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

Optimal. Leaf size=105 \[ x \sqrt {a+\frac {b}{x}+\frac {c}{x^2}}+\frac {b \tanh ^{-1}\left (\frac {2 a+\frac {b}{x}}{2 \sqrt {a} \sqrt {a+\frac {b}{x}+\frac {c}{x^2}}}\right )}{2 \sqrt {a}}-\sqrt {c} \tanh ^{-1}\left (\frac {b+\frac {2 c}{x}}{2 \sqrt {c} \sqrt {a+\frac {b}{x}+\frac {c}{x^2}}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {1342, 732, 843, 621, 206, 724} \[ x \sqrt {a+\frac {b}{x}+\frac {c}{x^2}}+\frac {b \tanh ^{-1}\left (\frac {2 a+\frac {b}{x}}{2 \sqrt {a} \sqrt {a+\frac {b}{x}+\frac {c}{x^2}}}\right )}{2 \sqrt {a}}-\sqrt {c} \tanh ^{-1}\left (\frac {b+\frac {2 c}{x}}{2 \sqrt {c} \sqrt {a+\frac {b}{x}+\frac {c}{x^2}}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + c/x^2 + b/x],x]

[Out]

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 732

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 1)), x] - Dist[p/(e*(m + 1)), Int[(d + e*x)^(m + 1)*(b + 2*c*x)*(a + b*x + c*x^2)^
(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ
[2*c*d - b*e, 0] && GtQ[p, 0] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m + 2*p + 1, 0] && IntQua
draticQ[a, b, c, d, e, m, p, x]

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 1342

Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n + c/x^(2*n))^p/x^2,
x], x, 1/x] /; FreeQ[{a, b, c, p}, x] && EqQ[n2, 2*n] && ILtQ[n, 0]

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 128, normalized size = 1.22 \[ \frac {x \sqrt {a+\frac {b x+c}{x^2}} \left (b \tanh ^{-1}\left (\frac {2 a x+b}{2 \sqrt {a} \sqrt {x (a x+b)+c}}\right )+2 \sqrt {a} \left (\sqrt {x (a x+b)+c}-\sqrt {c} \tanh ^{-1}\left (\frac {b x+2 c}{2 \sqrt {c} \sqrt {x (a x+b)+c}}\right )\right )\right )}{2 \sqrt {a} \sqrt {x (a x+b)+c}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + c/x^2 + b/x],x]

[Out]

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

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fricas [A]  time = 1.31, size = 590, normalized size = 5.62 \[ \left [\frac {4 \, a x \sqrt {\frac {a x^{2} + b x + c}{x^{2}}} + \sqrt {a} b \log \left (-8 \, a^{2} x^{2} - 8 \, a b x - b^{2} - 4 \, a c - 4 \, {\left (2 \, a x^{2} + b x\right )} \sqrt {a} \sqrt {\frac {a x^{2} + b x + c}{x^{2}}}\right ) + 2 \, a \sqrt {c} \log \left (-\frac {8 \, b c x + {\left (b^{2} + 4 \, a c\right )} x^{2} + 8 \, c^{2} - 4 \, {\left (b x^{2} + 2 \, c x\right )} \sqrt {c} \sqrt {\frac {a x^{2} + b x + c}{x^{2}}}}{x^{2}}\right )}{4 \, a}, \frac {2 \, a x \sqrt {\frac {a x^{2} + b x + c}{x^{2}}} - \sqrt {-a} b \arctan \left (\frac {{\left (2 \, a x^{2} + b x\right )} \sqrt {-a} \sqrt {\frac {a x^{2} + b x + c}{x^{2}}}}{2 \, {\left (a^{2} x^{2} + a b x + a c\right )}}\right ) + a \sqrt {c} \log \left (-\frac {8 \, b c x + {\left (b^{2} + 4 \, a c\right )} x^{2} + 8 \, c^{2} - 4 \, {\left (b x^{2} + 2 \, c x\right )} \sqrt {c} \sqrt {\frac {a x^{2} + b x + c}{x^{2}}}}{x^{2}}\right )}{2 \, a}, \frac {4 \, a x \sqrt {\frac {a x^{2} + b x + c}{x^{2}}} + 4 \, a \sqrt {-c} \arctan \left (\frac {{\left (b x^{2} + 2 \, c x\right )} \sqrt {-c} \sqrt {\frac {a x^{2} + b x + c}{x^{2}}}}{2 \, {\left (a c x^{2} + b c x + c^{2}\right )}}\right ) + \sqrt {a} b \log \left (-8 \, a^{2} x^{2} - 8 \, a b x - b^{2} - 4 \, a c - 4 \, {\left (2 \, a x^{2} + b x\right )} \sqrt {a} \sqrt {\frac {a x^{2} + b x + c}{x^{2}}}\right )}{4 \, a}, \frac {2 \, a x \sqrt {\frac {a x^{2} + b x + c}{x^{2}}} - \sqrt {-a} b \arctan \left (\frac {{\left (2 \, a x^{2} + b x\right )} \sqrt {-a} \sqrt {\frac {a x^{2} + b x + c}{x^{2}}}}{2 \, {\left (a^{2} x^{2} + a b x + a c\right )}}\right ) + 2 \, a \sqrt {-c} \arctan \left (\frac {{\left (b x^{2} + 2 \, c x\right )} \sqrt {-c} \sqrt {\frac {a x^{2} + b x + c}{x^{2}}}}{2 \, {\left (a c x^{2} + b c x + c^{2}\right )}}\right )}{2 \, a}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(x)]index.cc index_m operator + Error: Bad Argument Value

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maple [A]  time = 0.01, size = 121, normalized size = 1.15 \[ \frac {\sqrt {\frac {a \,x^{2}+b x +c}{x^{2}}}\, \left (-2 \sqrt {a}\, \sqrt {c}\, \ln \left (\frac {b x +2 c +2 \sqrt {a \,x^{2}+b x +c}\, \sqrt {c}}{x}\right )+b \ln \left (\frac {2 a x +b +2 \sqrt {a \,x^{2}+b x +c}\, \sqrt {a}}{2 \sqrt {a}}\right )+2 \sqrt {a \,x^{2}+b x +c}\, \sqrt {a}\right ) x}{2 \sqrt {a \,x^{2}+b x +c}\, \sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a*x^2+b*x+c)/x^2)^(1/2)*x*(-2*c^(1/2)*ln((b*x+2*c+2*(a*x^2+b*x+c)^(1/2)*c^(1/2))/x)*a^(1/2)+b*ln(1/2*(2*
a*x+b+2*(a*x^2+b*x+c)^(1/2)*a^(1/2))/a^(1/2))+2*(a*x^2+b*x+c)^(1/2)*a^(1/2))/(a*x^2+b*x+c)^(1/2)/a^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + \frac {b}{x} + \frac {c}{x^{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a + b/x + c/x^2), x)

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mupad [B]  time = 0.13, size = 100, normalized size = 0.95 \[ x\,\sqrt {\frac {1}{x^2}}\,\sqrt {a\,x^2+b\,x+c}-\sqrt {c}\,x\,\ln \left (\frac {2\,c+2\,\sqrt {c}\,\sqrt {a\,x^2+b\,x+c}+b\,x}{x}\right )\,\sqrt {\frac {1}{x^2}}+\frac {b\,x\,\ln \left (\frac {\frac {b}{2}+\sqrt {a}\,\sqrt {a\,x^2+b\,x+c}+a\,x}{\sqrt {a}}\right )\,\sqrt {\frac {1}{x^2}}}{2\,\sqrt {a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x + c/x^2)^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a + b/x + c/x**2), x)

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