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

Optimal. Leaf size=91 \[ \frac {c (b c-4 a d) \tanh ^{-1}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )}{8 d^{3/2}}+\frac {\sqrt {c+\frac {d}{x^2}} (b c-4 a d)}{8 d x}-\frac {b \left (c+\frac {d}{x^2}\right )^{3/2}}{4 d x} \]

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Rubi [A]  time = 0.05, antiderivative size = 91, 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, 195, 217, 206} \begin {gather*} \frac {c (b c-4 a d) \tanh ^{-1}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )}{8 d^{3/2}}+\frac {\sqrt {c+\frac {d}{x^2}} (b c-4 a d)}{8 d x}-\frac {b \left (c+\frac {d}{x^2}\right )^{3/2}}{4 d x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.19, size = 100, normalized size = 1.10 \begin {gather*} -\frac {\sqrt {c+\frac {d}{x^2}} \left (\left (c x^2+d\right ) \left (4 a d x^2+b c x^2+2 b d\right )+c x^4 \sqrt {\frac {c x^2}{d}+1} (4 a d-b c) \tanh ^{-1}\left (\sqrt {\frac {c x^2}{d}+1}\right )\right )}{8 d x^3 \left (c x^2+d\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.18, size = 103, normalized size = 1.13 \begin {gather*} \frac {x \sqrt {c+\frac {d}{x^2}} \left (\frac {\left (b c^2-4 a c d\right ) \tanh ^{-1}\left (\frac {\sqrt {c x^2+d}}{\sqrt {d}}\right )}{8 d^{3/2}}+\frac {\sqrt {c x^2+d} \left (-4 a d x^2-b c x^2-2 b d\right )}{8 d 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^2,x]

[Out]

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

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fricas [A]  time = 0.44, size = 194, normalized size = 2.13 \begin {gather*} \left [-\frac {{\left (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 (b c d + 4 \, a d^{2}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{16 \, d^{2} x^{3}}, -\frac {{\left (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 (b c d + 4 \, a d^{2}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{8 \, d^{2} x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.24, size = 130, normalized size = 1.43 \begin {gather*} -\frac {\frac {{\left (b c^{3} \mathrm {sgn}\relax (x) - 4 \, a c^{2} d \mathrm {sgn}\relax (x)\right )} \arctan \left (\frac {\sqrt {c x^{2} + d}}{\sqrt {-d}}\right )}{\sqrt {-d} d} + \frac {{\left (c x^{2} + d\right )}^{\frac {3}{2}} b c^{3} \mathrm {sgn}\relax (x) + 4 \, {\left (c x^{2} + d\right )}^{\frac {3}{2}} a c^{2} d \mathrm {sgn}\relax (x) + \sqrt {c x^{2} + d} b c^{3} d \mathrm {sgn}\relax (x) - 4 \, \sqrt {c x^{2} + d} a c^{2} d^{2} \mathrm {sgn}\relax (x)}{c^{2} d x^{4}}}{8 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*((b*c^3*sgn(x) - 4*a*c^2*d*sgn(x))*arctan(sqrt(c*x^2 + d)/sqrt(-d))/(sqrt(-d)*d) + ((c*x^2 + d)^(3/2)*b*c
^3*sgn(x) + 4*(c*x^2 + d)^(3/2)*a*c^2*d*sgn(x) + sqrt(c*x^2 + d)*b*c^3*d*sgn(x) - 4*sqrt(c*x^2 + d)*a*c^2*d^2*
sgn(x))/(c^2*d*x^4))/c

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maple [B]  time = 0.06, size = 175, normalized size = 1.92 \begin {gather*} -\frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, \left (4 a c \,d^{\frac {3}{2}} x^{4} \ln \left (\frac {2 d +2 \sqrt {c \,x^{2}+d}\, \sqrt {d}}{x}\right )-b \,c^{2} \sqrt {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 c d \,x^{4}+\sqrt {c \,x^{2}+d}\, b \,c^{2} x^{4}+4 \left (c \,x^{2}+d \right )^{\frac {3}{2}} a d \,x^{2}-\left (c \,x^{2}+d \right )^{\frac {3}{2}} b c \,x^{2}+2 \left (c \,x^{2}+d \right )^{\frac {3}{2}} b d \right )}{8 \sqrt {c \,x^{2}+d}\, d^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.35, size = 193, normalized size = 2.12 \begin {gather*} -\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 )} a - \frac {1}{16} \, {\left (\frac {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 {3}{2}}} + \frac {2 \, {\left ({\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c^{2} x^{3} + \sqrt {c + \frac {d}{x^{2}}} c^{2} d x\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{2} d x^{4} - 2 \, {\left (c + \frac {d}{x^{2}}\right )} d^{2} x^{2} + d^{3}}\right )} b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 6.97, size = 144, normalized size = 1.58 \begin {gather*} - \frac {a \sqrt {c} \sqrt {1 + \frac {d}{c x^{2}}}}{2 x} - \frac {a c \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {c} x} \right )}}{2 \sqrt {d}} - \frac {b c^{\frac {3}{2}}}{8 d x \sqrt {1 + \frac {d}{c x^{2}}}} - \frac {3 b \sqrt {c}}{8 x^{3} \sqrt {1 + \frac {d}{c x^{2}}}} + \frac {b c^{2} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {c} x} \right )}}{8 d^{\frac {3}{2}}} - \frac {b d}{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)*(c+d/x**2)**(1/2)/x**2,x)

[Out]

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

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