\(\int \sqrt {e x} \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} \, dx\) [856]

Optimal result

Integrand size = 36, antiderivative size = 208 \[ \int \sqrt {e x} \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} \, dx=-\frac {3 c^2 \sqrt {e x} \sqrt {b c^2-b d^2 x^2}}{8 d \sqrt {c+d x}}+\frac {3 c (e x)^{3/2} \sqrt {b c^2-b d^2 x^2}}{4 e \sqrt {c+d x}}-\frac {(e x)^{3/2} \left (b c^2-b d^2 x^2\right )^{3/2}}{3 b e (c+d x)^{3/2}}-\frac {3 \sqrt {b} c^3 \sqrt {e} \arctan \left (\frac {\sqrt {e} \sqrt {b c^2-b d^2 x^2}}{\sqrt {b} \sqrt {d} \sqrt {e x} \sqrt {c+d x}}\right )}{8 d^{3/2}} \] Output:

-3/8*c^2*(e*x)^(1/2)*(-b*d^2*x^2+b*c^2)^(1/2)/d/(d*x+c)^(1/2)+3/4*c*(e*x)^ 
(3/2)*(-b*d^2*x^2+b*c^2)^(1/2)/e/(d*x+c)^(1/2)-1/3*(e*x)^(3/2)*(-b*d^2*x^2 
+b*c^2)^(3/2)/b/e/(d*x+c)^(3/2)-3/8*b^(1/2)*c^3*e^(1/2)*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^(3/2)
 

Mathematica [A] (verified)

Time = 2.25 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.62 \[ \int \sqrt {e x} \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} \, dx=\frac {\sqrt {e x} \sqrt {b \left (c^2-d^2 x^2\right )} \left (\sqrt {d} \sqrt {x} \sqrt {1-\frac {d x}{c}} \left (-9 c^2+10 c d x+8 d^2 x^2\right )+9 c^{5/2} \arcsin \left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}}\right )\right )}{24 d^{3/2} \sqrt {x} \sqrt {c+d x} \sqrt {1-\frac {d x}{c}}} \] Input:

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

Output:

(Sqrt[e*x]*Sqrt[b*(c^2 - d^2*x^2)]*(Sqrt[d]*Sqrt[x]*Sqrt[1 - (d*x)/c]*(-9* 
c^2 + 10*c*d*x + 8*d^2*x^2) + 9*c^(5/2)*ArcSin[(Sqrt[d]*Sqrt[x])/Sqrt[c]]) 
)/(24*d^(3/2)*Sqrt[x]*Sqrt[c + d*x]*Sqrt[1 - (d*x)/c])
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {574, 576, 586, 60, 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.

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

\(\Big \downarrow \) 574

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

\(\Big \downarrow \) 576

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

\(\Big \downarrow \) 586

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 65

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

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 = 0.26 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.78

method	result	size
default	\(-\frac {\sqrt {e x}\, \sqrt {b \left (-d^{2} x^{2}+c^{2}\right )}\, \left (9 \arctan \left (\frac {\sqrt {b d e}\, \left (-2 d x +c \right )}{2 d \sqrt {\left (-d x +c \right ) b e x}}\right ) b \,c^{3} e -16 d^{2} x^{2} \sqrt {\left (-d x +c \right ) b e x}\, \sqrt {b d e}-20 \sqrt {\left (-d x +c \right ) b e x}\, \sqrt {b d e}\, c d x +18 \sqrt {\left (-d x +c \right ) b e x}\, \sqrt {b d e}\, c^{2}\right )}{48 \sqrt {d x +c}\, d \sqrt {\left (-d x +c \right ) b e x}\, \sqrt {b d e}}\)	\(162\)
risch	\(-\frac {\left (-8 d^{2} x^{2}-10 c d x +9 c^{2}\right ) \left (-d x +c \right ) x \sqrt {-\frac {e x b \left (d^{2} x^{2}-c^{2}\right )}{d x +c}}\, \sqrt {d x +c}\, b e}{24 d \sqrt {-b e \left (d x -c \right ) x}\, \sqrt {e x}\, \sqrt {-b \left (d^{2} x^{2}-c^{2}\right )}}+\frac {3 c^{3} \arctan \left (\frac {\sqrt {b d e}\, \left (x -\frac {c}{2 d}\right )}{\sqrt {-b d \,x^{2} e +b e x c}}\right ) \sqrt {-\frac {e x b \left (d^{2} x^{2}-c^{2}\right )}{d x +c}}\, \sqrt {d x +c}\, b e}{16 d \sqrt {b d e}\, \sqrt {e x}\, \sqrt {-b \left (d^{2} x^{2}-c^{2}\right )}}\)	\(211\)

Input:

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

Output:

-1/48*(e*x)^(1/2)*(b*(-d^2*x^2+c^2))^(1/2)*(9*arctan(1/2*(b*d*e)^(1/2)*(-2 
*d*x+c)/d/((-d*x+c)*b*e*x)^(1/2))*b*c^3*e-16*d^2*x^2*((-d*x+c)*b*e*x)^(1/2 
)*(b*d*e)^(1/2)-20*((-d*x+c)*b*e*x)^(1/2)*(b*d*e)^(1/2)*c*d*x+18*((-d*x+c) 
*b*e*x)^(1/2)*(b*d*e)^(1/2)*c^2)/(d*x+c)^(1/2)/d/((-d*x+c)*b*e*x)^(1/2)/(b 
*d*e)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.56 \[ \int \sqrt {e x} \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} \, dx=\left [\frac {4 \, \sqrt {-b d^{2} x^{2} + b c^{2}} {\left (8 \, d^{2} x^{2} + 10 \, c d x - 9 \, c^{2}\right )} \sqrt {d x + c} \sqrt {e x} + 9 \, {\left (c^{3} d x + c^{4}\right )} \sqrt {-\frac {b e}{d}} \log \left (-\frac {8 \, b d^{3} e x^{3} - 7 \, b c^{2} d e x + b c^{3} e + 4 \, \sqrt {-b d^{2} x^{2} + b c^{2}} {\left (2 \, d^{2} x - c d\right )} \sqrt {d x + c} \sqrt {e x} \sqrt {-\frac {b e}{d}}}{d x + c}\right )}{96 \, {\left (d^{2} x + c d\right )}}, \frac {2 \, \sqrt {-b d^{2} x^{2} + b c^{2}} {\left (8 \, d^{2} x^{2} + 10 \, c d x - 9 \, c^{2}\right )} \sqrt {d x + c} \sqrt {e x} - 9 \, {\left (c^{3} d x + c^{4}\right )} \sqrt {\frac {b e}{d}} \arctan \left (\frac {2 \, \sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c} \sqrt {e x} d \sqrt {\frac {b e}{d}}}{2 \, b d^{2} e x^{2} + b c d e x - b c^{2} e}\right )}{48 \, {\left (d^{2} x + c d\right )}}\right ] \] Input:

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

Output:

[1/96*(4*sqrt(-b*d^2*x^2 + b*c^2)*(8*d^2*x^2 + 10*c*d*x - 9*c^2)*sqrt(d*x 
+ c)*sqrt(e*x) + 9*(c^3*d*x + c^4)*sqrt(-b*e/d)*log(-(8*b*d^3*e*x^3 - 7*b* 
c^2*d*e*x + b*c^3*e + 4*sqrt(-b*d^2*x^2 + b*c^2)*(2*d^2*x - c*d)*sqrt(d*x 
+ c)*sqrt(e*x)*sqrt(-b*e/d))/(d*x + c)))/(d^2*x + c*d), 1/48*(2*sqrt(-b*d^ 
2*x^2 + b*c^2)*(8*d^2*x^2 + 10*c*d*x - 9*c^2)*sqrt(d*x + c)*sqrt(e*x) - 9* 
(c^3*d*x + c^4)*sqrt(b*e/d)*arctan(2*sqrt(-b*d^2*x^2 + b*c^2)*sqrt(d*x + c 
)*sqrt(e*x)*d*sqrt(b*e/d)/(2*b*d^2*e*x^2 + b*c*d*e*x - b*c^2*e)))/(d^2*x + 
 c*d)]
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate((e*x)^(1/2)*(d*x+c)^(1/2)*(-b*d^2*x^2+b*c^2)^(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 = 0.18 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.02 \[ \int \sqrt {e x} \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} \, dx=-\frac {{\left (\frac {b c^{2} e^{3} \log \left ({\left | -\sqrt {-b d e} \sqrt {e x} + \sqrt {-b d e^{2} x + b c e^{2}} \right |}\right )}{\sqrt {-b d e} d} - \sqrt {-b d e^{2} x + b c e^{2}} {\left (2 \, e x - \frac {c e}{d}\right )} \sqrt {e x}\right )} c {\left | e \right |}}{4 \, e^{3}} - \frac {{\left (\frac {3 \, b c^{3} e^{4} \log \left ({\left | -\sqrt {-b d e} \sqrt {e x} + \sqrt {-b d e^{2} x + b c e^{2}} \right |}\right )}{\sqrt {-b d e} d^{2}} - \sqrt {-b d e^{2} x + b c e^{2}} {\left (2 \, {\left (4 \, e x - \frac {c e}{d}\right )} e x - \frac {3 \, c^{2} e^{2}}{d^{2}}\right )} \sqrt {e x}\right )} d {\left | e \right |}}{24 \, e^{4}} \] Input:

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

Output:

-1/4*(b*c^2*e^3*log(abs(-sqrt(-b*d*e)*sqrt(e*x) + sqrt(-b*d*e^2*x + b*c*e^ 
2)))/(sqrt(-b*d*e)*d) - sqrt(-b*d*e^2*x + b*c*e^2)*(2*e*x - c*e/d)*sqrt(e* 
x))*c*abs(e)/e^3 - 1/24*(3*b*c^3*e^4*log(abs(-sqrt(-b*d*e)*sqrt(e*x) + sqr 
t(-b*d*e^2*x + b*c*e^2)))/(sqrt(-b*d*e)*d^2) - sqrt(-b*d*e^2*x + b*c*e^2)* 
(2*(4*e*x - c*e/d)*e*x - 3*c^2*e^2/d^2)*sqrt(e*x))*d*abs(e)/e^4
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.41 \[ \int \sqrt {e x} \sqrt {c+d x} \sqrt {b c^2-b d^2 x^2} \, dx=\frac {\sqrt {e}\, \sqrt {b}\, \left (-9 \sqrt {x}\, \sqrt {-d x +c}\, c^{2} d +10 \sqrt {x}\, \sqrt {-d x +c}\, c \,d^{2} x +8 \sqrt {x}\, \sqrt {-d x +c}\, d^{3} x^{2}-9 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {-d x +c}+\sqrt {x}\, \sqrt {d}\, i}{\sqrt {c}}\right ) c^{3} i \right )}{24 d^{2}} \] Input:

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

Output:

(sqrt(e)*sqrt(b)*( - 9*sqrt(x)*sqrt(c - d*x)*c**2*d + 10*sqrt(x)*sqrt(c - 
d*x)*c*d**2*x + 8*sqrt(x)*sqrt(c - d*x)*d**3*x**2 - 9*sqrt(d)*log((sqrt(c 
- d*x) + sqrt(x)*sqrt(d)*i)/sqrt(c))*c**3*i))/(24*d**2)