\(\int \frac {\sqrt {a+b \sqrt {x}}}{x^3} \, dx\) [128]

Optimal result

Integrand size = 17, antiderivative size = 133 \[ \int \frac {\sqrt {a+b \sqrt {x}}}{x^3} \, dx=-\frac {\sqrt {a+b \sqrt {x}}}{2 x^2}-\frac {b \sqrt {a+b \sqrt {x}}}{12 a x^{3/2}}+\frac {5 b^2 \sqrt {a+b \sqrt {x}}}{48 a^2 x}-\frac {5 b^3 \sqrt {a+b \sqrt {x}}}{32 a^3 \sqrt {x}}+\frac {5 b^4 \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {x}}}{\sqrt {a}}\right )}{32 a^{7/2}} \] Output:

-1/2*(a+b*x^(1/2))^(1/2)/x^2-1/12*b*(a+b*x^(1/2))^(1/2)/a/x^(3/2)+5/48*b^2 
*(a+b*x^(1/2))^(1/2)/a^2/x-5/32*b^3*(a+b*x^(1/2))^(1/2)/a^3/x^(1/2)+5/32*b 
^4*arctanh((a+b*x^(1/2))^(1/2)/a^(1/2))/a^(7/2)
 

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.68 \[ \int \frac {\sqrt {a+b \sqrt {x}}}{x^3} \, dx=\frac {\sqrt {a+b \sqrt {x}} \left (-48 a^3-8 a^2 b \sqrt {x}+10 a b^2 x-15 b^3 x^{3/2}\right )}{96 a^3 x^2}+\frac {5 b^4 \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {x}}}{\sqrt {a}}\right )}{32 a^{7/2}} \] Input:

Integrate[Sqrt[a + b*Sqrt[x]]/x^3,x]
 

Output:

(Sqrt[a + b*Sqrt[x]]*(-48*a^3 - 8*a^2*b*Sqrt[x] + 10*a*b^2*x - 15*b^3*x^(3 
/2)))/(96*a^3*x^2) + (5*b^4*ArcTanh[Sqrt[a + b*Sqrt[x]]/Sqrt[a]])/(32*a^(7 
/2))
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {798, 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.

\(\displaystyle \int \frac {\sqrt {a+b \sqrt {x}}}{x^3} \, dx\)

\(\Big \downarrow \) 798

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

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[Sqrt[a + b*Sqrt[x]]/x^3,x]
 

Output:

2*(-1/4*Sqrt[a + b*Sqrt[x]]/x^2 + (b*(-1/3*Sqrt[a + b*Sqrt[x]]/(a*x^(3/2)) 
 - (5*b*(-1/2*Sqrt[a + b*Sqrt[x]]/(a*x) - (3*b*(-(Sqrt[a + b*Sqrt[x]]/(a*S 
qrt[x])) + (b*ArcTanh[Sqrt[a + b*Sqrt[x]]/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 798

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n   Subst 
[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, 
b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
 

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.66

method	result	size
derivativedivides	\(4 b^{4} \left (-\frac {\frac {5 \left (a +b \sqrt {x}\right )^{\frac {7}{2}}}{128 a^{3}}-\frac {55 \left (a +b \sqrt {x}\right )^{\frac {5}{2}}}{384 a^{2}}+\frac {73 \left (a +b \sqrt {x}\right )^{\frac {3}{2}}}{384 a}+\frac {5 \sqrt {a +b \sqrt {x}}}{128}}{b^{4} x^{2}}+\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {a +b \sqrt {x}}}{\sqrt {a}}\right )}{128 a^{\frac {7}{2}}}\right )\)	\(88\)
default	\(4 b^{4} \left (-\frac {\frac {5 \left (a +b \sqrt {x}\right )^{\frac {7}{2}}}{128 a^{3}}-\frac {55 \left (a +b \sqrt {x}\right )^{\frac {5}{2}}}{384 a^{2}}+\frac {73 \left (a +b \sqrt {x}\right )^{\frac {3}{2}}}{384 a}+\frac {5 \sqrt {a +b \sqrt {x}}}{128}}{b^{4} x^{2}}+\frac {5 \,\operatorname {arctanh}\left (\frac {\sqrt {a +b \sqrt {x}}}{\sqrt {a}}\right )}{128 a^{\frac {7}{2}}}\right )\)	\(88\)

Input:

int((a+b*x^(1/2))^(1/2)/x^3,x,method=_RETURNVERBOSE)
 

Output:

4*b^4*(-(5/128/a^3*(a+b*x^(1/2))^(7/2)-55/384/a^2*(a+b*x^(1/2))^(5/2)+73/3 
84/a*(a+b*x^(1/2))^(3/2)+5/128*(a+b*x^(1/2))^(1/2))/b^4/x^2+5/128/a^(7/2)* 
arctanh((a+b*x^(1/2))^(1/2)/a^(1/2)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.56 \[ \int \frac {\sqrt {a+b \sqrt {x}}}{x^3} \, dx=\left [\frac {15 \, \sqrt {a} b^{4} x^{2} \log \left (\frac {b x + 2 \, \sqrt {b \sqrt {x} + a} \sqrt {a} \sqrt {x} + 2 \, a \sqrt {x}}{x}\right ) + 2 \, {\left (10 \, a^{2} b^{2} x - 48 \, a^{4} - {\left (15 \, a b^{3} x + 8 \, a^{3} b\right )} \sqrt {x}\right )} \sqrt {b \sqrt {x} + a}}{192 \, a^{4} x^{2}}, -\frac {15 \, \sqrt {-a} b^{4} x^{2} \arctan \left (\frac {{\left (\sqrt {-a} b \sqrt {x} - \sqrt {-a} a\right )} \sqrt {b \sqrt {x} + a}}{b^{2} x - a^{2}}\right ) - {\left (10 \, a^{2} b^{2} x - 48 \, a^{4} - {\left (15 \, a b^{3} x + 8 \, a^{3} b\right )} \sqrt {x}\right )} \sqrt {b \sqrt {x} + a}}{96 \, a^{4} x^{2}}\right ] \] Input:

integrate((a+b*x^(1/2))^(1/2)/x^3,x, algorithm="fricas")
 

Output:

[1/192*(15*sqrt(a)*b^4*x^2*log((b*x + 2*sqrt(b*sqrt(x) + a)*sqrt(a)*sqrt(x 
) + 2*a*sqrt(x))/x) + 2*(10*a^2*b^2*x - 48*a^4 - (15*a*b^3*x + 8*a^3*b)*sq 
rt(x))*sqrt(b*sqrt(x) + a))/(a^4*x^2), -1/96*(15*sqrt(-a)*b^4*x^2*arctan(( 
sqrt(-a)*b*sqrt(x) - sqrt(-a)*a)*sqrt(b*sqrt(x) + a)/(b^2*x - a^2)) - (10* 
a^2*b^2*x - 48*a^4 - (15*a*b^3*x + 8*a^3*b)*sqrt(x))*sqrt(b*sqrt(x) + a))/ 
(a^4*x^2)]
 

Sympy [A] (verification not implemented)

Time = 23.05 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.28 \[ \int \frac {\sqrt {a+b \sqrt {x}}}{x^3} \, dx=- \frac {a}{2 \sqrt {b} x^{\frac {9}{4}} \sqrt {\frac {a}{b \sqrt {x}} + 1}} - \frac {7 \sqrt {b}}{12 x^{\frac {7}{4}} \sqrt {\frac {a}{b \sqrt {x}} + 1}} + \frac {b^{\frac {3}{2}}}{48 a x^{\frac {5}{4}} \sqrt {\frac {a}{b \sqrt {x}} + 1}} - \frac {5 b^{\frac {5}{2}}}{96 a^{2} x^{\frac {3}{4}} \sqrt {\frac {a}{b \sqrt {x}} + 1}} - \frac {5 b^{\frac {7}{2}}}{32 a^{3} \sqrt [4]{x} \sqrt {\frac {a}{b \sqrt {x}} + 1}} + \frac {5 b^{4} \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt [4]{x}} \right )}}{32 a^{\frac {7}{2}}} \] Input:

integrate((a+b*x**(1/2))**(1/2)/x**3,x)
 

Output:

-a/(2*sqrt(b)*x**(9/4)*sqrt(a/(b*sqrt(x)) + 1)) - 7*sqrt(b)/(12*x**(7/4)*s 
qrt(a/(b*sqrt(x)) + 1)) + b**(3/2)/(48*a*x**(5/4)*sqrt(a/(b*sqrt(x)) + 1)) 
 - 5*b**(5/2)/(96*a**2*x**(3/4)*sqrt(a/(b*sqrt(x)) + 1)) - 5*b**(7/2)/(32* 
a**3*x**(1/4)*sqrt(a/(b*sqrt(x)) + 1)) + 5*b**4*asinh(sqrt(a)/(sqrt(b)*x** 
(1/4)))/(32*a**(7/2))
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.25 \[ \int \frac {\sqrt {a+b \sqrt {x}}}{x^3} \, dx=-\frac {5 \, b^{4} \log \left (\frac {\sqrt {b \sqrt {x} + a} - \sqrt {a}}{\sqrt {b \sqrt {x} + a} + \sqrt {a}}\right )}{64 \, a^{\frac {7}{2}}} - \frac {15 \, {\left (b \sqrt {x} + a\right )}^{\frac {7}{2}} b^{4} - 55 \, {\left (b \sqrt {x} + a\right )}^{\frac {5}{2}} a b^{4} + 73 \, {\left (b \sqrt {x} + a\right )}^{\frac {3}{2}} a^{2} b^{4} + 15 \, \sqrt {b \sqrt {x} + a} a^{3} b^{4}}{96 \, {\left ({\left (b \sqrt {x} + a\right )}^{4} a^{3} - 4 \, {\left (b \sqrt {x} + a\right )}^{3} a^{4} + 6 \, {\left (b \sqrt {x} + a\right )}^{2} a^{5} - 4 \, {\left (b \sqrt {x} + a\right )} a^{6} + a^{7}\right )}} \] Input:

integrate((a+b*x^(1/2))^(1/2)/x^3,x, algorithm="maxima")
 

