3.7.39 \(\int \frac {a+\frac {b}{x^2}}{(c+\frac {d}{x^2})^{3/2} x^6} \, dx\)

Optimal. Leaf size=123 \[ -\frac {3 c (5 b c-4 a d) \tanh ^{-1}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )}{8 d^{7/2}}+\frac {3 \sqrt {c+\frac {d}{x^2}} (5 b c-4 a d)}{8 d^3 x}-\frac {5 b c-4 a d}{4 d^2 x^3 \sqrt {c+\frac {d}{x^2}}}-\frac {b}{4 d x^5 \sqrt {c+\frac {d}{x^2}}} \]

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Rubi [A]  time = 0.07, 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, 288, 321, 217, 206} \begin {gather*} \frac {3 \sqrt {c+\frac {d}{x^2}} (5 b c-4 a d)}{8 d^3 x}-\frac {5 b c-4 a d}{4 d^2 x^3 \sqrt {c+\frac {d}{x^2}}}-\frac {3 c (5 b c-4 a d) \tanh ^{-1}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )}{8 d^{7/2}}-\frac {b}{4 d x^5 \sqrt {c+\frac {d}{x^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)/((c + d/x^2)^(3/2)*x^6),x]

[Out]

-b/(4*d*Sqrt[c + d/x^2]*x^5) - (5*b*c - 4*a*d)/(4*d^2*Sqrt[c + d/x^2]*x^3) + (3*(5*b*c - 4*a*d)*Sqrt[c + d/x^2
])/(8*d^3*x) - (3*c*(5*b*c - 4*a*d)*ArcTanh[Sqrt[d]/(Sqrt[c + d/x^2]*x)])/(8*d^(7/2))

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 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 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 {a+\frac {b}{x^2}}{\left (c+\frac {d}{x^2}\right )^{3/2} x^6} \, dx &=-\frac {b}{4 d \sqrt {c+\frac {d}{x^2}} x^5}+\frac {(-5 b c+4 a d) \int \frac {1}{\left (c+\frac {d}{x^2}\right )^{3/2} x^6} \, dx}{4 d}\\ &=-\frac {b}{4 d \sqrt {c+\frac {d}{x^2}} x^5}-\frac {(-5 b c+4 a d) \operatorname {Subst}\left (\int \frac {x^4}{\left (c+d x^2\right )^{3/2}} \, dx,x,\frac {1}{x}\right )}{4 d}\\ &=-\frac {b}{4 d \sqrt {c+\frac {d}{x^2}} x^5}-\frac {5 b c-4 a d}{4 d^2 \sqrt {c+\frac {d}{x^2}} x^3}+\frac {(3 (5 b c-4 a d)) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {c+d x^2}} \, dx,x,\frac {1}{x}\right )}{4 d^2}\\ &=-\frac {b}{4 d \sqrt {c+\frac {d}{x^2}} x^5}-\frac {5 b c-4 a d}{4 d^2 \sqrt {c+\frac {d}{x^2}} x^3}+\frac {3 (5 b c-4 a d) \sqrt {c+\frac {d}{x^2}}}{8 d^3 x}-\frac {(3 c (5 b c-4 a d)) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c+d x^2}} \, dx,x,\frac {1}{x}\right )}{8 d^3}\\ &=-\frac {b}{4 d \sqrt {c+\frac {d}{x^2}} x^5}-\frac {5 b c-4 a d}{4 d^2 \sqrt {c+\frac {d}{x^2}} x^3}+\frac {3 (5 b c-4 a d) \sqrt {c+\frac {d}{x^2}}}{8 d^3 x}-\frac {(3 c (5 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^3}\\ &=-\frac {b}{4 d \sqrt {c+\frac {d}{x^2}} x^5}-\frac {5 b c-4 a d}{4 d^2 \sqrt {c+\frac {d}{x^2}} x^3}+\frac {3 (5 b c-4 a d) \sqrt {c+\frac {d}{x^2}}}{8 d^3 x}-\frac {3 c (5 b c-4 a d) \tanh ^{-1}\left (\frac {\sqrt {d}}{\sqrt {c+\frac {d}{x^2}} x}\right )}{8 d^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 60, normalized size = 0.49 \begin {gather*} \frac {c x^4 (5 b c-4 a d) \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};\frac {c x^2}{d}+1\right )-b d^2}{4 d^3 x^5 \sqrt {c+\frac {d}{x^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)/((c + d/x^2)^(3/2)*x^6),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.25, size = 126, normalized size = 1.02 \begin {gather*} \frac {x \sqrt {c+\frac {d}{x^2}} \left (\frac {-12 a c d x^4-4 a d^2 x^2+15 b c^2 x^4+5 b c d x^2-2 b d^2}{8 d^3 x^4 \sqrt {c x^2+d}}-\frac {3 \left (5 b c^2-4 a c d\right ) \tanh ^{-1}\left (\frac {\sqrt {c x^2+d}}{\sqrt {d}}\right )}{8 d^{7/2}}\right )}{\sqrt {c x^2+d}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b/x^2)/((c + d/x^2)^(3/2)*x^6),x]

