\(\int \frac {(e x)^{5/2} \sqrt {b c^2-b d^2 x^2}}{\sqrt {c+d x}} \, dx\) [1088]

Optimal result

Integrand size = 36, antiderivative size = 254 \[ \int \frac {(e x)^{5/2} \sqrt {b c^2-b d^2 x^2}}{\sqrt {c+d x}} \, dx=-\frac {5 c^3 e^2 \sqrt {e x} \sqrt {b c^2-b d^2 x^2}}{64 d^3 \sqrt {c+d x}}-\frac {5 c^2 e (e x)^{3/2} \sqrt {b c^2-b d^2 x^2}}{96 d^2 \sqrt {c+d x}}-\frac {c (e x)^{5/2} \sqrt {b c^2-b d^2 x^2}}{24 d \sqrt {c+d x}}+\frac {(e x)^{7/2} \sqrt {b c^2-b d^2 x^2}}{4 e \sqrt {c+d x}}-\frac {5 \sqrt {b} c^4 e^{5/2} \arctan \left (\frac {\sqrt {e} \sqrt {b c^2-b d^2 x^2}}{\sqrt {b} \sqrt {d} \sqrt {e x} \sqrt {c+d x}}\right )}{64 d^{7/2}} \] Output:

-5/64*c^3*e^2*(e*x)^(1/2)*(-b*d^2*x^2+b*c^2)^(1/2)/d^3/(d*x+c)^(1/2)-5/96* 
c^2*e*(e*x)^(3/2)*(-b*d^2*x^2+b*c^2)^(1/2)/d^2/(d*x+c)^(1/2)-1/24*c*(e*x)^ 
(5/2)*(-b*d^2*x^2+b*c^2)^(1/2)/d/(d*x+c)^(1/2)+1/4*(e*x)^(7/2)*(-b*d^2*x^2 
+b*c^2)^(1/2)/e/(d*x+c)^(1/2)-5/64*b^(1/2)*c^4*e^(5/2)*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^(7/2)
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.56 \[ \int \frac {(e x)^{5/2} \sqrt {b c^2-b d^2 x^2}}{\sqrt {c+d x}} \, dx=\frac {(e x)^{5/2} \sqrt {b \left (c^2-d^2 x^2\right )} \left (-\sqrt {d} \sqrt {x} \sqrt {1-\frac {d x}{c}} \left (15 c^3+10 c^2 d x+8 c d^2 x^2-48 d^3 x^3\right )+15 c^{7/2} \arcsin \left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}}\right )\right )}{192 d^{7/2} x^{5/2} \sqrt {c+d x} \sqrt {1-\frac {d x}{c}}} \] Input:

Integrate[((e*x)^(5/2)*Sqrt[b*c^2 - b*d^2*x^2])/Sqrt[c + d*x],x]
 

Output:

((e*x)^(5/2)*Sqrt[b*(c^2 - d^2*x^2)]*(-(Sqrt[d]*Sqrt[x]*Sqrt[1 - (d*x)/c]* 
(15*c^3 + 10*c^2*d*x + 8*c*d^2*x^2 - 48*d^3*x^3)) + 15*c^(7/2)*ArcSin[(Sqr 
t[d]*Sqrt[x])/Sqrt[c]]))/(192*d^(7/2)*x^(5/2)*Sqrt[c + d*x]*Sqrt[1 - (d*x) 
/c])
 

Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {576, 586, 60, 60, 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.

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

\(\Big \downarrow \) 576

\(\displaystyle \frac {1}{8} b c \int \frac {(e x)^{5/2} \sqrt {c+d x}}{\sqrt {b c^2-b d^2 x^2}}dx+\frac {(e x)^{7/2} \sqrt {b c^2-b d^2 x^2}}{4 e \sqrt {c+d x}}\)

