3.3.66 \(\int \frac {\sqrt {\frac {e (a+b x^2)}{c+d x^2}}}{x} \, dx\) [266]

Optimal. Leaf size=112 \[ -\frac {\sqrt {a} \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {a} \sqrt {e}}\right )}{\sqrt {c}}+\frac {\sqrt {b} \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{\sqrt {d}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.14, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {1981, 1980, 492, 214} \begin {gather*} \frac {\sqrt {b} \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{\sqrt {d}}-\frac {\sqrt {a} \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {a} \sqrt {e}}\right )}{\sqrt {c}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 214

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

Rule 492

Int[((e_.)*(x_))^(m_.)/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(-a)*(e^n/(b*c -
a*d)), Int[(e*x)^(m - n)/(a + b*x^n), x], x] + Dist[c*(e^n/(b*c - a*d)), Int[(e*x)^(m - n)/(c + d*x^n), x], x
] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LeQ[n, m, 2*n - 1]

Rule 1980

Int[(x_)^(m_.)*(((e_.)*((a_.) + (b_.)*(x_)))/((c_) + (d_.)*(x_)))^(p_), x_Symbol] :> With[{q = Denominator[p]}
, Dist[q*e*(b*c - a*d), Subst[Int[x^(q*(p + 1) - 1)*(((-a)*e + c*x^q)^m/(b*e - d*x^q)^(m + 2)), x], x, (e*((a
+ b*x)/(c + d*x)))^(1/q)], x]] /; FreeQ[{a, b, c, d, e, m}, x] && FractionQ[p] && IntegerQ[m]

Rule 1981

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.47, size = 147, normalized size = 1.31 \begin {gather*} \frac {\sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}} \sqrt {c+d x^2} \left (-\sqrt {a} \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x^2}}{\sqrt {a} \sqrt {c+d x^2}}\right )+\sqrt {b} \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b} \sqrt {c+d x^2}}\right )\right )}{\sqrt {c} \sqrt {d} \sqrt {a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(178\) vs. \(2(84)=168\).
time = 0.06, size = 179, normalized size = 1.60

method	result	size
default	\(\frac {\sqrt {\frac {e \left (b \,x^{2}+a \right )}{d \,x^{2}+c}}\, \left (d \,x^{2}+c \right ) \left (b \ln \left (\frac {2 b d \,x^{2}+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}+a d +b c}{2 \sqrt {b d}}\right ) \sqrt {a c}-a \ln \left (\frac {a d \,x^{2}+b c \,x^{2}+2 \sqrt {a c}\, \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}+2 a c}{x^{2}}\right ) \sqrt {b d}\right )}{2 \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {b d}\, \sqrt {a c}}\)	\(179\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(e*(b*x^2+a)/(d*x^2+c))^(1/2)*(d*x^2+c)*(b*ln(1/2*(2*b*d*x^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(
1/2)+a*d+b*c)/(b*d)^(1/2))*(a*c)^(1/2)-a*ln((a*d*x^2+b*c*x^2+2*(a*c)^(1/2)*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)
+2*a*c)/x^2)*(b*d)^(1/2))/((d*x^2+c)*(b*x^2+a))^(1/2)/(b*d)^(1/2)/(a*c)^(1/2)

