3.960 \(\int \frac {(a+\frac {b}{x^2}) (c+\frac {d}{x^2})^{3/2}}{x^2} \, dx\)

Optimal. Leaf size=123 \[ \frac {c^2 (b c-6 a d) \tanh ^{-1}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )}{16 d^{3/2}}+\frac {\left (c+\frac {d}{x^2}\right )^{3/2} (b c-6 a d)}{24 d x}+\frac {c \sqrt {c+\frac {d}{x^2}} (b c-6 a d)}{16 d x}-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{6 d x} \]

[Out]

1/24*(-6*a*d+b*c)*(c+d/x^2)^(3/2)/d/x-1/6*b*(c+d/x^2)^(5/2)/d/x+1/16*c^2*(-6*a*d+b*c)*arctanh(d^(1/2)/x/(c+d/x
^2)^(1/2))/d^(3/2)+1/16*c*(-6*a*d+b*c)*(c+d/x^2)^(1/2)/d/x

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Rubi [A]  time = 0.06, antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ \frac {c^2 (b c-6 a d) \tanh ^{-1}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )}{16 d^{3/2}}+\frac {\left (c+\frac {d}{x^2}\right )^{3/2} (b c-6 a d)}{24 d x}+\frac {c \sqrt {c+\frac {d}{x^2}} (b c-6 a d)}{16 d x}-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{6 d x} \]

Antiderivative was successfully verified.

[In]

Int[((a + b/x^2)*(c + d/x^2)^(3/2))/x^2,x]

[Out]

(c*(b*c - 6*a*d)*Sqrt[c + d/x^2])/(16*d*x) + ((b*c - 6*a*d)*(c + d/x^2)^(3/2))/(24*d*x) - (b*(c + d/x^2)^(5/2)
)/(6*d*x) + (c^2*(b*c - 6*a*d)*ArcTanh[Sqrt[d]/(Sqrt[c + d/x^2]*x)])/(16*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 ) \left (c+\frac {d}{x^2}\right )^{3/2}}{x^2} \, dx &=-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{6 d x}+\frac {(-b c+6 a d) \int \frac {\left (c+\frac {d}{x^2}\right )^{3/2}}{x^2} \, dx}{6 d}\\ &=-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{6 d x}-\frac {(-b c+6 a d) \operatorname {Subst}\left (\int \left (c+d x^2\right )^{3/2} \, dx,x,\frac {1}{x}\right )}{6 d}\\ &=\frac {(b c-6 a d) \left (c+\frac {d}{x^2}\right )^{3/2}}{24 d x}-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{6 d x}+\frac {(c (b c-6 a d)) \operatorname {Subst}\left (\int \sqrt {c+d x^2} \, dx,x,\frac {1}{x}\right )}{8 d}\\ &=\frac {c (b c-6 a d) \sqrt {c+\frac {d}{x^2}}}{16 d x}+\frac {(b c-6 a d) \left (c+\frac {d}{x^2}\right )^{3/2}}{24 d x}-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{6 d x}+\frac {\left (c^2 (b c-6 a d)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c+d x^2}} \, dx,x,\frac {1}{x}\right )}{16 d}\\ &=\frac {c (b c-6 a d) \sqrt {c+\frac {d}{x^2}}}{16 d x}+\frac {(b c-6 a d) \left (c+\frac {d}{x^2}\right )^{3/2}}{24 d x}-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{6 d x}+\frac {\left (c^2 (b c-6 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}\\ &=\frac {c (b c-6 a d) \sqrt {c+\frac {d}{x^2}}}{16 d x}+\frac {(b c-6 a d) \left (c+\frac {d}{x^2}\right )^{3/2}}{24 d x}-\frac {b \left (c+\frac {d}{x^2}\right )^{5/2}}{6 d x}+\frac {c^2 (b c-6 a d) \tanh ^{-1}\left (\frac {\sqrt {d}}{\sqrt {c+\frac {d}{x^2}} x}\right )}{16 d^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 126, normalized size = 1.02 \[ -\frac {\sqrt {c+\frac {d}{x^2}} \left (\left (c x^2+d\right ) \left (6 a d x^2 \left (5 c x^2+2 d\right )+b \left (3 c^2 x^4+14 c d x^2+8 d^2\right )\right )+3 c^2 x^6 \sqrt {\frac {c x^2}{d}+1} (6 a d-b c) \tanh ^{-1}\left (\sqrt {\frac {c x^2}{d}+1}\right )\right )}{48 d x^5 \left (c x^2+d\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b/x^2)*(c + d/x^2)^(3/2))/x^2,x]

