3.1.42 \(\int \frac {x^9 (A+B x^2)}{b x^2+c x^4} \, dx\) [42]

Optimal. Leaf size=96 \[ -\frac {b^2 (b B-A c) x^2}{2 c^4}+\frac {b (b B-A c) x^4}{4 c^3}-\frac {(b B-A c) x^6}{6 c^2}+\frac {B x^8}{8 c}+\frac {b^3 (b B-A c) \log \left (b+c x^2\right )}{2 c^5} \]

[Out]

-1/2*b^2*(-A*c+B*b)*x^2/c^4+1/4*b*(-A*c+B*b)*x^4/c^3-1/6*(-A*c+B*b)*x^6/c^2+1/8*B*x^8/c+1/2*b^3*(-A*c+B*b)*ln(
c*x^2+b)/c^5

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Rubi [A]
time = 0.09, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1598, 457, 78} \begin {gather*} \frac {b^3 (b B-A c) \log \left (b+c x^2\right )}{2 c^5}-\frac {b^2 x^2 (b B-A c)}{2 c^4}+\frac {b x^4 (b B-A c)}{4 c^3}-\frac {x^6 (b B-A c)}{6 c^2}+\frac {B x^8}{8 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^9*(A + B*x^2))/(b*x^2 + c*x^4),x]

[Out]

-1/2*(b^2*(b*B - A*c)*x^2)/c^4 + (b*(b*B - A*c)*x^4)/(4*c^3) - ((b*B - A*c)*x^6)/(6*c^2) + (B*x^8)/(8*c) + (b^
3*(b*B - A*c)*Log[b + c*x^2])/(2*c^5)

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 457

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

Rule 1598

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

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Mathematica [A]
time = 0.03, size = 92, normalized size = 0.96 \begin {gather*} \frac {c x^2 \left (-12 b^3 B+6 b^2 c \left (2 A+B x^2\right )-2 b c^2 x^2 \left (3 A+2 B x^2\right )+c^3 x^4 \left (4 A+3 B x^2\right )\right )+12 b^3 (b B-A c) \log \left (b+c x^2\right )}{24 c^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^9*(A + B*x^2))/(b*x^2 + c*x^4),x]

[Out]

(c*x^2*(-12*b^3*B + 6*b^2*c*(2*A + B*x^2) - 2*b*c^2*x^2*(3*A + 2*B*x^2) + c^3*x^4*(4*A + 3*B*x^2)) + 12*b^3*(b
*B - A*c)*Log[b + c*x^2])/(24*c^5)

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Maple [A]
time = 0.38, size = 98, normalized size = 1.02

method	result	size
norman	\(\frac {\frac {B \,x^{9}}{8 c}+\frac {\left (A c -B b \right ) x^{7}}{6 c^{2}}-\frac {b \left (A c -B b \right ) x^{5}}{4 c^{3}}+\frac {b^{2} \left (A c -B b \right ) x^{3}}{2 c^{4}}}{x}-\frac {b^{3} \left (A c -B b \right ) \ln \left (c \,x^{2}+b \right )}{2 c^{5}}\)	\(92\)
default	\(\frac {\frac {1}{4} B \,c^{3} x^{8}+\frac {1}{3} A \,c^{3} x^{6}-\frac {1}{3} x^{6} B b \,c^{2}-\frac {1}{2} A b \,c^{2} x^{4}+\frac {1}{2} x^{4} B \,b^{2} c +A \,b^{2} c \,x^{2}-x^{2} B \,b^{3}}{2 c^{4}}-\frac {b^{3} \left (A c -B b \right ) \ln \left (c \,x^{2}+b \right )}{2 c^{5}}\)	\(98\)
risch	\(\frac {B \,x^{8}}{8 c}+\frac {A \,x^{6}}{6 c}-\frac {x^{6} B b}{6 c^{2}}-\frac {A b \,x^{4}}{4 c^{2}}+\frac {x^{4} B \,b^{2}}{4 c^{3}}+\frac {A \,b^{2} x^{2}}{2 c^{3}}-\frac {x^{2} B \,b^{3}}{2 c^{4}}-\frac {b^{3} \ln \left (c \,x^{2}+b \right ) A}{2 c^{4}}+\frac {b^{4} \ln \left (c \,x^{2}+b \right ) B}{2 c^{5}}\)	\(110\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9*(B*x^2+A)/(c*x^4+b*x^2),x,method=_RETURNVERBOSE)

[Out]

1/2/c^4*(1/4*B*c^3*x^8+1/3*A*c^3*x^6-1/3*x^6*B*b*c^2-1/2*A*b*c^2*x^4+1/2*x^4*B*b^2*c+A*b^2*c*x^2-x^2*B*b^3)-1/
2*b^3*(A*c-B*b)/c^5*ln(c*x^2+b)

