\(\int \frac {\sqrt {a+b \sqrt {c x^2}}}{x^5} \, dx\) [29]

Optimal result

Integrand size = 21, antiderivative size = 171 \[ \int \frac {\sqrt {a+b \sqrt {c x^2}}}{x^5} \, dx=-\frac {\sqrt {a+b \sqrt {c x^2}}}{4 x^4}+\frac {5 b^2 c \sqrt {a+b \sqrt {c x^2}}}{96 a^2 x^2}-\frac {b c^2 \sqrt {a+b \sqrt {c x^2}}}{24 a \left (c x^2\right )^{3/2}}-\frac {5 b^3 c^2 \sqrt {a+b \sqrt {c x^2}}}{64 a^3 \sqrt {c x^2}}+\frac {5 b^4 c^2 \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {c x^2}}}{\sqrt {a}}\right )}{64 a^{7/2}} \] Output:

-1/4*(a+b*(c*x^2)^(1/2))^(1/2)/x^4+5/96*b^2*c*(a+b*(c*x^2)^(1/2))^(1/2)/a^ 
2/x^2-1/24*b*c^2*(a+b*(c*x^2)^(1/2))^(1/2)/a/(c*x^2)^(3/2)-5/64*b^3*c^2*(a 
+b*(c*x^2)^(1/2))^(1/2)/a^3/(c*x^2)^(1/2)+5/64*b^4*c^2*arctanh((a+b*(c*x^2 
)^(1/2))^(1/2)/a^(1/2))/a^(7/2)
 

Mathematica [A] (verified)

Time = 1.49 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.65 \[ \int \frac {\sqrt {a+b \sqrt {c x^2}}}{x^5} \, dx=-\frac {\sqrt {a+b \sqrt {c x^2}} \left (48 a^3-10 a b^2 c x^2+8 a^2 b \sqrt {c x^2}+15 b^3 \left (c x^2\right )^{3/2}\right )}{192 a^3 x^4}+\frac {5 b^4 c^2 \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {c x^2}}}{\sqrt {a}}\right )}{64 a^{7/2}} \] Input:

Integrate[Sqrt[a + b*Sqrt[c*x^2]]/x^5,x]
 

Output:

-1/192*(Sqrt[a + b*Sqrt[c*x^2]]*(48*a^3 - 10*a*b^2*c*x^2 + 8*a^2*b*Sqrt[c* 
x^2] + 15*b^3*(c*x^2)^(3/2)))/(a^3*x^4) + (5*b^4*c^2*ArcTanh[Sqrt[a + b*Sq 
rt[c*x^2]]/Sqrt[a]])/(64*a^(7/2))
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {892, 51, 52, 52, 52, 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.

\(\displaystyle \int \frac {\sqrt {a+b \sqrt {c x^2}}}{x^5} \, dx\)

\(\Big \downarrow \) 892

\(\displaystyle c^2 \int \frac {\sqrt {a+b \sqrt {c x^2}}}{\left (c x^2\right )^{5/2}}d\sqrt {c x^2}\)

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 52

\(\displaystyle c^2 \left (\frac {1}{8} b \left (-\frac {5 b \int \frac {1}{\left (c x^2\right )^{3/2} \sqrt {a+b \sqrt {c x^2}}}d\sqrt {c x^2}}{6 a}-\frac {\sqrt {a+b \sqrt {c x^2}}}{3 a \left (c x^2\right )^{3/2}}\right )-\frac {\sqrt {a+b \sqrt {c x^2}}}{4 c^2 x^4}\right )\)

\(\Big \downarrow \) 52

\(\displaystyle c^2 \left (\frac {1}{8} b \left (-\frac {5 b \left (-\frac {3 b \int \frac {1}{c x^2 \sqrt {a+b \sqrt {c x^2}}}d\sqrt {c x^2}}{4 a}-\frac {\sqrt {a+b \sqrt {c x^2}}}{2 a c x^2}\right )}{6 a}-\frac {\sqrt {a+b \sqrt {c x^2}}}{3 a \left (c x^2\right )^{3/2}}\right )-\frac {\sqrt {a+b \sqrt {c x^2}}}{4 c^2 x^4}\right )\)

