3.5.50 \(\int (a+\frac {c}{x^2}+\frac {b}{x})^{3/2} \, dx\) [450]

Optimal. Leaf size=145 \[ -\frac {3}{4} \sqrt {a+\frac {c}{x^2}+\frac {b}{x}} \left (3 b+\frac {2 c}{x}\right )+\left (a+\frac {c}{x^2}+\frac {b}{x}\right )^{3/2} x+\frac {3}{2} \sqrt {a} b \tanh ^{-1}\left (\frac {2 a+\frac {b}{x}}{2 \sqrt {a} \sqrt {a+\frac {c}{x^2}+\frac {b}{x}}}\right )-\frac {3 \left (b^2+4 a c\right ) \tanh ^{-1}\left (\frac {b+\frac {2 c}{x}}{2 \sqrt {c} \sqrt {a+\frac {c}{x^2}+\frac {b}{x}}}\right )}{8 \sqrt {c}} \]

[Out]

(a+c/x^2+b/x)^(3/2)*x+3/2*b*arctanh(1/2*(2*a+b/x)/a^(1/2)/(a+c/x^2+b/x)^(1/2))*a^(1/2)-3/8*(4*a*c+b^2)*arctanh
(1/2*(b+2*c/x)/c^(1/2)/(a+c/x^2+b/x)^(1/2))/c^(1/2)-3/4*(3*b+2*c/x)*(a+c/x^2+b/x)^(1/2)

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Rubi [A]
time = 0.09, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {1356, 746, 828, 857, 635, 212, 738} \begin {gather*} -\frac {3 \left (4 a c+b^2\right ) \tanh ^{-1}\left (\frac {b+\frac {2 c}{x}}{2 \sqrt {c} \sqrt {a+\frac {b}{x}+\frac {c}{x^2}}}\right )}{8 \sqrt {c}}+x \left (a+\frac {b}{x}+\frac {c}{x^2}\right )^{3/2}-\frac {3}{4} \left (3 b+\frac {2 c}{x}\right ) \sqrt {a+\frac {b}{x}+\frac {c}{x^2}}+\frac {3}{2} \sqrt {a} b \tanh ^{-1}\left (\frac {2 a+\frac {b}{x}}{2 \sqrt {a} \sqrt {a+\frac {b}{x}+\frac {c}{x^2}}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + c/x^2 + b/x)^(3/2),x]

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
 /; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 635

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

Rule 738

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 746

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((
a + b*x + c*x^2)^p/(e*(m + 1))), x] - Dist[p/(e*(m + 1)), Int[(d + e*x)^(m + 1)*(b + 2*c*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] && GtQ[p, 0] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m + 2*p + 1, 0] && IntQua
draticQ[a, b, c, d, e, m, p, x]

Rule 828

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[(d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*((a + b*x + c*x^
2)^p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2))), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
 + b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
 c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*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] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 857

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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]
&&  !IGtQ[m, 0]

Rule 1356

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]

Rubi steps

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

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Mathematica [A]
time = 0.54, size = 158, normalized size = 1.09 \begin {gather*} \frac {\sqrt {a+\frac {c+b x}{x^2}} \left (3 \left (b^2+4 a c\right ) x^2 \tanh ^{-1}\left (\frac {\sqrt {a} x-\sqrt {c+x (b+a x)}}{\sqrt {c}}\right )-\sqrt {c} \left ((2 c+x (5 b-4 a x)) \sqrt {c+x (b+a x)}+6 \sqrt {a} b x^2 \log \left (b+2 a x-2 \sqrt {a} \sqrt {c+x (b+a x)}\right )\right )\right )}{4 \sqrt {c} x \sqrt {c+x (b+a x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + c/x^2 + b/x)^(3/2),x]

[Out]

