3.6.98 \(\int \frac {(a+\frac {b}{x^2}) \sqrt {c+\frac {d}{x^2}}}{x^4} \, dx\)

Optimal. Leaf size=123 \[ -\frac {c^2 (b c-2 a d) \tanh ^{-1}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )}{16 d^{5/2}}+\frac {c \sqrt {c+\frac {d}{x^2}} (b c-2 a d)}{16 d^2 x}+\frac {\sqrt {c+\frac {d}{x^2}} (b c-2 a d)}{8 d x^3}-\frac {b \left (c+\frac {d}{x^2}\right )^{3/2}}{6 d x^3} \]

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Rubi [A]  time = 0.08, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {459, 335, 279, 321, 217, 206} \begin {gather*} -\frac {c^2 (b c-2 a d) \tanh ^{-1}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )}{16 d^{5/2}}+\frac {c \sqrt {c+\frac {d}{x^2}} (b c-2 a d)}{16 d^2 x}+\frac {\sqrt {c+\frac {d}{x^2}} (b c-2 a d)}{8 d x^3}-\frac {b \left (c+\frac {d}{x^2}\right )^{3/2}}{6 d x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b/x^2)*Sqrt[c + d/x^2])/x^4,x]

[Out]

((b*c - 2*a*d)*Sqrt[c + d/x^2])/(8*d*x^3) - (b*(c + d/x^2)^(3/2))/(6*d*x^3) + (c*(b*c - 2*a*d)*Sqrt[c + d/x^2]
)/(16*d^2*x) - (c^2*(b*c - 2*a*d)*ArcTanh[Sqrt[d]/(Sqrt[c + d/x^2]*x)])/(16*d^(5/2))

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 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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^4} \, dx &=-\frac {b \left (c+\frac {d}{x^2}\right )^{3/2}}{6 d x^3}+\frac {(-3 b c+6 a d) \int \frac {\sqrt {c+\frac {d}{x^2}}}{x^4} \, dx}{6 d}\\ &=-\frac {b \left (c+\frac {d}{x^2}\right )^{3/2}}{6 d x^3}-\frac {(-3 b c+6 a d) \operatorname {Subst}\left (\int x^2 \sqrt {c+d x^2} \, dx,x,\frac {1}{x}\right )}{6 d}\\ &=\frac {(b c-2 a d) \sqrt {c+\frac {d}{x^2}}}{8 d x^3}-\frac {b \left (c+\frac {d}{x^2}\right )^{3/2}}{6 d x^3}+\frac {(c (b c-2 a d)) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {c+d x^2}} \, dx,x,\frac {1}{x}\right )}{8 d}\\ &=\frac {(b c-2 a d) \sqrt {c+\frac {d}{x^2}}}{8 d x^3}-\frac {b \left (c+\frac {d}{x^2}\right )^{3/2}}{6 d x^3}+\frac {c (b c-2 a d) \sqrt {c+\frac {d}{x^2}}}{16 d^2 x}-\frac {\left (c^2 (b c-2 a d)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c+d x^2}} \, dx,x,\frac {1}{x}\right )}{16 d^2}\\ &=\frac {(b c-2 a d) \sqrt {c+\frac {d}{x^2}}}{8 d x^3}-\frac {b \left (c+\frac {d}{x^2}\right )^{3/2}}{6 d x^3}+\frac {c (b c-2 a d) \sqrt {c+\frac {d}{x^2}}}{16 d^2 x}-\frac {\left (c^2 (b c-2 a d)\right ) \operatorname {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {1}{\sqrt {c+\frac {d}{x^2}} x}\right )}{16 d^2}\\ &=\frac {(b c-2 a d) \sqrt {c+\frac {d}{x^2}}}{8 d x^3}-\frac {b \left (c+\frac {d}{x^2}\right )^{3/2}}{6 d x^3}+\frac {c (b c-2 a d) \sqrt {c+\frac {d}{x^2}}}{16 d^2 x}-\frac {c^2 (b c-2 a d) \tanh ^{-1}\left (\frac {\sqrt {d}}{\sqrt {c+\frac {d}{x^2}} x}\right )}{16 d^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 68, normalized size = 0.55 \begin {gather*} \frac {\sqrt {c+\frac {d}{x^2}} \left (c x^2+d\right ) \left (c^2 x^6 (b c-2 a d) \, _2F_1\left (\frac {3}{2},3;\frac {5}{2};\frac {c x^2}{d}+1\right )-b d^3\right )}{6 d^4 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b/x^2)*Sqrt[c + d/x^2])/x^4,x]

