\(\int \frac {(c^2-d^2 x^2)^{5/2}}{x^2 (c+d x)^2} \, dx\) [300]

Optimal result

Integrand size = 27, antiderivative size = 105 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{x^2 (c+d x)^2} \, dx=-\frac {1}{2} d (4 c+d x) \sqrt {c^2-d^2 x^2}-\frac {\left (c^2-d^2 x^2\right )^{3/2}}{x}-\frac {1}{2} c^2 d \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )+2 c^2 d \text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{c}\right ) \] Output:

-1/2*d*(d*x+4*c)*(-d^2*x^2+c^2)^(1/2)-(-d^2*x^2+c^2)^(3/2)/x-1/2*c^2*d*arc 
tan(d*x/(-d^2*x^2+c^2)^(1/2))+2*c^2*d*arctanh((-d^2*x^2+c^2)^(1/2)/c)
 

Mathematica [A] (verified)

Time = 0.33 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.24 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{x^2 (c+d x)^2} \, dx=\frac {\sqrt {c^2-d^2 x^2} \left (-2 c^2-4 c d x+d^2 x^2\right )}{2 x}+c^2 d \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )+2 c \sqrt {c^2} d \log (x)-2 c \sqrt {c^2} d \log \left (\sqrt {c^2}-\sqrt {c^2-d^2 x^2}\right ) \] Input:

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

Output:

(Sqrt[c^2 - d^2*x^2]*(-2*c^2 - 4*c*d*x + d^2*x^2))/(2*x) + c^2*d*ArcTan[(d 
*x)/(Sqrt[c^2] - Sqrt[c^2 - d^2*x^2])] + 2*c*Sqrt[c^2]*d*Log[x] - 2*c*Sqrt 
[c^2]*d*Log[Sqrt[c^2] - Sqrt[c^2 - d^2*x^2]]
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.99, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {570, 540, 27, 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.

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

\(\Big \downarrow \) 570

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

\(\Big \downarrow \) 540

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 535

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

\(\Big \downarrow \) 538

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

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 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 540

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol 
] :> With[{Qx = PolynomialQuotient[(c + d*x)^n, x, x], R = PolynomialRemain 
der[(c + d*x)^n, x, x]}, Simp[R*x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))) 
, x] + Simp[1/(a*(m + 1))   Int[x^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*(m + 
1)*Qx - b*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d, p}, x] && IG 
tQ[n, 1] && ILtQ[m, -1] && GtQ[p, -1] && IntegerQ[2*p]
 

rule 570

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

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.33

method	result	size
risch	\(-\frac {c^{2} \sqrt {-d^{2} x^{2}+c^{2}}}{x}+\frac {d^{2} x \sqrt {-d^{2} x^{2}+c^{2}}}{2}-\frac {d^{2} c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}+\frac {2 d \,c^{3} \ln \left (\frac {2 c^{2}+2 \sqrt {c^{2}}\, \sqrt {-d^{2} x^{2}+c^{2}}}{x}\right )}{\sqrt {c^{2}}}-2 c d \sqrt {-d^{2} x^{2}+c^{2}}\)	\(140\)
default	\(\frac {-\frac {\left (-d^{2} x^{2}+c^{2}\right )^{\frac {7}{2}}}{c^{2} x}-\frac {6 d^{2} \left (\frac {x \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}{6}+\frac {5 c^{2} \left (\frac {x \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{4}+\frac {3 c^{2} \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )}{4}\right )}{6}\right )}{c^{2}}}{c^{2}}+\frac {\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {7}{2}}}{3 c d \left (x +\frac {c}{d}\right )^{2}}+\frac {5 d \left (\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 )\right )}{3 c}}{c^{2}}-\frac {2 d \left (\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 )\right )}{c^{3}}+\frac {2 d \left (\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 )\right )}{c^{3}}\)	\(674\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.06 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{x^2 (c+d x)^2} \, dx=\frac {2 \, c^{2} d x \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) - 4 \, c^{2} d x \log \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{x}\right ) - 4 \, c^{2} d x + {\left (d^{2} x^{2} - 4 \, c d x - 2 \, c^{2}\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{2 \, x} \] Input:

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

Output:

1/2*(2*c^2*d*x*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x)) - 4*c^2*d*x*log(- 
(c - sqrt(-d^2*x^2 + c^2))/x) - 4*c^2*d*x + (d^2*x^2 - 4*c*d*x - 2*c^2)*sq 
rt(-d^2*x^2 + c^2))/x
 

