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

Optimal. Leaf size=98 \[ \frac {b^{5/2} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{c^{9/2}}-\frac {b^2 x (b B-A c)}{c^4}+\frac {b x^3 (b B-A c)}{3 c^3}-\frac {x^5 (b B-A c)}{5 c^2}+\frac {B x^7}{7 c} \]

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 98, 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} \begin {gather*} -\frac {b^2 x (b B-A c)}{c^4}+\frac {b^{5/2} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{c^{9/2}}-\frac {x^5 (b B-A c)}{5 c^2}+\frac {b x^3 (b B-A c)}{3 c^3}+\frac {B x^7}{7 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^2*(b*B - A*c)*x)/c^4) + (b*(b*B - A*c)*x^3)/(3*c^3) - ((b*B - A*c)*x^5)/(5*c^2) + (B*x^7)/(7*c) + (b^(5/2
)*(b*B - A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/c^(9/2)

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^8 \left (A+B x^2\right )}{b x^2+c x^4} \, dx &=\int \frac {x^6 \left (A+B x^2\right )}{b+c x^2} \, dx\\ &=\frac {B x^7}{7 c}-\frac {(7 b B-7 A c) \int \frac {x^6}{b+c x^2} \, dx}{7 c}\\ &=\frac {B x^7}{7 c}-\frac {(7 b B-7 A c) \int \left (\frac {b^2}{c^3}-\frac {b x^2}{c^2}+\frac {x^4}{c}-\frac {b^3}{c^3 \left (b+c x^2\right )}\right ) \, dx}{7 c}\\ &=-\frac {b^2 (b B-A c) x}{c^4}+\frac {b (b B-A c) x^3}{3 c^3}-\frac {(b B-A c) x^5}{5 c^2}+\frac {B x^7}{7 c}+\frac {\left (b^3 (b B-A c)\right ) \int \frac {1}{b+c x^2} \, dx}{c^4}\\ &=-\frac {b^2 (b B-A c) x}{c^4}+\frac {b (b B-A c) x^3}{3 c^3}-\frac {(b B-A c) x^5}{5 c^2}+\frac {B x^7}{7 c}+\frac {b^{5/2} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{c^{9/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.07, size = 98, normalized size = 1.00 \begin {gather*} \frac {b^{5/2} (b B-A c) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b}}\right )}{c^{9/2}}-\frac {b^2 x (b B-A c)}{c^4}+\frac {b x^3 (b B-A c)}{3 c^3}+\frac {x^5 (A c-b B)}{5 c^2}+\frac {B x^7}{7 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((b^2*(b*B - A*c)*x)/c^4) + (b*(b*B - A*c)*x^3)/(3*c^3) + ((-(b*B) + A*c)*x^5)/(5*c^2) + (B*x^7)/(7*c) + (b^(
5/2)*(b*B - A*c)*ArcTan[(Sqrt[c]*x)/Sqrt[b]])/c^(9/2)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.40, size = 228, normalized size = 2.33 \begin {gather*} \left [\frac {30 \, B c^{3} x^{7} - 42 \, {\left (B b c^{2} - A c^{3}\right )} x^{5} + 70 \, {\left (B b^{2} c - A b c^{2}\right )} x^{3} - 105 \, {\left (B b^{3} - A b^{2} 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 ) - 210 \, {\left (B b^{3} - A b^{2} c\right )} x}{210 \, c^{4}}, \frac {15 \, B c^{3} x^{7} - 21 \, {\left (B b c^{2} - A c^{3}\right )} x^{5} + 35 \, {\left (B b^{2} c - A b c^{2}\right )} x^{3} + 105 \, {\left (B b^{3} - A b^{2} c\right )} \sqrt {\frac {b}{c}} \arctan \left (\frac {c x \sqrt {\frac {b}{c}}}{b}\right ) - 105 \, {\left (B b^{3} - A b^{2} c\right )} x}{105 \, c^{4}}\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),x, algorithm="fricas")

[Out]

[1/210*(30*B*c^3*x^7 - 42*(B*b*c^2 - A*c^3)*x^5 + 70*(B*b^2*c - A*b*c^2)*x^3 - 105*(B*b^3 - A*b^2*c)*sqrt(-b/c
)*log((c*x^2 - 2*c*x*sqrt(-b/c) - b)/(c*x^2 + b)) - 210*(B*b^3 - A*b^2*c)*x)/c^4, 1/105*(15*B*c^3*x^7 - 21*(B*
b*c^2 - A*c^3)*x^5 + 35*(B*b^2*c - A*b*c^2)*x^3 + 105*(B*b^3 - A*b^2*c)*sqrt(b/c)*arctan(c*x*sqrt(b/c)/b) - 10
5*(B*b^3 - A*b^2*c)*x)/c^4]

________________________________________________________________________________________

giac [A]  time = 0.18, size = 108, normalized size = 1.10 \begin {gather*} \frac {{\left (B b^{4} - A b^{3} c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c} c^{4}} + \frac {15 \, B c^{6} x^{7} - 21 \, B b c^{5} x^{5} + 21 \, A c^{6} x^{5} + 35 \, B b^{2} c^{4} x^{3} - 35 \, A b c^{5} x^{3} - 105 \, B b^{3} c^{3} x + 105 \, A b^{2} c^{4} x}{105 \, c^{7}} \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),x, algorithm="giac")