[Out]

(Sqrt[c + d/x^2]*(d + c*x^2)*(-(b*d^3) + c^2*(b*c - 2*a*d)*x^6*Hypergeometric2F1[3/2, 3, 5/2, 1 + (c*x^2)/d]))
/(6*d^4*x^5)

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IntegrateAlgebraic [A]  time = 0.24, size = 128, normalized size = 1.04 \begin {gather*} \frac {x \sqrt {c+\frac {d}{x^2}} \left (\frac {\sqrt {c x^2+d} \left (-6 a c d x^4-12 a d^2 x^2+3 b c^2 x^4-2 b c d x^2-8 b d^2\right )}{48 d^2 x^6}+\frac {\left (2 a c^2 d-b c^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c x^2+d}}{\sqrt {d}}\right )}{16 d^{5/2}}\right )}{\sqrt {c x^2+d}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((a + b/x^2)*Sqrt[c + d/x^2])/x^4,x]

[Out]

(Sqrt[c + d/x^2]*x*((Sqrt[d + c*x^2]*(-8*b*d^2 - 2*b*c*d*x^2 - 12*a*d^2*x^2 + 3*b*c^2*x^4 - 6*a*c*d*x^4))/(48*
d^2*x^6) + ((-(b*c^3) + 2*a*c^2*d)*ArcTanh[Sqrt[d + c*x^2]/Sqrt[d]])/(16*d^(5/2))))/Sqrt[d + c*x^2]

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fricas [A]  time = 0.46, size = 244, normalized size = 1.98 \begin {gather*} \left [-\frac {3 \, {\left (b c^{3} - 2 \, a c^{2} d\right )} \sqrt {d} x^{5} \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 (3 \, {\left (b c^{2} d - 2 \, a c d^{2}\right )} x^{4} - 8 \, b d^{3} - 2 \, {\left (b c d^{2} + 6 \, a d^{3}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{96 \, d^{3} x^{5}}, \frac {3 \, {\left (b c^{3} - 2 \, a c^{2} d\right )} \sqrt {-d} x^{5} \arctan \left (\frac {\sqrt {-d} x \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) + {\left (3 \, {\left (b c^{2} d - 2 \, a c d^{2}\right )} x^{4} - 8 \, b d^{3} - 2 \, {\left (b c d^{2} + 6 \, a d^{3}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{48 \, d^{3} x^{5}}\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^4,x, algorithm="fricas")

[Out]

[-1/96*(3*(b*c^3 - 2*a*c^2*d)*sqrt(d)*x^5*log(-(c*x^2 + 2*sqrt(d)*x*sqrt((c*x^2 + d)/x^2) + 2*d)/x^2) - 2*(3*(
b*c^2*d - 2*a*c*d^2)*x^4 - 8*b*d^3 - 2*(b*c*d^2 + 6*a*d^3)*x^2)*sqrt((c*x^2 + d)/x^2))/(d^3*x^5), 1/48*(3*(b*c
^3 - 2*a*c^2*d)*sqrt(-d)*x^5*arctan(sqrt(-d)*x*sqrt((c*x^2 + d)/x^2)/(c*x^2 + d)) + (3*(b*c^2*d - 2*a*c*d^2)*x
^4 - 8*b*d^3 - 2*(b*c*d^2 + 6*a*d^3)*x^2)*sqrt((c*x^2 + d)/x^2))/(d^3*x^5)]

