3.7.38 \(\int \frac {a+\frac {b}{x^2}}{(c+\frac {d}{x^2})^{3/2} x^4} \, dx\)

Optimal. Leaf size=92 \[ \frac {(3 b c-2 a d) \tanh ^{-1}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )}{2 d^{5/2}}-\frac {3 b c-2 a d}{2 d^2 x \sqrt {c+\frac {d}{x^2}}}-\frac {b}{2 d x^3 \sqrt {c+\frac {d}{x^2}}} \]

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Rubi [A]  time = 0.05, antiderivative size = 92, 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, 288, 217, 206} \begin {gather*} -\frac {3 b c-2 a d}{2 d^2 x \sqrt {c+\frac {d}{x^2}}}+\frac {(3 b c-2 a d) \tanh ^{-1}\left (\frac {\sqrt {d}}{x \sqrt {c+\frac {d}{x^2}}}\right )}{2 d^{5/2}}-\frac {b}{2 d x^3 \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^4),x]

[Out]

-b/(2*d*Sqrt[c + d/x^2]*x^3) - (3*b*c - 2*a*d)/(2*d^2*Sqrt[c + d/x^2]*x) + ((3*b*c - 2*a*d)*ArcTanh[Sqrt[d]/(S
qrt[c + d/x^2]*x)])/(2*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 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 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^4} \, dx &=-\frac {b}{2 d \sqrt {c+\frac {d}{x^2}} x^3}+\frac {(-3 b c+2 a d) \int \frac {1}{\left (c+\frac {d}{x^2}\right )^{3/2} x^4} \, dx}{2 d}\\ &=-\frac {b}{2 d \sqrt {c+\frac {d}{x^2}} x^3}-\frac {(-3 b c+2 a d) \operatorname {Subst}\left (\int \frac {x^2}{\left (c+d x^2\right )^{3/2}} \, dx,x,\frac {1}{x}\right )}{2 d}\\ &=-\frac {b}{2 d \sqrt {c+\frac {d}{x^2}} x^3}-\frac {3 b c-2 a d}{2 d^2 \sqrt {c+\frac {d}{x^2}} x}+\frac {(3 b c-2 a d) \operatorname {Subst}\left (\int \frac {1}{\sqrt {c+d x^2}} \, dx,x,\frac {1}{x}\right )}{2 d^2}\\ &=-\frac {b}{2 d \sqrt {c+\frac {d}{x^2}} x^3}-\frac {3 b c-2 a d}{2 d^2 \sqrt {c+\frac {d}{x^2}} x}+\frac {(3 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 )}{2 d^2}\\ &=-\frac {b}{2 d \sqrt {c+\frac {d}{x^2}} x^3}-\frac {3 b c-2 a d}{2 d^2 \sqrt {c+\frac {d}{x^2}} x}+\frac {(3 b c-2 a d) \tanh ^{-1}\left (\frac {\sqrt {d}}{\sqrt {c+\frac {d}{x^2}} x}\right )}{2 d^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 57, normalized size = 0.62 \begin {gather*} \frac {x^2 (2 a d-3 b c) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {c x^2}{d}+1\right )-b d}{2 d^2 x^3 \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^4),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.24, size = 101, normalized size = 1.10 \begin {gather*} \frac {x \sqrt {c+\frac {d}{x^2}} \left (\frac {(3 b c-2 a d) \tanh ^{-1}\left (\frac {\sqrt {c x^2+d}}{\sqrt {d}}\right )}{2 d^{5/2}}+\frac {2 a d x^2-3 b c x^2-b d}{2 d^2 x^2 \sqrt {c x^2+d}}\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^4),x]

[Out]

(Sqrt[c + d/x^2]*x*((-(b*d) - 3*b*c*x^2 + 2*a*d*x^2)/(2*d^2*x^2*Sqrt[d + c*x^2]) + ((3*b*c - 2*a*d)*ArcTanh[Sq
rt[d + c*x^2]/Sqrt[d]])/(2*d^(5/2))))/Sqrt[d + c*x^2]

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fricas [A]  time = 0.44, size = 248, normalized size = 2.70 \begin {gather*} \left [-\frac {{\left ({\left (3 \, b c^{2} - 2 \, a c d\right )} x^{3} + {\left (3 \, b c d - 2 \, a d^{2}\right )} x\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 (b d^{2} + {\left (3 \, b c d - 2 \, a d^{2}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{4 \, {\left (c d^{3} x^{3} + d^{4} x\right )}}, -\frac {{\left ({\left (3 \, b c^{2} - 2 \, a c d\right )} x^{3} + {\left (3 \, b c d - 2 \, a d^{2}\right )} x\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) + {\left (b d^{2} + {\left (3 \, b c d - 2 \, a d^{2}\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{2 \, {\left (c d^{3} x^{3} + d^{4} x\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^4,x, algorithm="fricas")

[Out]

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

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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^4,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 = 132, normalized size = 1.43 \begin {gather*} \frac {\left (c \,x^{2}+d \right ) \left (-2 \sqrt {c \,x^{2}+d}\, a \,d^{2} x^{2} \ln \left (\frac {2 d +2 \sqrt {c \,x^{2}+d}\, \sqrt {d}}{x}\right )+3 \sqrt {c \,x^{2}+d}\, b c d \,x^{2} \ln \left (\frac {2 d +2 \sqrt {c \,x^{2}+d}\, \sqrt {d}}{x}\right )+2 a \,d^{\frac {5}{2}} x^{2}-3 b c \,d^{\frac {3}{2}} x^{2}-b \,d^{\frac {5}{2}}\right )}{2 \left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} d^{\frac {7}{2}} x^{5}} \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^4,x)

[Out]

1/2*(c*x^2+d)*(2*d^(5/2)*x^2*a-3*d^(3/2)*x^2*b*c-2*ln(2*(d+(c*x^2+d)^(1/2)*d^(1/2))/x)*(c*x^2+d)^(1/2)*x^2*a*d
^2+3*ln(2*(d+(c*x^2+d)^(1/2)*d^(1/2))/x)*(c*x^2+d)^(1/2)*x^2*b*c*d-d^(5/2)*b)/((c*x^2+d)/x^2)^(3/2)/x^5/d^(7/2
)

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maxima [B]  time = 1.22, size = 162, normalized size = 1.76 \begin {gather*} -\frac {1}{4} \, b {\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 )} + \frac {1}{2} \, a {\left (\frac {\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}{\sqrt {c + \frac {d}{x^{2}}} d x}\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^4,x, algorithm="maxima")

[Out]

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

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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^4\,{\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^4*(c + d/x^2)^(3/2)),x)

[Out]

int((a + b/x^2)/(x^4*(c + d/x^2)^(3/2)), x)

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sympy [B]  time = 18.44, size = 262, normalized size = 2.85 \begin {gather*} a \left (\frac {c d^{2} x^{2} \log {\left (\frac {c x^{2}}{d} \right )}}{2 c d^{\frac {7}{2}} x^{2} + 2 d^{\frac {9}{2}}} - \frac {2 c d^{2} x^{2} \log {\left (\sqrt {\frac {c x^{2}}{d} + 1} + 1 \right )}}{2 c d^{\frac {7}{2}} x^{2} + 2 d^{\frac {9}{2}}} + \frac {2 d^{3} \sqrt {\frac {c x^{2}}{d} + 1}}{2 c d^{\frac {7}{2}} x^{2} + 2 d^{\frac {9}{2}}} + \frac {d^{3} \log {\left (\frac {c x^{2}}{d} \right )}}{2 c d^{\frac {7}{2}} x^{2} + 2 d^{\frac {9}{2}}} - \frac {2 d^{3} \log {\left (\sqrt {\frac {c x^{2}}{d} + 1} + 1 \right )}}{2 c d^{\frac {7}{2}} x^{2} + 2 d^{\frac {9}{2}}}\right ) + b \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 ) \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**4,x)

[Out]

a*(c*d**2*x**2*log(c*x**2/d)/(2*c*d**(7/2)*x**2 + 2*d**(9/2)) - 2*c*d**2*x**2*log(sqrt(c*x**2/d + 1) + 1)/(2*c
*d**(7/2)*x**2 + 2*d**(9/2)) + 2*d**3*sqrt(c*x**2/d + 1)/(2*c*d**(7/2)*x**2 + 2*d**(9/2)) + d**3*log(c*x**2/d)
/(2*c*d**(7/2)*x**2 + 2*d**(9/2)) - 2*d**3*log(sqrt(c*x**2/d + 1) + 1)/(2*c*d**(7/2)*x**2 + 2*d**(9/2))) + b*(
-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))))

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