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

Optimal. Leaf size=151 \[ -\frac{a^{3/2} e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{a} \sqrt{e}}\right )}{c^{3/2}}+\frac{b^{3/2} e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{b} \sqrt{e}}\right )}{d^{3/2}}-\frac{e (b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{c d} \]

[Out]

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

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Rubi [A]  time = 0.19189, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {1960, 479, 522, 208} \[ -\frac{a^{3/2} e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{a} \sqrt{e}}\right )}{c^{3/2}}+\frac{b^{3/2} e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{b} \sqrt{e}}\right )}{d^{3/2}}-\frac{e (b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{c d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((b*c - a*d)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(c*d)) - (a^(3/2)*e^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[(e*(a + b*
x^2))/(c + d*x^2)])/(Sqrt[a]*Sqrt[e])])/c^(3/2) + (b^(3/2)*e^(3/2)*ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2))/(c +
d*x^2)])/(Sqrt[b]*Sqrt[e])])/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 479

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(2*n
- 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*d*(m + n*(p + q) + 1)), x] - Dist[e^(2*n)
/(b*d*(m + n*(p + q) + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) + (a*d*(m +
n*(q - 1) + 1) + b*c*(m + n*(p - 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d
, 0] && IGtQ[n, 0] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, 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{\left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{x} \, dx &=((b c-a d) e) \operatorname{Subst}\left (\int \frac{x^4}{\left (-a e+c x^2\right ) \left (b e-d x^2\right )} \, dx,x,\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}\right )\\ &=-\frac{(b c-a d) e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{c d}+\frac{((b c-a d) e) \operatorname{Subst}\left (\int \frac{-a b e^2+(b c+a d) e x^2}{\left (-a e+c x^2\right ) \left (b e-d x^2\right )} \, dx,x,\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}\right )}{c d}\\ &=-\frac{(b c-a d) e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{c d}+\frac{\left (a^2 e^2\right ) \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 )}{c}+\frac{\left (b^2 e^2\right ) \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 )}{d}\\ &=-\frac{(b c-a d) e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{c d}-\frac{a^{3/2} e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{a} \sqrt{e}}\right )}{c^{3/2}}+\frac{b^{3/2} e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{b} \sqrt{e}}\right )}{d^{3/2}}\\ \end{align*}

Mathematica [A]  time = 1.28275, size = 193, normalized size = 1.28 \[ \frac{e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (\sqrt{d} \left (-\frac{a^{3/2} d \sqrt{c+d x^2} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x^2}}{\sqrt{a} \sqrt{c+d x^2}}\right )}{c^{3/2} \sqrt{a+b x^2}}+\frac{a d}{c}-b\right )+\frac{b \sqrt{b c-a d} \sqrt{\frac{b \left (c+d x^2\right )}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b c-a d}}\right )}{\sqrt{a+b x^2}}\right )}{d^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.015, size = 401, normalized size = 2.7 \begin{align*}{\frac{d{x}^{2}+c}{2\,cd \left ( b{x}^{2}+a \right ) } \left ( \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 ) \sqrt{ac}{x}^{2}{b}^{2}cd-\sqrt{bd}\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 ){x}^{2}{a}^{2}{d}^{2}+\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 ) \sqrt{ac}{b}^{2}{c}^{2}-\sqrt{bd}\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 ){a}^{2}cd+2\,\sqrt{bd}\sqrt{ac}\sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }ad-2\,\sqrt{bd}\sqrt{ac}\sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }bc \right ) \left ({\frac{ \left ( b{x}^{2}+a \right ) e}{d{x}^{2}+c}} \right ) ^{{\frac{3}{2}}}{\frac{1}{\sqrt{ac}}}{\frac{1}{\sqrt{bd}}}{\frac{1}{\sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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))^(3/2)/x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 6.45608, size = 2226, normalized size = 14.74 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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))**(3/2)/x,x)

[Out]

Timed out

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

[Out]

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