3.7.29 \(\int \frac {a+\frac {b}{x^2}}{(c+\frac {d}{x^2})^{3/2} x} \, dx\)

Optimal. Leaf size=52 \[ \frac {b c-a d}{c d \sqrt {c+\frac {d}{x^2}}}+\frac {a \tanh ^{-1}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{c^{3/2}} \]

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Rubi [A]  time = 0.04, antiderivative size = 52, 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 = {446, 78, 63, 208} \begin {gather*} \frac {b c-a d}{c d \sqrt {c+\frac {d}{x^2}}}+\frac {a \tanh ^{-1}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{c^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)/((c + d/x^2)^(3/2)*x),x]

[Out]

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

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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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 446

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

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 73, normalized size = 1.40 \begin {gather*} \frac {\sqrt {c} x (b c-a d)+a d^{3/2} \sqrt {\frac {c x^2}{d}+1} \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {d}}\right )}{c^{3/2} d x \sqrt {c+\frac {d}{x^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)/((c + d/x^2)^(3/2)*x),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.11, size = 73, normalized size = 1.40 \begin {gather*} \frac {a \tanh ^{-1}\left (\frac {\sqrt {\frac {c x^2+d}{x^2}}}{\sqrt {c}}\right )}{c^{3/2}}-\frac {x^2 \sqrt {\frac {c x^2+d}{x^2}} (a d-b c)}{c d \left (c x^2+d\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b/x^2)/((c + d/x^2)^(3/2)*x),x]

[Out]

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

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fricas [B]  time = 0.46, size = 200, normalized size = 3.85 \begin {gather*} \left [\frac {2 \, {\left (b c^{2} - a c d\right )} x^{2} \sqrt {\frac {c x^{2} + d}{x^{2}}} + {\left (a c d x^{2} + a d^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c} x^{2} \sqrt {\frac {c x^{2} + d}{x^{2}}} - d\right )}{2 \, {\left (c^{3} d x^{2} + c^{2} d^{2}\right )}}, \frac {{\left (b c^{2} - a c d\right )} x^{2} \sqrt {\frac {c x^{2} + d}{x^{2}}} - {\left (a c d x^{2} + a d^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x^{2} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right )}{c^{3} d x^{2} + c^{2} d^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)/(c+d/x^2)^(3/2)/x,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)]Unable to divide, perhaps due to rounding error%%%{%%%{-2,[1]%%%},[2,1,2]%%%}+%%%{%%{[4,0]:[1,0,%%
%{-1,[1]%%%}]%%},[1,1,3]%%%}+%%%{-2,[0,1,4]%%%} / %%%{%%%{1,[2]%%%},[2,0,0]%%%}+%%%{%%{[%%%{-2,[1]%%%},0]:[1,0
,%%%{-1,[1]%%%}]%%},[1,0,1]%%%}+%%%{%%%{1,[1]%%%},[0,0,2]%%%} Error: Bad Argument Value

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maple [A]  time = 0.06, size = 75, normalized size = 1.44 \begin {gather*} \frac {\left (c \,x^{2}+d \right ) \left (-a \,c^{\frac {3}{2}} d x +b \,c^{\frac {5}{2}} x +\sqrt {c \,x^{2}+d}\, a c d \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right )\right )}{\left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} c^{\frac {5}{2}} d \,x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.26, size = 69, normalized size = 1.33 \begin {gather*} -\frac {1}{2} \, a {\left (\frac {\log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} - \sqrt {c}}{\sqrt {c + \frac {d}{x^{2}}} + \sqrt {c}}\right )}{c^{\frac {3}{2}}} + \frac {2}{\sqrt {c + \frac {d}{x^{2}}} c}\right )} + \frac {b}{\sqrt {c + \frac {d}{x^{2}}} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.06, size = 54, normalized size = 1.04 \begin {gather*} \frac {a\,\mathrm {atanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{c^{3/2}}-\frac {a}{c\,\sqrt {c+\frac {d}{x^2}}}+\frac {b\,\sqrt {x^2}}{d\,\sqrt {c\,x^2+d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 20.33, size = 49, normalized size = 0.94 \begin {gather*} - \frac {a \operatorname {atan}{\left (\frac {\sqrt {c + \frac {d}{x^{2}}}}{\sqrt {- c}} \right )}}{c \sqrt {- c}} - \frac {a d - b c}{c d \sqrt {c + \frac {d}{x^{2}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a*atan(sqrt(c + d/x**2)/sqrt(-c))/(c*sqrt(-c)) - (a*d - b*c)/(c*d*sqrt(c + d/x**2))

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