\(\int \frac {(c^2-d^2 x^2)^{3/2}}{x^4 (c+d x)^2} \, dx\) [231]

Optimal result

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

-1/3*(-d^2*x^2+c^2)^(1/2)/x^3+d*(-d^2*x^2+c^2)^(1/2)/c/x^2-5/3*d^2*(-d^2*x 
^2+c^2)^(1/2)/c^2/x+d^3*arctanh((-d^2*x^2+c^2)^(1/2)/c)/c^2
 

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.94 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{x^4 (c+d x)^2} \, dx=-\frac {\frac {c \sqrt {c^2-d^2 x^2} \left (c^2-3 c d x+5 d^2 x^2\right )}{x^3}-3 \sqrt {c^2} d^3 \log (x)+3 \sqrt {c^2} d^3 \log \left (\sqrt {c^2}-\sqrt {c^2-d^2 x^2}\right )}{3 c^3} \] Input:

Integrate[(c^2 - d^2*x^2)^(3/2)/(x^4*(c + d*x)^2),x]
 

Output:

-1/3*((c*Sqrt[c^2 - d^2*x^2]*(c^2 - 3*c*d*x + 5*d^2*x^2))/x^3 - 3*Sqrt[c^2 
]*d^3*Log[x] + 3*Sqrt[c^2]*d^3*Log[Sqrt[c^2] - Sqrt[c^2 - d^2*x^2]])/c^3
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.07, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {570, 540, 27, 539, 27, 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.

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

\(\Big \downarrow \) 570

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

\(\Big \downarrow \) 540

\(\displaystyle -\frac {\int \frac {c^2 d (6 c-5 d x)}{x^3 \sqrt {c^2-d^2 x^2}}dx}{3 c^2}-\frac {\sqrt {c^2-d^2 x^2}}{3 x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {1}{3} d \int \frac {6 c-5 d x}{x^3 \sqrt {c^2-d^2 x^2}}dx-\frac {\sqrt {c^2-d^2 x^2}}{3 x^3}\)

\(\Big \downarrow \) 539

\(\displaystyle -\frac {1}{3} d \left (-\frac {\int \frac {2 c d (5 c-3 d x)}{x^2 \sqrt {c^2-d^2 x^2}}dx}{2 c^2}-\frac {3 \sqrt {c^2-d^2 x^2}}{c x^2}\right )-\frac {\sqrt {c^2-d^2 x^2}}{3 x^3}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {1}{3} d \left (-\frac {d \int \frac {5 c-3 d x}{x^2 \sqrt {c^2-d^2 x^2}}dx}{c}-\frac {3 \sqrt {c^2-d^2 x^2}}{c x^2}\right )-\frac {\sqrt {c^2-d^2 x^2}}{3 x^3}\)

\(\Big \downarrow \) 534

\(\displaystyle -\frac {1}{3} d \left (-\frac {d \left (-3 d \int \frac {1}{x \sqrt {c^2-d^2 x^2}}dx-\frac {5 \sqrt {c^2-d^2 x^2}}{c x}\right )}{c}-\frac {3 \sqrt {c^2-d^2 x^2}}{c x^2}\right )-\frac {\sqrt {c^2-d^2 x^2}}{3 x^3}\)

\(\Big \downarrow \) 243

\(\displaystyle -\frac {1}{3} d \left (-\frac {d \left (-\frac {3}{2} d \int \frac {1}{x^2 \sqrt {c^2-d^2 x^2}}dx^2-\frac {5 \sqrt {c^2-d^2 x^2}}{c x}\right )}{c}-\frac {3 \sqrt {c^2-d^2 x^2}}{c x^2}\right )-\frac {\sqrt {c^2-d^2 x^2}}{3 x^3}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {1}{3} d \left (-\frac {d \left (\frac {3 \int \frac {1}{\frac {c^2}{d^2}-\frac {x^4}{d^2}}d\sqrt {c^2-d^2 x^2}}{d}-\frac {5 \sqrt {c^2-d^2 x^2}}{c x}\right )}{c}-\frac {3 \sqrt {c^2-d^2 x^2}}{c x^2}\right )-\frac {\sqrt {c^2-d^2 x^2}}{3 x^3}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {1}{3} d \left (-\frac {d \left (\frac {3 d \text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{c}\right )}{c}-\frac {5 \sqrt {c^2-d^2 x^2}}{c x}\right )}{c}-\frac {3 \sqrt {c^2-d^2 x^2}}{c x^2}\right )-\frac {\sqrt {c^2-d^2 x^2}}{3 x^3}\)

Input:

Int[(c^2 - d^2*x^2)^(3/2)/(x^4*(c + d*x)^2),x]
 

Output:

-1/3*Sqrt[c^2 - d^2*x^2]/x^3 - (d*((-3*Sqrt[c^2 - d^2*x^2])/(c*x^2) - (d*( 
(-5*Sqrt[c^2 - d^2*x^2])/(c*x) + (3*d*ArcTanh[Sqrt[c^2 - d^2*x^2]/c])/c))/ 
c))/3
 

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 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 539

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

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 = 0.30 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.81

method	result	size
risch	\(-\frac {\sqrt {-d^{2} x^{2}+c^{2}}\, \left (5 d^{2} x^{2}-3 c d x +c^{2}\right )}{3 x^{3} c^{2}}+\frac {d^{3} \ln \left (\frac {2 c^{2}+2 \sqrt {c^{2}}\, \sqrt {-d^{2} x^{2}+c^{2}}}{x}\right )}{c \sqrt {c^{2}}}\)	\(85\)
default	\(\frac {-\frac {\left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}{3 c^{2} x^{3}}-\frac {2 d^{2} \left (-\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}}\right )}{3 c^{2}}}{c^{2}}+\frac {d^{2} \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^{4}}-\frac {2 d \left (-\frac {\left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}{2 c^{2} x^{2}}-\frac {3 d^{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 )}{2 c^{2}}\right )}{c^{3}}+\frac {3 d^{2} \left (-\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}}\right )}{c^{4}}-\frac {4 d^{3} \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^{5}}+\frac {4 d^{3} \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^{5}}\)	\(782\)

Input:

int((-d^2*x^2+c^2)^(3/2)/x^4/(d*x+c)^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.70 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{x^4 (c+d x)^2} \, dx=-\frac {3 \, d^{3} x^{3} \log \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{x}\right ) + {\left (5 \, d^{2} x^{2} - 3 \, c d x + c^{2}\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{3 \, c^{2} x^{3}} \] Input:

integrate((-d^2*x^2+c^2)^(3/2)/x^4/(d*x+c)^2,x, algorithm="fricas")
 

Output:

-1/3*(3*d^3*x^3*log(-(c - sqrt(-d^2*x^2 + c^2))/x) + (5*d^2*x^2 - 3*c*d*x 
+ c^2)*sqrt(-d^2*x^2 + c^2))/(c^2*x^3)
 

Sympy [F]

\[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{x^4 (c+d x)^2} \, dx=\int \frac {\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {3}{2}}}{x^{4} \left (c + d x\right )^{2}}\, dx \] Input:

integrate((-d**2*x**2+c**2)**(3/2)/x**4/(d*x+c)**2,x)
 

Output:

Integral((-(-c + d*x)*(c + d*x))**(3/2)/(x**4*(c + d*x)**2), x)
 

Maxima [F]

\[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{x^4 (c+d x)^2} \, dx=\int { \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}}}{{\left (d x + c\right )}^{2} x^{4}} \,d x } \] Input:

integrate((-d^2*x^2+c^2)^(3/2)/x^4/(d*x+c)^2,x, algorithm="maxima")
 

Output:

integrate((-d^2*x^2 + c^2)^(3/2)/((d*x + c)^2*x^4), x)
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.16 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.06 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{x^4 (c+d x)^2} \, dx=\frac {1}{12} \, {\left (\frac {12 \, d^{2} \log \left (\sqrt {\frac {2 \, c}{d x + c} - 1} + 1\right ) \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )}{c^{2}} - \frac {12 \, d^{2} \log \left ({\left | \sqrt {\frac {2 \, c}{d x + c} - 1} - 1 \right |}\right ) \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )}{c^{2}} + \frac {2 \, {\left (3 \, d^{2} \log \left (2\right ) - 6 \, d^{2} \log \left (i + 1\right ) + 10 i \, d^{2}\right )} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )}{c^{2}} + \frac {9 \, d^{2} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {5}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) - 8 \, d^{2} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) + 3 \, d^{2} \sqrt {\frac {2 \, c}{d x + c} - 1} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )}{c^{2} {\left (\frac {c}{d x + c} - 1\right )}^{3}}\right )} {\left | d \right |} \] Input:

integrate((-d^2*x^2+c^2)^(3/2)/x^4/(d*x+c)^2,x, algorithm="giac")
 

Output:

1/12*(12*d^2*log(sqrt(2*c/(d*x + c) - 1) + 1)*sgn(1/(d*x + c))*sgn(d)/c^2 
- 12*d^2*log(abs(sqrt(2*c/(d*x + c) - 1) - 1))*sgn(1/(d*x + c))*sgn(d)/c^2 
 + 2*(3*d^2*log(2) - 6*d^2*log(I + 1) + 10*I*d^2)*sgn(1/(d*x + c))*sgn(d)/ 
c^2 + (9*d^2*(2*c/(d*x + c) - 1)^(5/2)*sgn(1/(d*x + c))*sgn(d) - 8*d^2*(2* 
c/(d*x + c) - 1)^(3/2)*sgn(1/(d*x + c))*sgn(d) + 3*d^2*sqrt(2*c/(d*x + c) 
- 1)*sgn(1/(d*x + c))*sgn(d))/(c^2*(c/(d*x + c) - 1)^3))*abs(d)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{x^4 (c+d x)^2} \, dx=\int \frac {{\left (c^2-d^2\,x^2\right )}^{3/2}}{x^4\,{\left (c+d\,x\right )}^2} \,d x \] Input:

int((c^2 - d^2*x^2)^(3/2)/(x^4*(c + d*x)^2),x)
 

Output:

int((c^2 - d^2*x^2)^(3/2)/(x^4*(c + d*x)^2), x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.81 \[ \int \frac {\left (c^2-d^2 x^2\right )^{3/2}}{x^4 (c+d x)^2} \, dx=\frac {-\sqrt {-d^{2} x^{2}+c^{2}}\, c^{2}+3 \sqrt {-d^{2} x^{2}+c^{2}}\, c d x -5 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{2} x^{2}-3 \,\mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )\right ) d^{3} x^{3}}{3 c^{2} x^{3}} \] Input:

int((-d^2*x^2+c^2)^(3/2)/x^4/(d*x+c)^2,x)
 

Output:

( - sqrt(c**2 - d**2*x**2)*c**2 + 3*sqrt(c**2 - d**2*x**2)*c*d*x - 5*sqrt( 
c**2 - d**2*x**2)*d**2*x**2 - 3*log(tan(asin((d*x)/c)/2))*d**3*x**3)/(3*c* 
*2*x**3)