\(\Big \downarrow \) 586

\(\displaystyle \frac {b c \sqrt {c+d x} \sqrt {b c-b d x} \int \frac {(e x)^{5/2}}{\sqrt {b c-b d x}}dx}{8 \sqrt {b c^2-b d^2 x^2}}+\frac {(e x)^{7/2} \sqrt {b c^2-b d^2 x^2}}{4 e \sqrt {c+d x}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {b c \sqrt {c+d x} \sqrt {b c-b d x} \left (\frac {5 c e \int \frac {(e x)^{3/2}}{\sqrt {b c-b d x}}dx}{6 d}-\frac {(e x)^{5/2} \sqrt {b c-b d x}}{3 b d}\right )}{8 \sqrt {b c^2-b d^2 x^2}}+\frac {(e x)^{7/2} \sqrt {b c^2-b d^2 x^2}}{4 e \sqrt {c+d x}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {b c \sqrt {c+d x} \sqrt {b c-b d x} \left (\frac {5 c e \left (\frac {3 c e \int \frac {\sqrt {e x}}{\sqrt {b c-b d x}}dx}{4 d}-\frac {(e x)^{3/2} \sqrt {b c-b d x}}{2 b d}\right )}{6 d}-\frac {(e x)^{5/2} \sqrt {b c-b d x}}{3 b d}\right )}{8 \sqrt {b c^2-b d^2 x^2}}+\frac {(e x)^{7/2} \sqrt {b c^2-b d^2 x^2}}{4 e \sqrt {c+d x}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {b c \sqrt {c+d x} \sqrt {b c-b d x} \left (\frac {5 c e \left (\frac {3 c e \left (\frac {c e \int \frac {1}{\sqrt {e x} \sqrt {b c-b d x}}dx}{2 d}-\frac {\sqrt {e x} \sqrt {b c-b d x}}{b d}\right )}{4 d}-\frac {(e x)^{3/2} \sqrt {b c-b d x}}{2 b d}\right )}{6 d}-\frac {(e x)^{5/2} \sqrt {b c-b d x}}{3 b d}\right )}{8 \sqrt {b c^2-b d^2 x^2}}+\frac {(e x)^{7/2} \sqrt {b c^2-b d^2 x^2}}{4 e \sqrt {c+d x}}\)

\(\Big \downarrow \) 65

\(\displaystyle \frac {b c \sqrt {c+d x} \sqrt {b c-b d x} \left (\frac {5 c e \left (\frac {3 c e \left (\frac {c e \int \frac {1}{\frac {b d x e}{b c-b d x}+e}d\frac {\sqrt {e x}}{\sqrt {b c-b d x}}}{d}-\frac {\sqrt {e x} \sqrt {b c-b d x}}{b d}\right )}{4 d}-\frac {(e x)^{3/2} \sqrt {b c-b d x}}{2 b d}\right )}{6 d}-\frac {(e x)^{5/2} \sqrt {b c-b d x}}{3 b d}\right )}{8 \sqrt {b c^2-b d^2 x^2}}+\frac {(e x)^{7/2} \sqrt {b c^2-b d^2 x^2}}{4 e \sqrt {c+d x}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {b c \sqrt {c+d x} \sqrt {b c-b d x} \left (\frac {5 c e \left (\frac {3 c e \left (\frac {c \sqrt {e} \arctan \left (\frac {\sqrt {b} \sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {b c-b d x}}\right )}{\sqrt {b} d^{3/2}}-\frac {\sqrt {e x} \sqrt {b c-b d x}}{b d}\right )}{4 d}-\frac {(e x)^{3/2} \sqrt {b c-b d x}}{2 b d}\right )}{6 d}-\frac {(e x)^{5/2} \sqrt {b c-b d x}}{3 b d}\right )}{8 \sqrt {b c^2-b d^2 x^2}}+\frac {(e x)^{7/2} \sqrt {b c^2-b d^2 x^2}}{4 e \sqrt {c+d x}}\)

Input:

Int[((e*x)^(5/2)*Sqrt[b*c^2 - b*d^2*x^2])/Sqrt[c + d*x],x]
 

Output:

((e*x)^(7/2)*Sqrt[b*c^2 - b*d^2*x^2])/(4*e*Sqrt[c + d*x]) + (b*c*Sqrt[c + 
d*x]*Sqrt[b*c - b*d*x]*(-1/3*((e*x)^(5/2)*Sqrt[b*c - b*d*x])/(b*d) + (5*c* 
e*(-1/2*((e*x)^(3/2)*Sqrt[b*c - b*d*x])/(b*d) + (3*c*e*(-((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*d)))/(6*d)))/(8*Sqrt[b*c^2 
 - b*d^2*x^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 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 = 0.29 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.76

method	result	size
default	\(-\frac {e^{2} \sqrt {e x}\, \sqrt {b \left (-d^{2} x^{2}+c^{2}\right )}\, \left (-96 d^{3} x^{3} \sqrt {\left (-d x +c \right ) b e x}\, \sqrt {b d e}+15 \arctan \left (\frac {\sqrt {b d e}\, \left (-2 d x +c \right )}{2 d \sqrt {\left (-d x +c \right ) b e x}}\right ) b \,c^{4} e +16 c \,d^{2} x^{2} \sqrt {\left (-d x +c \right ) b e x}\, \sqrt {b d e}+20 \sqrt {\left (-d x +c \right ) b e x}\, \sqrt {b d e}\, c^{2} d x +30 \sqrt {\left (-d x +c \right ) b e x}\, \sqrt {b d e}\, c^{3}\right )}{384 \sqrt {d x +c}\, \sqrt {\left (-d x +c \right ) b e x}\, d^{3} \sqrt {b d e}}\)	\(194\)
risch	\(-\frac {\left (-48 d^{3} x^{3}+8 c \,d^{2} x^{2}+10 c^{2} d x +15 c^{3}\right ) \left (-d x +c \right ) x \sqrt {-\frac {e x b \left (d^{2} x^{2}-c^{2}\right )}{d x +c}}\, \sqrt {d x +c}\, b \,e^{3}}{192 d^{3} \sqrt {-b e \left (d x -c \right ) x}\, \sqrt {e x}\, \sqrt {-b \left (d^{2} x^{2}-c^{2}\right )}}+\frac {5 c^{4} \arctan \left (\frac {\sqrt {b d e}\, \left (x -\frac {c}{2 d}\right )}{\sqrt {-b d \,x^{2} e +b c e x}}\right ) \sqrt {-\frac {e x b \left (d^{2} x^{2}-c^{2}\right )}{d x +c}}\, \sqrt {d x +c}\, b \,e^{3}}{128 d^{3} \sqrt {b d e}\, \sqrt {e x}\, \sqrt {-b \left (d^{2} x^{2}-c^{2}\right )}}\)	\(226\)

