3.1014 \(\int \frac{x^2}{\sqrt{a+b x^2+(2+2 c-2 (1+c)) x^4}} \, dx\)

Optimal. Leaf size=49 \[ \frac{x \sqrt{a+b x^2}}{2 b}-\frac{a \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 b^{3/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0128613, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {5, 321, 217, 206} \[ \frac{x \sqrt{a+b x^2}}{2 b}-\frac{a \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0179933, size = 49, normalized size = 1. \[ \frac{x \sqrt{a+b x^2}}{2 b}-\frac{a \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.045, size = 39, normalized size = 0.8 \begin{align*}{\frac{x}{2\,b}\sqrt{b{x}^{2}+a}}-{\frac{a}{2}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*x*(b*x^2+a)^(1/2)/b-1/2*a/b^(3/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))

________________________________________________________________________________________

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^2/(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.64332, size = 238, normalized size = 4.86 \begin{align*} \left [\frac{2 \, \sqrt{b x^{2} + a} b x + a \sqrt{b} \log \left (-2 \, b x^{2} + 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right )}{4 \, b^{2}}, \frac{\sqrt{b x^{2} + a} b x + a \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right )}{2 \, b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*sqrt(b*x^2 + a)*b*x + a*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a))/b^2, 1/2*(sqrt(b*x^2
+ a)*b*x + a*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)))/b^2]

________________________________________________________________________________________

Sympy [A]  time = 2.11753, size = 42, normalized size = 0.86 \begin{align*} \frac{\sqrt{a} x \sqrt{1 + \frac{b x^{2}}{a}}}{2 b} - \frac{a \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{2 b^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(a)*x*sqrt(1 + b*x**2/a)/(2*b) - a*asinh(sqrt(b)*x/sqrt(a))/(2*b**(3/2))

________________________________________________________________________________________

Giac [A]  time = 1.13947, size = 54, normalized size = 1.1 \begin{align*} \frac{\sqrt{b x^{2} + a} x}{2 \, b} + \frac{a \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{2 \, b^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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