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

Optimal. Leaf size=59 \[ -a \sqrt {c+\frac {d}{x^2}}+a \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )-\frac {b \left (c+\frac {d}{x^2}\right )^{3/2}}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 59, 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 = {446, 80, 50, 63, 208} \[ -a \sqrt {c+\frac {d}{x^2}}+a \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )-\frac {b \left (c+\frac {d}{x^2}\right )^{3/2}}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 80

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

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

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Mathematica [A]  time = 0.15, size = 82, normalized size = 1.39 \[ \frac {\sqrt {c+\frac {d}{x^2}} \left (\frac {3 a \sqrt {c} \sqrt {d} x^3 \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {d}}\right )}{\sqrt {\frac {c x^2}{d}+1}}-3 a d x^2-b \left (c x^2+d\right )\right )}{3 d x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.90, size = 166, normalized size = 2.81 \[ \left [\frac {3 \, a \sqrt {c} d x^{2} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c} x^{2} \sqrt {\frac {c x^{2} + d}{x^{2}}} - d\right ) - 2 \, {\left ({\left (b c + 3 \, a d\right )} x^{2} + b d\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{6 \, d x^{2}}, -\frac {3 \, a \sqrt {-c} d x^{2} \arctan \left (\frac {\sqrt {-c} x^{2} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) + {\left ({\left (b c + 3 \, a d\right )} x^{2} + b d\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{3 \, d x^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*a*sqrt(c)*d*x^2*log(-2*c*x^2 - 2*sqrt(c)*x^2*sqrt((c*x^2 + d)/x^2) - d) - 2*((b*c + 3*a*d)*x^2 + b*d)*
sqrt((c*x^2 + d)/x^2))/(d*x^2), -1/3*(3*a*sqrt(-c)*d*x^2*arctan(sqrt(-c)*x^2*sqrt((c*x^2 + d)/x^2)/(c*x^2 + d)
) + ((b*c + 3*a*d)*x^2 + b*d)*sqrt((c*x^2 + d)/x^2))/(d*x^2)]

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giac [B]  time = 0.49, size = 163, normalized size = 2.76 \[ -\frac {1}{2} \, a \sqrt {c} \log \left ({\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2}\right ) \mathrm {sgn}\relax (x) + \frac {2 \, {\left (3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{4} b c^{\frac {3}{2}} \mathrm {sgn}\relax (x) + 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{4} a \sqrt {c} d \mathrm {sgn}\relax (x) - 6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} a \sqrt {c} d^{2} \mathrm {sgn}\relax (x) + b c^{\frac {3}{2}} d^{2} \mathrm {sgn}\relax (x) + 3 \, a \sqrt {c} d^{3} \mathrm {sgn}\relax (x)\right )}}{3 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} - d\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a*sqrt(c)*log((sqrt(c)*x - sqrt(c*x^2 + d))^2)*sgn(x) + 2/3*(3*(sqrt(c)*x - sqrt(c*x^2 + d))^4*b*c^(3/2)*
sgn(x) + 3*(sqrt(c)*x - sqrt(c*x^2 + d))^4*a*sqrt(c)*d*sgn(x) - 6*(sqrt(c)*x - sqrt(c*x^2 + d))^2*a*sqrt(c)*d^
2*sgn(x) + b*c^(3/2)*d^2*sgn(x) + 3*a*sqrt(c)*d^3*sgn(x))/((sqrt(c)*x - sqrt(c*x^2 + d))^2 - d)^3

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maple [B]  time = 0.06, size = 109, normalized size = 1.85 \[ -\frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, \left (-3 a c d \,x^{3} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right )-3 \sqrt {c \,x^{2}+d}\, a \,c^{\frac {3}{2}} x^{4}+3 \left (c \,x^{2}+d \right )^{\frac {3}{2}} a \sqrt {c}\, x^{2}+\left (c \,x^{2}+d \right )^{\frac {3}{2}} b \sqrt {c}\right )}{3 \sqrt {c \,x^{2}+d}\, \sqrt {c}\, d \,x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.23, size = 67, normalized size = 1.14 \[ -\frac {1}{2} \, {\left (\sqrt {c} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} - \sqrt {c}}{\sqrt {c + \frac {d}{x^{2}}} + \sqrt {c}}\right ) + 2 \, \sqrt {c + \frac {d}{x^{2}}}\right )} a - \frac {b {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.21, size = 57, normalized size = 0.97 \[ a\,\sqrt {c}\,\mathrm {atanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )-a\,\sqrt {c+\frac {d}{x^2}}-\frac {b\,\sqrt {c+\frac {d}{x^2}}\,\left (c\,x^2+d\right )}{3\,d\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 23.44, size = 75, normalized size = 1.27 \[ \frac {a \left (- \frac {2 c \operatorname {atan}{\left (\frac {\sqrt {c + \frac {d}{x^{2}}}}{\sqrt {- c}} \right )}}{\sqrt {- c}} - 2 \sqrt {c + \frac {d}{x^{2}}}\right )}{2} + \frac {b \left (\begin {cases} - \frac {\sqrt {c}}{x^{2}} & \text {for}\: d = 0 \\- \frac {2 \left (c + \frac {d}{x^{2}}\right )^{\frac {3}{2}}}{3 d} & \text {otherwise} \end {cases}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(-2*c*atan(sqrt(c + d/x**2)/sqrt(-c))/sqrt(-c) - 2*sqrt(c + d/x**2))/2 + b*Piecewise((-sqrt(c)/x**2, Eq(d, 0
)), (-2*(c + d/x**2)**(3/2)/(3*d), True))/2

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