3.75 \(\int \frac{x^2 (d+e x^2)}{\sqrt{a^2+2 a b x^2+b^2 x^4}} \, dx\)

Optimal. Leaf size=145 \[ \frac{x \left (a+b x^2\right ) (b d-a e)}{b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\sqrt{a} \left (a+b x^2\right ) (b d-a e) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{b^{5/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{e x^3 \left (a+b x^2\right )}{3 b \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

[Out]

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

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Rubi [A]  time = 0.0901167, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {1250, 459, 321, 205} \[ \frac{x \left (a+b x^2\right ) (b d-a e)}{b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\sqrt{a} \left (a+b x^2\right ) (b d-a e) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{b^{5/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{e x^3 \left (a+b x^2\right )}{3 b \sqrt{a^2+2 a b x^2+b^2 x^4}} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(d + e*x^2))/Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

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

Rule 1250

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dis
t[(a + b*x^2 + c*x^4)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(f*x)^m*(d + e*x^2)^q*(b/2
+ c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, m, p, q}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p]

Rule 459

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(b*e*(m + n*(p + 1) + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 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 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{x^2 \left (d+e x^2\right )}{\sqrt{a^2+2 a b x^2+b^2 x^4}} \, dx &=\frac{\left (a b+b^2 x^2\right ) \int \frac{x^2 \left (d+e x^2\right )}{a b+b^2 x^2} \, dx}{\sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{e x^3 \left (a+b x^2\right )}{3 b \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\left (\left (-3 b^2 d+3 a b e\right ) \left (a b+b^2 x^2\right )\right ) \int \frac{x^2}{a b+b^2 x^2} \, dx}{3 b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{(b d-a e) x \left (a+b x^2\right )}{b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{e x^3 \left (a+b x^2\right )}{3 b \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{\left (a \left (-3 b^2 d+3 a b e\right ) \left (a b+b^2 x^2\right )\right ) \int \frac{1}{a b+b^2 x^2} \, dx}{3 b^3 \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ &=\frac{(b d-a e) x \left (a+b x^2\right )}{b^2 \sqrt{a^2+2 a b x^2+b^2 x^4}}+\frac{e x^3 \left (a+b x^2\right )}{3 b \sqrt{a^2+2 a b x^2+b^2 x^4}}-\frac{\sqrt{a} (b d-a e) \left (a+b x^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{b^{5/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}\\ \end{align*}

Mathematica [A]  time = 0.0493694, size = 80, normalized size = 0.55 \[ \frac{\left (a+b x^2\right ) \left (\sqrt{b} x \left (-3 a e+3 b d+b e x^2\right )+3 \sqrt{a} (a e-b d) \tan ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right )}{3 b^{5/2} \sqrt{\left (a+b x^2\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(d + e*x^2))/Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4],x]

[Out]

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

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Maple [A]  time = 0.046, size = 90, normalized size = 0.6 \begin{align*}{\frac{b{x}^{2}+a}{3\,{b}^{2}} \left ( \sqrt{ab}{x}^{3}be+3\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ){a}^{2}e-3\,\arctan \left ({\frac{bx}{\sqrt{ab}}} \right ) abd-3\,\sqrt{ab}xae+3\,\sqrt{ab}xbd \right ){\frac{1}{\sqrt{ \left ( b{x}^{2}+a \right ) ^{2}}}}{\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.55687, size = 277, normalized size = 1.91 \begin{align*} \left [\frac{2 \, b e x^{3} - 3 \,{\left (b d - a e\right )} \sqrt{-\frac{a}{b}} \log \left (\frac{b x^{2} + 2 \, b x \sqrt{-\frac{a}{b}} - a}{b x^{2} + a}\right ) + 6 \,{\left (b d - a e\right )} x}{6 \, b^{2}}, \frac{b e x^{3} - 3 \,{\left (b d - a e\right )} \sqrt{\frac{a}{b}} \arctan \left (\frac{b x \sqrt{\frac{a}{b}}}{a}\right ) + 3 \,{\left (b d - a e\right )} x}{3 \, b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.511312, size = 90, normalized size = 0.62 \begin{align*} - \frac{\sqrt{- \frac{a}{b^{5}}} \left (a e - b d\right ) \log{\left (- b^{2} \sqrt{- \frac{a}{b^{5}}} + x \right )}}{2} + \frac{\sqrt{- \frac{a}{b^{5}}} \left (a e - b d\right ) \log{\left (b^{2} \sqrt{- \frac{a}{b^{5}}} + x \right )}}{2} + \frac{e x^{3}}{3 b} - \frac{x \left (a e - b d\right )}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(-a/b**5)*(a*e - b*d)*log(-b**2*sqrt(-a/b**5) + x)/2 + sqrt(-a/b**5)*(a*e - b*d)*log(b**2*sqrt(-a/b**5) +
x)/2 + e*x**3/(3*b) - x*(a*e - b*d)/b**2

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Giac [A]  time = 1.10465, size = 136, normalized size = 0.94 \begin{align*} -\frac{{\left (a b d \mathrm{sgn}\left (b x^{2} + a\right ) - a^{2} e \mathrm{sgn}\left (b x^{2} + a\right )\right )} \arctan \left (\frac{b x}{\sqrt{a b}}\right )}{\sqrt{a b} b^{2}} + \frac{b^{2} x^{3} e \mathrm{sgn}\left (b x^{2} + a\right ) + 3 \, b^{2} d x \mathrm{sgn}\left (b x^{2} + a\right ) - 3 \, a b x e \mathrm{sgn}\left (b x^{2} + a\right )}{3 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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