\(\int \frac {\sqrt {b c^2-b d^2 x^2}}{x^3 (c+d x)^{5/2}} \, dx\) [662]

Optimal result

Integrand size = 32, antiderivative size = 238 \[ \int \frac {\sqrt {b c^2-b d^2 x^2}}{x^3 (c+d x)^{5/2}} \, dx=\frac {13 d^2 \sqrt {b c^2-b d^2 x^2}}{4 c^3 (c+d x)^{3/2}}-\frac {\sqrt {b c^2-b d^2 x^2}}{2 c x^2 (c+d x)^{3/2}}+\frac {7 d \sqrt {b c^2-b d^2 x^2}}{4 c^2 x (c+d x)^{3/2}}-\frac {31 \sqrt {b} d^2 \text {arctanh}\left (\frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {b} \sqrt {c} \sqrt {c+d x}}\right )}{4 c^{7/2}}+\frac {11 \sqrt {b} d^2 \text {arctanh}\left (\frac {\sqrt {b c^2-b d^2 x^2}}{\sqrt {2} \sqrt {b} \sqrt {c} \sqrt {c+d x}}\right )}{\sqrt {2} c^{7/2}} \] Output:

13/4*d^2*(-b*d^2*x^2+b*c^2)^(1/2)/c^3/(d*x+c)^(3/2)-1/2*(-b*d^2*x^2+b*c^2) 
^(1/2)/c/x^2/(d*x+c)^(3/2)+7/4*d*(-b*d^2*x^2+b*c^2)^(1/2)/c^2/x/(d*x+c)^(3 
/2)-31/4*b^(1/2)*d^2*arctanh((-b*d^2*x^2+b*c^2)^(1/2)/b^(1/2)/c^(1/2)/(d*x 
+c)^(1/2))/c^(7/2)+11/2*b^(1/2)*d^2*arctanh(1/2*(-b*d^2*x^2+b*c^2)^(1/2)*2 
^(1/2)/b^(1/2)/c^(1/2)/(d*x+c)^(1/2))*2^(1/2)/c^(7/2)
 

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.86 \[ \int \frac {\sqrt {b c^2-b d^2 x^2}}{x^3 (c+d x)^{5/2}} \, dx=-\frac {\sqrt {b \left (c^2-d^2 x^2\right )} \left (\sqrt {c} \left (2 c^2-7 c d x-13 d^2 x^2\right ) \sqrt {c^2-d^2 x^2}+31 d^2 x^2 (c+d x)^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )-22 \sqrt {2} d^2 x^2 (c+d x)^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {c+d x}}{\sqrt {c^2-d^2 x^2}}\right )\right )}{4 c^{7/2} x^2 (c+d x)^{3/2} \sqrt {c^2-d^2 x^2}} \] Input:

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

Output:

-1/4*(Sqrt[b*(c^2 - d^2*x^2)]*(Sqrt[c]*(2*c^2 - 7*c*d*x - 13*d^2*x^2)*Sqrt 
[c^2 - d^2*x^2] + 31*d^2*x^2*(c + d*x)^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[c + d*x 
])/Sqrt[c^2 - d^2*x^2]] - 22*Sqrt[2]*d^2*x^2*(c + d*x)^(3/2)*ArcTanh[(Sqrt 
[2]*Sqrt[c]*Sqrt[c + d*x])/Sqrt[c^2 - d^2*x^2]]))/(c^(7/2)*x^2*(c + d*x)^( 
3/2)*Sqrt[c^2 - d^2*x^2])
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.01, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {586, 110, 27, 168, 27, 168, 27, 174, 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.

\(\displaystyle \int \frac {\sqrt {b c^2-b d^2 x^2}}{x^3 (c+d x)^{5/2}} \, dx\)

\(\Big \downarrow \) 586

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

\(\Big \downarrow \) 110

\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (\frac {\int -\frac {b d (7 c-5 d x)}{2 x^2 (c+d x)^2 \sqrt {b c-b d x}}dx}{2 c}-\frac {\sqrt {b c-b d x}}{2 c x^2 (c+d x)}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (-\frac {b d \int \frac {7 c-5 d x}{x^2 (c+d x)^2 \sqrt {b c-b d x}}dx}{4 c}-\frac {\sqrt {b c-b d x}}{2 c x^2 (c+d x)}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (-\frac {b d \left (-\frac {\int \frac {b c d (31 c-21 d x)}{2 x (c+d x)^2 \sqrt {b c-b d x}}dx}{b c^2}-\frac {7 \sqrt {b c-b d x}}{b c x (c+d x)}\right )}{4 c}-\frac {\sqrt {b c-b d x}}{2 c x^2 (c+d x)}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (-\frac {b d \left (-\frac {d \int \frac {31 c-21 d x}{x (c+d x)^2 \sqrt {b c-b d x}}dx}{2 c}-\frac {7 \sqrt {b c-b d x}}{b c x (c+d x)}\right )}{4 c}-\frac {\sqrt {b c-b d x}}{2 c x^2 (c+d x)}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\)

