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

Optimal. Leaf size=161 \[ \frac{\sqrt{e} (b c-a d) (a d+3 b c) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{b} \sqrt{e}}\right )}{8 b^{3/2} d^{5/2}}+\frac{\left (c+d x^2\right )^2 \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{4 d^2}-\frac{\left (c+d x^2\right ) (5 b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{8 b d^2} \]

[Out]

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

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Rubi [A]  time = 0.162943, antiderivative size = 161, 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} \[ \frac{\sqrt{e} (b c-a d) (a d+3 b c) \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{b} \sqrt{e}}\right )}{8 b^{3/2} d^{5/2}}+\frac{\left (c+d x^2\right )^2 \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{4 d^2}-\frac{\left (c+d x^2\right ) (5 b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{8 b d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((5*b*c - a*d)*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*(c + d*x^2))/(8*b*d^2) + (Sqrt[(e*(a + b*x^2))/(c + d*x^2)]*
(c + d*x^2)^2)/(4*d^2) + ((b*c - a*d)*(3*b*c + a*d)*Sqrt[e]*ArcTanh[(Sqrt[d]*Sqrt[(e*(a + b*x^2))/(c + d*x^2)]
)/(Sqrt[b]*Sqrt[e])])/(8*b^(3/2)*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 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 x^3 \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \, dx &=((b c-a d) e) \operatorname{Subst}\left (\int \frac{x^2 \left (-a e+c x^2\right )}{\left (b e-d x^2\right )^3} \, dx,x,\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}\right )\\ &=\frac{\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{4 d^2}-\frac{((b c-a d) e) \operatorname{Subst}\left (\int \frac{(b c-a d) e+4 c d 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 )}{4 d^2}\\ &=-\frac{(5 b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{8 b d^2}+\frac{\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{4 d^2}+\frac{((b c-a d) (3 b c+a d) e) \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 )}{8 b d^2}\\ &=-\frac{(5 b c-a d) \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )}{8 b d^2}+\frac{\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (c+d x^2\right )^2}{4 d^2}+\frac{(b c-a d) (3 b c+a d) \sqrt{e} \tanh ^{-1}\left (\frac{\sqrt{d} \sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}}}{\sqrt{b} \sqrt{e}}\right )}{8 b^{3/2} d^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.38612, size = 149, normalized size = 0.93 \[ \frac{\sqrt{\frac{e \left (a+b x^2\right )}{c+d x^2}} \left (b \sqrt{d} \left (c+d x^2\right ) \left (a d-3 b c+2 b d x^2\right )+\frac{(a d+3 b c) (b c-a d)^{3/2} \sqrt{\frac{b \left (c+d x^2\right )}{b c-a d}} \sinh ^{-1}\left (\frac{\sqrt{d} \sqrt{a+b x^2}}{\sqrt{b c-a d}}\right )}{\sqrt{a+b x^2}}\right )}{8 b^2 d^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.012, size = 342, normalized size = 2.1 \begin{align*}{\frac{d{x}^{2}+c}{16\,b{d}^{2}}\sqrt{{\frac{e \left ( b{x}^{2}+a \right ) }{d{x}^{2}+c}}} \left ( 4\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}{x}^{2}bd-{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 ){a}^{2}-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 ) 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}ad-6\,\sqrt{bd{x}^{4}+ad{x}^{2}+bc{x}^{2}+ac}\sqrt{bd}bc \right ){\frac{1}{\sqrt{ \left ( d{x}^{2}+c \right ) \left ( b{x}^{2}+a \right ) }}}{\frac{1}{\sqrt{bd}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(e*(b*x^2+a)/(d*x^2+c))^(1/2)*(d*x^2+c)*(4*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*x^2*b*d-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^2-2*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*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)*a*d-6*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)*(b*d)^(1/2)*b*c)/((d*x^2+c)*(b*x^2+a))^(1/2)/d^2/b/(b*d)
^(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(x^3*(e*(b*x^2+a)/(d*x^2+c))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.32398, size = 849, normalized size = 5.27 \begin{align*} \left [-\frac{{\left (3 \, b^{2} c^{2} - 2 \, a b c d - a^{2} d^{2}\right )} \sqrt{\frac{e}{b d}} \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^{2} d^{3} x^{4} + b^{2} c^{2} d + a b c d^{2} +{\left (3 \, b^{2} c d^{2} + a b d^{3}\right )} x^{2}\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}} \sqrt{\frac{e}{b d}}\right ) - 4 \,{\left (2 \, b d^{2} x^{4} - 3 \, b c^{2} + a c d -{\left (b c d - a d^{2}\right )} x^{2}\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{32 \, b d^{2}}, -\frac{{\left (3 \, b^{2} c^{2} - 2 \, a b c d - a^{2} d^{2}\right )} \sqrt{-\frac{e}{b d}} \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{e}{b d}}}{2 \,{\left (b e x^{2} + a e\right )}}\right ) - 2 \,{\left (2 \, b d^{2} x^{4} - 3 \, b c^{2} + a c d -{\left (b c d - a d^{2}\right )} x^{2}\right )} \sqrt{\frac{b e x^{2} + a e}{d x^{2} + c}}}{16 \, b d^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*(b*x^2+a)/(d*x^2+c))^(1/2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.4132, size = 257, normalized size = 1.6 \begin{align*} \frac{1}{16} \,{\left (2 \, \sqrt{b d x^{4} e + b c x^{2} e + a d x^{2} e + a c e}{\left (\frac{2 \, x^{2}}{d} - \frac{3 \, b c - a d}{b d^{2}}\right )} - \frac{{\left (3 \, b^{2} c^{2} e - 2 \, a b c d e - a^{2} d^{2} e\right )} \sqrt{b d} e^{\left (-\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^{2} d^{3}}\right )} \mathrm{sgn}\left (d x^{2} + c\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(e*(b*x^2+a)/(d*x^2+c))^(1/2),x, algorithm="giac")

[Out]

1/16*(2*sqrt(b*d*x^4*e + b*c*x^2*e + a*d*x^2*e + a*c*e)*(2*x^2/d - (3*b*c - a*d)/(b*d^2)) - (3*b^2*c^2*e - 2*a
*b*c*d*e - a^2*d^2*e)*sqrt(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^2*d^3))*sgn(d*x^2 + c)