\(\int \frac {\sqrt {e x} (c^2-d^2 x^2)^{3/2}}{(c+d x)^{5/2}} \, dx\) [977]

Optimal result

Integrand size = 33, antiderivative size = 210 \[ \int \frac {\sqrt {e x} \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{5/2}} \, dx=\frac {7 c \sqrt {e x} \sqrt {c^2-d^2 x^2}}{4 d \sqrt {c+d x}}+\frac {\sqrt {e x} \left (c^2-d^2 x^2\right )^{3/2}}{2 d (c+d x)^{3/2}}-\frac {23 c^2 \sqrt {e} \arctan \left (\frac {\sqrt {e} \sqrt {c^2-d^2 x^2}}{\sqrt {d} \sqrt {e x} \sqrt {c+d x}}\right )}{4 d^{3/2}}+\frac {4 \sqrt {2} c^2 \sqrt {e} \arctan \left (\frac {\sqrt {e} \sqrt {c^2-d^2 x^2}}{\sqrt {2} \sqrt {d} \sqrt {e x} \sqrt {c+d x}}\right )}{d^{3/2}} \] Output:

7/4*c*(e*x)^(1/2)*(-d^2*x^2+c^2)^(1/2)/d/(d*x+c)^(1/2)+1/2*(e*x)^(1/2)*(-d 
^2*x^2+c^2)^(3/2)/d/(d*x+c)^(3/2)-23/4*c^2*e^(1/2)*arctan(e^(1/2)*(-d^2*x^ 
2+c^2)^(1/2)/d^(1/2)/(e*x)^(1/2)/(d*x+c)^(1/2))/d^(3/2)+4*2^(1/2)*c^2*e^(1 
/2)*arctan(1/2*e^(1/2)*(-d^2*x^2+c^2)^(1/2)*2^(1/2)/d^(1/2)/(e*x)^(1/2)/(d 
*x+c)^(1/2))/d^(3/2)
 

Mathematica [A] (verified)

Time = 8.99 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt {e x} \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{5/2}} \, dx=\frac {\sqrt {e x} \sqrt {c^2-d^2 x^2} \left (23 c^{3/2} (c-d x) \arcsin \left (\frac {\sqrt {d} \sqrt {x}}{\sqrt {c}}\right )+\sqrt {1-\frac {d x}{c}} \left (\sqrt {d} \sqrt {x} \left (9 c^2-11 c d x+2 d^2 x^2\right )-16 \sqrt {2} c^2 \sqrt {c-d x} \arctan \left (\frac {\sqrt {2} \sqrt {d} \sqrt {x}}{\sqrt {c-d x}}\right )\right )\right )}{4 d^{3/2} \sqrt {x} (c-d x) \sqrt {c+d x} \sqrt {1-\frac {d x}{c}}} \] Input:

Integrate[(Sqrt[e*x]*(c^2 - d^2*x^2)^(3/2))/(c + d*x)^(5/2),x]
 

Output:

(Sqrt[e*x]*Sqrt[c^2 - d^2*x^2]*(23*c^(3/2)*(c - d*x)*ArcSin[(Sqrt[d]*Sqrt[ 
x])/Sqrt[c]] + Sqrt[1 - (d*x)/c]*(Sqrt[d]*Sqrt[x]*(9*c^2 - 11*c*d*x + 2*d^ 
2*x^2) - 16*Sqrt[2]*c^2*Sqrt[c - d*x]*ArcTan[(Sqrt[2]*Sqrt[d]*Sqrt[x])/Sqr 
t[c - d*x]])))/(4*d^(3/2)*Sqrt[x]*(c - d*x)*Sqrt[c + d*x]*Sqrt[1 - (d*x)/c 
])
 

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.91, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {586, 112, 27, 171, 27, 175, 65, 104, 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.

