\(\int \frac {(c^2-d^2 x^2)^{5/2}}{x (c+d x)} \, dx\) [285]

Optimal result

Integrand size = 27, antiderivative size = 113 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{x (c+d x)} \, dx=\frac {1}{8} c^2 (8 c-3 d x) \sqrt {c^2-d^2 x^2}+\frac {1}{12} (4 c-3 d x) \left (c^2-d^2 x^2\right )^{3/2}-\frac {3}{8} c^4 \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )-c^4 \text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{c}\right ) \] Output:

1/8*c^2*(-3*d*x+8*c)*(-d^2*x^2+c^2)^(1/2)+1/12*(-3*d*x+4*c)*(-d^2*x^2+c^2) 
^(3/2)-3/8*c^4*arctan(d*x/(-d^2*x^2+c^2)^(1/2))-c^4*arctanh((-d^2*x^2+c^2) 
^(1/2)/c)
 

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.26 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{x (c+d x)} \, dx=\frac {1}{24} \sqrt {c^2-d^2 x^2} \left (32 c^3-15 c^2 d x-8 c d^2 x^2+6 d^3 x^3\right )+\frac {3}{4} c^4 \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )-c^3 \sqrt {c^2} \log (x)+c^3 \sqrt {c^2} \log \left (\sqrt {c^2}-\sqrt {c^2-d^2 x^2}\right ) \] Input:

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

Output:

(Sqrt[c^2 - d^2*x^2]*(32*c^3 - 15*c^2*d*x - 8*c*d^2*x^2 + 6*d^3*x^3))/24 + 
 (3*c^4*ArcTan[(d*x)/(Sqrt[c^2] - Sqrt[c^2 - d^2*x^2])])/4 - c^3*Sqrt[c^2] 
*Log[x] + c^3*Sqrt[c^2]*Log[Sqrt[c^2] - Sqrt[c^2 - d^2*x^2]]
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {566, 535, 535, 538, 224, 216, 243, 73, 221}

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.

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

\(\Big \downarrow \) 566

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

\(\Big \downarrow \) 535

\(\displaystyle \frac {1}{4} c^2 \int \frac {(4 c-3 d x) \sqrt {c^2-d^2 x^2}}{x}dx+\frac {1}{12} (4 c-3 d x) \left (c^2-d^2 x^2\right )^{3/2}\)

\(\Big \downarrow \) 535

\(\displaystyle \frac {1}{4} c^2 \left (\frac {1}{2} c^2 \int \frac {8 c-3 d x}{x \sqrt {c^2-d^2 x^2}}dx+\frac {1}{2} (8 c-3 d x) \sqrt {c^2-d^2 x^2}\right )+\frac {1}{12} (4 c-3 d x) \left (c^2-d^2 x^2\right )^{3/2}\)

\(\Big \downarrow \) 538

\(\displaystyle \frac {1}{4} c^2 \left (\frac {1}{2} c^2 \left (8 c \int \frac {1}{x \sqrt {c^2-d^2 x^2}}dx-3 d \int \frac {1}{\sqrt {c^2-d^2 x^2}}dx\right )+\frac {1}{2} (8 c-3 d x) \sqrt {c^2-d^2 x^2}\right )+\frac {1}{12} (4 c-3 d x) \left (c^2-d^2 x^2\right )^{3/2}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {1}{4} c^2 \left (\frac {1}{2} c^2 \left (8 c \int \frac {1}{x \sqrt {c^2-d^2 x^2}}dx-3 d \int \frac {1}{\frac {d^2 x^2}{c^2-d^2 x^2}+1}d\frac {x}{\sqrt {c^2-d^2 x^2}}\right )+\frac {1}{2} (8 c-3 d x) \sqrt {c^2-d^2 x^2}\right )+\frac {1}{12} (4 c-3 d x) \left (c^2-d^2 x^2\right )^{3/2}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{4} c^2 \left (\frac {1}{2} c^2 \left (8 c \int \frac {1}{x \sqrt {c^2-d^2 x^2}}dx-3 \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )\right )+\frac {1}{2} (8 c-3 d x) \sqrt {c^2-d^2 x^2}\right )+\frac {1}{12} (4 c-3 d x) \left (c^2-d^2 x^2\right )^{3/2}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {1}{4} c^2 \left (\frac {1}{2} c^2 \left (4 c \int \frac {1}{x^2 \sqrt {c^2-d^2 x^2}}dx^2-3 \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )\right )+\frac {1}{2} (8 c-3 d x) \sqrt {c^2-d^2 x^2}\right )+\frac {1}{12} (4 c-3 d x) \left (c^2-d^2 x^2\right )^{3/2}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {1}{4} c^2 \left (\frac {1}{2} c^2 \left (-\frac {8 c \int \frac {1}{\frac {c^2}{d^2}-\frac {x^4}{d^2}}d\sqrt {c^2-d^2 x^2}}{d^2}-3 \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )\right )+\frac {1}{2} (8 c-3 d x) \sqrt {c^2-d^2 x^2}\right )+\frac {1}{12} (4 c-3 d x) \left (c^2-d^2 x^2\right )^{3/2}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {1}{4} c^2 \left (\frac {1}{2} c^2 \left (-3 \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )-8 \text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{c}\right )\right )+\frac {1}{2} (8 c-3 d x) \sqrt {c^2-d^2 x^2}\right )+\frac {1}{12} (4 c-3 d x) \left (c^2-d^2 x^2\right )^{3/2}\)

