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

Optimal. Leaf size=86 \[ \frac {(2 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{2 c^{5/2}}-\frac {2 b c-3 a d}{2 c^2 \sqrt {c+\frac {d}{x^2}}}+\frac {a x^2}{2 c \sqrt {c+\frac {d}{x^2}}} \]

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Rubi [A]  time = 0.06, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {446, 78, 51, 63, 208} \begin {gather*} -\frac {2 b c-3 a d}{2 c^2 \sqrt {c+\frac {d}{x^2}}}+\frac {(2 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{2 c^{5/2}}+\frac {a x^2}{2 c \sqrt {c+\frac {d}{x^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

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

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Mathematica [A]  time = 0.22, size = 89, normalized size = 1.03 \begin {gather*} \frac {\sqrt {c} x \left (a c x^2+3 a d-2 b c\right )-\sqrt {d} \sqrt {\frac {c x^2}{d}+1} (3 a d-2 b c) \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {d}}\right )}{2 c^{5/2} x \sqrt {c+\frac {d}{x^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.18, size = 93, normalized size = 1.08 \begin {gather*} \frac {(2 b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {\frac {c x^2+d}{x^2}}}{\sqrt {c}}\right )}{2 c^{5/2}}+\frac {\sqrt {\frac {c x^2+d}{x^2}} \left (a c x^4+3 a d x^2-2 b c x^2\right )}{2 c^2 \left (c x^2+d\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 249, normalized size = 2.90 \begin {gather*} \left [-\frac {{\left (2 \, b c d - 3 \, a d^{2} + {\left (2 \, b c^{2} - 3 \, a c d\right )} x^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} + 2 \, \sqrt {c} x^{2} \sqrt {\frac {c x^{2} + d}{x^{2}}} - d\right ) - 2 \, {\left (a c^{2} x^{4} - {\left (2 \, b c^{2} - 3 \, a c d\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{4 \, {\left (c^{4} x^{2} + c^{3} d\right )}}, -\frac {{\left (2 \, b c d - 3 \, a d^{2} + {\left (2 \, b c^{2} - 3 \, a c d\right )} x^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x^{2} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{c x^{2} + d}\right ) - {\left (a c^{2} x^{4} - {\left (2 \, b c^{2} - 3 \, a c d\right )} x^{2}\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{2 \, {\left (c^{4} x^{2} + c^{3} d\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)*x/(c+d/x^2)^(3/2),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)]Unable to divide, perhaps due to rounding error%%%{%%%{-2,[1]%%%},[2,1,2]%%%}+%%%{%%{[4,0]:[1,0,%%
%{-1,[1]%%%}]%%},[1,1,3]%%%}+%%%{-2,[0,1,4]%%%} / %%%{%%%{1,[2]%%%},[2,0,0]%%%}+%%%{%%{[%%%{-2,[1]%%%},0]:[1,0
,%%%{-1,[1]%%%}]%%},[1,0,1]%%%}+%%%{%%%{1,[1]%%%},[0,0,2]%%%} Error: Bad Argument Value

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maple [A]  time = 0.06, size = 115, normalized size = 1.34 \begin {gather*} -\frac {\left (c \,x^{2}+d \right ) \left (-a \,c^{\frac {5}{2}} x^{3}-3 a \,c^{\frac {3}{2}} d x +2 b \,c^{\frac {5}{2}} x +3 \sqrt {c \,x^{2}+d}\, a c d \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right )-2 \sqrt {c \,x^{2}+d}\, b \,c^{2} \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+d}\right )\right )}{2 \left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} c^{\frac {7}{2}} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.20, size = 144, normalized size = 1.67 \begin {gather*} \frac {1}{4} \, a {\left (\frac {2 \, {\left (3 \, {\left (c + \frac {d}{x^{2}}\right )} d - 2 \, c d\right )}}{{\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} c^{2} - \sqrt {c + \frac {d}{x^{2}}} c^{3}} + \frac {3 \, d \log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} - \sqrt {c}}{\sqrt {c + \frac {d}{x^{2}}} + \sqrt {c}}\right )}{c^{\frac {5}{2}}}\right )} - \frac {1}{2} \, b {\left (\frac {\log \left (\frac {\sqrt {c + \frac {d}{x^{2}}} - \sqrt {c}}{\sqrt {c + \frac {d}{x^{2}}} + \sqrt {c}}\right )}{c^{\frac {3}{2}}} + \frac {2}{\sqrt {c + \frac {d}{x^{2}}} c}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.61, size = 90, normalized size = 1.05 \begin {gather*} \frac {b\,\mathrm {atanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{c^{3/2}}-\frac {b}{c\,\sqrt {c+\frac {d}{x^2}}}+\frac {a\,x^2}{2\,c\,\sqrt {c+\frac {d}{x^2}}}-\frac {3\,a\,d\,\mathrm {atanh}\left (\frac {\sqrt {c+\frac {d}{x^2}}}{\sqrt {c}}\right )}{2\,c^{5/2}}+\frac {3\,a\,d}{2\,c^2\,\sqrt {c+\frac {d}{x^2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 52.42, size = 264, normalized size = 3.07 \begin {gather*} a \left (\frac {x^{3}}{2 c \sqrt {d} \sqrt {\frac {c x^{2}}{d} + 1}} + \frac {3 \sqrt {d} x}{2 c^{2} \sqrt {\frac {c x^{2}}{d} + 1}} - \frac {3 d \operatorname {asinh}{\left (\frac {\sqrt {c} x}{\sqrt {d}} \right )}}{2 c^{\frac {5}{2}}}\right ) + b \left (- \frac {2 c^{3} x^{2} \sqrt {1 + \frac {d}{c x^{2}}}}{2 c^{\frac {9}{2}} x^{2} + 2 c^{\frac {7}{2}} d} - \frac {c^{3} x^{2} \log {\left (\frac {d}{c x^{2}} \right )}}{2 c^{\frac {9}{2}} x^{2} + 2 c^{\frac {7}{2}} d} + \frac {2 c^{3} x^{2} \log {\left (\sqrt {1 + \frac {d}{c x^{2}}} + 1 \right )}}{2 c^{\frac {9}{2}} x^{2} + 2 c^{\frac {7}{2}} d} - \frac {c^{2} d \log {\left (\frac {d}{c x^{2}} \right )}}{2 c^{\frac {9}{2}} x^{2} + 2 c^{\frac {7}{2}} d} + \frac {2 c^{2} d \log {\left (\sqrt {1 + \frac {d}{c x^{2}}} + 1 \right )}}{2 c^{\frac {9}{2}} x^{2} + 2 c^{\frac {7}{2}} d}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(x**3/(2*c*sqrt(d)*sqrt(c*x**2/d + 1)) + 3*sqrt(d)*x/(2*c**2*sqrt(c*x**2/d + 1)) - 3*d*asinh(sqrt(c)*x/sqrt(
d))/(2*c**(5/2))) + b*(-2*c**3*x**2*sqrt(1 + d/(c*x**2))/(2*c**(9/2)*x**2 + 2*c**(7/2)*d) - c**3*x**2*log(d/(c
*x**2))/(2*c**(9/2)*x**2 + 2*c**(7/2)*d) + 2*c**3*x**2*log(sqrt(1 + d/(c*x**2)) + 1)/(2*c**(9/2)*x**2 + 2*c**(
7/2)*d) - c**2*d*log(d/(c*x**2))/(2*c**(9/2)*x**2 + 2*c**(7/2)*d) + 2*c**2*d*log(sqrt(1 + d/(c*x**2)) + 1)/(2*
c**(9/2)*x**2 + 2*c**(7/2)*d))

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