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

Optimal. Leaf size=165 \[ -\frac{3 \sqrt{a} e^{3/2} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{a} \sqrt{e}}\right )}{2 c^{5/2}}+\frac{3 e (b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{2 c^2}+\frac{(b c-a d) \left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{2 c \left (a-\frac{c \left (a+b x^2\right )}{c+d x^2}\right )} \]

[Out]

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

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Rubi [A]  time = 0.103718, antiderivative size = 165, 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, 288, 321, 208} \[ -\frac{3 \sqrt{a} e^{3/2} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{a} \sqrt{e}}\right )}{2 c^{5/2}}+\frac{3 e (b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{2 c^2}+\frac{(b c-a d) \left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{2 c \left (a-\frac{c \left (a+b x^2\right )}{c+d x^2}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(b*c - a*d)*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(2*c^2) + ((b*c - a*d)*((e*(a + b*x^2))/(c + d*x^2))^(3/2)
)/(2*c*(a - (c*(a + b*x^2))/(c + d*x^2))) - (3*Sqrt[a]*(b*c - a*d)*e^(3/2)*ArcTanh[(Sqrt[c]*Sqrt[(e*(a + b*x^2
))/(c + d*x^2)])/(Sqrt[a]*Sqrt[e])])/(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 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 321

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*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 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{\left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{x^3} \, dx &=((b c-a d) e) \operatorname{Subst}\left (\int \frac{x^4}{\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) \left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{2 c \left (a-\frac{c \left (a+b x^2\right )}{c+d x^2}\right )}+\frac{(3 (b c-a d) e) \operatorname{Subst}\left (\int \frac{x^2}{-a e+c x^2} \, dx,x,\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}\right )}{2 c}\\ &=\frac{3 (b c-a d) e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{2 c^2}+\frac{(b c-a d) \left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{2 c \left (a-\frac{c \left (a+b x^2\right )}{c+d x^2}\right )}+\frac{\left (3 a (b c-a d) 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 )}{2 c^2}\\ &=\frac{3 (b c-a d) e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{2 c^2}+\frac{(b c-a d) \left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{2 c \left (a-\frac{c \left (a+b x^2\right )}{c+d x^2}\right )}-\frac{3 \sqrt{a} (b c-a d) 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 )}{2 c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0906491, size = 146, normalized size = 0.88 \[ \frac{e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (\sqrt{c} \sqrt{a+b x^2} \left (2 b c x^2-a \left (c+3 d x^2\right )\right )-3 \sqrt{a} x^2 \sqrt{c+d x^2} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{a+b x^2}}{\sqrt{a} \sqrt{c+d x^2}}\right )\right )}{2 c^{5/2} x^2 \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.015, size = 641, normalized size = 3.9 \begin{align*} -{\frac{d{x}^{2}+c}{4\,{x}^{2}{c}^{3} \left ( b{x}^{2}+a \right ) } \left ( -2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{ac}{x}^{6}b{d}^{2}-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 ){x}^{4}{a}^{2}c{d}^{2}+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 ){x}^{4}ab{c}^{2}d-2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{ac}{x}^{4}a{d}^{2}-4\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{ac}{x}^{4}bcd-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 ){x}^{2}{a}^{2}{c}^{2}d+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 ){x}^{2}ab{c}^{3}+2\, \left ( bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac \right ) ^{3/2}\sqrt{ac}{x}^{2}d-2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{ac}{x}^{2}acd-2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{ac}{x}^{2}b{c}^{2}+4\,\sqrt{ac}\sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }{x}^{2}acd-4\,\sqrt{ac}\sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }{x}^{2}b{c}^{2}+2\, \left ( bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac \right ) ^{3/2}\sqrt{ac}c \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{ \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^3,x)

[Out]

-1/4*(-2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*x^6*b*d^2-3*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)*x^4*a^2*c*d^2+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)*x^4*a*b*c^2*d-2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*x^4*a*d^2-4*(b*d*
x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*x^4*b*c*d-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)*x^2*a^2*c^2*d+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)*x^2*a*b*c^3+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(3/2)*(a*c)^(1/2)*x^2*d-2*(b*d*x^4+a*d*x^2+b*c*x^2+
a*c)^(1/2)*(a*c)^(1/2)*x^2*a*c*d-2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(a*c)^(1/2)*x^2*b*c^2+4*(a*c)^(1/2)*((d
*x^2+c)*(b*x^2+a))^(1/2)*x^2*a*c*d-4*(a*c)^(1/2)*((d*x^2+c)*(b*x^2+a))^(1/2)*x^2*b*c^2+2*(b*d*x^4+a*d*x^2+b*c*
x^2+a*c)^(3/2)*(a*c)^(1/2)*c)*(d*x^2+c)*(e*(b*x^2+a)/(d*x^2+c))^(3/2)/x^2/(a*c)^(1/2)/c^3/(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^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 7.50565, size = 748, normalized size = 4.53 \begin{align*} \left [-\frac{3 \,{\left (b c - a d\right )} \sqrt{\frac{a e}{c}} e 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^{2} d + a c d^{2}\right )} x^{4} + 2 \, a c^{3} +{\left (b c^{3} + 3 \, a c^{2} d\right )} x^{2}\right )} \sqrt{\frac{a e}{c}} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{x^{4}}\right ) - 4 \,{\left ({\left (2 \, b c - 3 \, a d\right )} e x^{2} - a c e\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{8 \, c^{2} x^{2}}, \frac{3 \,{\left (b c - a d\right )} \sqrt{-\frac{a e}{c}} e x^{2} \arctan \left (\frac{{\left ({\left (b c + a d\right )} x^{2} + 2 \, a c\right )} \sqrt{-\frac{a e}{c}} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{2 \,{\left (a b e x^{2} + a^{2} e\right )}}\right ) + 2 \,{\left ({\left (2 \, b c - 3 \, a d\right )} e x^{2} - a c e\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{4 \, c^{2} 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))^(3/2)/x^3,x, algorithm="fricas")

[Out]

[-1/8*(3*(b*c - a*d)*sqrt(a*e/c)*e*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^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*((2*b*c - 3*a*d)*e*x^2 - a*c*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(c^2*x^2), 1/4*
(3*(b*c - a*d)*sqrt(-a*e/c)*e*x^2*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)) + 2*((2*b*c - 3*a*d)*e*x^2 - a*c*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/(c^2*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))**(3/2)/x**3,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^{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))^(3/2)/x^3,x, algorithm="giac")

[Out]

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