[Out]

-1/48*(Sqrt[c + d/x^2]*((d + c*x^2)*(6*a*d*x^2*(2*d + 5*c*x^2) + b*(8*d^2 + 14*c*d*x^2 + 3*c^2*x^4)) + 3*c^2*(
-(b*c) + 6*a*d)*x^6*Sqrt[1 + (c*x^2)/d]*ArcTanh[Sqrt[1 + (c*x^2)/d]]))/(d*x^5*(d + c*x^2))

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fricas [A]  time = 0.88, size = 246, normalized size = 2.00 \[ \left [-\frac {3 \, {\left (b c^{3} - 6 \, a c^{2} d\right )} \sqrt {d} x^{5} \log \left (-\frac {c x^{2} - 2 \, \sqrt {d} x \sqrt {\frac {c x^{2} + d}{x^{2}}} + 2 \, d}{x^{2}}\right ) + 2 \, {\left (3 \, {\left (b c^{2} d + 10 \, a c d^{2}\right )} x^{4} + 8 \, b d^{3} + 2 \, {\left (7 \, b c d^{2} + 6 \, a d^{3}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{96 \, d^{2} x^{5}}, -\frac {3 \, {\left (b c^{3} - 6 \, a c^{2} d\right )} \sqrt {-d} x^{5} \arctan \left (\frac {\sqrt {-d} x \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) + {\left (3 \, {\left (b c^{2} d + 10 \, a c d^{2}\right )} x^{4} + 8 \, b d^{3} + 2 \, {\left (7 \, b c d^{2} + 6 \, a d^{3}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{48 \, d^{2} x^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)*(c+d/x^2)^(3/2)/x^2,x, algorithm="fricas")

[Out]

[-1/96*(3*(b*c^3 - 6*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 + 10*a*c*d^2)*x^4 + 8*b*d^3 + 2*(7*b*c*d^2 + 6*a*d^3)*x^2)*sqrt((c*x^2 + d)/x^2))/(d^2*x^5), -1/48*(3*
(b*c^3 - 6*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 + 10*a*c*d
^2)*x^4 + 8*b*d^3 + 2*(7*b*c*d^2 + 6*a*d^3)*x^2)*sqrt((c*x^2 + d)/x^2))/(d^2*x^5)]

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giac [A]  time = 0.30, size = 173, normalized size = 1.41 \[ -\frac {\frac {3 \, {\left (b c^{4} \mathrm {sgn}\relax (x) - 6 \, a c^{3} d \mathrm {sgn}\relax (x)\right )} \arctan \left (\frac {\sqrt {c x^{2} + d}}{\sqrt {-d}}\right )}{\sqrt {-d} d} + \frac {3 \, {\left (c x^{2} + d\right )}^{\frac {5}{2}} b c^{4} \mathrm {sgn}\relax (x) + 30 \, {\left (c x^{2} + d\right )}^{\frac {5}{2}} a c^{3} d \mathrm {sgn}\relax (x) + 8 \, {\left (c x^{2} + d\right )}^{\frac {3}{2}} b c^{4} d \mathrm {sgn}\relax (x) - 48 \, {\left (c x^{2} + d\right )}^{\frac {3}{2}} a c^{3} d^{2} \mathrm {sgn}\relax (x) - 3 \, \sqrt {c x^{2} + d} b c^{4} d^{2} \mathrm {sgn}\relax (x) + 18 \, \sqrt {c x^{2} + d} a c^{3} d^{3} \mathrm {sgn}\relax (x)}{c^{3} d x^{6}}}{48 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)*(c+d/x^2)^(3/2)/x^2,x, algorithm="giac")