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giac [A]  time = 0.30, size = 153, normalized size = 1.24 \begin {gather*} \frac {\frac {3 \, {\left (b c^{4} \mathrm {sgn}\relax (x) - 2 \, a c^{3} d \mathrm {sgn}\relax (x)\right )} \arctan \left (\frac {\sqrt {c x^{2} + d}}{\sqrt {-d}}\right )}{\sqrt {-d} d^{2}} + \frac {3 \, {\left (c x^{2} + d\right )}^{\frac {5}{2}} b c^{4} \mathrm {sgn}\relax (x) - 6 \, {\left (c x^{2} + d\right )}^{\frac {5}{2}} a c^{3} d \mathrm {sgn}\relax (x) - 8 \, {\left (c x^{2} + d\right )}^{\frac {3}{2}} b c^{4} d \mathrm {sgn}\relax (x) - 3 \, \sqrt {c x^{2} + d} b c^{4} d^{2} \mathrm {sgn}\relax (x) + 6 \, \sqrt {c x^{2} + d} a c^{3} d^{3} \mathrm {sgn}\relax (x)}{c^{3} d^{2} x^{6}}}{48 \, 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^4,x, algorithm="giac")

[Out]

1/48*(3*(b*c^4*sgn(x) - 2*a*c^3*d*sgn(x))*arctan(sqrt(c*x^2 + d)/sqrt(-d))/(sqrt(-d)*d^2) + (3*(c*x^2 + d)^(5/
2)*b*c^4*sgn(x) - 6*(c*x^2 + d)^(5/2)*a*c^3*d*sgn(x) - 8*(c*x^2 + d)^(3/2)*b*c^4*d*sgn(x) - 3*sqrt(c*x^2 + d)*
b*c^4*d^2*sgn(x) + 6*sqrt(c*x^2 + d)*a*c^3*d^3*sgn(x))/(c^3*d^2*x^6))/c

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maple [B]  time = 0.06, size = 220, normalized size = 1.79 \begin {gather*} \frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, \left (6 a \,c^{2} d^{\frac {3}{2}} x^{6} \ln \left (\frac {2 d +2 \sqrt {c \,x^{2}+d}\, \sqrt {d}}{x}\right )-3 b \,c^{3} \sqrt {d}\, x^{6} \ln \left (\frac {2 d +2 \sqrt {c \,x^{2}+d}\, \sqrt {d}}{x}\right )-6 \sqrt {c \,x^{2}+d}\, a \,c^{2} d \,x^{6}+3 \sqrt {c \,x^{2}+d}\, b \,c^{3} x^{6}+6 \left (c \,x^{2}+d \right )^{\frac {3}{2}} a c d \,x^{4}-3 \left (c \,x^{2}+d \right )^{\frac {3}{2}} b \,c^{2} x^{4}-12 \left (c \,x^{2}+d \right )^{\frac {3}{2}} a \,d^{2} x^{2}+6 \left (c \,x^{2}+d \right )^{\frac {3}{2}} b c d \,x^{2}-8 \left (c \,x^{2}+d \right )^{\frac {3}{2}} b \,d^{2}\right )}{48 \sqrt {c \,x^{2}+d}\, d^{3} x^{5}} \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^4,x)

[Out]

1/48*((c*x^2+d)/x^2)^(1/2)/x^5*(6*d^(3/2)*ln(2*(d+(c*x^2+d)^(1/2)*d^(1/2))/x)*x^6*a*c^2-3*d^(1/2)*ln(2*(d+(c*x
^2+d)^(1/2)*d^(1/2))/x)*x^6*b*c^3-6*(c*x^2+d)^(1/2)*x^6*a*c^2*d+3*(c*x^2+d)^(1/2)*x^6*b*c^3+6*(c*x^2+d)^(3/2)*
x^4*a*c*d-3*(c*x^2+d)^(3/2)*x^4*b*c^2-12*(c*x^2+d)^(3/2)*x^2*a*d^2+6*(c*x^2+d)^(3/2)*x^2*b*c*d-8*(c*x^2+d)^(3/
2)*b*d^2)/(c*x^2+d)^(1/2)/d^3

