3 c x \sqrt {c x}\, \sqrt {c \left (b x +a \right )}\, b^{2} a \sqrt {a c}}\)	\(176\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.82 \[ \int \frac {A+B x+C x^2+D x^3}{(c x)^{3/2} \sqrt {a x+b x^2}} \, dx=\left [-\frac {3 \, {\left (2 \, B a b^{2} - A b^{3}\right )} \sqrt {a c} x^{2} \log \left (-\frac {b c x^{2} + 2 \, a c x + 2 \, \sqrt {b x^{2} + a x} \sqrt {a c} \sqrt {c x}}{x^{2}}\right ) - 2 \, {\left (2 \, D a^{2} b x^{2} - 3 \, A a b^{2} - 2 \, {\left (2 \, D a^{3} - 3 \, C a^{2} b\right )} x\right )} \sqrt {b x^{2} + a x} \sqrt {c x}}{6 \, a^{2} b^{2} c^{2} x^{2}}, \frac {3 \, {\left (2 \, B a b^{2} - A b^{3}\right )} \sqrt {-a c} x^{2} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-a c} \sqrt {c x}}{a c x}\right ) + {\left (2 \, D a^{2} b x^{2} - 3 \, A a b^{2} - 2 \, {\left (2 \, D a^{3} - 3 \, C a^{2} b\right )} x\right )} \sqrt {b x^{2} + a x} \sqrt {c x}}{3 \, a^{2} b^{2} c^{2} x^{2}}\right ] \] Input:

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

Output:

[-1/6*(3*(2*B*a*b^2 - A*b^3)*sqrt(a*c)*x^2*log(-(b*c*x^2 + 2*a*c*x + 2*sqr 
t(b*x^2 + a*x)*sqrt(a*c)*sqrt(c*x))/x^2) - 2*(2*D*a^2*b*x^2 - 3*A*a*b^2 - 
2*(2*D*a^3 - 3*C*a^2*b)*x)*sqrt(b*x^2 + a*x)*sqrt(c*x))/(a^2*b^2*c^2*x^2), 
 1/3*(3*(2*B*a*b^2 - A*b^3)*sqrt(-a*c)*x^2*arctan(sqrt(b*x^2 + a*x)*sqrt(- 
a*c)*sqrt(c*x)/(a*c*x)) + (2*D*a^2*b*x^2 - 3*A*a*b^2 - 2*(2*D*a^3 - 3*C*a^ 
2*b)*x)*sqrt(b*x^2 + a*x)*sqrt(c*x))/(a^2*b^2*c^2*x^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.97 \[ \int \frac {A+B x+C x^2+D x^3}{(c x)^{3/2} \sqrt {a x+b x^2}} \, dx=-\frac {b {\left (\frac {3 \, \sqrt {b c x + a c} A c}{a b x} - \frac {3 \, {\left (2 \, B a c^{2} - A b c^{2}\right )} \arctan \left (\frac {\sqrt {b c x + a c}}{\sqrt {-a c}}\right )}{\sqrt {-a c} a b} + \frac {2 \, {\left (3 \, \sqrt {b c x + a c} D a b^{6} c - 3 \, \sqrt {b c x + a c} C b^{7} c - {\left (b c x + a c\right )}^{\frac {3}{2}} D b^{6}\right )}}{b^{9}}\right )}}{3 \, c^{2} {\left | c \right |}} \] Input:

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

Output:

-1/3*b*(3*sqrt(b*c*x + a*c)*A*c/(a*b*x) - 3*(2*B*a*c^2 - A*b*c^2)*arctan(s 
qrt(b*c*x + a*c)/sqrt(-a*c))/(sqrt(-a*c)*a*b) + 2*(3*sqrt(b*c*x + a*c)*D*a 
*b^6*c - 3*sqrt(b*c*x + a*c)*C*b^7*c - (b*c*x + a*c)^(3/2)*D*b^6)/b^9)/(c^ 
2*abs(c))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.75 \[ \int \frac {A+B x+C x^2+D x^3}{(c x)^{3/2} \sqrt {a x+b x^2}} \, dx=\frac {\sqrt {c}\, \left (-8 \sqrt {b x +a}\, a^{2} d x -6 \sqrt {b x +a}\, a \,b^{2}+12 \sqrt {b x +a}\, a b c x +4 \sqrt {b x +a}\, a b d \,x^{2}+3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{3} x -3 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{3} x \right )}{6 a \,b^{2} c^{2} x} \] Input:

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

Output:

(sqrt(c)*( - 8*sqrt(a + b*x)*a**2*d*x - 6*sqrt(a + b*x)*a*b**2 + 12*sqrt(a 
 + b*x)*a*b*c*x + 4*sqrt(a + b*x)*a*b*d*x**2 + 3*sqrt(a)*log(sqrt(a + b*x) 
 - sqrt(a))*b**3*x - 3*sqrt(a)*log(sqrt(a + b*x) + sqrt(a))*b**3*x))/(6*a* 
b**2*c**2*x)