\(\int \frac {x^{3/2}}{(a+b x)^{3/2} (a c-b c x)^{3/2}} \, dx\) [177]

Optimal result

Integrand size = 28, antiderivative size = 108 \[ \int \frac {x^{3/2}}{(a+b x)^{3/2} (a c-b c x)^{3/2}} \, dx=\frac {\sqrt {x}}{b^2 c \sqrt {a+b x} \sqrt {a c-b c x}}-\frac {\sqrt {a} \sqrt {1-\frac {b^2 x^2}{a^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right ),-1\right )}{b^{5/2} c \sqrt {a+b x} \sqrt {a c-b c x}} \] Output:

x^(1/2)/b^2/c/(b*x+a)^(1/2)/(-b*c*x+a*c)^(1/2)-a^(1/2)*(1-b^2*x^2/a^2)^(1/ 
2)*EllipticF(b^(1/2)*x^(1/2)/a^(1/2),I)/b^(5/2)/c/(b*x+a)^(1/2)/(-b*c*x+a* 
c)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 6.46 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.72 \[ \int \frac {x^{3/2}}{(a+b x)^{3/2} (a c-b c x)^{3/2}} \, dx=\frac {\sqrt {x}-\sqrt {x} \sqrt {1-\frac {b^2 x^2}{a^2}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\frac {b^2 x^2}{a^2}\right )}{b^2 c \sqrt {c (a-b x)} \sqrt {a+b x}} \] Input:

Integrate[x^(3/2)/((a + b*x)^(3/2)*(a*c - b*c*x)^(3/2)),x]
 

Output:

(Sqrt[x] - Sqrt[x]*Sqrt[1 - (b^2*x^2)/a^2]*Hypergeometric2F1[1/4, 1/2, 5/4 
, (b^2*x^2)/a^2])/(b^2*c*Sqrt[c*(a - b*x)]*Sqrt[a + b*x])
 

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {109, 27, 35, 127, 126}

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.

\(\displaystyle \int \frac {x^{3/2}}{(a+b x)^{3/2} (a c-b c x)^{3/2}} \, dx\)

\(\Big \downarrow \) 109

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 35

\(\displaystyle \frac {\sqrt {x}}{b^2 c \sqrt {a+b x} \sqrt {a c-b c x}}-\frac {\int \frac {1}{\sqrt {x} \sqrt {a+b x} \sqrt {a c-b c x}}dx}{2 b^2 c}\)

\(\Big \downarrow \) 127

\(\displaystyle \frac {\sqrt {x}}{b^2 c \sqrt {a+b x} \sqrt {a c-b c x}}-\frac {\sqrt {1-\frac {b x}{a}} \sqrt {\frac {b x}{a}+1} \int \frac {1}{\sqrt {x} \sqrt {1-\frac {b x}{a}} \sqrt {\frac {b x}{a}+1}}dx}{2 b^2 c \sqrt {a+b x} \sqrt {a c-b c x}}\)

\(\Big \downarrow \) 126

\(\displaystyle \frac {\sqrt {x}}{b^2 c \sqrt {a+b x} \sqrt {a c-b c x}}-\frac {\sqrt {a} \sqrt {1-\frac {b x}{a}} \sqrt {\frac {b x}{a}+1} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right ),-1\right )}{b^{5/2} c \sqrt {a+b x} \sqrt {a c-b c x}}\)

Input:

Int[x^(3/2)/((a + b*x)^(3/2)*(a*c - b*c*x)^(3/2)),x]
 

Output:

Sqrt[x]/(b^2*c*Sqrt[a + b*x]*Sqrt[a*c - b*c*x]) - (Sqrt[a]*Sqrt[1 - (b*x)/ 
a]*Sqrt[1 + (b*x)/a]*EllipticF[ArcSin[(Sqrt[b]*Sqrt[x])/Sqrt[a]], -1])/(b^ 
(5/2)*c*Sqrt[a + b*x]*Sqrt[a*c - b*c*x])
 

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 35

Int[(u_.)*((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.), x_Symbol] :> 
 Simp[(b/d)^m   Int[u*(c + d*x)^(m + n), x], x] /; FreeQ[{a, b, c, d, m, n} 
, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] &&  !(IntegerQ[n] && SimplerQ[a + 
b*x, c + d*x])
 

rule 109

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(b*c - a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((e + f 
*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] + Simp[1/(b*(b*e - a*f)*(m + 1)) 
 Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) 
+ c*f*(p + 1)) + b*c*(d*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) 
 + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /; FreeQ[{a, b, c, 
d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || 
IntegersQ[m, n + p] || IntegersQ[p, m + n])
 

rule 126

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(e_) + (f_.)*(x_)]), x 
_] :> Simp[(2/(b*Sqrt[e]))*Rt[-b/d, 2]*EllipticF[ArcSin[Sqrt[b*x]/(Sqrt[c]* 
Rt[-b/d, 2])], c*(f/(d*e))], x] /; FreeQ[{b, c, d, e, f}, x] && GtQ[c, 0] & 
& GtQ[e, 0] && (PosQ[-b/d] || NegQ[-b/f])
 

rule 127

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(e_) + (f_.)*(x_)]), x 
_] :> Simp[Sqrt[1 + d*(x/c)]*(Sqrt[1 + f*(x/e)]/(Sqrt[c + d*x]*Sqrt[e + f*x 
]))   Int[1/(Sqrt[b*x]*Sqrt[1 + d*(x/c)]*Sqrt[1 + f*(x/e)]), x], x] /; Free 
Q[{b, c, d, e, f}, x] &&  !(GtQ[c, 0] && GtQ[e, 0])
 

