3.30.82 \(\int \sqrt {a+b \sqrt {\frac {c}{x}}} x \, dx\) [2982]

3.30.82.1 Optimal result

Integrand size = 19, antiderivative size = 169 \[ \int \sqrt {a+b \sqrt {\frac {c}{x}}} x \, dx=\frac {b c^2 \sqrt {a+b \sqrt {\frac {c}{x}}}}{12 a \left (\frac {c}{x}\right )^{3/2}}+\frac {5 b^3 c^2 \sqrt {a+b \sqrt {\frac {c}{x}}}}{32 a^3 \sqrt {\frac {c}{x}}}-\frac {5 b^2 c \sqrt {a+b \sqrt {\frac {c}{x}}} x}{48 a^2}+\frac {1}{2} \sqrt {a+b \sqrt {\frac {c}{x}}} x^2-\frac {5 b^4 c^2 \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {\frac {c}{x}}}}{\sqrt {a}}\right )}{32 a^{7/2}} \]

output

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

3.30.82.2 Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.66 \[ \int \sqrt {a+b \sqrt {\frac {c}{x}}} x \, dx=\frac {\sqrt {a+b \sqrt {\frac {c}{x}}} \left (48 a^3+8 a^2 b \sqrt {\frac {c}{x}}+15 b^3 \left (\frac {c}{x}\right )^{3/2}-\frac {10 a b^2 c}{x}\right ) x^2}{96 a^3}-\frac {5 b^4 c^2 \text {arctanh}\left (\frac {\sqrt {a+b \sqrt {\frac {c}{x}}}}{\sqrt {a}}\right )}{32 a^{7/2}} \]

input

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

output

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

3.30.82.3 Rubi [A] (warning: unable to verify)

Time = 0.27 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.11, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {893, 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.

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

\(\Big \downarrow \) 893

\(\displaystyle \int x \sqrt {a+\frac {b \sqrt {c}}{\sqrt {x}}}dx\)

\(\Big \downarrow \) 798

\(\displaystyle -2 \int \sqrt {a+\frac {b \sqrt {c}}{\sqrt {x}}} x^{5/2}d\frac {1}{\sqrt {x}}\)

\(\Big \downarrow \) 51

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 52

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

input

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

output

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

3.30.82.3.1 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]]
 

rule 893

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

3.30.82.4 Maple [A] (verified)

Time = 4.00 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.25

method	result	size
default	\(\frac {\sqrt {a +b \sqrt {\frac {c}{x}}}\, \sqrt {x}\, \left (30 a^{\frac {3}{2}} \sqrt {a x +b \sqrt {\frac {c}{x}}\, x}\, \left (\frac {c}{x}\right )^{\frac {3}{2}} x^{\frac {3}{2}} b^{3}-15 c^{2} \ln \left (\frac {b \sqrt {\frac {c}{x}}\, \sqrt {x}+2 \sqrt {a x +b \sqrt {\frac {c}{x}}\, x}\, \sqrt {a}+2 a \sqrt {x}}{2 \sqrt {a}}\right ) a \,b^{4}+60 c \,a^{\frac {5}{2}} \sqrt {a x +b \sqrt {\frac {c}{x}}\, x}\, \sqrt {x}\, b^{2}-80 a^{\frac {5}{2}} \left (a x +b \sqrt {\frac {c}{x}}\, x \right )^{\frac {3}{2}} \sqrt {\frac {c}{x}}\, \sqrt {x}\, b +96 \sqrt {x}\, \left (a x +b \sqrt {\frac {c}{x}}\, x \right )^{\frac {3}{2}} a^{\frac {7}{2}}\right )}{192 \sqrt {x \left (a +b \sqrt {\frac {c}{x}}\right )}\, a^{\frac {9}{2}}}\)	\(211\)

input

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

output

1/192*(a+b*(c/x)^(1/2))^(1/2)*x^(1/2)*(30*a^(3/2)*(a*x+b*(c/x)^(1/2)*x)^(1 
/2)*(c/x)^(3/2)*x^(3/2)*b^3-15*c^2*ln(1/2*(b*(c/x)^(1/2)*x^(1/2)+2*(a*x+b* 
(c/x)^(1/2)*x)^(1/2)*a^(1/2)+2*a*x^(1/2))/a^(1/2))*a*b^4+60*c*a^(5/2)*(a*x 
+b*(c/x)^(1/2)*x)^(1/2)*x^(1/2)*b^2-80*a^(5/2)*(a*x+b*(c/x)^(1/2)*x)^(3/2) 
*(c/x)^(1/2)*x^(1/2)*b+96*x^(1/2)*(a*x+b*(c/x)^(1/2)*x)^(3/2)*a^(7/2))/(x* 
(a+b*(c/x)^(1/2)))^(1/2)/a^(9/2)
 

3.30.82.5 Fricas [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.33 \[ \int \sqrt {a+b \sqrt {\frac {c}{x}}} x \, dx=\left [\frac {15 \, \sqrt {a} b^{4} c^{2} \log \left (-2 \, \sqrt {b \sqrt {\frac {c}{x}} + a} \sqrt {a} x \sqrt {\frac {c}{x}} + 2 \, a x \sqrt {\frac {c}{x}} + b c\right ) - 2 \, {\left (10 \, a^{2} b^{2} c x - 48 \, a^{4} x^{2} - {\left (15 \, a b^{3} c x + 8 \, a^{3} b x^{2}\right )} \sqrt {\frac {c}{x}}\right )} \sqrt {b \sqrt {\frac {c}{x}} + a}}{192 \, a^{4}}, \frac {15 \, \sqrt {-a} b^{4} c^{2} \arctan \left (\frac {\sqrt {b \sqrt {\frac {c}{x}} + a} \sqrt {-a}}{a}\right ) - {\left (10 \, a^{2} b^{2} c x - 48 \, a^{4} x^{2} - {\left (15 \, a b^{3} c x + 8 \, a^{3} b x^{2}\right )} \sqrt {\frac {c}{x}}\right )} \sqrt {b \sqrt {\frac {c}{x}} + a}}{96 \, a^{4}}\right ] \]

input

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

output

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

3.30.82.6 Sympy [F]

\[ \int \sqrt {a+b \sqrt {\frac {c}{x}}} x \, dx=\int x \sqrt {a + b \sqrt {\frac {c}{x}}}\, dx \]

input

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

output

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

3.30.82.7 Maxima [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.25 \[ \int \sqrt {a+b \sqrt {\frac {c}{x}}} x \, dx=\frac {1}{192} \, {\left (\frac {15 \, b^{4} \log \left (\frac {\sqrt {b \sqrt {\frac {c}{x}} + a} - \sqrt {a}}{\sqrt {b \sqrt {\frac {c}{x}} + a} + \sqrt {a}}\right )}{a^{\frac {7}{2}}} + \frac {2 \, {\left (15 \, {\left (b \sqrt {\frac {c}{x}} + a\right )}^{\frac {7}{2}} b^{4} - 55 \, {\left (b \sqrt {\frac {c}{x}} + a\right )}^{\frac {5}{2}} a b^{4} + 73 \, {\left (b \sqrt {\frac {c}{x}} + a\right )}^{\frac {3}{2}} a^{2} b^{4} + 15 \, \sqrt {b \sqrt {\frac {c}{x}} + a} a^{3} b^{4}\right )}}{{\left (b \sqrt {\frac {c}{x}} + a\right )}^{4} a^{3} - 4 \, {\left (b \sqrt {\frac {c}{x}} + a\right )}^{3} a^{4} + 6 \, {\left (b \sqrt {\frac {c}{x}} + a\right )}^{2} a^{5} - 4 \, {\left (b \sqrt {\frac {c}{x}} + a\right )} a^{6} + a^{7}}\right )} c^{2} \]

input

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

output

1/192*(15*b^4*log((sqrt(b*sqrt(c/x) + a) - sqrt(a))/(sqrt(b*sqrt(c/x) + a) 
 + sqrt(a)))/a^(7/2) + 2*(15*(b*sqrt(c/x) + a)^(7/2)*b^4 - 55*(b*sqrt(c/x) 
 + a)^(5/2)*a*b^4 + 73*(b*sqrt(c/x) + a)^(3/2)*a^2*b^4 + 15*sqrt(b*sqrt(c/ 
x) + a)*a^3*b^4)/((b*sqrt(c/x) + a)^4*a^3 - 4*(b*sqrt(c/x) + a)^3*a^4 + 6* 
(b*sqrt(c/x) + a)^2*a^5 - 4*(b*sqrt(c/x) + a)*a^6 + a^7))*c^2
 

3.30.82.8 Giac [F(-2)]

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

input

integrate(x*(a+b*(c/x)^(1/2))^(1/2),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 ValueMinimal poly. in r 
ootof mus
 

3.30.82.9 Mupad [F(-1)]

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

input

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

output

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