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

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

[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 = 0.13, 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 = {1342, 732, 814, 843, 621, 206, 724} \[ -\frac {3 \left (4 a c+b^2\right ) \tanh ^{-1}\left (\frac {b+\frac {2 c}{x}}{2 \sqrt {c} \sqrt {a+\frac {b}{x}+\frac {c}{x^2}}}\right )}{8 \sqrt {c}}+x \left (a+\frac {b}{x}+\frac {c}{x^2}\right )^{3/2}-\frac {3}{4} \left (3 b+\frac {2 c}{x}\right ) \sqrt {a+\frac {b}{x}+\frac {c}{x^2}}+\frac {3}{2} \sqrt {a} b \tanh ^{-1}\left (\frac {2 a+\frac {b}{x}}{2 \sqrt {a} \sqrt {a+\frac {b}{x}+\frac {c}{x^2}}}\right ) \]

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 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])

Rule 621

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 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 732

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 814

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 843

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 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]

Rubi steps

\begin {align*} \int \left (a+\frac {c}{x^2}+\frac {b}{x}\right )^{3/2} \, dx &=-\operatorname {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} \operatorname {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 \operatorname {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) \operatorname {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 ) \operatorname {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) \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 )-\frac {1}{4} \left (3 \left (b^2+4 a c\right )\right ) \operatorname {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 = 0.25, size = 163, normalized size = 1.12 \[ \frac {\sqrt {a+\frac {b x+c}{x^2}} \left (-3 x^2 \left (4 a c+b^2\right ) \tanh ^{-1}\left (\frac {b x+2 c}{2 \sqrt {c} \sqrt {x (a x+b)+c}}\right )+12 \sqrt {a} b \sqrt {c} x^2 \tanh ^{-1}\left (\frac {2 a x+b}{2 \sqrt {a} \sqrt {x (a x+b)+c}}\right )-2 \sqrt {c} (x (5 b-4 a x)+2 c) \sqrt {x (a x+b)+c}\right )}{8 \sqrt {c} x \sqrt {x (a x+b)+c}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + c/x^2 + b/x)^(3/2),x]

[Out]

(Sqrt[a + (c + b*x)/x^2]*(-2*Sqrt[c]*(2*c + x*(5*b - 4*a*x))*Sqrt[c + x*(b + a*x)] + 12*Sqrt[a]*b*Sqrt[c]*x^2*
ArcTanh[(b + 2*a*x)/(2*Sqrt[a]*Sqrt[c + x*(b + a*x)])] - 3*(b^2 + 4*a*c)*x^2*ArcTanh[(2*c + b*x)/(2*Sqrt[c]*Sq
rt[c + x*(b + a*x)])]))/(8*Sqrt[c]*x*Sqrt[c + x*(b + a*x)])

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fricas [A]  time = 1.39, size = 709, normalized size = 4.89 \[ \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 ] \]

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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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maple [B]  time = 0.01, size = 334, normalized size = 2.30 \[ -\frac {\left (\frac {a \,x^{2}+b x +c}{x^{2}}\right )^{\frac {3}{2}} \left (12 a^{\frac {5}{2}} c^{\frac {5}{2}} x^{2} \ln \left (\frac {b x +2 c +2 \sqrt {a \,x^{2}+b x +c}\, \sqrt {c}}{x}\right )-12 a^{2} b \,c^{2} x^{2} \ln \left (\frac {2 a x +b +2 \sqrt {a \,x^{2}+b x +c}\, \sqrt {a}}{2 \sqrt {a}}\right )+3 a^{\frac {3}{2}} b^{2} c^{\frac {3}{2}} x^{2} \ln \left (\frac {b x +2 c +2 \sqrt {a \,x^{2}+b x +c}\, \sqrt {c}}{x}\right )-6 \sqrt {a \,x^{2}+b x +c}\, a^{\frac {5}{2}} b c \,x^{3}-2 \left (a \,x^{2}+b x +c \right )^{\frac {3}{2}} a^{\frac {5}{2}} b \,x^{3}-12 \sqrt {a \,x^{2}+b x +c}\, a^{\frac {5}{2}} c^{2} x^{2}-6 \sqrt {a \,x^{2}+b x +c}\, a^{\frac {3}{2}} b^{2} c \,x^{2}-4 \left (a \,x^{2}+b x +c \right )^{\frac {3}{2}} a^{\frac {5}{2}} c \,x^{2}-2 \left (a \,x^{2}+b x +c \right )^{\frac {3}{2}} a^{\frac {3}{2}} b^{2} x^{2}+2 \left (a \,x^{2}+b x +c \right )^{\frac {5}{2}} a^{\frac {3}{2}} b x +4 \left (a \,x^{2}+b x +c \right )^{\frac {5}{2}} a^{\frac {3}{2}} c \right ) x}{8 \left (a \,x^{2}+b x +c \right )^{\frac {3}{2}} a^{\frac {3}{2}} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*((a*x^2+b*x+c)/x^2)^(3/2)*x*(12*a^(5/2)*c^(5/2)*ln((b*x+2*c+2*(a*x^2+b*x+c)^(1/2)*c^(1/2))/x)*x^2-2*a^(5/
2)*(a*x^2+b*x+c)^(3/2)*x^3*b-4*a^(5/2)*(a*x^2+b*x+c)^(3/2)*x^2*c-6*a^(5/2)*(a*x^2+b*x+c)^(1/2)*x^3*b*c+3*a^(3/
2)*c^(3/2)*ln((b*x+2*c+2*(a*x^2+b*x+c)^(1/2)*c^(1/2))/x)*x^2*b^2-12*a^(5/2)*(a*x^2+b*x+c)^(1/2)*x^2*c^2+2*a^(3
/2)*(a*x^2+b*x+c)^(5/2)*x*b-2*a^(3/2)*(a*x^2+b*x+c)^(3/2)*x^2*b^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)*x^2*b^2*c-12*a^2*ln(1/2*(2*a*x+b+2*(a*x^2+b*x+c)^(1/2)*a^(1/2))/a^(1/2))*x^2*b*c^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 \[ \int {\left (a + \frac {b}{x} + \frac {c}{x^{2}}\right )}^{\frac {3}{2}}\,{d x} \]

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

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

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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