(Sqrt[a + (c + b*x)/x^2]*(3*(b^2 + 4*a*c)*x^2*ArcTanh[(Sqrt[a]*x - Sqrt[c + x*(b + a*x)])/Sqrt[c]] - Sqrt[c]*(
(2*c + x*(5*b - 4*a*x))*Sqrt[c + x*(b + a*x)] + 6*Sqrt[a]*b*x^2*Log[b + 2*a*x - 2*Sqrt[a]*Sqrt[c + x*(b + a*x)
]])))/(4*Sqrt[c]*x*Sqrt[c + x*(b + a*x)])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(333\) vs. \(2(119)=238\).
time = 0.04, size = 334, normalized size = 2.30

method result size
risch \(-\frac {\left (5 b x +2 c \right ) \sqrt {\frac {a \,x^{2}+b x +c}{x^{2}}}}{4 x}+\frac {\left (a \sqrt {a \,x^{2}+b x +c}+\frac {3 \sqrt {a}\, b \ln \left (\frac {\frac {b}{2}+a x}{\sqrt {a}}+\sqrt {a \,x^{2}+b x +c}\right )}{2}-\frac {3 \sqrt {c}\, \ln \left (\frac {2 c +b x +2 \sqrt {c}\, \sqrt {a \,x^{2}+b x +c}}{x}\right ) a}{2}-\frac {3 \ln \left (\frac {2 c +b x +2 \sqrt {c}\, \sqrt {a \,x^{2}+b x +c}}{x}\right ) b^{2}}{8 \sqrt {c}}\right ) \sqrt {\frac {a \,x^{2}+b x +c}{x^{2}}}\, x}{\sqrt {a \,x^{2}+b x +c}}\) \(179\)
default \(-\frac {\left (\frac {a \,x^{2}+b x +c}{x^{2}}\right )^{\frac {3}{2}} x \left (12 a^{\frac {5}{2}} c^{\frac {5}{2}} \ln \left (\frac {2 c +b x +2 \sqrt {c}\, \sqrt {a \,x^{2}+b x +c}}{x}\right ) x^{2}-2 a^{\frac {5}{2}} \left (a \,x^{2}+b x +c \right )^{\frac {3}{2}} b \,x^{3}-4 a^{\frac {5}{2}} \left (a \,x^{2}+b x +c \right )^{\frac {3}{2}} c \,x^{2}-6 a^{\frac {5}{2}} \sqrt {a \,x^{2}+b x +c}\, b c \,x^{3}+3 a^{\frac {3}{2}} c^{\frac {3}{2}} \ln \left (\frac {2 c +b x +2 \sqrt {c}\, \sqrt {a \,x^{2}+b x +c}}{x}\right ) b^{2} x^{2}-12 a^{\frac {5}{2}} \sqrt {a \,x^{2}+b x +c}\, c^{2} x^{2}+2 a^{\frac {3}{2}} \left (a \,x^{2}+b x +c \right )^{\frac {5}{2}} b x -2 a^{\frac {3}{2}} \left (a \,x^{2}+b x +c \right )^{\frac {3}{2}} b^{2} x^{2}+4 \left (a \,x^{2}+b x +c \right )^{\frac {5}{2}} c \,a^{\frac {3}{2}}-6 a^{\frac {3}{2}} \sqrt {a \,x^{2}+b x +c}\, b^{2} c \,x^{2}-12 a^{2} \ln \left (\frac {2 \sqrt {a \,x^{2}+b x +c}\, \sqrt {a}+2 a x +b}{2 \sqrt {a}}\right ) b \,c^{2} x^{2}\right )}{8 \left (a \,x^{2}+b x +c \right )^{\frac {3}{2}} c^{2} a^{\frac {3}{2}}}\) \(334\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+c/x^2+b/x)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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 = 0.42, size = 709, normalized size = 4.89 \begin {gather*} \left [\frac {12 \, \sqrt {a} b c x \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 ) + 3 \, {\left (b^{2} + 4 \, a c\right )} \sqrt {c} x \log \left (-\frac {8 \, b c x + {\left (b^{2} + 4 \, a c\right )} x^{2} + 8 \, c^{2} - 4 \, {\left (b x^{2} + 2 \, c x\right )} \sqrt {c} \sqrt {\frac {a x^{2} + b x + c}{x^{2}}}}{x^{2}}\right ) + 4 \, {\left (4 \, a c x^{2} - 5 \, b c x - 2 \, c^{2}\right )} \sqrt {\frac {a x^{2} + b x + c}{x^{2}}}}{16 \, c x}, -\frac {24 \, \sqrt {-a} b c