Maple [A] (verified)

Time = 1.04 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.94

method	result	size
default	\(-\frac {\sqrt {b x +a}\, \sqrt {c \left (-b x +a \right )}\, \left (\sqrt {2}\, \operatorname {EllipticF}\left (\sqrt {\frac {b x +a}{a}}, \frac {\sqrt {2}}{2}\right ) a \sqrt {\frac {b x +a}{a}}\, \sqrt {\frac {-b x +a}{a}}\, \sqrt {-\frac {b x}{a}}-2 b x \right )}{2 \sqrt {x}\, c^{2} \left (-b^{2} x^{2}+a^{2}\right ) b^{3}}\)	\(102\)
elliptic	\(\frac {\sqrt {c x \left (-b^{2} x^{2}+a^{2}\right )}\, \left (\frac {x}{b^{2} c \sqrt {-\left (x^{2}-\frac {a^{2}}{b^{2}}\right ) b^{2} c x}}-\frac {a \sqrt {\frac {\left (x +\frac {a}{b}\right ) b}{a}}\, \sqrt {-\frac {2 \left (x -\frac {a}{b}\right ) b}{a}}\, \sqrt {-\frac {b x}{a}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {a}{b}\right ) b}{a}}, \frac {\sqrt {2}}{2}\right )}{2 b^{3} c \sqrt {-b^{2} c \,x^{3}+a^{2} c x}}\right )}{\sqrt {x}\, \sqrt {b x +a}\, \sqrt {c \left (-b x +a \right )}}\)	\(155\)

Input:

int(x^(3/2)/(b*x+a)^(3/2)/(-b*c*x+a*c)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/2*(b*x+a)^(1/2)*(c*(-b*x+a))^(1/2)*(2^(1/2)*EllipticF(((b*x+a)/a)^(1/2) 
,1/2*2^(1/2))*a*((b*x+a)/a)^(1/2)*((-b*x+a)/a)^(1/2)*(-b*x/a)^(1/2)-2*b*x) 
/x^(1/2)/c^2/(-b^2*x^2+a^2)/b^3
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.80 \[ \int \frac {x^{3/2}}{(a+b x)^{3/2} (a c-b c x)^{3/2}} \, dx=-\frac {\sqrt {-b c x + a c} \sqrt {b x + a} b^{2} \sqrt {x} - {\left (b^{2} x^{2} - a^{2}\right )} \sqrt {-b^{2} c} {\rm weierstrassPInverse}\left (\frac {4 \, a^{2}}{b^{2}}, 0, x\right )}{b^{6} c^{2} x^{2} - a^{2} b^{4} c^{2}} \] Input:

integrate(x^(3/2)/(b*x+a)^(3/2)/(-b*c*x+a*c)^(3/2),x, algorithm="fricas")
 

Output:

-(sqrt(-b*c*x + a*c)*sqrt(b*x + a)*b^2*sqrt(x) - (b^2*x^2 - a^2)*sqrt(-b^2 
*c)*weierstrassPInverse(4*a^2/b^2, 0, x))/(b^6*c^2*x^2 - a^2*b^4*c^2)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{3/2}}{(a+b x)^{3/2} (a c-b c x)^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(x**(3/2)/(b*x+a)**(3/2)/(-b*c*x+a*c)**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

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

integrate(x^(3/2)/(b*x+a)^(3/2)/(-b*c*x+a*c)^(3/2),x, algorithm="maxima")
 

Output:

integrate(x^(3/2)/((-b*c*x + a*c)^(3/2)*(b*x + a)^(3/2)), x)
 

Giac [F(-1)]

Timed out. \[ \int \frac {x^{3/2}}{(a+b x)^{3/2} (a c-b c x)^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(x^(3/2)/(b*x+a)^(3/2)/(-b*c*x+a*c)^(3/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

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

int(x^(3/2)/((a*c - b*c*x)^(3/2)*(a + b*x)^(3/2)),x)
 

Output:

int(x^(3/2)/((a*c - b*c*x)^(3/2)*(a + b*x)^(3/2)), x)
 

Reduce [F]

\[ \int \frac {x^{3/2}}{(a+b x)^{3/2} (a c-b c x)^{3/2}} \, dx=\frac {\sqrt {c}\, \left (2 \sqrt {x}\, \sqrt {b x +a}\, \sqrt {-b x +a}-\left (\int \frac {\sqrt {x}\, \sqrt {b x +a}\, \sqrt {-b x +a}}{b^{4} x^{5}-2 a^{2} b^{2} x^{3}+a^{4} x}d x \right ) a^{4}+\left (\int \frac {\sqrt {x}\, \sqrt {b x +a}\, \sqrt {-b x +a}}{b^{4} x^{5}-2 a^{2} b^{2} x^{3}+a^{4} x}d x \right ) a^{2} b^{2} x^{2}\right )}{b^{2} c^{2} \left (-b^{2} x^{2}+a^{2}\right )} \] Input:

int(x^(3/2)/(b*x+a)^(3/2)/(-b*c*x+a*c)^(3/2),x)
 

Output:

(sqrt(c)*(2*sqrt(x)*sqrt(a + b*x)*sqrt(a - b*x) - int((sqrt(x)*sqrt(a + b* 
x)*sqrt(a - b*x))/(a**4*x - 2*a**2*b**2*x**3 + b**4*x**5),x)*a**4 + int((s 
qrt(x)*sqrt(a + b*x)*sqrt(a - b*x))/(a**4*x - 2*a**2*b**2*x**3 + b**4*x**5 
),x)*a**2*b**2*x**2))/(b**2*c**2*(a**2 - b**2*x**2))