\(\int \frac {\sqrt {c^2-d^2 x^2}}{x (c+d x)^4} \, dx\) [150]

Optimal result

Integrand size = 27, antiderivative size = 111 \[ \int \frac {\sqrt {c^2-d^2 x^2}}{x (c+d x)^4} \, dx=-\frac {4 d x}{5 c \left (c^2-d^2 x^2\right )^{3/2}}+\frac {8 c}{5 (c+d x) \left (c^2-d^2 x^2\right )^{3/2}}+\frac {5 c-8 d x}{5 c^3 \sqrt {c^2-d^2 x^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{c}\right )}{c^3} \] Output:

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

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {c^2-d^2 x^2}}{x (c+d x)^4} \, dx=\frac {\frac {\sqrt {c^2-d^2 x^2} \left (13 c^2+19 c d x+8 d^2 x^2\right )}{(c+d x)^3}+10 \text {arctanh}\left (\frac {\sqrt {-d^2} x-\sqrt {c^2-d^2 x^2}}{c}\right )}{5 c^3} \] Input:

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

Output:

((Sqrt[c^2 - d^2*x^2]*(13*c^2 + 19*c*d*x + 8*d^2*x^2))/(c + d*x)^3 + 10*Ar 
cTanh[(Sqrt[-d^2]*x - Sqrt[c^2 - d^2*x^2])/c])/(5*c^3)
 

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {570, 532, 25, 2336, 27, 532, 27, 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.

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

\(\Big \downarrow \) 570

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

\(\Big \downarrow \) 532

\(\displaystyle \frac {8 c (c-d x)}{5 \left (c^2-d^2 x^2\right )^{5/2}}-\frac {\int -\frac {5 c^4-12 d x c^3-5 d^2 x^2 c^2}{x \left (c^2-d^2 x^2\right )^{5/2}}dx}{5 c^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {5 c^4-12 d x c^3-5 d^2 x^2 c^2}{x \left (c^2-d^2 x^2\right )^{5/2}}dx}{5 c^2}+\frac {8 c (c-d x)}{5 \left (c^2-d^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 2336

\(\displaystyle \frac {-\frac {\int -\frac {3 c^3 (5 c-8 d x)}{x \left (c^2-d^2 x^2\right )^{3/2}}dx}{3 c^2}-\frac {4 c d x}{\left (c^2-d^2 x^2\right )^{3/2}}}{5 c^2}+\frac {8 c (c-d x)}{5 \left (c^2-d^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {c \int \frac {5 c-8 d x}{x \left (c^2-d^2 x^2\right )^{3/2}}dx-\frac {4 c d x}{\left (c^2-d^2 x^2\right )^{3/2}}}{5 c^2}+\frac {8 c (c-d x)}{5 \left (c^2-d^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 532

\(\displaystyle \frac {c \left (\frac {5 c-8 d x}{c^2 \sqrt {c^2-d^2 x^2}}-\frac {\int -\frac {5 c}{x \sqrt {c^2-d^2 x^2}}dx}{c^2}\right )-\frac {4 c d x}{\left (c^2-d^2 x^2\right )^{3/2}}}{5 c^2}+\frac {8 c (c-d x)}{5 \left (c^2-d^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {c \left (\frac {5 \int \frac {1}{x \sqrt {c^2-d^2 x^2}}dx}{c}+\frac {5 c-8 d x}{c^2 \sqrt {c^2-d^2 x^2}}\right )-\frac {4 c d x}{\left (c^2-d^2 x^2\right )^{3/2}}}{5 c^2}+\frac {8 c (c-d x)}{5 \left (c^2-d^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {c \left (\frac {5 \int \frac {1}{x^2 \sqrt {c^2-d^2 x^2}}dx^2}{2 c}+\frac {5 c-8 d x}{c^2 \sqrt {c^2-d^2 x^2}}\right )-\frac {4 c d x}{\left (c^2-d^2 x^2\right )^{3/2}}}{5 c^2}+\frac {8 c (c-d x)}{5 \left (c^2-d^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {c \left (\frac {5 c-8 d x}{c^2 \sqrt {c^2-d^2 x^2}}-\frac {5 \int \frac {1}{\frac {c^2}{d^2}-\frac {x^4}{d^2}}d\sqrt {c^2-d^2 x^2}}{c d^2}\right )-\frac {4 c d x}{\left (c^2-d^2 x^2\right )^{3/2}}}{5 c^2}+\frac {8 c (c-d x)}{5 \left (c^2-d^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {c \left (\frac {5 c-8 d x}{c^2 \sqrt {c^2-d^2 x^2}}-\frac {5 \text {arctanh}\left (\frac {\sqrt {c^2-d^2 x^2}}{c}\right )}{c^2}\right )-\frac {4 c d x}{\left (c^2-d^2 x^2\right )^{3/2}}}{5 c^2}+\frac {8 c (c-d x)}{5 \left (c^2-d^2 x^2\right )^{5/2}}\)