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

\(\Big \downarrow \) 586

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

\(\Big \downarrow \) 112

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 171

\(\displaystyle \frac {\sqrt {c^2-d^2 x^2} \left (\frac {\sqrt {e x} (c-d x)^{3/2}}{2 d}-\frac {c e \left (\frac {\int \frac {c d e (9 c-23 d x)}{2 \sqrt {e x} \sqrt {c-d x} (c+d x)}dx}{d e}-\frac {7 \sqrt {e x} \sqrt {c-d x}}{e}\right )}{4 d}\right )}{\sqrt {c-d x} \sqrt {c+d x}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {c^2-d^2 x^2} \left (\frac {\sqrt {e x} (c-d x)^{3/2}}{2 d}-\frac {c e \left (\frac {1}{2} c \int \frac {9 c-23 d x}{\sqrt {e x} \sqrt {c-d x} (c+d x)}dx-\frac {7 \sqrt {e x} \sqrt {c-d x}}{e}\right )}{4 d}\right )}{\sqrt {c-d x} \sqrt {c+d x}}\)

\(\Big \downarrow \) 175

\(\displaystyle \frac {\sqrt {c^2-d^2 x^2} \left (\frac {\sqrt {e x} (c-d x)^{3/2}}{2 d}-\frac {c e \left (\frac {1}{2} c \left (32 c \int \frac {1}{\sqrt {e x} \sqrt {c-d x} (c+d x)}dx-23 \int \frac {1}{\sqrt {e x} \sqrt {c-d x}}dx\right )-\frac {7 \sqrt {e x} \sqrt {c-d x}}{e}\right )}{4 d}\right )}{\sqrt {c-d x} \sqrt {c+d x}}\)

\(\Big \downarrow \) 65

\(\displaystyle \frac {\sqrt {c^2-d^2 x^2} \left (\frac {\sqrt {e x} (c-d x)^{3/2}}{2 d}-\frac {c e \left (\frac {1}{2} c \left (32 c \int \frac {1}{\sqrt {e x} \sqrt {c-d x} (c+d x)}dx-46 \int \frac {1}{\frac {d x e}{c-d x}+e}d\frac {\sqrt {e x}}{\sqrt {c-d x}}\right )-\frac {7 \sqrt {e x} \sqrt {c-d x}}{e}\right )}{4 d}\right )}{\sqrt {c-d x} \sqrt {c+d x}}\)

\(\Big \downarrow \) 104

\(\displaystyle \frac {\sqrt {c^2-d^2 x^2} \left (\frac {\sqrt {e x} (c-d x)^{3/2}}{2 d}-\frac {c e \left (\frac {1}{2} c \left (64 c \int \frac {1}{c e+\frac {2 c d x e}{c-d x}}d\frac {\sqrt {e x}}{\sqrt {c-d x}}-46 \int \frac {1}{\frac {d x e}{c-d x}+e}d\frac {\sqrt {e x}}{\sqrt {c-d x}}\right )-\frac {7 \sqrt {e x} \sqrt {c-d x}}{e}\right )}{4 d}\right )}{\sqrt {c-d x} \sqrt {c+d x}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\sqrt {c^2-d^2 x^2} \left (\frac {\sqrt {e x} (c-d x)^{3/2}}{2 d}-\frac {c e \left (\frac {1}{2} c \left (\frac {32 \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c-d x}}\right )}{\sqrt {d} \sqrt {e}}-\frac {46 \arctan \left (\frac {\sqrt {d} \sqrt {e x}}{\sqrt {e} \sqrt {c-d x}}\right )}{\sqrt {d} \sqrt {e}}\right )-\frac {7 \sqrt {e x} \sqrt {c-d x}}{e}\right )}{4 d}\right )}{\sqrt {c-d x} \sqrt {c+d x}}\)

Input:

Int[(Sqrt[e*x]*(c^2 - d^2*x^2)^(3/2))/(c + d*x)^(5/2),x]
 

Output:

(Sqrt[c^2 - d^2*x^2]*((Sqrt[e*x]*(c - d*x)^(3/2))/(2*d) - (c*e*((-7*Sqrt[e 
*x]*Sqrt[c - d*x])/e + (c*((-46*ArcTan[(Sqrt[d]*Sqrt[e*x])/(Sqrt[e]*Sqrt[c 
 - d*x])])/(Sqrt[d]*Sqrt[e]) + (32*Sqrt[2]*ArcTan[(Sqrt[2]*Sqrt[d]*Sqrt[e* 
x])/(Sqrt[e]*Sqrt[c - d*x])])/(Sqrt[d]*Sqrt[e])))/2))/(4*d)))/(Sqrt[c - d* 
x]*Sqrt[c + d*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 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 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 112

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(a + b*x)^m*(c + d*x)^n*((e + f*x)^(p + 1)/(f*(m + n + 
p + 1))), x] - Simp[1/(f*(m + n + p + 1))   Int[(a + b*x)^(m - 1)*(c + d*x) 
^(n - 1)*(e + f*x)^p*Simp[c*m*(b*e - a*f) + a*n*(d*e - c*f) + (d*m*(b*e - a 
*f) + b*n*(d*e - c*f))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && 
GtQ[m, 0] && GtQ[n, 0] && NeQ[m + n + p + 1, 0] && (IntegersQ[2*m, 2*n, 2*p 
] || (IntegersQ[m, n + p] || IntegersQ[p, m + n]))
 

rule 171

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*(( 
e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Simp[1/(d*f*(m + n + p + 2)) 
  Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2 
) - h*(b*c*e*m + a*(d*e*(n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) 
+ h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x], x], x] /; Fre 
eQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] 
 && IntegersQ[2*m, 2*n, 2*p]
 

rule 175

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_ 
)))/((a_.) + (b_.)*(x_)), x_] :> Simp[h/b   Int[(c + d*x)^n*(e + f*x)^p, x] 
, x] + Simp[(b*g - a*h)/b   Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)), x], x] 
 /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]
 

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 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 = 0.31 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.16

method	result	size
default	\(\frac {\sqrt {e x}\, \sqrt {-d^{2} x^{2}+c^{2}}\, \left (-4 \sqrt {\left (-d x +c \right ) e x}\, \sqrt {d e}\, \sqrt {2}\, \sqrt {-\frac {e \,c^{2}}{d}}\, d^{2} x +32 \sqrt {d e}\, \ln \left (\frac {3 c d x e +2 \sqrt {2}\, \sqrt {-\frac {e \,c^{2}}{d}}\, \sqrt {\left (-d x +c \right ) e x}\, d -e \,c^{2}}{d x +c}\right ) c^{3} e -23 \arctan \left (\frac {\sqrt {d e}\, \left (-2 d x +c \right )}{2 d \sqrt {\left (-d x +c \right ) e x}}\right ) \sqrt {2}\, \sqrt {-\frac {e \,c^{2}}{d}}\, c^{2} d e +18 \sqrt {d e}\, \sqrt {2}\, \sqrt {-\frac {e \,c^{2}}{d}}\, \sqrt {\left (-d x +c \right ) e x}\, c d \right ) \sqrt {2}}{16 \sqrt {d x +c}\, \sqrt {\left (-d x +c \right ) e x}\, d^{2} \sqrt {d e}\, \sqrt {-\frac {e \,c^{2}}{d}}}\)	\(243\)
risch	\(\frac {\left (-2 d x +9 c \right ) \left (-d x +c \right ) x \sqrt {\frac {e x \left (-d^{2} x^{2}+c^{2}\right )}{d x +c}}\, \sqrt {d x +c}\, e}{4 d \sqrt {-e \left (d x -c \right ) x}\, \sqrt {e x}\, \sqrt {-d^{2} x^{2}+c^{2}}}+\frac {\left (\frac {23 c^{2} \arctan \left (\frac {\sqrt {d e}\, \left (x -\frac {c}{2 d}\right )}{\sqrt {-d e \,x^{2}+c e x}}\right )}{8 d \sqrt {d e}}+\frac {4 c^{3} \ln \left (\frac {-\frac {4 e \,c^{2}}{d}+3 c e \left (x +\frac {c}{d}\right )+2 \sqrt {-\frac {2 e \,c^{2}}{d}}\, \sqrt {-d e \left (x +\frac {c}{d}\right )^{2}+3 c e \left (x +\frac {c}{d}\right )-\frac {2 e \,c^{2}}{d}}}{x +\frac {c}{d}}\right )}{d^{2} \sqrt {-\frac {2 e \,c^{2}}{d}}}\right ) \sqrt {\frac {e x \left (-d^{2} x^{2}+c^{2}\right )}{d x +c}}\, \sqrt {d x +c}\, e}{\sqrt {e x}\, \sqrt {-d^{2} x^{2}+c^{2}}}\)	\(281\)

Input:

int((e*x)^(1/2)*(-d^2*x^2+c^2)^(3/2)/(d*x+c)^(5/2),x,method=_RETURNVERBOSE 
)
 

Output:

1/16*(e*x)^(1/2)*(-d^2*x^2+c^2)^(1/2)/(d*x+c)^(1/2)*(-4*((-d*x+c)*e*x)^(1/ 
2)*(d*e)^(1/2)*2^(1/2)*(-1/d*e*c^2)^(1/2)*d^2*x+32*(d*e)^(1/2)*ln((3*c*d*x 
*e+2*2^(1/2)*(-1/d*e*c^2)^(1/2)*((-d*x+c)*e*x)^(1/2)*d-e*c^2)/(d*x+c))*c^3 
*e-23*arctan(1/2*(d*e)^(1/2)*(-2*d*x+c)/d/((-d*x+c)*e*x)^(1/2))*2^(1/2)*(- 
1/d*e*c^2)^(1/2)*c^2*d*e+18*(d*e)^(1/2)*2^(1/2)*(-1/d*e*c^2)^(1/2)*((-d*x+ 
c)*e*x)^(1/2)*c*d)/((-d*x+c)*e*x)^(1/2)/d^2/(d*e)^(1/2)*2^(1/2)/(-1/d*e*c^ 
2)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 504, normalized size of antiderivative = 2.40 \[ \int \frac {\sqrt {e x} \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{5/2}} \, dx=\left [\frac {16 \, \sqrt {2} {\left (c^{2} d x + c^{3}\right )} \sqrt {-\frac {e}{d}} \log \left (-\frac {17 \, d^{3} e x^{3} + 3 \, c d^{2} e x^{2} - 13 \, c^{2} d e x + c^{3} e - 4 \, \sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} {\left (3 \, d^{2} x - c d\right )} \sqrt {d x + c} \sqrt {e x} \sqrt {-\frac {e}{d}}}{d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}}\right ) - 4 \, \sqrt {-d^{2} x^{2} + c^{2}} {\left (2 \, d x - 9 \, c\right )} \sqrt {d x + c} \sqrt {e x} + 23 \, {\left (c^{2} d x + c^{3}\right )} \sqrt {-\frac {e}{d}} \log \left (-\frac {8 \, d^{3} e x^{3} - 7 \, c^{2} d e x + c^{3} e + 4 \, \sqrt {-d^{2} x^{2} + c^{2}} {\left (2 \, d^{2} x - c d\right )} \sqrt {d x + c} \sqrt {e x} \sqrt {-\frac {e}{d}}}{d x + c}\right )}{16 \, {\left (d^{2} x + c d\right )}}, \frac {16 \, \sqrt {2} {\left (c^{2} d x + c^{3}\right )} \sqrt {\frac {e}{d}} \arctan \left (\frac {2 \, \sqrt {2} \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {e x} d \sqrt {\frac {e}{d}}}{3 \, d^{2} e x^{2} + 2 \, c d e x - c^{2} e}\right ) - 2 \, \sqrt {-d^{2} x^{2} + c^{2}} {\left (2 \, d x - 9 \, c\right )} \sqrt {d x + c} \sqrt {e x} - 23 \, {\left (c^{2} d x + c^{3}\right )} \sqrt {\frac {e}{d}} \arctan \left (\frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} \sqrt {d x + c} \sqrt {e x} d \sqrt {\frac {e}{d}}}{2 \, d^{2} e x^{2} + c d e x - c^{2} e}\right )}{8 \, {\left (d^{2} x + c d\right )}}\right ] \] Input:

integrate((e*x)^(1/2)*(-d^2*x^2+c^2)^(3/2)/(d*x+c)^(5/2),x, algorithm="fri 
cas")
 

Output:

[1/16*(16*sqrt(2)*(c^2*d*x + c^3)*sqrt(-e/d)*log(-(17*d^3*e*x^3 + 3*c*d^2* 
e*x^2 - 13*c^2*d*e*x + c^3*e - 4*sqrt(2)*sqrt(-d^2*x^2 + c^2)*(3*d^2*x - c 
*d)*sqrt(d*x + c)*sqrt(e*x)*sqrt(-e/d))/(d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x 
 + c^3)) - 4*sqrt(-d^2*x^2 + c^2)*(2*d*x - 9*c)*sqrt(d*x + c)*sqrt(e*x) + 
23*(c^2*d*x + c^3)*sqrt(-e/d)*log(-(8*d^3*e*x^3 - 7*c^2*d*e*x + c^3*e + 4* 
sqrt(-d^2*x^2 + c^2)*(2*d^2*x - c*d)*sqrt(d*x + c)*sqrt(e*x)*sqrt(-e/d))/( 
d*x + c)))/(d^2*x + c*d), 1/8*(16*sqrt(2)*(c^2*d*x + c^3)*sqrt(e/d)*arctan 
(2*sqrt(2)*sqrt(-d^2*x^2 + c^2)*sqrt(d*x + c)*sqrt(e*x)*d*sqrt(e/d)/(3*d^2 
*e*x^2 + 2*c*d*e*x - c^2*e)) - 2*sqrt(-d^2*x^2 + c^2)*(2*d*x - 9*c)*sqrt(d 
*x + c)*sqrt(e*x) - 23*(c^2*d*x + c^3)*sqrt(e/d)*arctan(2*sqrt(-d^2*x^2 + 
c^2)*sqrt(d*x + c)*sqrt(e*x)*d*sqrt(e/d)/(2*d^2*e*x^2 + c*d*e*x - c^2*e))) 
/(d^2*x + c*d)]
 

Sympy [F]

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

integrate((e*x)**(1/2)*(-d**2*x**2+c**2)**(3/2)/(d*x+c)**(5/2),x)
 

Output:

Integral(sqrt(e*x)*(-(-c + d*x)*(c + d*x))**(3/2)/(c + d*x)**(5/2), x)
 

Maxima [F]

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

integrate((e*x)^(1/2)*(-d^2*x^2+c^2)^(3/2)/(d*x+c)^(5/2),x, algorithm="max 
ima")
 

Output:

integrate((-d^2*x^2 + c^2)^(3/2)*sqrt(e*x)/(d*x + c)^(5/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sqrt {e x} \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((e*x)^(1/2)*(-d^2*x^2+c^2)^(3/2)/(d*x+c)^(5/2),x, algorithm="gia 
c")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E 
rror: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {e x} \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{5/2}} \, dx=\int \frac {{\left (c^2-d^2\,x^2\right )}^{3/2}\,\sqrt {e\,x}}{{\left (c+d\,x\right )}^{5/2}} \,d x \] Input:

int(((c^2 - d^2*x^2)^(3/2)*(e*x)^(1/2))/(c + d*x)^(5/2),x)
                                                                                    
                                                                                    
 

Output:

int(((c^2 - d^2*x^2)^(3/2)*(e*x)^(1/2))/(c + d*x)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.45 \[ \int \frac {\sqrt {e x} \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^{5/2}} \, dx=\frac {\sqrt {e}\, \left (16 \sqrt {d}\, \sqrt {2}\, \mathit {atan} \left (\frac {\sqrt {x}\, \sqrt {d}\, \sqrt {-d x +c}\, \sqrt {2}}{2 d x}\right ) c^{2}+23 \sqrt {d}\, \mathit {atan} \left (\frac {\sqrt {x}\, \sqrt {d}\, \sqrt {-d x +c}}{-d x +c}\right ) c^{2}+9 \sqrt {x}\, \sqrt {-d x +c}\, c d -2 \sqrt {x}\, \sqrt {-d x +c}\, d^{2} x \right )}{4 d^{2}} \] Input:

int((e*x)^(1/2)*(-d^2*x^2+c^2)^(3/2)/(d*x+c)^(5/2),x)
 

Output:

(sqrt(e)*(16*sqrt(d)*sqrt(2)*atan((sqrt(x)*sqrt(d)*sqrt(c - d*x)*sqrt(2))/ 
(2*d*x))*c**2 + 23*sqrt(d)*atan((sqrt(x)*sqrt(d)*sqrt(c - d*x))/(c - d*x)) 
*c**2 + 9*sqrt(x)*sqrt(c - d*x)*c*d - 2*sqrt(x)*sqrt(c - d*x)*d**2*x))/(4* 
d**2)