3.45 \(\int \frac {x^6 (A+B x^2)}{b x^2+c x^4} \, dx\)

Optimal. Leaf size=77 \[ -\frac {b^{3/2} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{c^{7/2}}+\frac {b x (b B-A c)}{c^3}-\frac {x^3 (b B-A c)}{3 c^2}+\frac {B x^5}{5 c} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 77, 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, 459, 302, 205} \[ -\frac {b^{3/2} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{c^{7/2}}-\frac {x^3 (b B-A c)}{3 c^2}+\frac {b x (b B-A c)}{c^3}+\frac {B x^5}{5 c} \]

Antiderivative was successfully verified.

[In]

Int[(x^6*(A + B*x^2))/(b*x^2 + c*x^4),x]

[Out]

(b*(b*B - A*c)*x)/c^3 - ((b*B - A*c)*x^3)/(3*c^2) + (B*x^5)/(5*c) - (b^(3/2)*(b*B - A*c)*ArcTan[(Sqrt[c]*x)/Sq
rt[b]])/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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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^6 \left (A+B x^2\right )}{b x^2+c x^4} \, dx &=\int \frac {x^4 \left (A+B x^2\right )}{b+c x^2} \, dx\\ &=\frac {B x^5}{5 c}-\frac {(5 b B-5 A c) \int \frac {x^4}{b+c x^2} \, dx}{5 c}\\ &=\frac {B x^5}{5 c}-\frac {(5 b B-5 A c) \int \left (-\frac {b}{c^2}+\frac {x^2}{c}+\frac {b^2}{c^2 \left (b+c x^2\right )}\right ) \, dx}{5 c}\\ &=\frac {b (b B-A c) x}{c^3}-\frac {(b B-A c) x^3}{3 c^2}+\frac {B x^5}{5 c}-\frac {\left (b^2 (b B-A c)\right ) \int \frac {1}{b+c x^2} \, dx}{c^3}\\ &=\frac {b (b B-A c) x}{c^3}-\frac {(b B-A c) x^3}{3 c^2}+\frac {B x^5}{5 c}-\frac {b^{3/2} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{c^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 77, normalized size = 1.00 \[ -\frac {b^{3/2} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{c^{7/2}}+\frac {b x (b B-A c)}{c^3}+\frac {x^3 (A c-b B)}{3 c^2}+\frac {B x^5}{5 c} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^6*(A + B*x^2))/(b*x^2 + c*x^4),x]

[Out]

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

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fricas [A]  time = 0.96, size = 178, normalized size = 2.31 \[ \left [\frac {6 \, B c^{2} x^{5} - 10 \, {\left (B b c - A c^{2}\right )} x^{3} - 15 \, {\left (B b^{2} - A b c\right )} \sqrt {-\frac {b}{c}} \log \left (\frac {c x^{2} + 2 \, c x \sqrt {-\frac {b}{c}} - b}{c x^{2} + b}\right ) + 30 \, {\left (B b^{2} - A b c\right )} x}{30 \, c^{3}}, \frac {3 \, B c^{2} x^{5} - 5 \, {\left (B b c - A c^{2}\right )} x^{3} - 15 \, {\left (B b^{2} - A b c\right )} \sqrt {\frac {b}{c}} \arctan \left (\frac {c x \sqrt {\frac {b}{c}}}{b}\right ) + 15 \, {\left (B b^{2} - A b c\right )} x}{15 \, c^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="fricas")

[Out]

[1/30*(6*B*c^2*x^5 - 10*(B*b*c - A*c^2)*x^3 - 15*(B*b^2 - A*b*c)*sqrt(-b/c)*log((c*x^2 + 2*c*x*sqrt(-b/c) - b)
/(c*x^2 + b)) + 30*(B*b^2 - A*b*c)*x)/c^3, 1/15*(3*B*c^2*x^5 - 5*(B*b*c - A*c^2)*x^3 - 15*(B*b^2 - A*b*c)*sqrt
(b/c)*arctan(c*x*sqrt(b/c)/b) + 15*(B*b^2 - A*b*c)*x)/c^3]

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giac [A]  time = 0.18, size = 85, normalized size = 1.10 \[ -\frac {{\left (B b^{3} - A b^{2} c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c} c^{3}} + \frac {3 \, B c^{4} x^{5} - 5 \, B b c^{3} x^{3} + 5 \, A c^{4} x^{3} + 15 \, B b^{2} c^{2} x - 15 \, A b c^{3} x}{15 \, c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="giac")

[Out]

-(B*b^3 - A*b^2*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c^3) + 1/15*(3*B*c^4*x^5 - 5*B*b*c^3*x^3 + 5*A*c^4*x^3 + 1
5*B*b^2*c^2*x - 15*A*b*c^3*x)/c^5

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maple [A]  time = 0.05, size = 92, normalized size = 1.19 \[ \frac {B \,x^{5}}{5 c}+\frac {A \,x^{3}}{3 c}-\frac {B b \,x^{3}}{3 c^{2}}+\frac {A \,b^{2} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c}\, c^{2}}-\frac {B \,b^{3} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c}\, c^{3}}-\frac {A b x}{c^{2}}+\frac {B \,b^{2} x}{c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*(B*x^2+A)/(c*x^4+b*x^2),x)

[Out]

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

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maxima [A]  time = 2.86, size = 78, normalized size = 1.01 \[ -\frac {{\left (B b^{3} - A b^{2} c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c} c^{3}} + \frac {3 \, B c^{2} x^{5} - 5 \, {\left (B b c - A c^{2}\right )} x^{3} + 15 \, {\left (B b^{2} - A b c\right )} x}{15 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="maxima")

[Out]

-(B*b^3 - A*b^2*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c^3) + 1/15*(3*B*c^2*x^5 - 5*(B*b*c - A*c^2)*x^3 + 15*(B*b
^2 - A*b*c)*x)/c^3

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mupad [B]  time = 0.07, size = 96, normalized size = 1.25 \[ x^3\,\left (\frac {A}{3\,c}-\frac {B\,b}{3\,c^2}\right )+\frac {B\,x^5}{5\,c}-\frac {b^{3/2}\,\mathrm {atan}\left (\frac {b^{3/2}\,\sqrt {c}\,x\,\left (A\,c-B\,b\right )}{B\,b^3-A\,b^2\,c}\right )\,\left (A\,c-B\,b\right )}{c^{7/2}}-\frac {b\,x\,\left (\frac {A}{c}-\frac {B\,b}{c^2}\right )}{c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^6*(A + B*x^2))/(b*x^2 + c*x^4),x)

[Out]

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

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sympy [B]  time = 0.37, size = 153, normalized size = 1.99 \[ \frac {B x^{5}}{5 c} + x^{3} \left (\frac {A}{3 c} - \frac {B b}{3 c^{2}}\right ) + x \left (- \frac {A b}{c^{2}} + \frac {B b^{2}}{c^{3}}\right ) + \frac {\sqrt {- \frac {b^{3}}{c^{7}}} \left (- A c + B b\right ) \log {\left (- \frac {c^{3} \sqrt {- \frac {b^{3}}{c^{7}}} \left (- A c + B b\right )}{- A b c + B b^{2}} + x \right )}}{2} - \frac {\sqrt {- \frac {b^{3}}{c^{7}}} \left (- A c + B b\right ) \log {\left (\frac {c^{3} \sqrt {- \frac {b^{3}}{c^{7}}} \left (- A c + B b\right )}{- A b c + B b^{2}} + x \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6*(B*x**2+A)/(c*x**4+b*x**2),x)

[Out]

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

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