\(\Big \downarrow \) 168

\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (-\frac {b d \left (-\frac {d \left (\frac {\int \frac {2 b c d (31 c-13 d x)}{x (c+d x) \sqrt {b c-b d x}}dx}{2 b c^2 d}+\frac {26 \sqrt {b c-b d x}}{b c (c+d x)}\right )}{2 c}-\frac {7 \sqrt {b c-b d x}}{b c x (c+d x)}\right )}{4 c}-\frac {\sqrt {b c-b d x}}{2 c x^2 (c+d x)}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (-\frac {b d \left (-\frac {d \left (\frac {\int \frac {31 c-13 d x}{x (c+d x) \sqrt {b c-b d x}}dx}{c}+\frac {26 \sqrt {b c-b d x}}{b c (c+d x)}\right )}{2 c}-\frac {7 \sqrt {b c-b d x}}{b c x (c+d x)}\right )}{4 c}-\frac {\sqrt {b c-b d x}}{2 c x^2 (c+d x)}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\)

\(\Big \downarrow \) 174

\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (-\frac {b d \left (-\frac {d \left (\frac {31 \int \frac {1}{x \sqrt {b c-b d x}}dx-44 d \int \frac {1}{(c+d x) \sqrt {b c-b d x}}dx}{c}+\frac {26 \sqrt {b c-b d x}}{b c (c+d x)}\right )}{2 c}-\frac {7 \sqrt {b c-b d x}}{b c x (c+d x)}\right )}{4 c}-\frac {\sqrt {b c-b d x}}{2 c x^2 (c+d x)}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (-\frac {b d \left (-\frac {d \left (\frac {\frac {88 \int \frac {1}{2 c-\frac {b c-b d x}{b}}d\sqrt {b c-b d x}}{b}-\frac {62 \int \frac {1}{\frac {c}{d}-\frac {b c-b d x}{b d}}d\sqrt {b c-b d x}}{b d}}{c}+\frac {26 \sqrt {b c-b d x}}{b c (c+d x)}\right )}{2 c}-\frac {7 \sqrt {b c-b d x}}{b c x (c+d x)}\right )}{4 c}-\frac {\sqrt {b c-b d x}}{2 c x^2 (c+d x)}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\sqrt {b c^2-b d^2 x^2} \left (-\frac {b d \left (-\frac {d \left (\frac {\frac {44 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {b c-b d x}}{\sqrt {2} \sqrt {b} \sqrt {c}}\right )}{\sqrt {b} \sqrt {c}}-\frac {62 \text {arctanh}\left (\frac {\sqrt {b c-b d x}}{\sqrt {b} \sqrt {c}}\right )}{\sqrt {b} \sqrt {c}}}{c}+\frac {26 \sqrt {b c-b d x}}{b c (c+d x)}\right )}{2 c}-\frac {7 \sqrt {b c-b d x}}{b c x (c+d x)}\right )}{4 c}-\frac {\sqrt {b c-b d x}}{2 c x^2 (c+d x)}\right )}{\sqrt {c+d x} \sqrt {b c-b d x}}\)

Input:

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

Output:

(Sqrt[b*c^2 - b*d^2*x^2]*(-1/2*Sqrt[b*c - b*d*x]/(c*x^2*(c + d*x)) - (b*d* 
((-7*Sqrt[b*c - b*d*x])/(b*c*x*(c + d*x)) - (d*((26*Sqrt[b*c - b*d*x])/(b* 
c*(c + d*x)) + ((-62*ArcTanh[Sqrt[b*c - b*d*x]/(Sqrt[b]*Sqrt[c])])/(Sqrt[b 
]*Sqrt[c]) + (44*Sqrt[2]*ArcTanh[Sqrt[b*c - b*d*x]/(Sqrt[2]*Sqrt[b]*Sqrt[c 
])])/(Sqrt[b]*Sqrt[c]))/c))/(2*c)))/(4*c)))/(Sqrt[c + d*x]*Sqrt[b*c - b*d* 
x])
 