[Out]

(Sqrt[c + d/x^2]*x*((-2*b*d^2 + 5*b*c*d*x^2 - 4*a*d^2*x^2 + 15*b*c^2*x^4 - 12*a*c*d*x^4)/(8*d^3*x^4*Sqrt[d + c
*x^2]) - (3*(5*b*c^2 - 4*a*c*d)*ArcTanh[Sqrt[d + c*x^2]/Sqrt[d]])/(8*d^(7/2))))/Sqrt[d + c*x^2]

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fricas [A]  time = 0.45, size = 314, normalized size = 2.55 \begin {gather*} \left [-\frac {3 \, {\left ({\left (5 \, b c^{3} - 4 \, a c^{2} d\right )} x^{5} + {\left (5 \, b c^{2} d - 4 \, a c d^{2}\right )} x^{3}\right )} \sqrt {d} \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 (5 \, b c^{2} d - 4 \, a c d^{2}\right )} x^{4} - 2 \, b d^{3} + {\left (5 \, b c d^{2} - 4 \, a d^{3}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{16 \, {\left (c d^{4} x^{5} + d^{5} x^{3}\right )}}, \frac {3 \, {\left ({\left (5 \, b c^{3} - 4 \, a c^{2} d\right )} x^{5} + {\left (5 \, b c^{2} d - 4 \, a c d^{2}\right )} x^{3}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) + {\left (3 \, {\left (5 \, b c^{2} d - 4 \, a c d^{2}\right )} x^{4} - 2 \, b d^{3} + {\left (5 \, b c d^{2} - 4 \, a d^{3}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{8 \, {\left (c d^{4} x^{5} + d^{5} x^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)/(c+d/x^2)^(3/2)/x^6,x, algorithm="fricas")

[Out]

[-1/16*(3*((5*b*c^3 - 4*a*c^2*d)*x^5 + (5*b*c^2*d - 4*a*c*d^2)*x^3)*sqrt(d)*log(-(c*x^2 + 2*sqrt(d)*x*sqrt((c*
x^2 + d)/x^2) + 2*d)/x^2) - 2*(3*(5*b*c^2*d - 4*a*c*d^2)*x^4 - 2*b*d^3 + (5*b*c*d^2 - 4*a*d^3)*x^2)*sqrt((c*x^
2 + d)/x^2))/(c*d^4*x^5 + d^5*x^3), 1/8*(3*((5*b*c^3 - 4*a*c^2*d)*x^5 + (5*b*c^2*d - 4*a*c*d^2)*x^3)*sqrt(-d)*
arctan(sqrt(-d)*x*sqrt((c*x^2 + d)/x^2)/(c*x^2 + d)) + (3*(5*b*c^2*d - 4*a*c*d^2)*x^4 - 2*b*d^3 + (5*b*c*d^2 -
4*a*d^3)*x^2)*sqrt((c*x^2 + d)/x^2))/(c*d^4*x^5 + d^5*x^3)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)/(c+d/x^2)^(3/2)/x^6,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is
real):Check [sign(t_nostep),sign(t_nostep+sqrt(d)/c*sign(t_nostep))]sym2poly/r2sym(const gen & e,const index_
m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.06, size = 157, normalized size = 1.28 \begin {gather*} \frac {\left (c \,x^{2}+d \right ) \left (12 \sqrt {c \,x^{2}+d}\, a c \,d^{2} x^{4} \ln \left (\frac {2 d +2 \sqrt {c \,x^{2}+d}\, \sqrt {d}}{x}\right )-15 \sqrt {c \,x^{2}+d}\, b \,c^{2} d \,x^{4} \ln \left (\frac {2 d +2 \sqrt {c \,x^{2}+d}\, \sqrt {d}}{x}\right )-12 a c \,d^{\frac {5}{2}} x^{4}+15 b \,c^{2} d^{\frac {3}{2}} x^{4}-4 a \,d^{\frac {7}{2}} x^{2}+5 b c \,d^{\frac {5}{2}} x^{2}-2 b \,d^{\frac {7}{2}}\right )}{8 \left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} d^{\frac {9}{2}} x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x^2)/(c+d/x^2)^(3/2)/x^6,x)

[Out]

