3.1.62 \(\int \frac {x^8 (A+B x^2)}{(b x^2+c x^4)^2} \, dx\)

Optimal. Leaf size=89 \[ \frac {\sqrt {b} (5 b B-3 A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 c^{7/2}}-\frac {b x (b B-A c)}{2 c^3 \left (b+c x^2\right )}-\frac {x (2 b B-A c)}{c^3}+\frac {B x^3}{3 c^2} \]

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Rubi [A]  time = 0.09, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {1584, 455, 1153, 205} \begin {gather*} -\frac {b x (b B-A c)}{2 c^3 \left (b+c x^2\right )}-\frac {x (2 b B-A c)}{c^3}+\frac {\sqrt {b} (5 b B-3 A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 c^{7/2}}+\frac {B x^3}{3 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^8*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

-(((2*b*B - A*c)*x)/c^3) + (B*x^3)/(3*c^2) - (b*(b*B - A*c)*x)/(2*c^3*(b + c*x^2)) + (Sqrt[b]*(5*b*B - 3*A*c)*
ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(2*c^(7/2))

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 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 1153

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -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]

Rubi steps

\begin {align*} \int \frac {x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx &=\int \frac {x^4 \left (A+B x^2\right )}{\left (b+c x^2\right )^2} \, dx\\ &=-\frac {b (b B-A c) x}{2 c^3 \left (b+c x^2\right )}-\frac {\int \frac {-b (b B-A c)+2 c (b B-A c) x^2-2 B c^2 x^4}{b+c x^2} \, dx}{2 c^3}\\ &=-\frac {b (b B-A c) x}{2 c^3 \left (b+c x^2\right )}-\frac {\int \left (2 (2 b B-A c)-2 B c x^2+\frac {-5 b^2 B+3 A b c}{b+c x^2}\right ) \, dx}{2 c^3}\\ &=-\frac {(2 b B-A c) x}{c^3}+\frac {B x^3}{3 c^2}-\frac {b (b B-A c) x}{2 c^3 \left (b+c x^2\right )}+\frac {(b (5 b B-3 A c)) \int \frac {1}{b+c x^2} \, dx}{2 c^3}\\ &=-\frac {(2 b B-A c) x}{c^3}+\frac {B x^3}{3 c^2}-\frac {b (b B-A c) x}{2 c^3 \left (b+c x^2\right )}+\frac {\sqrt {b} (5 b B-3 A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 89, normalized size = 1.00 \begin {gather*} \frac {x \left (A b c-b^2 B\right )}{2 c^3 \left (b+c x^2\right )}+\frac {\sqrt {b} (5 b B-3 A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{2 c^{7/2}}+\frac {x (A c-2 b B)}{c^3}+\frac {B x^3}{3 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^8*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

((-2*b*B + A*c)*x)/c^3 + (B*x^3)/(3*c^2) + ((-(b^2*B) + A*b*c)*x)/(2*c^3*(b + c*x^2)) + (Sqrt[b]*(5*b*B - 3*A*
c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/(2*c^(7/2))

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^8 \left (A+B x^2\right )}{\left (b x^2+c x^4\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(x^8*(A + B*x^2))/(b*x^2 + c*x^4)^2,x]

[Out]

IntegrateAlgebraic[(x^8*(A + B*x^2))/(b*x^2 + c*x^4)^2, x]

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fricas [A]  time = 0.41, size = 240, normalized size = 2.70 \begin {gather*} \left [\frac {4 \, B c^{2} x^{5} - 4 \, {\left (5 \, B b c - 3 \, A c^{2}\right )} x^{3} - 3 \, {\left (5 \, B b^{2} - 3 \, A b c + {\left (5 \, B b c - 3 \, A c^{2}\right )} x^{2}\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 (5 \, B b^{2} - 3 \, A b c\right )} x}{12 \, {\left (c^{4} x^{2} + b c^{3}\right )}}, \frac {2 \, B c^{2} x^{5} - 2 \, {\left (5 \, B b c - 3 \, A c^{2}\right )} x^{3} + 3 \, {\left (5 \, B b^{2} - 3 \, A b c + {\left (5 \, B b c - 3 \, A c^{2}\right )} x^{2}\right )} \sqrt {\frac {b}{c}} \arctan \left (\frac {c x \sqrt {\frac {b}{c}}}{b}\right ) - 3 \, {\left (5 \, B b^{2} - 3 \, A b c\right )} x}{6 \, {\left (c^{4} x^{2} + b c^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="fricas")

[Out]

[1/12*(4*B*c^2*x^5 - 4*(5*B*b*c - 3*A*c^2)*x^3 - 3*(5*B*b^2 - 3*A*b*c + (5*B*b*c - 3*A*c^2)*x^2)*sqrt(-b/c)*lo
g((c*x^2 - 2*c*x*sqrt(-b/c) - b)/(c*x^2 + b)) - 6*(5*B*b^2 - 3*A*b*c)*x)/(c^4*x^2 + b*c^3), 1/6*(2*B*c^2*x^5 -
2*(5*B*b*c - 3*A*c^2)*x^3 + 3*(5*B*b^2 - 3*A*b*c + (5*B*b*c - 3*A*c^2)*x^2)*sqrt(b/c)*arctan(c*x*sqrt(b/c)/b)
- 3*(5*B*b^2 - 3*A*b*c)*x)/(c^4*x^2 + b*c^3)]

