3.267 \(\int \frac{\sqrt{\frac{e (a+b x^2)}{c+d x^2}}}{x^3} \, dx\)

Optimal. Leaf size=127 \[ \frac{(b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{2 c \left (a-\frac{c \left (a+b x^2\right )}{c+d x^2}\right )}-\frac{\sqrt{e} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{a} \sqrt{e}}\right )}{2 \sqrt{a} c^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.086561, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {1960, 288, 208} \[ \frac{(b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{2 c \left (a-\frac{c \left (a+b x^2\right )}{c+d x^2}\right )}-\frac{\sqrt{e} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{a} \sqrt{e}}\right )}{2 \sqrt{a} c^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.100646, size = 133, normalized size = 1.05 \[ \frac{\sqrt{c+d x^2} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (-\frac{(b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x^2}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{\sqrt{a} c^{3/2}}-\frac{\sqrt{a+b x^2} \sqrt{c+d x^2}}{c x^2}\right )}{2 \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.02, size = 326, normalized size = 2.6 \begin{align*} -{\frac{d{x}^{2}+c}{4\,a{c}^{2}{x}^{2}}\sqrt{{\frac{e \left ( b{x}^{2}+a \right ) }{d{x}^{2}+c}}} \left ( -2\,bd\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}{x}^{4}\sqrt{ac}-{a}^{2}\ln \left ({\frac{1}{{x}^{2}} \left ( ad{x}^{2}+bc{x}^{2}+2\,\sqrt{ac}\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}+2\,ac \right ) } \right ) dc{x}^{2}+{c}^{2}\ln \left ({\frac{1}{{x}^{2}} \left ( ad{x}^{2}+bc{x}^{2}+2\,\sqrt{ac}\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}+2\,ac \right ) } \right ) ba{x}^{2}-2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}da{x}^{2}\sqrt{ac}-2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}bc{x}^{2}\sqrt{ac}+2\, \left ( bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac \right ) ^{3/2}\sqrt{ac} \right ){\frac{1}{\sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }}}{\frac{1}{\sqrt{ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*(b*x^2+a)/(d*x^2+c))^(1/2)/x^3,x)

[Out]

-1/4*(e*(b*x^2+a)/(d*x^2+c))^(1/2)*(d*x^2+c)*(-2*b*d*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*x^4*(a*c)^(1/2)-a^2*l
n((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)*d*c*x^2+c^2*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*a*x^2-2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)
*d*a*x^2*(a*c)^(1/2)-2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*b*c*x^2*(a*c)^(1/2)+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)
^(3/2)*(a*c)^(1/2))/((d*x^2+c)*(b*x^2+a))^(1/2)/c^2/(a*c)^(1/2)/a/x^2

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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((e*(b*x^2+a)/(d*x^2+c))^(1/2)/x^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.48498, size = 697, normalized size = 5.49 \begin{align*} \left [-\frac{{\left (b c - a d\right )} x^{2} \sqrt{\frac{e}{a c}} \log \left (\frac{{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e x^{4} + 8 \, a^{2} c^{2} e + 8 \,{\left (a b c^{2} + a^{2} c d\right )} e x^{2} + 4 \,{\left (2 \, a^{2} c^{3} +{\left (a b c^{2} d + a^{2} c d^{2}\right )} x^{4} +{\left (a b c^{3} + 3 \, a^{2} c^{2} d\right )} x^{2}\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}} \sqrt{\frac{e}{a c}}}{x^{4}}\right ) + 4 \,{\left (d x^{2} + c\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{8 \, c x^{2}}, \frac{{\left (b c - a d\right )} x^{2} \sqrt{-\frac{e}{a c}} \arctan \left (\frac{{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}} \sqrt{-\frac{e}{a c}}}{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 \, c x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*(b*x^2+a)/(d*x^2+c))^(1/2)/x^3,x, algorithm="fricas")

[Out]

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

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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((e*(b*x**2+a)/(d*x**2+c))**(1/2)/x**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\frac{{\left (b x^{2} + a\right )} e}{d x^{2} + c}}}{x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*(b*x^2+a)/(d*x^2+c))^(1/2)/x^3,x, algorithm="giac")

[Out]

integrate(sqrt((b*x^2 + a)*e/(d*x^2 + c))/x^3, x)