Input:

int((e*x)^(5/2)*(-b*d^2*x^2+b*c^2)^(1/2)/(d*x+c)^(1/2),x,method=_RETURNVER 
BOSE)
 

Output:

-1/384*e^2*(e*x)^(1/2)*(b*(-d^2*x^2+c^2))^(1/2)*(-96*d^3*x^3*((-d*x+c)*b*e 
*x)^(1/2)*(b*d*e)^(1/2)+15*arctan(1/2*(b*d*e)^(1/2)*(-2*d*x+c)/d/((-d*x+c) 
*b*e*x)^(1/2))*b*c^4*e+16*c*d^2*x^2*((-d*x+c)*b*e*x)^(1/2)*(b*d*e)^(1/2)+2 
0*((-d*x+c)*b*e*x)^(1/2)*(b*d*e)^(1/2)*c^2*d*x+30*((-d*x+c)*b*e*x)^(1/2)*( 
b*d*e)^(1/2)*c^3)/(d*x+c)^(1/2)/((-d*x+c)*b*e*x)^(1/2)/d^3/(b*d*e)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 388, normalized size of antiderivative = 1.53 \[ \int \frac {(e x)^{5/2} \sqrt {b c^2-b d^2 x^2}}{\sqrt {c+d x}} \, dx=\left [\frac {4 \, {\left (48 \, d^{3} e^{2} x^{3} - 8 \, c d^{2} e^{2} x^{2} - 10 \, c^{2} d e^{2} x - 15 \, c^{3} e^{2}\right )} \sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c} \sqrt {e x} + 15 \, {\left (c^{4} d e^{2} x + c^{5} e^{2}\right )} \sqrt {-\frac {b e}{d}} \log \left (-\frac {8 \, b d^{3} e x^{3} - 7 \, b c^{2} d e x + b c^{3} e + 4 \, \sqrt {-b d^{2} x^{2} + b c^{2}} {\left (2 \, d^{2} x - c d\right )} \sqrt {d x + c} \sqrt {e x} \sqrt {-\frac {b e}{d}}}{d x + c}\right )}{768 \, {\left (d^{4} x + c d^{3}\right )}}, \frac {2 \, {\left (48 \, d^{3} e^{2} x^{3} - 8 \, c d^{2} e^{2} x^{2} - 10 \, c^{2} d e^{2} x - 15 \, c^{3} e^{2}\right )} \sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c} \sqrt {e x} - 15 \, {\left (c^{4} d e^{2} x + c^{5} e^{2}\right )} \sqrt {\frac {b e}{d}} \arctan \left (\frac {2 \, \sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c} \sqrt {e x} d \sqrt {\frac {b e}{d}}}{2 \, b d^{2} e x^{2} + b c d e x - b c^{2} e}\right )}{384 \, {\left (d^{4} x + c d^{3}\right )}}\right ] \] Input:

integrate((e*x)^(5/2)*(-b*d^2*x^2+b*c^2)^(1/2)/(d*x+c)^(1/2),x, algorithm= 
"fricas")
 

Output:

[1/768*(4*(48*d^3*e^2*x^3 - 8*c*d^2*e^2*x^2 - 10*c^2*d*e^2*x - 15*c^3*e^2) 
*sqrt(-b*d^2*x^2 + b*c^2)*sqrt(d*x + c)*sqrt(e*x) + 15*(c^4*d*e^2*x + c^5* 
e^2)*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^4*x + c*d^3), 1/384*(2*(48*d^3*e^2*x^3 - 8*c*d^2*e^2*x^2 - 1 
0*c^2*d*e^2*x - 15*c^3*e^2)*sqrt(-b*d^2*x^2 + b*c^2)*sqrt(d*x + c)*sqrt(e* 
x) - 15*(c^4*d*e^2*x + c^5*e^2)*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^4*x + c*d^3)]
 

