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

Optimal. Leaf size=106 \[ \frac {(b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{2 b^{3/2} \sqrt {d} \sqrt {e}}+\frac {\left (c+d x^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 b e} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1960, 199, 208} \[ \frac {(b c-a d) \tanh ^{-1}\left (\frac {\sqrt {d} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt {b} \sqrt {e}}\right )}{2 b^{3/2} \sqrt {d} \sqrt {e}}+\frac {\left (c+d x^2\right ) \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}}{2 b e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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 1960

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

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 152, normalized size = 1.43 \[ \frac {\sqrt {a+b x^2} \left (\sqrt {d} \sqrt {a+b x^2} \sqrt {\frac {b \left (c+d x^2\right )}{b c-a d}}+\sqrt {b c-a d} \sinh ^{-1}\left (\frac {\sqrt {d} \sqrt {a+b x^2}}{\sqrt {b c-a d}}\right )\right )}{2 b \sqrt {d} \sqrt {\frac {b \left (c+d x^2\right )}{b c-a d}} \sqrt {\frac {e \left (a+b x^2\right )}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.44, size = 313, normalized size = 2.95 \[ \left [-\frac {\sqrt {b d e} {\left (b c - a d\right )} \log \left (8 \, b^{2} d^{2} e x^{4} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} e x^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e - 4 \, {\left (2 \, b d^{2} x^{4} + b c^{2} + a c d + {\left (3 \, b c d + a d^{2}\right )} x^{2}\right )} \sqrt {b d e} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}\right ) - 4 \, {\left (b d^{2} x^{2} + b c d\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{8 \, b^{2} d e}, -\frac {\sqrt {-b d e} {\left (b c - a d\right )} \arctan \left (\frac {{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt {-b d e} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{2 \, {\left (b^{2} d e x^{2} + a b d e\right )}}\right ) - 2 \, {\left (b d^{2} x^{2} + b c d\right )} \sqrt {\frac {b e x^{2} + a e}{d x^{2} + c}}}{4 \, b^{2} d e}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/8*(sqrt(b*d*e)*(b*c - a*d)*log(8*b^2*d^2*e*x^4 + 8*(b^2*c*d + a*b*d^2)*e*x^2 + (b^2*c^2 + 6*a*b*c*d + a^2*
d^2)*e - 4*(2*b*d^2*x^4 + b*c^2 + a*c*d + (3*b*c*d + a*d^2)*x^2)*sqrt(b*d*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))
) - 4*(b*d^2*x^2 + b*c*d)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(b^2*d*e), -1/4*(sqrt(-b*d*e)*(b*c - a*d)*arctan(
1/2*(2*b*d*x^2 + b*c + a*d)*sqrt(-b*d*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))/(b^2*d*e*x^2 + a*b*d*e)) - 2*(b*d^2
*x^2 + b*c*d)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(b^2*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(x/(e*(b*x^2+a)/(d*x^2+c))^(1/2),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(t_nostep*d+c)]Unable to divide, perhaps due to rounding error%%%{%%%{1,[0,1,0]%%%},[2,0]%%%}+%%%{%%{[-2,0]:[
1,0,%%%{-1,[1,1,1]%%%}]%%},[1,1]%%%}+%%%{%%%{1,[1,0,1]%%%},[0,2]%%%} / %%%{%%%{1,[0,2,0]%%%},[2,0]%%%}+%%%{%%{
[%%%{-2,[0,1,0]%%%},0]:[1,0,%%%{-1,[1,1,1]%%%}]%%},[1,1]%%%}+%%%{%%%{1,[1,1,1]%%%},[0,2]%%%} Error: Bad Argume
nt Value

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maple [B]  time = 0.03, size = 200, normalized size = 1.89 \[ \frac {\left (b \,x^{2}+a \right ) \left (-a d \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+b c \ln \left (\frac {2 b d \,x^{2}+a d +b c +2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}}{2 \sqrt {b d}}\right )+2 \sqrt {b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}\, \sqrt {b d}\right )}{4 \sqrt {\frac {\left (b \,x^{2}+a \right ) e}{d \,x^{2}+c}}\, \sqrt {\left (d \,x^{2}+c \right ) \left (b \,x^{2}+a \right )}\, \sqrt {b d}\, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.18, size = 153, normalized size = 1.44 \[ \frac {1}{4} \, e {\left (\frac {2 \, {\left (b c - a d\right )} \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}}}{b^{2} e^{2} - \frac {{\left (b x^{2} + a\right )} b d e^{2}}{d x^{2} + c}} - \frac {{\left (b c - a d\right )} \log \left (\frac {d \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} - \sqrt {b d e}}{d \sqrt {\frac {{\left (b x^{2} + a\right )} e}{d x^{2} + c}} + \sqrt {b d e}}\right )}{\sqrt {b d e} b e}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*e*(2*(b*c - a*d)*sqrt((b*x^2 + a)*e/(d*x^2 + c))/(b^2*e^2 - (b*x^2 + a)*b*d*e^2/(d*x^2 + c)) - (b*c - a*d)
*log((d*sqrt((b*x^2 + a)*e/(d*x^2 + c)) - sqrt(b*d*e))/(d*sqrt((b*x^2 + a)*e/(d*x^2 + c)) + sqrt(b*d*e)))/(sqr
t(b*d*e)*b*e))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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