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

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

[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))

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Rubi [A]  time = 0.0583173, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {446, 78, 51, 63, 208} \[ -\frac{2 b c-3 a d}{2 c^2 \sqrt{c+\frac{d}{x^2}}}+\frac{(2 b c-3 a d) \tanh ^{-1}\left (\frac{\sqrt{c+\frac{d}{x^2}}}{\sqrt{c}}\right )}{2 c^{5/2}}+\frac{a x^2}{2 c \sqrt{c+\frac{d}{x^2}}} \]

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 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]]

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 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 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]

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*}

Mathematica [A]  time = 0.115579, size = 89, normalized size = 1.03 \[ \frac{\sqrt{c} x \left (a c x^2+3 a d-2 b c\right )-\sqrt{d} \sqrt{\frac{c x^2}{d}+1} (3 a d-2 b c) \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{d}}\right )}{2 c^{5/2} x \sqrt{c+\frac{d}{x^2}}} \]

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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Maple [A]  time = 0.009, size = 114, normalized size = 1.3 \begin{align*}{\frac{c{x}^{2}+d}{2\,{x}^{3}} \left ({c}^{{\frac{5}{2}}}{x}^{3}a+3\,{c}^{3/2}xad-2\,{c}^{5/2}xb-3\,\ln \left ( \sqrt{c}x+\sqrt{c{x}^{2}+d} \right ) \sqrt{c{x}^{2}+d}acd+2\,\ln \left ( \sqrt{c}x+\sqrt{c{x}^{2}+d} \right ) \sqrt{c{x}^{2}+d}b{c}^{2} \right ) \left ({\frac{c{x}^{2}+d}{{x}^{2}}} \right ) ^{-{\frac{3}{2}}}{c}^{-{\frac{7}{2}}}} \end{align*}

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*ln(c^(1/2)*x+(c*x^2+d)^(1/2))*(c*x^2+d)^(1/2)*a*c
*d+2*ln(c^(1/2)*x+(c*x^2+d)^(1/2))*(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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.62802, size = 543, normalized size = 6.31 \begin{align*} \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{align*}

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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Sympy [B]  time = 27.2976, size = 264, normalized size = 3.07 \begin{align*} 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{align*}

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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Giac [A]  time = 1.17672, size = 182, normalized size = 2.12 \begin{align*} -\frac{1}{2} \, d{\left (\frac{{\left (2 \, b c - 3 \, a d\right )} \arctan \left (\frac{\sqrt{\frac{c x^{2} + d}{x^{2}}}}{\sqrt{-c}}\right )}{\sqrt{-c} c^{2} d} + \frac{2 \, b c^{2} - 2 \, a c d - \frac{2 \,{\left (c x^{2} + d\right )} b c}{x^{2}} + \frac{3 \,{\left (c x^{2} + d\right )} a d}{x^{2}}}{{\left (c \sqrt{\frac{c x^{2} + d}{x^{2}}} - \frac{{\left (c x^{2} + d\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{x^{2}}\right )} c^{2} d}\right )} \end{align*}

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]

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