3.454 \(\int \frac{1}{(a+\frac{c}{x^2}+\frac{b}{x})^{5/2}} \, dx\)

Optimal. Leaf size=220 \[ -\frac{2 x \left (32 a^2 c^2+\frac{b c \left (5 b^2-28 a c\right )}{x}-32 a b^2 c+5 b^4\right )}{3 a^2 \left (b^2-4 a c\right )^2 \sqrt{a+\frac{b}{x}+\frac{c}{x^2}}}+\frac{x \left (128 a^2 c^2-100 a b^2 c+15 b^4\right ) \sqrt{a+\frac{b}{x}+\frac{c}{x^2}}}{3 a^3 \left (b^2-4 a c\right )^2}-\frac{5 b \tanh ^{-1}\left (\frac{2 a+\frac{b}{x}}{2 \sqrt{a} \sqrt{a+\frac{b}{x}+\frac{c}{x^2}}}\right )}{2 a^{7/2}}-\frac{2 x \left (-2 a c+b^2+\frac{b c}{x}\right )}{3 a \left (b^2-4 a c\right ) \left (a+\frac{b}{x}+\frac{c}{x^2}\right )^{3/2}} \]

[Out]

((15*b^4 - 100*a*b^2*c + 128*a^2*c^2)*Sqrt[a + c/x^2 + b/x]*x)/(3*a^3*(b^2 - 4*a*c)^2) - (2*(b^2 - 2*a*c + (b*
c)/x)*x)/(3*a*(b^2 - 4*a*c)*(a + c/x^2 + b/x)^(3/2)) - (2*(5*b^4 - 32*a*b^2*c + 32*a^2*c^2 + (b*c*(5*b^2 - 28*
a*c))/x)*x)/(3*a^2*(b^2 - 4*a*c)^2*Sqrt[a + c/x^2 + b/x]) - (5*b*ArcTanh[(2*a + b/x)/(2*Sqrt[a]*Sqrt[a + c/x^2
+ b/x])])/(2*a^(7/2))

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Rubi [A]  time = 0.193687, antiderivative size = 220, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {1342, 740, 822, 806, 724, 206} \[ -\frac{2 x \left (32 a^2 c^2+\frac{b c \left (5 b^2-28 a c\right )}{x}-32 a b^2 c+5 b^4\right )}{3 a^2 \left (b^2-4 a c\right )^2 \sqrt{a+\frac{b}{x}+\frac{c}{x^2}}}+\frac{x \left (128 a^2 c^2-100 a b^2 c+15 b^4\right ) \sqrt{a+\frac{b}{x}+\frac{c}{x^2}}}{3 a^3 \left (b^2-4 a c\right )^2}-\frac{5 b \tanh ^{-1}\left (\frac{2 a+\frac{b}{x}}{2 \sqrt{a} \sqrt{a+\frac{b}{x}+\frac{c}{x^2}}}\right )}{2 a^{7/2}}-\frac{2 x \left (-2 a c+b^2+\frac{b c}{x}\right )}{3 a \left (b^2-4 a c\right ) \left (a+\frac{b}{x}+\frac{c}{x^2}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + c/x^2 + b/x)^(-5/2),x]

[Out]

((15*b^4 - 100*a*b^2*c + 128*a^2*c^2)*Sqrt[a + c/x^2 + b/x]*x)/(3*a^3*(b^2 - 4*a*c)^2) - (2*(b^2 - 2*a*c + (b*
c)/x)*x)/(3*a*(b^2 - 4*a*c)*(a + c/x^2 + b/x)^(3/2)) - (2*(5*b^4 - 32*a*b^2*c + 32*a^2*c^2 + (b*c*(5*b^2 - 28*
a*c))/x)*x)/(3*a^2*(b^2 - 4*a*c)^2*Sqrt[a + c/x^2 + b/x]) - (5*b*ArcTanh[(2*a + b/x)/(2*Sqrt[a]*Sqrt[a + c/x^2
+ b/x])])/(2*a^(7/2))

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 822

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

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 )^{5/2}} \, dx &=-\operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b x+c x^2\right )^{5/2}} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{2 \left (b^2-2 a c+\frac{b c}{x}\right ) x}{3 a \left (b^2-4 a c\right ) \left (a+\frac{c}{x^2}+\frac{b}{x}\right )^{3/2}}+\frac{2 \operatorname{Subst}\left (\int \frac{\frac{1}{2} \left (-5 b^2+16 a c\right )-3 b c x}{x^2 \left (a+b x+c x^2\right )^{3/2}} \, dx,x,\frac{1}{x}\right )}{3 a \left (b^2-4 a c\right )}\\ &=-\frac{2 \left (b^2-2 a c+\frac{b c}{x}\right ) x}{3 a \left (b^2-4 a c\right ) \left (a+\frac{c}{x^2}+\frac{b}{x}\right )^{3/2}}-\frac{2 \left (5 b^4-32 a b^2 c+32 a^2 c^2+\frac{b c \left (5 b^2-28 a c\right )}{x}\right ) x}{3 a^2 \left (b^2-4 a c\right )^2 \sqrt{a+\frac{c}{x^2}+\frac{b}{x}}}-\frac{4 \operatorname{Subst}\left (\int \frac{\frac{1}{4} \left (15 b^4-100 a b^2 c+128 a^2 c^2\right )+\frac{1}{2} b c \left (5 b^2-28 a c\right ) x}{x^2 \sqrt{a+b x+c x^2}} \, dx,x,\frac{1}{x}\right )}{3 a^2 \left (b^2-4 a c\right )^2}\\ &=\frac{\left (15 b^4-100 a b^2 c+128 a^2 c^2\right ) \sqrt{a+\frac{c}{x^2}+\frac{b}{x}} x}{3 a^3 \left (b^2-4 a c\right )^2}-\frac{2 \left (b^2-2 a c+\frac{b c}{x}\right ) x}{3 a \left (b^2-4 a c\right ) \left (a+\frac{c}{x^2}+\frac{b}{x}\right )^{3/2}}-\frac{2 \left (5 b^4-32 a b^2 c+32 a^2 c^2+\frac{b c \left (5 b^2-28 a c\right )}{x}\right ) x}{3 a^2 \left (b^2-4 a c\right )^2 \sqrt{a+\frac{c}{x^2}+\frac{b}{x}}}+\frac{(5 b) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x+c x^2}} \, dx,x,\frac{1}{x}\right )}{2 a^3}\\ &=\frac{\left (15 b^4-100 a b^2 c+128 a^2 c^2\right ) \sqrt{a+\frac{c}{x^2}+\frac{b}{x}} x}{3 a^3 \left (b^2-4 a c\right )^2}-\frac{2 \left (b^2-2 a c+\frac{b c}{x}\right ) x}{3 a \left (b^2-4 a c\right ) \left (a+\frac{c}{x^2}+\frac{b}{x}\right )^{3/2}}-\frac{2 \left (5 b^4-32 a b^2 c+32 a^2 c^2+\frac{b c \left (5 b^2-28 a c\right )}{x}\right ) x}{3 a^2 \left (b^2-4 a c\right )^2 \sqrt{a+\frac{c}{x^2}+\frac{b}{x}}}-\frac{(5 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^3}\\ &=\frac{\left (15 b^4-100 a b^2 c+128 a^2 c^2\right ) \sqrt{a+\frac{c}{x^2}+\frac{b}{x}} x}{3 a^3 \left (b^2-4 a c\right )^2}-\frac{2 \left (b^2-2 a c+\frac{b c}{x}\right ) x}{3 a \left (b^2-4 a c\right ) \left (a+\frac{c}{x^2}+\frac{b}{x}\right )^{3/2}}-\frac{2 \left (5 b^4-32 a b^2 c+32 a^2 c^2+\frac{b c \left (5 b^2-28 a c\right )}{x}\right ) x}{3 a^2 \left (b^2-4 a c\right )^2 \sqrt{a+\frac{c}{x^2}+\frac{b}{x}}}-\frac{5 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^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.400968, size = 256, normalized size = 1.16 \[ \frac{2 \sqrt{a} \left (3 b^4 \left (a^2 x^4-30 a c x^2+5 c^2\right )-4 a b^2 c \left (6 a^2 x^4-12 a c x^2+25 c^2\right )+8 a^2 b c^2 x \left (32 a x^2+39 c\right )+16 a^2 c^2 \left (3 a^2 x^4+12 a c x^2+8 c^2\right )+10 b^5 \left (2 a x^3+3 c x\right )-2 a b^3 c x \left (74 a x^2+105 c\right )+15 b^6 x^2\right )-15 b \left (b^2-4 a c\right )^2 (x (a x+b)+c)^{3/2} \tanh ^{-1}\left (\frac{2 a x+b}{2 \sqrt{a} \sqrt{x (a x+b)+c}}\right )}{6 a^{7/2} x \left (b^2-4 a c\right )^2 (x (a x+b)+c) \sqrt{a+\frac{b x+c}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + c/x^2 + b/x)^(-5/2),x]

