3.453 \(\int \frac{1}{(a+\frac{c}{x^2}+\frac{b}{x})^{3/2}} \, dx\)

Optimal. Leaf size=133 \[ \frac{x \left (3 b^2-8 a c\right ) \sqrt{a+\frac{b}{x}+\frac{c}{x^2}}}{a^2 \left (b^2-4 a c\right )}-\frac{3 b \tanh ^{-1}\left (\frac{2 a+\frac{b}{x}}{2 \sqrt{a} \sqrt{a+\frac{b}{x}+\frac{c}{x^2}}}\right )}{2 a^{5/2}}-\frac{2 x \left (-2 a c+b^2+\frac{b c}{x}\right )}{a \left (b^2-4 a c\right ) \sqrt{a+\frac{b}{x}+\frac{c}{x^2}}} \]

[Out]

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

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Rubi [A]  time = 0.0992594, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {1342, 740, 806, 724, 206} \[ \frac{x \left (3 b^2-8 a c\right ) \sqrt{a+\frac{b}{x}+\frac{c}{x^2}}}{a^2 \left (b^2-4 a c\right )}-\frac{3 b \tanh ^{-1}\left (\frac{2 a+\frac{b}{x}}{2 \sqrt{a} \sqrt{a+\frac{b}{x}+\frac{c}{x^2}}}\right )}{2 a^{5/2}}-\frac{2 x \left (-2 a c+b^2+\frac{b c}{x}\right )}{a \left (b^2-4 a c\right ) \sqrt{a+\frac{b}{x}+\frac{c}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[(a + c/x^2 + b/x)^(-3/2),x]

[Out]

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

Rule 1342

Int[((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n + c/x^(2*n))^p/x^2,
x], x, 1/x] /; FreeQ[{a, b, c, p}, x] && EqQ[n2, 2*n] && ILtQ[n, 0]

Rule 740

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e
+ a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*Simp[b*c*d*e*(2*p - m
+ 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x
, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b
*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[p, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x],
x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[Sim
plify[m + 2*p + 3], 0]

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

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])

Rubi steps

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

Mathematica [A]  time = 0.174029, size = 138, normalized size = 1.04 \[ -\frac{2 \sqrt{a} \left (-b^2 \left (a x^2+3 c\right )+10 a b c x+4 a c \left (a x^2+2 c\right )-3 b^3 x\right )+3 b \left (b^2-4 a c\right ) \sqrt{x (a x+b)+c} \tanh ^{-1}\left (\frac{2 a x+b}{2 \sqrt{a} \sqrt{x (a x+b)+c}}\right )}{2 a^{5/2} x \left (b^2-4 a c\right ) \sqrt{a+\frac{b x+c}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + c/x^2 + b/x)^(-3/2),x]

[Out]

-(2*Sqrt[a]*(-3*b^3*x + 10*a*b*c*x + 4*a*c*(2*c + a*x^2) - b^2*(3*c + a*x^2)) + 3*b*(b^2 - 4*a*c)*Sqrt[c + x*(
b + a*x)]*ArcTanh[(b + 2*a*x)/(2*Sqrt[a]*Sqrt[c + x*(b + a*x)])])/(2*a^(5/2)*(b^2 - 4*a*c)*x*Sqrt[a + (c + b*x
)/x^2])

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Maple [A]  time = 0.009, size = 197, normalized size = 1.5 \begin{align*}{\frac{a{x}^{2}+bx+c}{2\,{x}^{3} \left ( 4\,ac-{b}^{2} \right ) } \left ( 8\,{a}^{7/2}{x}^{2}c-2\,{a}^{5/2}{x}^{2}{b}^{2}+20\,{a}^{5/2}xbc-6\,{a}^{3/2}x{b}^{3}+16\,{a}^{5/2}{c}^{2}-6\,{a}^{3/2}{b}^{2}c-12\,\ln \left ( 1/2\,{\frac{2\,\sqrt{a{x}^{2}+bx+c}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ) \sqrt{a{x}^{2}+bx+c}{a}^{2}bc+3\,\ln \left ( 1/2\,{\frac{2\,\sqrt{a{x}^{2}+bx+c}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ) \sqrt{a{x}^{2}+bx+c}a{b}^{3} \right ){a}^{-{\frac{7}{2}}} \left ({\frac{a{x}^{2}+bx+c}{{x}^{2}}} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+c/x^2+b/x)^(3/2),x)

[Out]

1/2*(a*x^2+b*x+c)/a^(7/2)*(8*a^(7/2)*x^2*c-2*a^(5/2)*x^2*b^2+20*a^(5/2)*x*b*c-6*a^(3/2)*x*b^3+16*a^(5/2)*c^2-6
*a^(3/2)*b^2*c-12*ln(1/2*(2*(a*x^2+b*x+c)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*(a*x^2+b*x+c)^(1/2)*a^2*b*c+3*ln(1/2
*(2*(a*x^2+b*x+c)^(1/2)*a^(1/2)+2*a*x+b)/a^(1/2))*(a*x^2+b*x+c)^(1/2)*a*b^3)/((a*x^2+b*x+c)/x^2)^(3/2)/x^3/(4*
a*c-b^2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a + \frac{b}{x} + \frac{c}{x^{2}}\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+c/x^2+b/x)^(3/2),x, algorithm="maxima")

