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

Optimal. Leaf size=146 \[ \frac{3 \sqrt{d} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{b} \sqrt{e}}\right )}{2 b^{5/2} e^{3/2}}-\frac{3 (b c-a d)}{2 b^2 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{c+d x^2}{2 b e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}} \]

[Out]

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

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Rubi [A]  time = 0.0983798, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {1960, 290, 325, 208} \[ \frac{3 \sqrt{d} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{b} \sqrt{e}}\right )}{2 b^{5/2} e^{3/2}}-\frac{3 (b c-a d)}{2 b^2 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{c+d x^2}{2 b e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(b*c - a*d))/(2*b^2*e*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]) + (c + d*x^2)/(2*b*e*Sqrt[(e*(a + b*x^2))/(c + d*
x^2)]) + (3*Sqrt[d]*(b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)])/(Sqrt[b]*Sqrt[e])])/(2*b^(
5/2)*e^(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 290

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

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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{x}{\left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \, dx &=((b c-a d) e) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (b e-d x^2\right )^2} \, dx,x,\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}\right )\\ &=\frac{c+d x^2}{2 b e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{(3 (b c-a d)) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (b e-d x^2\right )} \, dx,x,\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}\right )}{2 b}\\ &=-\frac{3 (b c-a d)}{2 b^2 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{c+d x^2}{2 b e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{(3 d (b c-a d)) \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 )}{2 b^2 e}\\ &=-\frac{3 (b c-a d)}{2 b^2 e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{c+d x^2}{2 b e \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}+\frac{3 \sqrt{d} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{b} \sqrt{e}}\right )}{2 b^{5/2} e^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.0702331, size = 86, normalized size = 0.59 \[ -\frac{\left (a+b x^2\right ) \, _2F_1\left (-\frac{3}{2},-\frac{1}{2};\frac{1}{2};\frac{d \left (b x^2+a\right )}{a d-b c}\right )}{b \left (\frac{b \left (c+d x^2\right )}{b c-a d}\right )^{3/2} \left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a + b*x^2)*Hypergeometric2F1[-3/2, -1/2, 1/2, (d*(a + b*x^2))/(-(b*c) + a*d)])/(b*((e*(a + b*x^2))/(c + d*
x^2))^(3/2)*((b*(c + d*x^2))/(b*c - a*d))^(3/2)))

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Maple [B]  time = 0.01, size = 432, normalized size = 3. \begin{align*}{\frac{b{x}^{2}+a}{4\,{b}^{2} \left ( d{x}^{2}+c \right ) } \left ( -3\,\ln \left ( 1/2\,{\frac{2\,bd{x}^{2}+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){x}^{2}ab{d}^{2}+3\,\ln \left ( 1/2\,{\frac{2\,bd{x}^{2}+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){x}^{2}{b}^{2}cd+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}{x}^{2}bd-3\,{d}^{2}\ln \left ( 1/2\,{\frac{2\,bd{x}^{2}+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ){a}^{2}+3\,\ln \left ( 1/2\,{\frac{2\,bd{x}^{2}+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}+ad+bc}{\sqrt{bd}}} \right ) acbd+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}ad+4\,\sqrt{bd}\sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }ad-4\,\sqrt{bd}\sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }bc \right ){\frac{1}{\sqrt{bd}}}{\frac{1}{\sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }}} \left ({\frac{e \left ( b{x}^{2}+a \right ) }{d{x}^{2}+c}} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.93855, size = 914, normalized size = 6.26 \begin{align*} \left [-\frac{3 \,{\left ({\left (b^{2} c - a b d\right )} e x^{2} +{\left (a b c - a^{2} d\right )} e\right )} \sqrt{\frac{d}{b e}} \log \left (8 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} + 8 \,{\left (b^{2} c d + a b d^{2}\right )} x^{2} - 4 \,{\left (2 \, b^{2} d^{2} x^{4} + b^{2} c^{2} + a b c d +{\left (3 \, b^{2} c d + a b d^{2}\right )} x^{2}\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}} \sqrt{\frac{d}{b e}}\right ) - 4 \,{\left (b d^{2} x^{4} - 2 \, b c^{2} + 3 \, a c d -{\left (b c d - 3 \, a d^{2}\right )} x^{2}\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{8 \,{\left (b^{3} e^{2} x^{2} + a b^{2} e^{2}\right )}}, -\frac{3 \,{\left ({\left (b^{2} c - a b d\right )} e x^{2} +{\left (a b c - a^{2} d\right )} e\right )} \sqrt{-\frac{d}{b e}} \arctan \left (\frac{{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}} \sqrt{-\frac{d}{b e}}}{2 \,{\left (b d x^{2} + a d\right )}}\right ) - 2 \,{\left (b d^{2} x^{4} - 2 \, b c^{2} + 3 \, a c d -{\left (b c d - 3 \, a d^{2}\right )} x^{2}\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{4 \,{\left (b^{3} e^{2} x^{2} + a b^{2} e^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError