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

Optimal. Leaf size=84 \[ -\frac {\sqrt {c+\frac {d}{x^2}} (a d+2 b c)}{2 c}+\frac {(a d+2 b c) \tanh ^{-1}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{2 \sqrt {c}}+\frac {a x^2 \left (c+\frac {d}{x^2}\right )^{3/2}}{2 c} \]

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Rubi [A]  time = 0.06, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {446, 78, 50, 63, 208} \begin {gather*} -\frac {\sqrt {c+\frac {d}{x^2}} (a d+2 b c)}{2 c}+\frac {(a d+2 b c) \tanh ^{-1}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{2 \sqrt {c}}+\frac {a x^2 \left (c+\frac {d}{x^2}\right )^{3/2}}{2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)*Sqrt[c + d/x^2]*x,x]

[Out]

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

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Mathematica [A]  time = 0.19, size = 71, normalized size = 0.85 \begin {gather*} \frac {1}{2} \sqrt {c+\frac {d}{x^2}} \left (\frac {x (a d+2 b c) \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {d}}\right )}{\sqrt {c} \sqrt {d} \sqrt {\frac {c x^2}{d}+1}}+a x^2-2 b\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.11, size = 68, normalized size = 0.81 \begin {gather*} \frac {1}{2} \left (a x^2-2 b\right ) \sqrt {\frac {c x^2+d}{x^2}}+\frac {(a d+2 b c) \tanh ^{-1}\left (\frac {\sqrt {\frac {c x^2+d}{x^2}}}{\sqrt {c}}\right )}{2 \sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 155, normalized size = 1.85 \begin {gather*} \left [\frac {{\left (2 \, b c + a d\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 (a c x^{2} - 2 \, b c\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{4 \, c}, -\frac {{\left (2 \, b c + a d\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x^{2} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) - {\left (a c x^{2} - 2 \, b c\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{2 \, c}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.29, size = 92, normalized size = 1.10 \begin {gather*} \frac {1}{2} \, \sqrt {c x^{2} + d} a x \mathrm {sgn}\relax (x) + \frac {2 \, b \sqrt {c} d \mathrm {sgn}\relax (x)}{{\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2} - d} - \frac {{\left (2 \, b c^{\frac {3}{2}} \mathrm {sgn}\relax (x) + a \sqrt {c} d \mathrm {sgn}\relax (x)\right )} \log \left ({\left (\sqrt {c} x - \sqrt {c x^{2} + d}\right )}^{2}\right )}{4 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 129, normalized size = 1.54 \begin {gather*} -\frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, \left (-a \,d^{2} x \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right )-2 b c d x \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right )-\sqrt {c \,x^{2}+d}\, a \sqrt {c}\, d \,x^{2}-2 \sqrt {c \,x^{2}+d}\, b \,c^{\frac {3}{2}} x^{2}+2 \left (c \,x^{2}+d \right )^{\frac {3}{2}} b \sqrt {c}\right )}{2 \sqrt {c \,x^{2}+d}\, \sqrt {c}\, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.21, size = 108, normalized size = 1.29 \begin {gather*} \frac {1}{4} \, {\left (2 \, \sqrt {c + \frac {d}{x^{2}}} x^{2} - \frac {d \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} - \sqrt {c}}{\sqrt {c + \frac {d}{x^{2}}} + \sqrt {c}}\right )}{\sqrt {c}}\right )} a - \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 )} b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.10, size = 68, normalized size = 0.81 \begin {gather*} \frac {a\,x^2\,\sqrt {c+\frac {d}{x^2}}}{2}-b\,\sqrt {c+\frac {d}{x^2}}+b\,\sqrt {c}\,\mathrm {atanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )+\frac {a\,d\,\mathrm {atanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{2\,\sqrt {c}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 60.43, size = 107, normalized size = 1.27 \begin {gather*} \frac {a \sqrt {d} x \sqrt {\frac {c x^{2}}{d} + 1}}{2} + \frac {a d \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {d}} \right )}}{2 \sqrt {c}} + b \sqrt {c} \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {d}} \right )} - \frac {b c x}{\sqrt {d} \sqrt {\frac {c x^{2}}{d} + 1}} - \frac {b \sqrt {d}}{x \sqrt {\frac {c x^{2}}{d} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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