________________________________________________________________________________________

Maxima [A]
time = 0.52, size = 140, normalized size = 1.25 \begin {gather*} \frac {1}{2} \, {\left (\frac {a \log \left (\frac {c \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} - \sqrt {a c}}{c \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} + \sqrt {a c}}\right )}{\sqrt {a c}} - \frac {b \log \left (\frac {d \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} - \sqrt {b d}}{d \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} + \sqrt {b d}}\right )}{\sqrt {b d}}\right )} e^{\frac {1}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 0.46, size = 815, normalized size = 7.28 \begin {gather*} \left [\frac {1}{4} \, \sqrt {\frac {b}{d}} e^{\frac {1}{2}} \log \left (8 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x^{2} + 4 \, {\left (2 \, b d^{3} x^{4} + b c^{2} d + a c d^{2} + {\left (3 \, b c d^{2} + a d^{3}\right )} x^{2}\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} \sqrt {\frac {b}{d}}\right ) + \frac {1}{4} \, \sqrt {\frac {a}{c}} e^{\frac {1}{2}} \log \left (\frac {{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x^{2} - 4 \, {\left ({\left (b c^{2} d + a c d^{2}\right )} x^{4} + 2 \, a c^{3} + {\left (b c^{3} + 3 \, a c^{2} d\right )} x^{2}\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} \sqrt {\frac {a}{c}}}{x^{4}}\right ), -\frac {1}{2} \, \sqrt {-\frac {b}{d}} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} \sqrt {-\frac {b}{d}}}{2 \, {\left (b^{2} x^{2} + a b\right )}}\right ) e^{\frac {1}{2}} + \frac {1}{4} \, \sqrt {\frac {a}{c}} e^{\frac {1}{2}} \log \left (\frac {{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{4} + 8 \, a^{2} c^{2} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x^{2} - 4 \, {\left ({\left (b c^{2} d + a c d^{2}\right )} x^{4} + 2 \, a c^{3} + {\left (b c^{3} + 3 \, a c^{2} d\right )} x^{2}\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} \sqrt {\frac {a}{c}}}{x^{4}}\right ), \frac {1}{2} \, \sqrt {-\frac {a}{c}} \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} \sqrt {-\frac {a}{c}}}{2 \, {\left (a b x^{2} + a^{2}\right )}}\right ) e^{\frac {1}{2}} + \frac {1}{4} \, \sqrt {\frac {b}{d}} e^{\frac {1}{2}} \log \left (8 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x^{2} + 4 \, {\left (2 \, b d^{3} x^{4} + b c^{2} d + a c d^{2} + {\left (3 \, b c d^{2} + a d^{3}\right )} x^{2}\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} \sqrt {\frac {b}{d}}\right ), \frac {1}{2} \, \sqrt {-\frac {a}{c}} \arctan \left (\frac {{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} \sqrt {-\frac {a}{c}}}{2 \, {\left (a b x^{2} + a^{2}\right )}}\right ) e^{\frac {1}{2}} - \frac {1}{2} \, \sqrt {-\frac {b}{d}} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {\frac {b x^{2} + a}{d x^{2} + c}} \sqrt {-\frac {b}{d}}}{2 \, {\left (b^{2} x^{2} + a b\right )}}\right ) e^{\frac {1}{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*sqrt(b/d)*e^(1/2)*log(8*b^2*d^2*x^4 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 8*(b^2*c*d + a*b*d^2)*x^2 + 4*(2*b*
d^3*x^4 + b*c^2*d + a*c*d^2 + (3*b*c*d^2 + a*d^3)*x^2)*sqrt((b*x^2 + a)/(d*x^2 + c))*sqrt(b/d)) + 1/4*sqrt(a/c
)*e^(1/2)*log(((b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^4 + 8*a^2*c^2 + 8*(a*b*c^2 + a^2*c*d)*x^2 - 4*((b*c^2*d + a*c
*d^2)*x^4 + 2*a*c^3 + (b*c^3 + 3*a*c^2*d)*x^2)*sqrt((b*x^2 + a)/(d*x^2 + c))*sqrt(a/c))/x^4), -1/2*sqrt(-b/d)*
arctan(1/2*(2*b*d*x^2 + b*c + a*d)*sqrt((b*x^2 + a)/(d*x^2 + c))*sqrt(-b/d)/(b^2*x^2 + a*b))*e^(1/2) + 1/4*sqr
t(a/c)*e^(1/2)*log(((b^2*c^2 + 6*a*b*c*d + a^2*d^2)*x^4 + 8*a^2*c^2 + 8*(a*b*c^2 + a^2*c*d)*x^2 - 4*((b*c^2*d
+ a*c*d^2)*x^4 + 2*a*c^3 + (b*c^3 + 3*a*c^2*d)*x^2)*sqrt((b*x^2 + a)/(d*x^2 + c))*sqrt(a/c))/x^4), 1/2*sqrt(-a
/c)*arctan(1/2*((b*c + a*d)*x^2 + 2*a*c)*sqrt((b*x^2 + a)/(d*x^2 + c))*sqrt(-a/c)/(a*b*x^2 + a^2))*e^(1/2) + 1
/4*sqrt(b/d)*e^(1/2)*log(8*b^2*d^2*x^4 + b^2*c^2 + 6*a*b*c*d + a^2*d^2 + 8*(b^2*c*d + a*b*d^2)*x^2 + 4*(2*b*d^
3*x^4 + b*c^2*d + a*c*d^2 + (3*b*c*d^2 + a*d^3)*x^2)*sqrt((b*x^2 + a)/(d*x^2 + c))*sqrt(b/d)), 1/2*sqrt(-a/c)*
arctan(1/2*((b*c + a*d)*x^2 + 2*a*c)*sqrt((b*x^2 + a)/(d*x^2 + c))*sqrt(-a/c)/(a*b*x^2 + a^2))*e^(1/2) - 1/2*s
qrt(-b/d)*arctan(1/2*(2*b*d*x^2 + b*c + a*d)*sqrt((b*x^2 + a)/(d*x^2 + c))*sqrt(-b/d)/(b^2*x^2 + a*b))*e^(1/2)
]

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(sa

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________