[Out]

(B*b^4 - A*b^3*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c^4) + 1/105*(15*B*c^6*x^7 - 21*B*b*c^5*x^5 + 21*A*c^6*x^5
+ 35*B*b^2*c^4*x^3 - 35*A*b*c^5*x^3 - 105*B*b^3*c^3*x + 105*A*b^2*c^4*x)/c^7

________________________________________________________________________________________

maple [A]  time = 0.05, size = 116, normalized size = 1.18 \begin {gather*} \frac {B \,x^{7}}{7 c}+\frac {A \,x^{5}}{5 c}-\frac {B b \,x^{5}}{5 c^{2}}-\frac {A b \,x^{3}}{3 c^{2}}+\frac {B \,b^{2} x^{3}}{3 c^{3}}-\frac {A \,b^{3} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c}\, c^{3}}+\frac {B \,b^{4} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c}\, c^{4}}+\frac {A \,b^{2} x}{c^{3}}-\frac {B \,b^{3} x}{c^{4}} \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),x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 2.94, size = 100, normalized size = 1.02 \begin {gather*} \frac {{\left (B b^{4} - A b^{3} c\right )} \arctan \left (\frac {c x}{\sqrt {b c}}\right )}{\sqrt {b c} c^{4}} + \frac {15 \, B c^{3} x^{7} - 21 \, {\left (B b c^{2} - A c^{3}\right )} x^{5} + 35 \, {\left (B b^{2} c - A b c^{2}\right )} x^{3} - 105 \, {\left (B b^{3} - A b^{2} c\right )} x}{105 \, c^{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),x, algorithm="maxima")

[Out]

(B*b^4 - A*b^3*c)*arctan(c*x/sqrt(b*c))/(sqrt(b*c)*c^4) + 1/105*(15*B*c^3*x^7 - 21*(B*b*c^2 - A*c^3)*x^5 + 35*
(B*b^2*c - A*b*c^2)*x^3 - 105*(B*b^3 - A*b^2*c)*x)/c^4

________________________________________________________________________________________

mupad [B]  time = 0.04, size = 118, normalized size = 1.20 \begin {gather*} x^5\,\left (\frac {A}{5\,c}-\frac {B\,b}{5\,c^2}\right )+\frac {B\,x^7}{7\,c}+\frac {b^{5/2}\,\mathrm {atan}\left (\frac {b^{5/2}\,\sqrt {c}\,x\,\left (A\,c-B\,b\right )}{B\,b^4-A\,b^3\,c}\right )\,\left (A\,c-B\,b\right )}{c^{9/2}}-\frac {b\,x^3\,\left (\frac {A}{c}-\frac {B\,b}{c^2}\right )}{3\,c}+\frac {b^2\,x\,\left (\frac {A}{c}-\frac {B\,b}{c^2}\right )}{c^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),x)

[Out]

x^5*(A/(5*c) - (B*b)/(5*c^2)) + (B*x^7)/(7*c) + (b^(5/2)*atan((b^(5/2)*c^(1/2)*x*(A*c - B*b))/(B*b^4 - A*b^3*c
))*(A*c - B*b))/c^(9/2) - (b*x^3*(A/c - (B*b)/c^2))/(3*c) + (b^2*x*(A/c - (B*b)/c^2))/c^2

________________________________________________________________________________________

sympy [B]  time = 0.41, size = 180, normalized size = 1.84 \begin {gather*} \frac {B x^{7}}{7 c} + x^{5} \left (\frac {A}{5 c} - \frac {B b}{5 c^{2}}\right ) + x^{3} \left (- \frac {A b}{3 c^{2}} + \frac {B b^{2}}{3 c^{3}}\right ) + x \left (\frac {A b^{2}}{c^{3}} - \frac {B b^{3}}{c^{4}}\right ) - \frac {\sqrt {- \frac {b^{5}}{c^{9}}} \left (- A c + B b\right ) \log {\left (- \frac {c^{4} \sqrt {- \frac {b^{5}}{c^{9}}} \left (- A c + B b\right )}{- A b^{2} c + B b^{3}} + x \right )}}{2} + \frac {\sqrt {- \frac {b^{5}}{c^{9}}} \left (- A c + B b\right ) \log {\left (\frac {c^{4} \sqrt {- \frac {b^{5}}{c^{9}}} \left (- A c + B b\right )}{- A b^{2} c + B b^{3}} + x \right )}}{2} \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),x)

[Out]

B*x**7/(7*c) + x**5*(A/(5*c) - B*b/(5*c**2)) + x**3*(-A*b/(3*c**2) + B*b**2/(3*c**3)) + x*(A*b**2/c**3 - B*b**
3/c**4) - sqrt(-b**5/c**9)*(-A*c + B*b)*log(-c**4*sqrt(-b**5/c**9)*(-A*c + B*b)/(-A*b**2*c + B*b**3) + x)/2 +
sqrt(-b**5/c**9)*(-A*c + B*b)*log(c**4*sqrt(-b**5/c**9)*(-A*c + B*b)/(-A*b**2*c + B*b**3) + x)/2

________________________________________________________________________________________