\(\Big \downarrow \) 52

\(\displaystyle c^2 \left (\frac {1}{8} b \left (-\frac {5 b \left (-\frac {3 b \left (-\frac {b \int \frac {1}{\sqrt {c x^2} \sqrt {a+b \sqrt {c x^2}}}d\sqrt {c x^2}}{2 a}-\frac {\sqrt {a+b \sqrt {c x^2}}}{a \sqrt {c x^2}}\right )}{4 a}-\frac {\sqrt {a+b \sqrt {c x^2}}}{2 a c x^2}\right )}{6 a}-\frac {\sqrt {a+b \sqrt {c x^2}}}{3 a \left (c x^2\right )^{3/2}}\right )-\frac {\sqrt {a+b \sqrt {c x^2}}}{4 c^2 x^4}\right )\)

\(\Big \downarrow \) 73

\(\displaystyle c^2 \left (\frac {1}{8} b \left (-\frac {5 b \left (-\frac {3 b \left (-\frac {\int \frac {1}{\frac {c x^2}{b}-\frac {a}{b}}d\sqrt {a+b \sqrt {c x^2}}}{a}-\frac {\sqrt {a+b \sqrt {c x^2}}}{a \sqrt {c x^2}}\right )}{4 a}-\frac {\sqrt {a+b \sqrt {c x^2}}}{2 a c x^2}\right )}{6 a}-\frac {\sqrt {a+b \sqrt {c x^2}}}{3 a \left (c x^2\right )^{3/2}}\right )-\frac {\sqrt {a+b \sqrt {c x^2}}}{4 c^2 x^4}\right )\)

\(\Big \downarrow \) 221

\(\displaystyle c^2 \left (\frac {1}{8} b \left (-\frac {5 b \left (-\frac {3 b \left (\frac {b \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {c x^2}}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\sqrt {a+b \sqrt {c x^2}}}{a \sqrt {c x^2}}\right )}{4 a}-\frac {\sqrt {a+b \sqrt {c x^2}}}{2 a c x^2}\right )}{6 a}-\frac {\sqrt {a+b \sqrt {c x^2}}}{3 a \left (c x^2\right )^{3/2}}\right )-\frac {\sqrt {a+b \sqrt {c x^2}}}{4 c^2 x^4}\right )\)

Input:

Int[Sqrt[a + b*Sqrt[c*x^2]]/x^5,x]
 

Output:

c^2*(-1/4*Sqrt[a + b*Sqrt[c*x^2]]/(c^2*x^4) + (b*(-1/3*Sqrt[a + b*Sqrt[c*x 
^2]]/(a*(c*x^2)^(3/2)) - (5*b*(-1/2*Sqrt[a + b*Sqrt[c*x^2]]/(a*c*x^2) - (3 
*b*(-(Sqrt[a + b*Sqrt[c*x^2]]/(a*Sqrt[c*x^2])) + (b*ArcTanh[Sqrt[a + b*Sqr 
t[c*x^2]]/Sqrt[a]])/a^(3/2)))/(4*a)))/(6*a)))/8)
 

Defintions of rubi rules used

rule 51

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

rule 52

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

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 892

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbo 
l] :> Simp[(d*x)^(m + 1)/(d*((c*x^q)^(1/q))^(m + 1))   Subst[Int[x^m*(a + b 
*x^(n*q))^p, x], x, (c*x^q)^(1/q)], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x 
] && IntegerQ[n*q] && NeQ[x, (c*x^q)^(1/q)]
 

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.67

method	result	size
default	\(-\frac {15 \left (a +b \sqrt {c \,x^{2}}\right )^{\frac {7}{2}} a^{\frac {7}{2}}-15 \,\operatorname {arctanh}\left (\frac {\sqrt {a +b \sqrt {c \,x^{2}}}}{\sqrt {a}}\right ) a^{3} b^{4} c^{2} x^{4}-55 \left (a +b \sqrt {c \,x^{2}}\right )^{\frac {5}{2}} a^{\frac {9}{2}}+73 \left (a +b \sqrt {c \,x^{2}}\right )^{\frac {3}{2}} a^{\frac {11}{2}}+15 \sqrt {a +b \sqrt {c \,x^{2}}}\, a^{\frac {13}{2}}}{192 a^{\frac {13}{2}} x^{4}}\)	\(114\)

