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

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

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Rubi [A]  time = 0.04, antiderivative size = 97, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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 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]

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)]

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*}

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Mathematica [C]  time = 0.02, size = 52, normalized size = 0.54 \begin {gather*} -\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} \end {gather*}

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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IntegrateAlgebraic [B]  time = 1.64, size = 200, normalized size = 2.06 \begin {gather*} \frac {\frac {b^2 c^{3/2} x \sqrt {c x^2} \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {c x^2}}}{\sqrt {a}}\right )}{2 a^{3/2}}+\frac {b^2 c^2 x^2 \tanh ^{-1}\left (\frac {\sqrt {a+b \sqrt {c x^2}}}{\sqrt {a}}\right )}{2 a^{3/2}}-\frac {b c^{3/2} x \sqrt {a+b \sqrt {c x^2}}}{2 a}-\frac {b c \sqrt {c x^2} \sqrt {a+b \sqrt {c x^2}}}{2 a}-2 c \sqrt {a+b \sqrt {c x^2}}}{\left (\sqrt {c x^2}+\sqrt {c} x\right )^2} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [A]  time = 0.58, size = 174, normalized size = 1.79 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.17, size = 90, normalized size = 0.93 \begin {gather*} -\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 {gather*}

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))

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maple [A]  time = 0.01, size = 72, normalized size = 0.74 \begin {gather*} -\frac {-a \,b^{2} c \,x^{2} \arctanh \left (\frac {\sqrt {a +\sqrt {c \,x^{2}}\, b}}{\sqrt {a}}\right )+\sqrt {a +\sqrt {c \,x^{2}}\, b}\, a^{\frac {5}{2}}+\left (a +\sqrt {c \,x^{2}}\, b \right )^{\frac {3}{2}} a^{\frac {3}{2}}}{4 a^{\frac {5}{2}} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.21, size = 126, normalized size = 1.30 \begin {gather*} -\frac {1}{8} \, {\left (\frac {b^{2} \log \left (\frac {\sqrt {\sqrt {c x^{2}} b + a} - \sqrt {a}}{\sqrt {\sqrt {c x^{2}} b + a} + \sqrt {a}}\right )}{a^{\frac {3}{2}}} + \frac {2 \, {\left ({\left (\sqrt {c x^{2}} b + a\right )}^{\frac {3}{2}} b^{2} + \sqrt {\sqrt {c x^{2}} b + a} a b^{2}\right )}}{{\left (\sqrt {c x^{2}} b + a\right )}^{2} a - 2 \, {\left (\sqrt {c x^{2}} b + a\right )} a^{2} + a^{3}}\right )} c \end {gather*}

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]

-1/8*(b^2*log((sqrt(sqrt(c*x^2)*b + a) - sqrt(a))/(sqrt(sqrt(c*x^2)*b + a) + sqrt(a)))/a^(3/2) + 2*((sqrt(c*x^
2)*b + a)^(3/2)*b^2 + sqrt(sqrt(c*x^2)*b + a)*a*b^2)/((sqrt(c*x^2)*b + a)^2*a - 2*(sqrt(c*x^2)*b + a)*a^2 + a^
3))*c

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {a+b\,\sqrt {c\,x^2}}}{x^3} \,d x \end {gather*}

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]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {a + b \sqrt {c x^{2}}}}{x^{3}}\, dx \end {gather*}

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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