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 110

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/((m + 
1)*(b*e - a*f))), x] - Simp[1/((m + 1)*(b*e - a*f))   Int[(a + b*x)^(m + 1) 
*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + 
p + 2)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && Gt 
Q[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, 
 m + n])
 

rule 168

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + 
 d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S 
imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f))   Int[(a + b*x)^(m + 1)*(c + d*x)^n 
*(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* 
h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], 
 x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]
 

rule 174

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* 
((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d)   Int[(e + f*x)^ 
p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d)   Int[(e + f*x)^p/(c + d 
*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, 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 586

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

Maple [A] (verified)

Time = 0.30 (sec) , antiderivative size = 229, normalized size of antiderivative = 0.96

method	result	size
risch	\(-\frac {\left (-d x +c \right ) \left (-9 d x +2 c \right ) \sqrt {-\frac {b \left (d^{2} x^{2}-c^{2}\right )}{d x +c}}\, \sqrt {d x +c}\, b}{4 c^{3} x^{2} \sqrt {-b \left (d x -c \right )}\, \sqrt {-b \left (d^{2} x^{2}-c^{2}\right )}}-\frac {d^{2} \left (\frac {8 \sqrt {-b d x +b c}}{-b d x -b c}-\frac {44 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-b d x +b c}\, \sqrt {2}}{2 \sqrt {b c}}\right )}{\sqrt {b c}}+\frac {62 \,\operatorname {arctanh}\left (\frac {\sqrt {-b d x +b c}}{\sqrt {b c}}\right )}{\sqrt {b c}}\right ) \sqrt {-\frac {b \left (d^{2} x^{2}-c^{2}\right )}{d x +c}}\, \sqrt {d x +c}\, b}{8 c^{3} \sqrt {-b \left (d^{2} x^{2}-c^{2}\right )}}\)	\(229\)
default	\(\frac {\sqrt {b \left (-d^{2} x^{2}+c^{2}\right )}\, \left (22 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {\left (-d x +c \right ) b}\, \sqrt {2}}{2 \sqrt {b c}}\right ) b \,d^{3} x^{3}+22 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {\left (-d x +c \right ) b}\, \sqrt {2}}{2 \sqrt {b c}}\right ) b c \,d^{2} x^{2}-31 \,\operatorname {arctanh}\left (\frac {\sqrt {\left (-d x +c \right ) b}}{\sqrt {b c}}\right ) b \,d^{3} x^{3}-31 \,\operatorname {arctanh}\left (\frac {\sqrt {\left (-d x +c \right ) b}}{\sqrt {b c}}\right ) b c \,d^{2} x^{2}+13 d^{2} x^{2} \sqrt {\left (-d x +c \right ) b}\, \sqrt {b c}+7 c d x \sqrt {\left (-d x +c \right ) b}\, \sqrt {b c}-2 \sqrt {\left (-d x +c \right ) b}\, \sqrt {b c}\, c^{2}\right )}{4 \left (d x +c \right )^{\frac {3}{2}} \sqrt {\left (-d x +c \right ) b}\, c^{3} x^{2} \sqrt {b c}}\)	\(231\)

Input:

int((-b*d^2*x^2+b*c^2)^(1/2)/x^3/(d*x+c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/4*(-d*x+c)*(-9*d*x+2*c)/c^3/x^2/(-b*(d*x-c))^(1/2)*(-1/(d*x+c)*b*(d^2*x 
^2-c^2))^(1/2)*(d*x+c)^(1/2)/(-b*(d^2*x^2-c^2))^(1/2)*b-1/8/c^3*d^2*(8*(-b 
*d*x+b*c)^(1/2)/(-b*d*x-b*c)-44*2^(1/2)/(b*c)^(1/2)*arctanh(1/2*(-b*d*x+b* 
c)^(1/2)*2^(1/2)/(b*c)^(1/2))+62/(b*c)^(1/2)*arctanh((-b*d*x+b*c)^(1/2)/(b 
*c)^(1/2)))*(-1/(d*x+c)*b*(d^2*x^2-c^2))^(1/2)*(d*x+c)^(1/2)/(-b*(d^2*x^2- 
c^2))^(1/2)*b
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 562, normalized size of antiderivative = 2.36 \[ \int \frac {\sqrt {b c^2-b d^2 x^2}}{x^3 (c+d x)^{5/2}} \, dx=\left [\frac {44 \, \sqrt {\frac {1}{2}} {\left (d^{4} x^{4} + 2 \, c d^{3} x^{3} + c^{2} d^{2} x^{2}\right )} \sqrt {\frac {b}{c}} \log \left (-\frac {b d^{2} x^{2} - 2 \, b c d x - 3 \, b c^{2} - 4 \, \sqrt {\frac {1}{2}} \sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c} c \sqrt {\frac {b}{c}}}{d^{2} x^{2} + 2 \, c d x + c^{2}}\right ) + 31 \, {\left (d^{4} x^{4} + 2 \, c d^{3} x^{3} + c^{2} d^{2} x^{2}\right )} \sqrt {\frac {b}{c}} \log \left (-\frac {b d^{2} x^{2} - b c d x - 2 \, b c^{2} + 2 \, \sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c} c \sqrt {\frac {b}{c}}}{d x^{2} + c x}\right ) + 2 \, \sqrt {-b d^{2} x^{2} + b c^{2}} {\left (13 \, d^{2} x^{2} + 7 \, c d x - 2 \, c^{2}\right )} \sqrt {d x + c}}{8 \, {\left (c^{3} d^{2} x^{4} + 2 \, c^{4} d x^{3} + c^{5} x^{2}\right )}}, \frac {44 \, \sqrt {\frac {1}{2}} {\left (d^{4} x^{4} + 2 \, c d^{3} x^{3} + c^{2} d^{2} x^{2}\right )} \sqrt {-\frac {b}{c}} \arctan \left (\frac {2 \, \sqrt {\frac {1}{2}} \sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c} c \sqrt {-\frac {b}{c}}}{b d^{2} x^{2} - b c^{2}}\right ) - 31 \, {\left (d^{4} x^{4} + 2 \, c d^{3} x^{3} + c^{2} d^{2} x^{2}\right )} \sqrt {-\frac {b}{c}} \arctan \left (\frac {\sqrt {-b d^{2} x^{2} + b c^{2}} \sqrt {d x + c} c \sqrt {-\frac {b}{c}}}{b d^{2} x^{2} - b c^{2}}\right ) + \sqrt {-b d^{2} x^{2} + b c^{2}} {\left (13 \, d^{2} x^{2} + 7 \, c d x - 2 \, c^{2}\right )} \sqrt {d x + c}}{4 \, {\left (c^{3} d^{2} x^{4} + 2 \, c^{4} d x^{3} + c^{5} x^{2}\right )}}\right ] \] Input:

integrate((-b*d^2*x^2+b*c^2)^(1/2)/x^3/(d*x+c)^(5/2),x, algorithm="fricas" 
)
 

Output:

[1/8*(44*sqrt(1/2)*(d^4*x^4 + 2*c*d^3*x^3 + c^2*d^2*x^2)*sqrt(b/c)*log(-(b 
*d^2*x^2 - 2*b*c*d*x - 3*b*c^2 - 4*sqrt(1/2)*sqrt(-b*d^2*x^2 + b*c^2)*sqrt 
(d*x + c)*c*sqrt(b/c))/(d^2*x^2 + 2*c*d*x + c^2)) + 31*(d^4*x^4 + 2*c*d^3* 
x^3 + c^2*d^2*x^2)*sqrt(b/c)*log(-(b*d^2*x^2 - b*c*d*x - 2*b*c^2 + 2*sqrt( 
-b*d^2*x^2 + b*c^2)*sqrt(d*x + c)*c*sqrt(b/c))/(d*x^2 + c*x)) + 2*sqrt(-b* 
d^2*x^2 + b*c^2)*(13*d^2*x^2 + 7*c*d*x - 2*c^2)*sqrt(d*x + c))/(c^3*d^2*x^ 
4 + 2*c^4*d*x^3 + c^5*x^2), 1/4*(44*sqrt(1/2)*(d^4*x^4 + 2*c*d^3*x^3 + c^2 
*d^2*x^2)*sqrt(-b/c)*arctan(2*sqrt(1/2)*sqrt(-b*d^2*x^2 + b*c^2)*sqrt(d*x 
+ c)*c*sqrt(-b/c)/(b*d^2*x^2 - b*c^2)) - 31*(d^4*x^4 + 2*c*d^3*x^3 + c^2*d 
^2*x^2)*sqrt(-b/c)*arctan(sqrt(-b*d^2*x^2 + b*c^2)*sqrt(d*x + c)*c*sqrt(-b 
/c)/(b*d^2*x^2 - b*c^2)) + sqrt(-b*d^2*x^2 + b*c^2)*(13*d^2*x^2 + 7*c*d*x 
- 2*c^2)*sqrt(d*x + c))/(c^3*d^2*x^4 + 2*c^4*d*x^3 + c^5*x^2)]
 

