3.268 \(\int \frac{\sqrt{\frac{e (a+b x^2)}{c+d x^2}}}{x^5} \, dx\)

Optimal. Leaf size=208 \[ \frac{\sqrt{e} (3 a d+b c) (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{a} \sqrt{e}}\right )}{8 a^{3/2} c^{5/2}}-\frac{(b c-a d)^2 \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{4 c^2 \left (a-\frac{c \left (a+b x^2\right )}{c+d x^2}\right )^2}+\frac{(b c-5 a d) (b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{8 a c^2 \left (a-\frac{c \left (a+b x^2\right )}{c+d x^2}\right )} \]

[Out]

-((b*c - a*d)^2*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(4*c^2*(a - (c*(a + b*x^2))/(c + d*x^2))^2) + ((b*c - 5*a*d
)*(b*c - a*d)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(8*a*c^2*(a - (c*(a + b*x^2))/(c + d*x^2))) + ((b*c - a*d)*(b
*c + 3*a*d)*Sqrt[e]*ArcTanh[(Sqrt[c]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[a]*Sqrt[e])])/(8*a^(3/2)*c^(5/2)
)

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Rubi [A]  time = 0.169509, antiderivative size = 208, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {1960, 455, 385, 208} \[ \frac{\sqrt{e} (3 a d+b c) (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{a} \sqrt{e}}\right )}{8 a^{3/2} c^{5/2}}-\frac{(b c-a d)^2 \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{4 c^2 \left (a-\frac{c \left (a+b x^2\right )}{c+d x^2}\right )^2}+\frac{(b c-5 a d) (b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{8 a c^2 \left (a-\frac{c \left (a+b x^2\right )}{c+d x^2}\right )} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[(e*(a + b*x^2))/(c + d*x^2)]/x^5,x]

[Out]

-((b*c - a*d)^2*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(4*c^2*(a - (c*(a + b*x^2))/(c + d*x^2))^2) + ((b*c - 5*a*d
)*(b*c - a*d)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(8*a*c^2*(a - (c*(a + b*x^2))/(c + d*x^2))) + ((b*c - a*d)*(b
*c + 3*a*d)*Sqrt[e]*ArcTanh[(Sqrt[c]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[a]*Sqrt[e])])/(8*a^(3/2)*c^(5/2)
)

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 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

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^5} \, dx &=((b c-a d) e) \operatorname{Subst}\left (\int \frac{x^2 \left (b e-d x^2\right )}{\left (-a e+c x^2\right )^3} \, dx,x,\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}\right )\\ &=-\frac{(b c-a d)^2 \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{4 c^2 \left (a-\frac{c \left (a+b x^2\right )}{c+d x^2}\right )^2}-\frac{((b c-a d) e) \operatorname{Subst}\left (\int \frac{-(b c-a d) e+4 c d 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 )}{4 c^2}\\ &=-\frac{(b c-a d)^2 \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{4 c^2 \left (a-\frac{c \left (a+b x^2\right )}{c+d x^2}\right )^2}+\frac{(b c-5 a d) (b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{8 a c^2 \left (a-\frac{c \left (a+b x^2\right )}{c+d x^2}\right )}-\frac{((b c-a d) (b c+3 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 )}{8 a c^2}\\ &=-\frac{(b c-a d)^2 \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{4 c^2 \left (a-\frac{c \left (a+b x^2\right )}{c+d x^2}\right )^2}+\frac{(b c-5 a d) (b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{8 a c^2 \left (a-\frac{c \left (a+b x^2\right )}{c+d x^2}\right )}+\frac{(b c-a d) (b c+3 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 )}{8 a^{3/2} c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.110395, size = 174, normalized size = 0.84 \[ \frac{\sqrt{c+d x^2} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (x^4 \left (-3 a^2 d^2+2 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x^2}}{\sqrt{a} \sqrt{c+d x^2}}\right )+\sqrt{a} \sqrt{c} \sqrt{a+b x^2} \sqrt{c+d x^2} \left (-2 a c+3 a d x^2-b c x^2\right )\right )}{8 a^{3/2} c^{5/2} x^4 \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[(e*(a + b*x^2))/(c + d*x^2)]/x^5,x]

[Out]

(Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*Sqrt[c + d*x^2]*(Sqrt[a]*Sqrt[c]*Sqrt[a + b*x^2]*Sqrt[c + d*x^2]*(-2*a*c -
b*c*x^2 + 3*a*d*x^2) + (b^2*c^2 + 2*a*b*c*d - 3*a^2*d^2)*x^4*ArcTanh[(Sqrt[c]*Sqrt[a + b*x^2])/(Sqrt[a]*Sqrt[c
+ d*x^2])]))/(8*a^(3/2)*c^(5/2)*x^4*Sqrt[a + b*x^2])

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Maple [B]  time = 0.023, size = 558, normalized size = 2.7 \begin{align*}{\frac{d{x}^{2}+c}{16\,{a}^{2}{c}^{3}{x}^{4}}\sqrt{{\frac{e \left ( b{x}^{2}+a \right ) }{d{x}^{2}+c}}} \left ( -10\,b{d}^{2}\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}{x}^{6}a\sqrt{ac}-2\,{b}^{2}d\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}{x}^{6}c\sqrt{ac}-3\,{a}^{3}\ln \left ({\frac{ad{x}^{2}+bc{x}^{2}+2\,\sqrt{ac}\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}+2\,ac}{{x}^{2}}} \right ){d}^{2}c{x}^{4}+2\,\ln \left ({\frac{ad{x}^{2}+bc{x}^{2}+2\,\sqrt{ac}\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}+2\,ac}{{x}^{2}}} \right ) db{a}^{2}{c}^{2}{x}^{4}+{c}^{3}\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 ){b}^{2}a{x}^{4}-10\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}{d}^{2}{a}^{2}{x}^{4}\sqrt{ac}-8\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{ac}{x}^{4}abcd-2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}{b}^{2}{c}^{2}{x}^{4}\sqrt{ac}+10\, \left ( bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac \right ) ^{3/2}da{x}^{2}\sqrt{ac}+2\, \left ( bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac \right ) ^{3/2}bc{x}^{2}\sqrt{ac}-4\, \left ( bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac \right ) ^{3/2}ac\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^5,x)

[Out]

1/16*(e*(b*x^2+a)/(d*x^2+c))^(1/2)*(d*x^2+c)*(-10*b*d^2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*x^6*a*(a*c)^(1/2)-
2*b^2*d*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*x^6*c*(a*c)^(1/2)-3*a^3*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^2*c*x^4+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*b*a^2*c^2*x^4+c^3*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^2*a*x^4-10*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*d^2*a^2*x^4*(a*c)^(1/2)-8*(b*d*x^4+a*d*x^2+b*
c*x^2+a*c)^(1/2)*(a*c)^(1/2)*x^4*a*b*c*d-2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*b^2*c^2*x^4*(a*c)^(1/2)+10*(b*d
*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*d*a*x^2*(a*c)^(1/2)+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*b*c*x^2*(a*c)^(1/2)-
4*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*a*c*(a*c)^(1/2))/((d*x^2+c)*(b*x^2+a))^(1/2)/c^3/(a*c)^(1/2)/a^2/x^4

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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^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 7.78701, size = 890, normalized size = 4.28 \begin{align*} \left [-\frac{{\left (b^{2} c^{2} + 2 \, a b c d - 3 \, a^{2} d^{2}\right )} x^{4} \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 ({\left (b c d - 3 \, a d^{2}\right )} x^{4} + 2 \, a c^{2} +{\left (b c^{2} - a c d\right )} x^{2}\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{32 \, a c^{2} x^{4}}, -\frac{{\left (b^{2} c^{2} + 2 \, a b c d - 3 \, a^{2} d^{2}\right )} x^{4} \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 ({\left (b c d - 3 \, a d^{2}\right )} x^{4} + 2 \, a c^{2} +{\left (b c^{2} - a c d\right )} x^{2}\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{16 \, a c^{2} x^{4}}\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^5,x, algorithm="fricas")

[Out]

[-1/32*((b^2*c^2 + 2*a*b*c*d - 3*a^2*d^2)*x^4*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*((b*c*d - 3*a*d^2)*x^4 + 2*a*c^2 + (b*c^2 - a*c*d
)*x^2)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(a*c^2*x^4), -1/16*((b^2*c^2 + 2*a*b*c*d - 3*a^2*d^2)*x^4*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*((b*c*d - 3*a*d^2)*x^4 + 2*a*c^2 + (b*c^2 - a*c*d)*x^2)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(a*c^2*x^4)]

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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**5,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^{5}}\,{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^5,x, algorithm="giac")

[Out]

integrate(sqrt((b*x^2 + a)*e/(d*x^2 + c))/x^5, x)