3.2982 \(\int \sqrt{a+b \sqrt{\frac{c}{x}}} x \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.101512, 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 = {369, 266, 47, 51, 63, 208} \[ \frac{5 b^3 c^2 \sqrt{a+b \sqrt{\frac{c}{x}}}}{32 a^3 \sqrt{\frac{c}{x}}}-\frac{5 b^4 c^2 \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{\frac{c}{x}}}}{\sqrt{a}}\right )}{32 a^{7/2}}-\frac{5 b^2 c x \sqrt{a+b \sqrt{\frac{c}{x}}}}{48 a^2}+\frac{b x^3 \left (\frac{c}{x}\right )^{3/2} \sqrt{a+b \sqrt{\frac{c}{x}}}}{12 a c}+\frac{1}{2} x^2 \sqrt{a+b \sqrt{\frac{c}{x}}} \]

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 369

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]

Rule 266

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 47

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},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 51

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] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \sqrt{a+b \sqrt{\frac{c}{x}}} x \, dx &=\operatorname{Subst}\left (\int \sqrt{a+\frac{b \sqrt{c}}{\sqrt{x}}} x \, dx,\sqrt{x},\frac{\sqrt{\frac{c}{x}} x}{\sqrt{c}}\right )\\ &=-\operatorname{Subst}\left (2 \operatorname{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-\operatorname{Subst}\left (\frac{1}{4} \left (b \sqrt{c}\right ) \operatorname{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}+\operatorname{Subst}\left (\frac{\left (5 b^2 c\right ) \operatorname{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}-\operatorname{Subst}\left (\frac{\left (5 b^3 c^{3/2}\right ) \operatorname{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}+\operatorname{Subst}\left (\frac{\left (5 b^4 c^2\right ) \operatorname{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}+\operatorname{Subst}\left (\frac{\left (5 b^3 c^{3/2}\right ) \operatorname{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 = 0.128274, size = 111, normalized size = 0.66 \[ \frac{\sqrt{a} x \sqrt{a+b \sqrt{\frac{c}{x}}} \left (8 a^2 b x \sqrt{\frac{c}{x}}+48 a^3 x-10 a b^2 c+15 b^3 c \sqrt{\frac{c}{x}}\right )-15 b^4 c^2 \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{\frac{c}{x}}}}{\sqrt{a}}\right )}{96 a^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*Sqrt[c/x]]*x,x]

[Out]

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

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Maple [A]  time = 0.028, size = 211, normalized size = 1.3 \begin{align*} -{\frac{1}{192}\sqrt{a+b\sqrt{{\frac{c}{x}}}}\sqrt{x} \left ( 15\,\ln \left ( 1/2\,{\frac{1}{\sqrt{a}} \left ( b\sqrt{{\frac{c}{x}}}\sqrt{x}+2\,\sqrt{ax+b\sqrt{{\frac{c}{x}}}x}\sqrt{a}+2\,a\sqrt{x} \right ) } \right ){c}^{2}a{b}^{4}-30\,{a}^{3/2}\sqrt{ax+b\sqrt{{\frac{c}{x}}}x} \left ({\frac{c}{x}} \right ) ^{3/2}{x}^{3/2}{b}^{3}-60\,{a}^{5/2}\sqrt{ax+b\sqrt{{\frac{c}{x}}}x}c\sqrt{x}{b}^{2}+80\,{a}^{5/2} \left ( ax+b\sqrt{{\frac{c}{x}}}x \right ) ^{3/2}\sqrt{{\frac{c}{x}}}\sqrt{x}b-96\,\sqrt{x} \left ( ax+b\sqrt{{\frac{c}{x}}}x \right ) ^{3/2}{a}^{7/2} \right ){\frac{1}{\sqrt{x \left ( a+b\sqrt{{\frac{c}{x}}} \right ) }}}{a}^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.51508, size = 506, normalized size = 2.99 \begin{align*} \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{align*}

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]

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

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)

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out