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

Optimal. Leaf size=141 \[ -\frac{3 \sqrt{b} e^{3/2} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{b} \sqrt{e}}\right )}{2 d^{5/2}}+\frac{3 e (b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{2 d^2}+\frac{\left (c+d x^2\right ) \left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0899479, antiderivative size = 141, 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, 288, 321, 208} \[ -\frac{3 \sqrt{b} e^{3/2} (b c-a d) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{b} \sqrt{e}}\right )}{2 d^{5/2}}+\frac{3 e (b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{2 d^2}+\frac{\left (c+d x^2\right ) \left (\frac{e \left (a+b x^2\right )}{c+d x^2}\right )^{3/2}}{2 d} \]

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

Mathematica [C]  time = 0.0600596, size = 96, normalized size = 0.68 \[ \frac{e \left (a+b x^2\right )^2 \sqrt{\frac{b \left (c+d x^2\right )}{b c-a d}} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \, _2F_1\left (\frac{3}{2},\frac{5}{2};\frac{7}{2};\frac{d \left (b x^2+a\right )}{a d-b c}\right )}{5 b c-5 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*(a + b*x^2)^2*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*Sqrt[(b*(c + d*x^2))/(b*c - a*d)]*Hypergeometric2F1[3/2, 5/
2, 7/2, (d*(a + b*x^2))/(-(b*c) + a*d)])/(5*b*c - 5*a*d)

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Maple [B]  time = 0.012, size = 432, normalized size = 3.1 \begin{align*}{\frac{d{x}^{2}+c}{4\,{d}^{2} \left ( b{x}^{2}+a \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\,\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-3\,{b}^{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 ){c}^{2}+2\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}bc-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 ) \left ({\frac{ \left ( b{x}^{2}+a \right ) e}{d{x}^{2}+c}} \right ) ^{{\frac{3}{2}}}{\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(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*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-3*b^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))*c^2+2*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*b*c-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)/d^2*(d*x^2+c)*(e*(b*x^2+a)/(d*x^2+c))^(3/2
)/(b*d)^(1/2)/((d*x^2+c)*(b*x^2+a))^(1/2)/(b*x^2+a)

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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 = 4.25471, size = 707, normalized size = 5.01 \begin{align*} \left [-\frac{3 \,{\left (b c - a d\right )} \sqrt{\frac{b e}{d}} e \log \left (8 \, b^{2} d^{2} e x^{4} + 8 \,{\left (b^{2} c d + a b d^{2}\right )} e x^{2} +{\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} e + 4 \,{\left (2 \, b d^{3} x^{4} + b c^{2} d + a c d^{2} +{\left (3 \, b c d^{2} + a d^{3}\right )} x^{2}\right )} \sqrt{\frac{b e}{d}} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}\right ) - 4 \,{\left (b d e x^{2} +{\left (3 \, b c - 2 \, a d\right )} e\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{8 \, d^{2}}, \frac{3 \,{\left (b c - a d\right )} \sqrt{-\frac{b e}{d}} e \arctan \left (\frac{{\left (2 \, b d x^{2} + b c + a d\right )} \sqrt{-\frac{b e}{d}} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{2 \,{\left (b^{2} e x^{2} + a b e\right )}}\right ) + 2 \,{\left (b d e x^{2} +{\left (3 \, b c - 2 \, a d\right )} e\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{4 \, d^{2}}\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*c - a*d)*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*d*e*x^2 + (3*b*c - 2*a*d)*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/d^2, 1/4*(3*(b*c - a*d)*sqr
t(-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*d*e*x^2 + (3*b*c - 2*a*d)*e)*sqrt((b*e*x^2 + a*e)/(d*x^2 + c)))/d^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 [B]  time = 1.38977, size = 339, normalized size = 2.4 \begin{align*} \frac{1}{4} \,{\left (\frac{2 \, \sqrt{b d x^{4} e + b c x^{2} e + a d x^{2} e + a c e} b}{d^{2}} + \frac{4 \,{\left (b^{2} c^{2} e - 2 \, a b c d e + a^{2} d^{2} e\right )}}{{\left (\sqrt{b d} c e^{\frac{1}{2}} +{\left (\sqrt{b d} x^{2} e^{\frac{1}{2}} - \sqrt{b d x^{4} e + b c x^{2} e + a d x^{2} e + a c e}\right )} d\right )} d^{2}} + \frac{3 \,{\left (\sqrt{b d} b^{2} c e^{\frac{1}{2}} - \sqrt{b d} a b d e^{\frac{1}{2}}\right )} \log \left ({\left | -\sqrt{b d} b c e^{\frac{1}{2}} - \sqrt{b d} a d e^{\frac{1}{2}} - 2 \,{\left (\sqrt{b d} x^{2} e^{\frac{1}{2}} - \sqrt{b d x^{4} e + b c x^{2} e + a d x^{2} e + a c e}\right )} b d \right |}\right )}{b d^{3}}\right )} e \mathrm{sgn}\left (d x^{2} + c\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="giac")

[Out]

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