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.52 (sec) , antiderivative size = 330, normalized size of antiderivative = 3.14 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{x^2 (c+d x)^2} \, dx=c^{2} \left (\begin {cases} \frac {i c}{x \sqrt {-1 + \frac {d^{2} x^{2}}{c^{2}}}} + i d \operatorname {acosh}{\left (\frac {d x}{c} \right )} - \frac {i d^{2} x}{c \sqrt {-1 + \frac {d^{2} x^{2}}{c^{2}}}} & \text {for}\: \left |{\frac {d^{2} x^{2}}{c^{2}}}\right | > 1 \\- \frac {c}{x \sqrt {1 - \frac {d^{2} x^{2}}{c^{2}}}} - d \operatorname {asin}{\left (\frac {d x}{c} \right )} + \frac {d^{2} x}{c \sqrt {1 - \frac {d^{2} x^{2}}{c^{2}}}} & \text {otherwise} \end {cases}\right ) - 2 c d \left (\begin {cases} \frac {c^{2}}{d x \sqrt {\frac {c^{2}}{d^{2} x^{2}} - 1}} - c \operatorname {acosh}{\left (\frac {c}{d x} \right )} - \frac {d x}{\sqrt {\frac {c^{2}}{d^{2} x^{2}} - 1}} & \text {for}\: \left |{\frac {c^{2}}{d^{2} x^{2}}}\right | > 1 \\- \frac {i c^{2}}{d x \sqrt {- \frac {c^{2}}{d^{2} x^{2}} + 1}} + i c \operatorname {asin}{\left (\frac {c}{d x} \right )} + \frac {i d x}{\sqrt {- \frac {c^{2}}{d^{2} x^{2}} + 1}} & \text {otherwise} \end {cases}\right ) + d^{2} \left (\begin {cases} \frac {c^{2} \left (\begin {cases} \frac {\log {\left (- 2 d^{2} x + 2 \sqrt {- d^{2}} \sqrt {c^{2} - d^{2} x^{2}} \right )}}{\sqrt {- d^{2}}} & \text {for}\: c^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- d^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {c^{2} - d^{2} x^{2}}}{2} & \text {for}\: d^{2} \neq 0 \\x \sqrt {c^{2}} & \text {otherwise} \end {cases}\right ) \] Input:

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

Output:

c**2*Piecewise((I*c/(x*sqrt(-1 + d**2*x**2/c**2)) + I*d*acosh(d*x/c) - I*d 
**2*x/(c*sqrt(-1 + d**2*x**2/c**2)), Abs(d**2*x**2/c**2) > 1), (-c/(x*sqrt 
(1 - d**2*x**2/c**2)) - d*asin(d*x/c) + d**2*x/(c*sqrt(1 - d**2*x**2/c**2) 
), True)) - 2*c*d*Piecewise((c**2/(d*x*sqrt(c**2/(d**2*x**2) - 1)) - c*aco 
sh(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/sqr 
t(-c**2/(d**2*x**2) + 1), True)) + d**2*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)/sqrt(-d**2*x**2), True))/2 + x*sqrt(c**2 - d**2*x**2)/2, Ne(d**2 
, 0)), (x*sqrt(c**2), True))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.07 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{x^2 (c+d x)^2} \, dx=-\frac {1}{2} \, c^{2} d \arcsin \left (\frac {d x}{c}\right ) + 2 \, c^{2} d \log \left (\frac {2 \, c^{2}}{{\left | x \right |}} + \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} c}{{\left | x \right |}}\right ) + \frac {1}{2} \, \sqrt {-d^{2} x^{2} + c^{2}} d^{2} x - 2 \, \sqrt {-d^{2} x^{2} + c^{2}} c d - \frac {\sqrt {-d^{2} x^{2} + c^{2}} c^{2}}{x} \] Input:

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

Output:

-1/2*c^2*d*arcsin(d*x/c) + 2*c^2*d*log(2*c^2/abs(x) + 2*sqrt(-d^2*x^2 + c^ 
2)*c/abs(x)) + 1/2*sqrt(-d^2*x^2 + c^2)*d^2*x - 2*sqrt(-d^2*x^2 + c^2)*c*d 
 - sqrt(-d^2*x^2 + c^2)*c^2/x
 

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Limit: Max order reached or unable 
to make series expansion Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.96 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{x^2 (c+d x)^2} \, dx=\frac {-\mathit {asin} \left (\frac {d x}{c}\right ) c^{2} d x -2 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{2}-4 \sqrt {-d^{2} x^{2}+c^{2}}\, c d x +\sqrt {-d^{2} x^{2}+c^{2}}\, d^{2} x^{2}-4 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )\right ) c^{2} d x +4 c^{2} d x}{2 x} \] Input:

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

Output:

( - asin((d*x)/c)*c**2*d*x - 2*sqrt(c**2 - d**2*x**2)*c**2 - 4*sqrt(c**2 - 
 d**2*x**2)*c*d*x + sqrt(c**2 - d**2*x**2)*d**2*x**2 - 4*log(tan(asin((d*x 
)/c)/2))*c**2*d*x + 4*c**2*d*x)/(2*x)