\(\int \frac {\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}}{\sqrt {e x}} \, dx\) [857]

Optimal result

Integrand size = 36, antiderivative size = 163 \[ \int \frac {\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}}{\sqrt {e x}} \, dx=\frac {5 c \sqrt {e x} \sqrt {b c^2-b d^2 x^2}}{4 e \sqrt {c+d x}}-\frac {\sqrt {e x} \left (b c^2-b d^2 x^2\right )^{3/2}}{2 b e (c+d x)^{3/2}}-\frac {5 \sqrt {b} c^2 \arctan \left (\frac {\sqrt {e} \sqrt {b c^2-b d^2 x^2}}{\sqrt {b} \sqrt {d} \sqrt {e x} \sqrt {c+d x}}\right )}{4 \sqrt {d} \sqrt {e}} \] Output:

5/4*c*(e*x)^(1/2)*(-b*d^2*x^2+b*c^2)^(1/2)/e/(d*x+c)^(1/2)-1/2*(e*x)^(1/2) 
*(-b*d^2*x^2+b*c^2)^(3/2)/b/e/(d*x+c)^(3/2)-5/4*b^(1/2)*c^2*arctan(e^(1/2) 
*(-b*d^2*x^2+b*c^2)^(1/2)/b^(1/2)/d^(1/2)/(e*x)^(1/2)/(d*x+c)^(1/2))/d^(1/ 
2)/e^(1/2)
 

Mathematica [A] (verified)

Time = 2.48 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.72 \[ \int \frac {\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}}{\sqrt {e x}} \, dx=\frac {\sqrt {x} \sqrt {b \left (c^2-d^2 x^2\right )} \left (\sqrt {d} \sqrt {x} (3 c+2 d x) \sqrt {1-\frac {d x}{c}}+5 c^{3/2} \arcsin \left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}}\right )\right )}{4 \sqrt {d} \sqrt {e x} \sqrt {c+d x} \sqrt {1-\frac {d x}{c}}} \] Input:

Integrate[(Sqrt[c + d*x]*Sqrt[b*c^2 - b*d^2*x^2])/Sqrt[e*x],x]
 

Output:

(Sqrt[x]*Sqrt[b*(c^2 - d^2*x^2)]*(Sqrt[d]*Sqrt[x]*(3*c + 2*d*x)*Sqrt[1 - ( 
d*x)/c] + 5*c^(3/2)*ArcSin[(Sqrt[d]*Sqrt[x])/Sqrt[c]]))/(4*Sqrt[d]*Sqrt[e* 
x]*Sqrt[c + d*x]*Sqrt[1 - (d*x)/c])
 

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.14, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {574, 576, 586, 65, 218}

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.

\(\displaystyle \int \frac {\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}}{\sqrt {e x}} \, dx\)

\(\Big \downarrow \) 574

\(\displaystyle \frac {5}{4} c \int \frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {e x} \sqrt {c+d x}}dx-\frac {\sqrt {e x} \left (b c^2-b d^2 x^2\right )^{3/2}}{2 b e (c+d x)^{3/2}}\)

\(\Big \downarrow \) 576

\(\displaystyle \frac {5}{4} c \left (\frac {1}{2} b c \int \frac {\sqrt {c+d x}}{\sqrt {e x} \sqrt {b c^2-b d^2 x^2}}dx+\frac {\sqrt {e x} \sqrt {b c^2-b d^2 x^2}}{e \sqrt {c+d x}}\right )-\frac {\sqrt {e x} \left (b c^2-b d^2 x^2\right )^{3/2}}{2 b e (c+d x)^{3/2}}\)

\(\Big \downarrow \) 586

\(\displaystyle \frac {5}{4} c \left (\frac {b c \sqrt {c+d x} \sqrt {b c-b d x} \int \frac {1}{\sqrt {e x} \sqrt {b c-b d x}}dx}{2 \sqrt {b c^2-b d^2 x^2}}+\frac {\sqrt {e x} \sqrt {b c^2-b d^2 x^2}}{e \sqrt {c+d x}}\right )-\frac {\sqrt {e x} \left (b c^2-b d^2 x^2\right )^{3/2}}{2 b e (c+d x)^{3/2}}\)

