3.1012 \(\int \frac{x^4}{\sqrt{a+b x^2+(2+2 c-2 (1+c)) x^4}} \, dx\)

Optimal. Leaf size=73 \[ \frac{3 a^2 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{8 b^{5/2}}-\frac{3 a x \sqrt{a+b x^2}}{8 b^2}+\frac{x^3 \sqrt{a+b x^2}}{4 b} \]

[Out]

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

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Rubi [A]  time = 0.022045, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {5, 321, 217, 206} \[ \frac{3 a^2 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{8 b^{5/2}}-\frac{3 a x \sqrt{a+b x^2}}{8 b^2}+\frac{x^3 \sqrt{a+b x^2}}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[x^4/Sqrt[a + b*x^2 + (2 + 2*c - 2*(1 + c))*x^4],x]

[Out]

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

Rule 5

Int[(u_.)*((a_.) + (c_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[u*(a + b*x^n)^p, x] /; FreeQ[{
a, b, c, n, p}, x] && EqQ[j, 2*n] && EqQ[c, 0]

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0240006, size = 62, normalized size = 0.85 \[ \frac{3 a^2 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )+\sqrt{b} x \sqrt{a+b x^2} \left (2 b x^2-3 a\right )}{8 b^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/Sqrt[a + b*x^2 + (2 + 2*c - 2*(1 + c))*x^4],x]

[Out]

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

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Maple [A]  time = 0.045, size = 59, normalized size = 0.8 \begin{align*}{\frac{{x}^{3}}{4\,b}\sqrt{b{x}^{2}+a}}-{\frac{3\,ax}{8\,{b}^{2}}\sqrt{b{x}^{2}+a}}+{\frac{3\,{a}^{2}}{8}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(b*x^2+a)^(1/2),x)

[Out]

1/4*x^3*(b*x^2+a)^(1/2)/b-3/8*a*x*(b*x^2+a)^(1/2)/b^2+3/8/b^(5/2)*a^2*ln(x*b^(1/2)+(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(x^4/(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.6038, size = 300, normalized size = 4.11 \begin{align*} \left [\frac{3 \, a^{2} \sqrt{b} \log \left (-2 \, b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) + 2 \,{\left (2 \, b^{2} x^{3} - 3 \, a b x\right )} \sqrt{b x^{2} + a}}{16 \, b^{3}}, -\frac{3 \, a^{2} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) -{\left (2 \, b^{2} x^{3} - 3 \, a b x\right )} \sqrt{b x^{2} + a}}{8 \, b^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[1/16*(3*a^2*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(2*b^2*x^3 - 3*a*b*x)*sqrt(b*x^2 + a)
)/b^3, -1/8*(3*a^2*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (2*b^2*x^3 - 3*a*b*x)*sqrt(b*x^2 + a))/b^3]

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Sympy [A]  time = 3.776, size = 95, normalized size = 1.3 \begin{align*} - \frac{3 a^{\frac{3}{2}} x}{8 b^{2} \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{\sqrt{a} x^{3}}{8 b \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{3 a^{2} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{8 b^{\frac{5}{2}}} + \frac{x^{5}}{4 \sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/(b*x**2+a)**(1/2),x)

[Out]

-3*a**(3/2)*x/(8*b**2*sqrt(1 + b*x**2/a)) - sqrt(a)*x**3/(8*b*sqrt(1 + b*x**2/a)) + 3*a**2*asinh(sqrt(b)*x/sqr
t(a))/(8*b**(5/2)) + x**5/(4*sqrt(a)*sqrt(1 + b*x**2/a))

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Giac [A]  time = 1.22533, size = 73, normalized size = 1. \begin{align*} \frac{1}{8} \, \sqrt{b x^{2} + a} x{\left (\frac{2 \, x^{2}}{b} - \frac{3 \, a}{b^{2}}\right )} - \frac{3 \, a^{2} \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{8 \, b^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

1/8*sqrt(b*x^2 + a)*x*(2*x^2/b - 3*a/b^2) - 3/8*a^2*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2)