\(\int \frac {A+B x+C x^2+D x^3}{(c x)^{5/2} \sqrt {a x+b x^2}} \, dx\) [7]

Optimal result

Integrand size = 36, antiderivative size = 161 \[ \int \frac {A+B x+C x^2+D x^3}{(c x)^{5/2} \sqrt {a x+b x^2}} \, dx=-\frac {A \sqrt {a x+b x^2}}{2 a (c x)^{5/2}}+\frac {(3 A b-4 a B) \sqrt {a x+b x^2}}{4 a^2 c (c x)^{3/2}}+\frac {2 D \sqrt {a x+b x^2}}{b c^2 \sqrt {c x}}-\frac {\left (3 A b^2-4 a b B+8 a^2 C\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {a x+b x^2}}{\sqrt {a} \sqrt {c x}}\right )}{4 a^{5/2} c^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.74 \[ \int \frac {A+B x+C x^2+D x^3}{(c x)^{5/2} \sqrt {a x+b x^2}} \, dx=\frac {x \left (\sqrt {a} (a+b x) \left (3 A b^2 x+8 a^2 D x^2-2 a b (A+2 B x)\right )-b \left (3 A b^2-4 a b B+8 a^2 C\right ) x^2 \sqrt {a+b x} \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )\right )}{4 a^{5/2} b (c x)^{5/2} \sqrt {x (a+b x)}} \] Input:

Integrate[(A + B*x + C*x^2 + D*x^3)/((c*x)^(5/2)*Sqrt[a*x + b*x^2]),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.95 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.22, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2169, 27, 2169, 27, 1220, 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.

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

\(\Big \downarrow \) 2169

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2169

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1220

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

\(\Big \downarrow \) 1135

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

\(\Big \downarrow \) 1136

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

\(\Big \downarrow \) 221

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

Input:

Int[(A + B*x + C*x^2 + D*x^3)/((c*x)^(5/2)*Sqrt[a*x + b*x^2]),x]
 

Output:

(2*D*Sqrt[a*x + b*x^2])/(b*c^2*Sqrt[c*x]) + ((-2*c^2*C*Sqrt[a*x + b*x^2])/ 
(c*x)^(3/2) + c^3*(-1/2*(A*b*Sqrt[a*x + b*x^2])/(a*(c*x)^(5/2)) - ((3*A*b^ 
2 - 4*a*b*B + 8*a^2*C)*(-(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)))/(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 = 207, normalized size of antiderivative = 1.29

method	result	size
default	\(-\frac {\sqrt {x \left (b x +a \right )}\, \left (3 A \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (b x +a \right )}}{\sqrt {a c}}\right ) b^{3} c \,x^{2}-4 B \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (b x +a \right )}}{\sqrt {a c}}\right ) a \,b^{2} c \,x^{2}+8 C \,\operatorname {arctanh}\left (\frac {\sqrt {c \left (b x +a \right )}}{\sqrt {a c}}\right ) a^{2} b c \,x^{2}-8 D \sqrt {c \left (b x +a \right )}\, a^{2} x^{2} \sqrt {a c}-3 A \,b^{2} x \sqrt {c \left (b x +a \right )}\, \sqrt {a c}+4 B a b x \sqrt {c \left (b x +a \right )}\, \sqrt {a c}+2 A \sqrt {c \left (b x +a \right )}\, \sqrt {a c}\, a b \right )}{4 c^{2} x^{2} \sqrt {c x}\, b \sqrt {c \left (b x +a \right )}\, a^{2} \sqrt {a c}}\)	\(207\)

Input:

int((D*x^3+C*x^2+B*x+A)/(c*x)^(5/2)/(b*x^2+a*x)^(1/2),x,method=_RETURNVERB 
OSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.68 \[ \int \frac {A+B x+C x^2+D x^3}{(c x)^{5/2} \sqrt {a x+b x^2}} \, dx=\left [\frac {{\left (8 \, C a^{2} b - 4 \, B a b^{2} + 3 \, A b^{3}\right )} \sqrt {a c} x^{3} \log \left (-\frac {b c x^{2} + 2 \, a c x - 2 \, \sqrt {b x^{2} + a x} \sqrt {a c} \sqrt {c x}}{x^{2}}\right ) + 2 \, {\left (8 \, D a^{3} x^{2} - 2 \, A a^{2} b - {\left (4 \, B a^{2} b - 3 \, A a b^{2}\right )} x\right )} \sqrt {b x^{2} + a x} \sqrt {c x}}{8 \, a^{3} b c^{3} x^{3}}, \frac {{\left (8 \, C a^{2} b - 4 \, B a b^{2} + 3 \, A b^{3}\right )} \sqrt {-a c} x^{3} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-a c} \sqrt {c x}}{a c x}\right ) + {\left (8 \, D a^{3} x^{2} - 2 \, A a^{2} b - {\left (4 \, B a^{2} b - 3 \, A a b^{2}\right )} x\right )} \sqrt {b x^{2} + a x} \sqrt {c x}}{4 \, a^{3} b c^{3} x^{3}}\right ] \] Input:

integrate((D*x^3+C*x^2+B*x+A)/(c*x)^(5/2)/(b*x^2+a*x)^(1/2),x, algorithm=" 
fricas")
                                                                                    