\(\Big \downarrow \) 65

\(\displaystyle \frac {5}{4} c \left (\frac {b c \sqrt {c+d x} \sqrt {b c-b d x} \int \frac {1}{\frac {b d x e}{b c-b d x}+e}d\frac {\sqrt {e x}}{\sqrt {b c-b d x}}}{\sqrt {b c^2-b d^2 x^2}}+\frac {\sqrt {e x} \sqrt {b c^2-b d^2 x^2}}{e \sqrt {c+d x}}\right )-\frac {\sqrt {e x} \left (b c^2-b d^2 x^2\right )^{3/2}}{2 b e (c+d x)^{3/2}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {5}{4} c \left (\frac {\sqrt {b} c \sqrt {c+d x} \sqrt {b c-b d x} \arctan \left (\frac {\sqrt {b} \sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {b c-b d x}}\right )}{\sqrt {d} \sqrt {e} \sqrt {b c^2-b d^2 x^2}}+\frac {\sqrt {e x} \sqrt {b c^2-b d^2 x^2}}{e \sqrt {c+d x}}\right )-\frac {\sqrt {e x} \left (b c^2-b d^2 x^2\right )^{3/2}}{2 b e (c+d x)^{3/2}}\)

Input:

Int[(Sqrt[c + d*x]*Sqrt[b*c^2 - b*d^2*x^2])/Sqrt[e*x],x]
 

Output:

-1/2*(Sqrt[e*x]*(b*c^2 - b*d^2*x^2)^(3/2))/(b*e*(c + d*x)^(3/2)) + (5*c*(( 
Sqrt[e*x]*Sqrt[b*c^2 - b*d^2*x^2])/(e*Sqrt[c + d*x]) + (Sqrt[b]*c*Sqrt[c + 
 d*x]*Sqrt[b*c - b*d*x]*ArcTan[(Sqrt[b]*Sqrt[d]*Sqrt[e*x])/(Sqrt[e]*Sqrt[b 
*c - b*d*x])])/(Sqrt[d]*Sqrt[e]*Sqrt[b*c^2 - b*d^2*x^2])))/4
 

Defintions of rubi rules used

rule 65

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2   Sub 
st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d 
}, x] &&  !GtQ[c, 0]
 

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 574

Int[((e_.)*(x_))^(n_)*((c_) + (d_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), 
x_Symbol] :> Simp[d^2*(e*x)^(n + 1)*(c + d*x)^(m - 2)*((a + b*x^2)^(p + 1)/ 
(b*e*(n + p + 2))), x] + Simp[c*((2*n + p + 3)/(n + p + 2))   Int[(e*x)^n*( 
c + d*x)^(m - 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x 
] && EqQ[b*c^2 + a*d^2, 0] && EqQ[m + p - 1, 0] &&  !LtQ[n, -1] && IntegerQ 
[2*p]
 

rule 576

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), 
x_Symbol] :> Simp[(-(e*x)^(m + 1))*(c + d*x)^n*((a + b*x^2)^p/(e*(n - m - 1 
))), x] - Simp[b*c*(n/(d^2*(n - m - 1)))   Int[(e*x)^m*(c + d*x)^(n + 1)*(a 
 + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && EqQ[b*c^2 + a* 
d^2, 0] && EqQ[n + p, 0] && GtQ[p, 0] && NeQ[m - n + 1, 0] &&  !IGtQ[m, 0] 
&&  !(IntegerQ[m + p] && LtQ[m + p + 2, 0]) && RationalQ[m]
 