[Out]

-1/48*(3*(b*c^4*sgn(x) - 6*a*c^3*d*sgn(x))*arctan(sqrt(c*x^2 + d)/sqrt(-d))/(sqrt(-d)*d) + (3*(c*x^2 + d)^(5/2
)*b*c^4*sgn(x) + 30*(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) - 48*(c*x^2 + d)^(3/
2)*a*c^3*d^2*sgn(x) - 3*sqrt(c*x^2 + d)*b*c^4*d^2*sgn(x) + 18*sqrt(c*x^2 + d)*a*c^3*d^3*sgn(x))/(c^3*d*x^6))/c

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maple [B]  time = 0.07, size = 259, normalized size = 2.11 \[ -\frac {\left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} \left (18 a \,c^{2} d^{\frac {5}{2}} x^{6} \ln \left (\frac {2 d +2 \sqrt {c \,x^{2}+d}\, \sqrt {d}}{x}\right )-3 b \,c^{3} d^{\frac {3}{2}} x^{6} \ln \left (\frac {2 d +2 \sqrt {c \,x^{2}+d}\, \sqrt {d}}{x}\right )-18 \sqrt {c \,x^{2}+d}\, a \,c^{2} d^{2} x^{6}+3 \sqrt {c \,x^{2}+d}\, b \,c^{3} d \,x^{6}-6 \left (c \,x^{2}+d \right )^{\frac {3}{2}} a \,c^{2} d \,x^{6}+\left (c \,x^{2}+d \right )^{\frac {3}{2}} b \,c^{3} x^{6}+6 \left (c \,x^{2}+d \right )^{\frac {5}{2}} a c d \,x^{4}-\left (c \,x^{2}+d \right )^{\frac {5}{2}} b \,c^{2} x^{4}+12 \left (c \,x^{2}+d \right )^{\frac {5}{2}} a \,d^{2} x^{2}-2 \left (c \,x^{2}+d \right )^{\frac {5}{2}} b c d \,x^{2}+8 \left (c \,x^{2}+d \right )^{\frac {5}{2}} b \,d^{2}\right )}{48 \left (c \,x^{2}+d \right )^{\frac {3}{2}} d^{3} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x^2)*(c+d/x^2)^(3/2)/x^2,x)

[Out]

-1/48*((c*x^2+d)/x^2)^(3/2)/x^3*(18*d^(5/2)*ln(2*(d+(c*x^2+d)^(1/2)*d^(1/2))/x)*x^6*a*c^2-3*d^(3/2)*ln(2*(d+(c
*x^2+d)^(1/2)*d^(1/2))/x)*x^6*b*c^3-6*(c*x^2+d)^(3/2)*x^6*a*c^2*d+(c*x^2+d)^(3/2)*x^6*b*c^3+6*(c*x^2+d)^(5/2)*
x^4*a*c*d-(c*x^2+d)^(5/2)*x^4*b*c^2-18*(c*x^2+d)^(1/2)*x^6*a*c^2*d^2+3*(c*x^2+d)^(1/2)*x^6*b*c^3*d+12*(c*x^2+d
)^(5/2)*x^2*a*d^2-2*(c*x^2+d)^(5/2)*x^2*b*c*d+8*(c*x^2+d)^(5/2)*b*d^2)/(c*x^2+d)^(3/2)/d^3

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maxima [B]  time = 1.38, size = 275, normalized size = 2.24 \[ \frac {1}{16} \, {\left (\frac {3 \, c^{2} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} x - \sqrt {d}}{\sqrt {c + \frac {d}{x^{2}}} x + \sqrt {d}}\right )}{\sqrt {d}} - \frac {2 \, {\left (5 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c^{2} x^{3} - 3 \, \sqrt {c + \frac {d}{x^{2}}} c^{2} d x\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{2} x^{4} - 2 \, {\left (c + \frac {d}{x^{2}}\right )} d x^{2} + d^{2}}\right )} a - \frac {1}{96} \, {\left (\frac {3 \, c^{3} \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} x - \sqrt {d}}{\sqrt {c + \frac {d}{x^{2}}} x + \sqrt {d}}\right )}{d^{\frac {3}{2}}} + \frac {2 \, {\left (3 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} c^{3} x^{5} + 8 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c^{3} d x^{3} - 3 \, \sqrt {c + \frac {d}{x^{2}}} c^{3} d^{2} x\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{3} d x^{6} - 3 \, {\left (c + \frac {d}{x^{2}}\right )}^{2} d^{2} x^{4} + 3 \, {\left (c + \frac {d}{x^{2}}\right )} d^{3} x^{2} - d^{4}}\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)*(c+d/x^2)^(3/2)/x^2,x, algorithm="maxima")