                                                                                    
 

Output:

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

Sympy [F]

\[ \int \frac {A+B x+C x^2+D x^3}{(c x)^{5/2} \sqrt {a x+b x^2}} \, dx=\int \frac {A + B x + C x^{2} + D x^{3}}{\left (c x\right )^{\frac {5}{2}} \sqrt {x \left (a + b x\right )}}\, dx \] Input:

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

Output:

Integral((A + B*x + C*x**2 + D*x**3)/((c*x)**(5/2)*sqrt(x*(a + b*x))), x)
 

Maxima [F]

\[ \int \frac {A+B x+C x^2+D x^3}{(c x)^{5/2} \sqrt {a x+b x^2}} \, dx=\int { \frac {D x^{3} + C x^{2} + B x + A}{\sqrt {b x^{2} + a x} \left (c x\right )^{\frac {5}{2}}} \,d x } \] Input:

integrate((D*x^3+C*x^2+B*x+A)/(c*x)^(5/2)/(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)^(5/2)), x)
 

Giac [A] (verification not implemented)

Time = 0.40 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.05 \[ \int \frac {A+B x+C x^2+D x^3}{(c x)^{5/2} \sqrt {a x+b x^2}} \, dx=\frac {8 \, \sqrt {b c x + a c} D + \frac {{\left (8 \, C a^{2} b c - 4 \, B a b^{2} c + 3 \, A b^{3} c\right )} \arctan \left (\frac {\sqrt {b c x + a c}}{\sqrt {-a c}}\right )}{\sqrt {-a c} a^{2}} + \frac {4 \, \sqrt {b c x + a c} B a^{2} b^{2} c^{2} - 5 \, \sqrt {b c x + a c} A a b^{3} c^{2} - 4 \, {\left (b c x + a c\right )}^{\frac {3}{2}} B a b^{2} c + 3 \, {\left (b c x + a c\right )}^{\frac {3}{2}} A b^{3} c}{a^{2} b^{2} c^{2} x^{2}}}{4 \, b c^{2} {\left | c \right |}} \] Input:

integrate((D*x^3+C*x^2+B*x+A)/(c*x)^(5/2)/(b*x^2+a*x)^(1/2),x, algorithm=" 
giac")
 

Output:

1/4*(8*sqrt(b*c*x + a*c)*D + (8*C*a^2*b*c - 4*B*a*b^2*c + 3*A*b^3*c)*arcta 
n(sqrt(b*c*x + a*c)/sqrt(-a*c))/(sqrt(-a*c)*a^2) + (4*sqrt(b*c*x + a*c)*B* 
a^2*b^2*c^2 - 5*sqrt(b*c*x + a*c)*A*a*b^3*c^2 - 4*(b*c*x + a*c)^(3/2)*B*a* 
b^2*c + 3*(b*c*x + a*c)^(3/2)*A*b^3*c)/(a^2*b^2*c^2*x^2))/(b*c^2*abs(c))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{(c x)^{5/2} \sqrt {a x+b x^2}} \, dx=\int \frac {A+B\,x+C\,x^2+x^3\,D}{\sqrt {b\,x^2+a\,x}\,{\left (c\,x\right )}^{5/2}} \,d x \] Input:

int((A + B*x + C*x^2 + x^3*D)/((a*x + b*x^2)^(1/2)*(c*x)^(5/2)),x)
 

Output:

int((A + B*x + C*x^2 + x^3*D)/((a*x + b*x^2)^(1/2)*(c*x)^(5/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.87 \[ \int \frac {A+B x+C x^2+D x^3}{(c x)^{5/2} \sqrt {a x+b x^2}} \, dx=\frac {\sqrt {c}\, \left (-4 \sqrt {b x +a}\, a^{2} b +16 \sqrt {b x +a}\, a^{2} d \,x^{2}-2 \sqrt {b x +a}\, a \,b^{2} x +8 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) a b c \,x^{2}-\sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}-\sqrt {a}\right ) b^{3} x^{2}-8 \sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) a b c \,x^{2}+\sqrt {a}\, \mathrm {log}\left (\sqrt {b x +a}+\sqrt {a}\right ) b^{3} x^{2}\right )}{8 a^{2} b \,c^{3} x^{2}} \] Input:

int((D*x^3+C*x^2+B*x+A)/(c*x)^(5/2)/(b*x^2+a*x)^(1/2),x)
 

Output:

(sqrt(c)*( - 4*sqrt(a + b*x)*a**2*b + 16*sqrt(a + b*x)*a**2*d*x**2 - 2*sqr 
t(a + b*x)*a*b**2*x + 8*sqrt(a)*log(sqrt(a + b*x) - sqrt(a))*a*b*c*x**2 - 
sqrt(a)*log(sqrt(a + b*x) - sqrt(a))*b**3*x**2 - 8*sqrt(a)*log(sqrt(a + b* 
x) + sqrt(a))*a*b*c*x**2 + sqrt(a)*log(sqrt(a + b*x) + sqrt(a))*b**3*x**2) 
)/(8*a**2*b*c**3*x**2)