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

Optimal. Leaf size=115 \[ \frac{x^2 \left (c+\frac{d}{x^2}\right )^{3/2} (a d+4 b c)}{8 c}-\frac{3 d \sqrt{c+\frac{d}{x^2}} (a d+4 b c)}{8 c}+\frac{3 d (a d+4 b c) \tanh ^{-1}\left (\frac{\sqrt{c+\frac{d}{x^2}}}{\sqrt{c}}\right )}{8 \sqrt{c}}+\frac{a x^4 \left (c+\frac{d}{x^2}\right )^{5/2}}{4 c} \]

[Out]

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

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Rubi [A]  time = 0.0844772, antiderivative size = 115, normalized size of antiderivative = 1., 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 = {446, 78, 47, 50, 63, 208} \[ \frac{x^2 \left (c+\frac{d}{x^2}\right )^{3/2} (a d+4 b c)}{8 c}-\frac{3 d \sqrt{c+\frac{d}{x^2}} (a d+4 b c)}{8 c}+\frac{3 d (a d+4 b c) \tanh ^{-1}\left (\frac{\sqrt{c+\frac{d}{x^2}}}{\sqrt{c}}\right )}{8 \sqrt{c}}+\frac{a x^4 \left (c+\frac{d}{x^2}\right )^{5/2}}{4 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 47

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 \left (a+\frac{b}{x^2}\right ) \left (c+\frac{d}{x^2}\right )^{3/2} x^3 \, dx &=-\left (\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x) (c+d x)^{3/2}}{x^3} \, dx,x,\frac{1}{x^2}\right )\right )\\ &=\frac{a \left (c+\frac{d}{x^2}\right )^{5/2} x^4}{4 c}-\frac{(4 b c+a d) \operatorname{Subst}\left (\int \frac{(c+d x)^{3/2}}{x^2} \, dx,x,\frac{1}{x^2}\right )}{8 c}\\ &=\frac{(4 b c+a d) \left (c+\frac{d}{x^2}\right )^{3/2} x^2}{8 c}+\frac{a \left (c+\frac{d}{x^2}\right )^{5/2} x^4}{4 c}-\frac{(3 d (4 b c+a d)) \operatorname{Subst}\left (\int \frac{\sqrt{c+d x}}{x} \, dx,x,\frac{1}{x^2}\right )}{16 c}\\ &=-\frac{3 d (4 b c+a d) \sqrt{c+\frac{d}{x^2}}}{8 c}+\frac{(4 b c+a d) \left (c+\frac{d}{x^2}\right )^{3/2} x^2}{8 c}+\frac{a \left (c+\frac{d}{x^2}\right )^{5/2} x^4}{4 c}-\frac{1}{16} (3 d (4 b c+a d)) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+d x}} \, dx,x,\frac{1}{x^2}\right )\\ &=-\frac{3 d (4 b c+a d) \sqrt{c+\frac{d}{x^2}}}{8 c}+\frac{(4 b c+a d) \left (c+\frac{d}{x^2}\right )^{3/2} x^2}{8 c}+\frac{a \left (c+\frac{d}{x^2}\right )^{5/2} x^4}{4 c}-\frac{1}{8} (3 (4 b c+a d)) \operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+\frac{d}{x^2}}\right )\\ &=-\frac{3 d (4 b c+a d) \sqrt{c+\frac{d}{x^2}}}{8 c}+\frac{(4 b c+a d) \left (c+\frac{d}{x^2}\right )^{3/2} x^2}{8 c}+\frac{a \left (c+\frac{d}{x^2}\right )^{5/2} x^4}{4 c}+\frac{3 d (4 b c+a d) \tanh ^{-1}\left (\frac{\sqrt{c+\frac{d}{x^2}}}{\sqrt{c}}\right )}{8 \sqrt{c}}\\ \end{align*}