rule 586

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

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.80

method	result	size
default	\(-\frac {\sqrt {b \left (-d^{2} x^{2}+c^{2}\right )}\, x \left (5 b \,c^{2} e \arctan \left (\frac {\sqrt {b d e}\, \left (-2 d x +c \right )}{2 d \sqrt {\left (-d x +c \right ) b e x}}\right )-4 \sqrt {\left (-d x +c \right ) b e x}\, \sqrt {b d e}\, d x -6 \sqrt {\left (-d x +c \right ) b e x}\, \sqrt {b d e}\, c \right )}{8 \sqrt {d x +c}\, \sqrt {e x}\, \sqrt {\left (-d x +c \right ) b e x}\, \sqrt {b d e}}\)	\(131\)
risch	\(\frac {\left (2 d x +3 c \right ) \left (-d x +c \right ) x \sqrt {-\frac {e x b \left (d^{2} x^{2}-c^{2}\right )}{d x +c}}\, \sqrt {d x +c}\, b}{4 \sqrt {-b e \left (d x -c \right ) x}\, \sqrt {e x}\, \sqrt {-b \left (d^{2} x^{2}-c^{2}\right )}}+\frac {5 c^{2} \arctan \left (\frac {\sqrt {b d e}\, \left (x -\frac {c}{2 d}\right )}{\sqrt {-b d \,x^{2} e +b e x c}}\right ) \sqrt {-\frac {e x b \left (d^{2} x^{2}-c^{2}\right )}{d x +c}}\, \sqrt {d x +c}\, b}{8 \sqrt {b d e}\, \sqrt {e x}\, \sqrt {-b \left (d^{2} x^{2}-c^{2}\right )}}\)	\(192\)

Input:

int((d*x+c)^(1/2)*(-b*d^2*x^2+b*c^2)^(1/2)/(e*x)^(1/2),x,method=_RETURNVER 
BOSE)
 

Output:

-1/8*(b*(-d^2*x^2+c^2))^(1/2)*x*(5*b*c^2*e*arctan(1/2*(b*d*e)^(1/2)*(-2*d* 
x+c)/d/((-d*x+c)*b*e*x)^(1/2))-4*((-d*x+c)*b*e*x)^(1/2)*(b*d*e)^(1/2)*d*x- 
6*((-d*x+c)*b*e*x)^(1/2)*(b*d*e)^(1/2)*c)/(d*x+c)^(1/2)/(e*x)^(1/2)/((-d*x 
+c)*b*e*x)^(1/2)/(b*d*e)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.89 \[ \int \frac {\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}}{\sqrt {e x}} \, dx=\left [\frac {4 \, \sqrt {-b d^{2} x^{2} + b c^{2}} {\left (2 \, d x + 3 \, c\right )} \sqrt {d x + c} \sqrt {e x} + 5 \, {\left (c^{2} d e x + c^{3} e\right )} \sqrt {-\frac {b}{d e}} \log \left (-\frac {8 \, b d^{3} x^{3} - 7 \, b c^{2} d x + b c^{3} + 4 \, \sqrt {-b d^{2} x^{2} + b c^{2}} {\left (2 \, d^{2} x - c d\right )} \sqrt {d x + c} \sqrt {e x} \sqrt {-\frac {b}{d e}}}{d x + c}\right )}{16 \, {\left (d e x + c e\right )}}, \frac {2 \, \sqrt {-b d^{2} x^{2} + b c^{2}} {\left (2 \, d x + 3 \, c\right )} \sqrt {d x + c} \sqrt {e x} - 5 \, {\left (c^{2} d e x + c^{3} e\right )} \sqrt {\frac {b}{d e}} \arctan \left (\frac {2 \, \sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c} \sqrt {e x} d \sqrt {\frac {b}{d e}}}{2 \, b d^{2} x^{2} + b c d x - b c^{2}}\right )}{8 \, {\left (d e x + c e\right )}}\right ] \] Input:

integrate((d*x+c)^(1/2)*(-b*d^2*x^2+b*c^2)^(1/2)/(e*x)^(1/2),x, algorithm= 
"fricas")
 

Output:

[1/16*(4*sqrt(-b*d^2*x^2 + b*c^2)*(2*d*x + 3*c)*sqrt(d*x + c)*sqrt(e*x) + 
5*(c^2*d*e*x + c^3*e)*sqrt(-b/(d*e))*log(-(8*b*d^3*x^3 - 7*b*c^2*d*x + b*c 
^3 + 4*sqrt(-b*d^2*x^2 + b*c^2)*(2*d^2*x - c*d)*sqrt(d*x + c)*sqrt(e*x)*sq 
rt(-b/(d*e)))/(d*x + c)))/(d*e*x + c*e), 1/8*(2*sqrt(-b*d^2*x^2 + b*c^2)*( 
2*d*x + 3*c)*sqrt(d*x + c)*sqrt(e*x) - 5*(c^2*d*e*x + c^3*e)*sqrt(b/(d*e)) 
*arctan(2*sqrt(-b*d^2*x^2 + b*c^2)*sqrt(d*x + c)*sqrt(e*x)*d*sqrt(b/(d*e)) 
/(2*b*d^2*x^2 + b*c*d*x - b*c^2)))/(d*e*x + c*e)]
 