Sympy [F]

\[ \int \frac {\sqrt {b c^2-b d^2 x^2}}{x^3 (c+d x)^{5/2}} \, dx=\int \frac {\sqrt {- b \left (- c + d x\right ) \left (c + d x\right )}}{x^{3} \left (c + d x\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((-b*d**2*x**2+b*c**2)**(1/2)/x**3/(d*x+c)**(5/2),x)
 

Output:

Integral(sqrt(-b*(-c + d*x)*(c + d*x))/(x**3*(c + d*x)**(5/2)), x)
 

Maxima [F]

\[ \int \frac {\sqrt {b c^2-b d^2 x^2}}{x^3 (c+d x)^{5/2}} \, dx=\int { \frac {\sqrt {-b d^{2} x^{2} + b c^{2}}}{{\left (d x + c\right )}^{\frac {5}{2}} x^{3}} \,d x } \] Input:

integrate((-b*d^2*x^2+b*c^2)^(1/2)/x^3/(d*x+c)^(5/2),x, algorithm="maxima" 
)
 

Output:

integrate(sqrt(-b*d^2*x^2 + b*c^2)/((d*x + c)^(5/2)*x^3), x)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.71 \[ \int \frac {\sqrt {b c^2-b d^2 x^2}}{x^3 (c+d x)^{5/2}} \, dx=-\frac {1}{4} \, d^{2} {\left (\frac {22 \, \sqrt {2} b \arctan \left (\frac {\sqrt {2} \sqrt {-{\left (d x + c\right )} b + 2 \, b c}}{2 \, \sqrt {-b c}}\right )}{\sqrt {-b c} c^{3}} - \frac {31 \, b \arctan \left (\frac {\sqrt {-{\left (d x + c\right )} b + 2 \, b c}}{\sqrt {-b c}}\right )}{\sqrt {-b c} c^{3}} - \frac {4 \, \sqrt {-{\left (d x + c\right )} b + 2 \, b c}}{{\left (d x + c\right )} c^{3}} - \frac {7 \, \sqrt {-{\left (d x + c\right )} b + 2 \, b c} b^{2} c - 9 \, {\left (-{\left (d x + c\right )} b + 2 \, b c\right )}^{\frac {3}{2}} b}{{\left ({\left (d x + c\right )} b - b c\right )}^{2} c^{3}}\right )} \] Input:

integrate((-b*d^2*x^2+b*c^2)^(1/2)/x^3/(d*x+c)^(5/2),x, algorithm="giac")
 

Output:

-1/4*d^2*(22*sqrt(2)*b*arctan(1/2*sqrt(2)*sqrt(-(d*x + c)*b + 2*b*c)/sqrt( 
-b*c))/(sqrt(-b*c)*c^3) - 31*b*arctan(sqrt(-(d*x + c)*b + 2*b*c)/sqrt(-b*c 
))/(sqrt(-b*c)*c^3) - 4*sqrt(-(d*x + c)*b + 2*b*c)/((d*x + c)*c^3) - (7*sq 
rt(-(d*x + c)*b + 2*b*c)*b^2*c - 9*(-(d*x + c)*b + 2*b*c)^(3/2)*b)/(((d*x 
+ c)*b - b*c)^2*c^3))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {b c^2-b d^2 x^2}}{x^3 (c+d x)^{5/2}} \, dx=\int \frac {\sqrt {b\,c^2-b\,d^2\,x^2}}{x^3\,{\left (c+d\,x\right )}^{5/2}} \,d x \] Input:

int((b*c^2 - b*d^2*x^2)^(1/2)/(x^3*(c + d*x)^(5/2)),x)
 

Output:

int((b*c^2 - b*d^2*x^2)^(1/2)/(x^3*(c + d*x)^(5/2)), x)
 

Reduce [F]

\[ \int \frac {\sqrt {b c^2-b d^2 x^2}}{x^3 (c+d x)^{5/2}} \, dx=\int \frac {\sqrt {-b \,d^{2} x^{2}+b \,c^{2}}}{x^{3} \left (d x +c \right )^{\frac {5}{2}}}d x \] Input:

int((-b*d^2*x^2+b*c^2)^(1/2)/x^3/(d*x+c)^(5/2),x)
 

Output:

int((-b*d^2*x^2+b*c^2)^(1/2)/x^3/(d*x+c)^(5/2),x)