3.300 \(\int \frac{1}{x^3 \sqrt{\frac{e (a+b x^2)}{c+d x^2}}} \, dx\)

Optimal. Leaf size=130 \[ \frac{(b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{a} \sqrt{e}}\right )}{2 a^{3/2} \sqrt{c} \sqrt{e}}+\frac{(b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{2 a \left (a e-\frac{c e \left (a+b x^2\right )}{c+d x^2}\right )} \]

[Out]

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

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Rubi [A]  time = 0.08368, antiderivative size = 130, 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, 199, 208} \[ \frac{(b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{a} \sqrt{e}}\right )}{2 a^{3/2} \sqrt{c} \sqrt{e}}+\frac{(b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{2 a \left (a e-\frac{c e \left (a+b x^2\right )}{c+d x^2}\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]),x]

[Out]

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

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 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]

Rubi steps

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

Mathematica [A]  time = 0.122445, size = 133, normalized size = 1.02 \[ \frac{\sqrt{a+b x^2} \left (-\frac{(a d-b c) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x^2}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{a^{3/2} \sqrt{c}}-\frac{\sqrt{a+b x^2} \sqrt{c+d x^2}}{a x^2}\right )}{2 \sqrt{c+d x^2} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]),x]

[Out]

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

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Maple [B]  time = 0.012, size = 326, normalized size = 2.5 \begin{align*} -{\frac{b{x}^{2}+a}{4\,{a}^{2}c{x}^{2}} \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{{\frac{e \left ( b{x}^{2}+a \right ) }{d{x}^{2}+c}}}}}{\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(1/x^3/(e*(b*x^2+a)/(d*x^2+c))^(1/2),x)

[Out]

-1/4*(b*x^2+a)*(-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*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)*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))/(e*(b*x^2+
a)/(d*x^2+c))^(1/2)/((d*x^2+c)*(b*x^2+a))^(1/2)/a^2/(a*c)^(1/2)/c/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(1/x^3/(e*(b*x^2+a)/(d*x^2+c))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.42144, size = 714, normalized size = 5.49 \begin{align*} \left [-\frac{\sqrt{a c e}{\left (b c - a d\right )} x^{2} \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 ({\left (b c d + a d^{2}\right )} x^{4} + 2 \, a c^{2} +{\left (b c^{2} + 3 \, a c d\right )} x^{2}\right )} \sqrt{a c e} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{x^{4}}\right ) + 4 \,{\left (a c d x^{2} + a c^{2}\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{8 \, a^{2} c e x^{2}}, -\frac{\sqrt{-a c e}{\left (b c - a d\right )} x^{2} \arctan \left (\frac{\sqrt{-a c e}{\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}}}{2 \,{\left (a b c e x^{2} + a^{2} c e\right )}}\right ) + 2 \,{\left (a c d x^{2} + a c^{2}\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{4 \, a^{2} c e x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\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(1/x^3/(e*(b*x^2+a)/(d*x^2+c))^(1/2),x, algorithm="giac")

[Out]

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