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giac [A]  time = 0.16, size = 88, normalized size = 0.99 \begin {gather*} \frac {{\left (5 \, B b^{2} - 3 \, A b c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} c^{3}} - \frac {B b^{2} x - A b c x}{2 \, {\left (c x^{2} + b\right )} c^{3}} + \frac {B c^{4} x^{3} - 6 \, B b c^{3} x + 3 \, A c^{4} x}{3 \, c^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="giac")

[Out]

1/2*(5*B*b^2 - 3*A*b*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c^3) - 1/2*(B*b^2*x - A*b*c*x)/((c*x^2 + b)*c^3) + 1/
3*(B*c^4*x^3 - 6*B*b*c^3*x + 3*A*c^4*x)/c^6

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maple [A]  time = 0.06, size = 105, normalized size = 1.18 \begin {gather*} \frac {B \,x^{3}}{3 c^{2}}+\frac {A b x}{2 \left (c \,x^{2}+b \right ) c^{2}}-\frac {3 A b \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \sqrt {b c}\, c^{2}}-\frac {B \,b^{2} x}{2 \left (c \,x^{2}+b \right ) c^{3}}+\frac {5 B \,b^{2} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \sqrt {b c}\, c^{3}}+\frac {A x}{c^{2}}-\frac {2 B b x}{c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^8*(B*x^2+A)/(c*x^4+b*x^2)^2,x)

[Out]

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

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maxima [A]  time = 3.03, size = 85, normalized size = 0.96 \begin {gather*} -\frac {{\left (B b^{2} - A b c\right )} x}{2 \, {\left (c^{4} x^{2} + b c^{3}\right )}} + \frac {{\left (5 \, B b^{2} - 3 \, A b c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{2 \, \sqrt {b c} c^{3}} + \frac {B c x^{3} - 3 \, {\left (2 \, B b - A c\right )} x}{3 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^8*(B*x^2+A)/(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

-1/2*(B*b^2 - A*b*c)*x/(c^4*x^2 + b*c^3) + 1/2*(5*B*b^2 - 3*A*b*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c^3) + 1/3
*(B*c*x^3 - 3*(2*B*b - A*c)*x)/c^3

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mupad [B]  time = 0.11, size = 104, normalized size = 1.17 \begin {gather*} x\,\left (\frac {A}{c^2}-\frac {2\,B\,b}{c^3}\right )-\frac {x\,\left (\frac {B\,b^2}{2}-\frac {A\,b\,c}{2}\right )}{c^4\,x^2+b\,c^3}+\frac {B\,x^3}{3\,c^2}+\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {c}\,x\,\left (3\,A\,c-5\,B\,b\right )}{5\,B\,b^2-3\,A\,b\,c}\right )\,\left (3\,A\,c-5\,B\,b\right )}{2\,c^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^8*(A + B*x^2))/(b*x^2 + c*x^4)^2,x)

[Out]

x*(A/c^2 - (2*B*b)/c^3) - (x*((B*b^2)/2 - (A*b*c)/2))/(b*c^3 + c^4*x^2) + (B*x^3)/(3*c^2) + (b^(1/2)*atan((b^(
1/2)*c^(1/2)*x*(3*A*c - 5*B*b))/(5*B*b^2 - 3*A*b*c))*(3*A*c - 5*B*b))/(2*c^(7/2))

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sympy [A]  time = 0.69, size = 129, normalized size = 1.45 \begin {gather*} \frac {B x^{3}}{3 c^{2}} + x \left (\frac {A}{c^{2}} - \frac {2 B b}{c^{3}}\right ) + \frac {x \left (A b c - B b^{2}\right )}{2 b c^{3} + 2 c^{4} x^{2}} - \frac {\sqrt {- \frac {b}{c^{7}}} \left (- 3 A c + 5 B b\right ) \log {\left (- c^{3} \sqrt {- \frac {b}{c^{7}}} + x \right )}}{4} + \frac {\sqrt {- \frac {b}{c^{7}}} \left (- 3 A c + 5 B b\right ) \log {\left (c^{3} \sqrt {- \frac {b}{c^{7}}} + x \right )}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**8*(B*x**2+A)/(c*x**4+b*x**2)**2,x)

[Out]

B*x**3/(3*c**2) + x*(A/c**2 - 2*B*b/c**3) + x*(A*b*c - B*b**2)/(2*b*c**3 + 2*c**4*x**2) - sqrt(-b/c**7)*(-3*A*
c + 5*B*b)*log(-c**3*sqrt(-b/c**7) + x)/4 + sqrt(-b/c**7)*(-3*A*c + 5*B*b)*log(c**3*sqrt(-b/c**7) + x)/4

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