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

Optimal. Leaf size=103 \[ \frac{\left (c+d x^2\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{2 d}-\frac{\sqrt{e} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{b} \sqrt{e}}\right )}{2 \sqrt{b} d^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0698071, antiderivative size = 103, normalized size of antiderivative = 1., 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, 288, 208} \[ \frac{\left (c+d x^2\right ) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{2 d}-\frac{\sqrt{e} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{b} \sqrt{e}}\right )}{2 \sqrt{b} d^{3/2}} \]

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

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]]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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]

Rubi steps

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

Mathematica [A]  time = 0.284086, size = 143, normalized size = 1.39 \[ \frac{\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (b \sqrt{d} \left (a+b x^2\right ) \left (c+d x^2\right )-\sqrt{a+b x^2} (b c-a d)^{3/2} \sqrt{\frac{b \left (c+d x^2\right )}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b c-a d}}\right )\right )}{2 b d^{3/2} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.006, size = 200, normalized size = 1.9 \begin{align*}{\frac{d{x}^{2}+c}{4\,d}\sqrt{{\frac{e \left ( b{x}^{2}+a \right ) }{d{x}^{2}+c}}} \left ( a\ln \left ({\frac{1}{2} \left ( 2\,bd{x}^{2}+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc \right ){\frac{1}{\sqrt{bd}}}} \right ) d-b\ln \left ({\frac{1}{2} \left ( 2\,bd{x}^{2}+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc \right ){\frac{1}{\sqrt{bd}}}} \right ) c+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd} \right ){\frac{1}{\sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }}}{\frac{1}{\sqrt{bd}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.58203, size = 656, normalized size = 6.37 \begin{align*} \left [-\frac{{\left (b c - a d\right )} \sqrt{\frac{e}{b d}} \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^{2} d^{3} x^{4} + b^{2} c^{2} d + a b c d^{2} +{\left (3 \, b^{2} c d^{2} + a b d^{3}\right )} x^{2}\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}} \sqrt{\frac{e}{b d}}\right ) - 4 \,{\left (d x^{2} + c\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{8 \, d}, \frac{{\left (b c - a d\right )} \sqrt{-\frac{e}{b d}} \arctan \left (\frac{{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}} \sqrt{-\frac{e}{b d}}}{2 \,{\left (b e x^{2} + a e\right )}}\right ) + 2 \,{\left (d x^{2} + c\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{4 \, d}\right ] \end{align*}

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*((b*c - a*d)*sqrt(e/(b*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^2*d^3*x^4 + b^2*c^2*d + a*b*c*d^2 + (3*b^2*c*d^2 + a*b*d^3)*x^2)*sqrt((b*e*x^2 + a*e)/(d*x^2
+ c))*sqrt(e/(b*d))) - 4*(d*x^2 + c)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/d, 1/4*((b*c - a*d)*sqrt(-e/(b*d))*ar
ctan(1/2*(2*b*d*x^2 + b*c + a*d)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c))*sqrt(-e/(b*d))/(b*e*x^2 + a*e)) + 2*(d*x^2
+ c)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/d]

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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

________________________________________________________________________________________

Giac [A]  time = 1.3491, size = 201, normalized size = 1.95 \begin{align*} \frac{1}{4} \,{\left (\frac{{\left (b c e - a d e\right )} \sqrt{b d} e^{\left (-\frac{1}{2}\right )} \log \left ({\left | -\sqrt{b d} b c e^{\frac{1}{2}} - \sqrt{b d} a d e^{\frac{1}{2}} - 2 \,{\left (\sqrt{b d} x^{2} e^{\frac{1}{2}} - \sqrt{b d x^{4} e + b c x^{2} e + a d x^{2} e + a c e}\right )} b d \right |}\right )}{b d^{2}} + \frac{2 \, \sqrt{b d x^{4} e + b c x^{2} e + a d x^{2} e + a c e}}{d}\right )} \mathrm{sgn}\left (d x^{2} + c\right ) \end{align*}

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]

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