[Out]

(2*Sqrt[a]*(15*b^6*x^2 + 8*a^2*b*c^2*x*(39*c + 32*a*x^2) - 2*a*b^3*c*x*(105*c + 74*a*x^2) + 10*b^5*(3*c*x + 2*
a*x^3) + 3*b^4*(5*c^2 - 30*a*c*x^2 + a^2*x^4) + 16*a^2*c^2*(8*c^2 + 12*a*c*x^2 + 3*a^2*x^4) - 4*a*b^2*c*(25*c^
2 - 12*a*c*x^2 + 6*a^2*x^4)) - 15*b*(b^2 - 4*a*c)^2*(c + x*(b + a*x))^(3/2)*ArcTanh[(b + 2*a*x)/(2*Sqrt[a]*Sqr
t[c + x*(b + a*x)])])/(6*a^(7/2)*(b^2 - 4*a*c)^2*x*(c + x*(b + a*x))*Sqrt[a + (c + b*x)/x^2])

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Maple [A]  time = 0.01, size = 376, normalized size = 1.7 \begin{align*}{\frac{a{x}^{2}+bx+c}{6\,{x}^{5} \left ( 4\,ac-{b}^{2} \right ) ^{2}} \left ( 96\,{a}^{13/2}{x}^{4}{c}^{2}-48\,{a}^{11/2}{x}^{4}{b}^{2}c+512\,{a}^{11/2}{x}^{3}b{c}^{2}+6\,{a}^{9/2}{x}^{4}{b}^{4}+384\,{a}^{11/2}{x}^{2}{c}^{3}-296\,{a}^{9/2}{x}^{3}{b}^{3}c+96\,{a}^{9/2}{x}^{2}{b}^{2}{c}^{2}+40\,{a}^{7/2}{x}^{3}{b}^{5}+624\,{a}^{9/2}xb{c}^{3}-180\,{a}^{7/2}{x}^{2}{b}^{4}c+256\,{a}^{9/2}{c}^{4}-420\,{a}^{7/2}x{b}^{3}{c}^{2}+30\,{a}^{5/2}{x}^{2}{b}^{6}-200\,{a}^{7/2}{b}^{2}{c}^{3}+60\,{a}^{5/2}x{b}^{5}c+30\,{a}^{5/2}{b}^{4}{c}^{2}-240\,\ln \left ( 1/2\,{\frac{2\,\sqrt{a{x}^{2}+bx+c}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ) \left ( a{x}^{2}+bx+c \right ) ^{3/2}{a}^{4}b{c}^{2}+120\,\ln \left ( 1/2\,{\frac{2\,\sqrt{a{x}^{2}+bx+c}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ) \left ( a{x}^{2}+bx+c \right ) ^{3/2}{a}^{3}{b}^{3}c-15\,\ln \left ( 1/2\,{\frac{2\,\sqrt{a{x}^{2}+bx+c}\sqrt{a}+2\,ax+b}{\sqrt{a}}} \right ) \left ( a{x}^{2}+bx+c \right ) ^{3/2}{a}^{2}{b}^{5} \right ){a}^{-{\frac{11}{2}}} \left ({\frac{a{x}^{2}+bx+c}{{x}^{2}}} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+c/x^2+b/x)^(5/2),x)

[Out]

1/6*(a*x^2+b*x+c)*(96*a^(13/2)*x^4*c^2-48*a^(11/2)*x^4*b^2*c+512*a^(11/2)*x^3*b*c^2+6*a^(9/2)*x^4*b^4+384*a^(1
1/2)*x^2*c^3-296*a^(9/2)*x^3*b^3*c+96*a^(9/2)*x^2*b^2*c^2+40*a^(7/2)*x^3*b^5+624*a^(9/2)*x*b*c^3-180*a^(7/2)*x
^2*b^4*c+256*a^(9/2)*c^4-420*a^(7/2)*x*b^3*c^2+30*a^(5/2)*x^2*b^6-200*a^(7/2)*b^2*c^3+60*a^(5/2)*x*b^5*c+30*a^
(5/2)*b^4*c^2-240*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)^(3/2)*a^4*b*c^2+120*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)^(3/2)*a^3*b^3*c-15*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)^(3/2)*a^2*b^5)/a^(11/2)/((a*x^2+b*x+c)/x^2)^(5/2)/x^5/(4*a*c-b^
2)^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{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+c/x^2+b/x)^(5/2),x, algorithm="maxima")