Sympy [F]

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

integrate((e*x)**(5/2)*(-b*d**2*x**2+b*c**2)**(1/2)/(d*x+c)**(1/2),x)
 

Output:

Integral((e*x)**(5/2)*sqrt(-b*(-c + d*x)*(c + d*x))/sqrt(c + d*x), x)
 

Maxima [F]

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

integrate((e*x)^(5/2)*(-b*d^2*x^2+b*c^2)^(1/2)/(d*x+c)^(1/2),x, algorithm= 
"maxima")
 

Output:

integrate(sqrt(-b*d^2*x^2 + b*c^2)*(e*x)^(5/2)/sqrt(d*x + c), x)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.50 \[ \int \frac {(e x)^{5/2} \sqrt {b c^2-b d^2 x^2}}{\sqrt {c+d x}} \, dx=-\frac {1}{192} \, {\left (\frac {15 \, b c^{4} e^{2} \log \left ({\left | -\sqrt {-b d e} \sqrt {e x} + \sqrt {-b d e^{2} x + b c e^{2}} \right |}\right )}{\sqrt {-b d e} d^{3}} - \sqrt {-b d e^{2} x + b c e^{2}} {\left (2 \, {\left (4 \, e x {\left (\frac {6 \, x}{e^{2}} - \frac {c}{d e^{2}}\right )} - \frac {5 \, c^{2}}{d^{2} e}\right )} e x - \frac {15 \, c^{3}}{d^{3}}\right )} \sqrt {e x}\right )} {\left | e \right |} \] Input:

integrate((e*x)^(5/2)*(-b*d^2*x^2+b*c^2)^(1/2)/(d*x+c)^(1/2),x, algorithm= 
"giac")
 

Output:

-1/192*(15*b*c^4*e^2*log(abs(-sqrt(-b*d*e)*sqrt(e*x) + sqrt(-b*d*e^2*x + b 
*c*e^2)))/(sqrt(-b*d*e)*d^3) - sqrt(-b*d*e^2*x + b*c*e^2)*(2*(4*e*x*(6*x/e 
^2 - c/(d*e^2)) - 5*c^2/(d^2*e))*e*x - 15*c^3/d^3)*sqrt(e*x))*abs(e)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^{5/2} \sqrt {b c^2-b d^2 x^2}}{\sqrt {c+d x}} \, dx=\int \frac {{\left (e\,x\right )}^{5/2}\,\sqrt {b\,c^2-b\,d^2\,x^2}}{\sqrt {c+d\,x}} \,d x \] Input:

int(((e*x)^(5/2)*(b*c^2 - b*d^2*x^2)^(1/2))/(c + d*x)^(1/2),x)
 

Output:

int(((e*x)^(5/2)*(b*c^2 - b*d^2*x^2)^(1/2))/(c + d*x)^(1/2), x)
 

Reduce [F]

\[ \int \frac {(e x)^{5/2} \sqrt {b c^2-b d^2 x^2}}{\sqrt {c+d x}} \, dx=\frac {\sqrt {e}\, \sqrt {b}\, e^{2} \left (46 \sqrt {x}\, \sqrt {d x +c}\, \sqrt {-d^{2} x^{2}+c^{2}}\, c^{2}-56 \sqrt {x}\, \sqrt {d x +c}\, \sqrt {-d^{2} x^{2}+c^{2}}\, c d x +48 \sqrt {x}\, \sqrt {d x +c}\, \sqrt {-d^{2} x^{2}+c^{2}}\, d^{2} x^{2}-23 \left (\int \frac {\sqrt {-d x +c}}{\sqrt {x}\, c -\sqrt {x}\, d x}d x \right ) c^{4}+61 \left (\int \frac {\sqrt {x}\, \sqrt {d x +c}\, \sqrt {-d^{2} x^{2}+c^{2}}}{-d^{2} x^{2}+c^{2}}d x \right ) c^{3} d \right )}{192 d^{3}} \] Input:

int((e*x)^(5/2)*(-b*d^2*x^2+b*c^2)^(1/2)/(d*x+c)^(1/2),x)
 

Output:

(sqrt(e)*sqrt(b)*e**2*(46*sqrt(x)*sqrt(c + d*x)*sqrt(c**2 - d**2*x**2)*c** 
2 - 56*sqrt(x)*sqrt(c + d*x)*sqrt(c**2 - d**2*x**2)*c*d*x + 48*sqrt(x)*sqr 
t(c + d*x)*sqrt(c**2 - d**2*x**2)*d**2*x**2 - 23*int(sqrt(c - d*x)/(sqrt(x 
)*c - sqrt(x)*d*x),x)*c**4 + 61*int((sqrt(x)*sqrt(c + d*x)*sqrt(c**2 - d** 
2*x**2))/(c**2 - d**2*x**2),x)*c**3*d))/(192*d**3)