x \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 ) - 3 \, {\left (b^{2} + 4 \, a c\right )} \sqrt {c} x \log \left (-\frac {8 \, b c x + {\left (b^{2} + 4 \, a c\right )} x^{2} + 8 \, c^{2} - 4 \, {\left (b x^{2} + 2 \, c x\right )} \sqrt {c} \sqrt {\frac {a x^{2} + b x + c}{x^{2}}}}{x^{2}}\right ) - 4 \, {\left (4 \, a c x^{2} - 5 \, b c x - 2 \, c^{2}\right )} \sqrt {\frac {a x^{2} + b x + c}{x^{2}}}}{16 \, c x}, \frac {6 \, \sqrt {a} b c x \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 ) + 3 \, {\left (b^{2} + 4 \, a c\right )} \sqrt {-c} x \arctan \left (\frac {{\left (b x^{2} + 2 \, c x\right )} \sqrt {-c} \sqrt {\frac {a x^{2} + b x + c}{x^{2}}}}{2 \, {\left (a c x^{2} + b c x + c^{2}\right )}}\right ) + 2 \, {\left (4 \, a c x^{2} - 5 \, b c x - 2 \, c^{2}\right )} \sqrt {\frac {a x^{2} + b x + c}{x^{2}}}}{8 \, c x}, -\frac {12 \, \sqrt {-a} b c x \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 ) - 3 \, {\left (b^{2} + 4 \, a c\right )} \sqrt {-c} x \arctan \left (\frac {{\left (b x^{2} + 2 \, c x\right )} \sqrt {-c} \sqrt {\frac {a x^{2} + b x + c}{x^{2}}}}{2 \, {\left (a c x^{2} + b c x + c^{2}\right )}}\right ) - 2 \, {\left (4 \, a c x^{2} - 5 \, b c x - 2 \, c^{2}\right )} \sqrt {\frac {a x^{2} + b x + c}{x^{2}}}}{8 \, c x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(12*sqrt(a)*b*c*x*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)) + 3*(b^2 + 4*a*c)*sqrt(c)*x*log(-(8*b*c*x + (b^2 + 4*a*c)*x^2 + 8*c^2 - 4*(b*x^2 + 2*c*x)*sqrt(c)*sq
rt((a*x^2 + b*x + c)/x^2))/x^2) + 4*(4*a*c*x^2 - 5*b*c*x - 2*c^2)*sqrt((a*x^2 + b*x + c)/x^2))/(c*x), -1/16*(2
4*sqrt(-a)*b*c*x*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)) - 3*
(b^2 + 4*a*c)*sqrt(c)*x*log(-(8*b*c*x + (b^2 + 4*a*c)*x^2 + 8*c^2 - 4*(b*x^2 + 2*c*x)*sqrt(c)*sqrt((a*x^2 + b*
x + c)/x^2))/x^2) - 4*(4*a*c*x^2 - 5*b*c*x - 2*c^2)*sqrt((a*x^2 + b*x + c)/x^2))/(c*x), 1/8*(6*sqrt(a)*b*c*x*l
og(-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)) + 3*(b^2 + 4*a*
c)*sqrt(-c)*x*arctan(1/2*(b*x^2 + 2*c*x)*sqrt(-c)*sqrt((a*x^2 + b*x + c)/x^2)/(a*c*x^2 + b*c*x + c^2)) + 2*(4*
a*c*x^2 - 5*b*c*x - 2*c^2)*sqrt((a*x^2 + b*x + c)/x^2))/(c*x), -1/8*(12*sqrt(-a)*b*c*x*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)) - 3*(b^2 + 4*a*c)*sqrt(-c)*x*arctan(1/2*(b*x
^2 + 2*c*x)*sqrt(-c)*sqrt((a*x^2 + b*x + c)/x^2)/(a*c*x^2 + b*c*x + c^2)) - 2*(4*a*c*x^2 - 5*b*c*x - 2*c^2)*sq
rt((a*x^2 + b*x + c)/x^2))/(c*x)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + \frac {b}{x} + \frac {c}{x^{2}}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (a+\frac {b}{x}+\frac {c}{x^2}\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x + c/x^2)^(3/2),x)

[Out]

int((a + b/x + c/x^2)^(3/2), x)

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