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

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

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Rubi [A]  time = 0.05, antiderivative size = 58, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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} \end {gather*}

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 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 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 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 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^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*}

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Mathematica [A]  time = 0.04, size = 57, normalized size = 0.98 \begin {gather*} \frac {\sqrt {b} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{c^{5/2}}+\frac {x (A c-b B)}{c^2}+\frac {B x^3}{3 c} \end {gather*}

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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^4 \left (A+B x^2\right )}{b x^2+c x^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 129, normalized size = 2.22 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.16, size = 57, normalized size = 0.98 \begin {gather*} \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 {gather*}

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

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maple [A]  time = 0.05, size = 68, normalized size = 1.17 \begin {gather*} \frac {B \,x^{3}}{3 c}-\frac {A b \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c}\, c}+\frac {B \,b^{2} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c}\, c^{2}}+\frac {A x}{c}-\frac {B b x}{c^{2}} \end {gather*}

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

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maxima [A]  time = 2.86, size = 53, normalized size = 0.91 \begin {gather*} \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 x^{3} - 3 \, {\left (B b - A c\right )} x}{3 \, c^{2}} \end {gather*}

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]

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

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mupad [B]  time = 0.11, size = 70, normalized size = 1.21 \begin {gather*} x\,\left (\frac {A}{c}-\frac {B\,b}{c^2}\right )+\frac {B\,x^3}{3\,c}+\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c}\,x\,\left (A\,c-B\,b\right )}{B\,b^2-A\,b\,c}\right )\,\left (A\,c-B\,b\right )}{c^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*(A/c - (B*b)/c^2) + (B*x^3)/(3*c) + (b^(1/2)*atan((b^(1/2)*c^(1/2)*x*(A*c - B*b))/(B*b^2 - A*b*c))*(A*c - B*
b))/c^(5/2)

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sympy [A]  time = 0.34, size = 90, normalized size = 1.55 \begin {gather*} \frac {B x^{3}}{3 c} + x \left (\frac {A}{c} - \frac {B b}{c^{2}}\right ) - \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} \end {gather*}

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) + x*(A/c - B*b/c**2) - 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

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