Sympy [F]

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

integrate((d*x+c)**(1/2)*(-b*d**2*x**2+b*c**2)**(1/2)/(e*x)**(1/2),x)
 

Output:

Integral(sqrt(-b*(-c + d*x)*(c + d*x))*sqrt(c + d*x)/sqrt(e*x), x)
 

Maxima [F]

\[ \int \frac {\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}}{\sqrt {e x}} \, dx=\int { \frac {\sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c}}{\sqrt {e x}} \,d x } \] Input:

integrate((d*x+c)^(1/2)*(-b*d^2*x^2+b*c^2)^(1/2)/(e*x)^(1/2),x, algorithm= 
"maxima")
 

Output:

integrate(sqrt(-b*d^2*x^2 + b*c^2)*sqrt(d*x + c)/sqrt(e*x), x)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.83 \[ \int \frac {\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}}{\sqrt {e x}} \, dx=\frac {{\left (\frac {5 \, c^{2} \log \left ({\left | -\sqrt {-b d e} \sqrt {-b d x + b c} + \sqrt {b^{2} c d e + {\left (b d x - b c\right )} b d e} \right |}\right )}{\sqrt {-b d e}} + \sqrt {b^{2} c d e + {\left (b d x - b c\right )} b d e} \sqrt {-b d x + b c} {\left (\frac {5 \, c}{b^{2} d e} + \frac {2 \, {\left (b d x - b c\right )}}{b^{3} d e}\right )}\right )} {\left | b \right |} {\left | d \right |}}{4 \, d} \] Input:

integrate((d*x+c)^(1/2)*(-b*d^2*x^2+b*c^2)^(1/2)/(e*x)^(1/2),x, algorithm= 
"giac")
 

Output:

1/4*(5*c^2*log(abs(-sqrt(-b*d*e)*sqrt(-b*d*x + b*c) + sqrt(b^2*c*d*e + (b* 
d*x - b*c)*b*d*e)))/sqrt(-b*d*e) + sqrt(b^2*c*d*e + (b*d*x - b*c)*b*d*e)*s 
qrt(-b*d*x + b*c)*(5*c/(b^2*d*e) + 2*(b*d*x - b*c)/(b^3*d*e)))*abs(b)*abs( 
d)/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}}{\sqrt {e x}} \, dx=\int \frac {\sqrt {b\,c^2-b\,d^2\,x^2}\,\sqrt {c+d\,x}}{\sqrt {e\,x}} \,d x \] Input:

int(((b*c^2 - b*d^2*x^2)^(1/2)*(c + d*x)^(1/2))/(e*x)^(1/2),x)
 

Output:

int(((b*c^2 - b*d^2*x^2)^(1/2)*(c + d*x)^(1/2))/(e*x)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.42 \[ \int \frac {\sqrt {c+d x} \sqrt {b c^2-b d^2 x^2}}{\sqrt {e x}} \, dx=\frac {\sqrt {e}\, \sqrt {b}\, \left (3 \sqrt {x}\, \sqrt {-d x +c}\, c d +2 \sqrt {x}\, \sqrt {-d x +c}\, d^{2} x -5 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {-d x +c}+\sqrt {x}\, \sqrt {d}\, i}{\sqrt {c}}\right ) c^{2} i \right )}{4 d e} \] Input:

int((d*x+c)^(1/2)*(-b*d^2*x^2+b*c^2)^(1/2)/(e*x)^(1/2),x)
 

Output:

(sqrt(e)*sqrt(b)*(3*sqrt(x)*sqrt(c - d*x)*c*d + 2*sqrt(x)*sqrt(c - d*x)*d* 
*2*x - 5*sqrt(d)*log((sqrt(c - d*x) + sqrt(x)*sqrt(d)*i)/sqrt(c))*c**2*i)) 
/(4*d*e)