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

Optimal. Leaf size=40 \[ \frac {B x}{c}-\frac {(b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{\sqrt {b} c^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1584, 388, 205} \[ \frac {B x}{c}-\frac {(b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{\sqrt {b} c^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 388

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 40, normalized size = 1.00 \[ \frac {B x}{c}-\frac {(b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{\sqrt {b} c^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.90, size = 99, normalized size = 2.48 \[ \left [\frac {2 \, B b c x + {\left (B b - A c\right )} \sqrt {-b c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-b c} x - b}{c x^{2} + b}\right )}{2 \, b c^{2}}, \frac {B b c x - {\left (B b - A c\right )} \sqrt {b c} \arctan \left (\frac {\sqrt {b c} x}{b}\right )}{b c^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 34, normalized size = 0.85 \[ \frac {B x}{c} - \frac {{\left (B b - A c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*x/c - (B*b - A*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c)

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maple [A]  time = 0.04, size = 45, normalized size = 1.12 \[ \frac {A \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c}}-\frac {B b \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c}\, c}+\frac {B x}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.92, size = 34, normalized size = 0.85 \[ \frac {B x}{c} - \frac {{\left (B b - A c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*x/c - (B*b - A*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c)

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mupad [B]  time = 0.05, size = 31, normalized size = 0.78 \[ \frac {B\,x}{c}+\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {b}}\right )\,\left (A\,c-B\,b\right )}{\sqrt {b}\,c^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 0.29, size = 82, normalized size = 2.05 \[ \frac {B x}{c} + \frac {\sqrt {- \frac {1}{b c^{3}}} \left (- A c + B b\right ) \log {\left (- b c \sqrt {- \frac {1}{b c^{3}}} + x \right )}}{2} - \frac {\sqrt {- \frac {1}{b c^{3}}} \left (- A c + B b\right ) \log {\left (b c \sqrt {- \frac {1}{b c^{3}}} + x \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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