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Maxima [A]
time = 0.28, size = 97, normalized size = 1.01 \begin {gather*} \frac {3 \, B c^{3} x^{8} - 4 \, {\left (B b c^{2} - A c^{3}\right )} x^{6} + 6 \, {\left (B b^{2} c - A b c^{2}\right )} x^{4} - 12 \, {\left (B b^{3} - A b^{2} c\right )} x^{2}}{24 \, c^{4}} + \frac {{\left (B b^{4} - A b^{3} c\right )} \log \left (c x^{2} + b\right )}{2 \, c^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="maxima")

[Out]

1/24*(3*B*c^3*x^8 - 4*(B*b*c^2 - A*c^3)*x^6 + 6*(B*b^2*c - A*b*c^2)*x^4 - 12*(B*b^3 - A*b^2*c)*x^2)/c^4 + 1/2*
(B*b^4 - A*b^3*c)*log(c*x^2 + b)/c^5

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Fricas [A]
time = 1.51, size = 98, normalized size = 1.02 \begin {gather*} \frac {3 \, B c^{4} x^{8} - 4 \, {\left (B b c^{3} - A c^{4}\right )} x^{6} + 6 \, {\left (B b^{2} c^{2} - A b c^{3}\right )} x^{4} - 12 \, {\left (B b^{3} c - A b^{2} c^{2}\right )} x^{2} + 12 \, {\left (B b^{4} - A b^{3} c\right )} \log \left (c x^{2} + b\right )}{24 \, c^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="fricas")

[Out]

1/24*(3*B*c^4*x^8 - 4*(B*b*c^3 - A*c^4)*x^6 + 6*(B*b^2*c^2 - A*b*c^3)*x^4 - 12*(B*b^3*c - A*b^2*c^2)*x^2 + 12*
(B*b^4 - A*b^3*c)*log(c*x^2 + b))/c^5

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Sympy [A]
time = 0.18, size = 94, normalized size = 0.98 \begin {gather*} \frac {B x^{8}}{8 c} + \frac {b^{3} \left (- A c + B b\right ) \log {\left (b + c x^{2} \right )}}{2 c^{5}} + x^{6} \left (\frac {A}{6 c} - \frac {B b}{6 c^{2}}\right ) + x^{4} \left (- \frac {A b}{4 c^{2}} + \frac {B b^{2}}{4 c^{3}}\right ) + x^{2} \left (\frac {A b^{2}}{2 c^{3}} - \frac {B b^{3}}{2 c^{4}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9*(B*x**2+A)/(c*x**4+b*x**2),x)

[Out]

B*x**8/(8*c) + b**3*(-A*c + B*b)*log(b + c*x**2)/(2*c**5) + x**6*(A/(6*c) - B*b/(6*c**2)) + x**4*(-A*b/(4*c**2
) + B*b**2/(4*c**3)) + x**2*(A*b**2/(2*c**3) - B*b**3/(2*c**4))

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Giac [A]
time = 0.63, size = 101, normalized size = 1.05 \begin {gather*} \frac {3 \, B c^{3} x^{8} - 4 \, B b c^{2} x^{6} + 4 \, A c^{3} x^{6} + 6 \, B b^{2} c x^{4} - 6 \, A b c^{2} x^{4} - 12 \, B b^{3} x^{2} + 12 \, A b^{2} c x^{2}}{24 \, c^{4}} + \frac {{\left (B b^{4} - A b^{3} c\right )} \log \left ({\left | c x^{2} + b \right |}\right )}{2 \, c^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9*(B*x^2+A)/(c*x^4+b*x^2),x, algorithm="giac")

[Out]

1/24*(3*B*c^3*x^8 - 4*B*b*c^2*x^6 + 4*A*c^3*x^6 + 6*B*b^2*c*x^4 - 6*A*b*c^2*x^4 - 12*B*b^3*x^2 + 12*A*b^2*c*x^
2)/c^4 + 1/2*(B*b^4 - A*b^3*c)*log(abs(c*x^2 + b))/c^5

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Mupad [B]
time = 0.06, size = 100, normalized size = 1.04 \begin {gather*} x^6\,\left (\frac {A}{6\,c}-\frac {B\,b}{6\,c^2}\right )+\frac {B\,x^8}{8\,c}+\frac {\ln \left (c\,x^2+b\right )\,\left (B\,b^4-A\,b^3\,c\right )}{2\,c^5}+\frac {b^2\,x^2\,\left (\frac {A}{c}-\frac {B\,b}{c^2}\right )}{2\,c^2}-\frac {b\,x^4\,\left (\frac {A}{c}-\frac {B\,b}{c^2}\right )}{4\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^9*(A + B*x^2))/(b*x^2 + c*x^4),x)

[Out]

x^6*(A/(6*c) - (B*b)/(6*c^2)) + (B*x^8)/(8*c) + (log(b + c*x^2)*(B*b^4 - A*b^3*c))/(2*c^5) + (b^2*x^2*(A/c - (
B*b)/c^2))/(2*c^2) - (b*x^4*(A/c - (B*b)/c^2))/(4*c)

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