[Out]

integrate((a + b/x + c/x^2)^(-3/2), x)

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Fricas [A]  time = 2.30451, size = 1002, normalized size = 7.53 \begin{align*} \left [\frac{3 \,{\left (b^{3} c - 4 \, a b c^{2} +{\left (a b^{3} - 4 \, a^{2} b c\right )} x^{2} +{\left (b^{4} - 4 \, a b^{2} c\right )} x\right )} \sqrt{a} \log \left (-8 \, a^{2} x^{2} - 8 \, a b x - b^{2} - 4 \, a c + 4 \,{\left (2 \, a x^{2} + b x\right )} \sqrt{a} \sqrt{\frac{a x^{2} + b x + c}{x^{2}}}\right ) + 4 \,{\left ({\left (a^{2} b^{2} - 4 \, a^{3} c\right )} x^{3} +{\left (3 \, a b^{3} - 10 \, a^{2} b c\right )} x^{2} +{\left (3 \, a b^{2} c - 8 \, a^{2} c^{2}\right )} x\right )} \sqrt{\frac{a x^{2} + b x + c}{x^{2}}}}{4 \,{\left (a^{3} b^{2} c - 4 \, a^{4} c^{2} +{\left (a^{4} b^{2} - 4 \, a^{5} c\right )} x^{2} +{\left (a^{3} b^{3} - 4 \, a^{4} b c\right )} x\right )}}, \frac{3 \,{\left (b^{3} c - 4 \, a b c^{2} +{\left (a b^{3} - 4 \, a^{2} b c\right )} x^{2} +{\left (b^{4} - 4 \, a b^{2} c\right )} x\right )} \sqrt{-a} \arctan \left (\frac{{\left (2 \, a x^{2} + b x\right )} \sqrt{-a} \sqrt{\frac{a x^{2} + b x + c}{x^{2}}}}{2 \,{\left (a^{2} x^{2} + a b x + a c\right )}}\right ) + 2 \,{\left ({\left (a^{2} b^{2} - 4 \, a^{3} c\right )} x^{3} +{\left (3 \, a b^{3} - 10 \, a^{2} b c\right )} x^{2} +{\left (3 \, a b^{2} c - 8 \, a^{2} c^{2}\right )} x\right )} \sqrt{\frac{a x^{2} + b x + c}{x^{2}}}}{2 \,{\left (a^{3} b^{2} c - 4 \, a^{4} c^{2} +{\left (a^{4} b^{2} - 4 \, a^{5} c\right )} x^{2} +{\left (a^{3} b^{3} - 4 \, a^{4} b c\right )} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+c/x^2+b/x)^(3/2),x, algorithm="fricas")

[Out]

[1/4*(3*(b^3*c - 4*a*b*c^2 + (a*b^3 - 4*a^2*b*c)*x^2 + (b^4 - 4*a*b^2*c)*x)*sqrt(a)*log(-8*a^2*x^2 - 8*a*b*x -
b^2 - 4*a*c + 4*(2*a*x^2 + b*x)*sqrt(a)*sqrt((a*x^2 + b*x + c)/x^2)) + 4*((a^2*b^2 - 4*a^3*c)*x^3 + (3*a*b^3
- 10*a^2*b*c)*x^2 + (3*a*b^2*c - 8*a^2*c^2)*x)*sqrt((a*x^2 + b*x + c)/x^2))/(a^3*b^2*c - 4*a^4*c^2 + (a^4*b^2
- 4*a^5*c)*x^2 + (a^3*b^3 - 4*a^4*b*c)*x), 1/2*(3*(b^3*c - 4*a*b*c^2 + (a*b^3 - 4*a^2*b*c)*x^2 + (b^4 - 4*a*b^
2*c)*x)*sqrt(-a)*arctan(1/2*(2*a*x^2 + b*x)*sqrt(-a)*sqrt((a*x^2 + b*x + c)/x^2)/(a^2*x^2 + a*b*x + a*c)) + 2*
((a^2*b^2 - 4*a^3*c)*x^3 + (3*a*b^3 - 10*a^2*b*c)*x^2 + (3*a*b^2*c - 8*a^2*c^2)*x)*sqrt((a*x^2 + b*x + c)/x^2)
)/(a^3*b^2*c - 4*a^4*c^2 + (a^4*b^2 - 4*a^5*c)*x^2 + (a^3*b^3 - 4*a^4*b*c)*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + \frac{b}{x} + \frac{c}{x^{2}}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+c/x**2+b/x)**(3/2),x)

[Out]

Integral((a + b/x + c/x**2)**(-3/2), x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+c/x^2+b/x)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError