3.921 \(\int \frac {\sqrt {a+\frac {c}{x^2}+\frac {b}{x}}}{d+e x} \, dx\)

Optimal. Leaf size=181 \[ -\frac {\sqrt {a d^2-e (b d-c e)} \tanh ^{-1}\left (\frac {2 a d+\frac {b d-2 c e}{x}-b e}{2 \sqrt {a+\frac {b}{x}+\frac {c}{x^2}} \sqrt {a d^2-e (b d-c e)}}\right )}{d e}-\frac {\sqrt {c} \tanh ^{-1}\left (\frac {b+\frac {2 c}{x}}{2 \sqrt {c} \sqrt {a+\frac {b}{x}+\frac {c}{x^2}}}\right )}{d}+\frac {\sqrt {a} \tanh ^{-1}\left (\frac {2 a+\frac {b}{x}}{2 \sqrt {a} \sqrt {a+\frac {b}{x}+\frac {c}{x^2}}}\right )}{e} \]

[Out]

arctanh(1/2*(2*a+b/x)/a^(1/2)/(a+c/x^2+b/x)^(1/2))*a^(1/2)/e-arctanh(1/2*(b+2*c/x)/c^(1/2)/(a+c/x^2+b/x)^(1/2)
)*c^(1/2)/d-arctanh(1/2*(2*a*d-b*e+(b*d-2*c*e)/x)/(a*d^2-e*(b*d-c*e))^(1/2)/(a+c/x^2+b/x)^(1/2))*(a*d^2-e*(b*d
-c*e))^(1/2)/d/e

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Rubi [A]  time = 0.27, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1443, 1474, 895, 724, 206, 843, 621} \[ -\frac {\sqrt {a d^2-e (b d-c e)} \tanh ^{-1}\left (\frac {2 a d+\frac {b d-2 c e}{x}-b e}{2 \sqrt {a+\frac {b}{x}+\frac {c}{x^2}} \sqrt {a d^2-e (b d-c e)}}\right )}{d e}-\frac {\sqrt {c} \tanh ^{-1}\left (\frac {b+\frac {2 c}{x}}{2 \sqrt {c} \sqrt {a+\frac {b}{x}+\frac {c}{x^2}}}\right )}{d}+\frac {\sqrt {a} \tanh ^{-1}\left (\frac {2 a+\frac {b}{x}}{2 \sqrt {a} \sqrt {a+\frac {b}{x}+\frac {c}{x^2}}}\right )}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a]*ArcTanh[(2*a + b/x)/(2*Sqrt[a]*Sqrt[a + c/x^2 + b/x])])/e - (Sqrt[c]*ArcTanh[(b + (2*c)/x)/(2*Sqrt[c]
*Sqrt[a + c/x^2 + b/x])])/d - (Sqrt[a*d^2 - e*(b*d - c*e)]*ArcTanh[(2*a*d - b*e + (b*d - 2*c*e)/x)/(2*Sqrt[a*d
^2 - e*(b*d - c*e)]*Sqrt[a + c/x^2 + b/x])])/(d*e)

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 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 895

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

Rule 1443

Int[((d_) + (e_.)*(x_)^(mn_.))^(q_.)*((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Int[((
e + d*x^n)^q*(a + b*x^n + c*x^(2*n))^p)/x^(n*q), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[n2, 2*n] && EqQ[
mn, -n] && IntegerQ[q] && (PosQ[n] ||  !IntegerQ[p])

Rule 1474

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :>
Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a,
b, c, d, e, m, n, p, q}, x] && EqQ[n2, 2*n] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

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

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Mathematica [A]  time = 0.28, size = 189, normalized size = 1.04 \[ -\frac {x \sqrt {a+\frac {b x+c}{x^2}} \left (\sqrt {a d^2-b d e+c e^2} \tanh ^{-1}\left (\frac {2 a d x+b d-b e x-2 c e}{2 \sqrt {x (a x+b)+c} \sqrt {a d^2-b d e+c e^2}}\right )-\sqrt {a} d \tanh ^{-1}\left (\frac {2 a x+b}{2 \sqrt {a} \sqrt {x (a x+b)+c}}\right )+\sqrt {c} e \tanh ^{-1}\left (\frac {b x+2 c}{2 \sqrt {c} \sqrt {x (a x+b)+c}}\right )\right )}{d e \sqrt {x (a x+b)+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((x*Sqrt[a + (c + b*x)/x^2]*(-(Sqrt[a]*d*ArcTanh[(b + 2*a*x)/(2*Sqrt[a]*Sqrt[c + x*(b + a*x)])]) + Sqrt[c]*e*
ArcTanh[(2*c + b*x)/(2*Sqrt[c]*Sqrt[c + x*(b + a*x)])] + Sqrt[a*d^2 - b*d*e + c*e^2]*ArcTanh[(b*d - 2*c*e + 2*
a*d*x - b*e*x)/(2*Sqrt[a*d^2 - b*d*e + c*e^2]*Sqrt[c + x*(b + a*x)])]))/(d*e*Sqrt[c + x*(b + a*x)]))

