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

Optimal. Leaf size=61 \[ \frac {(b c-2 a d) \tanh ^{-1}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )}{2 d^{3/2}}-\frac {b \sqrt {c+\frac {d}{x^2}}}{2 d x} \]

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Rubi [A]  time = 0.04, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {459, 335, 217, 206} \begin {gather*} \frac {(b c-2 a d) \tanh ^{-1}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )}{2 d^{3/2}}-\frac {b \sqrt {c+\frac {d}{x^2}}}{2 d x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)/(Sqrt[c + d/x^2]*x^2),x]

[Out]

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

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 217

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

Rule 335

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

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 80, normalized size = 1.31 \begin {gather*} \frac {x^2 \sqrt {c x^2+d} (b c-2 a d) \tanh ^{-1}\left (\frac {\sqrt {c x^2+d}}{\sqrt {d}}\right )-b \sqrt {d} \left (c x^2+d\right )}{2 d^{3/2} x^3 \sqrt {c+\frac {d}{x^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)/(Sqrt[c + d/x^2]*x^2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.12, size = 82, normalized size = 1.34 \begin {gather*} \frac {x \sqrt {c+\frac {d}{x^2}} \left (\frac {(b c-2 a d) \tanh ^{-1}\left (\frac {\sqrt {c x^2+d}}{\sqrt {d}}\right )}{2 d^{3/2}}-\frac {b \sqrt {c x^2+d}}{2 d x^2}\right )}{\sqrt {c x^2+d}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b/x^2)/(Sqrt[c + d/x^2]*x^2),x]

[Out]

(Sqrt[c + d/x^2]*x*(-1/2*(b*Sqrt[d + c*x^2])/(d*x^2) + ((b*c - 2*a*d)*ArcTanh[Sqrt[d + c*x^2]/Sqrt[d]])/(2*d^(
3/2))))/Sqrt[d + c*x^2]

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fricas [A]  time = 0.42, size = 144, normalized size = 2.36 \begin {gather*} \left [-\frac {{\left (b c - 2 \, a d\right )} \sqrt {d} x \log \left (-\frac {c x^{2} - 2 \, \sqrt {d} x \sqrt {\frac {c x^{2} + d}{x^{2}}} + 2 \, d}{x^{2}}\right ) + 2 \, b d \sqrt {\frac {c x^{2} + d}{x^{2}}}}{4 \, d^{2} x}, -\frac {{\left (b c - 2 \, a d\right )} \sqrt {-d} x \arctan \left (\frac {\sqrt {-d} x \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) + b d \sqrt {\frac {c x^{2} + d}{x^{2}}}}{2 \, d^{2} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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((a+b/x^2)/x^2/(c+d/x^2)^(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(t_nostep)]Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is
real):Check [sign(t_nostep),sign(t_nostep+sqrt(d)/c*sign(t_nostep))]sym2poly/r2sym(const gen & e,const index_
m & i,const vecteur & l) Error: Bad Argument Value

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maple [B]  time = 0.06, size = 105, normalized size = 1.72 \begin {gather*} -\frac {\sqrt {c \,x^{2}+d}\, \left (2 a \,d^{2} x^{2} \ln \left (\frac {2 d +2 \sqrt {c \,x^{2}+d}\, \sqrt {d}}{x}\right )-b c d \,x^{2} \ln \left (\frac {2 d +2 \sqrt {c \,x^{2}+d}\, \sqrt {d}}{x}\right )+\sqrt {c \,x^{2}+d}\, b \,d^{\frac {3}{2}}\right )}{2 \sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, d^{\frac {5}{2}} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(c*x^2+d)^(1/2)*(2*a*ln(2*(d+(c*x^2+d)^(1/2)*d^(1/2))/x)*x^2*d^2-ln(2*(d+(c*x^2+d)^(1/2)*d^(1/2))/x)*x^2*
b*c*d+(c*x^2+d)^(1/2)*d^(3/2)*b)/((c*x^2+d)/x^2)^(1/2)/x^3/d^(5/2)

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maxima [B]  time = 1.18, size = 121, normalized size = 1.98 \begin {gather*} -\frac {1}{4} \, {\left (\frac {2 \, \sqrt {c + \frac {d}{x^{2}}} c x}{{\left (c + \frac {d}{x^{2}}\right )} d x^{2} - d^{2}} + \frac {c \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} x - \sqrt {d}}{\sqrt {c + \frac {d}{x^{2}}} x + \sqrt {d}}\right )}{d^{\frac {3}{2}}}\right )} b + \frac {a \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} x - \sqrt {d}}{\sqrt {c + \frac {d}{x^{2}}} x + \sqrt {d}}\right )}{2 \, \sqrt {d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.53, size = 94, normalized size = 1.54 \begin {gather*} \left \{\begin {array}{cl} -\frac {3\,a\,x^2+b}{3\,\sqrt {c}\,x^3} & \text {\ if\ \ }d=0\\ \frac {b\,c\,\ln \left (2\,\sqrt {c+\frac {d}{x^2}}+\frac {2\,\sqrt {d}}{x}\right )}{2\,d^{3/2}}-\frac {b\,\sqrt {c+\frac {d}{x^2}}}{2\,d\,x}-\frac {a\,\ln \left (\sqrt {c+\frac {d}{x^2}}+\frac {\sqrt {d}}{x}\right )}{\sqrt {d}} & \text {\ if\ \ }d\neq 0 \end {array}\right . \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

piecewise(d == 0, -(b + 3*a*x^2)/(3*c^(1/2)*x^3), d ~= 0, - (a*log((c + d/x^2)^(1/2) + d^(1/2)/x))/d^(1/2) - (
b*(c + d/x^2)^(1/2))/(2*d*x) + (b*c*log(2*(c + d/x^2)^(1/2) + (2*d^(1/2))/x))/(2*d^(3/2)))

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sympy [A]  time = 4.92, size = 66, normalized size = 1.08 \begin {gather*} - \frac {a \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {c} x} \right )}}{\sqrt {d}} - \frac {b \sqrt {c} \sqrt {1 + \frac {d}{c x^{2}}}}{2 d x} + \frac {b c \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {c} x} \right )}}{2 d^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*asinh(sqrt(d)/(sqrt(c)*x))/sqrt(d) - b*sqrt(c)*sqrt(1 + d/(c*x**2))/(2*d*x) + b*c*asinh(sqrt(d)/(sqrt(c)*x)
)/(2*d**(3/2))

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