1/8*(c*x^2+d)*(12*(c*x^2+d)^(1/2)*ln(2*(d+(c*x^2+d)^(1/2)*d^(1/2))/x)*x^4*a*c*d^2-15*(c*x^2+d)^(1/2)*ln(2*(d+(
c*x^2+d)^(1/2)*d^(1/2))/x)*x^4*b*c^2*d-12*d^(5/2)*x^4*a*c+15*d^(3/2)*x^4*b*c^2-4*d^(7/2)*x^2*a+5*d^(5/2)*x^2*b
*c-2*d^(7/2)*b)/((c*x^2+d)/x^2)^(3/2)/x^7/d^(9/2)

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maxima [B]  time = 1.28, size = 243, normalized size = 1.98 \begin {gather*} \frac {1}{16} \, b {\left (\frac {2 \, {\left (15 \, {\left (c + \frac {d}{x^{2}}\right )}^{2} c^{2} x^{4} - 25 \, {\left (c + \frac {d}{x^{2}}\right )} c^{2} d x^{2} + 8 \, c^{2} d^{2}\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} d^{3} x^{5} - 2 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} d^{4} x^{3} + \sqrt {c + \frac {d}{x^{2}}} d^{5} x} + \frac {15 \, 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 {7}{2}}}\right )} - \frac {1}{4} \, a {\left (\frac {2 \, {\left (3 \, {\left (c + \frac {d}{x^{2}}\right )} c x^{2} - 2 \, c d\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} d^{2} x^{3} - \sqrt {c + \frac {d}{x^{2}}} d^{3} x} + \frac {3 \, c \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}}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)/(c+d/x^2)^(3/2)/x^6,x, algorithm="maxima")

[Out]

1/16*b*(2*(15*(c + d/x^2)^2*c^2*x^4 - 25*(c + d/x^2)*c^2*d*x^2 + 8*c^2*d^2)/((c + d/x^2)^(5/2)*d^3*x^5 - 2*(c
+ d/x^2)^(3/2)*d^4*x^3 + sqrt(c + d/x^2)*d^5*x) + 15*c^2*log((sqrt(c + d/x^2)*x - sqrt(d))/(sqrt(c + d/x^2)*x
+ sqrt(d)))/d^(7/2)) - 1/4*a*(2*(3*(c + d/x^2)*c*x^2 - 2*c*d)/((c + d/x^2)^(3/2)*d^2*x^3 - sqrt(c + d/x^2)*d^3
*x) + 3*c*log((sqrt(c + d/x^2)*x - sqrt(d))/(sqrt(c + d/x^2)*x + sqrt(d)))/d^(5/2))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+\frac {b}{x^2}}{x^6\,{\left (c+\frac {d}{x^2}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x^2)/(x^6*(c + d/x^2)^(3/2)),x)

[Out]

int((a + b/x^2)/(x^6*(c + d/x^2)^(3/2)), x)

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sympy [A]  time = 29.81, size = 180, normalized size = 1.46 \begin {gather*} a \left (- \frac {3 \sqrt {c}}{2 d^{2} x \sqrt {1 + \frac {d}{c x^{2}}}} + \frac {3 c \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {c} x} \right )}}{2 d^{\frac {5}{2}}} - \frac {1}{2 \sqrt {c} d x^{3} \sqrt {1 + \frac {d}{c x^{2}}}}\right ) + b \left (\frac {15 c^{\frac {3}{2}}}{8 d^{3} x \sqrt {1 + \frac {d}{c x^{2}}}} + \frac {5 \sqrt {c}}{8 d^{2} x^{3} \sqrt {1 + \frac {d}{c x^{2}}}} - \frac {15 c^{2} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {c} x} \right )}}{8 d^{\frac {7}{2}}} - \frac {1}{4 \sqrt {c} d x^{5} \sqrt {1 + \frac {d}{c x^{2}}}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x**2)/(c+d/x**2)**(3/2)/x**6,x)

[Out]

a*(-3*sqrt(c)/(2*d**2*x*sqrt(1 + d/(c*x**2))) + 3*c*asinh(sqrt(d)/(sqrt(c)*x))/(2*d**(5/2)) - 1/(2*sqrt(c)*d*x
**3*sqrt(1 + d/(c*x**2)))) + b*(15*c**(3/2)/(8*d**3*x*sqrt(1 + d/(c*x**2))) + 5*sqrt(c)/(8*d**2*x**3*sqrt(1 +
d/(c*x**2))) - 15*c**2*asinh(sqrt(d)/(sqrt(c)*x))/(8*d**(7/2)) - 1/(4*sqrt(c)*d*x**5*sqrt(1 + d/(c*x**2))))

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