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fricas [A]  time = 103.95, size = 2411, normalized size = 13.32 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(a)*d*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)
) + sqrt(c)*e*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) + sqrt(a*d^2 - b*d*e + c*e^2)*log((8*b*c*d*e - 8*c^2*e^2 - (b^2 + 4*a*c)*d^2 - (8*a^2*d^2 - 8*a*b*d*e
+ (b^2 + 4*a*c)*e^2)*x^2 - 2*(4*a*b*d^2 + 4*b*c*e^2 - (3*b^2 + 4*a*c)*d*e)*x + 4*sqrt(a*d^2 - b*d*e + c*e^2)*(
(2*a*d - b*e)*x^2 + (b*d - 2*c*e)*x)*sqrt((a*x^2 + b*x + c)/x^2))/(e^2*x^2 + 2*d*e*x + d^2)))/(d*e), -1/2*(2*s
qrt(-a)*d*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)) - sqrt(c)*e
*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) - sqr
t(a*d^2 - b*d*e + c*e^2)*log((8*b*c*d*e - 8*c^2*e^2 - (b^2 + 4*a*c)*d^2 - (8*a^2*d^2 - 8*a*b*d*e + (b^2 + 4*a*
c)*e^2)*x^2 - 2*(4*a*b*d^2 + 4*b*c*e^2 - (3*b^2 + 4*a*c)*d*e)*x + 4*sqrt(a*d^2 - b*d*e + c*e^2)*((2*a*d - b*e)
*x^2 + (b*d - 2*c*e)*x)*sqrt((a*x^2 + b*x + c)/x^2))/(e^2*x^2 + 2*d*e*x + d^2)))/(d*e), 1/2*(sqrt(a)*d*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)) + sqrt(c)*e*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) - 2*sqrt(-a*d^2
+ b*d*e - c*e^2)*arctan(-1/2*sqrt(-a*d^2 + b*d*e - c*e^2)*((2*a*d - b*e)*x^2 + (b*d - 2*c*e)*x)*sqrt((a*x^2 +
b*x + c)/x^2)/(a*c*d^2 - b*c*d*e + c^2*e^2 + (a^2*d^2 - a*b*d*e + a*c*e^2)*x^2 + (a*b*d^2 - b^2*d*e + b*c*e^2
)*x)))/(d*e), -1/2*(2*sqrt(-a)*d*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)) - sqrt(c)*e*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) + 2*sqrt(-a*d^2 + b*d*e - c*e^2)*arctan(-1/2*sqrt(-a*d^2 + b*d*e - c*e^2)*((2*a*d - b*e)*x^2
+ (b*d - 2*c*e)*x)*sqrt((a*x^2 + b*x + c)/x^2)/(a*c*d^2 - b*c*d*e + c^2*e^2 + (a^2*d^2 - a*b*d*e + a*c*e^2)*x
^2 + (a*b*d^2 - b^2*d*e + b*c*e^2)*x)))/(d*e), 1/2*(2*sqrt(-c)*e*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)) + sqrt(a)*d*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)) + sqrt(a*d^2 - b*d*e + c*e^2)*log((8*b*c*d*e - 8*c^2*e^2 - (b^2 + 4*
a*c)*d^2 - (8*a^2*d^2 - 8*a*b*d*e + (b^2 + 4*a*c)*e^2)*x^2 - 2*(4*a*b*d^2 + 4*b*c*e^2 - (3*b^2 + 4*a*c)*d*e)*x
+ 4*sqrt(a*d^2 - b*d*e + c*e^2)*((2*a*d - b*e)*x^2 + (b*d - 2*c*e)*x)*sqrt((a*x^2 + b*x + c)/x^2))/(e^2*x^2 +
2*d*e*x + d^2)))/(d*e), -1/2*(2*sqrt(-a)*d*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*sqrt(-c)*e*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)) - sqrt(a*d^2 - b*d*e + c*e^2)*log((8*b*c*d*e - 8*c^2*e^2 - (b^2 + 4*a*c)*d^2 - (8*a^2*d^2 - 8
*a*b*d*e + (b^2 + 4*a*c)*e^2)*x^2 - 2*(4*a*b*d^2 + 4*b*c*e^2 - (3*b^2 + 4*a*c)*d*e)*x + 4*sqrt(a*d^2 - b*d*e +
c*e^2)*((2*a*d - b*e)*x^2 + (b*d - 2*c*e)*x)*sqrt((a*x^2 + b*x + c)/x^2))/(e^2*x^2 + 2*d*e*x + d^2)))/(d*e),
1/2*(2*sqrt(-c)*e*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)) + s
qrt(a)*d*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)) - 2*s
qrt(-a*d^2 + b*d*e - c*e^2)*arctan(-1/2*sqrt(-a*d^2 + b*d*e - c*e^2)*((2*a*d - b*e)*x^2 + (b*d - 2*c*e)*x)*sqr
t((a*x^2 + b*x + c)/x^2)/(a*c*d^2 - b*c*d*e + c^2*e^2 + (a^2*d^2 - a*b*d*e + a*c*e^2)*x^2 + (a*b*d^2 - b^2*d*e
+ b*c*e^2)*x)))/(d*e), -(sqrt(-a)*d*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)) - sqrt(-c)*e*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)) + sqrt(-a*d^2 + b*d*e - c*e^2)*arctan(-1/2*sqrt(-a*d^2 + b*d*e - c*e^2)*((2*a*d - b*e)*x^2 + (b*d - 2*
c*e)*x)*sqrt((a*x^2 + b*x + c)/x^2)/(a*c*d^2 - b*c*d*e + c^2*e^2 + (a^2*d^2 - a*b*d*e + a*c*e^2)*x^2 + (a*b*d^
2 - b^2*d*e + b*c*e^2)*x)))/(d*e)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(x)]Error: Bad Argument Type

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maple [B]  time = 0.04, size = 397, normalized size = 2.19 \[ -\frac {\sqrt {\frac {a \,x^{2}+b x +c}{x^{2}}}\, \left (-a^{\frac {3}{2}} d^{2} \ln \left (\frac {-2 a d x +b e x -b d +2 c e +2 \sqrt {a \,x^{2}+b x +c}\, \sqrt {\frac {a \,d^{2}-b d e +c \,e^{2}}{e^{2}}}\, e}{e x +d}\right )+\sqrt {a}\, b d e \ln \left (\frac {-2 a d x +b e x -b d +2 c e +2 \sqrt {a \,x^{2}+b x +c}\, \sqrt {\frac {a \,d^{2}-b d e +c \,e^{2}}{e^{2}}}\, e}{e x +d}\right )-\sqrt {a}\, c \,e^{2} \ln \left (\frac {-2 a d x +b e x -b d +2 c e +2 \sqrt {a \,x^{2}+b x +c}\, \sqrt {\frac {a \,d^{2}-b d e +c \,e^{2}}{e^{2}}}\, e}{e x +d}\right )-\sqrt {\frac {a \,d^{2}-b d e +c \,e^{2}}{e^{2}}}\, a d e \ln \left (\frac {2 a x +b +2 \sqrt {a \,x^{2}+b x +c}\, \sqrt {a}}{2 \sqrt {a}}\right )+\sqrt {\frac {a \,d^{2}-b d e +c \,e^{2}}{e^{2}}}\, \sqrt {a}\, \sqrt {c}\, e^{2} \ln \left (\frac {b x +2 c +2 \sqrt {a \,x^{2}+b x +c}\, \sqrt {c}}{x}\right )\right ) x}{\sqrt {a \,x^{2}+b x +c}\, \sqrt {\frac {a \,d^{2}-b d e +c \,e^{2}}{e^{2}}}\, \sqrt {a}\, d \,e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+c/x^2+b/x)^(1/2)/(e*x+d),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a + \frac {b}{x} + \frac {c}{x^{2}}}}{e x + d}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(a + b/x + c/x^2)/(e*x + d), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a+\frac {b}{x}+\frac {c}{x^2}}}{d+e\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x + c/x^2)^(1/2)/(d + e*x),x)

[Out]

int((a + b/x + c/x^2)^(1/2)/(d + e*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a + \frac {b}{x} + \frac {c}{x^{2}}}}{d + e x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a + b/x + c/x**2)/(d + e*x), x)

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