3.7.26 \(\int \frac {a+\frac {b}{x^2}}{\sqrt {c+\frac {d}{x^2}} x^4} \, dx\)

Optimal. Leaf size=93 \[ -\frac {c (3 b c-4 a d) \tanh ^{-1}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )}{8 d^{5/2}}+\frac {\sqrt {c+\frac {d}{x^2}} (3 b c-4 a d)}{8 d^2 x}-\frac {b \sqrt {c+\frac {d}{x^2}}}{4 d x^3} \]

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {459, 335, 321, 217, 206} \begin {gather*} \frac {\sqrt {c+\frac {d}{x^2}} (3 b c-4 a d)}{8 d^2 x}-\frac {c (3 b c-4 a d) \tanh ^{-1}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )}{8 d^{5/2}}-\frac {b \sqrt {c+\frac {d}{x^2}}}{4 d x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)/(Sqrt[c + d/x^2]*x^4),x]

[Out]

-(b*Sqrt[c + d/x^2])/(4*d*x^3) + ((3*b*c - 4*a*d)*Sqrt[c + d/x^2])/(8*d^2*x) - (c*(3*b*c - 4*a*d)*ArcTanh[Sqrt
[d]/(Sqrt[c + d/x^2]*x)])/(8*d^(5/2))

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 321

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

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^4} \, dx &=-\frac {b \sqrt {c+\frac {d}{x^2}}}{4 d x^3}+\frac {(-3 b c+4 a d) \int \frac {1}{\sqrt {c+\frac {d}{x^2}} x^4} \, dx}{4 d}\\ &=-\frac {b \sqrt {c+\frac {d}{x^2}}}{4 d x^3}-\frac {(-3 b c+4 a d) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {c+d x^2}} \, dx,x,\frac {1}{x}\right )}{4 d}\\ &=-\frac {b \sqrt {c+\frac {d}{x^2}}}{4 d x^3}+\frac {(3 b c-4 a d) \sqrt {c+\frac {d}{x^2}}}{8 d^2 x}-\frac {(c (3 b c-4 a d)) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c+d x^2}} \, dx,x,\frac {1}{x}\right )}{8 d^2}\\ &=-\frac {b \sqrt {c+\frac {d}{x^2}}}{4 d x^3}+\frac {(3 b c-4 a d) \sqrt {c+\frac {d}{x^2}}}{8 d^2 x}-\frac {(c (3 b c-4 a d)) \operatorname {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {1}{\sqrt {c+\frac {d}{x^2}} x}\right )}{8 d^2}\\ &=-\frac {b \sqrt {c+\frac {d}{x^2}}}{4 d x^3}+\frac {(3 b c-4 a d) \sqrt {c+\frac {d}{x^2}}}{8 d^2 x}-\frac {c (3 b c-4 a d) \tanh ^{-1}\left (\frac {\sqrt {d}}{\sqrt {c+\frac {d}{x^2}} x}\right )}{8 d^{5/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.18, size = 107, normalized size = 1.15 \begin {gather*} -\frac {\left (c x^2+d\right ) \left (d \sqrt {\frac {c x^2}{d}+1} \left (4 a d x^2-3 b c x^2+2 b d\right )+c x^4 (3 b c-4 a d) \tanh ^{-1}\left (\sqrt {\frac {c x^2}{d}+1}\right )\right )}{8 d^3 x^5 \sqrt {c+\frac {d}{x^2}} \sqrt {\frac {c x^2}{d}+1}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)/(Sqrt[c + d/x^2]*x^4),x]

[Out]

-1/8*((d + c*x^2)*(d*Sqrt[1 + (c*x^2)/d]*(2*b*d - 3*b*c*x^2 + 4*a*d*x^2) + c*(3*b*c - 4*a*d)*x^4*ArcTanh[Sqrt[
1 + (c*x^2)/d]]))/(d^3*Sqrt[c + d/x^2]*x^5*Sqrt[1 + (c*x^2)/d])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.20, size = 104, normalized size = 1.12 \begin {gather*} \frac {x \sqrt {c+\frac {d}{x^2}} \left (\frac {\left (4 a c d-3 b c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c x^2+d}}{\sqrt {d}}\right )}{8 d^{5/2}}+\frac {\sqrt {c x^2+d} \left (-4 a d x^2+3 b c x^2-2 b d\right )}{8 d^2 x^4}\right )}{\sqrt {c x^2+d}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b/x^2)/(Sqrt[c + d/x^2]*x^4),x]

[Out]

(Sqrt[c + d/x^2]*x*((Sqrt[d + c*x^2]*(-2*b*d + 3*b*c*x^2 - 4*a*d*x^2))/(8*d^2*x^4) + ((-3*b*c^2 + 4*a*c*d)*Arc
Tanh[Sqrt[d + c*x^2]/Sqrt[d]])/(8*d^(5/2))))/Sqrt[d + c*x^2]

________________________________________________________________________________________

fricas [A]  time = 0.45, size = 201, normalized size = 2.16 \begin {gather*} \left [-\frac {{\left (3 \, b c^{2} - 4 \, a c d\right )} \sqrt {d} x^{3} \log \left (-\frac {c x^{2} + 2 \, \sqrt {d} x \sqrt {\frac {c x^{2} + d}{x^{2}}} + 2 \, d}{x^{2}}\right ) + 2 \, {\left (2 \, b d^{2} - {\left (3 \, b c d - 4 \, a d^{2}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{16 \, d^{3} x^{3}}, \frac {{\left (3 \, b c^{2} - 4 \, a c d\right )} \sqrt {-d} x^{3} \arctan \left (\frac {\sqrt {-d} x \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) - {\left (2 \, b d^{2} - {\left (3 \, b c d - 4 \, a d^{2}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{8 \, d^{3} x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)/x^4/(c+d/x^2)^(1/2),x, algorithm="fricas")

[Out]

[-1/16*((3*b*c^2 - 4*a*c*d)*sqrt(d)*x^3*log(-(c*x^2 + 2*sqrt(d)*x*sqrt((c*x^2 + d)/x^2) + 2*d)/x^2) + 2*(2*b*d
^2 - (3*b*c*d - 4*a*d^2)*x^2)*sqrt((c*x^2 + d)/x^2))/(d^3*x^3), 1/8*((3*b*c^2 - 4*a*c*d)*sqrt(-d)*x^3*arctan(s
qrt(-d)*x*sqrt((c*x^2 + d)/x^2)/(c*x^2 + d)) - (2*b*d^2 - (3*b*c*d - 4*a*d^2)*x^2)*sqrt((c*x^2 + d)/x^2))/(d^3
*x^3)]

________________________________________________________________________________________

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^4/(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

________________________________________________________________________________________

maple [A]  time = 0.06, size = 146, normalized size = 1.57 \begin {gather*} -\frac {\sqrt {c \,x^{2}+d}\, \left (-4 a c \,d^{2} x^{4} \ln \left (\frac {2 d +2 \sqrt {c \,x^{2}+d}\, \sqrt {d}}{x}\right )+3 b \,c^{2} d \,x^{4} \ln \left (\frac {2 d +2 \sqrt {c \,x^{2}+d}\, \sqrt {d}}{x}\right )+4 \sqrt {c \,x^{2}+d}\, a \,d^{\frac {5}{2}} x^{2}-3 \sqrt {c \,x^{2}+d}\, b c \,d^{\frac {3}{2}} x^{2}+2 \sqrt {c \,x^{2}+d}\, b \,d^{\frac {5}{2}}\right )}{8 \sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, d^{\frac {7}{2}} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x^2)/x^4/(c+d/x^2)^(1/2),x)

[Out]

-1/8*(c*x^2+d)^(1/2)*(-4*ln(2*(d+(c*x^2+d)^(1/2)*d^(1/2))/x)*x^4*a*c*d^2+3*ln(2*(d+(c*x^2+d)^(1/2)*d^(1/2))/x)
*x^4*b*c^2*d+4*(c*x^2+d)^(1/2)*d^(5/2)*x^2*a-3*(c*x^2+d)^(1/2)*d^(3/2)*x^2*b*c+2*(c*x^2+d)^(1/2)*d^(5/2)*b)/((
c*x^2+d)/x^2)^(1/2)/x^5/d^(7/2)

________________________________________________________________________________________

maxima [B]  time = 1.23, size = 200, normalized size = 2.15 \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 )} a + \frac {1}{16} \, b {\left (\frac {3 \, c^{2} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} x - \sqrt {d}}{\sqrt {c + \frac {d}{x^{2}}} x + \sqrt {d}}\right )}{d^{\frac {5}{2}}} + \frac {2 \, {\left (3 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c^{2} x^{3} - 5 \, \sqrt {c + \frac {d}{x^{2}}} c^{2} d x\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{2} d^{2} x^{4} - 2 \, {\left (c + \frac {d}{x^{2}}\right )} d^{3} x^{2} + d^{4}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)/x^4/(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))*a + 1/16*b*(3*c^2*log((sqrt(c + d/x^2)*x - sqrt(d))/(sqrt(c + d/x^2)*x + sqrt(d)))/d^(5/
2) + 2*(3*(c + d/x^2)^(3/2)*c^2*x^3 - 5*sqrt(c + d/x^2)*c^2*d*x)/((c + d/x^2)^2*d^2*x^4 - 2*(c + d/x^2)*d^3*x^
2 + d^4))

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+\frac {b}{x^2}}{x^4\,\sqrt {c+\frac {d}{x^2}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x^2)/(x^4*(c + d/x^2)^(1/2)),x)

[Out]

int((a + b/x^2)/(x^4*(c + d/x^2)^(1/2)), x)

________________________________________________________________________________________

sympy [A]  time = 8.46, size = 150, normalized size = 1.61 \begin {gather*} - \frac {a \sqrt {c} \sqrt {1 + \frac {d}{c x^{2}}}}{2 d x} + \frac {a c \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {c} x} \right )}}{2 d^{\frac {3}{2}}} + \frac {3 b c^{\frac {3}{2}}}{8 d^{2} x \sqrt {1 + \frac {d}{c x^{2}}}} + \frac {b \sqrt {c}}{8 d x^{3} \sqrt {1 + \frac {d}{c x^{2}}}} - \frac {3 b c^{2} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {c} x} \right )}}{8 d^{\frac {5}{2}}} - \frac {b}{4 \sqrt {c} x^{5} \sqrt {1 + \frac {d}{c x^{2}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x**2)/x**4/(c+d/x**2)**(1/2),x)

[Out]

-a*sqrt(c)*sqrt(1 + d/(c*x**2))/(2*d*x) + a*c*asinh(sqrt(d)/(sqrt(c)*x))/(2*d**(3/2)) + 3*b*c**(3/2)/(8*d**2*x
*sqrt(1 + d/(c*x**2))) + b*sqrt(c)/(8*d*x**3*sqrt(1 + d/(c*x**2))) - 3*b*c**2*asinh(sqrt(d)/(sqrt(c)*x))/(8*d*
*(5/2)) - b/(4*sqrt(c)*x**5*sqrt(1 + d/(c*x**2)))

________________________________________________________________________________________