[Out]

integrate((a + b/x + c/x^2)^(-5/2), x)

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Fricas [B]  time = 4.32664, size = 2286, normalized size = 10.39 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+c/x^2+b/x)^(5/2),x, algorithm="fricas")

[Out]

[1/12*(15*(b^5*c^2 - 8*a*b^3*c^3 + 16*a^2*b*c^4 + (a^2*b^5 - 8*a^3*b^3*c + 16*a^4*b*c^2)*x^4 + 2*(a*b^6 - 8*a^
2*b^4*c + 16*a^3*b^2*c^2)*x^3 + (b^7 - 6*a*b^5*c + 32*a^3*b*c^3)*x^2 + 2*(b^6*c - 8*a*b^4*c^2 + 16*a^2*b^2*c^3
)*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*(3*(a^3*b^4 - 8*a^4*b^2*c + 16*a^5*c^2)*x^5 + 4*(5*a^2*b^5 - 37*a^3*b^3*c + 64*a^4*b*c^2)*x^4 + 3*(5*a*b^6
- 30*a^2*b^4*c + 16*a^3*b^2*c^2 + 64*a^4*c^3)*x^3 + 6*(5*a*b^5*c - 35*a^2*b^3*c^2 + 52*a^3*b*c^3)*x^2 + (15*a*
b^4*c^2 - 100*a^2*b^2*c^3 + 128*a^3*c^4)*x)*sqrt((a*x^2 + b*x + c)/x^2))/(a^4*b^4*c^2 - 8*a^5*b^2*c^3 + 16*a^6
*c^4 + (a^6*b^4 - 8*a^7*b^2*c + 16*a^8*c^2)*x^4 + 2*(a^5*b^5 - 8*a^6*b^3*c + 16*a^7*b*c^2)*x^3 + (a^4*b^6 - 6*
a^5*b^4*c + 32*a^7*c^3)*x^2 + 2*(a^4*b^5*c - 8*a^5*b^3*c^2 + 16*a^6*b*c^3)*x), 1/6*(15*(b^5*c^2 - 8*a*b^3*c^3
+ 16*a^2*b*c^4 + (a^2*b^5 - 8*a^3*b^3*c + 16*a^4*b*c^2)*x^4 + 2*(a*b^6 - 8*a^2*b^4*c + 16*a^3*b^2*c^2)*x^3 + (
b^7 - 6*a*b^5*c + 32*a^3*b*c^3)*x^2 + 2*(b^6*c - 8*a*b^4*c^2 + 16*a^2*b^2*c^3)*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*(3*(a^3*b^4 - 8*a^4*b^2*c + 16*a^5*c
^2)*x^5 + 4*(5*a^2*b^5 - 37*a^3*b^3*c + 64*a^4*b*c^2)*x^4 + 3*(5*a*b^6 - 30*a^2*b^4*c + 16*a^3*b^2*c^2 + 64*a^
4*c^3)*x^3 + 6*(5*a*b^5*c - 35*a^2*b^3*c^2 + 52*a^3*b*c^3)*x^2 + (15*a*b^4*c^2 - 100*a^2*b^2*c^3 + 128*a^3*c^4
)*x)*sqrt((a*x^2 + b*x + c)/x^2))/(a^4*b^4*c^2 - 8*a^5*b^2*c^3 + 16*a^6*c^4 + (a^6*b^4 - 8*a^7*b^2*c + 16*a^8*
c^2)*x^4 + 2*(a^5*b^5 - 8*a^6*b^3*c + 16*a^7*b*c^2)*x^3 + (a^4*b^6 - 6*a^5*b^4*c + 32*a^7*c^3)*x^2 + 2*(a^4*b^
5*c - 8*a^5*b^3*c^2 + 16*a^6*b*c^3)*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{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+c/x**2+b/x)**(5/2),x)

[Out]

Integral((a + b/x + c/x**2)**(-5/2), 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)^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError