3.920 \(\int \frac {\sqrt {a+\frac {c}{x^2}}}{d+e x} \, dx\)
Optimal. Leaf size=121 \[ -\frac {\sqrt {a d^2+c e^2} \tanh ^{-1}\left (\frac {a d-\frac {c e}{x}}{\sqrt {a+\frac {c}{x^2}} \sqrt {a d^2+c e^2}}\right )}{d e}-\frac {\sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c}}{x \sqrt {a+\frac {c}{x^2}}}\right )}{d}+\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+\frac {c}{x^2}}}{\sqrt {a}}\right )}{e} \]
[Out]
arctanh((a+c/x^2)^(1/2)/a^(1/2))*a^(1/2)/e-arctanh(c^(1/2)/x/(a+c/x^2)^(1/2))*c^(1/2)/d-arctanh((a*d-c*e/x)/(a
*d^2+c*e^2)^(1/2)/(a+c/x^2)^(1/2))*(a*d^2+c*e^2)^(1/2)/d/e
________________________________________________________________________________________
Rubi [A] time = 0.16, antiderivative size = 121, normalized size of antiderivative = 1.00,
number of steps used = 11, number of rules used = 10, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used
= {1444, 1475, 896, 266, 63, 208, 844, 217, 206, 725} \[ -\frac {\sqrt {a d^2+c e^2} \tanh ^{-1}\left (\frac {a d-\frac {c e}{x}}{\sqrt {a+\frac {c}{x^2}} \sqrt {a d^2+c e^2}}\right )}{d e}-\frac {\sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c}}{x \sqrt {a+\frac {c}{x^2}}}\right )}{d}+\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+\frac {c}{x^2}}}{\sqrt {a}}\right )}{e} \]
Antiderivative was successfully verified.
[In]
Int[Sqrt[a + c/x^2]/(d + e*x),x]
[Out]
(Sqrt[a]*ArcTanh[Sqrt[a + c/x^2]/Sqrt[a]])/e - (Sqrt[a*d^2 + c*e^2]*ArcTanh[(a*d - (c*e)/x)/(Sqrt[a*d^2 + c*e^
2]*Sqrt[a + c/x^2])])/(d*e) - (Sqrt[c]*ArcTanh[Sqrt[c]/(Sqrt[a + c/x^2]*x)])/d
Rule 63
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]
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 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rule 217
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] && !GtQ[a, 0]
Rule 266
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
Rule 725
Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]
Rule 844
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && !IGtQ[m, 0]
Rule 896
Int[((a_) + (c_.)*(x_)^2)^(p_)/(((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))), x_Symbol] :> Dist[(c*d^2 + a*e^2)/
(e*(e*f - d*g)), Int[(a + c*x^2)^(p - 1)/(d + e*x), x], x] - Dist[1/(e*(e*f - d*g)), Int[(Simp[c*d*f + a*e*g -
c*(e*f - d*g)*x, x]*(a + c*x^2)^(p - 1))/(f + g*x), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g,
0] && NeQ[c*d^2 + a*e^2, 0] && FractionQ[p] && GtQ[p, 0]
Rule 1444
Int[((d_) + (e_.)*(x_)^(mn_.))^(q_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Int[x^(mn*q)*(e + d/x^mn)^
q*(a + c*x^n2)^p, x] /; FreeQ[{a, c, d, e, mn, p}, x] && EqQ[n2, -2*mn] && IntegerQ[q] && (PosQ[n2] || !Integ
erQ[p])
Rule 1475
Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, 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}}}{d+e x} \, dx &=\int \frac {\sqrt {a+\frac {c}{x^2}}}{\left (e+\frac {d}{x}\right ) x} \, dx\\ &=-\operatorname {Subst}\left (\int \frac {\sqrt {a+c x^2}}{x (e+d x)} \, dx,x,\frac {1}{x}\right )\\ &=\frac {\operatorname {Subst}\left (\int \frac {a d-c e x}{(e+d x) \sqrt {a+c x^2}} \, dx,x,\frac {1}{x}\right )}{e}-\frac {a \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+c x^2}} \, dx,x,\frac {1}{x}\right )}{e}\\ &=-\frac {c \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+c x^2}} \, dx,x,\frac {1}{x}\right )}{d}-\frac {a \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,\frac {1}{x^2}\right )}{2 e}+\left (\frac {a d}{e}+\frac {c e}{d}\right ) \operatorname {Subst}\left (\int \frac {1}{(e+d x) \sqrt {a+c x^2}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {c \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {1}{\sqrt {a+\frac {c}{x^2}} x}\right )}{d}-\frac {a \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{c}+\frac {x^2}{c}} \, dx,x,\sqrt {a+\frac {c}{x^2}}\right )}{c e}+\left (-\frac {a d}{e}-\frac {c e}{d}\right ) \operatorname {Subst}\left (\int \frac {1}{a d^2+c e^2-x^2} \, dx,x,\frac {a d-\frac {c e}{x}}{\sqrt {a+\frac {c}{x^2}}}\right )\\ &=\frac {\sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+\frac {c}{x^2}}}{\sqrt {a}}\right )}{e}-\frac {\sqrt {a d^2+c e^2} \tanh ^{-1}\left (\frac {a d-\frac {c e}{x}}{\sqrt {a d^2+c e^2} \sqrt {a+\frac {c}{x^2}}}\right )}{d e}-\frac {\sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c}}{\sqrt {a+\frac {c}{x^2}} x}\right )}{d}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 0.10, size = 136, normalized size = 1.12 \[ \frac {x \sqrt {a+\frac {c}{x^2}} \left (\sqrt {a d^2+c e^2} \tanh ^{-1}\left (\frac {c e-a d x}{\sqrt {a x^2+c} \sqrt {a d^2+c e^2}}\right )+\sqrt {a} d \tanh ^{-1}\left (\frac {\sqrt {a} x}{\sqrt {a x^2+c}}\right )-\sqrt {c} e \tanh ^{-1}\left (\frac {\sqrt {a x^2+c}}{\sqrt {c}}\right )\right )}{d e \sqrt {a x^2+c}} \]
Antiderivative was successfully verified.
[In]
Integrate[Sqrt[a + c/x^2]/(d + e*x),x]
[Out]
(Sqrt[a + c/x^2]*x*(Sqrt[a]*d*ArcTanh[(Sqrt[a]*x)/Sqrt[c + a*x^2]] + Sqrt[a*d^2 + c*e^2]*ArcTanh[(c*e - a*d*x)
/(Sqrt[a*d^2 + c*e^2]*Sqrt[c + a*x^2])] - Sqrt[c]*e*ArcTanh[Sqrt[c + a*x^2]/Sqrt[c]]))/(d*e*Sqrt[c + a*x^2])
________________________________________________________________________________________
fricas [A] time = 2.05, size = 1532, normalized size = 12.66 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+c/x^2)^(1/2)/(e*x+d),x, algorithm="fricas")
[Out]
[1/2*(sqrt(a)*d*log(-2*a*x^2 - 2*sqrt(a)*x^2*sqrt((a*x^2 + c)/x^2) - c) + sqrt(c)*e*log(-(a*x^2 - 2*sqrt(c)*x*
sqrt((a*x^2 + c)/x^2) + 2*c)/x^2) + sqrt(a*d^2 + c*e^2)*log((2*a*c*d*e*x - a*c*d^2 - 2*c^2*e^2 - (2*a^2*d^2 +
a*c*e^2)*x^2 + 2*(a*d*x^2 - c*e*x)*sqrt(a*d^2 + c*e^2)*sqrt((a*x^2 + c)/x^2))/(e^2*x^2 + 2*d*e*x + d^2)))/(d*e
), -1/2*(2*sqrt(-a)*d*arctan(sqrt(-a)*x^2*sqrt((a*x^2 + c)/x^2)/(a*x^2 + c)) - sqrt(c)*e*log(-(a*x^2 - 2*sqrt(
c)*x*sqrt((a*x^2 + c)/x^2) + 2*c)/x^2) - sqrt(a*d^2 + c*e^2)*log((2*a*c*d*e*x - a*c*d^2 - 2*c^2*e^2 - (2*a^2*d
^2 + a*c*e^2)*x^2 + 2*(a*d*x^2 - c*e*x)*sqrt(a*d^2 + c*e^2)*sqrt((a*x^2 + c)/x^2))/(e^2*x^2 + 2*d*e*x + d^2)))
/(d*e), 1/2*(sqrt(a)*d*log(-2*a*x^2 - 2*sqrt(a)*x^2*sqrt((a*x^2 + c)/x^2) - c) + sqrt(c)*e*log(-(a*x^2 - 2*sqr
t(c)*x*sqrt((a*x^2 + c)/x^2) + 2*c)/x^2) + 2*sqrt(-a*d^2 - c*e^2)*arctan((a*d*x^2 - c*e*x)*sqrt(-a*d^2 - c*e^2
)*sqrt((a*x^2 + c)/x^2)/(a*c*d^2 + c^2*e^2 + (a^2*d^2 + a*c*e^2)*x^2)))/(d*e), -1/2*(2*sqrt(-a)*d*arctan(sqrt(
-a)*x^2*sqrt((a*x^2 + c)/x^2)/(a*x^2 + c)) - sqrt(c)*e*log(-(a*x^2 - 2*sqrt(c)*x*sqrt((a*x^2 + c)/x^2) + 2*c)/
x^2) - 2*sqrt(-a*d^2 - c*e^2)*arctan((a*d*x^2 - c*e*x)*sqrt(-a*d^2 - c*e^2)*sqrt((a*x^2 + c)/x^2)/(a*c*d^2 + c
^2*e^2 + (a^2*d^2 + a*c*e^2)*x^2)))/(d*e), 1/2*(2*sqrt(-c)*e*arctan(sqrt(-c)*x*sqrt((a*x^2 + c)/x^2)/(a*x^2 +
c)) + sqrt(a)*d*log(-2*a*x^2 - 2*sqrt(a)*x^2*sqrt((a*x^2 + c)/x^2) - c) + sqrt(a*d^2 + c*e^2)*log((2*a*c*d*e*x
- a*c*d^2 - 2*c^2*e^2 - (2*a^2*d^2 + a*c*e^2)*x^2 + 2*(a*d*x^2 - c*e*x)*sqrt(a*d^2 + c*e^2)*sqrt((a*x^2 + c)/
x^2))/(e^2*x^2 + 2*d*e*x + d^2)))/(d*e), -1/2*(2*sqrt(-a)*d*arctan(sqrt(-a)*x^2*sqrt((a*x^2 + c)/x^2)/(a*x^2 +
c)) - 2*sqrt(-c)*e*arctan(sqrt(-c)*x*sqrt((a*x^2 + c)/x^2)/(a*x^2 + c)) - sqrt(a*d^2 + c*e^2)*log((2*a*c*d*e*
x - a*c*d^2 - 2*c^2*e^2 - (2*a^2*d^2 + a*c*e^2)*x^2 + 2*(a*d*x^2 - c*e*x)*sqrt(a*d^2 + c*e^2)*sqrt((a*x^2 + c)
/x^2))/(e^2*x^2 + 2*d*e*x + d^2)))/(d*e), 1/2*(2*sqrt(-c)*e*arctan(sqrt(-c)*x*sqrt((a*x^2 + c)/x^2)/(a*x^2 + c
)) + sqrt(a)*d*log(-2*a*x^2 - 2*sqrt(a)*x^2*sqrt((a*x^2 + c)/x^2) - c) + 2*sqrt(-a*d^2 - c*e^2)*arctan((a*d*x^
2 - c*e*x)*sqrt(-a*d^2 - c*e^2)*sqrt((a*x^2 + c)/x^2)/(a*c*d^2 + c^2*e^2 + (a^2*d^2 + a*c*e^2)*x^2)))/(d*e), -
(sqrt(-a)*d*arctan(sqrt(-a)*x^2*sqrt((a*x^2 + c)/x^2)/(a*x^2 + c)) - sqrt(-c)*e*arctan(sqrt(-c)*x*sqrt((a*x^2
+ c)/x^2)/(a*x^2 + c)) - sqrt(-a*d^2 - c*e^2)*arctan((a*d*x^2 - c*e*x)*sqrt(-a*d^2 - c*e^2)*sqrt((a*x^2 + c)/x
^2)/(a*c*d^2 + c^2*e^2 + (a^2*d^2 + a*c*e^2)*x^2)))/(d*e)]
________________________________________________________________________________________
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)^(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
________________________________________________________________________________________
maple [B] time = 0.03, size = 244, normalized size = 2.02 \[ \frac {\sqrt {\frac {a \,x^{2}+c}{x^{2}}}\, \left (a \,d^{2} \ln \left (\frac {-2 a d x +2 c e +2 \sqrt {a \,x^{2}+c}\, \sqrt {\frac {a \,d^{2}+c \,e^{2}}{e^{2}}}\, e}{e x +d}\right )+c \,e^{2} \ln \left (\frac {-2 a d x +2 c e +2 \sqrt {a \,x^{2}+c}\, \sqrt {\frac {a \,d^{2}+c \,e^{2}}{e^{2}}}\, e}{e x +d}\right )+\sqrt {\frac {a \,d^{2}+c \,e^{2}}{e^{2}}}\, \sqrt {a}\, d e \ln \left (\frac {a x +\sqrt {a \,x^{2}+c}\, \sqrt {a}}{\sqrt {a}}\right )-\sqrt {\frac {a \,d^{2}+c \,e^{2}}{e^{2}}}\, \sqrt {c}\, e^{2} \ln \left (\frac {2 c +2 \sqrt {a \,x^{2}+c}\, \sqrt {c}}{x}\right )\right ) x}{\sqrt {a \,x^{2}+c}\, \sqrt {\frac {a \,d^{2}+c \,e^{2}}{e^{2}}}\, d \,e^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+c/x^2)^(1/2)/(e*x+d),x)
[Out]
((a*x^2+c)/x^2)^(1/2)*x*(a^(1/2)*d*ln(((a*x^2+c)^(1/2)*a^(1/2)+a*x)/a^(1/2))*e*((a*d^2+c*e^2)/e^2)^(1/2)-((a*d
^2+c*e^2)/e^2)^(1/2)*c^(1/2)*ln(2*(c^(1/2)*(a*x^2+c)^(1/2)+c)/x)*e^2+ln(2*((a*x^2+c)^(1/2)*((a*d^2+c*e^2)/e^2)
^(1/2)*e-a*x*d+c*e)/(e*x+d))*a*d^2+ln(2*((a*x^2+c)^(1/2)*((a*d^2+c*e^2)/e^2)^(1/2)*e-a*x*d+c*e)/(e*x+d))*c*e^2
)/(a*x^2+c)^(1/2)/d/e^2/((a*d^2+c*e^2)/e^2)^(1/2)
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a + \frac {c}{x^{2}}}}{e x + d}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+c/x^2)^(1/2)/(e*x+d),x, algorithm="maxima")
[Out]
integrate(sqrt(a + c/x^2)/(e*x + d), x)
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a+\frac {c}{x^2}}}{d+e\,x} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + c/x^2)^(1/2)/(d + e*x),x)
[Out]
int((a + c/x^2)^(1/2)/(d + e*x), x)
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {a + \frac {c}{x^{2}}}}{d + e x}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+c/x**2)**(1/2)/(e*x+d),x)
[Out]
Integral(sqrt(a + c/x**2)/(d + e*x), x)
________________________________________________________________________________________