Input:

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

Output:

((4*c - 3*d*x)*(c^2 - d^2*x^2)^(3/2))/12 + (c^2*(((8*c - 3*d*x)*Sqrt[c^2 - 
 d^2*x^2])/2 + (c^2*(-3*ArcTan[(d*x)/Sqrt[c^2 - d^2*x^2]] - 8*ArcTanh[Sqrt 
[c^2 - d^2*x^2]/c]))/2))/4
 

Defintions of rubi rules used

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 221

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

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 535

Int[(((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_))/(x_), x_Symbol] :> Sim 
p[(c*(2*p + 1) + 2*d*p*x)*((a + b*x^2)^p/(2*p*(2*p + 1))), x] + Simp[a/(2*p 
 + 1)   Int[(c*(2*p + 1) + 2*d*p*x)*((a + b*x^2)^(p - 1)/x), x], x] /; Free 
Q[{a, b, c, d}, x] && GtQ[p, 0] && IntegerQ[2*p]
 

rule 538

Int[((c_) + (d_.)*(x_))/((x_)*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Simp 
[c   Int[1/(x*Sqrt[a + b*x^2]), x], x] + Simp[d   Int[1/Sqrt[a + b*x^2], x] 
, x] /; FreeQ[{a, b, c, d}, x]
 

rule 566

Int[((x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)), x_Symbol] : 
> Int[x^m*(a/c + b*(x/d))*(a + b*x^2)^(p - 1), x] /; FreeQ[{a, b, c, d}, x] 
 && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(296\) vs. \(2(99)=198\).

Time = 0.22 (sec) , antiderivative size = 297, normalized size of antiderivative = 2.63

method	result	size
default	\(\frac {\frac {\left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}{5}+c^{2} \left (\frac {\left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{3}+c^{2} \left (\sqrt {-d^{2} x^{2}+c^{2}}-\frac {c^{2} \ln \left (\frac {2 c^{2}+2 \sqrt {c^{2}}\, \sqrt {-d^{2} x^{2}+c^{2}}}{x}\right )}{\sqrt {c^{2}}}\right )\right )}{c}-\frac {\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {5}{2}}}{5}+c d \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d \right ) \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{8 d^{2}}+\frac {3 c^{2} \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d \right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{4 d^{2}}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{2 \sqrt {d^{2}}}\right )}{4}\right )}{c}\)	\(297\)

Input:

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

Output:

1/c*(1/5*(-d^2*x^2+c^2)^(5/2)+c^2*(1/3*(-d^2*x^2+c^2)^(3/2)+c^2*((-d^2*x^2 
+c^2)^(1/2)-c^2/(c^2)^(1/2)*ln((2*c^2+2*(c^2)^(1/2)*(-d^2*x^2+c^2)^(1/2))/ 
x))))-1/c*(1/5*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(5/2)+c*d*(-1/8*(-2*d^2*(x+c 
/d)+2*c*d)/d^2*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(3/2)+3/4*c^2*(-1/4*(-2*d^2* 
(x+c/d)+2*c*d)/d^2*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(1/2)+1/2*c^2/(d^2)^(1/2 
)*arctan((d^2)^(1/2)*x/(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(1/2)))))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.95 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{x (c+d x)} \, dx=\frac {3}{4} \, c^{4} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + c^{4} \log \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{x}\right ) + \frac {1}{24} \, {\left (6 \, d^{3} x^{3} - 8 \, c d^{2} x^{2} - 15 \, c^{2} d x + 32 \, c^{3}\right )} \sqrt {-d^{2} x^{2} + c^{2}} \] Input:

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

Output:

3/4*c^4*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x)) + c^4*log(-(c - sqrt(-d^ 
2*x^2 + c^2))/x) + 1/24*(6*d^3*x^3 - 8*c*d^2*x^2 - 15*c^2*d*x + 32*c^3)*sq 
rt(-d^2*x^2 + c^2)
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 6.07 (sec) , antiderivative size = 400, normalized size of antiderivative = 3.54 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{x (c+d x)} \, dx=c^{3} \left (\begin {cases} \frac {c^{2}}{d x \sqrt {\frac {c^{2}}{d^{2} x^{2}} - 1}} - c \operatorname {acosh}{\left (\frac {c}{d x} \right )} - \frac {d x}{\sqrt {\frac {c^{2}}{d^{2} x^{2}} - 1}} & \text {for}\: \left |{\frac {c^{2}}{d^{2} x^{2}}}\right | > 1 \\- \frac {i c^{2}}{d x \sqrt {- \frac {c^{2}}{d^{2} x^{2}} + 1}} + i c \operatorname {asin}{\left (\frac {c}{d x} \right )} + \frac {i d x}{\sqrt {- \frac {c^{2}}{d^{2} x^{2}} + 1}} & \text {otherwise} \end {cases}\right ) - c^{2} d \left (\begin {cases} \frac {c^{2} \left (\begin {cases} \frac {\log {\left (- 2 d^{2} x + 2 \sqrt {- d^{2}} \sqrt {c^{2} - d^{2} x^{2}} \right )}}{\sqrt {- d^{2}}} & \text {for}\: c^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- d^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {c^{2} - d^{2} x^{2}}}{2} & \text {for}\: d^{2} \neq 0 \\x \sqrt {c^{2}} & \text {otherwise} \end {cases}\right ) - c d^{2} \left (\begin {cases} - \frac {c^{2} \sqrt {c^{2} - d^{2} x^{2}}}{3 d^{2}} + \frac {x^{2} \sqrt {c^{2} - d^{2} x^{2}}}{3} & \text {for}\: d^{2} \neq 0 \\\frac {x^{2} \sqrt {c^{2}}}{2} & \text {otherwise} \end {cases}\right ) + d^{3} \left (\begin {cases} \frac {c^{4} \left (\begin {cases} \frac {\log {\left (- 2 d^{2} x + 2 \sqrt {- d^{2}} \sqrt {c^{2} - d^{2} x^{2}} \right )}}{\sqrt {- d^{2}}} & \text {for}\: c^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- d^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{8 d^{2}} - \frac {c^{2} x \sqrt {c^{2} - d^{2} x^{2}}}{8 d^{2}} + \frac {x^{3} \sqrt {c^{2} - d^{2} x^{2}}}{4} & \text {for}\: d^{2} \neq 0 \\\frac {x^{3} \sqrt {c^{2}}}{3} & \text {otherwise} \end {cases}\right ) \] Input:

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

Output:

c**3*Piecewise((c**2/(d*x*sqrt(c**2/(d**2*x**2) - 1)) - c*acosh(c/(d*x)) - 
 d*x/sqrt(c**2/(d**2*x**2) - 1), Abs(c**2/(d**2*x**2)) > 1), (-I*c**2/(d*x 
*sqrt(-c**2/(d**2*x**2) + 1)) + I*c*asin(c/(d*x)) + I*d*x/sqrt(-c**2/(d**2 
*x**2) + 1), True)) - c**2*d*Piecewise((c**2*Piecewise((log(-2*d**2*x + 2* 
sqrt(-d**2)*sqrt(c**2 - d**2*x**2))/sqrt(-d**2), Ne(c**2, 0)), (x*log(x)/s 
qrt(-d**2*x**2), True))/2 + x*sqrt(c**2 - d**2*x**2)/2, Ne(d**2, 0)), (x*s 
qrt(c**2), True)) - c*d**2*Piecewise((-c**2*sqrt(c**2 - d**2*x**2)/(3*d**2 
) + x**2*sqrt(c**2 - d**2*x**2)/3, Ne(d**2, 0)), (x**2*sqrt(c**2)/2, True) 
) + d**3*Piecewise((c**4*Piecewise((log(-2*d**2*x + 2*sqrt(-d**2)*sqrt(c** 
2 - d**2*x**2))/sqrt(-d**2), Ne(c**2, 0)), (x*log(x)/sqrt(-d**2*x**2), Tru 
e))/(8*d**2) - c**2*x*sqrt(c**2 - d**2*x**2)/(8*d**2) + x**3*sqrt(c**2 - d 
**2*x**2)/4, Ne(d**2, 0)), (x**3*sqrt(c**2)/3, True))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.10 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{x (c+d x)} \, dx=-\frac {3}{8} \, c^{4} \arcsin \left (\frac {d x}{c}\right ) - c^{4} \log \left (\frac {2 \, c^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} c}{{\left | x \right |}}\right ) - \frac {3}{8} \, \sqrt {-d^{2} x^{2} + c^{2}} c^{2} d x + \sqrt {-d^{2} x^{2} + c^{2}} c^{3} - \frac {1}{4} \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d x + \frac {1}{3} \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c \] Input:

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

Output:

-3/8*c^4*arcsin(d*x/c) - c^4*log(2*c^2/abs(x) + 2*sqrt(-d^2*x^2 + c^2)*c/a 
bs(x)) - 3/8*sqrt(-d^2*x^2 + c^2)*c^2*d*x + sqrt(-d^2*x^2 + c^2)*c^3 - 1/4 
*(-d^2*x^2 + c^2)^(3/2)*d*x + 1/3*(-d^2*x^2 + c^2)^(3/2)*c
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.01 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{x (c+d x)} \, dx=-\frac {3 \, c^{4} d \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{8 \, {\left | d \right |}} - \frac {c^{4} d \log \left (\frac {{\left | -2 \, c d - 2 \, \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |} \right |}}{2 \, d^{2} {\left | x \right |}}\right )}{{\left | d \right |}} + \frac {1}{24} \, \sqrt {-d^{2} x^{2} + c^{2}} {\left (32 \, c^{3} - {\left (15 \, c^{2} d - 2 \, {\left (3 \, d^{3} x - 4 \, c d^{2}\right )} x\right )} x\right )} \] Input:

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

Output:

-3/8*c^4*d*arcsin(d*x/c)*sgn(c)*sgn(d)/abs(d) - c^4*d*log(1/2*abs(-2*c*d - 
 2*sqrt(-d^2*x^2 + c^2)*abs(d))/(d^2*abs(x)))/abs(d) + 1/24*sqrt(-d^2*x^2 
+ c^2)*(32*c^3 - (15*c^2*d - 2*(3*d^3*x - 4*c*d^2)*x)*x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.01 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{x (c+d x)} \, dx=-\frac {3 \mathit {asin} \left (\frac {d x}{c}\right ) c^{4}}{8}+\frac {4 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{3}}{3}-\frac {5 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{2} d x}{8}-\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, c \,d^{2} x^{2}}{3}+\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, d^{3} x^{3}}{4}+\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )\right ) c^{4}-\frac {4 c^{4}}{3} \] Input:

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

Output:

( - 9*asin((d*x)/c)*c**4 + 32*sqrt(c**2 - d**2*x**2)*c**3 - 15*sqrt(c**2 - 
 d**2*x**2)*c**2*d*x - 8*sqrt(c**2 - d**2*x**2)*c*d**2*x**2 + 6*sqrt(c**2 
- d**2*x**2)*d**3*x**3 + 24*log(tan(asin((d*x)/c)/2))*c**4 - 32*c**4)/24