Input:

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

Output:

-1/192*(15*(a+b*(c*x^2)^(1/2))^(7/2)*a^(7/2)-15*arctanh((a+b*(c*x^2)^(1/2) 
)^(1/2)/a^(1/2))*a^3*b^4*c^2*x^4-55*(a+b*(c*x^2)^(1/2))^(5/2)*a^(9/2)+73*( 
a+b*(c*x^2)^(1/2))^(3/2)*a^(11/2)+15*(a+b*(c*x^2)^(1/2))^(1/2)*a^(13/2))/a 
^(13/2)/x^4
 

Fricas [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.57 \[ \int \frac {\sqrt {a+b \sqrt {c x^2}}}{x^5} \, dx=\left [\frac {15 \, \sqrt {a} b^{4} c^{2} x^{4} \log \left (\frac {b c x^{2} + 2 \, \sqrt {c x^{2}} \sqrt {\sqrt {c x^{2}} b + a} \sqrt {a} + 2 \, \sqrt {c x^{2}} a}{x^{2}}\right ) + 2 \, {\left (10 \, a^{2} b^{2} c x^{2} - 48 \, a^{4} - {\left (15 \, a b^{3} c x^{2} + 8 \, a^{3} b\right )} \sqrt {c x^{2}}\right )} \sqrt {\sqrt {c x^{2}} b + a}}{384 \, a^{4} x^{4}}, -\frac {15 \, \sqrt {-a} b^{4} c^{2} x^{4} \arctan \left (\frac {{\left (\sqrt {c x^{2}} \sqrt {-a} b - \sqrt {-a} a\right )} \sqrt {\sqrt {c x^{2}} b + a}}{b^{2} c x^{2} - a^{2}}\right ) - {\left (10 \, a^{2} b^{2} c x^{2} - 48 \, a^{4} - {\left (15 \, a b^{3} c x^{2} + 8 \, a^{3} b\right )} \sqrt {c x^{2}}\right )} \sqrt {\sqrt {c x^{2}} b + a}}{192 \, a^{4} x^{4}}\right ] \] Input:

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

Output:

[1/384*(15*sqrt(a)*b^4*c^2*x^4*log((b*c*x^2 + 2*sqrt(c*x^2)*sqrt(sqrt(c*x^ 
2)*b + a)*sqrt(a) + 2*sqrt(c*x^2)*a)/x^2) + 2*(10*a^2*b^2*c*x^2 - 48*a^4 - 
 (15*a*b^3*c*x^2 + 8*a^3*b)*sqrt(c*x^2))*sqrt(sqrt(c*x^2)*b + a))/(a^4*x^4 
), -1/192*(15*sqrt(-a)*b^4*c^2*x^4*arctan((sqrt(c*x^2)*sqrt(-a)*b - sqrt(- 
a)*a)*sqrt(sqrt(c*x^2)*b + a)/(b^2*c*x^2 - a^2)) - (10*a^2*b^2*c*x^2 - 48* 
a^4 - (15*a*b^3*c*x^2 + 8*a^3*b)*sqrt(c*x^2))*sqrt(sqrt(c*x^2)*b + a))/(a^ 
4*x^4)]
 

Sympy [F]

\[ \int \frac {\sqrt {a+b \sqrt {c x^2}}}{x^5} \, dx=\int \frac {\sqrt {a + b \sqrt {c x^{2}}}}{x^{5}}\, dx \] Input:

integrate((a+b*(c*x**2)**(1/2))**(1/2)/x**5,x)
 

Output:

Integral(sqrt(a + b*sqrt(c*x**2))/x**5, x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.23 \[ \int \frac {\sqrt {a+b \sqrt {c x^2}}}{x^5} \, dx=-\frac {1}{384} \, {\left (\frac {15 \, b^{4} \log \left (\frac {\sqrt {\sqrt {c x^{2}} b + a} - \sqrt {a}}{\sqrt {\sqrt {c x^{2}} b + a} + \sqrt {a}}\right )}{a^{\frac {7}{2}}} + \frac {2 \, {\left (15 \, {\left (\sqrt {c x^{2}} b + a\right )}^{\frac {7}{2}} b^{4} - 55 \, {\left (\sqrt {c x^{2}} b + a\right )}^{\frac {5}{2}} a b^{4} + 73 \, {\left (\sqrt {c x^{2}} b + a\right )}^{\frac {3}{2}} a^{2} b^{4} + 15 \, \sqrt {\sqrt {c x^{2}} b + a} a^{3} b^{4}\right )}}{{\left (\sqrt {c x^{2}} b + a\right )}^{4} a^{3} - 4 \, {\left (\sqrt {c x^{2}} b + a\right )}^{3} a^{4} + 6 \, {\left (\sqrt {c x^{2}} b + a\right )}^{2} a^{5} - 4 \, {\left (\sqrt {c x^{2}} b + a\right )} a^{6} + a^{7}}\right )} c^{2} \] Input:

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

Output:

-1/384*(15*b^4*log((sqrt(sqrt(c*x^2)*b + a) - sqrt(a))/(sqrt(sqrt(c*x^2)*b 
 + a) + sqrt(a)))/a^(7/2) + 2*(15*(sqrt(c*x^2)*b + a)^(7/2)*b^4 - 55*(sqrt 
(c*x^2)*b + a)^(5/2)*a*b^4 + 73*(sqrt(c*x^2)*b + a)^(3/2)*a^2*b^4 + 15*sqr 
t(sqrt(c*x^2)*b + a)*a^3*b^4)/((sqrt(c*x^2)*b + a)^4*a^3 - 4*(sqrt(c*x^2)* 
b + a)^3*a^4 + 6*(sqrt(c*x^2)*b + a)^2*a^5 - 4*(sqrt(c*x^2)*b + a)*a^6 + a 
^7))*c^2
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a+b \sqrt {c x^2}}}{x^5} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {a+b \sqrt {c x^2}}}{x^5} \, dx=\int \frac {\sqrt {a+b\,\sqrt {c\,x^2}}}{x^5} \,d x \] Input:

int((a + b*(c*x^2)^(1/2))^(1/2)/x^5,x)
 

Output:

int((a + b*(c*x^2)^(1/2))^(1/2)/x^5, x)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt {a+b \sqrt {c x^2}}}{x^5} \, dx=\frac {-16 \sqrt {c}\, \sqrt {\sqrt {c}\, b x +a}\, a^{3} b x -30 \sqrt {c}\, \sqrt {\sqrt {c}\, b x +a}\, a \,b^{3} c \,x^{3}-96 \sqrt {\sqrt {c}\, b x +a}\, a^{4}+20 \sqrt {\sqrt {c}\, b x +a}\, a^{2} b^{2} c \,x^{2}-15 \sqrt {a}\, \mathrm {log}\left (\sqrt {\sqrt {c}\, b x +a}-\sqrt {a}\right ) b^{4} c^{2} x^{4}+15 \sqrt {a}\, \mathrm {log}\left (\sqrt {\sqrt {c}\, b x +a}+\sqrt {a}\right ) b^{4} c^{2} x^{4}}{384 a^{4} x^{4}} \] Input:

int((a+b*(c*x^2)^(1/2))^(1/2)/x^5,x)
 

Output:

( - 16*sqrt(c)*sqrt(sqrt(c)*b*x + a)*a**3*b*x - 30*sqrt(c)*sqrt(sqrt(c)*b* 
x + a)*a*b**3*c*x**3 - 96*sqrt(sqrt(c)*b*x + a)*a**4 + 20*sqrt(sqrt(c)*b*x 
 + a)*a**2*b**2*c*x**2 - 15*sqrt(a)*log(sqrt(sqrt(c)*b*x + a) - sqrt(a))*b 
**4*c**2*x**4 + 15*sqrt(a)*log(sqrt(sqrt(c)*b*x + a) + sqrt(a))*b**4*c**2* 
x**4)/(384*a**4*x**4)