[Out]

1/16*(3*c^2*log((sqrt(c + d/x^2)*x - sqrt(d))/(sqrt(c + d/x^2)*x + sqrt(d)))/sqrt(d) - 2*(5*(c + d/x^2)^(3/2)*
c^2*x^3 - 3*sqrt(c + d/x^2)*c^2*d*x)/((c + d/x^2)^2*x^4 - 2*(c + d/x^2)*d*x^2 + d^2))*a - 1/96*(3*c^3*log((sqr
t(c + d/x^2)*x - sqrt(d))/(sqrt(c + d/x^2)*x + sqrt(d)))/d^(3/2) + 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*x^6 - 3*(c + d/x^2)^2*d^2*x^4 + 3*(c + d/x
^2)*d^3*x^2 - d^4))*b

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (a+\frac {b}{x^2}\right )\,{\left (c+\frac {d}{x^2}\right )}^{3/2}}{x^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b/x^2)*(c + d/x^2)^(3/2))/x^2,x)

[Out]

int(((a + b/x^2)*(c + d/x^2)^(3/2))/x^2, x)

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sympy [B]  time = 18.95, size = 253, normalized size = 2.06 \[ - \frac {a c^{\frac {3}{2}} \sqrt {1 + \frac {d}{c x^{2}}}}{2 x} - \frac {a c^{\frac {3}{2}}}{8 x \sqrt {1 + \frac {d}{c x^{2}}}} - \frac {3 a \sqrt {c} d}{8 x^{3} \sqrt {1 + \frac {d}{c x^{2}}}} - \frac {3 a c^{2} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {c} x} \right )}}{8 \sqrt {d}} - \frac {a d^{2}}{4 \sqrt {c} x^{5} \sqrt {1 + \frac {d}{c x^{2}}}} - \frac {b c^{\frac {5}{2}}}{16 d x \sqrt {1 + \frac {d}{c x^{2}}}} - \frac {17 b c^{\frac {3}{2}}}{48 x^{3} \sqrt {1 + \frac {d}{c x^{2}}}} - \frac {11 b \sqrt {c} d}{24 x^{5} \sqrt {1 + \frac {d}{c x^{2}}}} + \frac {b c^{3} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {c} x} \right )}}{16 d^{\frac {3}{2}}} - \frac {b d^{2}}{6 \sqrt {c} x^{7} \sqrt {1 + \frac {d}{c x^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x**2)*(c+d/x**2)**(3/2)/x**2,x)

[Out]

-a*c**(3/2)*sqrt(1 + d/(c*x**2))/(2*x) - a*c**(3/2)/(8*x*sqrt(1 + d/(c*x**2))) - 3*a*sqrt(c)*d/(8*x**3*sqrt(1
+ d/(c*x**2))) - 3*a*c**2*asinh(sqrt(d)/(sqrt(c)*x))/(8*sqrt(d)) - a*d**2/(4*sqrt(c)*x**5*sqrt(1 + d/(c*x**2))
) - b*c**(5/2)/(16*d*x*sqrt(1 + d/(c*x**2))) - 17*b*c**(3/2)/(48*x**3*sqrt(1 + d/(c*x**2))) - 11*b*sqrt(c)*d/(
24*x**5*sqrt(1 + d/(c*x**2))) + b*c**3*asinh(sqrt(d)/(sqrt(c)*x))/(16*d**(3/2)) - b*d**2/(6*sqrt(c)*x**7*sqrt(
1 + d/(c*x**2)))

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