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maxima [B]  time = 1.27, size = 277, normalized size = 2.25 \begin {gather*} -\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 )} a + \frac {1}{96} \, {\left (\frac {3 \, c^{3} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} x - \sqrt {d}}{\sqrt {c + \frac {d}{x^{2}}} x + \sqrt {d}}\right )}{d^{\frac {5}{2}}} + \frac {2 \, {\left (3 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} c^{3} x^{5} - 8 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c^{3} d x^{3} - 3 \, \sqrt {c + \frac {d}{x^{2}}} c^{3} d^{2} x\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{3} d^{2} x^{6} - 3 \, {\left (c + \frac {d}{x^{2}}\right )}^{2} d^{3} x^{4} + 3 \, {\left (c + \frac {d}{x^{2}}\right )} d^{4} x^{2} - d^{5}}\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^4,x, algorithm="maxima")

[Out]

-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))*a + 1/96*(3*c^3*log((sqrt
(c + d/x^2)*x - sqrt(d))/(sqrt(c + d/x^2)*x + sqrt(d)))/d^(5/2) + 2*(3*(c + d/x^2)^(5/2)*c^3*x^5 - 8*(c + d/x^
2)^(3/2)*c^3*d*x^3 - 3*sqrt(c + d/x^2)*c^3*d^2*x)/((c + d/x^2)^3*d^2*x^6 - 3*(c + d/x^2)^2*d^3*x^4 + 3*(c + d/
x^2)*d^4*x^2 - d^5))*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^4} \,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^4,x)

[Out]

int(((a + b/x^2)*(c + d/x^2)^(1/2))/x^4, x)

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sympy [B]  time = 11.42, size = 226, normalized size = 1.84 \begin {gather*} - \frac {a c^{\frac {3}{2}}}{8 d x \sqrt {1 + \frac {d}{c x^{2}}}} - \frac {3 a \sqrt {c}}{8 x^{3} \sqrt {1 + \frac {d}{c x^{2}}}} + \frac {a c^{2} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {c} x} \right )}}{8 d^{\frac {3}{2}}} - \frac {a d}{4 \sqrt {c} x^{5} \sqrt {1 + \frac {d}{c x^{2}}}} + \frac {b c^{\frac {5}{2}}}{16 d^{2} x \sqrt {1 + \frac {d}{c x^{2}}}} + \frac {b c^{\frac {3}{2}}}{48 d x^{3} \sqrt {1 + \frac {d}{c x^{2}}}} - \frac {5 b \sqrt {c}}{24 x^{5} \sqrt {1 + \frac {d}{c x^{2}}}} - \frac {b c^{3} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {c} x} \right )}}{16 d^{\frac {5}{2}}} - \frac {b d}{6 \sqrt {c} x^{7} \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**4,x)

[Out]

-a*c**(3/2)/(8*d*x*sqrt(1 + d/(c*x**2))) - 3*a*sqrt(c)/(8*x**3*sqrt(1 + d/(c*x**2))) + a*c**2*asinh(sqrt(d)/(s
qrt(c)*x))/(8*d**(3/2)) - a*d/(4*sqrt(c)*x**5*sqrt(1 + d/(c*x**2))) + b*c**(5/2)/(16*d**2*x*sqrt(1 + d/(c*x**2
))) + b*c**(3/2)/(48*d*x**3*sqrt(1 + d/(c*x**2))) - 5*b*sqrt(c)/(24*x**5*sqrt(1 + d/(c*x**2))) - b*c**3*asinh(
sqrt(d)/(sqrt(c)*x))/(16*d**(5/2)) - b*d/(6*sqrt(c)*x**7*sqrt(1 + d/(c*x**2)))

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