Output:

-5/64*b^4*log((sqrt(b*sqrt(x) + a) - sqrt(a))/(sqrt(b*sqrt(x) + a) + sqrt( 
a)))/a^(7/2) - 1/96*(15*(b*sqrt(x) + a)^(7/2)*b^4 - 55*(b*sqrt(x) + a)^(5/ 
2)*a*b^4 + 73*(b*sqrt(x) + a)^(3/2)*a^2*b^4 + 15*sqrt(b*sqrt(x) + a)*a^3*b 
^4)/((b*sqrt(x) + a)^4*a^3 - 4*(b*sqrt(x) + a)^3*a^4 + 6*(b*sqrt(x) + a)^2 
*a^5 - 4*(b*sqrt(x) + a)*a^6 + a^7)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.73 \[ \int \frac {\sqrt {a+b \sqrt {x}}}{x^3} \, dx=-\frac {1}{96} \, b^{5} {\left (\frac {15 \, \arctan \left (\frac {\sqrt {b \sqrt {x} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3} b} + \frac {15 \, {\left (b \sqrt {x} + a\right )}^{\frac {7}{2}} - 55 \, {\left (b \sqrt {x} + a\right )}^{\frac {5}{2}} a + 73 \, {\left (b \sqrt {x} + a\right )}^{\frac {3}{2}} a^{2} + 15 \, \sqrt {b \sqrt {x} + a} a^{3}}{a^{3} b^{5} x^{2}}\right )} \] Input:

integrate((a+b*x^(1/2))^(1/2)/x^3,x, algorithm="giac")
 

Output:

-1/96*b^5*(15*arctan(sqrt(b*sqrt(x) + a)/sqrt(-a))/(sqrt(-a)*a^3*b) + (15* 
(b*sqrt(x) + a)^(7/2) - 55*(b*sqrt(x) + a)^(5/2)*a + 73*(b*sqrt(x) + a)^(3 
/2)*a^2 + 15*sqrt(b*sqrt(x) + a)*a^3)/(a^3*b^5*x^2))
 

Mupad [B] (verification not implemented)

Time = 0.69 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.68 \[ \int \frac {\sqrt {a+b \sqrt {x}}}{x^3} \, dx=\frac {55\,{\left (a+b\,\sqrt {x}\right )}^{5/2}}{96\,a^2\,x^2}-\frac {73\,{\left (a+b\,\sqrt {x}\right )}^{3/2}}{96\,a\,x^2}-\frac {5\,\sqrt {a+b\,\sqrt {x}}}{32\,x^2}-\frac {5\,{\left (a+b\,\sqrt {x}\right )}^{7/2}}{32\,a^3\,x^2}-\frac {b^4\,\mathrm {atan}\left (\frac {\sqrt {a+b\,\sqrt {x}}\,1{}\mathrm {i}}{\sqrt {a}}\right )\,5{}\mathrm {i}}{32\,a^{7/2}} \] Input:

int((a + b*x^(1/2))^(1/2)/x^3,x)
 

Output:

(55*(a + b*x^(1/2))^(5/2))/(96*a^2*x^2) - (b^4*atan(((a + b*x^(1/2))^(1/2) 
*1i)/a^(1/2))*5i)/(32*a^(7/2)) - (73*(a + b*x^(1/2))^(3/2))/(96*a*x^2) - ( 
5*(a + b*x^(1/2))^(1/2))/(32*x^2) - (5*(a + b*x^(1/2))^(7/2))/(32*a^3*x^2)
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.84 \[ \int \frac {\sqrt {a+b \sqrt {x}}}{x^3} \, dx=\frac {-16 \sqrt {x}\, \sqrt {\sqrt {x}\, b +a}\, a^{3} b -30 \sqrt {x}\, \sqrt {\sqrt {x}\, b +a}\, a \,b^{3} x -96 \sqrt {\sqrt {x}\, b +a}\, a^{4}+20 \sqrt {\sqrt {x}\, b +a}\, a^{2} b^{2} x -15 \sqrt {a}\, \mathrm {log}\left (\sqrt {\sqrt {x}\, b +a}-\sqrt {a}\right ) b^{4} x^{2}+15 \sqrt {a}\, \mathrm {log}\left (\sqrt {\sqrt {x}\, b +a}+\sqrt {a}\right ) b^{4} x^{2}}{192 a^{4} x^{2}} \] Input:

int((a+b*x^(1/2))^(1/2)/x^3,x)
 

Output:

( - 16*sqrt(x)*sqrt(sqrt(x)*b + a)*a**3*b - 30*sqrt(x)*sqrt(sqrt(x)*b + a) 
*a*b**3*x - 96*sqrt(sqrt(x)*b + a)*a**4 + 20*sqrt(sqrt(x)*b + a)*a**2*b**2 
*x - 15*sqrt(a)*log(sqrt(sqrt(x)*b + a) - sqrt(a))*b**4*x**2 + 15*sqrt(a)* 
log(sqrt(sqrt(x)*b + a) + sqrt(a))*b**4*x**2)/(192*a**4*x**2)