3.47 \(\int \frac{x^4 (A+B x^2)}{b x^2+c x^4} \, dx\)

Optimal. Leaf size=58 \[ -\frac{x (b B-A c)}{c^2}+\frac{\sqrt{b} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{c^{5/2}}+\frac{B x^3}{3 c} \]

[Out]

-(((b*B - A*c)*x)/c^2) + (B*x^3)/(3*c) + (Sqrt[b]*(b*B - A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/c^(5/2)

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Rubi [A]  time = 0.0498338, antiderivative size = 58, 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 = {1584, 459, 321, 205} \[ -\frac{x (b B-A c)}{c^2}+\frac{\sqrt{b} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{c^{5/2}}+\frac{B x^3}{3 c} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(A + B*x^2))/(b*x^2 + c*x^4),x]

[Out]

-(((b*B - A*c)*x)/c^2) + (B*x^3)/(3*c) + (Sqrt[b]*(b*B - A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/c^(5/2)

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - 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^4 \left (A+B x^2\right )}{b x^2+c x^4} \, dx &=\int \frac{x^2 \left (A+B x^2\right )}{b+c x^2} \, dx\\ &=\frac{B x^3}{3 c}-\frac{(3 b B-3 A c) \int \frac{x^2}{b+c x^2} \, dx}{3 c}\\ &=-\frac{(b B-A c) x}{c^2}+\frac{B x^3}{3 c}+\frac{(b (b B-A c)) \int \frac{1}{b+c x^2} \, dx}{c^2}\\ &=-\frac{(b B-A c) x}{c^2}+\frac{B x^3}{3 c}+\frac{\sqrt{b} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{c^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0384189, size = 57, normalized size = 0.98 \[ \frac{x (A c-b B)}{c^2}+\frac{\sqrt{b} (b B-A c) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b}}\right )}{c^{5/2}}+\frac{B x^3}{3 c} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(A + B*x^2))/(b*x^2 + c*x^4),x]

[Out]

((-(b*B) + A*c)*x)/c^2 + (B*x^3)/(3*c) + (Sqrt[b]*(b*B - A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/c^(5/2)

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Maple [A]  time = 0.004, size = 68, normalized size = 1.2 \begin{align*}{\frac{B{x}^{3}}{3\,c}}+{\frac{Ax}{c}}-{\frac{Bbx}{{c}^{2}}}-{\frac{Ab}{c}\arctan \left ({cx{\frac{1}{\sqrt{bc}}}} \right ){\frac{1}{\sqrt{bc}}}}+{\frac{B{b}^{2}}{{c}^{2}}\arctan \left ({cx{\frac{1}{\sqrt{bc}}}} \right ){\frac{1}{\sqrt{bc}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(B*x^2+A)/(c*x^4+b*x^2),x)

[Out]

1/3*B*x^3/c+1/c*A*x-1/c^2*B*b*x-b/c/(b*c)^(1/2)*arctan(x*c/(b*c)^(1/2))*A+b^2/c^2/(b*c)^(1/2)*arctan(x*c/(b*c)
^(1/2))*B

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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)/(c*x^4+b*x^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.589389, size = 277, normalized size = 4.78 \begin{align*} \left [\frac{2 \, B c x^{3} - 3 \,{\left (B b - A c\right )} \sqrt{-\frac{b}{c}} \log \left (\frac{c x^{2} - 2 \, c x \sqrt{-\frac{b}{c}} - b}{c x^{2} + b}\right ) - 6 \,{\left (B b - A c\right )} x}{6 \, c^{2}}, \frac{B c x^{3} + 3 \,{\left (B b - A c\right )} \sqrt{\frac{b}{c}} \arctan \left (\frac{c x \sqrt{\frac{b}{c}}}{b}\right ) - 3 \,{\left (B b - A c\right )} x}{3 \, c^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="fricas")

[Out]

[1/6*(2*B*c*x^3 - 3*(B*b - A*c)*sqrt(-b/c)*log((c*x^2 - 2*c*x*sqrt(-b/c) - b)/(c*x^2 + b)) - 6*(B*b - A*c)*x)/
c^2, 1/3*(B*c*x^3 + 3*(B*b - A*c)*sqrt(b/c)*arctan(c*x*sqrt(b/c)/b) - 3*(B*b - A*c)*x)/c^2]

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Sympy [A]  time = 0.461501, size = 90, normalized size = 1.55 \begin{align*} \frac{B x^{3}}{3 c} - \frac{\sqrt{- \frac{b}{c^{5}}} \left (- A c + B b\right ) \log{\left (- c^{2} \sqrt{- \frac{b}{c^{5}}} + x \right )}}{2} + \frac{\sqrt{- \frac{b}{c^{5}}} \left (- A c + B b\right ) \log{\left (c^{2} \sqrt{- \frac{b}{c^{5}}} + x \right )}}{2} - \frac{x \left (- A c + B b\right )}{c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(B*x**2+A)/(c*x**4+b*x**2),x)

[Out]

B*x**3/(3*c) - sqrt(-b/c**5)*(-A*c + B*b)*log(-c**2*sqrt(-b/c**5) + x)/2 + sqrt(-b/c**5)*(-A*c + B*b)*log(c**2
*sqrt(-b/c**5) + x)/2 - x*(-A*c + B*b)/c**2

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Giac [A]  time = 1.34879, size = 77, normalized size = 1.33 \begin{align*} \frac{{\left (B b^{2} - A b c\right )} \arctan \left (\frac{c x}{\sqrt{b c}}\right )}{\sqrt{b c} c^{2}} + \frac{B c^{2} x^{3} - 3 \, B b c x + 3 \, A c^{2} x}{3 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="giac")

[Out]

(B*b^2 - A*b*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c^2) + 1/3*(B*c^2*x^3 - 3*B*b*c*x + 3*A*c^2*x)/c^3