Input:

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

Output:

(8*c*(c - d*x))/(5*(c^2 - d^2*x^2)^(5/2)) + ((-4*c*d*x)/(c^2 - d^2*x^2)^(3 
/2) + c*((5*c - 8*d*x)/(c^2*Sqrt[c^2 - d^2*x^2]) - (5*ArcTanh[Sqrt[c^2 - d 
^2*x^2]/c])/c^2))/(5*c^2)
 

Defintions of rubi rules used

rule 25

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

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 532

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe 
ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol 
ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) 
*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[x^m 
*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*(Qx/x^m) + e*((2*p + 3)/x^m), 
x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && ILtQ[m, 0] && LtQ[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])
 

rule 2336

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{Q = PolynomialQuotient[(c*x)^m*Pq, a + b*x^2, x], f = Coeff[PolynomialRema 
inder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[(c*x) 
^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a* 
b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[(c*x)^m*(a + b*x^2)^(p + 1)*Ex 
pandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x], x]] /; F 
reeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(414\) vs. \(2(97)=194\).

Time = 0.24 (sec) , antiderivative size = 415, normalized size of antiderivative = 3.74

method	result	size
default	\(\frac {\sqrt {-d^{2} x^{2}+c^{2}}-\frac {c^{2} \ln \left (\frac {2 c^{2}+2 \sqrt {c^{2}}\, \sqrt {-d^{2} x^{2}+c^{2}}}{x}\right )}{\sqrt {c^{2}}}}{c^{4}}+\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{3 d^{3} c^{3} \left (x +\frac {c}{d}\right )^{3}}-\frac {-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{5 c d \left (x +\frac {c}{d}\right )^{4}}-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{15 c^{2} \left (x +\frac {c}{d}\right )^{3}}}{d^{3} c}-\frac {\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}+\frac {c d \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{\sqrt {d^{2}}}}{c^{4}}-\frac {-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{c d \left (x +\frac {c}{d}\right )^{2}}-\frac {d \left (\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}+\frac {c d \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{\sqrt {d^{2}}}\right )}{c}}{c^{3} d}\)	\(415\)

Input:

int((-d^2*x^2+c^2)^(1/2)/x/(d*x+c)^4,x,method=_RETURNVERBOSE)
 

Output:

1/c^4*((-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))+1/3/d^3/c^3/(x+c/d)^3*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^( 
3/2)-1/d^3/c*(-1/5/c/d/(x+c/d)^4*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(3/2)-1/15 
/c^2/(x+c/d)^3*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(3/2))-1/c^4*((-d^2*(x+c/d)^ 
2+2*c*d*(x+c/d))^(1/2)+c*d/(d^2)^(1/2)*arctan((d^2)^(1/2)*x/(-d^2*(x+c/d)^ 
2+2*c*d*(x+c/d))^(1/2)))-1/c^3/d*(-1/c/d/(x+c/d)^2*(-d^2*(x+c/d)^2+2*c*d*( 
x+c/d))^(3/2)-d/c*((-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(1/2)+c*d/(d^2)^(1/2)*ar 
ctan((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 = 153, normalized size of antiderivative = 1.38 \[ \int \frac {\sqrt {c^2-d^2 x^2}}{x (c+d x)^4} \, dx=\frac {13 \, d^{3} x^{3} + 39 \, c d^{2} x^{2} + 39 \, c^{2} d x + 13 \, c^{3} + 5 \, {\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}\right )} \log \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{x}\right ) + {\left (8 \, d^{2} x^{2} + 19 \, c d x + 13 \, c^{2}\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{5 \, {\left (c^{3} d^{3} x^{3} + 3 \, c^{4} d^{2} x^{2} + 3 \, c^{5} d x + c^{6}\right )}} \] Input:

integrate((-d^2*x^2+c^2)^(1/2)/x/(d*x+c)^4,x, algorithm="fricas")
 

Output:

1/5*(13*d^3*x^3 + 39*c*d^2*x^2 + 39*c^2*d*x + 13*c^3 + 5*(d^3*x^3 + 3*c*d^ 
2*x^2 + 3*c^2*d*x + c^3)*log(-(c - sqrt(-d^2*x^2 + c^2))/x) + (8*d^2*x^2 + 
 19*c*d*x + 13*c^2)*sqrt(-d^2*x^2 + c^2))/(c^3*d^3*x^3 + 3*c^4*d^2*x^2 + 3 
*c^5*d*x + c^6)
 

Sympy [F]

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

integrate((-d**2*x**2+c**2)**(1/2)/x/(d*x+c)**4,x)
 

Output:

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

Maxima [F]

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

integrate((-d^2*x^2+c^2)^(1/2)/x/(d*x+c)^4,x, algorithm="maxima")
 

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 212 vs. \(2 (97) = 194\).

Time = 0.14 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.91 \[ \int \frac {\sqrt {c^2-d^2 x^2}}{x (c+d x)^4} \, dx=-\frac {d \log \left (\frac {{\left | -2 \, c d - 2 \, \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |} \right |}}{2 \, d^{2} {\left | x \right |}}\right )}{c^{3} {\left | d \right |}} - \frac {2 \, {\left (13 \, d + \frac {45 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}}{d x} + \frac {75 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2}}{d^{3} x^{2}} + \frac {55 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{3}}{d^{5} x^{3}} + \frac {20 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{4}}{d^{7} x^{4}}\right )}}{5 \, c^{3} {\left (\frac {c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}}{d^{2} x} + 1\right )}^{5} {\left | d \right |}} \] Input:

integrate((-d^2*x^2+c^2)^(1/2)/x/(d*x+c)^4,x, algorithm="giac")
 

Output:

-d*log(1/2*abs(-2*c*d - 2*sqrt(-d^2*x^2 + c^2)*abs(d))/(d^2*abs(x)))/(c^3* 
abs(d)) - 2/5*(13*d + 45*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))/(d*x) + 75*(c 
*d + sqrt(-d^2*x^2 + c^2)*abs(d))^2/(d^3*x^2) + 55*(c*d + sqrt(-d^2*x^2 + 
c^2)*abs(d))^3/(d^5*x^3) + 20*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^4/(d^7*x 
^4))/(c^3*((c*d + sqrt(-d^2*x^2 + c^2)*abs(d))/(d^2*x) + 1)^5*abs(d))
 

Mupad [F(-1)]

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

int((c^2 - d^2*x^2)^(1/2)/(x*(c + d*x)^4),x)
 

Output:

int((c^2 - d^2*x^2)^(1/2)/(x*(c + d*x)^4), x)
 

Reduce [F]

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

int((-d^2*x^2+c^2)^(1/2)/x/(d*x+c)^4,x)
 

Output:

int((-d^2*x^2+c^2)^(1/2)/x/(d*x+c)^4,x)