3.2934 \(\int \frac{\sqrt{a+b \sqrt{c x^2}}}{x^3} \, dx\)

Optimal. Leaf size=97 \[ \frac{b^2 c \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{c x^2}}}{\sqrt{a}}\right )}{4 a^{3/2}}-\frac{b c \sqrt{a+b \sqrt{c x^2}}}{4 a \sqrt{c x^2}}-\frac{\sqrt{a+b \sqrt{c x^2}}}{2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.0427247, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {368, 47, 51, 63, 208} \[ \frac{b^2 c \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{c x^2}}}{\sqrt{a}}\right )}{4 a^{3/2}}-\frac{b c \sqrt{a+b \sqrt{c x^2}}}{4 a \sqrt{c x^2}}-\frac{\sqrt{a+b \sqrt{c x^2}}}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Sqrt[c*x^2]]/x^3,x]

[Out]

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

Rule 368

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

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 \frac{\sqrt{a+b \sqrt{c x^2}}}{x^3} \, dx &=c \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x^3} \, dx,x,\sqrt{c x^2}\right )\\ &=-\frac{\sqrt{a+b \sqrt{c x^2}}}{2 x^2}+\frac{1}{4} (b c) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x}} \, dx,x,\sqrt{c x^2}\right )\\ &=-\frac{\sqrt{a+b \sqrt{c x^2}}}{2 x^2}-\frac{b c \sqrt{a+b \sqrt{c x^2}}}{4 a \sqrt{c x^2}}-\frac{\left (b^2 c\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\sqrt{c x^2}\right )}{8 a}\\ &=-\frac{\sqrt{a+b \sqrt{c x^2}}}{2 x^2}-\frac{b c \sqrt{a+b \sqrt{c x^2}}}{4 a \sqrt{c x^2}}-\frac{(b c) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sqrt{c x^2}}\right )}{4 a}\\ &=-\frac{\sqrt{a+b \sqrt{c x^2}}}{2 x^2}-\frac{b c \sqrt{a+b \sqrt{c x^2}}}{4 a \sqrt{c x^2}}+\frac{b^2 c \tanh ^{-1}\left (\frac{\sqrt{a+b \sqrt{c x^2}}}{\sqrt{a}}\right )}{4 a^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.0115763, size = 52, normalized size = 0.54 \[ -\frac{2 b^2 c \left (a+b \sqrt{c x^2}\right )^{3/2} \, _2F_1\left (\frac{3}{2},3;\frac{5}{2};\frac{\sqrt{c x^2} b}{a}+1\right )}{3 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*Sqrt[c*x^2]]/x^3,x]

[Out]

(-2*b^2*c*(a + b*Sqrt[c*x^2])^(3/2)*Hypergeometric2F1[3/2, 3, 5/2, 1 + (b*Sqrt[c*x^2])/a])/(3*a^3)

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Maple [A]  time = 0.013, size = 72, normalized size = 0.7 \begin{align*} -{\frac{1}{4\,{x}^{2}} \left ( -{\it Artanh} \left ({\sqrt{a+b\sqrt{c{x}^{2}}}{\frac{1}{\sqrt{a}}}} \right ) c{x}^{2}a{b}^{2}+ \left ( a+b\sqrt{c{x}^{2}} \right ) ^{{\frac{3}{2}}}{a}^{{\frac{3}{2}}}+\sqrt{a+b\sqrt{c{x}^{2}}}{a}^{{\frac{5}{2}}} \right ){a}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.31517, size = 419, normalized size = 4.32 \begin{align*} \left [\frac{\sqrt{a} b^{2} c x^{2} \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 (\sqrt{c x^{2}} a b + 2 \, a^{2}\right )} \sqrt{\sqrt{c x^{2}} b + a}}{8 \, a^{2} x^{2}}, -\frac{\sqrt{-a} b^{2} c x^{2} \arctan \left (\frac{\sqrt{\sqrt{c x^{2}} b + a} \sqrt{-a}}{a}\right ) +{\left (\sqrt{c x^{2}} a b + 2 \, a^{2}\right )} \sqrt{\sqrt{c x^{2}} b + a}}{4 \, a^{2} x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(sqrt(a)*b^2*c*x^2*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*(sqrt(c*x^2)*a*b + 2*a^2)*sqrt(sqrt(c*x^2)*b + a))/(a^2*x^2), -1/4*(sqrt(-a)*b^2*c*x^2*arctan(sqrt(sqrt(c*x
^2)*b + a)*sqrt(-a)/a) + (sqrt(c*x^2)*a*b + 2*a^2)*sqrt(sqrt(c*x^2)*b + a))/(a^2*x^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a + b \sqrt{c x^{2}}}}{x^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a + b*sqrt(c*x**2))/x**3, x)

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Giac [A]  time = 1.22736, size = 122, normalized size = 1.26 \begin{align*} -\frac{\frac{b^{3} c^{\frac{3}{2}} \arctan \left (\frac{\sqrt{b \sqrt{c} x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a} + \frac{{\left (b \sqrt{c} x + a\right )}^{\frac{3}{2}} b^{3} c^{\frac{3}{2}} + \sqrt{b \sqrt{c} x + a} a b^{3} c^{\frac{3}{2}}}{a b^{2} c x^{2}}}{4 \, b \sqrt{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(b^3*c^(3/2)*arctan(sqrt(b*sqrt(c)*x + a)/sqrt(-a))/(sqrt(-a)*a) + ((b*sqrt(c)*x + a)^(3/2)*b^3*c^(3/2) +
sqrt(b*sqrt(c)*x + a)*a*b^3*c^(3/2))/(a*b^2*c*x^2))/(b*sqrt(c))