Mathematica [A]  time = 0.176663, size = 89, normalized size = 0.77 \[ \frac{1}{8} \sqrt{c+\frac{d}{x^2}} \left (\frac{3 \sqrt{d} x (a d+4 b c) \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{d}}\right )}{\sqrt{c} \sqrt{\frac{c x^2}{d}+1}}+2 a c x^4+5 a d x^2+4 b c x^2-8 b d\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.013, size = 174, normalized size = 1.5 \begin{align*}{\frac{{x}^{2}}{8\,d} \left ({\frac{c{x}^{2}+d}{{x}^{2}}} \right ) ^{{\frac{3}{2}}} \left ( 8\,{c}^{3/2} \left ( c{x}^{2}+d \right ) ^{3/2}{x}^{2}b+12\,{c}^{3/2}\sqrt{c{x}^{2}+d}{x}^{2}bd+2\,\sqrt{c} \left ( c{x}^{2}+d \right ) ^{3/2}{x}^{2}ad-8\,\sqrt{c} \left ( c{x}^{2}+d \right ) ^{5/2}b+3\,\sqrt{c}\sqrt{c{x}^{2}+d}{x}^{2}a{d}^{2}+3\,\ln \left ( \sqrt{c}x+\sqrt{c{x}^{2}+d} \right ) xa{d}^{3}+12\,\ln \left ( \sqrt{c}x+\sqrt{c{x}^{2}+d} \right ) xbc{d}^{2} \right ) \left ( c{x}^{2}+d \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.36592, size = 462, normalized size = 4.02 \begin{align*} \left [\frac{3 \,{\left (4 \, b c d + a d^{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^{2} x^{4} - 8 \, b c d +{\left (4 \, b c^{2} + 5 \, a c d\right )} x^{2}\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{16 \, c}, -\frac{3 \,{\left (4 \, b c d + a d^{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^{2} x^{4} - 8 \, b c d +{\left (4 \, b c^{2} + 5 \, a c d\right )} x^{2}\right )} \sqrt{\frac{c x^{2} + d}{x^{2}}}}{8 \, c}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 89.8581, size = 216, normalized size = 1.88 \begin{align*} \frac{a c^{2} x^{5}}{4 \sqrt{d} \sqrt{\frac{c x^{2}}{d} + 1}} + \frac{3 a c \sqrt{d} x^{3}}{8 \sqrt{\frac{c x^{2}}{d} + 1}} + \frac{a d^{\frac{3}{2}} x \sqrt{\frac{c x^{2}}{d} + 1}}{2} + \frac{a d^{\frac{3}{2}} x}{8 \sqrt{\frac{c x^{2}}{d} + 1}} + \frac{3 a d^{2} \operatorname{asinh}{\left (\frac{\sqrt{c} x}{\sqrt{d}} \right )}}{8 \sqrt{c}} + \frac{3 b \sqrt{c} d \operatorname{asinh}{\left (\frac{\sqrt{c} x}{\sqrt{d}} \right )}}{2} + \frac{b c \sqrt{d} x \sqrt{\frac{c x^{2}}{d} + 1}}{2} - \frac{b c \sqrt{d} x}{\sqrt{\frac{c x^{2}}{d} + 1}} - \frac{b d^{\frac{3}{2}}}{x \sqrt{\frac{c x^{2}}{d} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.29828, size = 170, normalized size = 1.48 \begin{align*} \frac{2 \, b \sqrt{c} d^{2} \mathrm{sgn}\left (x\right )}{{\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{2} - d} + \frac{1}{8} \,{\left (2 \, a c x^{2} \mathrm{sgn}\left (x\right ) + \frac{4 \, b c^{3} \mathrm{sgn}\left (x\right ) + 5 \, a c^{2} d \mathrm{sgn}\left (x\right )}{c^{2}}\right )} \sqrt{c x^{2} + d} x - \frac{3 \,{\left (4 \, b c^{\frac{3}{2}} d \mathrm{sgn}\left (x\right ) + a \sqrt{c} d^{2} \mathrm{sgn}\left (x\right )\right )} \log \left ({\left (\sqrt{c} x - \sqrt{c x^{2} + d}\right )}^{2}\right )}{16 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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