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

Optimal. Leaf size=118 \[ \frac{x^2 (4 b c-5 a d)}{8 c^2 \sqrt{c+\frac{d}{x^2}}}+\frac{3 d (4 b c-5 a d)}{8 c^3 \sqrt{c+\frac{d}{x^2}}}-\frac{3 d (4 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{c+\frac{d}{x^2}}}{\sqrt{c}}\right )}{8 c^{7/2}}+\frac{a x^4}{4 c \sqrt{c+\frac{d}{x^2}}} \]

[Out]

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

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Rubi [A]  time = 0.0848099, antiderivative size = 120, normalized size of antiderivative = 1.02, 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 = {446, 78, 51, 63, 208} \[ \frac{3 x^2 \sqrt{c+\frac{d}{x^2}} (4 b c-5 a d)}{8 c^3}-\frac{x^2 (4 b c-5 a d)}{4 c^2 \sqrt{c+\frac{d}{x^2}}}-\frac{3 d (4 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{c+\frac{d}{x^2}}}{\sqrt{c}}\right )}{8 c^{7/2}}+\frac{a x^4}{4 c \sqrt{c+\frac{d}{x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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^3}{\left (c+\frac{d}{x^2}\right )^{3/2}} \, dx &=-\left (\frac{1}{2} \operatorname{Subst}\left (\int \frac{a+b x}{x^3 (c+d x)^{3/2}} \, dx,x,\frac{1}{x^2}\right )\right )\\ &=\frac{a x^4}{4 c \sqrt{c+\frac{d}{x^2}}}-\frac{\left (2 b c-\frac{5 a d}{2}\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 (c+d x)^{3/2}} \, dx,x,\frac{1}{x^2}\right )}{4 c}\\ &=-\frac{(4 b c-5 a d) x^2}{4 c^2 \sqrt{c+\frac{d}{x^2}}}+\frac{a x^4}{4 c \sqrt{c+\frac{d}{x^2}}}-\frac{(3 (4 b c-5 a d)) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{c+d x}} \, dx,x,\frac{1}{x^2}\right )}{8 c^2}\\ &=-\frac{(4 b c-5 a d) x^2}{4 c^2 \sqrt{c+\frac{d}{x^2}}}+\frac{3 (4 b c-5 a d) \sqrt{c+\frac{d}{x^2}} x^2}{8 c^3}+\frac{a x^4}{4 c \sqrt{c+\frac{d}{x^2}}}+\frac{(3 d (4 b c-5 a d)) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+d x}} \, dx,x,\frac{1}{x^2}\right )}{16 c^3}\\ &=-\frac{(4 b c-5 a d) x^2}{4 c^2 \sqrt{c+\frac{d}{x^2}}}+\frac{3 (4 b c-5 a d) \sqrt{c+\frac{d}{x^2}} x^2}{8 c^3}+\frac{a x^4}{4 c \sqrt{c+\frac{d}{x^2}}}+\frac{(3 (4 b c-5 a d)) \operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+\frac{d}{x^2}}\right )}{8 c^3}\\ &=-\frac{(4 b c-5 a d) x^2}{4 c^2 \sqrt{c+\frac{d}{x^2}}}+\frac{3 (4 b c-5 a d) \sqrt{c+\frac{d}{x^2}} x^2}{8 c^3}+\frac{a x^4}{4 c \sqrt{c+\frac{d}{x^2}}}-\frac{3 d (4 b c-5 a d) \tanh ^{-1}\left (\frac{\sqrt{c+\frac{d}{x^2}}}{\sqrt{c}}\right )}{8 c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.14451, size = 111, normalized size = 0.94 \[ \frac{\sqrt{c} x \left (a \left (2 c^2 x^4-5 c d x^2-15 d^2\right )+4 b c \left (c x^2+3 d\right )\right )+3 d^{3/2} \sqrt{\frac{c x^2}{d}+1} (5 a d-4 b c) \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{d}}\right )}{8 c^{7/2} x \sqrt{c+\frac{d}{x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c]*x*(4*b*c*(3*d + c*x^2) + a*(-15*d^2 - 5*c*d*x^2 + 2*c^2*x^4)) + 3*d^(3/2)*(-4*b*c + 5*a*d)*Sqrt[1 + (
c*x^2)/d]*ArcSinh[(Sqrt[c]*x)/Sqrt[d]])/(8*c^(7/2)*Sqrt[c + d/x^2]*x)

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Maple [A]  time = 0.012, size = 140, normalized size = 1.2 \begin{align*}{\frac{c{x}^{2}+d}{8\,{x}^{3}} \left ( 2\,{c}^{7/2}{x}^{5}a-5\,{c}^{5/2}{x}^{3}ad+4\,{c}^{7/2}{x}^{3}b-15\,{c}^{3/2}xa{d}^{2}+12\,{c}^{5/2}xbd+15\,\ln \left ( \sqrt{c}x+\sqrt{c{x}^{2}+d} \right ) \sqrt{c{x}^{2}+d}ac{d}^{2}-12\,\ln \left ( \sqrt{c}x+\sqrt{c{x}^{2}+d} \right ) \sqrt{c{x}^{2}+d}b{c}^{2}d \right ) \left ({\frac{c{x}^{2}+d}{{x}^{2}}} \right ) ^{-{\frac{3}{2}}}{c}^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^3/(c+d/x^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.88141, size = 662, normalized size = 5.61 \begin{align*} \left [-\frac{3 \,{\left (4 \, b c d^{2} - 5 \, a d^{3} +{\left (4 \, b c^{2} d - 5 \, a c d^{2}\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 (2 \, a c^{3} x^{6} +{\left (4 \, b c^{3} - 5 \, a c^{2} d\right )} x^{4} + 3 \,{\left (4 \, b c^{2} d - 5 \, a c d^{2}\right )} x^{2}\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{16 \,{\left (c^{5} x^{2} + c^{4} d\right )}}, \frac{3 \,{\left (4 \, b c d^{2} - 5 \, a d^{3} +{\left (4 \, b c^{2} d - 5 \, a c d^{2}\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 (2 \, a c^{3} x^{6} +{\left (4 \, b c^{3} - 5 \, a c^{2} d\right )} x^{4} + 3 \,{\left (4 \, b c^{2} d - 5 \, a c d^{2}\right )} x^{2}\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{8 \,{\left (c^{5} x^{2} + c^{4} d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*(3*(4*b*c*d^2 - 5*a*d^3 + (4*b*c^2*d - 5*a*c*d^2)*x^2)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c)*x^2*sqrt((c*x^2
+ d)/x^2) - d) - 2*(2*a*c^3*x^6 + (4*b*c^3 - 5*a*c^2*d)*x^4 + 3*(4*b*c^2*d - 5*a*c*d^2)*x^2)*sqrt((c*x^2 + d)
/x^2))/(c^5*x^2 + c^4*d), 1/8*(3*(4*b*c*d^2 - 5*a*d^3 + (4*b*c^2*d - 5*a*c*d^2)*x^2)*sqrt(-c)*arctan(sqrt(-c)*
x^2*sqrt((c*x^2 + d)/x^2)/(c*x^2 + d)) + (2*a*c^3*x^6 + (4*b*c^3 - 5*a*c^2*d)*x^4 + 3*(4*b*c^2*d - 5*a*c*d^2)*
x^2)*sqrt((c*x^2 + d)/x^2))/(c^5*x^2 + c^4*d)]

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Sympy [A]  time = 52.4764, size = 177, normalized size = 1.5 \begin{align*} a \left (\frac{x^{5}}{4 c \sqrt{d} \sqrt{\frac{c x^{2}}{d} + 1}} - \frac{5 \sqrt{d} x^{3}}{8 c^{2} \sqrt{\frac{c x^{2}}{d} + 1}} - \frac{15 d^{\frac{3}{2}} x}{8 c^{3} \sqrt{\frac{c x^{2}}{d} + 1}} + \frac{15 d^{2} \operatorname{asinh}{\left (\frac{\sqrt{c} x}{\sqrt{d}} \right )}}{8 c^{\frac{7}{2}}}\right ) + b \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.1696, size = 261, normalized size = 2.21 \begin{align*} \frac{1}{8} \, d^{2}{\left (\frac{3 \,{\left (4 \, b c - 5 \, a d\right )} \arctan \left (\frac{\sqrt{\frac{c x^{2} + d}{x^{2}}}}{\sqrt{-c}}\right )}{\sqrt{-c} c^{3} d} + \frac{8 \,{\left (b c - a d\right )}}{c^{3} d \sqrt{\frac{c x^{2} + d}{x^{2}}}} - \frac{4 \, b c^{2} \sqrt{\frac{c x^{2} + d}{x^{2}}} - 9 \, a c d \sqrt{\frac{c x^{2} + d}{x^{2}}} - \frac{4 \,{\left (c x^{2} + d\right )} b c \sqrt{\frac{c x^{2} + d}{x^{2}}}}{x^{2}} + \frac{7 \,{\left (c x^{2} + d\right )} a d \sqrt{\frac{c x^{2} + d}{x^{2}}}}{x^{2}}}{{\left (c - \frac{c x^{2} + d}{x^{2}}\right )}^{2} c^{3} d}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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