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

Optimal. Leaf size=169 \[ \frac {b c^2 \sqrt {a+b \sqrt {\frac {c}{x}}}}{12 a \left (\frac {c}{x}\right )^{3/2}}+\frac {5 b^3 c^2 \sqrt {a+b \sqrt {\frac {c}{x}}}}{32 a^3 \sqrt {\frac {c}{x}}}-\frac {5 b^2 c \sqrt {a+b \sqrt {\frac {c}{x}}} x}{48 a^2}+\frac {1}{2} \sqrt {a+b \sqrt {\frac {c}{x}}} x^2-\frac {5 b^4 c^2 \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {\frac {c}{x}}}}{\sqrt {a}}\right )}{32 a^{7/2}} \]

[Out]

-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)

________________________________________________________________________________________

Rubi [A]
time = 0.07, antiderivative size = 172, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {376, 272, 43, 44, 65, 214} \begin {gather*} -\frac {5 b^4 c^2 \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {\frac {c}{x}}}}{\sqrt {a}}\right )}{32 a^{7/2}}+\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 x \sqrt {a+b \sqrt {\frac {c}{x}}}}{48 a^2}+\frac {b x^3 \left (\frac {c}{x}\right )^{3/2} \sqrt {a+b \sqrt {\frac {c}{x}}}}{12 a c}+\frac {1}{2} x^2 \sqrt {a+b \sqrt {\frac {c}{x}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Sqrt[c/x]]*x,x]

[Out]

(5*b^3*c^2*Sqrt[a + b*Sqrt[c/x]])/(32*a^3*Sqrt[c/x]) - (5*b^2*c*Sqrt[a + b*Sqrt[c/x]]*x)/(48*a^2) + (Sqrt[a +
b*Sqrt[c/x]]*x^2)/2 + (b*Sqrt[a + b*Sqrt[c/x]]*(c/x)^(3/2)*x^3)/(12*a*c) - (5*b^4*c^2*ArcTanh[Sqrt[a + b*Sqrt[
c/x]]/Sqrt[a]])/(32*a^(7/2))

Rule 43

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] - Dist[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] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 44

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] - Dist[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] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[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 376

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*((c_.)*(x_)^(q_))^(n_))^(p_.), x_Symbol] :> With[{k = Denominator[n]}, Su
bst[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]

Rubi steps

\begin {align*} \int \sqrt {a+b \sqrt {\frac {c}{x}}} x \, dx &=\text {Subst}\left (\int \sqrt {a+\frac {b \sqrt {c}}{\sqrt {x}}} x \, dx,\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right )\\ &=-\text {Subst}\left (2 \text {Subst}\left (\int \frac {\sqrt {a+b \sqrt {c} x}}{x^5} \, dx,x,\frac {1}{\sqrt {x}}\right ),\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right )\\ &=\frac {1}{2} \sqrt {a+b \sqrt {\frac {c}{x}}} x^2-\text {Subst}\left (\frac {1}{4} \left (b \sqrt {c}\right ) \text {Subst}\left (\int \frac {1}{x^4 \sqrt {a+b \sqrt {c} x}} \, dx,x,\frac {1}{\sqrt {x}}\right ),\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right )\\ &=\frac {1}{2} \sqrt {a+b \sqrt {\frac {c}{x}}} x^2+\frac {b \sqrt {a+b \sqrt {\frac {c}{x}}} \left (\frac {c}{x}\right )^{3/2} x^3}{12 a c}+\text {Subst}\left (\frac {\left (5 b^2 c\right ) \text {Subst}\left (\int \frac {1}{x^3 \sqrt {a+b \sqrt {c} x}} \, dx,x,\frac {1}{\sqrt {x}}\right )}{24 a},\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right )\\ &=-\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 {b \sqrt {a+b \sqrt {\frac {c}{x}}} \left (\frac {c}{x}\right )^{3/2} x^3}{12 a c}-\text {Subst}\left (\frac {\left (5 b^3 c^{3/2}\right ) \text {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b \sqrt {c} x}} \, dx,x,\frac {1}{\sqrt {x}}\right )}{32 a^2},\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right )\\ &=\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 {b \sqrt {a+b \sqrt {\frac {c}{x}}} \left (\frac {c}{x}\right )^{3/2} x^3}{12 a c}+\text {Subst}\left (\frac {\left (5 b^4 c^2\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+b \sqrt {c} x}} \, dx,x,\frac {1}{\sqrt {x}}\right )}{64 a^3},\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right )\\ &=\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 {b \sqrt {a+b \sqrt {\frac {c}{x}}} \left (\frac {c}{x}\right )^{3/2} x^3}{12 a c}+\text {Subst}\left (\frac {\left (5 b^3 c^{3/2}\right ) \text {Subst}\left (\int \frac {1}{-\frac {a}{b \sqrt {c}}+\frac {x^2}{b \sqrt {c}}} \, dx,x,\sqrt {a+\frac {b \sqrt {c}}{\sqrt {x}}}\right )}{32 a^3},\sqrt {x},\frac {\sqrt {\frac {c}{x}} x}{\sqrt {c}}\right )\\ &=\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 {b \sqrt {a+b \sqrt {\frac {c}{x}}} \left (\frac {c}{x}\right )^{3/2} x^3}{12 a c}-\frac {5 b^4 c^2 \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {\frac {c}{x}}}}{\sqrt {a}}\right )}{32 a^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.14, size = 112, normalized size = 0.66 \begin {gather*} \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 \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {\frac {c}{x}}}}{\sqrt {a}}\right )}{32 a^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*Sqrt[c/x]]*x,x]

[Out]

(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]]/Sqrt[a]])/(32*a^(7/2))

________________________________________________________________________________________

Maple [A]
time = 0.26, size = 211, normalized size = 1.25

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*(a+b*(c/x)^(1/2))^(1/2)*x^(1/2)*(15*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))*c^2*a*b^4-30*a^(3/2)*(a*x+b*(c/x)^(1/2)*x)^(1/2)*(c/x)^(3/2)*x^(3/2)*b^3-60*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)

________________________________________________________________________________________

Maxima [A]
time = 0.49, size = 211, normalized size = 1.25 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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*sqr
t(c/x) + a)*a^6 + a^7))*c^2

________________________________________________________________________________________

Fricas [A]
time = 0.42, size = 224, normalized size = 1.33 \begin {gather*} \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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[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]

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x \sqrt {a + b \sqrt {\frac {c}{x}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*sqrt(a + b*sqrt(c/x)), x)

________________________________________________________________________________________

Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x\,\sqrt {a+b\,\sqrt {\frac {c}{x}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*(c/x)^(1/2))^(1/2),x)

[Out]

int(x*(a + b*(c/x)^(1/2))^(1/2), x)

________________________________________________________________________________________