\(\int \frac {(c^2-d^2 x^2)^{3/2}}{x^2 (c+d x)^3} \, dx\) [239]

Optimal result

Integrand size = 27, antiderivative size = 81 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{x^2 (c+d x)^3} \, dx=-\frac {\sqrt {c^2-d^2 x^2}}{c x}-\frac {4 d \sqrt {c^2-d^2 x^2}}{c (c+d x)}+\frac {3 d \text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{c}\right )}{c} \] Output:

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

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.09 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{x^2 (c+d x)^3} \, dx=\frac {(-c-5 d x) \sqrt {c^2-d^2 x^2}}{c x (c+d x)}+\frac {3 d \log (x)}{\sqrt {c^2}}-\frac {3 d \log \left (\sqrt {c^2}-\sqrt {c^2-d^2 x^2}\right )}{\sqrt {c^2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {564, 25, 534, 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.

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

\(\Big \downarrow \) 564

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 534

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 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 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 534

Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[(-c)*x^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[d   Int[ 
x^(m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[m, 
0] && GtQ[p, -1] && EqQ[m + 2*p + 3, 0]
 

rule 564

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol 
] :> Simp[(-(-c)^(m - n - 2))*d^(2*n - m + 3)*(Sqrt[a + b*x^2]/(2^(n + 1)*b 
^(n + 2)*(c + d*x))), x] - Simp[d^(2*n + 2)/b^(n + 1)   Int[(x^m/Sqrt[a + b 
*x^2])*ExpandToSum[((2^(-n - 1)*(-c)^(m - n - 1))/(d^m*x^m) - (-c + d*x)^(- 
n - 1))/(c + d*x), x], x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^ 
2, 0] && ILtQ[m, 0] && ILtQ[n, 0] && EqQ[n + p, -3/2]
 

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.31

method	result	size
risch	\(-\frac {\sqrt {-d^{2} x^{2}+c^{2}}}{c x}+\frac {3 d \ln \left (\frac {2 c^{2}+2 \sqrt {c^{2}}\, \sqrt {-d^{2} x^{2}+c^{2}}}{x}\right )}{\sqrt {c^{2}}}-\frac {4 \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{c \left (x +\frac {c}{d}\right )}\)	\(106\)
default	\(\frac {-\frac {\left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}{c^{2} x}-\frac {4 d^{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 )}{c^{2}}}{c^{3}}+\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}}}{c d \left (x +\frac {c}{d}\right )^{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}}}{c d \left (x +\frac {c}{d}\right )^{2}}+\frac {3 d \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{3}+c d \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 )\right )}{c}\right )}{c}}{c^{2} d}-\frac {3 d \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^{4}}+\frac {3 d \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{3}+c d \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 )\right )}{c^{4}}+\frac {\frac {2 \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {5}{2}}}{c d \left (x +\frac {c}{d}\right )^{2}}+\frac {6 d \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{3}+c d \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 )\right )}{c}}{c^{3}}\)	\(759\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.10 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{x^2 (c+d x)^3} \, dx=-\frac {4 \, d^{2} x^{2} + 4 \, c d x + 3 \, {\left (d^{2} x^{2} + c d x\right )} \log \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{x}\right ) + \sqrt {-d^{2} x^{2} + c^{2}} {\left (5 \, d x + c\right )}}{c d x^{2} + c^{2} x} \] Input:

integrate((-d^2*x^2+c^2)^(3/2)/x^2/(d*x+c)^3,x, algorithm="fricas")
 

Output:

-(4*d^2*x^2 + 4*c*d*x + 3*(d^2*x^2 + c*d*x)*log(-(c - sqrt(-d^2*x^2 + c^2) 
)/x) + sqrt(-d^2*x^2 + c^2)*(5*d*x + c))/(c*d*x^2 + c^2*x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate((-d^2*x^2+c^2)^(3/2)/x^2/(d*x+c)^3,x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 178 vs. \(2 (75) = 150\).

Time = 0.14 (sec) , antiderivative size = 178, normalized size of antiderivative = 2.20 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{x^2 (c+d x)^3} \, dx=\frac {3 \, d^{2} \log \left (\frac {{\left | -2 \, c d - 2 \, \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |} \right |}}{2 \, d^{2} {\left | x \right |}}\right )}{c {\left | d \right |}} + \frac {{\left (d^{2} + \frac {17 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}}{x}\right )} d^{2} x}{2 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} c {\left (\frac {c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}}{d^{2} x} + 1\right )} {\left | d \right |}} - \frac {c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}}{2 \, c x {\left | d \right |}} \] Input:

integrate((-d^2*x^2+c^2)^(3/2)/x^2/(d*x+c)^3,x, algorithm="giac")
 

Output:

3*d^2*log(1/2*abs(-2*c*d - 2*sqrt(-d^2*x^2 + c^2)*abs(d))/(d^2*abs(x)))/(c 
*abs(d)) + 1/2*(d^2 + 17*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))/x)*d^2*x/((c* 
d + sqrt(-d^2*x^2 + c^2)*abs(d))*c*((c*d + sqrt(-d^2*x^2 + c^2)*abs(d))/(d 
^2*x) + 1)*abs(d)) - 1/2*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))/(c*x*abs(d))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.33 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{x^2 (c+d x)^3} \, dx=\frac {d \left (-6 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )\right ) \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{2}-6 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )\right ) \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )+\tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{3}+18 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{2}-1\right )}{2 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right ) c \left (\tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )+1\right )} \] Input:

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

Output:

(d*( - 6*log(tan(asin((d*x)/c)/2))*tan(asin((d*x)/c)/2)**2 - 6*log(tan(asi 
n((d*x)/c)/2))*tan(asin((d*x)/c)/2) + tan(asin((d*x)/c)/2)**3 + 18*tan(asi 
n((d*x)/c)/2)**2 - 1))/(2*tan(asin((d*x)/c)/2)*c*(tan(asin((d*x)/c)/2) + 1 
))