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

Optimal. Leaf size=85 \[ -\frac {\sqrt {c+\frac {d}{x^2}} (2 a d+b c)}{2 c x}-\frac {(2 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )}{2 \sqrt {d}}+\frac {a x \left (c+\frac {d}{x^2}\right )^{3/2}}{c} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {375, 453, 195, 217, 206} \[ -\frac {\sqrt {c+\frac {d}{x^2}} (2 a d+b c)}{2 c x}-\frac {(2 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )}{2 \sqrt {d}}+\frac {a x \left (c+\frac {d}{x^2}\right )^{3/2}}{c} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)*Sqrt[c + d/x^2],x]

[Out]

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

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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 375

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Subst[Int[((a + b/x^n)^p*(c +
d/x^n)^q)/x^2, x], x, 1/x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && ILtQ[n, 0]

Rule 453

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

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 75, normalized size = 0.88 \[ \frac {\sqrt {c+\frac {d}{x^2}} \left (-\frac {x^2 (2 a d+b c) \tanh ^{-1}\left (\frac {\sqrt {c x^2+d}}{\sqrt {d}}\right )}{\sqrt {d} \sqrt {c x^2+d}}+2 a x^2-b\right )}{2 x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.72, size = 164, normalized size = 1.93 \[ \left [\frac {{\left (b c + 2 \, a d\right )} \sqrt {d} x \log \left (-\frac {c x^{2} - 2 \, \sqrt {d} x \sqrt {\frac {c x^{2} + d}{x^{2}}} + 2 \, d}{x^{2}}\right ) + 2 \, {\left (2 \, a d x^{2} - b d\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{4 \, d x}, \frac {{\left (b c + 2 \, a d\right )} \sqrt {-d} x \arctan \left (\frac {\sqrt {-d} x \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) + {\left (2 \, a d x^{2} - b d\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{2 \, d x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/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*(2*a*d*x^2 - b*d
)*sqrt((c*x^2 + d)/x^2))/(d*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)) + (2*a*d*x^2 - b*d)*sqrt((c*x^2 + d)/x^2))/(d*x)]

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giac [A]  time = 0.25, size = 76, normalized size = 0.89 \[ \frac {2 \, \sqrt {c x^{2} + d} a c \mathrm {sgn}\relax (x) + \frac {{\left (b c^{2} \mathrm {sgn}\relax (x) + 2 \, a c d \mathrm {sgn}\relax (x)\right )} \arctan \left (\frac {\sqrt {c x^{2} + d}}{\sqrt {-d}}\right )}{\sqrt {-d}} - \frac {\sqrt {c x^{2} + d} b c \mathrm {sgn}\relax (x)}{x^{2}}}{2 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 135, normalized size = 1.59 \[ -\frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, \left (2 a \,d^{\frac {3}{2}} x^{2} \ln \left (\frac {2 d +2 \sqrt {c \,x^{2}+d}\, \sqrt {d}}{x}\right )+b c \sqrt {d}\, x^{2} \ln \left (\frac {2 d +2 \sqrt {c \,x^{2}+d}\, \sqrt {d}}{x}\right )-2 \sqrt {c \,x^{2}+d}\, a d \,x^{2}-\sqrt {c \,x^{2}+d}\, b c \,x^{2}+\left (c \,x^{2}+d \right )^{\frac {3}{2}} b \right )}{2 \sqrt {c \,x^{2}+d}\, d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.22, size = 133, normalized size = 1.56 \[ \frac {1}{2} \, {\left (2 \, \sqrt {c + \frac {d}{x^{2}}} x + \sqrt {d} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} x - \sqrt {d}}{\sqrt {c + \frac {d}{x^{2}}} x + \sqrt {d}}\right )\right )} a - \frac {1}{4} \, {\left (\frac {2 \, \sqrt {c + \frac {d}{x^{2}}} c x}{{\left (c + \frac {d}{x^{2}}\right )} x^{2} - d} - \frac {c \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} x - \sqrt {d}}{\sqrt {c + \frac {d}{x^{2}}} x + \sqrt {d}}\right )}{\sqrt {d}}\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.11, size = 97, normalized size = 1.14 \[ a\,x\,\sqrt {c+\frac {d}{x^2}}-\frac {b\,\sqrt {c+\frac {d}{x^2}}}{2\,x}-\frac {b\,c\,\ln \left (\sqrt {c+\frac {d}{x^2}}+\frac {\sqrt {d}}{x}\right )}{2\,\sqrt {d}}+\frac {a\,\sqrt {d}\,\mathrm {asin}\left (\frac {\sqrt {d}\,1{}\mathrm {i}}{\sqrt {c}\,x}\right )\,\sqrt {c+\frac {d}{x^2}}\,1{}\mathrm {i}}{\sqrt {c}\,\sqrt {\frac {d}{c\,x^2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x^2)*(c + d/x^2)^(1/2),x)

[Out]

a*x*(c + d/x^2)^(1/2) - (b*(c + d/x^2)^(1/2))/(2*x) - (b*c*log((c + d/x^2)^(1/2) + d^(1/2)/x))/(2*d^(1/2)) + (
a*d^(1/2)*asin((d^(1/2)*1i)/(c^(1/2)*x))*(c + d/x^2)^(1/2)*1i)/(c^(1/2)*(d/(c*x^2) + 1)^(1/2))

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sympy [A]  time = 4.48, size = 107, normalized size = 1.26 \[ \frac {a \sqrt {c} x}{\sqrt {1 + \frac {d}{c x^{2}}}} - a \sqrt {d} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {c} x} \right )} + \frac {a d}{\sqrt {c} x \sqrt {1 + \frac {d}{c x^{2}}}} - \frac {b \sqrt {c} \sqrt {1 + \frac {d}{c x^{2}}}}{2 x} - \frac {b c \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {c} x} \right )}}{2 \sqrt {d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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